Chapter 8
Quantities in Chemical Reactions
Products are
carbon dioxide
and water
Octane in gas tank
Octane mixes
with oxygen
2006, Prentice Hall
CHAPTER OUTLINE
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Stoichiometry
Molar Ratios
Mole-Mole Calculations
Mass-Mole Calculations
Mass-Mass Calculations
Limiting Reactant
Percent Yield
2
Global Warming: Too Much Carbon Dioxide
• The combustion of
fossil fuels such as
octane (shown here)
produces water and
carbon dioxide as
products.
• Carbon dioxide is a
greenhouse gas that
is believed to be
responsible for global
warming.
The greenhouse effect
• Greenhouse gases
act like glass in a
greenhouse, allowing
visible-light energy to
enter the atmosphere
but preventing heat
energy from
escaping.
• Outgoing heat is
trapped by
greenhouse gases.
Combustion of fossil fuels produces CO2.
• Consider the combustion of octane (C8H18), a
component of gasoline:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
• The balanced chemical equation shows that
16 mol of CO2 are produced for every 2 mol of
octane burned.
Global Warming
• scientists have measured an average
0.6°C rise in atmospheric temperature
since 1860
• during the same period atmospheric
CO2 levels have risen 25%
The Source of Increased CO2
• the primary source of the increased CO2
levels are combustion reactions of fossil fuels
we use to get energy (methane and octane)
– 1860 corresponds to the beginning of the
Industrial Revolution in the US and Europe
CH 4 (g)  2 O 2 (g)  CO 2 (g)  2 H 2 O (g)
2 C 8 H 18 (l)  25 O 2 (g)  16 CO 2 (g)  18 H 2 O (g)
STOICHIOMETRY
 Stoichiometry is the quantitative relationship
between the reactants and products in a
balanced chemical equation.
 A balanced chemical equation provides several
important information about the reactants and
products in a chemical reaction.
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MOLAR
RATIOS
For example:
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
This
is the molar 2 molecules
3 molecules
ratios between
100 molecules
300 molecules
the
reactants and200 molecules
products
6 molecules
106 molecules
3x10
2x106 molecules
1 molecule
1 mole
3 moles
2 moles
9
Examples:
Determine each mole ratio below based on the
reaction shown:
2 C4H10 + 13 O2  8 CO2 + 10 H2O
m ol O 2
13
=
m ol C O 2
m ol C 4 H 10
m ol H 2 O
8
=
2
10
10
Stoichiometry
Stoichiometry - Allows us to predict products that
form in a reaction based on amount of reactants.
The amount of each element must be the same throughout the
overall reaction.
For example, the amount of element H or O on the reactant side must
equal the amount of element H or O on the product side.
2H2 + O2
2H2O
STOICHIOMETRIC
CALCULATIONS
 Stoichiometric
calculations can be classified as
Mass-mass
one of the following:
calculations
MASS of
compound A
Mass-mole
MASS of
Mole-mole
calculations
compound B
calculations
MM
MM
MOLES of
compound A
molar ratio
MOLES of
compound B
12
MOLE-MOLE
CALCULATIONS
 Relates moles of reactants and products in a
balanced chemical equation
MOLES of
compound A
molar ratio
MOLES of
compound B
13
Example 1:
How many moles of ammonia can be produced from
32 moles of hydrogen? (Assume excess N2 present)
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
32 mol H2
x
2
3
m ol N H 3
m ol H 2
= 21 mol NH3
Mole ratio
14
Example 2:
In one experiment, 6.80 mol of ammonia are
prepared. How many moles of hydrogen were used
up in this experiment?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
6.80 mol NH3 x
3 m ol H 2
2 m ol N H 3 = 10.2 mol H2
Mole ratio
15
MASS-MOLE
CALCULATIONS
 Relates moles and mass of reactants or products
in a balanced chemical equation
MASS of
compound A
MM
MOLES of
compound A
molar ratio
MOLES of
compound B
16
Example 1:
How many moles of ammonia can be produced from
the reaction of 125 g of nitrogen?
1 N2 (g) + 3 H2 (g)  2 NH3 (g)
125 g N2
1
m ol N 2
28.0
g N2
x
Molar mass
x
2
1
m ol N H 3
m ol N 2
= 8.93 mol NH3
Mole ratio
17
MASS -MASS
CALCULATIONS
 Relates mass of reactants and products in
a balanced chemical equation
MASS of
compound A
MASS of
compound B
MM
MM
MOLES of
compound A
molar ratio
MOLES of
compound B
18
Example 1:
What mass of carbon dioxide will be produced from
the reaction of 175 g of propane, as shown?
1 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Mass of
propane
Moles of
propane
Moles of
carbon
dioxide
Mass of
carbon dioxide
19
Example 1:
1 C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
175 g C3H8
1 m ol C 3 H 8
x
44.1 g C 3 H 8
x
Molar mass
Molar mass
x
3
m ol C O 2
1
m ol C 3 H 8
44.0 g C O 2
1 m ol C O 2 = 524 g CO2
Mole ratio
20
LIMITING
REACTANT
 When 2 or more reactants are combined in nonstoichiometric ratios, the amount of product
produced is limited by the reactant that is not in
excess.
 This reactant is referred to as limiting reactant.
 When doing stoichiometric problems of this
type, the limiting reactant must be determined
first before proceeding with the calculations.
21
LIMITING REACTANT
ANALOGY
Consider the following recipe for a sundae:
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LIMITING REACTANT
ANALOGY
The number
How
many sundaes
of sundaes
can possible
be prepared
is limited
from the
by the
followingofingredients:
amount
syrup, the limiting reactant.
Limiting
reactant
Excess
reactants
23
LIMITING
REACTANT
 When
Compare
solving
yourlimiting
answersreactant
for eachproblems,
assumption;
assume
the
lower
each
value
reactant
is the correct
is limiting
assumption.
reactant,
and
Lower
calculate the desired quantity based onvalue
that is
assumption.
correct
A+BC
A is LR
Calculate
amount of C
B is LR
Calculate
amount of C
24
Example 1:
A fuel mixture used in the early days of rocketry
was a mixture of N2H4 and N2O4, as shown below.
How many grams of N2 gas is produced when 100 g
of N2H4 and 200 g of N2O4 are mixed?
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Limiting
reactant
Mass-mass
calculations
25
Example 1:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Assume N2H4 is LR
100 g N2H4
x
1
m ol N 2 H 4
32.04
g N 2H 4
x
3 m ol N 2
2 m ol N 2 H 4
=
4.68 mol N2
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Example 1:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Assume N2O4 is LR
200 g N2O4
x
1
m ol N 2 O 4
92.00
g N 2O 4
x
3 m ol N 2
1 m ol N 2 O 4
=
6.52 mol N2
27
Example 1:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Assume N2H4 is LR
4.68 mol N2
Assume N2O4 is LR
6.52 mol N2
N2H4 is
LR
Correct
amount
28
Example 1:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
Calculate mass of N2
4.68 mol N2
x
28.0 g N 2
1 m ol N 2 = 131 g N2
29
Example 2:
How many grams of AgBr can be produced when
50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as
shown below:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Limiting
Reactant
30
Example 2:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Assume MgBr2 is LR
50.0 g MgBr2
x
1 m ol M gB r x 2
184.1 g M gB r 1
2
2
x
m ol A gB r
m ol M gB r2
187.8 g A gB r =
1 m ol A gB r 102 g AgBr
31
Example 2:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Assume AgNO3 is LR
100.0 g AgNO3
x
1 m ol A gN O
169.9 g A gN O
3
x
x
3
2
2
m ol A gB r
m ol A gN O 3
187.8 g A gB r =
1 m ol A gB r 111 g AgBr
32
Example 2:
MgBr2 + 2 AgNO3  2 AgBr + Mg(NO3)2
Assume MgBr2 is LR
102 g AgBr
Assume AgNO3 is LR
111 g AgBr
MgBr2
is LR
Correct
amount
33
PERCENT YIELD
 The amount of product calculated through
stoichiometric ratios are the maximum amount
product that can be produced during the
reaction, and is thus called theoretical yield.
 The actual yield of a product in a chemical
reaction is the actual amount obtained from the
reaction.
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PERCENT YIELD
 The percent yield of a reaction is obtained as
follows:
A ctual y ield
x100 = P ercent y ield
T heoretical y ield
35
Example 1:
In an experiment forming ethanol, the theoretical
yield is 50.0 g and the actual yield is 46.8 g. What is
the percent yield for this reaction?
% y ield =
A ctual y ield
T heoretical y ield
x100 =
46.8 g
50.0 g
x100 =
92.7 %
36
Example 2:
Silicon carbide can be formed from the reaction of
sand (SiO2) with carbon as shown below:
1 SiO2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g)
When 100 g of sand are processed, 51.4g of SiC is
produced. What is the percent yield of SiC in this
reaction?
Actual
yield
37
Example 2:
1 SiO2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g)
Calculate theoretical yield
100 g SiO2
x
1 m ol S iO 2
60.1 g S iO 2
x
1 m ol S iC 40.1 g S iC
x
=
1 m ol S iO 2
1 m ol S iC
66.7 g SiC
38
Example 2:
Calculate percent yield
% y ield =
A ctual y ield
T heoretical y ield
x100 =
51.4 g
66.7 g
x100 =
77.1 %
39
Theoretical and Actual Yield
• In order to determine the theoretical yield, we
use reaction stoichiometry to determine the
amount of product each of our reactants
could make.
• The theoretical yield will always be the least
possible amount of product.
– The theoretical yield will always come from the
limiting reactant.
• Because of both controllable and
uncontrollable factors, the actual yield of
product will always be less than the
theoretical yield.
Chap. 8 terms you should know
1. Limiting reactant - the reactant that limits the amount of product produced in
a chemical reaction. The reactant that makes the least amount of product.
2. Theoretical yield - the amount of product that can be made in a chemical
reaction based on the amount of limiting reactant.
3. Actual yield - the amount of product actually produced by a chemical reaction.
4. Percent yield - The percent of the theoretical yield that was actually obtained.
actual yield
% yield =
x 100
theoretical yield
THE END
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