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:

Fe2 O3 + 3 CO  2 Fe + 3 CO 2
4
Chemical Equations
• Look at the information an equation provides:

Fe O + 3 CO yields
 2 Fe products
+ 3 CO 2
2 3
reactants
5
Chemical Equations
• Look at the information an equation provides:

Fe O + 3 CO yields
 2 Fe products
+ 3 CO 2
2 3
reactants
1 formula unit
molecules
3 molecules
2 atoms
3
6
Chemical Equations
• Look at the information an equation provides:

Fe O + 3 CO yields
 2 Fe products
+ 3 CO 2
2 3
reactants
1 formula unit
molecules
1 mole
3 molecules
3 moles
2 atoms
2 moles
3
3 moles
7
Chemical Equations
• Look at the information an equation provides:

Fe
O
+
3
CO



2
Fe
+
3
CO
2
3
2
reactants
yields
products
1 formula unit
molecules
1 mole
159.7 g
3 molecules
3 moles
84.0 g
2 atoms
2 moles
111.7 g
3
3 moles
132g
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,C3H8, burns in oxygen to give carbon dioxide
and water.

C3 H 8  5 O 2 
 3 CO2  4 H 2O
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).

Fe 2 O 3 + 3 CO  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
Fe2O3?
3 CO molecules
? CO molecules = 25 formula units Fe2O3 
1 Fe2O3 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 105
formula units of iron (III) oxide with excess
carbon monoxide?
? Fe atoms = 2.50 10 formula units Fe2O3
5
12
Calculations Based on Chemical
Equations
• Example 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 atoms = 2.50 10 formula units Fe2O3
5
2 Fe atoms


1 formula units Fe2O3
13
Calculations Based on Chemical
Equations
• Example 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 atoms = 2.50 10 formula units Fe2O3
5
2 Fe atoms
5

 5.00 10 Fe atoms
1 formula units Fe2O3
14
Calculations Based on Chemical
Equations
• Example 3-3: What mass of CO is required to
react with 146 g of iron (III) oxide?
1 mol Fe2O3
? g CO = 146 g Fe2O3 
159.7 g Fe2O3
15
Calculations Based on Chemical
Equations
• Example 3: What mass of CO is required to
react with 146 g of iron (III) oxide?
1 mol Fe2O3
3 mol CO
? g CO = 146 g Fe2O3 

159.7 g Fe2 O3 1 mol Fe2O3
16
Calculations Based on Chemical
Equations
• Example 3-3: What mass of CO is required to
react with 146 g of iron (III) oxide?
1 mol Fe2O3
3 mol CO
? g CO = 146 g Fe2O3 

159.7 g Fe2 O3 1 mol Fe2O3
28.0 g CO

 76.8 g CO
1 mol 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?
3 mol CO2
? g CO2  0.540 mol Fe2O3 
1 mol Fe2O3
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?
3 mol CO 2
44.0 g CO 2
? g CO 2  0.540 mol Fe 2 O 3 

1 mol Fe 2 O 3 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?
3 mol CO 2
44.0 g CO 2
? g CO 2  0.540 mol Fe 2 O 3 

1 mol Fe 2 O 3 1 mol CO 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?
1 molCO 2 1mol Fe2O3
? g Fe2O3  8.65 g CO2 


44.0 g CO2 3 mol CO2
159.7 g Fe2O3
 10.5 g Fe2O3
1 mol Fe2O3
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
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 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?
CS2  3 O2  CO2  2 SO2
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?
CS2  3 O 2  CO2  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?
CS2  3 O 2  CO 2  2 SO 2
1 mol
3 mol
1 mol
76.2 g
3(32.0 g) 44.0 g
2 mol
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?
CS2  3 O 2  CO2  2 SO 2
1 mol CS2
? mol SO 2  95.6 g CS2 
76.2 g
28
Limiting Reactant Concept
CS2  3 O 2  CO 2  2 SO 2
1 mol CS2
? g SO 2  95.6 g CS2 
76.2 g
2 mol SO 2 64.1 g SO 2


 161 g SO 2
1 mol CS2
1 mol SO 2
What do we do next?
You do it!
29
Limiting Reactant Concept
CS2  3 O 2  CO 2  2 SO 2
1 mol CS2 2 mol SO 2 64.1 g SO 2
? g SO 2  95.6 g CS2 


 161 g SO 2
76.2 g
1 mol CS2 1 mol SO 2
? g SO 2  110 g O 2 
•
•
•
•
1 mol O 2 2 mol SO 2 64.1 g SO 2


 147 g SO 2
32.0 g O 2 3 mol O 2
1 mol SO 2
Which is limiting reactant?
Limiting reactant is O2.
What is maximum mass of sulfur dioxide?
Maximum mass is 147 g.
30
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 =
 100%
theoretical yield
31
Percent Yields from Reactions
• Example 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?
32
Percent Yields from Reactions
CH3COOH + C 2 H 5OH  CH3COOC2 H 5  H 2 O
1. Calculate the theoretic al yield
88.0 g CH3COOC2 H 5
? g CH3COOC2 H 5 = 10.0 g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH3COOC2 H 5
33
Percent Yields from Reactions
CH3COOH + C 2 H 5OH  CH3COOC2 H 5  H 2 O
1. Calculate the theoretic al yield
88.0 g CH3COOC2 H 5
? g CH3COOC2 H 5 = 10.0 g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH3COOC2 H 5
34
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH  CH 3COOC2 H 5  H 2 O
1. Calculate the theoretic al yield
88.0 g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0 g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH 3COOC2 H 5
2. Calculate the percent yield.
35
Percent Yields from Reactions
CH 3COOH + C 2 H 5OH  CH 3COOC2 H 5  H 2 O
1. Calculate the theoretic al yield
88.0 g CH 3COOC2 H 5
? g CH 3COOC2 H 5 = 10.0 g C 2 H 5OH 
46.0 g C 2 H 5OH
 19.1 g CH 3COOC2 H 5
2. Calculate the percent yield.
14.8 g CH 3COOC2 H 5
% yield =
100%  77.5%
19.1 g CH 3COOC2 H 5
36