Chapter3 (Reaction Stoichiometry)

3Chemical Equations &

Reaction

Stoichiometry

3

Chapter Three Goals

1. Chemical Equations

2. Calculations Based on Chemical Equations

3. The Limiting Reactant Concept

4. Percent Yields from Chemical Reactions

5. Sequential Reactions

6. Concentrations of Solutions

7. Dilution of Solutions

8. Using Solutions in Chemical Reactions

9. Synthesis Question

4

Chemical Equations

• Symbolic representation of a chemical

reaction that shows:

1. reactants on left side of reaction

2. products on right side of equation

3. relative amounts of each using

stoichiometric coefficients

5

Chemical Equations

• Attempt to show on paper what is happening at

the laboratory and molecular levels.

6

Chemical Equations

• Look at the information an equation

provides:

Fe O + 3 CO 2 Fe + 3 CO2 3 2

Reactants Yields Products

7

Chemical Equations


provides:

Fe O + 3 CO 2 Fe + 3 CO2 3 2


1 formula

unit

3

molecules

2 atoms 3

molecules

8

Chemical Equations


provides:

Fe O + 3 CO 2 Fe + 3 CO2 3 2


1 formula

unit

3

molecules

2 atoms 3

molecules

1 mole 3 moles 2 moles 3 moles

9

Chemical Equations


provides:

Fe O + 3 CO 2 Fe + 3 CO2 3 2


1 formula

unit

3

molecules

2 atoms 3

molecules

1 mole 3 moles 2 moles 3 moles

159.7 g 84.0 g 111.7 g 132 g

10

Chemical Equations

• Law of Conservation of Matter

– There is no detectable change in quantity of matter in

an ordinary chemical reaction.

– Balanced chemical equations must always include the

same number of each kind of atom on both sides of

the equation.

– This law was determined by Antoine Lavoisier.

11

Chemical Equations

Propane,C3H8, burns in oxygen to give carbon

dioxide and water.

12

Law of Conservation of Matter

NH3 burns in oxygen to form NO & water

You do it!

13

Law of Conservation of Matter

C7H16 burns in oxygen to form carbon dioxide and

water.

You do it!

14

Calculations Based on

Chemical Equations• Can work in moles, formula units, etc.

• Frequently, we work in mass or weight

(grams or kg or pounds or tons).

Fe O + 3 CO 2 Fe + 3 CO2 3 2

15


Chemical EquationsExample 3-1: How many CO molecules are

required to react with 25 formula units of Fe2O3?

Fe O + 3 CO 2 Fe + 3 CO2 3 2

16


Chemical EquationsExample 3-2: How many iron atoms can be

produced by the reaction of 2.50 x 105 formula

units of iron (III) oxide with excess carbon

monoxide?

Fe O + 3 CO 2 Fe + 3 CO2 3 2

17


Chemical EquationsExample 3-3: What mass of CO is required to

react with 146 g of iron (III) oxide?

Fe O + 3 CO 2 Fe + 3 CO2 3 2

18


Chemical EquationsExample 3-4: What mass of carbon dioxide can be

produced by the reaction of 0.540 mole of iron (III)

oxide with excess carbon monoxide?

Fe O + 3 CO 2 Fe + 3 CO2 3 2

19


Chemical EquationsExample 3-5: What mass of iron (III) oxide reacted

with excess carbon monoxide if the carbon dioxide

produced by the reaction had a mass of 8.65

grams?

You do it!

Fe O + 3 CO 2 Fe + 3 CO2 3 2

20


Chemical EquationsExample 3-6: How many pounds of carbon

monoxide would react with 125 pounds of iron (III)

oxide?

You do it!

Fe O + 3 CO 2 Fe + 3 CO2 3 2

21

Limiting Reactant Concept

• Kitchen example of limiting reactant

concept.

1 packet of muffin mix + 2 eggs + 1 cup of milk

12 muffins

• How many muffins can we make with the

following amounts of mix, eggs, and milk?

22


• Mix Packets Eggs Milk

1 1 dozen 1 gallon

limiting reactant is the muffin mix

2 1 dozen 1 gallon

3 1 dozen 1 gallon

4 1 dozen 1 gallon

5 1 dozen 1 gallon

6 1 dozen 1 gallon

7 1 dozen 1 gallon

limiting reactant is the eggs

23


Example 3-7: Suppose a box contains 87 bolts,

110 washers, and 99 nuts. How many sets, each

consisting of one bolt, two washers, and one nut,

can you construct from the contents of one box?

24


• Look at a chemical limiting reactant

situation.

Zn + 2 HCl ZnCl2 + H2

MEDIA/Movies/05M05VD1.MOV

25


Example 3-8: What is the maximum mass of sulfur

dioxide that can be produced by the reaction of

95.6 g of carbon disulfide with 110. g of oxygen?

2222 SO 2 CO O 3 CS

26

Percent Yields from Reactions

• Theoretical yield is calculated by assuming that the reaction goes to completion.– Determined from the limiting reactant calculation.

• Actual yield is the amount of a specified pure product made in a given reaction.– In the laboratory, this is the amount of product that is

formed in your beaker, after it is purified and dried.

• Percent yield indicates how much of the product is obtained from a reaction.

% yield = actual yield

theoretical yield100%

27

Percent Yields from Reactions

Example 3-9: A 10.0 g sample of ethanol, C2H5OH,

was boiled with excess acetic acid, CH3COOH, to

produce 14.8 g of ethyl acetate, CH3COOC2H5.

What is the percent yield?

3 2 5 3 2 5 2CH COOH + C H OH CH COOC H H O

28

Sequential Reactions

Example 3-10: Starting with 10.0 g of benzene

(C6H6), calculate the theoretical yield of

nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).1 mol benzene 1 mol nitrobenzene

? g nitrobenzene = 10.0 g benzene78.0 g benzene 1 mol benzene

123.0 g nitrobenzene

1 mol nitrobenzene

15.8 g nitrobenzene

N O2NH

2

HNO3

H2SO4

Sn

Conc. HCl

benzene nitrobenzene aniline

29


Example 3-10: Starting with 10.0 g of benzene

(C6H6), calculate the theoretical yield of

nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).

N O2NH

2

HNO3

H2SO4

Sn

Conc. HCl

benzene nitrobenzene aniline

1 mol nitrobenzene 1 mol aniline? g aniline = 15.8 g nitrobenzene

123.0 g nitrobenzene 1 mol nitrobenzene

93.0 g aniline

1 mol aniline

11.9 g aniline

30


Example 3-10: If 6.7 g of aniline is prepared from

10.0 g of benzene, what is the percentage yield?

You do it!

31

Concentration of Solutions

• Solution is a mixture of two or more substances dissolved in another.– Solute is the substance present in the smaller

amount.

– Solvent is the substance present in the larger amount.

– In aqueous solutions, the solvent is water.

• The concentration of a solution defines the amount of solute dissolved in the solvent.– The amount of sugar in sweet tea can be defined by

its concentration.

32


One common unit of concentration is:

% by mass of solute has the symbol %w/w

mass of solute% by mass of solute = 100%

mass of solution

mass of solution = mass of solute + mass of solvent

33


Example 3-11: What mass of NaOH is required to

prepare 250.0 g of solution that is 8.00% w/w

NaOH?

34


Example 3-12: Calculate the mass of 8.00% w/w

NaOH solution that contains 32.0 g of NaOH.

35


Example 3-13: Calculate the mass of NaOH in

300.0 mL of an 8.00% w/w NaOH solution.

Density is 1.09 g/mL.

You do it!

36

Concentrations of Solutions

Example 3-14: What volume of 12.0% w/w KOH

contains 40.0 g of KOH? The density of the

solution is 1.11 g/mL.

You do it!

37


Second common unit of concentration:

mL

mmol

L

moles

solution of liters ofnumber

solute of moles ofnumber molarity

M

M


38


Example 3-15: Calculate the molarity of a solution

that contains 12.5 g of sulfuric acid in 1.75 L of

solution.

You do it!

39


Example 3-16: Determine the mass of calcium

nitrate required to prepare 3.50 L of a 0.800 M

Ca(NO3)2 .

You do it!

40


• One of the reasons that molarity is

commonly used is because:

M x L = moles solute

and

M x mL = mmol solute

41


Example 3-17: The specific gravity of a

concentrated HCl is 1.185 and it is 36.31% w/w

HCl. What is its molarity?

42

Dilution of Solutions

• To dilute a solution, add solvent to a

concentrated solution.

– One method to make tea “less sweet.”

– How fountain drinks are made from syrup.

• The number of moles of solute in the two

solutions remains constant.

• The relationship M1V1 = M2V2 is

appropriate for dilutions, but not for

chemical reactions.

43


• Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.


44


Example 3-18: If 10.0 mL of 12.0 M HCl is added

to enough water to give 100. mL of solution, what

is the concentration of the solution?

45


Example 3-19: What volume of an 18.0 M sulfuric

acid is required to make 2.50 L of a 2.40 M sulfuric

acid solution?

You do it!

46

Using Solutions in Chemical

Reactions

Combine the concepts of molarity and

stoichiometry to determine the amounts of

reactants and products involved in reactions

in solution.

47


ReactionsExample 3-20: What volume of a 0.500 M BaCl2 is

required to completely react with 4.32 g of

Na2SO4?

NaCl 2 + BaSO BaCl + SONa 4242

48


ReactionsExample 3-21: (a) What volume of a 0.200 M

NaOH will react with 50.0 mL of a 0.200 M

aluminum nitrate, Al(NO3)3?

You Do It!!

3 33 3Al NO 3 NaOH Al OH 3 NaNO

49


Reactions(b) What mass of Al(OH)3 precipitates in (a)?

You do it!

50


Reactions• Titrations are a method of determining the

concentration of an unknown solutions

from the known concentration of a solution

and solution reaction stoichiometry.

– Requires special lab glassware

• Buret, pipet, and flasks

– Must have an indicator also

Media/Movies/05M14VD1.MOV

51


ReactionsExample 3-22: What is the molarity of a KOH

solution if 38.7 mL of the KOH solution is required

to react with 43.2 mL of a 0.223 M HCl?

2KOH + HCl KCl + H O

52


ReactionsExample 3-23: What is the molarity of a barium

hydroxide solution if 44.1 mL of a 0.103 M HCl is

required to react with 38.3 mL of the Ba(OH)2

solution?

2 2 2Ba(OH) + 2 HCl BaCl + 2 H O

53

Synthesis Question

Nylon is made by the reaction of hexamethylene

diamine

with adipic acid.

CH2

CH2

CH2

CH2

CH2

CH2

NH2

NH2

C

OH

OCH2

C

H2

CH2

C

H2

C

OH

O

54

Synthesis Question

in a 1 to 1 mole ratio. The structure of nylon is:

CNH

O

C

H2

CH2

CH2

C

H2

CH2

C

H2

C

O

C

H2

CH2

C

H2

NH

** n

where the value of n is typically 450,000. On a daily

basis, a DuPont factory makes 1.5 million pounds of

nylon. How many pounds of hexamethylene diamine and

adipic acid must they have available in the plant each

day?

55

Synthesis Question

# of unitsC atoms H atoms N atoms O atoms

8

Molar mass of 1 nylon molecule [ (12 12) (1 20) (2 14) (2 16)] 450,000

[224 g/mol] 450,000

1.01 10 g/mol nylon molecules

56

Synthesis Question


8


[224 g/mol] 450,000


1.5 million pou 6

8

454 gnds (1.5 10 lb)

lb

6.81 10 g

57

Synthesis Question


8


[224 g/mol] 450,000


1.5 million pou 6

8

8

8

454 gnds (1.5 10 lb)

lb

6.81 10 g

1 mol nylon# mol of nylon molecules 6.81 10 g

1.01 10 g

6.76 mol of nylon

58

Synthesis Question

Because the nylon formation reaction uses 1 mole of adipic acid 450,000

plus 1 mole of hexamethylene diamine 450,000 per mole of nylon formed,

to make 6.76 mol of nylon requires:

59

Synthesis Question

Because the nylon formation reaction uses 1 mole of adipic acid 450,000

plus 1 mole of hexamethylene diamine 450,000 per mole of nylon formed,

to make 6.76 mol of nylon requires:

adipic acid - 6.68 8 51 lb 450,000 146 g/mol 3.53 10 g 7.78 10 lb

454 g

60

Synthesis Question

lb 1068.7g 454

lb 1g 1049.3 g/mol 116450,000 6.68 - diamine enehexamethyl

lb 1066.9g 454

lb 1g 104.39 g/mol 146450,000 6.68 - acid adipic

:requiresnylon of mol 6.68 make to

formed,nylon of moleper 450,000 diamine enehexamethyl of mole 1 plus

450,000 acid adipic of mole 1 usesreaction formation nylon theBecause

58

58

61

Group Activity

Manganese dioxide, potassium hydroxide

and oxygen react in the following fashion:

OH 2 KMnO 4 O 3 + KOH 4 + MnO 4 2422

A mixture of 272.9 g of MnO2, 26.6 L of 0.250

M KOH, and 41.92 g of O2 is allowed to react

as shown above. After the reaction is finished,

234.6 g of KMnO4 is separated from the

reaction mixture. What is the per cent yield of

this reaction?

3Chemical Equations &

Reaction

Stoichimoetry

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Chapter3 (Reaction Stoichiometry)