Chemical Reactions and Stoichiometrymsbotticelli.weebly.com/uploads/1/6/7/6/16766594/chapter_7.pdfChemistry 11 Solutions 978‐0‐07‐105107‐1 Chapter 7 Chemical Reactions and

Chemistry 11 Solutions

978‐0‐07‐105107‐1 Chapter 7 Chemical Reactions and Stoichiometry • MHR | 1

Chapter 7

Chemical Reactions and Stoichiometry

Section 7.1 What Is Stoichiometry? Solutions for Practice Problems Student Edition page 298 Write ratios of coefficients for the equations in questions 1–4. 1. Practice Problem (page 298)

Balanced chemical equation: 2Mg(s) + O2(g) → 2MgO(s) Ratio of coefficients: 2 1 2

2. Practice Problem (page 298)

Balanced chemical equation: 2NO(g) + O2(g) → 2NO2(g) Ratio of coefficients: 2 1 2


Balanced chemical equation: Ca(s) + 2H2O(ℓ) → Ca(OH)2(s) + H2(g) Ratio of coefficients: 1 2 1 1


Balanced chemical equation: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) Ratio of coefficients: 2 7 4 6

5. Practice Problem (page 298) How many molecules of nitrogen, N2(g), produce 10 molecules of ammonia, NH3(g), in the following reaction? N2(g) + 3H2(g) →2NH3(g) What Is Required? You need to determine the number of molecules of nitrogen that react. What Is Given? You know the balanced chemical equation: N2(g) + 3H2(g) →2NH3(g) You know the number of molecules of ammonia that are formed: 10



Plan Your Strategy Use the balanced chemical equation to determine the ratio of coefficients for nitrogen molecules to ammonia molecules. Let x be the unknown number of nitrogen molecules. Equate the known ratio of coefficients for nitrogen to ammonia to the unknown ratio. Solve for the unknown number of nitrogen molecules. Act on Your Strategy Balanced chemical equation: N2(g) + 3H2(g) → 2NH3(g) Particle ratio: 1 molecule 2 molecules Known ratio equated to the unknown ratio:

2

3 3

1 molecule of N 10 molecules of NH 2 molecules of NH

x=

Solving for x:

2

3 3

3

3 3 2 2 molecul

1 molecule of N 10

es of NH 10 molecules of NH 1 mo

molecules of NH 2 molecules of NH

10 molecules of NH

lecule of N

x

x

x=

=

× = ×

2

3

× 1 molecule of N 2 molecules of NH

25 molecules of N=

Five molecules of nitrogen will react. Check Your Solution The units are correct. The ratio 5:10 is equivalent to the ratio 1:2. The answer is reasonable.


Aluminum reacts with chlorine gas to form aluminum chloride: 2Al(s) + 3Cl2(g) → 2AlCl3(s) How many molecules of aluminum chloride form when 155 atoms of aluminum react with an excess of chlorine gas? What Is Required? You need to determine the number of molecules of aluminum chloride that form.



What Is Given? You know the balanced chemical equation: 2Al(s) + 3Cl2(g) → 2AlCl3(s) You know the number of atoms of aluminum that react: 155 Plan Your Strategy Use the balanced chemical equation to determine the ratio of coefficients for aluminum chloride molecules to aluminum atoms. Let x be the unknown number of aluminum chloride molecules. Equate the known ratio of coefficients for aluminum chloride to aluminum to the unknown ratio. Solve for the unknown number of aluminum chloride molecules. Act on Your Strategy Balanced chemical equation: 2Al(s) + 3Cl2(g) → 2AlCl3(s) Particle ratio: 2 atoms 2 molecules Known ratio equated to the unknown ratio:

3

3

2 molecules of AlCl 155 atoms of Al 2 atoms of AlCl

x=

Solving for x:

3

3 2 atoms of Al 155

2 m

atom

olecules

s of Al

of AlCl 155 atoms of Al 2 atoms

2 molecules of AlC of Al

155 atoms of

l

Al

x

x

x

=

= ×

=

×

3× 2 molecules of AlCl 2 atoms of Al

3155 molecules of AlCl=

155 molecules of aluminum chloride will form. Check Your Solution The units are correct. The ratio 155:155 is equivalent to the ratio 2:2. The answer is reasonable and correctly shows three significant digits.

7. Practice Problem (page 298) How many formula units of calcium chloride are produced by 6.7 × 1023

molecules of hydrochloric acid in the following reaction? Ca(OH)2(aq) + 2HCl(aq) → CaCl2(s) + 2H2O(ℓ) What Is Required? You need to determine the number of formula units of calcium chloride produced.



What Is Given? You know the balanced chemical equation: Ca(OH)2(aq) + 2HCl(aq) → CaCl2(s) + 2H2O(ℓ) You know the number of molecules of hydrochloric acid: 6.7 × 1023 Plan Your Strategy Use the balanced chemical equation to determine the ratio of coefficients for formula units of calcium chloride to molecules of hydrochloric acid. Let x be the unknown number of formula units of calcium chloride. Equate the known ratio of coefficients for calcium chloride to hydrochloric acid to the unknown ratio. Solve for the unknown number of formula units of calcium chloride. Act on Your Strategy Balanced chemical equation: Ca(OH)2(aq) + 2HCl(aq) → CaCl2(s) + 2H2O(ℓ) Particle ratio: 2 molecules 1 formula unit Known ratio equated to the unknown ratio:

323

1 formula unit of CaCl 6.7 ×10 molecules of HCl 2 molecules o Cl f H

x=

Solving for x:

323

232

23

× 2 molecules of HCl 6.7 × 10 molec

1 formula unit

ules of HCl ×

of CaCl 6.7 ×10 molecules of HCl 2 molecules o

1 formula unit of Caf HCl

6.7 ×

Cl

10 molecules of HCl

x

x

x

=

=

=

3 × 1 formula unit of CaCl 2 molecules of HCl

233

233

3.35 10 formula units

3.4 10 formula units

of CaCl

of CaCl

= ×

= ×

Therefore, 3.4 × 1023 formula units of calcium chloride will be produced. Check Your Solution The units are correct. The rounded ratio of 3.4 × 1023:6.7 × 1023 is equivalent to the ratio 1:2. The answer is reasonable and correctly shows two significant digits.



8. Practice Problem (page 298) How many formula units of magnesium chloride are produced by 7.7 × 1024

molecules of hydrochloric acid in this reaction? Mg(OH)2(aq) + 2HCl(aq) → MgCl2(s) + 2H2O(ℓ) What Is Required? You need to determine the number of formula units of magnesium chloride produced. What Is Given? You know the balanced chemical equation: Mg(OH)2(aq) + 2HCl(aq) → MgCl2(s) + 2H2O(ℓ) You know the number of molecules of hydrochloric acid: 7.7 × 1024 Plan Your Strategy Use the balanced chemical equation to determine the ratio of coefficients for formula units of magnesium chloride to molecules of hydrochloric acid. Let x be the unknown number of formula units of magnesium chloride. Equate the known ratio of coefficients for magnesium chloride to hydrochloric acid to the unknown ratio. Solve for the unknown number of formula units of magnesium chloride. Act on Your Strategy Balanced chemical equation: Mg(OH)2(aq) + 2HCl(aq) → MgCl2(s) + 2H2O(ℓ) Particle ratio: 2 molecules 1 formula unit Known ratio equated to the unknown ratio:

224

1 formula unit of MgCl 7.7 10 molecules of HCl 2 molecules of HCl

x=

×

Solving for x:

2

2

2

42

24

4

× 2 molecules of HCl 7.7 × 10 mole

1 formula u

cules of HC

nit of MgCl 7.7 × 10 molecules of HCl 2 molecule

l × 1 formula unit of MgCs of HCl

7.7 ×10 molecules of Cl

l

H

x

x

x

=

=

=

2× 1 formula unit of MgCl 2 molecules of HCl

242

242

3.85 × 10 formula units of MgCl

3.8 × 10 formula units of MgCl

=

=

3.8 × 1024 formula units of magnesium chloride will be produced.



Check Your Solution The units are correct. The rounded ratio of 3.8 × 1024:7.7 × 1024 is equivalent to the ratio 1:2. The answer is reasonable and correctly shows two significant digits.


The combustion of ethanol, C2H5OH(ℓ), is represented by the following equation: C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O(ℓ) How many molecules of oxygen, O2(g), produce 1.81× 1024 molecules of carbon dioxide, CO2(g), if an excess of ethanol is present? What Is Required? You need to determine the number of molecules of oxygen that react. What Is Given? You know the balanced chemical equation: C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O(ℓ) You know the number of molecules of carbon dioxide: 1.81× 1024 Plan Your Strategy Use the balanced chemical equation to determine the ratio of coefficients for oxygen molecules to carbon dioxide molecules. Let x be the unknown number of oxygen molecules. Equate the known ratio of coefficients for oxygen to carbon dioxide to the unknown ratio. Solve for the unknown number of oxygen molecules. Act on Your Strategy Balanced chemical equation: C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O(ℓ) Particle ratio: 3 molecules 2 molecules Known ratio equated to the unknown ratio:

224

2 2

3 molecules of O 1.81 × 10 molecules of CO 2 molecules of CO

x=



Solving for x: 2

24

242 2

2 2

2

2

42

3 molecules of O 1.81 × 10 molecules of CO 2 molecules of CO

molecules of molecules of molecules of 2 CO 1.81 × 10 CO × 3

1.81 ×10 molecules of CO

Ox

x

x

× =

=

=

2

2

× 3 molecules of O 2 molecules of CO

2

2

24

24

molecules of O2.715 10

2.72 1 molecule0 s of O

=

=

×

×

Therefore, 2.72 × 1024 molecules of oxygen are needed. Check Your Solution The units are correct. The rounded ratio of 2.72 × 1024:1.81 × 1024 is equivalent to the ratio 3:2. The answer is reasonable and correctly shows three significant digits.


Iron reacts with chlorine gas to form iron(III) chloride. How many atoms of iron react with three molecules of chlorine? What Is Required? You need to determine the number of atoms of iron, Fe(s), that react.

What Is Given? You know the product: iron(III) chloride You know the number of molecules of chlorine reacted: 3 Plan Your Strategy The chemical formula for chlorine gas is Cl2(g). Write the balanced chemical equation for this reaction. Use the balanced equation to determine the ratio of coefficients for atoms of iron to molecules of chlorine. Let x be the unknown number of atoms of iron. Equate the known ratio of coefficients for iron to chlorine to the unknown ratio. Solve for the unknown number of atoms of iron. Act on Your Strategy Balanced chemical equation: 2Fe(s) + 3Cl2(g) → 2FeCl3(s) Particle ratio: 2 atoms 3 molecules



Known ratio equated to the unknown ratio:

2 2

2 atoms of Fe3 molecules of Cl 3 molecules of Cl

x=

Solving for x:

2 2

2 2

2

2 atoms of Fe3 molecules of Cl 3 molecules of Cl 3 molecu × les of Cl 3 molecules of Cl 2 atoms of Fe

3 molecules of Cl

x

x

x

× =

=

=

2

2 atoms of Fe3 molecules of

× Cl

2 atoms of Fe=

Two atoms of iron react. Check Your Solution The units are correct. The ratio of iron to chlorine is identical to the ratio in the balanced chemical equation. The answer is reasonable and correctly shows one significant digit.

Section 7.1 What Is Stoichiometry? Solutions for Practice Problems Student Edition page 300-301

11. Practice Problem (page 300) What amount in moles of silver chromate, Ag2CrO4(s), is produced from 0.50 mol of silver nitrate, AgNO3(aq)? 2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq) What Is Required? You need to determine the amount in moles of silver chromate that is produced. What Is Given? You know the balanced chemical equation for the reaction: 2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq) You know the amount in moles of silver nitrate: 0.50 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of Ag2CrO4(s) to AgNO3(aq). Let

2 4Ag CrOn be the unknown amount in moles of Ag2CrO4(s).



Equate the known mole ratio of Ag2CrO4(s) to AgNO3(aq) to the unknown mole ratio. Solve for the unknown amount of Ag2CrO4(s). Act on Your Strategy Balanced chemical equation: 2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq) Mole ratio: 2 moles 1 mole Known mole ratio equated to the unknown ratio:

2 4Ag CrO 2 4

3 3

1 mol Ag CrO0.50 mol AgNO 2 mol AgNO

n=

Solving for

2 4Ag CrOn :

2 4

2 4

2 4

3 3

Ag CrO 2 4

3 3

Ag 2 4

A3

rO

C

g C

rO 2 mol AgNO 0.50 mol AgNO 1 mol Ag

1 mol Ag CrO0.50 mol AgNO 2 mol AgNO

0.50

C

m

rO

ol AgNO

n

n

n

× = ×

=

= 2 4

3

× 1 mol Ag CrO 2 mol AgNO

2 4mol Ag0.25 CrO=

The reaction produces 0.25 mol of silver chromate. Check Your Solution The amount in moles of Ag2CrO4 produced is half the amount of AgNO3. This is the same mole ratio as in the balanced chemical equation. The answer is reasonable and correctly shows two significant digits.


What amount in moles of water forms when 6.00 mol of carbon dioxide is consumed in the following reaction? 2NH3(g) + CO2(g) → NH2CONH2(s) + H2O(g) What Is Required? You need to determine the amount in moles of water that forms. What Is Given? You know the balanced chemical equation for the reaction: 2NH3(g) + CO2(g) → NH2CONH2(s) + H2O(g) You know the amount in moles of carbon dioxide: 6.00 mol



Plan Your Strategy Use the balanced chemical equation to write the mole ratio of water to carbon dioxide. Let

2H On be the unknown amount in moles of H2O(g). Equate the known mole ratio of H2O(g) to CO2(g) to the unknown ratio. Solve for the unknown amount of H2O(g). Act on Your Strategy Balanced chemical equation: 2NH3(g) + CO2(g) → NH2CONH2(s) + H2O(g) Mole ratio: 1 mole 1 mole Known mole ratio equated to the unknown ratio:

2H O 2

2 2

1 mol H O6.00 mol CO 1 mol CO

n=

Solving for

2H On :

2

2

2

2 2 2

2

H O 2

2 2

H O

H O

× 1 mol CO 6.00 mol CO ×

1

1

mol H O6.00 mol CO 1 mol C

mol H O

6.00 mol CO

O

n

n

n

=

=

=

2

2

× 1 mol H O1 mol CO

26.00 mol H O=

Therefore, 6.00 mol of water forms in the reaction. Check Your Solution The units are correct. The ratio 6.00:6.00 is equivalent to the ratio 1:1. The answer is reasonable and correctly shows three significant digits.


Calculate the amount in moles of ammonia, NH3(g), that is needed to prepare 22 500 mol of the fertilizer ammonium sulfate, (NH4)2SO4(s). 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s) What Is Required? You need to determine the amount in moles of ammonia that is needed to prepare a fertilizer.



What Is Given? You know the balanced chemical equation for the reaction: 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s) You know the amount in moles of ammonium sulfate: 22 500 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of ammonia to ammonium sulfate. Let

3NHn be the unknown amount in moles of NH3(g). Equate the known mole ratio of NH3(g) to (NH4)2SO4(s) to the unknown ratio. Solve for the unknown amount in moles of NH3(g). Act on Your Strategy Balanced chemical equation: 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s) Mole ratio: 2 moles 1 mole Known ratio equated to the unknown ratio:

3NH 3

4 2 4 4 2 4

2 mol NH22 500 mol (NH ) SO 1 mol (NH ) SO

n=

Solving for

3NHn :

( )

3

3

3

N

NH 3

4 2 4 4 2 4

4 2 4

4 2 4NH

H 4 4 32

2 mol NH22 500 mol (NH ) SO 1 mol (NH ) SO

(NH 1 mol 22 500 mol NH SO 2 mol) SO

22 500 mol (N

H ) S

NH

O

n

n

n

× = ×

=

= 3

4 2 4

× 2 mol NH 1 mol (NH ) SO

345 000 mol NH=

Therefore, 4.5 × 104 mol of ammonia is needed. Check Your Solution The units are correct. The ratio 45 000:22 500 is equivalent to the ratio 2:1. The answer is reasonable and correctly shows two significant digits.



14. Practice Problem (page 300) Calculate the amount in moles of oxygen that is needed to react with 2.4 mol of ammonia to produce poisonous hydrogen cyanide, HCN(g). 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g) What Is Required? You need to determine the amount in moles of oxygen, O2(g), that is needed to react. What Is Given? You know the balanced chemical equation for the reaction: 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g) You know the amount in moles of ammonia: 2.4 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of oxygen to ammonia. Let

2On be the unknown amount in moles of O2(g). Equate the known mole ratio of O2(g) to NH3(g) to the unknown ratio. Solve for the unknown amount in moles of O2(g). Act on Your Strategy Balanced chemical equation: 2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g) Mole ratio: 2 moles 3 moles Known mole ratio equated to the unknown ratio:

2

3

O 2

3

3 mol O 2.4 mol NH 2 mol NH

n=

Solving for

2On :

2

2

2

O

O 3

2

3 3

3

3 2

O

2 mol NH 2.4 mol NH

3 mol O 2.4 m

3 mol Ool NH 2 mol NH

2.4 mol NH

n

n

n

=

× = ×

= 2

3

3 mol O 2 mol NH

×

23.6 mol O =

Therefore, 3.6 mol of oxygen is needed.



Check Your Solution The units are correct. The ratio 3.6:2.4 is equivalent to the ratio 3:2. The answer is reasonable and correctly shows two significant digits.


What amount in moles of fluorine, F2(g), yields 2.35 mol of xenon tetrafluoride, XeF4(s)? Xe(g) + 2F2(g) → XeFe4(s) What Is Required? You need to determine the amount in moles of fluorine, F2(g), that is needed to react. What Is Given? You know the balanced chemical equation for the reaction: Xe(g) + 2F2(g) → XeFe4(s) You know the amount in moles of xenon tetrafluoride: 2.35 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of fluorine to xenon tetrafluoride. Let

2Fn be the unknown amount in moles of F2(g). Equate the known mole ratio of F2(g) to XeFe4(s) to the unknown ratio. Solve for the unknown amount in moles of F2(g). Act on Your Strategy Balanced chemical equation: Xe(g) + 2F2(g) → XeFe4(s) Mole ratio: 2 moles 1 mole Known mole ratio equated to the unknown ratio:

2F 2

4 4

2 mol F2.35 mol XeF 1 mol XeF

n=

Solving for

2Fn :

2

2

2

F 2

4 4

4 4 2

4

F

F

× 1 mol XeF 2.35 mol XeF

2 mol F2.35

× 2 mol Fmol XeF 1 mol

2.35 mol XeF

XeFn

n

n =

=

=

2

4

× 2 mol F1 mol XeF

24.70 mol F=



Therefore, 4.70 mol of fluorine is needed. Check Your Solution The units are correct. The ratio 4.70:2.35 is equivalent to the ratio 2:1. The answer is reasonable and correctly shows three significant digits.


These equations show two possible reactions: 2N2(g) + O2(g) → 2N2O(g) N2(g) + 2O2(g) → 2NO2(g) a. What amount in moles of oxygen reacts with 93.5 mol of nitrogen to form dinitrogen monoxide, N2O(g)? b. What amount in moles of nitrogen dioxide, NO2(g), forms in the other reaction? a. amount in moles of oxygen What Is Required? You need to determine the amount in moles of oxygen that is needed to react. What Is Given? You know the balanced chemical equation for the reaction: 2N2(g) + O2(g) → 2N2O(g) You know the amount in moles of nitrogen: 93.5 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of oxygen to nitrogen. Let

2On be the unknown amount in moles of O2(g) . Equate the known mole ratio of O2(g) to N2(g) to the unknown ratio. Solve for the unknown amount in moles of O2(g). Act on Your Strategy Balanced chemical equation: 2N2(g) + O2(g) → 2N2O(g) Mole ratio: 2 moles 1 mole Known mole ratio equated to the unknown ratio:

2O 2

2 2

1 mol O93.5 mol N 2 m

ol N

n=

Solving for

2On :



2

2

2O 2

O

2 2

O

2 2 2

2

2 mol N 93.5 mol

1 mol O93.5

N 1 mol Omol N 2 mol

93.5 mol N

Nn

n

n

=

=

× = ×

2

2

1 mol O2 mol N

×

2

2

mol Omol O

46.75 46.8 =

=

Therefore, 46.8 mol of oxygen is needed. Check Your Solution The units are correct. Rounded, the ratio 46.8:93.5 is equivalent to the ratio 1:2. The answer is reasonable and correctly shows three significant digits. b. amount in moles of nitrogen dioxide What Is Required? You need to determine the amount in moles of nitrogen dioxide that is formed. What Is Given? You know the balanced chemical equation for the reaction: N2(g) + 2O2(g) → 2NO2(g) You know the amount in moles of nitrogen: 93.5 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of nitrogen to nitrogen dioxide. Let

2NOn be the unknown amount in moles of NO2(g) . Equate the known mole ratio of NO2(g) to N2(g) to the unknown ratio. Solve for the unknown amount in moles of NO2(g). Act on Your Strategy Balanced chemical equation: N2(g) + 2O2(g) → 2NO2(g) Mole ratio: 1 mole 2 moles Known mole ratio equated to the unknown ratio:

2NO 2

2 2

2 mol NO93.5 mol N 1 m

ol N

n=

Solving for

2NOn :



2

2

2

2 2 2

2

NO 2

2 2

NO

NO

1 mol N 93.5 m

2 mol NO93.5 m

ol N 2 mol NOol N 1 mol

93.5 mol

N

N

n

n

n =

×

=

= ×

2

2

2 mol NO2 mol N

×

2mol187 NO=

Therefore, 187 mol of nitrogen dioxide is produced.

Check Your Solution The units are correct. The unknown ratio solves to match the ratio in the balanced chemical equation for each result. The answers are reasonable and correctly show three significant digits.


What amount in moles of oxygen reacts with 11.3 mol of propane gas, C3H8(g), during the combustion of propane? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) What Is Required? You need to determine the amount in moles of oxygen that is needed to react. What Is Given? You know the balanced chemical equation for the reaction: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) You know the amount in moles of propane: 11.3 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of oxygen to propane. Let

2On be the unknown amount in moles of O2(g). Equate the known mole ratio of O2(g) to C3H8(g) to the unknown ratio. Solve for the unknown amount in moles of O2(g). Act on Your Strategy Balanced chemical equation: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Mole ratio: 1 mole 5 moles Known mole ratio equated to the unknown ratio:

2O 2

3 8 3 8

5 mol O 11.3 mol C H 1 mol C H

n=



Solving for O2n :

2

2

2

O 2

3

3 8 3 8 2

3

8 8

O

O8

3

× 1 mol C H 11.3 mol

5 mol O 11.3

C H × 5 mol O

11.3 mol C

mol C H 1 mol C H

H

n

n

n =

=

=

2

3 8

× 5 mol O1 mol C H

2mo56.5 l O=

Therefore, 56.5 mol of oxygen is needed. Check Your Solution The units are correct. The rounded ratio of 56.5:11.3 is equivalent to the ratio 5:1. The answer is reasonable and correctly shows three significant digits.


What amount in moles of phosphorus produces 6.45 mol of tetraphosphorus hexoxide, P4O6(s)? P4(s) + 3O2(g) → P4O6(s)e Pro What Is Required? You need to determine the amount in moles of phosphorus. What Is Given? You know the balanced chemical equation for the reaction: P4(s) + 3O2(g) → P4O6(s)e Pro You know the amount in moles of tetraphosphorus hexoxide: 6.45 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of phosphorus to tetraphosphorus hexoxide. Let

4Pn be the unknown amount in moles of P4(s). Equate the known mole ratio of P4(s) to P4O6(s) to the unknown ratio. Solve for the unknown amount in moles of P4(s). Act on Your Strategy Balanced chemical equation: P4(s) + 3O2(g) → P4O6(s)e Pro Mole ratio: 1 mole 1 mole



Known mole ratio equated to the unknown ratio: 4P 4

4 6 4 6

1 mol P 6.45 mol P O 1 mol P O

n=

Solving for

4Pn :

4

4

4

4 1

P 4

4 6 4 6

0 4 10

4 1P

4P

0

1 mol P O 6.45 mol

1 mol of P 6.45 mo

P O × 1 mol l P O 1 mol P O

P

6.45 mol P O

n

n

n

=

=

× =

4

4 10

× 1 mol P1 mol P O

46.45 mol of P=

Therefore, 6.45 mol of phosphorus is needed. Check Your Solution The units are correct. The ratio 6.45:6.45 is equivalent to the ratio 1:1. The answer is reasonable and correctly shows three significant digits.


Silver tarnishes when it is exposed to small amounts of hydrogen sulfide, H2S(g), in the air. 4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s) + 2H2O(ℓ) How many molecules of hydrogen sulfide react with 1.7 mol of silver? What Is Required? You need to determine the number of molecules of hydrogen sulfide that react. What Is Given? You know the balanced chemical equation for the reaction: 4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s) + 2H2O(ℓ) You know the amount in moles of silver: 1.7 mol Plan Your Strategy Use the balanced chemical equation to write the mole ratio of silver to hydrogen sulfide. Let

2H Sn be the unknown amount in moles of H2S(g). Equate the known mole ratio of H2S(g) to Ag(s) to the unknown ratio. Solve for the unknown amount in moles of H2S(g).



Use Avogadro’s constant: NA = 6.02 × 1023 molecules/mol Calculate the number of molecules, N, of H2S(g) using

2H S A N n N= × .

Act on Your Strategy Balanced chemical equation: 4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s) + 2H2O(ℓ) Mole ratio: 4 moles 2 moles Known mole ratio equated to the unknown ratio:

2H S 22 mol H S1 .7 mol Ag 4 mol Ag

n=

Solving for

2H Sn :

2

2

2

2

H S 2

H S

H S

4 mol Ag =1.7 mol Ag

2 mol H S1 .7 mol Ag 4 mol Ag

× 2 mol H S

1.7 mol Ag

n

n

n

×

=

= 2× 2 mol H S4 mol of Ag

20.85 mol H S=

Number, N, of molecules of H2S(g):

2H S A

0.85 mol

N n N= ×

= 23 6.02 10 molecules/ mol× ×23

23

5.117 10 molecules5.1 10 molecules

= ×

= ×

Therefore, 5.1 × 1023 molecules of hydrogen sulfide are needed. Check Your Solution The units are correct. The answer is reasonable and correctly shows two significant digits.



20. Practice Problem (page 301) When heated, magnesium hydrogen carbonate, Mg(HCO3)2(s), decomposes and forms magnesium carbonate, MgCO3(s), carbon dioxide, and water vapour. What amount in moles of water is produced from 7.24 × 105 mol of magnesium hydrogen carbonate? What Is Required? You need to determine the amount in moles of water that is produced when magnesium hydrogen carbonate decomposes. What Is Given? You know the reactant: magnesium hydrogen carbonate You know the products: magnesium carbonate, carbon dioxide, and water vapour. Plan Your Strategy Write the balanced chemical equation for this decomposition. Determine the mole ratio from the balanced equation. Use the balanced chemical equation to write the mole ratio of water to magnesium hydrogen carbonate. Let

2H On be the unknown amount in moles of water. Equate the known mole ratio of H2O(g) to Mg(HCO3)2 to the unknown ratio. Solve for the unknown amount in moles of H2O(g). Act on Your Strategy Balanced chemical equation: Mg(HCO3)2 → MgCO3(s) + CO2(g) + H2O(g) Mole ratio: 1 mole 1 mole Known mole ratio equated to the unknown ratio:

2 H O 25

3 2 3 2

1 mol H O7.24 × 10 mol Mg(HCO ) 1 mol Mg(HCO )

n=

Solving for

2H On :

( ) ( )( )

2

2

2

53 3 22

H O 25

3 2 3 2

H 2

53 2

O

H O

1 mol Mg HCO 7.24 10 mol Mg HCO

1 mol H O7.24 × 10

1 mol H O

mol Mg(HCO ) 1 mol Mg(HCO )

7.24 10 mol Mg HCO

n

n

n×

=

× = × ×

=

( )2

3 2

1 mol H O

1 mol Mg HCO

×

527.24 10 mol H O = ×



Therefore, 7.24 × 105 mol of water is produced. Check Your Solution The units are correct. The ratio 7.24 × 105:7.24 × 105 is equivalent to the ratio 1:1. The answer is reasonable and correctly shows three significant digits.

Section 7.1 What Is Stoichiometry? Solutions for Practice Problems Student Edition page 304

21. Practice Problem (page 304) The production of acetic acid, CH3COOH(ℓ), is represented by the following chemical equation: CH3OH(ℓ) + CO(g) → CH3COOH(ℓ) Calculate the mass of acetic acid that is produced by the reaction of 6.0 × 104 g of carbon monoxide with sufficient methanol, CH3OH(ℓ). What Is Required? You need to determine the mass of acetic acid that is produced. What Is Given? You know the balanced chemical equation for the reaction: CH3OH(ℓ) + CO(g) → CH3COOH(ℓ) You know the mass of the carbon monoxide, CO(g): 6.0 × 104 g Plan Your Strategy Determine the mole ratio from the balanced chemical equation. Calculate the molar mass, M, of CO(g).

Calculate the amount in moles of the CO(g) using mnM

= .

Use the mole ratio in the balanced equation and the calculated amount in moles of CO(g) to determine the amount in moles of CH3COOH(ℓ). Determine the molar mass, M, of CH3COOH(ℓ). Calculate the mass of the CH3COOH(ℓ) using m n M= × . Act on Your Strategy Balanced chemical equation: CH3OH(ℓ) + CO(g) → CH3COOH(ℓ) Mole ratio: 1 mole 1 mole



Molar mass, M, of CO(g): CO C O1 1

1(12.01 g/mol) + 1(16.00 g/mol)28.01 g/mol

M M M= +

==

Amount in moles, n, of CO(g):

4

CO

6.0 × 10 g

mnM

=

=28.01 g

2142.092

/m

l

ol

mo= From the balanced chemical equation, the mole ratio of CO(g) to CH3COOH(ℓ) is 1:1, so the amount in moles of CH3COOH(ℓ) is also 2142.092 mol. Molar mass, M, of CH3COOH(ℓ):

3CH COOH C H O2 4 2

2(12.01 g/mol) + 4(1.01 g/mol) + 2(16.00 g/mol)60.06 g/mol

M M M M= + +

==

Mass, m, of the CH3COOH(ℓ):

3CH COOH

2214.2092 mol

m n M= ×

= 60.06 g/ mol×5

5

1.28654 10 g1.3 10 g

= ×

= ×

The mass of acetic acid produced is 1.3 × 105 g. Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of acetic acid.

3

4CH COOH 6.0 × 10 g COm =

1 mo × l CO28.01 g CO

31 mol CH C OOH×

1 mol CO

3

3

60.06 g CH COOH ×1 mol CH COOH

51.3 × 10 g=



22. Practice Problem (page 304) Calculate the mass of silver nitrate, AgNO3(aq), that must react with solid copper to provide 475 kg of copper nitrate, Cu(NO3)2(aq). Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq) What Is Required? You need to determine the mass of silver nitrate that reacts. What Is Given? You know the balanced chemical equation for the reaction: Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq) You know the mass of copper(II) nitrate: 475 kg Plan Your Strategy Determine the mole ratio from the balanced chemical equation. Calculate the molar mass, M, of Cu(NO3)2(aq). Convert the mass of Cu(NO3)2(aq) from kilograms to grams: 1 kg = 1 × 103 g

Calculate the amount in moles of the Cu(NO3)2(aq) using mnM

= .

Use the mole ratio in the balanced equation and the calculated amount in moles of Cu(NO3)2(aq) to determine the amount in moles of AgNO3(aq). Determine the molar mass, M, of AgNO3(aq). Calculate the mass of the AgNO3(aq) using m n M= × . Convert the mass from grams to kilograms: 1 g = 1 × 10–3 kg Act on Your Strategy Balanced chemical equation: Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq) Mole ratio: 2 moles 1 mole Molar mass, M, of Cu(NO3)2(aq):

( ) 33 2Cu NOCu NO

Cu N O

1 2

1 2[1( ) 3( )]1(63/55 g/mol) + 2[1(14.01 g/mol) + 3(16.00 g/mol)]187.57 g/mol

M M M

M M M

= +

= + +

==

Mass (in kilograms), m, of the Cu(NO3)2(aq):

( )3 2Cu NO 475 kg m = 3× 1 × 10 g/ kg54.75 10 g= ×



Amount in moles, n, of Cu(NO3)2(aq):

( )3 2

5

Cu NO

4.75 × 10 g187.57 g/mol2532.387 mol

mnM

=

=

=

Known mole ratio equated to the unknown ratio:

3AgNO 3

3 2 3 2

2 mol AgNO 2532.387 mol Cu(NO ) 1 mol Cu(NO )

n=

Amount in moles, n, of silver nitrate:

( ) ( )

3

3

3

3

AgNO 3

3 2 3 2

AgNO

3 2AgNO

3 32 2 1 mol Cu NO 2532.387 mol Cu NO

2 mo

2 mo

l AgNO 2532.387 mol Cu(NO ) 1 mol Cu(NO )

2532.387 mo

l

l

Ag

Cu(NO )

NO

n

n

n

×

=

= ×

= 3

3 2

× 2 mol AgNO 1 mol Cu(NO )

35064.7758 Amol gNO=

Molar mass, M, of AgNO3(aq):

3AgNO Ag N O1 1 3


M M M M= + +

==

Mass, m, of the AgNO3(aq):

3AgNO

5064.7758 mol

m n M= ×

= × 169.88 g/ mol58.6040 10 g= ×

Mass (in grams), m, of the AgNO3(aq):

3

5AgNO 8.6040 × 10 gm = –3 × 1 × 10 kg/ g

860 kg=

The mass of silver nitrate that reacts is 860 kg, Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of silver nitrate.



3

3AgNO 475 kg × 10 gm = / kg ( )3 2

Cu NO 3 21 mol Cu(N

O

)×

3 2187.57 g Cu(NO )

32 mol AgNO ×

3 21 mol Cu(NO )3169.88 g AgNO

× 3

1 mol AgNO

–3 1 × 10 × kg / g

860 kg=


What mass of oxygen is produced if 22.7 mol of carbon dioxide is consumed in a controlled photosynthesis reaction? 6CO2(g) + 6H2O(ℓ) → C6H12O6(s) + 6O2(g) What Is Required? You need to determine the mass of oxygen produced. What Is Given? You know the balanced chemical equation for the reaction: 6CO2(g) + 6H2O(ℓ) → C6H12O6(s) + 6O2(g) You know the amount in moles of carbon dioxide: 22.7 mol Plan Your Strategy Use the mole ratio in the balanced chemical equation and the given amount in moles of CO2(g) to calculate the amount in moles of O2(g). Determine the molar mass, M, of O2(g). Calculate the mass of O2(g) using the relationship m n M= × . Act Your Strategy Balanced chemical equation: 6CO2(g) + 6H2O(ℓ) → C6H12O6(s) + 6O2(g) Mole ratio: 6 moles 6 moles Known mole ratio equated to the unknown ratio:

2O 2

2 2

6 mol O22.7 mol CO 6 mol CO

n=



Amount in moles, n, of O2(g): 2

2

2

2O

O

2 2

2

2 2 2

O

6 mol CO 22.7 mol CO

6 mol O22.7

6 mol O mol CO 6 mol CO

22.7 mol CO

n

n

n

×

=

= ×

=

2

2

× 6 mol O 6 mol CO

222.7 mol O=

Molar mass, M, of O2(g):

2O O2

2(16.00 g/mol)32.00 g/mol

M M=

==

Mass, m, of the O2(g):

2O

22.7 mol

m n M= ×

= 32.00 g/ mol×726.4 g726 g

==

The mass of oxygen produced is 726 g. Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of oxygen.

2O 222.7 mol COm = 26 mol O×

26 mol CO2

2

32.00 g O ×1 mol O

726 g=


Sodium phosphate, Na3PO4(aq), is an all-purpose cleaner that can be used to clean walls before painting. It is often referred to as trisodium phosphate, or TSP, and it must be handled with care because it is corrosive. It is prepared by the following reaction: 3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3H2O(ℓ) What amount in moles of TSP is produced if 14.7 g of sodium hydroxide reacts with phosphoric acid, H3PO4(aq)?



What Is Required? You need to determine the amount in moles of triosodium phosphate produced in a reaction. What Is Given? You know the balanced chemical equation for the reaction: 3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3H2O(ℓ) You know the mass of sodium hydroxide, NaOH(aq): 14.7 g Plan Your Strategy Calculate the molar mass, M, of NaOH(aq).

Calculate the amount in moles of NaOH(aq) using the relationship mnM

= .

Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced chemical equation and the calculated amount in moles of NaOH(aq) to calculate the amount in moles of Na3PO4(aq). Act on Your Strategy Molar mass, M, of NaOH(aq):

NaOH N O H1 1 11(22.99 g/mol) + 1(16.00 g/mol) + 1(1.01 g/mol)40.00 g/mol

M M M M= + +==

Amount in moles, n, of NaOH(aq):

NaOH

14.7 g

mnM

=

=40.00 g /mol

0.3675 mol=

Balanced chemical equation: 3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3H2O(ℓ) Mole ratio: 3 moles 1 mole Known mole ratio equated to the unknown ratio:

3 4Na PO 3 41 mol Na PO0.3675 mol NaOH 3 mol NaOH

n=



Amount in moles, n, of Na3PO4(aq): 3 4

3 4

3 4

3

Na PO 3 4

Na PO

Na PO

4

3 mol NaOH 0.3675 mol NaOH 1

1 mol Na PO0.3675 mol NaOH 3 mol NaOH

0.3675 mol NaOH

mol Na PO

n

n

n

× = ×

=

= 3 4× 1 mol Na PO3 mol NaOH

3 4

3 4

0.1225 mol Na PO0.122 mol Na PO

==

The amount in moles of trisodium phosphate is 0.122 mol. Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the amount in moles of trisodium phosphate.

3 4Na PO 14.7 g NaOHm =1 mol × NaOH

40.00 g NaOH3 41 mol Na PO ×

3 mol NaOH

0.122 mol =


What mass of hydrogen is produced when 3.75 g of aluminum reacts with sulfuric acid, H2SO4(aq)? 2Al(s) + 3H2SO4(aq) → 3H2(g) + Al2(SO4)3(aq) What Is Required? You need to determine the mass of hydrogen produced in a reaction. What Is Given? You know the balanced chemical equation for the reaction: 2Al(s) + 3H2SO4(aq) → 3H2(g) + Al2(SO4)3(aq) You know the mass of aluminum: 3.75 g Plan Your Strategy Use the periodic table to find the atomic molar mass of Al(s).

Calculate the amount in moles of Al(s) using the relationship mnM

= .

Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of Al(s) to calculate the amount in moles of H2(g). Determine the molar mass, M, of H2(g). Calculate the mass of the H2(g) using the relationship m n M= × .



Act on Your Strategy Molar mass, M, of Al(s): M = 26.98 g/mol (from the periodic table) Amount in moles, n, of Al(s):

Al

3.75 g

mnM

=

=26.98 g

0.13899

/mo

l

l

mo=

Balanced chemical equation: 2Al(s) + 3H2SO4(aq) → 3H2(g) + Al2(SO4)3(aq) Mole ratio: 2 moles 3 moles Known mole ratio equated to the unknown ratio:

2H 23 mol H0.13899 mol Al 2 mol Al

n=

Amount in moles, n, of H2(g):

2

2

2

H 2

H

H 2

2 mol Al 0.13899 mol

3 mol H0.13899 mol Al 2 mol A

Al 3 mol l

0.13899 mol A

H

l

n

n

n

=

=

=

× ×

2× 3 mol H2 mol Al

20.208487 mol H =

Molar mass, m, of H2(g):

2H H2

2(1.01 g/mol)2.02 g/mol

M M=

==

Mass, m, of the H2(g):

2H

0.208487 mol

m n M= ×

= × 2.02 g/ mol0.421145 g0.421 g

==

The mass of hydrogen produced is 0.421 g.



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of hydrogen.

2H 3.75 g Alm =1 m × ol Al

26.98 g Al23 mol H

×

2 mol Al2

2

2.02 g H × 1 mol H

0.421 g=


Nitrogen monoxide, NO(g), reacts with oxygen gas to form nitrogen dioxide, NO2(g). What mass of nitrogen dioxide is produced from 2.84 g of nitrogen monoxide? What Is Required? You need to determine the mass of nitrogen dioxide produced in a reaction. What Is Given? You know the formula for the reactant, nitrogen monoxide: NO You know the formula for the product, nitrogen dioxide: NO2 You know the mass of nitrogen monoxide: 2.84 g Plan Your Strategy Calculate the molar mass, M, of NO(g).

Calculate the amount in moles of NO(g) using the relationship mnM

= .

Write the balanced chemical equation for the reaction. Determine the mole ratio from the balanced equation. Use the mole ratio in the balanced equation and the calculated amount in moles of NO(g) to calculate the amount in moles of NO2(g). Determine the molar mass, M, of NO2(g). Calculate the mass of the NO2(g) using the relationship m n M= × . Act on Your Strategy Molar mass, M, of NO(g):

NO N O1 11(14.01 g/mol) + 1(16.00 g/mol)30.01 g/mol

M M M= +==



Amount in moles, n, of NO(g):

NO

2.84 g

mnM

=

=30.01 g

0.094635

/mol

1 mol=

Balanced chemical equation: 2NO(g) + O2(g) → 2NO2(g) Mole ratio: 2 moles 2 moles Known mole ratio equated to the unknown ratio:

2NO 22 mol NO 0.0946351 mol NO 2 mol NO

n=

Amount in moles, n, of NO2(g):

2

2

2

N

NO

O 2

NO

2

2 mol NO 0.0946351 mol NO

2 mol NO 0.0946351 mol NO 2 mol NO

0.0946351 mol

2 mol

N

O

O

N

n

n

n

× = ×

=

=

2× 2 mol NO 2 mol NO

20.0946351 mol NO=

Molar mass, M, of NO2(g):

2NO N O1 2

1(14.01 g/mol) + 2(16.00 g/mol)46.01 g/mol

M M M= +

==

Mass, m, of the NO2(g):

2NO

0.0946351 mol

m n M= ×

= × 46.01 g/ mol4.354161 g4.35 g

==

The mass of nitrogen dioxide produced is 4.35 g



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of nitrogen dioxide.

2NO 2.84 g NOm =1 m × ol NO

30.01 g NO22 mol NO

×

2 mol NO2

2

46.01 g NO × 1 mol NO

4.35 g=


Iron(III) oxide, Fe2O3(s), reacts with carbon monoxide to form solid iron and carbon dioxide in the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) What mass (in grams) of carbon dioxide is produced from 12.4 g of iron(III) oxide? What Is Required? You need to determine the mass of carbon dioxide produced in a reaction. What Is Given? You know the balanced chemical equation for the reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) You know the mass of iron(III) oxide: 12.4 g Plan Your Strategy Calculate the molar mass of Fe2O3(s).

Calculate the amount in moles of Fe2O3(s) using the relationship mnM

= .

Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of Fe2O3(s) to calculate the amount in moles of the CO2(g). Determine the molar mass of CO2(g). Calculate the mass (in grams) of the CO2(g) using the relationship m n M= × . Act on Your Strategy Molar mass, M, of Fe2O3(s):

2 3Fe O Fe O2 3

2(55.85 g/mol) + 3(16.00 g/mol)159.7 g/mol

M M M= +

==



Amount in moles, n, of Fe2O3(s):

2 3Fe O

12.4 g

mnM

=

=159.7 g

0.0776455

/m

l

ol

mo=

Balanced chemical equation: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Mole ratio: 1 mole 3 moles Known mole ratio equated to the unknown ratio:

2CO 2

2 3 2 3

3 mol CO =0.0776455 mol Fe O 1 mol Fe O

n

Amount in moles, n, of CO2(g):

2

2

2

CO 2

CO 2

2 3 2

3 2 3 2

CO

3

2 3

1 mol Fe O 0.0776455 mol Fe O

3 mol CO =0.0776455

3 mol mol Fe O 1 mol Fe O

0.0776455 mol Fe O

COn

n

n

× = ×

= 2

2 3

× 3 mol CO 1 mol Fe O

20.232936 mol CO=

Molar mass, M, of CO2(g):

2CO C O1 2

1(12.01 g/mol) + 2(16.00 g/mol) 44.01 g/mol

M M M= +

==

Mass, m, of the CO2(g):

2CO

0.232936 mol

m n M= ×

= × 44.01 g/ mol10.2725 g10.3 g

==

The mass of carbon dioxide produced is 10.3 g



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of carbon dioxide.

2CO 2 312.4 g Fe Om = 2 31 mol

Fe

O×

2 3159.7 g Fe O23 mol CO

× 2 31 mol Fe O

2

2

44.01 g CO × 1 mol CO

10.3 g=


Methane, CH4(g), reacts with sulfur, S8(s), to produce carbon disulfide, CS2(ℓ), and hydrogen sulfide, H2S(g). Carbon disulfide is often used in the production of cellophane. 2CH4(g) + S8(s) → 2CS2(ℓ) + 4H2S(g) What mass of methane is required if 4.09 g of hydrogen sulfide is produced? What Is Required? You need to determine the mass of methane required for a reaction. What Is Given? You know the balanced chemical equation for the reaction: 2CH4(g) + S8(s) → 2CS2(ℓ) + 4H2S(g) You know the mass of hydrogen sulfide: 4.09 g Plan Your Strategy Calculate the molar mass of H2S(g).

Calculate the amount in moles of the H2S(g) using the relationship .mnM

=

Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of the H2S(g) to calculate the amount in moles of the CH4(g). Determine the molar mass of CH4(g). Calculate the mass of the CH4(g) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of H2S(g):

2H S H S2 1

1(1.01 g/mol) + 1(32.07 g/mol)34.09 g/mol

M M M= +

==



Amount in moles, n, of the H2S(g):

2H S

4.09 g

mnM

=

=34.09 g

0.119976 m

/ lmo

ol =

Balanced chemical equation: 2CH4(g) + S8(s) → 2CS2(ℓ) + 4H2S(g) Mole ratio: 2 moles 4 moles Known mole ratio equated to the unknown ratio:

4CH 4

2 2

2 mol CH 0.119976 mol H S 4 mol H S

n=

Amount in moles, n, of CH4(g):

4

4

4

2

CH 4

2 2

2 4CH

2CH

4 mol H S 0.119976 mo

2 mol CH 0.119976 mol H S 4 mol

l H S 2 mol C H S

0.119976 mol H S

H

n

n

n

× =

=

×

=

4

2

× 2 mol CH 4 mol H S

40.0599882 mol CH =

Molar mass, M, of CH4(g):

4CH C H1 4

1(12.01 g/mol) + 4(1.01 g/mol)16.05 g/mol

M M M= +

==

Mass, m, of the CH4(g):

4CH

0.0599882 mol

m n M= ×

= × 16.05 g/ mol0.96281 g0.963 g

==

The mass of methane that reacts is 0.963 g.



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of methane.

4CH 24.09 g H Sm = 21 mol H×

S

2

34.09 g H S

42 mol CH×

2

4 mol H S

4

4

16.05 g CH ×1 mol CH

0.963 g=


The addition of concentrated hydrochloric acid to manganese(IV) oxide, MnO2(s), produces chlorine gas, Cl2(g). 4HCl(aq) + MnO2(s) → MnCl2(aq) + Cl2(g) + 2H2O(ℓ) What mass of manganese(IV) oxide is needed to react with 8.65 × 10−2 g of hydrochloric acid? What Is Required? You need to determine the mass of manganese(IV) oxide required for a reaction. What Is Given? You know the balanced chemical equation for the reaction: 4HCl(aq) + MnO2(s) → MnCl2(aq) + Cl2(g) + 2H2O(ℓ) You know the mass of hydrochloric acid: 8.65 × 10−2 g Plan Your Strategy Calculate the molar mass of HCl(aq).

Calculate the amount in moles of HCl(aq) using the relationship .mnM

=

Determine the mole ratio from the balanced chemical equation. Using the mole ratio in the balanced chemical equation and the calculated amount in moles of the HCl(aq), calculate the amount in moles of the MnO2(s). Determine the molar mass of MnO2(s). Calculate the mass of the MnO2(s) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of HCl(aq):

HCl H Cl1 11(1.01 g/mol) + 1(35.45g/mol)36.46 g/mol

M M M= +==



Amount in moles, n, of the HCl(aq):

2

HCl

8.65 × 10 g

mnM

−

=

=36.46 g

–32.372462 10

/m

l

ol

mo= ×

Balanced chemical equation: 4HCl(aq) + MnO2(s) → MnCl2(aq) + Cl2(g) + 2H2O(ℓ) Mole ratio: 4 moles 1 mole Known mole ratio equated to the unknown ratio:

2MnO 23

1 mol MnO 2.372462 × 10 mol HCl 4 mol HCl

n− =

Amount in moles, n, of the MnO2(s):

2

2

2

MnO 23

MnO

3

M

–

n

32

O

× 4 mol HCl 2.372462 × 10 mol HCl

1 mol MnO 2.372462 × 10 mol HCl 4 mol HCl

2.372462 × 10 mol H

× 1 m

C

ol O

l

Mn

n

n

n

−

−

=

=

= 2× 1 mol MnO 4 mol HCl

–425.93115 10 mol MnO= ×

Molar mass, M, of MnO2(s):

2MnO N O1 2

1(54.94 g/mol) + 2(16.00 g/mol) 86.94 g/mol

M M M= +

==

Mass, m, of the MnO2(s):

2MnO

–4

5.93115 10 mol

m n M= ×

= × 86.94 g/ mol×–2

–2

5.15654 10 g5.16 10 g

= ×

= ×

The mass of manganese(IV) oxide is 5.16 × 10–2 g.



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of manganese(IV) oxide.

2

2MnO 8.65 × 10 g HClm −=

1 mo × l HCl 36.46 g HCl

21 mol MnO ×

4 mol HCl

2

2

86.94 g MnO × 1 mol MnO

–25.16 10 g×=


Aluminum carbide, Al4C3(s), is a yellow powder that reacts with water, H2O(ℓ), to produce aluminum hydroxide, Al(OH)3(s), and methane, CH4(g). Write a balanced chemical equation for the reaction and determine the mass of water required to react with 14.0 g of aluminum carbide. What Is Required? You need to write a balanced chemical equation for the reaction and calculate the mass of water that reacts. What Is Given? You know the formulas for the reactants: aluminum carbide, Al4C3(s); water, H2O(ℓ) You know the formulas for the products: aluminum hydroxide, Al(OH)3(s); methane, CH4(g) You know the mass of the aluminum carbide: 14.0 g Plan Your Strategy Write the balanced chemical equation for the reaction. Calculate the molar mass of Al4C3(s). Calculate the amount in moles of the Al4C3(s). Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of Al4C3(s) to calculate the amount in moles of the H2O(ℓ). Determine the molar mass of H2O(ℓ). Calculate the mass of the H2O(ℓ) using the relationship .m n M= × Act on Your Strategy Balanced chemical equation: Al4C3(s) + 12H2O(ℓ) → 4Al(OH)3(s) + 3CH4(g) Molar mass, M, of Al4C3(s):

4 3Al C Al C4 3

4(26.98 g/mol) + 3(12.01 g/mol) 143.95 g/mol

M M M= +

==



Amount in moles, n, of the Al4C3(s):

4 3Al C

14.0 g

mnM

=

=143.95 g

0.0972559

/m

l

ol

mo=

Balanced chemical equation: Al4C3(s) + 12H2O(ℓ) → 4Al(OH)3(s) + 3CH4(g) Mole ratio: 1 mole 12 moles Known mole ratio equated to the unknown ratio:

2H O 2

4 3 4 3

12 mol H O 0.0946351 mol Al C 1 mol Al C

n=

Amount in moles, n, of the H2O(ℓ):

2

2

2H O 2

4 3 4 3

4 3

H O 4 3 4 3 2

H O

× 1 mol Al C 0.0972559 mol Al C

12

12

mol H O 0.0946351 mol Al C 1 mol Al C

0.097255

mol H

9 mol Al C

On

n

n=

= ×

= 2

4 3

× 12 mol H O 1 mol Al C

1.16707 mol=

Molar mass, M, of H2O(ℓ):

2H O H O2 1

2(1.01 g/mol) + 1(16.00 g/mol) 18.02 g/mol

M M M= +

==

Mass, m, of the H2O(ℓ):

2H O

1.16707 mol

m n M= ×

= 18.02 g/ mol×21.03063 g21.0 g

==

The mass of water that reacts is 21.0 g



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of water.

2 4H O 314.0 g Al Cm = 4 31 mol

Al

C×

4 3

143.95 g Al C

212 mol H O ×

4 3

1 mol Al C

2

2

18.02 g H O × 1 mol H O

21.0 g=

Section 7.1 What Is Stoichiometry? Solutions for Selected Review Questions Student Edition page 305 4. Review Question (page 305)

The oxidation of aluminum is represented by the following chemical equation: 4Al(s) + 3O2(g) → 2Al2O3(aq) What mass of oxygen is required to oxidize 25 mol of aluminum? What Is Required? You need to determine the mass of oxygen that reacts. What Is Given? You know the balanced chemical equation: 4Al(s) + 3O2(g) → 2Al2O3(aq) You know the amount in moles of aluminum: 25 mol Plan Your Strategy Use the mole ratio in the balanced chemical equation and the given amount in moles of Al(s) to calculate the amount in moles of O2(g). Determine the molar mass of O2(g). Calculate the mass (in grams) of O2(g) using the relationship .m n M= × Act Your Strategy Balanced chemical equation: 4Al(s) + 3O2(g) → 2Al2O3(aq) Mole ratio: 4 moles 3 moles



Amount in moles, n, of O2(g): 2

2

2

2

O 2

O

O

4 mol Al 25 mol Al

3 mol O25 mol Al 4 mol Al

3 mol

25 mol Al

O

n

n

n

=

=

× = ×

2× 3 mol O 4 mol Al

218.75 mol O =


2O O2

2(16.00 g/mol) 32.00 g/mol

M M=

==

Mass, m, of the O2(g):

2O

18.75 mol

m n M= ×

= × 32.00 g/ mol

2

600 g6.0 10 g

=

= ×

The mass of oxygen produced is 6.0 × 102 g. Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of oxygen.

2O 25 mol Alm = 23 m

ol O×

4 mol Al2

2

32.00 g O × 1 mol O

226.0 10 O g ×=

5. Review Question (page 305)

The reaction of nitrogen gas with hydrogen gas is represented by the following chemical equation: N2(g) + 3H2(g) → 2NH3(g) What mass (in grams) of nitrogen reacts with 6.0 g of hydrogen? What Is Required? You need to determine the mass (in grams) of nitrogen used in a reaction.



What Is Given? You know the balanced chemical equation for the reaction: N2(g) + 3H2(g) → 2NH3(g) You know the mass of hydrogen: 6.0 g Plan Your Strategy Calculate the molar mass of H2(g). Calculate the amount in moles of H2(g). Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of H2(g) to calculate the amount in moles of N2(g). Determine the molar mass of N2(g) Calculate the mass (in grams) of N2(g) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of H2(g):

2H H2

2(1.01 g/mol) 2.01 g/mol

M M=

==

Amount in moles, n, of H2(g):

6.0 g

mnM

=

=2.02 g

2.97029

/mo

l

l

mo=

Balanced chemical equation: N2(g) + 3H2(g) → 2NH3(g) Mole ratio: 1 mole 3 moles Amount in moles, n, of N2(g) from the complete consumption of hydrogen gas:

2

2

2

N 2

2 2

N

2

2 2

N

2 3 mol H 2.97029 mol

1 mol N2.970

H 1 mol N29 mol H 3 mol H

2.97029 mol H

n

n

n =

× = ×

=

2

2

× 1 mol N3 mol H

20.990099 mol N=



Molar mass, M, of N2(g): 2N N2

2(14.01 g/mol)28.2 g/mol

M M=

==

Mass, m, of the N2(g):

2N

0.990099 mol

m n M= ×

= × 28.02 g/ mol27.7425 g28 g

==

The mass of nitrogen that reacts is 28 g. Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of nitrogen.

2N 2 6.0 g Hm = 21 m

ol H×

22.02 g H21 mo

l N

×

23 mol H2

2

28.02 g N 1 m

ol N

×

2 28 g N=


Iron ore, Fe2O3(s), is treated with carbon monoxide, CO(g), to extract and purify the iron. This reaction is represented by the following unbalanced chemical equation: ___Fe2O3(s) + ___CO(g) → ___Fe(s) + ___CO2(g) a. Balance the chemical equation. b. Calculate the minimum mass of carbon monoxide that must be ordered by a refining company for every metric tonne of iron ore that is processed a. equation Balanced chemical equation: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) b. mass of carbon monoxide What Is Required? You need to determine the minimum mass of carbon monoxide to process each metric tonne of ore.



What Is Given? You know the balanced chemical equation: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) You know the mass of iron ore: 1 metric tonne Plan Your Strategy Convert the mass of iron ore from metric tonnes to grams: 1 metric tonne = 1.000 × 103 kg = 1.000 × 106 g Determine the molar masses of Fe2O3(s) and CO(g). Determine the mole ratio in the balanced chemical equation. Use the factor-label method to set up an expression in which all the terms cancel except for the mass of CO(g). Express the answer in kilograms. Act on Your Strategy Mass conversion (metric tonne to grams):

3

6

1 metric tonne 1.000 10 kg 1.000 10 g

= ×

= ×

Molar mass, M, of Fe2O3(s):

2 3Fe O Fe O2 3

2(55.85 g/mol) + 3(16.00 g/mol) 159.70 g/mol

M M M= +

==

Molar mass, M, of CO(g):

CO C O1 11(12.01 g/mol) + 1(16.00 g/mol) 28.01 g/mol

M M M= +==

Balance chemical equation: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Mole ratio: 1 mole 3 moles Factor-label method to determine the mass, m, of the CO(g):

6CO 2 31.000 10 g Fe Om = × 2 31 mol

Fe

O

×2 3159.7 0 g Fe O

3 mo O l C×

2 31 mol Fe O28.01 g CO 1 mo

l CO

×

55.26174 10 g CO×= Mass conversion (grams to kilograms):

5CO

5

5.26174 10 g

5.26174 10 g

m = ×

= × –3 1 10 kg/ g× ×

526.2 kg=



The mass of carbon monoxide that must be ordered is 526.2 kg per metric tonne of ore. Check Your Solution The units have cancelled to give the answer in kilograms. The answer is reasonable and correctly shows four significant digits.


When heated, the orange crystals of ammonium dichromate, (NH4)2Cr2O7(s), slowly decompose to form green chromium(III) oxide, Cr2O3(s). Colourless nitrogen gas and water vapour are given off . (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) a. How many formula units of chromium(III) oxide are produced from the decomposition of 7.0 g of ammonium dichromate? b. How many formula units of ammonium dichromate are needed to produce 2.75 g of water vapour? a. formula units of chromium(III) oxide What Is Required? You need to determine the number of formula units of chromium(III) oxide that are produced in a reaction. What Is Given? You know the balanced chemical equation for the reaction: (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) You know the mass of ammonium dichromate: 7.0 g Plan Your Strategy Determine the molar mass of (NH4)2Cr2O7(s). Calculate the amount in moles of (NH4)2Cr2O7(s) using the relationship

.mnM

=

Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced chemical equation and the calculated amount in moles of (NH4)2Cr2O7(s) to determine the amount in moles of Cr2O3(s). Use the Avogadro constant: NA = 6.02 × 1023 Calculate the number of formula units, N, of Cr2O3(s) using the relationship

A .N n N= ×



Act on Your Strategy Molar mass, M, of (NH4)2Cr2O7(s):

( ) 44 2 72NH Cr ONH Cr O

N H Cr O

2 2 7

2[1 4 ] 2 72[1(14.01 g/mol) + 4(1.01 g/mol)] + 2(52.00 g/mol) + 7(16.00 g/mol) 252.1 g/mol

M M M M

M M M M

= + +

= + + +==

Amount in moles, n, of (NH4)2Cr2O7(s):

( )4 2 72NH Cr O

7.0 g

mnM

=

=252.1 g

0.0277667

/m

l

ol

mo=

Balanced chemical equation: (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) Mole ratio: 1 mole 1 mole Amount in moles, n, of Cr2O3(s):

2 3

2 3

2 3

Cr O 2 3

4 2 2 7 4 2 2 7

Cr O 4 2 2 7 4 2 2 7

4 2 2

2

7

3

Cr O

1 mol Cr O 0.0277667 mol (NH ) Cr O 1 mol (NH ) Cr O

(NH 1 mol 0.0277667 mol 1 mol C) Cr O (NH ) Cr O

0.0277667 mol (N ) r

r O

H C O

n

n

n

× =

=

×

=

2 3

4 2 2 7

× 1 mol Cr O1 mol (NH ) Cr O

2 30.0277667 Crmol O =

Number of formula units, N, of Cr2O3(s):

A

0.0277667 mol

N n N= ×

= 23× 6.02 × 10 formula units/ mol22

22

1.67155 × 10 formula units1.7 × 10 formula units

=

=

The number of formula units of chromium(III) oxide produced is 1.7 × 1022. b. formula units of ammonium dichromate What Is Required? You need to determine the number of formula units of ammonium dichromate.



What Is Given? You know the balanced chemical equation: (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) You know the mass of water vapour: 2.75 g Plan Your Strategy Determine the molar mass of H2O(g). Calculate the amount in moles of H2O(g). Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of H2O(g) to calculate the amount in moles of (NH4)2Cr2O7(s). Use the Avogadro constant: NA = 6.02 × 1023 Calculate the number of formula units of (NH4)2Cr2O7(s) using the relationship

A .N n N= × Act on Your Strategy Molar mass, M, of H2O(g):

2H O H O2 1

2(1.01 g/mol) + 1(16.00 g/mol) 18.02 g/mol

M M M= +

==

Amount in moles, n, of H2O(g):

2H O

2.75 g

mnM

=

=18.02 g

0.1526

/mol

08 mol=

Balanced chemical equation: (NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) Mole ratio: 1 mole 4 moles Amount in moles, n, of (NH4)2Cr2O7(s):

4 2 2 7

4 2 2 7

4 2 2 7

2 2 4 2 2 7

(NH ) Cr O 4 2 2 7

2 2

(NH ) Cr O

2(NH ) Cr O

4 mol H O 0.152608 mol H O 1 mol (NH ) C

1 mol (NH ) Cr O 0.152608 mol H O 4 mol H O

0.152608 mol H O

r

O

n

n

n

=

=

× = ×

4 2 2 7

2

× 1 mol (NH ) Cr O 4 mol H O

4 2 2 70.038152 mol (NH ) Cr O =



Number of formula units, N, of (NH4)2Cr2O7(s): A

0.038152 mol

N n N= ×

= 23× 6.02 × 10 formula units/ mol22

22

2.29675 10 formula units2.3 10 formula units

= ×

= ×

The number of formula units of ammonium dichromate needed is 2.3 × 1022. Check Your Solution The amounts in moles have been determined correctly and substituted into the mole ratio to give answers with the proper units. The answers seem reasonable and correctly show two significant digits.

Section 7.2 Limiting and Excess Reactants Solutions for Practice Problems Student Edition page 309

31. Practice Problem (page 309) Hydrogen fluoride, HF(g), is a highly toxic gas. It is produced according to the following balanced chemical equation: CaF2(s) + H2SO4 (aq) → 2HF(g) + CaSO4(s) Determine the limiting reactant when 1.00 g of calcium fluoride, CaF2(s), reacts with 15.5 g of sulfuric acid, H2SO4(aq). What Is Required? You need to determine whether calcium fluoride or sulfuric acid is the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: CaF2(s) + H2SO4 (aq) → 2HF(g) + CaSO4(s) You know the mass of calcium fluoride: 1.00 g You know the mass of sulfuric acid: 15.5 g Plan Your Strategy Calculate the molar masses of CaF2(s) and H2SO4(aq). Calculate the amounts in moles of CaF2(s) and H2SO4(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products.



Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The limiting reactant is the one that produces the lesser amount in moles of the product. Act on Your Strategy Molar mass, M, of CaF2(s):

2CaF Ca F1 2

1(40.08 g/mol) + 2(19.00 g/mol) 78.08 g/mol

M M M= +

==

Molar mass, M, of H2SO4(aq):

2 4H SO H S O2 1 4

2(1.01 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol) 98.09 g/mol

M M M M= + +

==

Amount in moles, n, of

CaF2(s):

2CaF

1.00 g

mnM

=

=78.08 g

0.0128073

/m

l

ol

mo=

Amount in moles, n, of H2SO4 (aq):

2 4H SO

15.5 g

mnM

=

=98.09 g

0.158018

/m

l

ol

mo=

Balanced chemical equation: CaF2(s) + H2SO4 (aq) → 2HF(g) + CaSO4(s) Mole ratio: 1 mole 1 mole 2 moles 1 mole Use the product HF(g) to determine the limiting reactant. Known mole ratio of HF to

CaF2 equated to the unknown ratio:

HF

2 2

2 mol HF0.0128073 mol CaF 1 mol CaF

n=



Amount in moles, n, of HF(g): HF

2 2

2

HF 2 2

HF

1 mol CaF 0.0128073 mol Ca

2 mol HF0.0128073 mol CaF 1 mol C

F 2 mol HFaF

0.0128073 mol CaF

n

n

n

× = ×

=

=

2

× 2 mol HF 1 mol CaF

0.0256146 mol HF0.026 mol HF

==

Known mole ratio of HF to

CaF2 equated to the unknown ratio:

HF

2 4 2 4

2 mol HF0.158018 mol H SO 1 mol H SO

n=

Amount in moles, n, of H2SO4(aq):

HF

2 4 2 4

2 4

HF 2 4 2 4

HF

1 mol H SO 0.158018 mol H SO

2 mol HF0.15

2 mol HF 8018 mol H SO 1 mol H SO

0.158018 mol H SO

n

n

n

× = ×

=

=

2 4

× 2 mol HF 1 mol H SO

0.316036 mol HF0.316 mol HF

==

CaF2(s) produces a lesser amount of HF. Calcium fluoride is the limiting reactant. Check Your Solution The mole ratio of CaF2(s) to H2SO4(aq) is 1:1. The reactant that is the lesser amount will produce the lesser amount of product. CaF2 is the reasonable answer.


An ester is an organic compound that forms when a carboxylic acid reacts with an alcohol. Esters often are used as essences or scents. One such ester is methyl salicylate, C8H8O3 (aq), which is oil of wintergreen. It is formed by the reaction of salicylic acid, C7H6O3(aq), and methanol, CH3OH(aq), as shown: C7H6O3(aq) + CH3OH(aq) → C8H8O3 (aq) + H2O(ℓ) If 100.11 g of salicylic acid and 90.4 g of methanol are used to produce oil of wintergreen, which is the limiting reactant?



What Is Required? You need to determine whether salicylic acid or methanol is the limiting reactant in the reaction. What Is Given? You know the balanced chemical equation for the reaction: C7H6O3(aq) + CH3OH(aq) → C8H8O3 (aq) + H2O(ℓ) You know the mass of salicylic acid: 100.11 g You know the mass of methanol: 90.4 g Plan Your Strategy Calculate the molar masses of C7H6O3(aq) and CH3OH(aq). Calculate the amounts in moles of C7H6O3(aq) and CH3OH(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The limiting reactant is the one that produces the lesser amount in moles of the product. Act on Your Strategy Molar mass, M, of C7H6O3(aq):

7 6 3C H O C H O7 6 3


M M M M= + +

==

Molar mass, M, of CH3OH(aq):

3CH OH C H O1 4 1


M M M M= + +

==

Amount in moles, n, of C7H6O3(aq):

7 6 3C H O

100.11 g

mnM

=

=138.13 g

7247

/mol

52 l0. mo=



Amount in moles, n, of CH3OH(aq):

3CH OH

90.4 g

mnM

=

=32.05 g

2.820592

/m

l

ol

mo=

Balanced chemical equation: C7H6O3(aq) + CH3OH(aq) → C8H8O3 (aq) + H2O(ℓ) Mole ratio: 1 mole 1 mole 1 mole 1 mole Use the product H2O(ℓ) to determine the limiting reactant. Amount in moles, n, of H2O(ℓ) from complete consumption of salicylic acid:

2

2

2

H O 2

7 6 3 7 6 3

H O 7 6 3 7 6 3 2

7 6 3H O

1 mol H O0. mol C H O 1 mol C H O

× 1 mol C H O 0. mol C H O 1 mol 724752

724752 ×

724752

H O

0. mol C H O

n

n

n

=

=

=

2

7 6 3

× 1 mol H O 1 mol C H O

2724752 mol. O 0 H=

Amount in moles, n, of H2O(ℓ) from complete consumption of methanol:

2

2

2

H O 2

3 3

H O 3 3 2

3H O

1 mol H O2.820592 mol CH OH 1 mol CH OH

× 1 mol CH OH 2.820592 mol CH OH 1 mol H O

2.820592 mol CH OH

×

n

n

n

=

=

= 2

3

× 1 mol H O 1 mol CH OH

22.820592 mol H O=

C7H6O3(aq) produces a lesser amount of H2O(ℓ). Salicylic acid is the limiting reactant. Check Your Solution The mole ratio of C7H6O3(aq) to CH3OH(aq) is 1:1. The reactant that is the lesser amount will produce the lesser amount of product. C7H6O3(aq) is the reasonable answer.



33. Practice Problem (page 309) Acetylene, C2H2(g), is used in welding. It forms when calcium carbide, CaC2(s), reacts with water, as shown below: CaC2(s) + 2H2O(ℓ) → Ca(OH)2(aq) + C2H2(g) If 5.50 mol of calcium carbide reacts with 3.75 mol of water, which is the limiting reactant? What Is Required? You need to determine whether calcium carbide or water is the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: CaC2(s) + 2H2O(ℓ) → Ca(OH)2(aq) + C2H2(g) You know the amount in moles of calcium carbide: 5.50 mol You know the amount in moles of water: 3.75 mol Plan Your Strategy Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The limiting reactant is the one that produces the lesser amount in moles of the product. Act on Your Strategy Balanced chemical equation: CaC2(s) + 2H2O(ℓ) → Ca(OH)2(aq) + C2H2(g) Mole ratio: 1 mole 2 moles 1 mole 1 mole Use the product C2H2(g) to determine the limiting reactant. Amount in moles, n, of the C2H2(g) from the complete consumption of calcium carbide:

2 2

2 2

2 2

2

C H 2 2

2 2

C H

2C H

2 2 2 1 mol CaC 5.50 mol CaC

1 mol

1 mol

C H5.50 mol

C HCaC 1 mol CaC

5.50 mol CaC

n

n

n

× = ×

=

= 2 2

2

× 1 mol C H 1 mol CaC

2 25.50 mo Cl H=



Amount in moles, n, of the C2H2(g) from the complete consumption of water: 2 2

2 2

2 2

2

C H 2 2

2 2

C H

2C H

2 2 2 2 mol H O 3.75 mol H O

1 mol

1 mo

C H3.75 mol H O 2 mol H O

3.75 mo

l C

l O

H

H

n

n

n

× = ×

=

= 2 2

2

× 1 mol C H 2 mol H O

2 2

2 2

1.875 mol 1.88 mo

HHl

CC

==

H2O(ℓ) produces a lesser amount of C2H2(g). Water is the limiting reactant. Check Your Solution The mole ratio of CaC2(s) to H2O(ℓ) is 1:2. For 5.50 mol of CaC2(s) to react

completely, 21

× 5.50 mol or 11.0 mol of H2O(ℓ) is required. Since the given

amount of H2O(ℓ) is 3.75 mol, water is the limiting reactant.

34. Practice Problem (page 309) Nickel(II) chloride, NiCl2(aq), reacts with sodium phosphate, Na3PO4(aq), according to the following balanced chemical equation: 3NiCl2(aq) + 2Na3PO4(aq) → Ni3(PO4)2(s) + 6NaCl(aq) If 10.0 g of each reactant is used, which is the limiting reactant? What Is Required? You need to determine whether nickel(II) chloride or sodium phosphate is the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: 3NiCl2(aq) + 2Na3PO4(aq) → Ni3(PO4)2(s) + 6NaCl(aq) You know the mass of nickel(II) chloride: 10.0 g You know the mass of sodium phosphate: 10.0 g Plan Your Strategy Calculate the molar masses of NiCl2(aq) and Na3PO4(aq). Calculate the amounts in moles of NiCl2(aq) and Na3PO4(aq) using the

relationship .mnM

=




Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The limiting reactant is the one that produces the lesser amount in moles of the product. Act on Your Strategy Molar mass, M, of NiCl2(aq):

2NiCl Ni Cl1 2

1(58.69 g/mol) + 2(35.45 g/mol)129.59 g/mol

M M M= +

==

Molar mass, M, of Na3PO4(aq):

3 4Na PO Na P O3 1 4


M M M M= + +

==

Amount in moles, n, of the NiCl2(aq):

2NiCl

10.0 g

mnM

=

=129.59 g

0.0771664

/m

l

ol

mo=

Amount in moles, n, of the Na3PO4(aq):

3 4Na PO

10.0 g

mnM

=

=169.94 g

0.0609979

/m

l

ol

mo=

Balanced chemical equation: 3NiCl2(aq) + 2Na3PO4(aq) → Ni3(PO4)2(s) + 6NaCl(aq) Mole ratio: 3 moles 2 moles 1 mole 6 moles Use the product NaCl(aq) to determine the limiting reactant.



Amount in moles, n, of the NaCl(aq) from the complete consumption of the nickel(II) chloride:

Na

NaCl

2 2

2

Cl 2 2

NaCl

3 mol NiCl 0.0771664 mol NiCl

6 mol NaCl0.0771

6 mol NaCl664 mol NiCl 3 mol NiCl

0.0771664 mol NiCl

n

n

n

× = ×

=

=

2

× 6 mol NaCl 3 mol NiCl

0.154332 mol NaCl=

Amount in moles, n, of the NaCl(aq) from the complete consumption of the sodium phosphate:

NaCl 3 4

NaCl

3 4 3 4

3 4

NaCl3 4

6 mol NaCl0

2 mol Na PO 0.0609979 mol Na PO 6 mol NaCl.0609979 mol Na P O 2 mol Na PO

0.0609979 mol Na PO

n

n

n

× =

=

=

×

3 4

× 6 mol NaCl 2 mol Na PO

0.1828837 mol Nacl=

NiCl2(aq) produces a lesser amount of NaCl(aq). Nickel(II) chloride is the limiting reactant. Check Your Solution The mole ratio of NiCl2(aq) to Na3PO4(aq) is 3:2. Using rounded numbers, for

0.077 mol of NiCl2(aq) to react completely, 23

× 0.077 mol or 0.051 mol of

Na3PO4(aq) is required. The Na3PO4(aq) is in excess since the given amount of this reactant is 0.061 mol. The NiCl2(aq) is the limiting reactant.


Copper metal reacts with nitric acid, HNO3(aq), as follows: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2 (aq) + 2NO(g) + 4H2O(ℓ) If 2.5 g of copper reacts with 25.0 g of nitric acid, which reactant is in excess? What Is Required? You need to determine whether copper metal or nitric acid is the reactant in excess.



What Is Given? You know the balanced chemical equation for the reaction: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2 (aq) + 2NO(g) + 4H2O(ℓ) You know the mass of copper: 2.5 g You know the mass of nitric acid: 25.0 g Plan Your Strategy Determine the atomic molar mass of Cu(s) using the periodic table. Calculate the molar mass of HNO3(aq). Calculate the amounts in moles of Cu(s) and HNO3(aq) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The reactant in excess is the one that produces the greater amount in moles of the product. Act on Your Strategy Molar mass, M, of Cu(s): M = 63.55 g/mol (from the periodic table) Molar mass, M, of HNO3(aq):

3HNO H N O1 1 3


M M M M= + +

==

Amount in moles, n, of Cu(s):

Cu

2.5 g

mnM

=

=63.55 g

0.039331

/m

l

ol

mo=

Amount in moles, n, of HNO3(aq):

3HNO

25.0 g

mnM

=

=63.02 g

0.396699

/m

l

ol

mo=



Balanced chemical equation: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2 (aq) + 2NO(g) + 4H2O(ℓ) Mole ratio: 3 moles 8 moles 3 moles 2 moles 4 moles Use the product NO(g) to determine the reactant in excess. Amount in moles, n, of the NO(g) from the complete consumption of the copper metal:

N

NO

O

NO

3 mol Cu 0.039331 mol

2 mol NO 0.039331 mol Cu 3

Cu 2 mol N mol Cu

0.039331 mol Cu

O

n

n

n

×

=

=

= ×

× 2 mol NO 3 mol Cu

0.026226 O mol N=

Amount in moles, n, of the NO(g) from the complete consumption of the nitric acid:

NO

3 3

3

NO 3 3

NO

8 mol HNO 0.396699 mol HN

2 mol NO0.396699 mol HNO 8 mol HN

O 2 mol NOO

0.396699 mol HNO

n

n

n

× = ×

=

=

3

× 2 mol NO 8 mol HNO

0.0495873 mol NO=

HNO3(aq) produces a greater amount of NO(g). Nitric acid is the reactant in excess. Check Your Solution The mole ratio of Cu(s) to HNO3(aq) is 3:8. Using rounded numbers, for 0.039

mol of Cu(s) to react completely, 83

× 0.039 mol or 0.10 mol of HNO3(aq) is

required. The HNO3(aq) is in excess since the given amount of this reactant is 0.40 mol.


Lithium reacts with oxygen to form lithium oxide, Li2O(s). 4Li(s) + O2(g) → 2Li2O(s) When 20.0 g of lithium metal reacts with 30.0 g of oxygen gas, which reactant is limiting and which reactant is in excess?



What Is Required? You need to determine the limiting reactant and the reactant is in excess when a sample of lithium reacts with oxygen. What Is Given? You know the balanced chemical equation for the reaction: 4Li(s) + O2(g) → 2Li2O(s) You know the mass of lithium: 20.0 g You know the mass of oxygen: 30.0 g Plan Your Strategy Determine the atomic molar mass of Li(s) using the periodic table. Calculate the molar mass of O2(g). Calculate the amounts in moles of Li(s) and O2(g) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of product, Li2O(s), that is produced if all of each reactant is completely consumed. The reactant that produces the greater amount in moles of Li2O(s) is in excess. The reactant that produces the lesser amount of Li2O(s) is the limiting reactant. Act on Your Strategy Molar mass, M, of Li(s): MLi = 6.94 g/mol (from the period table) Molar mass, M, of O2(g):

2O O2

2(16.00 g/mol)32.00 g/mol

M M=

==

Amount in moles, n, of Li(s):

Li

20.0 g

mnM

=

=6.94 g

2.88184

/mo

l

l

mo=



Amount in moles, n, of O2(g):

2O

30.0 g

mnM

=

=32.00 g

0.937500

/m

l

ol

mo=

Balanced chemical equation: 4Li(s) + O2(g) → 2Li2O(s) Mole ratio: 4 moles 1 mole 2 moles Amount in moles, n, of Li2O(s) from the complete consumption of lithium:

2

2

2

L

Li O 2

i O 2

Li O

4 mol Li 2.88184 mol Li

2 mol Li O2.88184 mol Li 4 mol Li

2.88184 mol L

2 mol O

i

Li

n

n

n

× = ×

=

=

2× 2 mol Li O 4 mol Li

2

2

Li O1.44092 mol 1.44 mo Ol Li

==

Amount in moles, n, of Li2O(s) from the complete consumption of oxygen gas:

2

2

2Li O 2

Li O

2 2

L

2

i

2

O

2

2

1 mol O 0.93750 m

2 mol Li O0.93750

ol O 2 mol 0 mol O 1 mol O

0.93750 ol O

Li O

m

n

n

n

=

× = ×

= 2

2

× 2 mol Li O 1 mol O

2

2

Li O1.875 mol 1.88 mol O Li

==

Li(s) produces a lesser amount of Li2O(s). Lithium is the limiting reactant. Oxygen is the reactant in excess. Check Your Solution The mole ratio of Li(s) to O2(g) is 4:1. Using rounded numbers, for 2.88 mol of

Li(s) to react completely, 14

× 2.88 mol or 0.72 mol of O2(g) is required. The

O2(g) is in excess since the given amount of this reactant is 0.94 mol.



37. Practice Problem (page 309) Chlorine gas is used in the textile industry to bleach fabric. Excess chlorine is removed by a reaction with sodium thiosulfate, Na2S2O3(aq), as shown below: Na2S2O3(aq) + 4Cl2(g) + 5H2O(ℓ) → 2NaHSO4(aq) + 8HCl(aq) If 42.5 g of sodium thiosulfate and 175 g of chlorine gas react with excess water, which is the limiting reactant? What Is Required? You need to determine whether sodium thiosulfate or chlorine gas is the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: Na2S2O3(aq) + 4Cl2(g) + 5H2O(ℓ) → 2NaHSO4(aq) + 8HCl(aq) You know the mass of sodium thiosulfate: 42.5 g You know the mass of chlorine gas: 175 g Plan Your Strategy Calculate the molar masses of Na2S2O3(aq) and Cl2(g). Calculate the amounts in moles of Na2S2O3(aq) and Cl2(g) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount in moles of the product is the limiting reactant Act on Your Strategy Molar mass, M, of Na2S2O3(aq):

2 2 3Na S O Na S O2 2 3


M M M M= + +

==

Molar mass, M, of Cl2(g):

2Cl Cl2

2(35.45 g/mol)70.9 g/mol

M M=

==



Amount in moles, n, of the Na2S2O3(aq):

2 2 3Na S O

42.5 g

mnM

=

=158.12 g

0.268783

/m

l

ol

mo=

Amount in moles, n, of the Cl2(g):

2Cl

175 g

mnM

=

=70.9 g

2.46826

/mo

l

l

mo=

Balanced chemical equation: Na2S2O3(aq) + 4Cl2(g) + 5H2O(ℓ) → 2NaHSO4(aq) + 8HCl(aq) Mole ratio: 1 mole 4 moles 5 moles 2 moles 8 moles Use the product HCl(aq) to determine the limiting reactant.

Amount in moles, n, of HCl(aq) from the complete consumption of sodium thiosulfate:

HCl 2 2 3

HCl

2 2 3 2

2 2 3

H

2 3

2 2C

3l

1 mol Na S O 0.268783 mol Na S

8 mol HCl0.268783

O 8 mol HClmol Na S O 1 mol Na S O

0.268783 mol Na S O

n

n

n

× = ×

=

=

2 2 3

× 8 mol HCl 1 mol Na S O

HCl2.15026 mol 2.15 m Clol H

==

Amount in moles, n, of HCl(aq) from the complete consumption of chlorine gas:

HCl

2 2

HCl 2 2

HCl2

4 mol Cl 2.46826 mol Cl

8 mol HCl2.468

8 mol HC26 mo

ll Cl 4 mol Cl

2.46826 mol Cl

n

n

n

× = ×

=

=

2

× 8 mol HCl 4 mol Cl

HCl4.93652 mol 4.94 m Clol H

==



Na2S2O3(aq) produces a lesser amount of HCl(aq). Sodium thiosulfate is the limiting reactant. Check Your Solution The mole ratio of Na2S2O3(aq) to Cl2(g) is 1:4. Using rounded numbers, for

0.27 mol of Na2S2O3(aq) to react completely, 41

× 0.27 mol or 1.1 mol of

Cl2(g) is required. The Cl2(g) is in excess since the given amount of this reactant is 2.5 mol. The limiting reactant is Na2S2O3(aq).


Acrylonitrile, C3H3N(g), is prepared by the reaction of propylene, C3H6(g), with nitric oxide, NO(g). 4C3H6(g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g) If 126 g of propylene reacts with 175 g of nitric oxide, which is the limiting reactant? What Is Required? You need to determine whether propylene or nitric oxide is the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: 4C3H6(g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g) You know the mass of propylene: 126 g You know the mass of nitric oxide: 175 g Plan Your Strategy Calculate the molar masses of C3H6(g) and NO(g). Calculate the amounts in moles of C3H6(g) and NO(g) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount in moles of the product is the limiting reactant.



Act on Your Strategy Molar mass, M, of C3H6(g):

3 6C H C H3 6

3(12.01 g/mol) + 6(1.01 g/mol)42.09 g/mol

M M M= +

==

Molar mass, M, of NO(g):

NO N O1 11(14.01 g/mol) + 1(16.00 g/mol) 30.01 g/mol

M M M= +==

Amount in moles, n, of C3H6(g):

3 6C H

126 g

mnM

=

=42.09 g

2.99358

/mo

l

l

mo=

Amount in moles, n, of NO(g):

NO

175 g

mnM

=

=30.01 g

5.83138

/mo

l

l

mo=

Balanced chemical equation: 4C3H6(g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g) Mole ratio: 4 moles 6 moles 4 moles 6 moles 1 mole Use the product N2(g) to determine the limiting reactant. Amount in moles, n, of N2(g) from the complete consumption of propylene:

2

2

2

3 6

N 2

3 6 3 6

N

3 6N

3 6 2 4 mol C H 2.99358 mol

1 mol N2.99358 m

C H 1 mol Nol C H 4 mol C H

2.99358 mol C H

n

n

n

× =

=

×

=

2

3 6

× 1 mol N 4 mol C H

2

2

N0.748395 mol 0.748 Nmol

==



Amount in moles, n, of N2(g) from the complete consumption of nitric acid: 2

2

2

N 2

N

N

2 6 mol NO 5.83138 mol N

1 mol N5.83138 mol NO 6 mol NO

5.83138 m

O 1 mo

o

l

l NO

N

n

n

n

× =

=

=

×

2× 1 mol N 6 mol NO

2

2

N0.971896 mol

0.972 Nmol ==

C3H6(g) produces a lesser amount of N2(g). Propylene is the limiting reactant. Check Your Solution The mole ratio of C3H6(g) to NO(g) is 4:6. Using rounded numbers, for 3.0

mol of C3H6(g) to react completely, 64

× 3.0 mol or 4.5 mol of NO(g) is

required. The NO(g) is in excess since the given amount of this reactant is 5.8 mol. The limiting reactant is C3H6(g).


Insoluble silver carbonate, Ag2CO3(s), forms in the following balanced chemical reaction: 2AgNO3(aq) + K2CO3 (aq) → Ag2CO3(s) + 2KNO3(aq) What mass of silver nitrate, AgNO3(aq), reacts with 25.0 g of potassium carbonate, K2CO3 (aq), if there is at least 5.5 g of silver nitrate in excess? What Is Required? You need to determine the mass of silver nitrate that reacts. What Is Given? You know the balanced chemical equation for the reaction: 2AgNO3(aq) + K2CO3 (aq) → Ag2CO3(s) + 2KNO3(aq) You know the mass of potassium carbonate: 25.0 g You know the minimum excess of silver nitrate: 5.5 g Plan Your Strategy Calculate the molar mass of K2CO3(aq).

Calculate the amount in moles of K2CO3(aq) using the relationship .mnM

=

All of the K2CO3(aq) must be consumed since the silver nitrate is in excess. Determine the mole ratio from the balanced chemical equation.



Use the mole ratio in the balanced chemical equation and the calculated amount in moles of K2CO3(aq) to calculate the amount in moles of AgNO3(aq). Determine the molar mass of AgNO3(aq). Calculate the mass of AgNO3(aq) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of K2CO3(aq):

2 3K CO K C O2 1 3


M M M M= + +

==

Amount in moles, n, of the K2CO3(aq):

2 3K CO

25.0 g

mnM

=

=138.21 g

0.180884

/m

l

ol

mo=

Balanced chemical equation: 2AgNO3(aq) + K2CO3 (aq) → Ag2CO3(s) + 2KNO3(aq) Mole ratio: 2 moles 1 mole Amount in moles, n, of the AgNO3(aq):

3

3

3

3

2 3 2 3

2 3 2 3

AgNO

AgNO

2

3

3AgNO

0.180884 mol K C

2 mol AgNO0.180884 mol K CO 1 mol K CO

1 mol K CO 0.180884 mol K CO 2 mo

O

l AgNO

n

n

n

=

× =

=

×

3

2 3

× 2 mol AgNO 1 mol K CO

30.361768 Amol gNO=

Molar mass, M, of AgNO3(aq):

3AgNO Ag N O1 1 3


M M M M= + +

==



Mass, m, of the AgNO3(aq): 3AgNO

0.361768 mol

m n M= ×

= 169.88 g/ mol×61.45714 g61.5 g

==

The mass of silver nitrate that reacts is 61.5 g. Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of silver nitrate.

3AgNO 2 325.0 g K COm = 2 31 mol

K

CO×

2 3138.21 g K CO32 mol AgNO

×2 3

1 mol K CO

3

3

169.88 g AgNO 1 mol AgNO

×

61.5 g=

Section 7.2 Limiting and Excess Reactants Solutions for Practice Problems Student Edition page 311

40. Practice Problem (page 311) The formation of water is represented by the following equation: 2H2(g) + O2(g) → 2H2O(g) a. What is the limiting reactant if 4 mol of oxygen reacts with 16 mol of hydrogen? b. What amount (in moles) of water is produced in this reaction? What Is Required? You need to determine whether oxygen or hydrogen is the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: 2H2(g) + O2(g) → 2H2O(g) You know the amount in moles of oxygen: 4 mol You know the amount in moles of hydrogen: 16 mol Plan Your Strategy a. limiting reactant Use the balanced chemical equation to determine the mole ratio between the reactants and products.



Calculate the amount in moles of H2O(g) that is produced if all of each reactant is completely consumed. The limiting reactant is the one that produces the lesser amount in moles of H2O(g). Act on Your Strategy Balanced chemical equation: 2H2(g) + O2(g) → 2H2O(g) Mole ratio: 2 moles 1 mole 2 moles Amount in moles, n, of H2O(g) from the complete consumption of hydrogen:

2

2

2

2

H O 2

2 2

H O

2H O

2 2 2 mol H 16 mo

2 mol H O16 mol

l H 2 mol H H 2 mol H

16 m

O

ol H

n

n

n =

×

=

= ×

2

2

× 2 mol H O 2 mol H

216 mol H O=

Amount in moles, n, of H2O(g) from the complete consumption of oxygen:

2

2

2

2

H O 2

2

2 2

2

H O

2H O

1 mol O 4 mo

2 mol H O4 mol

l O 2 mol HO 1 mol O

4 mol

O

O

n

n

n =

×

=

= ×

2

2

× 2 mol H O 1 mol O

28 mol H O=

O2(g) produces a lesser amount of H2O(g). Oxygen is the limiting reactant. Check Your Solution The mole ratio of H2(g) to O2(g) is 2:1. For 16 mol of H2(g) to react

completely, 12

× 16 mol or 8 mol of O2(g) is required. Since the given amount

of O2(g) is only 4 mol, oxygen is the limiting reactant. b. amount (in moles) of water What is Required? You need to determine the amount in moles of water that forms. What Is Given? You know the balanced chemical equation for the reaction: 2H2(g) + O2(g) → 2H2O(g)



You know from part a. the amount of O2(g) that reacts: 4 mol Plan Your Strategy Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced chemical equation and the given amount in moles of O2(g) to calculate the amount in moles of H2O(g). Act on Your Strategy Balanced chemical equation: 2H2(g) + O2(g) → 2H2O(g) Mole ratio: 1 mole 2 moles Amount in moles, n, of H2O(g) from the complete consumption of oxygen:

2

2

2

2

H O 2

2

2 2

2

H O

2H O

1 mol O 4 mol O

2 mol H O 4 mol

2 mol H OO 1 mol O

4 mol O

n

n

n

=

× = ×

= 2

2

× 2 mol H O1 mol O

28 mol H O=

The amount of water that forms is 8 moles. Check Your Solution The mole ratio of O2(g) to H2O(g) is 1:2. The answer matches this ratio and is reasonable.


Silver nitrate, AgNO3(aq), reacts with iron(III) chloride, FeCl3(aq), to produce silver chloride, AgCl(s), and iron(III) nitrate, Fe(NO3)3(aq). 3AgNO3(aq) + FeCl3(aq) → 3AgCl(s) + Fe(NO3)3(aq) a. If a solution containing 18.00 g of silver nitrate is mixed with a solution containing 32.4 g of iron(III) chloride, which is the limiting reactant? b. What amount in moles of iron(III) nitrate is produced in this reaction? a. limiting reactant What Is Required? You need to determine whether silver nitrate or iron(III) chloride is the limiting reactant.



What Is Given? You know the balanced chemical equation for the reaction: 3AgNO3(aq) + FeCl3(aq) → 3AgCl(s) + Fe(NO3)3(aq) You know the mass of silver nitrate: 18.00 g You know the mass of iron(III) chloride: 32.4 g Plan Your Strategy Calculate the molar masses of AgNO3(aq) and FeCl3(aq). Calculate the amounts in moles of AgNO3(aq) and FeCl3(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount in moles of the product is the limiting reactant. Act on Your Strategy Molar mass, M, of AgNO3(aq) :

3AgNO Ag N O1 1 3


M M M M= + +

==

Molar mass, M, of FeCl3(aq):

3FeCl Fe Cl1 3

1(55.85 g/mol) + 3(35.45 g/mol) 162.2 g/mol

M M M= +

==

Amount in moles, n, of AgNO3:

3AgNO

18.00 g

mnM

=

=169.88 g

0.105957

/m

l

ol

mo=



Amount in moles, n, of FeCl3:

3FeCl

32.4 g

mnM

=

=162.2 g

0.199753

/m

l

ol

mo=

Balanced chemical equation: 3AgNO3(aq) + FeCl3(aq) → 3AgCl(s) + Fe(NO3)3(aq) Mole ratio: 3 moles 1 mole 3 moles 1 mole Use the product AgCl(s) to determine the limiting reactant. Amount in moles, n, of AgCl(s) from the complete consumption of silver nitrate:

AgCl

3 3

Ag

3

Cl 3 3

AgCl

3 mol AgNO 0.105957 mol AgNO

3 mol AgCl0.1059

3 mol AgC57 mol

l AgNO 3 mol AgNO

0.105957 mol AgNO

n

n

n

× = ×

=

=

3

× 3 mol AgCl 3 mol AgNO

AgCl 0.105957 mol 0.106 mol

AgCl

==

Amount in moles, n, of AgCl(s) from the complete consumption of iron(III) chloride:

AgCl

3 3

Ag

3

Cl 3 3

AgCl

1 mol FeCl 0.199753 mol FeCl

3 mol AgCl0.1997

3 mol AgC53 mol

l FeCl 1 mol FeCl

0.199753 mol FeCl

n

n

n

× = ×

=

=

3

× 3 mol AgCl 1 mol FeCl

AgCl0.599260 mol 0.599 m Clol Ag

==

AgNO3(aq) produces a lesser amount of AgCl(s). Silver nitrate is the limiting reactant.



Check Your Solution The mole ratio in the balanced chemical equation of AgNO3(aq) to FeCl3(g) is 3:1. Using rounded numbers, for 0.10 mol of AgNO3(aq) to react completely, 13

× 0.10 mol or 0.03 mol of FeCl3(aq) is required. The FeCl3(aq) is in excess

since the given amount of this reactant is 0.20 mol. The limiting reactant is AgNO3(aq). b. amount in moles of iron(III) nitrate What Is Required? You need to determine the amount in moles of iron(III) nitrate that is produced. What Is Given? You know the balanced chemical equation: 3AgNO3(aq) + FeCl3(aq) → 3AgCl(s) + Fe(NO3)3(aq) You know from part a. the limiting reactant: AgNO3(aq). You know from part a. the amount in moles of AgNO3(aq): 0.105957 mol Plan Your Strategy Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced chemical equation and the amount in moles of AgNO3(aq) from part a. to calculate the amount in moles of Fe(NO3)3(aq). Act on Your Strategy Balanced chemical equation: 3AgNO3(aq) + FeCl3(aq) → 3AgCl(s) + Fe(NO3)3(aq) Mole ratio: 3 moles 1 mole Amount in moles, n, of Fe(NO3)3(aq):

3 3

3 3

3 3

3 3 3

Fe(NO ) 3 3

3 3

Fe 3(NO )

3Fe(NO )

3 mol AgNO 0.105957 mol AgNO 1 mol

1 mol Fe(NO ) 0.105957 mol AgNO 3 mol AgNO

0.1059

Fe

57 mol

N

A

O

gN

( )

O

n

n

n

=

×

=

= ×

3 3

3

× 1 mol Fe(NO ) 3 mol AgNO

3 3

3 3

mol Fe(NO ) 0.035319 0.035 mol Fe(NO )

==

The amount in moles of iron(III) nitrate produced is 0.035 mol.



Check Your Solution The mole ratio in the balanced chemical equation of AgNO3(aq) to Fe(NO3)3(aq) is 3:1. The answer matches this mole ratio and seems reasonable.


Barium sulfate, BaSO4(s), forms in the following reaction: Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq) If 75.00 g of barium nitrate, Ba(NO3)2(aq), reacts with 100.00 g of sodium sulfate, Na2SO4(aq), what mass of barium sulfate is produced? What Is Required? You need to determine the mass of barium sulfate that forms. What Is Given? You know the balanced chemical equation for the reaction: Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq) You know the mass of barium nitrate:75.00 g You know the mass of sodium sulfate: 100.00 g Plan Your Strategy Determine the limiting reactant: Calculate the molar masses of Ba(NO3)2(aq) and Na2SO4(aq). Calculate the amounts in moles of Ba(NO3)2(aq) and Na2SO4(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of BaSO4(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of barium sulfate: The lesser amount in moles of BaSO4(s) is the amount that is produced. Determine the molar mass of the BaSO4(s). Convert the amount in moles of BaSO4(s) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of Ba(NO3)2(aq):

( ) 33 2Ba NOBa NO

Ba N O

1 2

1 2[1 3 ]1(137.33 g/mol) + 2[1(14.01 g/mol) + 3(16.00 g/mol)]261.35 g/mol

M M M

M M M

= +

= + +

==



Molar mass, M, of Na2SO4(aq): 2 4Na SO Na S O2 1 4


M M M M= + +

==

Amount in moles, n, of Ba(NO3)2(aq):

( )3 2Ba NO

75.00 g

mnM

=

=261.35 g

0.286971

/m

l

ol

mo=

Amount in moles, n, of Na2SO4(aq):

2 4Na SO

100.00 g

mnM

=

=142.05 g

0.703977

/m

l

ol

mo=

Balanced chemical equation: Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq) Mole ratio: 1 mole 1 mole 1 mole 2 moles

Amount in moles, n, of BaSO4(s) from the complete consumption of barium nitrate:

( ) ( )

4

4

4

3

BaSO 4

3 2 3 2

BaSO

3 2BaS

32

O

42 1 mol Ba NO 0.286971 mol

1 mol BaSO0.286971

Ba NO 1 mol Bmol Ba(NO ) 1 mol Ba(NO )

0.286971 mol Ba(NO )

aSO

n

n

n

× = ×

=

= 4

3 2

× 1 mol BaSO1 mol Ba(NO )

40.286971 Bmol aSO=

Amount in moles, n, of BaSO4(s) from the complete consumption of sodium sulfate:

4

4

4

BaSO 2

BaSO 4

2 4

4 2 4 4

2 4

2 4BaSO

1 mol Na SO 0.703977 mol

1 mol BaSO0.703977 m

Na SO 1 mol Bol Na SO 1 mol Na SO

0. 7039

aS

77 mol Na S

O

O

n

n

n

× = ×

=

=

4

2 4

× 1 mol BaSO1 mol Na SO

40.703977 Bmol aSO=



The amount in moles of BaSO4(s) produced is the lesser of the two values: 0.286971 mol Molar mass, M, of BaSO4(s):

4BaSO Ba S O1 1 4


M M M M= + +

==

Mass, m, of the BaSO4(s):

0.286971 mol

m n M= ×

= 233.4 g/ mol×66.9790 g66.98 g

==

The mass of barium sulfate produced is 66.98 g. Check Your Solution The mole ratio of the reactants Ba(NO3)3(aq) and Na2SO4(aq) is 1:1. The units are correct and the limiting reactant seems reasonable. The mole ratio of the limiting reactant to BaSO4(s) is also 1:1. The calculated mass of BaSO4(s) seems reasonable and correctly shows four significant digits.


Zinc oxide, ZnO(s), is formed by the reaction of zinc sulfide, ZnS(s), with oxygen. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) If 16.7 g of zinc sulfide reacts with 6.70 g of oxygen, what mass of zinc oxide is produced? What Is Required? You need to determine the mass of zinc oxide that forms. What Is Given? You know the balanced chemical equation for the reaction: 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) You know the mass of zinc sulfide: 16.7 g You know the mass of oxygen: 6.70 g Plan Your Strategy Determine the limiting reactant: Calculate the molar masses of ZnS(s) and O2(g).



Calculate the amounts in moles of ZnS(s) and O2(g) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of ZnO(s) produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of zinc oxide: The lesser amount of ZnO(s) is the amount that is produced. Determine the molar mass of ZnO(s). Convert the amount in moles of ZnO(s) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of ZnS(s):

ZnS Zn S1 11(65.38 g/mol) + 1(32.07 g/mol)97.45 g/mol

M M M= +==


2O O2

2(16.00 g/mol)32.00 g/mol

M M=

==

Amount in moles, n, of ZnS(s):

ZnS

16.7 g

mnM

=

=97.45 g

0.171370

/m

l

ol

mo=


2O

6.70 g

mnM

=

=32.00 g

0.209375

/m

l

ol

mo=



Balanced chemical equation: 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) Mole ratio: 2 moles 3 moles 2 moles Amount in moles, n, of ZnO(s) from the complete consumption of zinc sulfide:

Zn

ZnO

O

ZnO

2 mol ZnS 0.171370 mol Z

2 mol ZnO0.171370 mol ZnS 2 mo

nS 2 mol l ZnS

0.171370 mol ZnS

ZnOn

n

n

=

=

× = ×

× 2 mol ZnO2 mol ZnS

0.171370 mol O Zn=

Amount in moles, n, of ZnO(s) from the complete consumption of oxygen gas:

ZnO

2 2

ZnO 2 2

ZnO2

3 mol O 0.209375 mol O ×

2 mol ZnO0.209375 mol O 3 m

2 mol ZnOol O

0.209375 mol O

n

n

n

× =

=

=

2

× 2 mol ZnO3 mol O

0.139583 mol O Zn=

The amount in moles of ZnO(s) produced is the lesser of the two values: 0.139583 mol Molar mass, M, of ZnO(s):

ZnO Zn O1 11(65.38 g/mol) + 1(16.00 g/mol) 81.38 g/mol

M M M= +==

Mass, m, of the ZnO(s):

ZnO

0.139583 mol

m n M= ×

= 81.38 g/ mol×11.36347 g11.4 g

==

The mass of zinc oxide produced is 11.4 g



Check Your Solution The mole ratio in the balanced chemical equation of the reactants ZnS(s) and O2(g) is 2:3. Using rounded numbers, for 0.17 mol of ZnS(s) to be completely

consumed, 32

× 0.17 mol or 0.25 mol of O2(g) is required. The O2(g) must be

the limiting reactant since only 0.21 mol is given. The mole ratio of O2(g) to ZnO(s) is 3:2. The calculated mass of ZnO(s) seems reasonable and correctly shows three significant digits.


The following balanced chemical equation represents the reaction of calcium carbonate, CaCO3(s), with hydrochloric acid: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(ℓ) If 155 g of calcium carbonate reacts with 245 g of hydrochloric acid, what mass of calcium chloride, CaCl2(s), is produced? What Is Required? You need to determine the mass of calcium chloride produced in a reaction. What Is Given? You know the balanced chemical equation for the reaction: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(ℓ) You know the mass of calcium carbonate: 155 g You know the mass of hydrochloric acid: 245 g Plan Your Strategy Determine the limiting reactant: Calculate the molar masses of CaCO3(s) and HCl(aq). Calculate the amounts in moles of CaCO3(s) and HCl(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of CaCl2(aq) produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of calcium chloride: The lesser amount in moles of CaCl2(aq) is the amount that is produced. Determine the molar mass of CaCl2(aq). Convert the amount in moles of CaCl2(aq) to mass (in grams) using the relationship .m n M= ×



Act on Your Strategy Molar mass, M, of CaCO3(s):

3CaCO Ca C O1 1 3


M M M M= + +

==

Molar mass, M, of HCl(aq):

HCl H Cl1 11(1.01 g/mol) + 1(35.45 g/mol) 36.46 g/mol

M M M= +==

Amount in moles, n, of CaCO3(s):

3CaCO

155 g

mnM

=

=100.09 g

1.54860

/mo

l

l

mo=

Amount in moles, n, of HCl(aq):

HCl

245 g

mnM

=

=36.46 g

6.71969

/mo

l

l

mo=

Balanced chemical equation: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(ℓ) Mole ratio: 1 mole 2 moles 1 mole Amount in moles, n, of CaCl2(aq) from the complete consumption of calcium carbonate:

2

2

2

3

CaCl 2

3 3

CaCl

3C C

3

a

2

l

1 mol CaCO 1.54860 mol CaCO 1

1

mol CaCl1.54860 mol CaCO 1 mol CaCO

1.54860 mol C

mol CaCl

aCO

n

n

n

×

=

=

= ×

2

3

× 1 mol CaCl 1 mol CaCO

21 .54860 Cmol aCl=



Amount in moles, n, of CaCl2(aq) from the complete consumption of hydrochloric acid:

2

2

2

2

CaCl 2

CaCl

CaCl

2 mol HCl 6.71969 mol HCl 1

1 mol CaCl6.71969 mol HCl 2 mol HCl

6.71969 mol HC

mol CaC

l

l

n

n

n

=

× = ×

= 2× 1 mol CaCl 2 mol HCl

23.35984 m Caol Cl =

The amount in moles of CaCl2(aq) produced is the lesser of the two values: 1.54860 mol Molar mass, M, of CaCl2(aq):

2CaCl Ca Cl1 2

1(40.08 g/mol) + 2(35.45 g/mol)110.98 g/mol

M M M= +

==

Mass, m, of the CaCl2(aq):

2CaCl

1.54860 mol

m n M= ×

= 110.98 g/ mol×171.8636 g172 g

==

The mass of calcium chloride produced is 172 g. Check Your Solution The mole ratio in the balanced chemical equation of the reactants CaCO3(s) and HCl(aq) is 1:2. Using rounded numbers, for 1.5 mol of HCl(aq) to be

completely consumed, 21

× 1.5 mol or 3 mol of HCl(aq) is required. The

HCl(aq) must be the reactant in excess since the given amount is 6.7 mol. The CaCO3(s) is the limiting reactant. The mole ratio of CaCO3(s) to CaCl2(aq) is 2:1. The calculated mass of CaCl2(aq) seems reasonable and correctly shows three significant digits.



45. Practice Problem (page 311) The reaction of aluminum hydroxide, Al(OH)3(aq), with hydrochloric acid produces water and aluminum chloride, AlCl3(s). 3HCl(aq) + Al(OH)3(aq) → 3H2O(ℓ) + AlCl3(s) What mass of aluminum chloride is produced when 8.0 g of hydrochloric acid reacts with an equal mass of aluminum hydroxide? What Is Required? You need to determine the mass of aluminum chloride produced in a reaction. What Is Given? You know the balanced chemical equation for the reaction: 3HCl(aq) + Al(OH)3(aq) → 3H2O(ℓ) + AlCl3(s) You know the mass of aluminum hydroxide: 8.0 g You know the mass of hydrochloric acid: 8.0 g Plan Your Strategy Determine the limiting reactant: Calculate the molar masses of HCl(aq) and Al(OH)3(aq). Calculate the amounts in moles of HCl(aq) and Al(OH)3(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of AlCl3(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of aluminum chloride: The lesser amount in moles of AlCl3(s) is the amount that is produced. Determine the molar mass of AlCl3(s). Convert the amount in moles of AlCl3(s) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of HCl(aq):

HCl H Cl1 11(1.01 g/mol) + 1(35.45 g/mol) 36.46 g/mol

M M M= +==



Molar mass, M, of Al(OH)3: ( )3 Al O HAl OH 1 3[1 1 ]

1(26.98 g/mol) + 3[1(16.00 g/mol) + 1(1.01 g/mol)]78.01 g/mol

M M M M= + +

==

Amount in moles, n, of HCl(aq):

HCl

8.0 g

mnM

=

=36.46 g

0.219418

/m

l

ol

mo=

Amount in moles, n, of Al(OH)3:

( )3Al OH

8.0 g

mnM

=

=78.01 g

0.102550

/m

l

ol

mo=

Balanced chemical equation: 3HCl(aq) + Al(OH)3(aq) → 3H2O(ℓ) + AlCl3(s) Mole ratio: 3 moles 1 mole 1 mole

Amount in moles, n, of AlCl3(s) from the complete consumption of hydrochloric acid: 3

3

3

AlCl 3

AlCl

Al l

3

C

3 mol HCl 0.219418 mol HCl 1

1 mol AlCl0.219418 mol HCl 3 mol HCl

0.219418 mol HC

mol AlC

l

l

n

n

n

=

=

× = ×

3× 1 mol AlCl 3 mol HCl

30.0731395 Amol lCl=

Amount in moles, n, of AlCl3(s) from the complete consumption of aluminum hydroxide:

( ) ( )

3

3

3

33

AlCl 3

3 3

AlCl

3AlC

3

l

1 mol Al OH 0.102550 mol Al O

1 mol AlCl0.102550

H 1 mol A mol Al(OH) 1 mol Al(OH)

0.102550 mol Al(OH)

lCl

n

n

n

× = ×

=

= 3

3

× 1 mol AlCl 1 mol Al(OH)

3 0.102550 Amol lCl=



The amount in moles of AlCl3(s) produced is the lesser of the two values: 0.0731395 mol Molar mass, M, of AlCl3(s):

3AlCl Al Cl1 3

1(26.98 g/mol) + 3(35.45 g/mol)133.33 g/mol

M M M= +

==

Mass, m, of the AlCl3(s):

3AlCl

0.0731395 mol

m n M= ×

= 133.33 g/ mol×9.751691 g9.8 g

==

The mass of aluminum chloride produced is 9.8 g. Check Your Solution The mole ratio in the balanced chemical equation of the reactants HCl(aq) and Al(OH)3(aq) is 3:1. Using rounded numbers, for 0.22 mol of HCl(aq) to be


× 0.22 mol or 0.07 mol of Al(OH)3(aq) is required.

The Al(OH)3(aq) must be the reactant in excess since 0.10 mol is given. The HCl(aq) is the limiting reactant. The mole ratio of HCl(aq) to AlCl3(s) is 3:1. The calculated mass of AlCl3(s) seems reasonable and correctly shows two significant digits.


The reaction between solid white phosphorus, P4(s), and oxygen gas produces solid tetraphosphorus decoxide, P4O10(s). Determine the mass of tetraphosphorus decoxide that is formed when 25.0 g of solid white phosphorus and 50.0 g of oxygen are combined. What Is Required? You need to determine the mass of tetraphosphorus decoxide produced in a reaction. What Is Given? You know the formula for the reactant phosphorus: P4(s). You know the formula for the product tetraphosphorus decoxide: P4O10(s) You know the mass of phosphorus: 25.0 g You know the mass of oxygen: 50.0 g



Plan Your Strategy Write the balanced chemical equation for the reaction. Determine the limiting reactant: Calculate the molar masses of P4(s) and O2(g). Calculate the amounts in moles of P4(s) and O2(g) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of P4O10(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of tetraphosphorus decoxide: The lesser amount in moles of P4O10(s) is the amount that is produced. Determine the molar mass of P4O10(s). Convert the amount in moles of P4O10(s) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Balanced chemical equation: P4(s) + 5O2(g) → P4O10(s) Molar mass, M, of P4(s):

4P P4

4(30.97 g/mol)123.88 g/mol

M M=

==


2O O2

2(16.00 g/mol)32.00 g/mol

M M=

==

Amount in moles, n, of P4(s):

4P

25.0 g

mnM

=

=123.88 g

0.201808

/m

l

ol

mo=




2O

50.0 g

mnM

=

=32.00 g

1.56250

/mo

l

l

mo=

Balanced chemical equation: P4(s) + 5O2(g) → P4O10(s) Mole ratio: 1 mole 5 moles 1 mole Amount in moles, n, of P4O10(s) from the complete consumption of solid white phosphorus:

4 10

4 10

4 10

4

P O 4 10

4 4

P O

4P O

4 4 10 1 mol P 0.201808 mol P 1 mol P

1 mol P O0.201808 mol P 1 mol P

0.201808 mol P

O

n

n

n

=

=

× = ×

4 10

4

× 1 mol P O 1 mol P

4 100.201808 m Pl Oo =

Amount in moles, n, of P4O10(s) from the complete consumption of oxygen gas:

4 10

4 10

4 10

2 2

P O 4 10

2 2

P

2

4

P O

10O 5 mol O 1.56250 mol O 1 mol P

1 mol P O1.56250 mol O 5 mol O

1.56250 mol O

O

n

n

n

=

× ×

=

=

4 10

2

× 1 mol P O 5 mol O

4 100.31250 m Pl Oo=

The amount in moles of P4O10(s) produced is the lesser of the two values: 0.201808 mol Molar mass, M, of P4O10(s):

4 10P O P O4 10

4(30.97 g/mol) + 10(16.00 g/mol)283.88 g/mol

M M M= +

==



Mass, m, of the P4O10(s): 4 10P O

0.201808 mol

m n M= ×

= 283.88 g/ mol×57.2892 g57.3 g

==

The mass of tetraphosphorus decoxide produced is 57.3 g. Check Your Solution The mole ratio in the balanced chemical equation of the reactants P4(s) and O2(g) is 1:5. Using rounded off numbers, for 0.2 mol of P4(s) to be completely

consumed, 51

× 0.2 mol or 1 mol of O2(g) is required. The O2(g) must be the

reactant in excess since 1.5 mol is given. The P4(s) is the limiting reactant. The mole ratio of P4(s) to P4O10(s) is 1:1. The calculated mass of P4O10(s) seems reasonable and correctly shows three significant digits.


A solution containing 14.0 g of silver nitrate, AgNO3(aq), is added to a solution containing 4.83 g of calcium chloride, CaCl2(aq). Find the mass of silver chloride, AgCl(s), produced. What Is Required? You need to determine the mass of silver chloride produced in a reaction. What Is Given? You know the names and chemical formulas for the two reactants: silver nitrate, AgNO3(aq); calcium chloride, CaCl2(aq) You know the name and chemical formula for one product: silver chloride, AgCl(s). You know the mass of silver nitrate: 14.0 g You know the mass of calcium chloride: 4.83 g Plan Your Strategy Write a balanced chemical equation for the reaction. The reaction is a double displacement reaction. The second product is calcium nitrate, Ca(NO3)2(aq). Determine the limiting reactant: Calculate the molar masses of AgNO3(aq) and CaCl2(aq). Calculate the amounts in moles of the AgNO3(aq) and the CaCl2(aq) using the

relationship .mnM

=




Calculate the amount in moles of AgCl(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of the silver chloride: The lesser amount in moles of AgCl(s) is the amount that is produced. Determine the molar mass, M, of AgCl(s). Convert the amount in moles of the AgCl(s) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Balanced chemical equation: 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) Molar mass, M, of AgNO3(aq):

3AgNO Ag N O1 1 3


M M M M= + +

==

Molar mass, M, of CaCl2(aq):

2CaCl Ca Cl1 2

4(40.08 g/mol) + 2(35.45 g/mol)110.98 g/mol

M M M= +

==

Amount in moles, n, of AgNO3(aq):

3AgNO

14.0 g

mnM

=

=169.88 g

0.0824111

/m

l

ol

mo=

Amount in moles, n, of CaCl2(aq):

2CaCl

4.83 g

mnM

=

=110.98 g

0.0435213

/m

l

ol

mo=



Balanced chemical equation: 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) Mole ratio: 2 moles 1 mole 2 moles 1 mole Amount in moles, n, of the AgCl(s) from the complete consumption of the silver nitrate:

AgCl

3 3

3

AgCl 3 3

AgCl


2 mol AgCl0.0824

2 mol AgC111 mol AgNO 2 mol AgNO

0.0824111 mol Ag

l

NO

n

n

n

× = ×

=

=

3


0.0824111 mol

l AgC=

Amount in moles, n, of the AgCl(s) from the complete consumption of the calcium chloride:

AgCl

2 2

AgCl

2AgCl

2 2 1 mol CaCl 0.0435213 mol CaCl

2 mol AgCl0.0435213 mol CaCl 1 mol CaCl

0.0435213 mol CaCl

2 mol AgCl

n

n

n

=

=

× = ×

2

× 2 mol AgCl 1 mol CaCl

0.0870426 mol

l AgC=

The amount in moles of the AgCl(s) produced is the lesser of the two values: 0.0824111 mol Molar mass, M, of AgCl(s):

AgCl Ag Cl1 1

1(107.87 g/mol) + 1(35.45 g/mol)143.32 g/mol

M M M= +

==

Mass, m, of the AgCl(s):

0.0824111 mol

m n M= ×

= 143.32 g/ mol×11.8111 g11.8 g

==

The mass of silver chloride produced is 11.8 g.



Check Your Solution The mole ratio of AgNO3(aq) to CaCl2(aq) is 2:1. Using rounded numbers, for

0.0824 mol of AgNO3(aq) to react completely, 12

× 0.0824 mol or 0.0412 mol

of CaCl2(aq) is required. The CaCl2(aq) is in excess since 0.0435 mol of this reactant is given. The limiting reactant is AgNO3(aq). The mole ratio of AgNO3(aq) to AgCl(s) is 2:2. The calculated mass of AgCl(s) is reasonable and correctly shows three significant digits.


The reaction between solid sodium and iron(III) oxide, Fe2O3(s), is one in a series of reactions that occurs when an automobile air bag inflates. 6Na(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) If 100.0 g of solid sodium and 100.0 g of iron(III) oxide are used in this reaction, what mass of solid iron will be produced? What Is Required? You need to determine the mass of iron, Fe(s), produced in a reaction. What Is Given? You know the balanced chemical equation: 6Na(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) You know the mass of sodium metal: 100.0 g You know the mass of iron(III) oxide: 100.0 g Plan Your Strategy Determine the limiting reactant: Determine the atomic molar mass of Na(s) using the periodic table. Calculate the molar mass of Fe2O3(s). Calculate the amounts in moles of Na(s) and Fe2O3(s) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of Fe(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the mass of iron: The lesser amount of Fe(s) is the amount that is produced. Determine the atomic molar mass of Fe(s) from the periodic table. Convert the amount in moles of the Fe(s) to mass (in grams) using the relationship m = n × M.



Act on Your Strategy Molar mass, M, of Na(s): M = 22.99 g/mol (from the periodic table) Molar mass, M, of Fe2O3(s):

2 3Fe O Fe O2 3

2(55.85 g/mol) + 3(16.00 g/mol)159.7 g/mol

M M M= +

==

Amount in moles, n, of Na(s):

Na

100.0 g

mnM

=

=22.99 g

4.3497

/mol

1mol=

Amount in moles, n, of Fe2O3(s):

2 3Fe O

100.0 g

mnM

=

=159.7 g

0.626174

/m

l

ol

mo=

Balanced chemical equation: 6Na(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) Mole ratio: 6 moles 1 mole 2 moles Amount in moles, n, of Fe(s) from the complete consumption of sodium metal:

Fe

e

Fe

F

6 mol Na 4.34971 mol

2 mol Fe4.34971 mol Na 6

Na 2 mol Fmol Na

4.34971 mol Na

e

n

n

n

× =

=

×

=× 2 mol Fe

6 mol Na1.44990 mol Fe=



Amount in moles, n, of Fe(s) from the complete consumption of iron(III) oxide:

Fe

2 3 2 3

Fe 2 3 2 3

2 3Fe

2 mol Fe0.626174 mol Fe O 1 mol Fe O

1 mol Fe O 0.626174 mol Fe O 2 mol Fe

0.626174 mol

Fe O

n

n

n

× = ×

=

=2 3

×2 mol Fe 1 mol Fe O

1.25234 mol Fe=

The Fe2O3(s) produces less Fe(s). Fe2O3(s) is the limiting reactant. The amount in moles of Fe(s) produced is the lesser of the two values: 1.25234 mol Molar mass, M, of Fe(s): M = 55.85 g/mol (from the periodic table) Mass, m, of the Fe(s):

1.25234 mol

m n M= ×

= 55.85 g/ mol×69.9436 g69.94 g

==

The mass of iron produced is 69.94 g. Check Your Solution The mole ratio of the reactants Na(s) and Fe2O3(s) in the balanced chemical equation is 6:1. Using rounded numbers, for 4.3 mol of Na(s) to be completely

consumed, 16

× 4.3 mol or 0.72 mol of Fe2O3(s) is required. The Fe2O3(s) must

be the limiting reactant since the given amount of iron(III) oxide is only 0.63 mol. The mole ratio of Fe2O3(s) to Fe(s) is 1:2. The calculated mass of Fe(s) seems reasonable and correctly shows four significant digits.


Manganese(III) fluoride, MnF3(s), is formed by the reaction of manganese(II) iodide, MnI2(s), with fluorine gas. 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ) a. If 1.23 g of manganese(II) iodide reacts with 25.0 g of fluorine, what mass of manganese(III) fluoride is produced?



b. Which reactant is in excess? How much of this reactant remains at the end of the reaction? a. mass of manganese(III) fluoride What Is Required? You need to determine the mass of manganese(III) fluoride produced in a reaction. What Is Given? You know the balanced chemical equation: 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ) You know the mass of manganese(II) iodide: 1.23 g You know the mass of fluorine: 25.0 g Plan Your Strategy Determine the limiting reactant: Calculate the molar masses of MnI2(s) and F2(g). Calculate the amounts in moles of MnI2(s) and F2(g) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of MnF3(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Mass of manganese(III) fluoride: The lesser amount of the MnF3(s) is the amount that is produced. Determine the molar mass of MnF3(s). Convert the amount in moles of the MnF3(s) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of MnI2(s):

2MnI Mn I1 2

1(54.94 g/mol) + 2(126.90 g/mol)308.74 g/mol

M M M= +

==

Molar mass, M, of F2(g):

2F F2

2(19.00 g/mol)38.00 g/mol

M M=

==



Amount in moles, n, of MnI2(s):

2MnI

1.23 g

mnM

=

=308.74 g

0.00398393

/ l

l

mo

mo=

Amount in moles, n, of the F2(g):

2F

25.0 g

mnM

=

=38.00 g

0.657894

/m

l

ol

mo=

Balanced chemical equation: 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ) Mole ratio: 2 moles 13 moles 2 moles Amount in moles, n, of the MnF3(s) from the complete consumption of manganese(II) iodide:

3

3

3

MnF 3

2 2

MnF

2MnF

2 2 3 2 mol MnI 0.00398393 mol MnI

2 mol MnF0

2 .00398393 mol MnI 2 mol MnI

0.00398393 mol

mol Mn

I

F

Mn

n

n

n

× = ×

=

= 3

2

× 2 mol MnF 2 mol MnI

30.00398393 mol MnF=

Amount in moles, n, of the MnF3(s) from the complete consumption of fluorine gas:

3

3

3

2

MnF 3

2 2

Mn 2 3F

2MnF

13 mol F 0.657894

2 mol MnF0.65789

mol F 2 mol 4 mol F 13 mol F

0.657894 mol F

MnF

n

n

n =

×

=

× =

3

2

× 2 mol MnF 13 mol F

30.101214 Mmol nF =

MnI2(s) produces the lesser amount in moles of MnF3(s). MnI2(s) is the limiting reactant.



The amount in moles of MnF3(s) produced is the lesser of the two values: 0.00398393 mol Molar mass, M, of MnF3(s):

3MnF Mn F1 3

1(54.94 g/mol) + 3(19.00 g/mol)111.94 g/mol

M M M= +

==

Mass. m, of the MnF3(s):

3MnF

0.00398393 mol

m n M= ×

= 111.94 g/ mol× 0.445961 g 0.446 g==

The mass of manganese(III) fluoride produced is 0.446 g. Check Your Solution The mole ratio of the reactants MnI2(s) and F2(g) in the balanced chemical equation is 2:13. Using rounded numbers, for 0.004 mol of MnI2(s) to be


× 0.004 mol or 0.026 mol of F2(g) is required. The

F2(g) must be the reactant in excess since the amount given is 0.66 mol. The mole ratio of MnI2(s) to MnF3(s) is 2:2. The calculated mass of MnF3(s) seems reasonable and correctly shows three significant digits. b. reactant in excess What is Required? You need to identify the reactant in excess and calculate the mass (in grams) of the reactant at the end of the reaction. What Is Given? You know the balanced chemical equation: 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ) From part a., you know the limiting reactant: MnI2(s) From part a., you know the amounts in moles of the reactants: 0.00398393 mol MnI2(s) 0.657894 mol F2(g) Plan Your Strategy Since the MnI2(s) is the limiting reactant, F2(g) is the reactant in excess. Use the mole ratio in the balanced chemical equation and the amount in moles of MnI2(s) from part a. to calculate the amount in moles of F2(g) that reacts.



Determine the molar mass of F2(g). Convert the amount in moles of F2(g) that reacts to mass (in grams) using the relationship .m n M= × Subtract the mass of the F2(g) that reacts from the initial mass of F2(g) to determine the mass of F2(g) that is in excess. Act on Your Strategy Amount in moles, n, of F2(g):

2

2

F 2

2 2

2F

13 mol F0.00398393 mol MnI 2 mol MnI

0.00398393 mol MnI

n

n

=

= 2

2

× 13 mol F 2 mol MnI

20.0258955 mol F =

Molar mass, M, of F2(g):

2F F2

2(19.00 g/mol) 38.00 g/mol

M M=

==

Mass, m, of the F2(g):

2F

0.0258955 mol

m n M= ×

= × 38.00 g/ mol0.984030 g0.984 g

==

mass of fluorine that remains initial mass mass that reacts

25.0 g 0.984 g24.0159 g24.0 g

= −= −==

Therefore, 24.0 g of F2(g) remains unreacted after the reaction. Check Your Strategy The units are correct and the substitution of the amounts in moles into the mole ratios is correct. The answer is reasonable and correctly shows three significant digits.



50. Practice Problem (page 311) Silver nitrate, AgNO3(aq), reacts with calcium chloride, CaCl2(aq), in the following reaction: 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) There are 7.000 mol of each reactant present. a. What is the mass of the excess reactant? b. What is the mass of the limiting reactant? c. What are the masses of each product that forms? a. mass of the excess reactant What Is Required? You need to determine the mass of the reactant that is in excess. What Is Given? You know the balanced chemical equation for the reaction: 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) You know the amount in moles of silver nitrate: 7.000 mol You know the amount in moles of calcium chloride: 7.000 mol Plan Your Strategy Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select a product to determine the reactant in excess. Calculate the amount in moles of the product produced if all of each reactant is completely consumed. The reactant in excess is the one that produces the greater amount in moles of product. Act on Your Strategy Balanced chemical equation: 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) Mole ratio: 2 moles 1 mole 2 moles Use the product AgCl(s) to determine the reactant in excess.

Amount in moles, n, of AgCl(s) from the complete consumption of silver nitrate:

AgCl 3

AgCl

3 3

3

3

AgCl


2 mol AgCl7.000

2 mol AgC mol A

lgNO 2 mol AgNO

7.000 mol AgNO

n

n

n

=

=

× = ×

3


7.000 mo l AgCl=



Amount in moles, n, of AgCl(s) from the complete consumption of calcium chloride:

AgCl 2

Ag

2

AgCl

Cl

2 2

2

1 mol CaCl 7.000 mol CaCl

2 mol AgCl7.000

2 mol AgClmol CaCl 1 mol CaCl

7.000 mol CaCl

n

n

n

× = ×

=

=

2


14.000 m lol AgC=

CaCl2(aq) produces a greater amount in moles of AgCl(s). CaCl2(aq) is the reactant in excess. Molar mass, M, of CaCl2(aq):

2CaCl Ca Cl1 2

1(40.08 g/mol) + 2(35.45 g/mol)110.98 g/mol

M M M= +

==

Mass, m, of the CaCl2(aq):

2CaCl

7.000 mol

m n M= ×

= 110.98 g/ mol×776.86 g=

The initial mass of calcium chloride is 776.9 g Since the mole ratio of AgNO3(aq) to CaCl2(aq) in the balanced chemical equation is 2:1, the amount in moles of CaCl2(aq) that reacts with 7.000 moles

of AgNO3(aq) is 12

× 7.000 mol or 3.5 mol.

mass of 3.5 mol CaCl2(aq) that reacts = 3.5 mol × 110.98 g/mol = 388.4 g b. mass of limiting agent What Is Required? You need to determine the mass of the limiting reactant. What is Given? You know from part a. which reactant is limiting: AgNO3(aq) You know the amount in moles of silver nitrate: 7.000 mol



Plan Your Strategy Since CaCl2(aq) has been identified as the reactant in excess, AgNO3(aq) must be the limiting reactant. Determine the molar mass of AgNO3(aq). Convert the amount in moles of AgNO3(aq) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of AgNO3(aq):

3AgNO Ag N O1 1 3


M M M M= + +

==

Mass, m, of the AgNO3(aq):

3AgNO

7.000 mol 169.88 g/mol1189.16 g1189 g

m n M= ×

= ×==

The mass of silver nitrate that reacts is 1189 g. c. mass of each product What Is Required? You need to determine the mass of each product. What is Given? You know the balanced chemical equation: 2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq) You know the limiting reactant: AgNO3(aq)You know the amount in moles of silver nitrate: 7.000 mol Plan Your Strategy Use the mole ratio in the balanced chemical equation and the given amount in moles of AgNO3(aq) to calculate the amount in moles of each product. Determine the molar masses of AgCl(s) and Ca(NO3)2(aq). Convert the amounts in moles of each product to masses (in grams) using the relationship .m n M= ×



Act on Your Strategy Amount in moles, n, of AgCl(s) from the complete consumption of silver nitrate:

AgCl

3 3

AgCl 3 3

AgCl3


2 mol AgCl7.000

2 mol AgCmol AgNO 2 mol AgNO

7.000 mol A

l

gNO

n

n

n

= ×

=

=

×

3

2 mol AgCl 2 mol AgNO

×

7.000 mol AgCl=

Molar mass, M, of AgCl(s):

AgCl Ag Cl1 1

1(107.87 g/mol) + 1(35.45 g/mol) 143.32 g/mol

M M M= +

==


AgCl

7.000 mol 143.32 g/mol1003.24 g1003 g

m n M= ×

= ×==

The mass of AgCl(s) is 1003 g.

Amount in moles, n, of Ca(NO3)2(aq) from the complete consumption of silver nitrate:

( )

3 2

3 2

3 2

3 3 3 2

Ca(NO ) 3 2

3 3

Ca(NO )

3Ca(NO )

2 mol AgNO 7.000 mol AgNO 1 mol

1 mol C

Ca NO

a(NO )7.000 mol AgNO 2 mol AgNO

7.000 mol AgNO

n

n

n

×

=

=

= ×

3 2

3

× 1 mol Ca(NO ) 2 mol AgNO

( )3 23.5 mol Ca NO

=

Molar mass, M, of Ca(NO3)2(aq):

( )

( ) ( )( ) ( ) ( )

33 2Ca NOCa NO

a N O

1 2

1 2 1 3

1 40.08 g/mol 2 1 14.01 g/mol 3 16.00 g/mol

164.10 g/mol

C

M M M

M M M

= +

⎡ ⎤= + +⎣ ⎦⎡ ⎤= + +⎣ ⎦

=



Mass, m, of the Ca(NO3)2(aq): ( )3 2Ca NO

3.5 mol 164.10 g/mol574.35 g574.4 g

m n M= ×

= ×==

The mass of Ca(NO3)2 is 574.3 g. Check Your Solution The units are correct and the substitutions into the mole ratio are correct. The answers match the mole ratio in the balanced chemical equation and seem reasonable. The answers should be expressed to four significant digits since the molar masses were determined to four significant digits.

Section 7.2 Limiting and Excess Reactants Solutions for Selected Review Questions Student Edition page 313 4. Review Question (page 313)

In an experiment, 57.4 g of iron(III) chloride in solution reacts with 45.3 g of sodium hydroxide in solution, as shown below. FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) a. Which reactant is the limiting reactant? b. How much of the excess reactant remains after the reaction? c. How much of each product forms? a. limiting reactant What Is Required? You need to determine the limiting reactant. What Is Given? You know the balanced chemical equation for the reaction: FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) You know the mass of iron(III) chloride: 57.4 g You know the mass of sodium hydroxide: 45.3 g Plan Your Strategy Calculate the molar masses of FeCl3(aq), and NaOH(aq). Calculate the amount in moles of FeCl3(aq) and NaOH(aq) using the

relationship .mnM

=



Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select one of the products. Calculate the amount in moles of this product that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount in moles of the product is the limiting reactant. Act on Your Strategy Molar mass, M, of FeCl3(aq):

3FeCl Fe Cl1 3

1(55.85 g/mol) + 3(35.45 g/mol) 162.2 g/mol

M M M= +

==

Molar mass, M, of NaOH(aq):

NaOH Na O H1 1 11(22.99 g/mol) + 1(16.00 g/mol) + 1(1.01 g/mol) 40.00 g/mol

M M M M= + +==

Amount in moles, n, of FeCl3(aq):

3FeCl

57.4 g

mnM

=

=162.2 g /mol

0.353884 mol=

Amount in moles, n, of NaOH(aq):

NaOH

45.3 g

mnM

=

=40.00 g

1.13250

/mo

l

l

mo=

Balanced chemical equation: FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) Mole ratio: 1 mole 3 moles 1 mole 3 moles Use the product NaCl(aq) to determine the limiting reactant.



Amount in moles, n, of NaCl(aq) from the complete consumption of iron(III) chloride:

NaCl 3 3

N

NaCl

3

3aC

3

l

353884 1 mol FeCl 0.353884 mo

3 mol NaCl0. mol

l FeCl 3 mol FeCl 1 mol FeCl

0.353884 mol Fe

NaCl

Cln

n

n

=

× = ×

=3

× 3 mol NaCl 1 mol FeCl

1.061652

mol NaCl =

Amount in moles, n, of NaCl(aq) from the complete consumption of sodium hydroxide:

Na

NaCl

Cl

NaCl

3 mol NaOH 1.13250 mol Na

3 mol NaCl1.13250 mol NaOH 3 mol

OH 3 mol Na NaOH

1.13250 mol Na

Cl

OH

n

n

n

=

=

× = ×

× 3 mol NaCl 3 mol NaOH

1.13250

mol NaCl=

FeCl3(aq) produces a lesser amount of NaCl(aq). Iron(III) chloride is the limiting reactant. Check Your Solution The mole ratio of FeCl3(aq) to NaOH(aq) is 1:3. Using rounded numbers, for

0.35 mol of FeCl3(aq) to react completely, 31

× 0.35 mol or 1.05 mol of

NaOH(aq) is required. The NaOH(aq) is in excess since the given amount of the reactant is 1.1 mol. The limiting reactant is FeCl3(aq). b. mass of excess reactant What Is Required? You need to calculate the mass that remains of the reactant in excess. What is Given? You know the balanced chemical equation: FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) You know the limiting reactant: FeCl3(aq) You know the amount in moles of FeCl3(aq): 0.353884 mol You know the total amount in moles of NaOH(aq): 1.13250 mol



Plan Your Strategy Since FeCl3(aq) is the limiting reactant, NaOH(aq) must be the reactant in excess. Use the mole ratio in the balanced chemical equation and the calculated amount in moles of FeCl3(aq) to calculate the amount in moles of NaOH(aq) that reacts. Subtract the amount in moles of NaOH(aq) that reacts from the initial amount in moles of NaOH(aq). Convert this amount in moles of NaOH(aq) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Amount in moles, n, of NaOH(aq):

NaOH

3 3

Na

3

OH 3 3

NaOH

1 mol FeCl 0.3353884 mol FeCl

3 mol NaOH0.3538

3 mol NaO84 mol

H FeCl 1 mol FeCl

0.353884 mol FeCl

n

n

n

× = ×

=

=

× 3 mol NaOH 1 mol FeCl 3

1 .06165 mol NaOH=

Amount in moles, n, of unreacted NaOH(aq):

unreacted NaOH initial amount in moles amount in moles that react1.13250 mol 1.06165 mol0.07848 mol

n = −= −=

Mass, m, of unreacted NaOH(aq):

NaOH

0.07848 mol

m n M= ×

= 40.00 g/ mol×2.83392 g2.83 g

==

Therefore, 2.83 g of sodium hydroxide remains. Check Your Solution The mole ratios have been substituted correctly. The answer seems reasonable and correctly shows three significant digits. c. amount of each product What Is Required? You need to determine the mass of each product.



What is Given? You know the balanced chemical equation: FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) You know the limiting reactant: FeCl3(aq). You know the amount in moles of FeCl3(aq)(from part a.): 0.353884 mol You know the amount in moles of NaCl(aq) produced (from part a.): 1.061652 mol Plan Your Strategy Determine the molar masses of Fe(OH)3(s) and NaCl(aq). Calculate the mass of the NaCl(aq) using the relationship .m n M= × Use the mole ratio in the balanced chemical equation and the amount in moles of the FeCl3(aq) to calculate the amount in moles of the Fe(OH)3(s). Convert each amounts in moles to mass (in grams) using the relationship

.m n M= × Act on Your Strategy Molar mass, M, of NaCl(aq):

NaCl Na Cl1 11(22.99 g/mol) + 1(35.99 g/mol) 58.44 g/mol

M M M= +==

Molar mass, M, of Fe(OH)3(s):

( )3 Fe OHFe OH

Fe O H

1 3

1 3[1 1 ]1(55.85 g/mol) + 3[1(16.00 g/mol) +1(1.01 g/mol)] 106.88 g/mol

M M M

M M M

= +

= + +==

Mass, m, of the NaCl(aq):

NaCl

1.061652 mol

m n M= ×

= × 58.44 g/ mol62.0429 g62.0 g

==

The mass of the NaCl(s) is 62.0 g. Balanced chemical equation: FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) Mole ratio: 1 mole 3 moles 1 mole 3 moles



Amount in moles, n, of the Fe(OH)3(s): 3

3

3

Fe(OH) 3

3 3

Fe(OH) 3 3 3

3Fe(OH)

1 mol Fe(OH)0.353884 mol FeCl 1 mol FeCl

1 mol FeCl 0.353884 mol FeCl 1 mol Fe(OH)

0.353884 mol FeCl

n

n

n

=

× = ×

= 3

3

× 1 mol Fe(OH)1 mol FeCl

30.0353884 mol Fe(OH)=

Mass, m, of the Fe(OH)3(s):

( )3Fe OH

0.353884 mol

m n M= ×

= × 106.88 g/ mol37.82312 g37.8 g

==

The mass of the iron(III) hydroxide is 37.8 g. Check Your Solution The units are correct and the substitutions into the mole ratio are correct. The answer matches the mole ratio in the balanced chemical equation and seems reasonable.


Solid sodium metal reacts with chlorine gas to form table salt. 2Na(s) + Cl2(g) → 2NaCl(s) What is the minimum mass of chlorine gas that is required to consume 2.25 g of solid sodium? What Is Required? You need to determine the mass of chlorine gas that reacts. What Is Given? You know the balanced chemical equation for the reaction: 2Na(s) + Cl2(g) → 2NaCl(s) You know the mass of the solid sodium metal: 2.25 g Plan Your Strategy Determine the atomic molar mass of Na(s) using the periodic table.

Calculate the amount in moles of Na(s) using the relationship .mnM

=



Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced chemical equation and the calculated amount in moles of the Na(s) to calculate the amount in moles of the Cl2(g). Determine the molar mass of Cl2(g). Calculate the mass (in grams) of the Cl2(g) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of Na(s): M = 22.99 g/mol (from the periodic table) Amount in moles, n, of Na(s):

2.25 g

mnM

=

=22.99 g /mol

0.0978686 mol=

Balanced chemical equation: 2Na(s) + Cl2(g) → 2NaCl(s) Mole ratio: 2 moles 1 mole Amount in moles, n, of the Cl2(g):

2

2

2

Cl 2

Cl 2

Cl

1 mol Cl0.0978686 mol Na 2 mol Na

2 mol Na 0.0978686 mol Na 1 mol Cl

0.0978686 mol Na

n

n

n

=

× = ×

= 2× 1 mol Cl2 mol Na

20.0489343 mol Cl=


2Cl Cl2

2(35.45 g/mol)70.90 g/mol

M M=

==

Mass, m, of the Cl2(g):

2Cl

0.0489

343 mol

m n M= ×

= 70.9 g/ mol×3.46944 g3.47 g

==

The mass of chlorine that reacts is 3.47 g



Check Your Solution Use the factor-label method to set up an expression in which all the terms cancel except for the mass of chlorine.

2Cl 2.25 g Nam =1 mol Na ×

22.99 g Na21 mol Cl

× 2 mol Na

2

2

70.9 g Cl × 1 mol Cl

3.47 g=


Copper reacts with nitric acid, HNO3(aq), as follows: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ) What mass of nitrogen monoxide, NO(g), is produced when 50.0 g of copper reacts with 150.0 g of nitric acid? What Is Required? You need to determine the mass of nitrogen monoxide produced in a reaction. What Is Given? You know the balanced chemical equation: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ) You know the mass of copper: 50.0 g You know the mass of nitric acid: 150.0 g Plan Your Strategy Determine the atomic molar mass of Cu(s) using the periodic table. Determine the molar mass of HNO3(aq). Calculate the amounts in moles of Cu(s) and HNO3(aq) using the relationship

.mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Calculate the amount in moles of NO(g) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. The lesser amount of NO(g) is the amount that is produced. Determine the molar mass of NO(g). Convert the amount in moles of NO(g) to mass (in grams) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of Cu(s): M = 63.55 g/mol (from the periodic table)



Molar mass, M, of the Fe2O3(s): 2 3Fe O Fe O2 3

2(55.85 g/mol) + 3(16.00 g/mol) 159.70 g/mol

M M M= +

==

Mass, m, of the Fe2O3(s):

2 3Fe O

9.43828 mol

m n M= ×

= 159.70 g/ mol×

3

1507.29 g1.51 10 g

=

= ×

Molar mass, M, of HNO3(aq):

3HNO H N O1 1 3


M M M M= + +

==

Amount in moles, n, of Cu(s):

Cu

50.0 g

mnM

=

=63.55 g /mol

0.786782 mol=

Amount in moles, n, of HNO3(aq):

3HNO

150.0 g

mnM

=

=63.02 g /mol

2.3801967 mol=

Balanced chemical equation: 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ) Mole ratio: 3 moles 8 moles 2 moles



Amount in moles, n, of the NO(g) from the complete consumption of the copper:

NO

O

2

NO

N

3 mol Cu 0.786782 mol C

2 mol NO0.786782 mol Cu 3 mo

u 2 mol l Cu

0.786782 mol

NO

Cu

n

n

n

=

=

× = ×

× 2 mol NO 3 mol Cu

0.524521 mol NO=

Amount in moles, n, of the NO(g) from the complete consumption of the nitric acid:

NO

3 3

3

NO 3 3

NO

8 mol HNO 2.38019 mol H

2 mol NO2.38019 mol HNO 8 mol HN

NO 2 mol NOO

2.38019 mol HNO

n

n

n

× = ×

=

=

3

× 2 mol NO 8 mol HNO

0.595047 O mol N=

Cu(s) produces the lesser amount in moles of NO(g). Cu(s) is the limiting reactant. The amount in moles of NO(g) produced is the lesser of the two values: 0.524521 mol Molar mass, M, of NO(g):

NO N O1 11(14.01 g/mol) + 1(16.00 g/mol) 30.01 g/mol

M M M= +==

Mass, m, of NO(g):

NO

0.524521 mol

m n M= ×

= 30.01 g/ mol×15.7408 g15.7 g

==

The mass of nitrogen monoxide produced is 15.7 g.



Check Your Solution The mole ratio of the reactants Cu(s) and HNO3(aq) in the balanced chemical equation is 3:8. Using rounded numbers, for 0.79 mol of Cu(s) to be


× 0.79 mol or 2.1 mol of HNO3(aq) is required. The

HNO3(aq) must be the reactant in excess since the given amount is 2.3 mol. The mole ratio of Cu(s) to NO(g) is 3:2. The calculated mass of NO(g) seems reasonable and correctly shows three significant digits.

Section 7.3 Reaction Yields Solutions for Practice Problems Student Edition page 319

51. Practice Problem (page 319) During an investigation, calcium carbide, CaC2(s), reacted with excess water to make calcium hydroxide, Ca(OH)2 (aq), and acetylene, C2H2(g). CaC2(s) + 2H2O(ℓ) → Ca(OH)2(aq) + C2H2(g) The data table for this investigation is given below.

Mass of Calcium Carbide That Reacted 2.38 g Mass of Acetylene That Was Produced 0.77 g

What was the theoretical yield and percentage yield of acetylene? What Is Required? You need to determine the theoretical yield and percentage yield of acetylene. What Is Given? You know the balanced chemical equation for the reaction: CaC2(s) + 2H2O(ℓ) → Ca(OH)2(aq) + C2H2(g) You know the mass of calcium carbide that reacts: 2.38 g You know the mass of acetylene produced: 0.77 g Plan Your Strategy Determine the molar mass of CaC2(s).

Calculate the amount in moles of CaC2(s) using the relationship .mnM

=

Use the mole ratio in the balanced chemical equation and the calculated amount in moles of CaC2(s) to determine the amount in moles of C2H2(g). Determine the molar mass of C2H2(g). Determine the theoretical yield by converting the amount in moles of the C2H2(g) to mass (in grams) using the relationship .m n M= ×



Act on Your Strategy Molar mass, M, of CaC2(s):

2CaC Ca C 1 2

1(40.08 g/mol) + 2(12.01 g/mol)64.10 g/mol

M M M= +

==

Amount in moles, n, of CaC2(s):

2CaC

2.38 g

mnM

=

=64.10 g

0.0371294

/m

l

ol

mo=

Balanced chemical equation: CaC2(s) + 2H2O(ℓ) → Ca(OH)2(aq) + C2H2(g) Mole ratio: 1 mole 1 mole Known ratio of acetylene to calcium carbide equated to the unknown ratio:

2 2C H 2 2

2 2

1 mol C H 0.0371294 mol CaC 1 mol CaC

n=

Amount in moles, n, of the C2H2(g):

2 2

2 2

2 2

C H 2 2

2 2

C H

2

2 2 2 2

C H

1 mol CaC 0.0371294 m

1 mol C H 0.037129

ol CaC 1 mol 4 mol CaC 1 mol CaC

0.0371294 mol CaC

C H

n

n

n

= ×

=

×

= 2 2

2

× 1 mol C H 1 mol CaC

0.0371294 mol=

Molar mass, M, of C2H2(g):

2 2C H C H2 2

1(12.01 g/mol) + 2(1.01 g/mol)26.04 g/mol

M M M= +

==

Mass, m, of the C2H2(g):

2 2C H

0.0371294 mol

m n M= ×

= 26.04 g/ mol×0.966849 g0.97 g

==



Percentage yield: actual yieldpercentage yield × 100%

theoretical

0.77 g

yield=

=0.97 g

× 100%

79.4%=

The theoretical yield is 0.97 g and the percentage yield is 79.4%. Check Your Solution The theoretical yield should be greater than the actual yield. The units are correct and the answer seems reasonable. The answer correctly shows three significant digits.


Suppose that 0.250 mol of potassium carbonate, K2CO3(s), reacts with excess hydrochloric acid as follows: K2CO3(s) + 2HCl(aq) → H2O(ℓ) + CO2(g) + 2KCl(aq) a. Calculate the theoretical yield of potassium chloride. b. Calculate the percentage yield of water if 0.189 mol of water is produced. What Is Required? You need to determine the theoretical yield of potassium chloride and the percentage yield of water. What Is Given? You know the balanced chemical equation for the reaction: K2CO3(s) + 2HCl(aq) → H2O(ℓ) + CO2(g) + 2KCl(aq) You know the amount in moles of potassium carbonate reacted: 0.250 mol You know the actual yield of water: 0.189 mol Plan Your Strategy a. theoretical yield of potassium chloride Use the mole ratio in the balanced equation and the given amount in moles of K2CO3(s) to calculate the amount in moles of KCl(aq). Determine the molar mass of KCl(aq). Calculate the mass (in grams) of KCl using the relationship .m n M= × This is the theoretical yield of KCl(aq).



b. percentage yield of water Use the mole ratio in the balanced equation and the given amount in moles of K2CO3(s) to calculate the amount in moles of H2O(ℓ). This is the theoretical yield of H2O(ℓ). The actual yield of H2O(ℓ) is 0.189 mol.

Calculate the percentage yield: actual yieldpercentage yield × 100%theoretical yield

=

Act on Your Strategy a. theoretical yield of potassium chloride Balanced equation: K2CO3(s) + 2HCl(aq) → H2O(ℓ) + CO2(g) + 2KCl(aq) Mole ratio: 1 mole 1 mole 2 moles Amount in moles, n, of KCl(aq) from the complete consumption of potassium carbonate:

KCl 2

KCl

2 3 2 3

2

3 2 3

KCl3

1 mol K CO 0.250 m

2 mol KCl 0.250 m

ol K CO 2 mol KClol K CO 1 mol K CO

0.250 mol K CO

n

n

n

×

=

=

= ×

2 3

× 2 mol KCl 1 mol K CO

0.500 m lol KC=

Molar mass, M, of KCl(aq):

KCl K Cl 1 11(39.10 g/mol) + 1(35.45 g/mol)74.55 g/mol

M M M= +==

Mass, m, of the KCl(aq):

0.500 mol

m n M= ×

= 74.55 g/ mol×37.275 g37.3 g

==

The theoretical yield of potassium chloride is 37.3 g. Check Your Solution Given the mole ratio K2CO3:KCl of 1:2 in the equation for this reaction, a theoretical yield of 37.3 g of KCl is reasonable. The answer correctly shows three significant digits.



b. percentage yield of water Balanced equation: K2CO3(s) + 2HCl(aq) → H2O(ℓ) + CO2(g) + 2KCl(aq) Mole ratio: 1 mole 1 mole 2 moles Amount in moles, n, of H2O(ℓ) from the complete consumption of potassium carbonate:

2

2

2

2

H O 2

2 3 2 3

3 2 3 2H O

2 3H O

1 mol K CO 0.250 mol

1 mol H O 0.250 m

K CO 1 mol ol K CO 1 mol K CO

0.250 mol K CO

H

O

n

n

n

=

=

× = ×

2

2 3

× 1 mol H O 1 mol K CO

20.250 mol H O=

The theoretical yield of H2O(ℓ) is 0.250 mol. The actual yield of H2O(ℓ) is 0.189 mol. Percentage yield of H2O(ℓ):

actual yieldpercentage yield × 100%theoretical

0.

yiel

189 o

d

m l

=

=0.250 mol

× 100%

75.6%=

The percentage yield of H2O(ℓ) is 75.6 %. Check Your Solution Given the mole ration K2CO3:H2O of 1:1 in the equation for this reaction and an actual yield of 0.189 mol of H2O, the calculated percentage yield is reasonable. The answer correctly shows three significant digits.


Phosphoric acid, H3PO4(aq), is neutralized by potassium hydroxide, KOH(aq), according to the following reaction: H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(ℓ) If 49.0 g of potassium phosphate, K3PO4(aq), is recovered after 49.0 g of phosphoric acid reacts with 49.0 g of potassium hydroxide, what is the percentage yield of the reaction?



What Is Required? You need to determine the percentage yield of potassium phosphate in the reaction. What Is Given? You know the balanced chemical equation for the reaction: H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(ℓ) You know the mass of phosphoric acid: 49.0 g You know the mass of potassium hydroxide: 49.0 g You know the actual yield of potassium phosphate: 49.0 g Plan Your Strategy Determine the limiting reactant in the reaction: Determine the molar masses of H3PO4(aq) and KOH(aq). Calculate the amounts in moles of H3PO4(aq) and KOH(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Identify the limiting reactant by calculate the amount in moles of K3PO4(aq) produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the percentage yield: The lesser amount of K3PO4(aq) is the theoretical yield. Determine the molar mass of K3PO4(aq). Convert the amount in moles of K3PO4(aq) to mass (in grams) using the relationship .m n M= ×


=

Act on Your Strategy Molar mass, M, of H3PO4(aq):

3 4H PO H P O3 1 4


M M M M= + +

==

Molar mass, M, of KOH(aq):

KOH K O H1 1 11(39.10 g/mol) + 1(16.00 g/mol) + 1(1.01 g/mol)56.11 g/mol

M M M M= + +==



Amount in moles, n, of H3PO4(aq):

3 4H PO

49.0 g

mnM

=

=98.00 g

0.5000

/mol

mol=

Amount in moles, n, of KOH(aq):

KOH

49.0 g

mnM

=

=56.11 g

0.873284

/m

l

ol

mo=

Balanced chemical equation: H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(ℓ) Mole ratio: 1 moles 3 moles 1 mole Amount in moles, n, of K3PO4(aq) from the complete consumption of phosphoric acid:

3 4

3 4

3 4

K PO 3 4

K PO 3 4

3 3

3 4 3 4

K P

3 3

O3 3

1 mol H PO 0.5000 mol H PO 1 mol

1 mol K PO 0.5000 mol H PO 1 mol H PO

0.5000 mol H PO

K PO

n

n

n =

=

× = ×

3 4

3 3

× 1 mol K PO 1 mol H PO

3 40.5000 mol K PO=

Amount in moles, n, of K3PO4(aq), from the complete consumption of potassium hydroxide:

4

4

3

3

3

4

K PO 3

K PO 3

4

O

4

K P

3 mol KOH 0.873284 mol KOH

1 mol K PO 0.873284 mol KOH

1 mol K PO3 mol KOH

0.873284 mol KOH

n

n

n

× = ×

=

=

3 4× 1 mol K PO 3 mol KOH

3 40.291094 ol Km PO=

KOH(aq) produces the lesser amount in moles of K3PO4(aq). Potassium hydroxide is the limiting reactant. The theoretical yield of K3PO4(aq) is 0.291094 mol.



Molar mass, M, of K3PO4(aq): 3 4K PO K P O3 1 4


M M M M= + +

==

Mass, m, of K3PO4(aq):

3 4K PO

0.291094 mol

m n M= ×

= × 212.27 g/ mol61.7907 g=

Percentage yield of K3PO4(aq):


49.0 g

yield=

=61.7907 g

× 100%

79.2999%79.3%

==

The percentage yield is 79.3%. Check Your Solution Using rounded values, estimate the amount in moles of each reactant and determine the limiting reactant using the mole ratio in the balanced chemical equation to estimate the amounts in moles of reactants that combine.

Estimate of amount in moles of H3PO4: 50 g

100 g0.5 mol

/mol≈

Estimate of amount in moles of KOH: 50 g

55 g1 mol

/mol≈

KOH

3 4 3 4

KOH

3 mol KOH0.5 mol H P O 1 mol H P O

1.5 mol

n

n

=

≈

This means that for all of the H3PO4 to be used up, about 1.5 mol of KOH is needed. Since only about 1 mol of KOH is given, KOH is the limiting reactant.



Use the mass of KOH and the factor-label method to check the theoretical yield and the percentage yield.

theoretical yield 49.0 g KOH=1 mol × KOH

56.11 g KOH3 41 mol K PO

×

3 mol KOH3 4

3 4

212.27 g K PO × 1 mol K PO

3 461.790 Kg PO=

percentage yield 49.0 g

=61.790 g

× 100%

79.3%=

Both methods give the same result.


The reaction of glucose, C6H12O6(s), with sulfuric acid, H2SO4(ℓ), produces carbon as follows: C6H12O6(s) + 2H2SO4(ℓ) → 6C(s) + 6H2O(ℓ) + 2H2SO4(aq) a. If 20.8 g of glucose reacts with excess sulfuric acid, what is the theoretical yield, in grams, of carbon? b. If the percentage yield is 72.0%, what mass of carbon is produced? What Is Required? You need to determine the theoretical yield of carbon and the actual mass of carbon produced in the reaction. What Is Given? You know the balanced chemical equation: C6H12O6(s) + 2H2SO4(ℓ) → 6C(s) + 6H2O(ℓ) + 2H2SO4(aq) You know the mass of glucose: 20.8 g You know the percentage yield: 72.0% Plan Your Strategy Determine the molar mass of C6H12O6(s).

Calculate the amount in moles of C6H12O6(s) using the relationship .mnM

=

Use the mole ratio in the balanced chemical equation and the calculated amount in moles of C6H12O6(s) to calculate the amount in moles of carbon produced in the reaction. This is the theoretical yield of carbon.



Determine the atomic molar mass of C(s) using the periodic table. Convert the amount in moles of C(s) to mass (in grams) using the relationship

.m n M= × Act on Your Strategy a. theoretical yield of carbon Molar mass, M, of C6H12O6(s):

6 12 6C H O C H O6 12 6


M M M M= + +

==

Amount in moles, n, of C6H12O6(s):

6 12 6C H O

20.8 g

mnM

=

=180.18 g

0.115440

/m

l

ol

mo=

Balanced chemical equation: C6H12O6(s) + 2H2SO4(ℓ) → 6C(s) + 6H2O(ℓ) + 2H2SO4(aq) Mole ratio: 1 mole 6 moles Known mole ratio of C(s) to C6H12O6(s) equated to the unknown ratio:

C

6 12 6 6 12 6

6 mol C0.115440 mol C 1 mol CH H OO

n=

Amount in moles, n, of C(s):

6 12 6

6 12 6

C

6 12 6

C 6 12 6

6 12 6C

0.115440 mol C H O 0.115440 mol C H O

6 mol C1 mol C H O

1 mol C H O 6×

0.115440 m

mo

ol

l

C

C

H O

n

n

n

=

× =

=6 12 6

6 mol C1 mol C H

O

×

0.692640 mol=

Molar mass of C(s): M = 12.01 g/mol (from the periodic table)



Mass, m, of the C(s): C

0.692640 mol

m n M= ×

= × 12.01 g/ mol8.31861 g8.32 g

==

The theoretical yield of carbon is 8.32 g. b. mass of carbon produced

actual yieldpercentage yield × 100%theoretical yieldactual yield72.0% × 100%

8.32 g

72.0 %actual yield

=

=

= 8.32 g

100 %×

5.99 g=

The actual yield of carbon is 5.99 g. Check Your Solution The actual yield should be about 70% of the theoretical yield. The answer is reasonable and correctly shows three significant digits.


Calcium chloride, CaCl2(aq), is mixed with silver nitrate, AgNO3(aq), to form calcium nitrate, Ca(NO3)2(aq), and silver chloride, AgCl(s). CaCl2(aq) + 2AgNO3(aq) → Ca(NO3)2(aq) + 2AgCl(s) If this reaction has an 81.5% yield, what mass of silver chloride is produced when 21.2 g of calcium chloride is added to excess silver nitrate? What Is Required? You need to determine the actual mass of silver chloride produced. What Is Given? You know the balanced chemical equation: CaCl2(aq) + 2AgNO3(aq) → Ca(NO3)2(aq) + 2AgCl(s) You know the mass of calcium chloride: 21.2 g You know the silver nitrate is in excess. You know the percentage yield: 81.5%



Plan Your Strategy Determine the molar mass of CaCl2(aq).

Calculate the amount in moles of CaCl2(aq) using the relationship .mnM

=

Use the mole ratio in the balanced chemical equation and the calculated amount in moles of CaCl2(aq) to calculate the amount in moles of AgCl(s) that forms. Determine the molar mass of AgCl(s). To determine the theoretical yield of AgCl(s), convert the amount in moles of the AgCl(s) to mass (in grams) using the relationship m n M= × . Use the expression for percentage yield to calculate the actual yield of

AgCl(s): actual yieldpercentage yield × 100%theoretical yield

=

Act on your Strategy Molar mass, M, of CaCl2(aq):

2CaCl Ca Cl1 2

1(40.08 g/mol) + 2(35.45 g/mol) 110.98 g/mol

M M M= +

==

Amount in moles, n, of the CaCl2(aq):

21.2 g

mnM

=

=110.98 g

0.191025

/m

l

ol

mo=

Balanced chemical equation: CaCl2(aq) + 2AgNO3(aq) → Ca(NO3)2(aq) + 2AgCl(s) Mole ratio: 1 mole 2 moles Amount in moles, n, of the AgCl(s):

AgCl

2 2

Ag

2

Cl 2 2

AgCl

1 mol CaCl 0.191025 mol CaCl

2 mol AgCl0.1910

2 mol AgCl25 mol CaCl 1 mol CaCl

0.191025 mol CaCl

n

n

n

× = ×

=

=

2


0.382050 mol AgCl=



Molar mass, M, of AgCl(s): AgCl Ag Cl1 1

1(107.87 g/mol) + 1(35.45 g/mol) 143.32 g/mol

M M M= +

==


AgCl

0.382050 mol

m n M= ×

= × 143.32 g/ mol54.7555 g54.8 g

==

The theoretical yield of silver chloride is 54.8 g. Actual yield of the AgCl(s):

actual yieldpercentage yield × 100%theoretical yieldactual yield81.5% × 100%

54.8 g

actual yield 8 5 1. %

=

=

= × 54.8 g

100 %44.6 g=

The actual yield of silver chloride is 44.6 g. Check Your Solution The actual yield should be about 80% of the theoretical yield. The answer is reasonable and correctly shows three significant digits.


The following reaction has a 68% yield. AlCl3(aq) + 4NaOH(aq) → NaAlO2(aq) + 3NaCl(aq) + 2H2O(ℓ) Calculate the actual mass of sodium chloride that is recovered if 18.2 g of aluminum chloride, AlCl3(aq), reacts with 16.00 g of sodium hydroxide. What Is Required? You need to determine the actual mass of sodium chloride produced in the reaction.



What Is Given? You know the balanced chemical equation: AlCl3(aq) + 4NaOH(aq) → NaAlO2(aq) + 3NaCl(aq) + 2H2O(ℓ) You know the mass of aluminum chloride: 18.2 g You know the mass of sodium hydroxide: 16.00 g You know the percentage yield: 68% Plan Your Strategy Determine the limiting reactant in the reaction: Determine the molar masses of AlCl3(aq) and NaOH(aq). Calculate the amounts in moles of AlCl3(aq) and NaOH(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Determine the limiting reactant by calculating the amount in moles of NaCl(aq) produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the actual yield of sodium chloride: The lesser amount of NaCl(aq) is the theoretical yield. Determine the molar mass of NaCl(aq). Convert the lesser amount in moles of NaCl(aq) to mass (in grams) using the relationship .m n M= × Use the expression for percentage yield to calculate the actual yield of

NaCl(aq): actual yieldpercentage yield × 100%theoretical yield

=

Act on Your Strategy Molar mass, M, of AlCl3(aq):

3AlCl Al Cl1 3

1(26.98 g/mol) + 3(35.45 g/mol) 133.33 g/mol

M M M= +

==

Molar mass, M, of NaOH(aq):

NaOH Na O H1 1 11(22.99 g/mol) + 1(16.00 g/mol) + 1(1.01 g/mol) 40.00 g/mol

M M M M= + +==



Amount in moles, n, of the AlCl3(aq):

3AlCl

18.2 g

mnM

=

=133.33 g

0.136503

/m

l

ol

mo=

Amount in moles, n, of the NaOH(aq):

NaOH

16.0 g

mnM

=

=40.00 g

0.4000

/mol

mol=

Balanced chemical equation: AlCl3(aq) + 4NaOH(aq) → NaAlO2(aq) + 3NaCl(aq) + 2H2O(ℓ) Mole ratio: 1 mole 4 moles 3 moles Amount in moles, n, of NaCl(aq), from the complete consumption of aluminum chloride:

NaCl

3 3

Na

3

Cl 3 3

NaCl

1 mol AlCl 0.136503 mol AlCl

3 mol NaCl0.1365

3 mol NaCl03 mol AlCl 1 mol AlCl

0.136503 mol AlCl

n

n

n

× = ×

=

=

3

× 3 mol NaCl 1 mol AlCl

0.409509 mol=

Amount in moles, n, of NaCl(aq), from the complete consumption of sodium hydroxide:

Na

NaCl

Cl

NaCl

4 mol NaOH 0.4000 mol NaO

3 mol NaCl0.4000 mol NaOH 4 mol Na

H 3 mol NaClOH

0.4000 mol NaOH

n

n

n

×

=

=

= ×

× 3 mol NaCl 4 mol NaOH

0.3000 m lol NaC=

NaOH(aq) produces the lesser amount in moles of NaCl(aq). Sodium hydroxide is the limiting reactant. The theoretical yield of NaCl(aq) is 0.3000 mol.



Molar mass, M, of NaCl(aq): NaCl Na Cl1 1

1(22.99 g/mol) + 1(35.45 g/mol) 58.44 g/mol

M M M= +==

Mass, m, of the NaCl(aq):

NaCl

0.30000 mol

m n M= ×

= × 58.44 g/ mol17.5320 g=

Calculation of actual yield:

actual yieldpercentage yield × 100%theoretical yieldactual yield68% × 100%17.5320 g

actual yield 68 %

=

=

= × 17.5320 g100 %

11.9217 g11.9 g

==

Check Your Solution The actual yield should be about 70% of the theoretical yield. The answer is reasonable and correctly shows three significant digits.


Ethyl butanoate, C6H13O2(ℓ), is an organic ester that has the flavour and scent of pineapple. It is prepared as follows: C4H9O2(ℓ) + C2H6O(ℓ) → C6H13O2(ℓ) + H2O(ℓ) During an investigation, 0.573 mol of butanoic acid, C4H9O2(ℓ), reacts with excess ethanol, C2H6O(ℓ). What mass of ethyl butanoate is produced if this reaction has a 92.0% yield? What Is Required? You need to determine the mass of ethyl butanoate produced. What Is Given? You know the balanced chemical equation for the reaction: C4H9O2(ℓ) + C2H6O(ℓ) → C6H13O2(ℓ) + H2O(ℓ)



You know the amount in moles of butanoic acid: 0.573 mol You know the percentage yield: 92.0% Plan Your Strategy Use the mole ratio in the balanced chemical equation and the given amount in moles of C4H9O2(ℓ) to calculate the amount in moles of C6H13O2(ℓ). Determine the molar mass of C6H13O2(ℓ). To determine the theoretical yield of C6H13O2(ℓ), convert the amount in moles of the C6H13O2(ℓ) to mass (in grams) using the relationship m n M= × . Use the expression for percentage yield to calculate the actual yield of C6H13O2(ℓ). Act on Your Strategy Balanced chemical equation: C4H9O2(ℓ) + C2H6O(ℓ) → C6H13O2(ℓ) + H2O(ℓ) Mole ratio: 1 mole 1 mole Amount in moles, n, of the C6H13O2(ℓ) from the complete consumption of butanoic acid:

6 13 2

6 13 2

6 13 2

4 9 2 4 9 2

C H O 6 13 2

4 9 2 4 9 2

C H O

4 9C O

6 3 2

2

1

H

1 mol C H O 0.573 mol C H O 1 mol C H

1 mol C H O 0.573 mol C H O 1 mol C H O

0.573 mol C H O

O

n

n

n

=

=

× = ×

6 13 2

4 9 2

× 1 mol C H O 1 mol C H O

6 13 20.573 mol C H O =

Molar mass, M, of C6H13O2(ℓ):

6 13 2C H O C H O6 13 2


M M M M= + +

==

Mass, m, of the C6H13O2(ℓ):

6 13 2C H O

0.573 mol

m n M= ×

= 117.19 g/ mol×67.14987 g=



Actual yield of the C6H13O2(ℓ): actual yieldpercentage yield × 100%

theoretical yieldactual yield92.0% × 100%67.14987 g

actual yiel 9 .d 2 0 %

=

=

= × 67.14987 g100 %

61.7778 g61.8 g

==

The actual yield of ethyl butanoate is 61.8 g. Check Your Solution The actual yield should be about 90% of the theoretical yield. The answer is reasonable and correctly shows three significant digits.


An impure sample of barium hydroxide, Ba(OH)2(aq), has a mass 0.540 g. It is dissolved in water and then treated with excess sulfuric acid, H2SO4(aq). This results in the formation of a precipitate of barium sulfate, BaSO4(s). H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O(ℓ) The barium sulfate is filtered, and any remaining sulfuric acid is washed away. Then the barium sulfate is dried and its mass is measured to be 0.62 g. What mass of barium hydroxide was in the original (impure) sample? What Is Required? You need to determine the mass of barium hydroxide in the original (impure) sample. What Is Given? You know the balanced chemical equation: H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O(ℓ) You know the mass of barium sulfate: 0.62 g You know the mass of the impure sample of barium hydroxide: 0.540 g Plan Your Strategy Determine the molar mass of BaSO4(s).

Calculate the amount in moles of BaSO4(s) using the relationship .mnM

=



Use the mole ratio in the balanced chemical equation and the calculated amount in moles of BaSO4(s) to calculate the amount in moles of Ba(OH)2(aq). Determine the molar mass of Ba(OH)2(aq). Calculate the mass of Ba(OH)2(aq) using the relationship .m n M= × Act on Your Strategy Molar mass, M, of BaSO4(s):

4BaSO Ba S O1 1 4


M M M M= + +

==

Amount in moles, n, of the BaSO4(s):

4BaSO

0.62 g

mnM

=

=233.40 g

0.00265638

/ l

l

mo

mo=

Balanced chemical equation: H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2H2O(ℓ) Mole ratio: 1 mole 1 mole Amount in moles, n, of Ba(OH)2:

( )

( )

2

2

2

4

Ba(OH) 2

4 4

4 2Ba OH

a4

B OH

× 1 mol BaSO 0.00265638 mol BaSO × 1 m

1 mol Ba(OH) 0.00265638 mol BaSO 1 mol BaSO

0.002656

ol Ba

38 mol BaSO

(OH)

n

n

n=

=

= 2

4

× 1 mol Ba(OH) 1 mol BaSO

0.00265638 mol=

Molar mass, M, of Ba(OH)2(aq):

( )2 Ba O HBa OH 1 2[1 1 ]


M M M M= + +

==



Mass, m, of the Ba(OH)2(aq): ( )2Ba OH

0.00265638 mol

m n M= ×

= × 171.35 g/ mol0.45517138 g 0.46 g

==

The mass of barium hydroxide in the original sample is 0.46 g. Check Your Solution The mass would be expected to be less than the mass of the impure sample. The answer is reasonable. The units are correct and the answer correctly shows two significant digits.


Iron pyrite, FeS2(s), reacts with oxygen as shown in the reaction below: 4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g) a. In a laboratory, 5.000 kg of an impure mineral, which contains 45.3% iron pyrite, reacts with oxygen. Calculate the mass of iron(III) oxide, Fe2O3(s), that forms. Assume that all the pyrite reacts. b. Suppose that the reaction has a 78.0% yield, due to an incomplete reaction. How many grams of iron(III) oxide are produced? What Is Required? a. You need to determine the mass of iron(III) oxide that forms. b. You need to determine how many grams of iron(III) oxide is produced in an incomplete reaction. What Is Given? You know the balanced chemical equation: 4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g) You know the mass of the mineral sample: 5.000 kg You know the mineral is 45.3% iron pyrite. Plan Your Strategy a. mass of iron(III) oxide Use the 45.3% mass percent of the iron pyrite in the mineral to find the mass of FeS2(s). Convert the mass of FeS2(s) from kilograms to grams: 1 kg = 1 × 103 g Determine the molar mass of FeS2(s).

Calculate the amount in moles of FeS2(s) using the relationship .mnM

=



Use the mole ratio in the balanced chemical equation and the calculated amount of FeS2(s) to calculate the amount in moles of Fe2O3(s). Determine the molar mass of Fe2O3(s). Convert the amount in moles of the Fe2O3(s) to mass (in grams) using the relationship .m n M= × Convert the mass from grams to kilograms: 1 g = 1 × 10–3 kg b. grams of iron(III) oxide produced Use the expression for percentage yield to calculate the actual

yield: actual yieldpercentage yield × 100%theoretical yield

=

Act on Your Strategy a. mass of iron(III) oxide Mass, m, of the FeS2(s):

45.3 %m =100 %

× 5.000 kg

2.265 kg=

Mass conversion:

2FeS 2.265 kg

2.265 kg

m =

= 3 × 1 × 10 g/ kg

2265 g=

Molar mass, M, of FeS2(s):

2FeS Fe S1 2

1(55.85 g/mol) + 2(32.07 g/mol) 119.99 g/mol

M M M= +

==

Amount in moles, n, of FeS2(s):

2FeS

2265 g

mnM

=

=119.99 g

18.87657

/m

l

ol

mo=

Balanced chemical equation: 4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g) Mole ratio: 4 moles 2 moles



Amount in moles, n, of the Fe2O3(s): 2 3

2 3

2 3

2

Fe O 2 3

2 2

Fe 2 2

2O

3O

Fe

4 mol FeS 18.87657 mol FeS

2

2

mol Fe O 1

mol Fe8.87657 mol FeS 4 mol FeS

18.87657 mol S

O

Fe

n

n

n

× = ×

=

= 2 3

2

× 2 mol Fe O 4 mol FeS

2 3mol9.43 F82 e8 O=

Molar mass, M, of the Fe2O3(s):

2 3Fe O Fe O2 3

2(55.85 g/mol) + 3(16.00 g/mol) 159.70 g/mol

M M M= +

==

Mass, m, of the Fe2O3(s):

2 3Fe O

9.43828 mol

m n M= ×

= 159.70 g/ mol×

3

1507.29 g1.51 10 g

=

= ×

Mass conversion:

2 3

3Fe O

3

1.51 10 g

1.51 10 g

m = ×

= × –3× 1 × 10 kg/ g

1.51 kg=

The mass of iron(III) oxide would be 1.51 × 103 g or 1.51 kg. b. grams of iron(III) oxide produced in an incomplete reaction:

actual yieldpercentage yield × 100%theoretical yieldactual yield78.0% × 100%1507.29 g

actual yiel 7 .d 8 0 %

=

=

= × 1507.29 g

100 %31.118 10 g= ×

The actual yield of iron(III) oxide is 1.18 × 103 g or 1.18 kg.



Check Your Solution The units are correct and show the correct number of significant digits. The answers are reasonable for the mole ratios in this equation.

60. Practice Problem (page 319) Sodium oxide, Na2O(s), reacts with water to form the base sodium hydroxide. Na2O(s) + H2O(ℓ)→ 2NaOH(aq) If this reaction has a 91% yield, what mass of sodium hydroxide is obtained when 0.483 mol of sodium oxide reacts with excess water? What Is Required? You need to determine the mass of sodium hydroxide produced. What Is Given? You know the balanced chemical equation for the reaction: Na2O(s) + H2O(ℓ) → 2NaOH(aq) You know the amount in moles of sodium oxide: 0.483 mol You know the percentage yield: 91% Plan Your Strategy Use the mole ratio in the balanced chemical equation and the given amount in moles of Na2O(s) to calculate the amount in moles of NaOH(aq). Determine the molar mass of NaOH(aq). Calculate the mass (in grams) of NaOH(aq) using the relationship .m n M= × This is the theoretical yield of NaOH(aq). Use the expression for percentage yield to calculate the actual yield of NaOH(aq):

actual yieldpercentage yield × 100%theoretical yield

=

Act on Your Strategy Balanced chemical equation: Na2O(s) + H2O(ℓ)→ 2NaOH(aq) Mole ratio: 1 mole 2 moles

Amount in moles, n, of NaOH(aq) from the complete consumption of sodium oxide:

NaOH 2

Na

2

NaOH

OH

2 2

2

1 mol Na O 0.483 mol Na O

2 mol NaOH 0.483 mol Na O 1 mo

2 mol Nal Na O

0.483 mol Na O

OHn

n

n

× = ×

=

=

2

× 2 mol NaOH1 mol Na O

0.9660 m Hol NaO=



Molar mass, M, of NaOH(aq): NaOH Na O H1 1 1

1(22.9 g/mol) + 1(16.00 g/mol)+ 1(1.01 g/mol) 40.00 g/mol

M M M M= + +==

Mass, m, of the NaOH(aq):

NaOH

0.9660 mol

m n M= ×

= × 40.00 g/ mol38.64 g=

Actual yield of NaOH(aq):

actual yieldpercentage yield × 100%theoretical yieldactual yield91% × 100%

38.64 g

91%actual yield

=

=

= × 38.64 g

100 %

35 g=

The actual yield of sodium hydroxide is 35 g Check Your Solution The actual yield should be about 90% of the theoretical yield. The answer is reasonable and correctly shows two significant digits.

Section 7.3 Reaction Yields Solutions for Selected Review Questions Student Edition page 321 4. Review Question (page 321)

Silver nitrate, AgNO3(aq), reacts with potassium bromide, KBr(aq), to produce silver bromide, AgBr(s). AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq) If 14.64 g of silver bromide is obtained when 14.00 g of silver nitrate reacts with excess potassium bromide, what is the percentage yield? What Is Required? You need to determine the percentage yield of silver bromide in a reaction.



What Is Given? You know the balanced chemical equation: AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq) You know the mass of silver nitrate that reacts: 14.00 g You know the actual yield of silver bromide: 14.64 g Plan Your Strategy Determine the molar mass of AgNO3(s).

Calculate the amount in moles of the AgNO3(s) using the relationship .mnM

=

Determine the mole ratio from the balanced chemical equation. Use the mole ratio in the balanced equation and the calculated amount in moles of silver nitrate to calculate the amount in moles of the AgBr(s). Determine the molar mass of AgBr(s). To determine the theoretical yield of the AgBr(s), convert the amount in moles of the AgBr(s) to mass (in grams) using the relationship .m n M= ×


=

Act on Your Strategy Molar mass, M, of AgNO3(s):

3AgNO Ag N O1 1 3


M M M M= + +

==

Amount in moles, n, of the AgNO3(s):

3AgNO

14.00 g

mnM

=

=169.88 g

0.0824111

/m

l

ol

mo=

Balanced chemical equation: AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq) Mole ratio: 1 mole 1 mole Known mole ratio equated to the unknown ratio:

AgBr

3 3

1 mol AgBr 0.0824111 mol AgNO 1 mol AgNO

n=



Amount in moles, n, of AgBr(s):

Ag

AgBr

3 3

3

Br 3 3

AgBr


1 mol AgBr 0.0824

1 mol A111 mol AgNO 1 mol AgNO

0.08

gBr

24111 mol AgNO

n

n

n

× = ×

=

=

3

× 1 mol AgBr 1 mol AgNO

0.0824111 r mol AgB=

Molar mass, M, of AgBr(s):

AgBr Ag Br1 1

1(107.87 g/mol) + 1(79.90 g/mol)187.77 g/mol

M M M= +

==

Mass, m, of the AgBr(s):

AgBr

0.0824111 mol

m n M= ×

= × 187.77 g/ mol15.4743 g15.47 g

==

The theoretical yield of silver bromide is 15.47 g. Percentage yield of the AgBr(s):


14.64 g

=

=15.47 g

× 100%

%

94.63=

The percentage yield is 94.63%.

Check Your Solution The units are correct and the answer seems reasonable. The answer correctly shows four significant digits.



5. Review Question (page 321) Sodium phosphate, Na3PO4(aq), reacts with barium nitrate, Ba(NO3)2(aq), as represented by 2Na3PO4(aq) + 3Ba(NO3)2(aq) → Ba3(PO4)2(s) + 6NaNO3(aq). If 5.00 g of sodium phosphate reacts with 10.90 g of barium nitrate and 7.69 g of solid precipitate is recovered, what is the percentage yield? What Is Required? You need to determine the percentage yield of barium phosphate in a reaction. What Is Given? You know the balanced chemical equation for the reaction: 2Na3PO4(aq) + 3Ba(NO3)2(aq) → Ba3(PO4)2(s) + 6NaNO3(aq) You know the mass of sodium phosphate: 5.00 g You know the mass of barium nitrate: 10.90 g You know the actual yield of the precipitate, barium phosphate: 7.69 g Plan Your Strategy Determine the limiting reactant in the reaction: Determine the molar masses of Na3PO4(aq) and Ba(NO3)2(aq). Calculate the amounts in moles of Na3PO4(aq) and Ba(NO3)2(aq) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio for the reactants and products. Identify the limiting reactant by calculating the amount in moles of the Ba3(PO4)2(s) produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the percentage yield: The lesser amount of Ba3(PO4)2(s) is the theoretical yield. Determine the molar mass of Ba3(PO4)2(s). Convert the lesser amount in moles of Ba3(PO4)2(s) to mass (in grams) using the relationship .m n M= ×


=

Act on Your Strategy Molar mass, M, of Na3PO4(aq):

3 4Na PO Na P O3 1 4


M M M M= + +

==



Molar mass, M, of Ba(NO3)2(aq): ( ) 33 2

Ba NOBa NO 1 2


M M M= +

==

Amount in moles, n, of Na3PO4(aq):

3 4Na PO

5.00 g

mnM

=

=163.94 g

0.0304989

/m

l

ol

mo=

Amount in moles, n, of Ba(NO3)2 (aq):

3 2Ba(NO )

10.9 g

mnM

=

=261.35 g

0.0417065

/m

l

ol

mo=

Balanced chemical equation: 2Na3PO4(aq) + 3Ba(NO3)2(aq) → Ba3(PO4)2(s) + 6NaNO3(aq) Mole ratio: 2 moles 3 moles 1 mole Amount in moles, n, of Ba3(PO4)2(s) from the complete consumption of sodium phosphate:

( )

3 4 2

3 4 2

3 4 2

2 4 3

Ba (PO ) 3 4 2

3 4 3 4

Ba (PO )

3 4Ba O

4 2

( )

4

P

3× 2 mol Na PO = 0.0304989 mol Na PO × 1 mol B

1 mol Ba (PO ) 0.0304989 mol

a PO

Na PO 2 mol Na PO

0.0304989 mol Na PO

n

n

n =

=

3 4 2

3 4

× 1 mol Ba (PO ) 2 mol Na PO

3 4 20.0152494 mo Ba (P Ol )=



Amount in moles, n, of Ba3(PO4)2(s) from the complete consumption of barium nitrate:

( )

3 4 2

3 4 2

3 4 2

Ba (PO ) 3 4 2

3 2 3 2

Ba (PO ) 3 2 3 3 4 22

3 2Ba (PO )

3 mol

1 mol Ba (PO ) 0.0417065 mol Ba(NO ) 3 mol Ba(NO )

Ba(NO ) B 0.0417065 mol a(NO )

0.0417065 mol Ba(

1 mol Ba

N )

PO

O

n

n

n

× = ×

=

=

3 4 2

3 2

× 1 mol Ba (PO ) 3 mol Ba(NO )

3 4 20.0139021 mo Ba (P Ol )=

Ba(NO3)2(aq) produces the lesser amount in moles of Ba3(PO4)2(s). Barium nitrate is the limiting reactant. The theoretical yield of Ba3(PO4)2(s) is 0.0139021 mol. Molar mass, M, of Ba3(PO4)2(s):

3 4 2 4Ba (PO ) Ba PO3 2


M M M= +

==

Mass, m, of Ba3(PO4)2(s):

( )3 4 2Ba PO

0.0139021 mol

m n M= ×

= × 601.93 g/ mol8.36813 g8.37 g

==

Percentage yield of Ba3(PO4)2(s):


7.69 g

yield=

=8.37 g

× 100%

91.8757%91.9%

==

The percentage yield is 91.9%. Check Your Solution Using rounded values, estimate the amount in moles of each reactant and determine the limiting reactant using the mole ratio in the balanced equation to estimate the amounts of reactants that combine.



Estimate of amount in moles of Na3PO4(aq): 5 g

160 g0.03 mol

/mol≈

Estimate of amount in moles of Ba(NO3)2(aq): 10 g

260 g0.04 mol

/mol≈

3 2

3 2

Ba(NO ) 3 2

3 4 3 4

Ba(NO )

3 mol Ba(NO )0.3 mol Na PO 2 mol Na PO

0.45 mol

n

n

=

≈

This means that for all of the Na3PO4(aq) to be used up, about 0.45 mol of Ba(NO3)2(aq) is needed. Since the given amount of Ba(NO3)2(aq) is only about 0.04 mol, Ba(NO3)2(aq) is the limiting reactant.

Use the mass of Ba(NO3)2 and the factor-label method to check the theoretical yield:

3 2theoretical yield g Ba(N0.9 O1 )= 3 21 mol Ba(N

O

)×

3 2261.35 g Ba(NO )

3 4 2

1 mol Ba (PO ) ×

3 2

3 mol Ba(NO )

3 4 2

3 4 2

601.93 g Ba (PO ) 1 mol Ba (PO )

×

3 4 28.37 g Ba (PO )=

actual yieldpercentage yield × 100%theoretical7.69 g8.3

yield

× 100%

91.9%7 g

=

=

=

Both methods give the same result.


Sodium iodide, NaI(s), and chlorine gas undergo a single displacement reaction, as represented by the following equation: 2NaI(s) + Cl2(g) → I2(s) + 2NaCl(s) a. If 4.0 g of sodium iodide and 4.0 g of chlorine gas react, what are the theoretical yields of both products? b. If the percentage yield of sodium chloride is 67%, what is the actual yield of sodium chloride?



c. Would it be appropriate to assume that the yield of the other product, iodine, I2(s), is also 67%? Explain. What Is Required? a. You need to determine the theoretical yields of iodine and sodium chloride. b. You need to calculate the actual yield of the sodium chloride. c. You need to explain if it is appropriate to assume that the yields of the products are equal. What Is Given? You know the balanced chemical equation: 2NaI(s) + Cl2(g) → I2(s) + 2NaCl(s) You know the mass of sodium iodide: 4.0 g You know the mass of chlorine gas: 4.0 g Plan Your Strategy Determine the limiting reactant: Determine the molar masses of NaI(s) and Cl2(g). Calculate the amounts in moles of NaI(s) and Cl2(g) using the

relationship .mnM

=

Use the balanced chemical equation to determine the mole ratio between the reactants and products. Select I2(s) to determine which reactant is limiting. Calculate the amount in moles of I2(s) that is produced if all of each reactant is completely consumed. The reactant that produces the lesser amount of product is the limiting reactant. Determine the theoretical yield of each product: The lesser amount of the I2(s) is the theoretical yield. Determine the molar mass of I2(s). Convert the amount in moles of the I2(s) to mass (in grams) using the relationship .m n M= × Use the limiting reactant to calculate the theoretical yield of the NaCl(s). Calculate the amount in moles of NaCl(s) that is produced if all of the limiting reactant is completely consumed. Calculate the molar mass of NaCl(s). Convert the amount in moles of the NaCl(s) to mass (in grams) using the relationship .m n M= × Calculate the actual yield of the NaCl(s) using the expression for percentage

yield:


=



Act on Your Strategy a. theoretical yields of iodine and sodium chloride Molar mass, M, of NaI(s):

NaI Na I1 11(22.99 g/mol) + 1(126.9 g/mol) 149.89 g/mol

M M M= +==


2Cl Cl2

2(35.45 g/mol)70.90 g/mol

M M=

==

Amount in moles, n, of NaI(s):

NaI

4.0 g

mnM

=

=149.89 g

0.0266862

/m

l

ol

mo=

Amount in moles, n, of Cl2(g):

2Cl

4.0 g

mnM

=

=70.90 g

0.0564174

/m

l

ol

mo=

Balanced chemical equation: 2NaI(s) + Cl2(g) → I2(s) + 2NaCl(s) Mole ratio: 2 moles 1 mole 1 mole 2 moles Amount in moles, n, of the I2(s) from the complete consumption of sodium iodide:

2

2

2

I 2

I

I

2 2 mol NaI 0.0266862 mol

1 mol I 0.0266862 mol NaI 2

NaI 1 mol I mol NaI

0.0266862 mol NaI

n

n

n

× ×

=

=

=

2× 1 mol I 2 mol NaI

20.0133431 mol I=



Amount in moles, n, of the I2(s) from the complete consumption of chlorine gas:

2

2

2

I 2

2 2

I 2 2

I

2

2

1 mol Cl 0.0564174 mol Cl

1 mol I 0.056

1 mol4174 mol Cl 1 mol Cl

0.0564174 m

I

ol Cl

n

n

n

× = ×

=

=

2

2

× 1 mol I 1 mol Cl

20.0564174 mol I=

NaI(aq) produces the lesser amount in moles of I2(s). Sodium iodide is the limiting reactant. The theoretical yield of I2(s) is 0.0133431 mol. Molar mass, M, of I2(s):

2I I2

2(126.9 g/mol)253.8 g/mol

M M=

==

Mass, m, of I2(s):

2I

0.0133431 mol

m n M= ×

= × 253.8 g/ mol3.38647 g3.4 g

==

The theoretical yield of iodine is 3.4 g

Amount in moles, n, of NaCl(s) from the complete consumption of the sodium iodide:

Na

NaCl

Cl

NaCl

2 mol NaI 0.0266862 mol NaI

2 mol NaCl 0.0266862 mol NaI 2 mol NaI

0.0266862 mol NaI

2 mol NaCln

n

n

× = ×

=

=

× 2 mol NaCl 2 mol NaI

0.0266862 mol NaCl=

Molar mass, M, of NaCl(s):

NaCl Na Cl1 11(22.99 g/mol) + 1(35.45 g/mol)58.44 g/mol

M M M= +==



Mass, m, of the NaCl(s): NaCl

0.0266862 mol

m n M= ×

= × 58.44 g/ mol1.55954 g1.6 g

==

The theoretical yield of sodium chloride is 1.6 g b. actual yield of sodium chloride

actual yieldpercentage yield × 100%theoretical yieldactual yield67% × 100%1.55954 g

67 %actual yield =

=

=

× 1.55954 g100 %

1.04453 g1.0 g

==

The actual yield of sodium chloride is 1.0 g. c. comparison of yield of products For this reaction, the percentage yield would be expected to be 67% for both products. The percentage yield is less than 100% because of factors that interfere with the reactants combining completely. This would affect the yield of both products in the same way. There are factors that can affect the measurement of the amount of product. For example, if the product is a gas, there may be experimental loss of product. If a product is partially soluble in water, there may be experimental error in measuring the amount of product.

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