Chapter 9

Chemical Equations & Reaction Stoichiometry

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 2

Chemical Equations • Attempt to show on paper what is happening at the laboratory and molecular levels.

3

Chemical Equations • Look at the information an equation provides:

2 Fe + 3 CO

2 4

Chemical Equations • Look at the information an equation provides:

Fe O + 3 CO

 yields

2 Fe + 3 CO

products 2 5

Chemical Equations • Look at the information an equation provides:

Fe O + 3 CO

 yields 1 formula unit 3 molecules molecules

2 Fe + 3 CO

products 2 2 atoms 3 6

Chemical Equations • Look at the information an equation provides:

Fe O + 3 CO

 yields 1 formula unit 3 molecules molecules 1 mole 3 moles

2 Fe + 3 CO

products 2 atoms 2 moles 3 moles 3 2 7

Chemical Equations • Look at the information an equation provides:

Fe O + 3 CO

reactants  yields 1 formula unit 3 molecules molecules 1 mole 3 moles 159.7 g 84.0 g

2 Fe + 3 CO

products 2 atoms 2 moles 111.7 g 3 moles 132g 3 2 8

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.

Propane,C 3 H 8 , burns in oxygen to give carbon dioxide and water.

C 3 H 8  5 O 2  3 CO 2  4 H 2 O 9

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).

2 Fe + 3 CO

2 10

Calculations Based on Chemical Equations • Example 1: How many CO molecules are required to react with 25 formula units of Fe 2 O 3 ?

?

CO molecules = 25 formula units Fe 2 O 3  3 CO molecules 1 Fe 2 O 3 formula unit  75 molecules of CO 11

Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide?

?

Fe atoms = 2.50

 10 5 formula units Fe 2 O 3 12

Calculations Based on Chemical Equations • Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide?

?

Fe atoms = 2.50

 10 5 formula units Fe 2 O 3  2 1 formula Fe atoms units Fe 2 O 3  13

Calculations Based on Chemical Equations • Example 2: How many iron atoms can be produced by the reaction of 2.50 x 10 5 formula units of iron (III) oxide with excess carbon monoxide?

?

Fe atoms = 2.50

 10 5 formula units Fe 2 O 3  2 1 formula Fe atoms units Fe 2 O 3  5.00

 10 5 Fe atoms 14

Calculations Based on Chemical Equations • Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

?

g CO = 146 g Fe 2 O 3  1 mol Fe 2 O 3 159 .

7 g Fe 2 O 3 15

Calculations Based on Chemical Equations • Example 3: What mass of CO is required to react with 146 g of iron (III) oxide?

?

g CO = 146 g Fe 2 O 3  1 mol Fe 2 O 3 159 .

7 g Fe 2 O 3  3 mol CO 1 mol Fe 2 O 3 16

Calculations Based on Chemical Equations • Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

?

g CO = 146 g Fe 2 O 3  1 mol Fe 2 O 3 159 .

7 g Fe 2 O 3  3 mol CO 1 mol Fe 2 O 3  28.0

g CO 1 mol CO  76 .

8 g CO 17

Calculations Based on Chemical Equations • Example 4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

?

g CO 2  0 .

540 mol Fe 2 O 3  3 mol CO 2 1 mol Fe 2 O 3 18

Calculations Based on Chemical Equations • Example 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?

?

g CO 2  0 .

540 mol Fe 2 O 3  3 mol CO 2 1 mol Fe 2 O 3  44 .

0 g CO 2 1 mol CO 2 19

Calculations Based on Chemical Equations • Example 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?

? g CO 2  .

mol Fe O 3  3 mol CO 2  .

g CO 1 mol CO 2 2 = 71.3 g CO 2 20

Calculations Based on Chemical Equations • Example 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!

21

Calculations Based on Chemical Equations • Example 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? ?

g Fe 2 O 3  8.65

g CO 2  1 molCO 2 44.0

g CO 2  1 mol Fe 2 O 3 3 mol CO 2  159 .

7 g Fe 1 mol Fe 2 2 O O 3 3  10 .

5 g Fe 2 O 3 22

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?

23

Limiting Reactant Concept • Mix Packets 1 2 Eggs 1 dozen 1 dozen Milk 1 gallon limiting reactant is the muffin mix 1 gallon 3 4 5 6 7 1 dozen 1 dozen 1 dozen 1 dozen 1 dozen 1 gallon 1 gallon 1 gallon 1 gallon 1 gallon limiting reactant is the dozen eggs 24

 Limiting Reactant Concept • Example 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?

CS 2  3 O 2  CO 2  2 SO 2 25

 Limiting Reactant Concept • Example 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?

CS

2 

3 O

2 

CO

2 

2 SO

2

1 mol 3 mol 1 mol 2 mol

26

Limiting Reactant Concept • Example 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?



1 mol 3 mol 1 mol 2 mol

2

76.2 g 3(32.0 g) 44.0 g 2(64.1 g)

27

Limiting Reactant Concept • Example 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?

CS 2  3 O 2  CO 2  2 SO 2 ?

mol SO 2  95 .

6 g CS 2  1 mol CS 2 76.2

g 28

Limiting Reactant Concept CS 2  3 O 2  CO 2  2 SO 2 ?

g SO 2  95 .

6 g CS 2  1 mol CS 2 76.2

g  2 mol SO 2 1 mol CS 2  64 .

1 g SO 2 1 mol SO 2  161 g SO 2 What do we do next?

You do it!

29

Limiting Reactant Concept • • • • CS 2  3 O 2  CO 2  2 SO 2 ?

g SO 2  95 .

6 g CS 2  1 mol CS 2 76.2

g  2 mol SO 2 1 mol CS 2  64 .

1 g SO 2 1 mol SO 2  161 g SO 2 ?

g SO 2  110 g O 2  1 mol O 2 32.0

g O 2  2 mol SO 2  3 mol O 2 64 .

1 g SO 2 1 mol SO 2  147 g SO 2 Which is limiting reactant?

Limiting reactant is O 2 .

What is maximum mass of sulfur dioxide?

Maximum mass is 147 g.

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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.

actual yield % yield = theoretical yield  100% 31

Percent Yields from Reactions • Example 9: A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5 . What is the percent yield?

32

Percent Yields from Reactions CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5  H 2 O 1.

Calculate the theoretic al yield ?

g CH 3 COOC 2 H 5 = 10.0

g C 2 H 5 OH  88.0

g CH 3 COOC 2 46 .

0 g C 2 H 5 OH  19 .

1 g CH 3 COOC 2 H 5 H 5 33

Percent Yields from Reactions CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5  H 2 O 1.

Calculate the theoretic al yield ?

g CH 3 COOC 2 H 5 = 10.0

g C 2 H 5 OH  88.0

g CH 3 COOC 2 46 .

0 g C 2 H 5 OH  19 .

1 g CH 3 COOC 2 H 5 H 5 34

Percent Yields from Reactions CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5  H 2 O 1.

Calculate the theoretic al yield ?

g CH 3 COOC 2 H 5 = 10.0

g C 2 H 5 OH  88.0

g CH 3 COOC 2 46 .

0 g C 2 H 5 OH  19 .

1 g CH 3 COOC 2 H 5 H 5 2 .

Calculate the percent yield.

35

Percent Yields from Reactions CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5  H 2 O 1.

Calculate the theoretic al yield ?

g CH 3 COOC 2 H 5 = 10.0

g C 2 H 5 OH  88.0

g CH 3 COOC 2 46 .

0 g C 2 H 5 OH  19 .

1 g CH 3 COOC 2 H 5 H 5 2 .

Calculate the percent yield.

% yield = 14.8

g CH 3 COOC 2 H 5  100 % 19.1

g CH 3 COOC 2 H 5  77 .

5 % 36