Prentice Hall Chemistry (c) 2005 Section Assessment Answers Chapter 12 By Daniel R. Barnes Init: 12/17/2008 WARNING: some images and content in this presentation

Prentice Hall Chemistry(c) 2005

Section Assessment Answers

Chapter 12By Daniel R. BarnesInit: 12/17/2008

WARNING: some images and content in this presentation may have been taken without permission from the world wide web. It is intended for use only by Mr. Barnes and his students. It is not meant to be copied or distributed.

SWBAT . . . . . . determine the mass of one substance involved in a chemical reaction if given the mass of any one of the other substances involved in the reaction.

in other words . . .

. . . convert grams of the known into grams of the unknown.

12.1 Section Assessment5. How is a balanced chemical equation similar to a recipe?

A recipe says what ingredients you need to make some kind of food. A balanced chemical equation says what reactants you need to make some kind of product.

As the teacher’s edition says, “Both a balanced equation and a recipe give quantitative information about the starting and end materials.

12.1 Section Assessment6. How do chemists use balanced equations?

Chemists use balanced equations as a basis to calculate how much reactant is needed or how much product is formed in a reaction.

If you run a factory that makes chemicals, knowing how much of each reactant you’re going to use lets you know how much it’s going to cost you, and knowing how much product will be formed tells you how much money you can expect to make when you sell the product.

Gross sales – fabrication costs =

PROFIT

For instance . . .

12.1 Section Assessment7. Chemical reactions can be described in terms of what quantities?

Chemical reactions can be described in terms of numbers atoms, molecules, or moles; in terms of mass; or in terms of volume.

12.1 Section Assessment8. What quantities are always conserved in chemical reactions?Mass and atoms are always conserved in chemical reactions.

If you start with 18 kg of reactants, you should end up with . . .

For example . . .

If 3 million carbon atoms react with 6 million oxygen atoms to form carbon dioxide, the carbon dioxide will consist of . . . 3 million carbon atoms and 6 million oxygen atoms.

Volume is NOT conserved. (Example: a small handful of nitrogen triiodide can explode to form several liters of nitrogen gas and iodine vapors.)Molecules are NOT conserved. (Example: two million hydrogen molecules and one million oxygen molecules become two million water molecules. 3 million molecules 2 million molecules.)

. . .18 kg of products. Chemical reactions do not create or destroy matter.

12.1 Section Assessment9. Interpret the given equation in terms of relative numbers of

representative particles , numbers of moles, and masses of reactants and products.

2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

In terms of “representative particles”:

Two atoms of K react with 2 molecules of H2O to form 2 formula units of KOH and one molecule of H2.

In terms of mass:

78.2 g K + 36.0 g H2O 112.2 g KOH + 2.0 g H2

Two moles of K

Two moles of H2O

Two moles of KOH

One mole of H2

No coefficient = an invisible coefficient of “1”

12.1 Section Assessment10. Balance this equation:C2H5OH(l) + O2(g) CO2(g) + H2O(g)

Show that the balanced equation obeys the law of conservation of mass.

C2H5OH(l) + O2(g) CO2(g) + H2O(g)

C = C =H = H = O = O =

263

123

2

2

3

65 7

3

7



First, I’d like to show how the unbalanced equation disobeys the law of conservation of mass

To do this, I’ll need to calculate the molar mass of each reactant and each product . . .

C2H5OH: C: 2 x 12 = 24 H: 6 x 1 = 6 O: 1 x 16 = 16----------------------- 46 g/mol

46 g





O2: O: 2 x 16 = 32----------------------- 32 g/mol

46 g 32 g





CO2: C: 1 x 12 = 12 O: 2 x 16 = 32----------------------- 44 g/mol

46 g 32 g 44 g





H2O: H: 2 x 1 = 2 O: 1 x 16 = 16----------------------- 18 g/mol

46 g 32 g 44 g 18 g





46 g 32 g 44 g 18 g

46 + 32 = 78 grams of reactants

44 + 18 = 62 grams of products

You can’t have 16 grams of matter just disappear like that.

In its unbalanced state, the equation violates the law of conservation of matter. It needs to be balanced!



C2H5OH(l) + O2(g) CO2(g) + H2O(g)2 33

46 g 32 g 44 g 18 gx 1

46 g

x 3

96 g

x 2

88 g

x 3

54 g

46 g + 96 g = 142 g 88 g + 54 g = 142 g

(in terms of mass, anyway)

Don’t forget that one “mole” of a material =

6.022 x 1023 molecules*

of that material.

*If the material is a noble gas or a metal, subsitute the word “atoms” for “molecules”. If the material is ionic, substitute “forumula units” for “molecules”.

12.2 Practice Problems13. Acetylene gas (C2H2) is produced by adding water to calcium

carbide (CaC2).

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)How many grams of acetylene are produced by adding water to

5.00 g CaC2?

5 g CaC2

1x

g CaC2

mol CaC2

CaC2: Ca: 1 x 40 = 40 C: 2 x 12 = 24

64 g/mol

64

1x

mol CaC2

mol C2H2

1

1x

mol C2H2

g C2H2

126

C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2

26 g/mol

5 g g

26x 5

130 ) 130 64

2.03

2.03

12.2 Practice Problems14. Using the same equation, determine how many moles of

CaC2 are needed to react completely with 49.0 g H2O.

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)

49 g H2O1

xg H2O

mol H2O18

1x

mol H2O

mol CaC22

1

H2O:

H: 2 x 1 = 2 O: 1 x 16 = 16---------------------- 18 g/mol

49

18 x 2 mol CaC2=

49

36 mol CaC2=

1.36 mol CaC2

= 4.82 x 1022 molecules O2

12.2 Practice Problems15. How many molecules of oxygen are produced by the

decomposition of 6.54 g of potassium chlorate (KClO3)?

2KClO3 2KCl + 3O2

6.54 g KClO3

1x

122 g KClO3

1 mol KClO3 x

KClO3:

K: 1 x 39 = 39 Cl: 1 x 35 = 35 O: 3 x 16 = 48 ---------------------- 122 g/mol

x2 mol KClO3

3 mol O21 mol O2

6 x 1023

molecules O2

6.54 x 3 x 6 x 1023

122 x 2117.72 x 1023

244= = 0.482459 x 1023

= 4.82459 x 1022

16. The last step in the production of nitric acid is the reaction of nitrogen dioxide with water:

3NO2 + H2O 2HNO3 + NO

How many grams of nitrogen dioxide must react with water to

produce 5.00 x 1022 molecules of nitrogen monoxide?

6 x 1023 molecules NO

12.2 Practice

Pro

blem

s

1

5 x 3 x 46 x 1022

1 mol NO

NO: N: 1 x 14 = 14 O: 2 x 16 = 32 ---------------------- 46 g/mol

5 x 1022 molecules NO 3 mol NO2

1 mol NO46 g NO2

1 mol NO2

6 x 1023= 690 x 1022

6 x 1023

= 115 x 10-1 g NO2

= 1.15 x 101 g NO2

= 11.5 g NO2

12.2 Practice Problems

Notice how the 22.4’s just ended up canceling each other out? It almost makes you wonder why I bothered putting them there in the first place . . .


Look back at my work-out for #18 and notice again how the 22.4’s just ended up canceling each other out. If I’d put those two fractions in my work-out for this one, the same thing would have happened, so I just didn’t bother putting them in. I could have also put in two fractions to turn mL to L and then L back to mL again, but the 1000’s would have also have just canceled each other out, so I didn’t bother. 12.2 Practice Problems


12.2 Section Assessment21. How are mole ratios used in chemical calculations?

Mole ratios are written using the coefficients from a balanced chemical equation. They are used to relate moles of reactants and products in stoichiometric calculations.

12.2 Section Assessment22. Outline the sequence of steps needed to solve a typical

stoichiometric problem.

i. Convert the given quantity to moles using the molar mass of the known chemical. ii. Use the mole ratio (from the appropriate coefficients in the balanced chemical equation for the reaction) to find moles of the desired chemical. iii. Using the molar mass of the desired chemical, convert moles of the desired chemical into grams of the desired chemical.

12.2 Section Assessment23.Write the 12 mole ratios that can be derived from the equation

for the combustion of isopropyl alcohol.

2C3H7OH(l) + 9O2(g) 6CO2(g) + 8H2O(g)

Holy crud. You ready? Here they come.

2 mol C3H7OH

9 mol O2

2 mol C3H7OH

6 mol CO2

2 mol C3H7OH

8 mol H2O

2 mol C3H7OH

9 mol O2

6 mol CO2 8 mol H2O

9 mol O2 9 mol O2

2 mol C3H7OH

6 mol CO2

8 mol H2O9 mol O2

6 mol CO2 6 mol CO2

2 mol C3H7OH

8 mol H2O

9 mol O2 6 mol CO2

8 mol H2O 8 mol H2O

24. The combustion of acetylene gas is represented by this equation:

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

How many grams of CO2 and H2O are produced when 52.0 g C2H2 burns in oxygen?

12.2 Section Assessment

52.0 g C2H2

1x

g C2H2

mol C2H2x

mol C2H2

mol CO2 xg CO2

mol CO2

C2H2:

C: 2 x 12 = 24 H: 2 x 1 = 2---------------------- 26 g/mol

26

1

2

4

CO2:

C: 1 x 12 = 12 O: 2 x 16 = 32---------------------- 44 g/mol

1

44

= 176 g CO2

17652.0 g g


24. The combustion of acetylene gas is represented by this equation:

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

How many grams of CO2 and H2O are produced when 52.0 g C2H2 burns in oxygen?

52.0 g C2H2

1x

g C2H2

mol C2H2x

mol C2H2

mol H2O xg H2O

mol H2O

C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2---------------------- 26 g/mol

26

1

2

2

H2O: H: 2 x 1 = 2 O: 1 x 16 = 16---------------------- 18 g/mol

1

18

= 36 g H2O

176 g 36 g


The math shows that 2.7 moles of C2H4 can create 5.4 moles of CO2, but 6.3 moles of O2 can only create 4.2 moles CO2. O2 is the weakest link, so it’s the limiting reactant. In this reaction, all the O2 would be used up, with leftover C2H4.



2.6 mol C2H4 can produce less H2O than 6.3 mol O2 can, so C2H4 is the limiting reactant. (I do this a little different from the book. I see how much product each amount of reactant can make. Whoever makes the least product is the limiting reactant.)


As with #27, I calculated how much product each amount of reactant could make. the 2.4 mol C2H2 produced the lesser amount of H2O, so C2H2 is the limiting reactant.


This is just an ordinary, g-this g-that, stoichiometry problem. The only difference is that now we’re referring to the results of such calculations as “theoretical yield”.


#30 is extra-evil because it doesn’t tell you what the formulas are of the chemicals in the reaction, and, since you have to make the equation from scratch, you have to balance it, too. You may have to go back to chapter 9 if you forget how to figure out the formula of a chemical when given only its name.


. . .


#32 = evil (no formulas or equation given AND 2 stoich calcs)


33. “In a chemical equation, an insufficient quantity of any of the reactants will limit the amount of product that forms.”

34. “The efficiency of a reaction carried out in a laboratory can be measured by calculating the percent yield.”


35. What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper (II) sulfate?

2Al + 3CuSO4 Al2(SO4)3 + 3Cu

1.87 g Al

2 mol Al

3 mol Cu

1 mol Cu

63.55 g Cu= 6.6 g Cu

27 g Al

1 mol Al

% yield =theoretical yield

actual yield =6.6 g Cu

4.65 g Cu = 0.705 = 70.5%

Chapter 12 Assessment

42.The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.

5F2(g) + 2NH3(g) N2F4(g) + 6HF(g)

a. If you have 66.6 g of NH3, how many grams of F2 are required

for complete reaction?



5F2(g) + 2NH3(g) N2F4(g) + 6HF(g)

a. If you have 66.6 g of NH3, how many grams of F2 are required

for complete reaction?

5F2 + 2NH3 N2F4 + 6HF

66.6 gg

66.6 g NH3

1x

17 g NH3

1 mol NH3

NH3: N: 1 x 14 = 14 H: 3 x 1 = 3 17 g/mol

x2 mol NH3

5 mol F2 x1 mol F2

38 g F2

F2: F: 2 x 19 = 38 38 g/mol

= 372 g F2

372



5F2(g) + 2NH3(g) N2F4(g) + 6HF(g)

b. How many grams of NH3 are required to produce 4.65 g HF?

5F2 + 2NH3 N2F4 + 6HF

4.65 gg



5F2(g) + 2NH3(g) N2F4(g) + 6HF(g)

b. How many grams of NH3 are required to produce 4.65 g HF?

5F2 + 2NH3 N2F4 + 6HF

4.65 gg

4.65 g HF

1x

20 g HF

1 mol HF

HF: H: 1 x 1 = 1 F: 1 x 19 = 19 20 g/mol

x6 mol HF

2 mol NH3 x1 mol NH3

17 g NH3 = 1.32 g NH3

1.32

NH3: N: 1 x 14 = 14 H: 3 x 1 = 3 17 g/mol



5F2(g) + 2NH3(g) N2F4(g) + 6HF(g)

c. How many grams of N2F4 can be produced from 225 g F2?

5F2 + 2NH3 N2F4 + 6HF

225 g g?



5F2(g) + 2NH3(g) N2F4(g) + 6HF(g)

c. How many grams of N2F4 can be produced from 225 g F2?

5F2 + 2NH3 N2F4 + 6HF

225 g g

225 g F21

x38 g F2

1 mol F2 x5 mol HF2

1 mol N2F4 x1 mol N2F4

104 g N2F4 = 123 g N2F4

123

N2F4:

N: 2 x 14 = 28 F: 4 x 19 = 76 104 g/mol

F2:

F: 2 x 19 = 38 38 g/mol


44. Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.

Li3N + 3H2O NH3 + 3LiOH

a. What mass of water is needed to react with 32.9 g Li3N?


32.9 g g?


44. Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide.


a. What mass of water is needed to react with 32.9 g Li3N?


32.9 g g

32.9 g Li3N1

x35 g Li3N

1 mol Li3N x1 mol Li3N

3 mol H2O x1 mol H2O

18 g H2O= 51 g H2O

51

Li3N:

Li: 3 x 7 = 21 N: 1 x 14 = 14 35 g/mol

H2O:

H: 2 x 1 = 2 O: 1 x 16 = 16 18 g/mol

This Power Point may soon again be . . .

10Q + “E”1,000,000,000,000,000,000,000,000,000,000,000,000,000,000

I’m working on the honors stuff, mostly . . .

Title Page

SWBATS


12.2 PracticeProblems

12.2 SectionAssessment

12.3 PracticeProblems

Ch 12Assessment


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Prentice Hall Chemistry (c) 2005 Section Assessment Answers Chapter 12 By Daniel R. Barnes Init: 12/17/2008 WARNING: some images and content in this presentation