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Chapter 7 Chemical Quantities
7.5 Mole Relationships in Chemical
Equations
7.6 Mass Calculations for Reactions
7.7 Percent Yield
7.8 Equilibrium Constants
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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7.5 Conservation of Mass

In a chemical reaction, the mass of the reactants
is equal to the mass of the products.
2Ag + S
Ag2S
mass reactants = mass products
2 moles Ag
+ 1 mole S = 1 mole Ag2S
2 (107.9 g)
+ 1(32.1 g) = 1 (247.9 g)
247.9 g reactants
= 247.9 g product
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Moles in Equations
 We can read the equation in “moles” by
placing the word “moles” between each
coefficient and formula.
4 Fe + 3 O2
2 Fe2O3
4 moles Fe + 3 moles O2
2 moles Fe2O3
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Writing Mole-Mole Factors
A mole-mole factor is a ratio of the coefficients
for two substances.
4 Fe + 3 O2
2 Fe2O3
Fe and O2
4 mole Fe and 3 mole O2
3 mole O2
4 mole Fe

Fe and Fe2O3
4 mole Fe and
2 mole Fe2O3
O2 and Fe2O3
3 mole O2
and 2 mole Fe2O3
2 mole Fe2O3
3 mole O2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
2 mole Fe2O3
4 mole Fe
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Learning Check
Consider the following equation:
3 H2 + N2
2 NH3
A. A mole factor for H2 and N2 is
1) 3 mole N2 2) 1 mole N2
1 mole H2
3 mole H2
B. A mole factor for NH3 and H2 is
1) 1 mole H2
2) 2 mole NH3
2 mole NH3
3 mole H2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
3) 1 mole N2
2 mole H2
3) 3 mole N2
2 mole NH3
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Solution
3 H2
+ N2
2 NH3
A. A mole factor for H2 and N2 is
2) 1 mole N2
3 mole H2
B. A mole factor for NH3 and H2 is
2) 2 mole NH3
3 mole H2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Calculations with Mole Factors
Consider the following reaction:
4 Fe + 3 O2
2 Fe2O3
How many moles of Fe2O3 are produced when 6.0
moles O2 react?
Use the appropriate mole factor to determine the
moles Fe2O3.
6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3
3 mole O2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Learning Check
Consider the following reaction:
4 Fe +
3 O2
2 Fe2O3
How many moles of Fe are needed to react
with 12.0 moles of O2?
1) 3.00 moles Fe
2) 9.00 moles Fe
3) 16.0 moles Fe
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Solution
3) 16.0 moles Fe
Consider the following reaction:
4 Fe +
3 O2
2 Fe2O3
How many moles of Fe are needed to react with
12.0 moles of O2?
12.0 mole O2 x 4 mole Fe = 16.0 moles Fe
3 mole O2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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7.6 Mass Calculations
Convert
1.
grams A
2.
moles A
3.
moles B
gA
gB
moles A
moles B (use mole factor)
grams B
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Learning Check
How many grams of O2 are needed to
produce 0.400 mole of Fe2O3?
4 Fe + 3 O2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
2 Fe2O3
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Solution
0.400 moles Fe2O3
1.
2.
? grams O2
moles Fe2O3 moles O2(use mole factor)
moles O2 grams O2
0.400 mole Fe2O3 x 3 mole O2
x 32.0 g O2
2 mole Fe2O3 1 mole O2
mole factor
molar mass
= 19.2 g O2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Calculating the Mass of a Reactant
The reaction between H2 and O2 produces 13.1 g
of water. How many grams of O2 reacted?
2H2 + O2
2H2O
?g
13.1 g
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Calculating the Mass of a Reactant
2H2
+
O2
1.
grams H2O
moles H2O
2.
moles H2O
moles O2
3.
moles O2
13.1 g H2O x
2H2O
grams O2
1 mole H2O x 1 mole O2 x 32.0 g O2
18.0 g H2O
2 mole H2O 1 mole O2
= 11.6 g O2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Learning Check
Acetylene gas C2H2 burns in the oxyactylene
torch for welding. How many grams of C2H2
are burned if the reaction produces 75.0 g of
CO2?
2 C2H2 + 5 O2
? grams
4 CO2 + 2 H2O
75.0 g
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Solution
1.
2.
3.
grams CO2 moles CO2
moles CO2 moles C2H2
moles C2H2 grams C2H2
2 C2H2 + 5 O2
4 CO2 + 2 H2O
75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H2
44.0 g CO2
4 moles CO2 1 mole C2H2
= 22.2 g C2H2
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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7.7 Percent Yield
You prepared cookie dough to make 5 dozen cookies.
The phone rings and you answer. While you talk, a
sheet of 12 cookies burns. You have to throw them out.
The rest of the cookies are okay. The results of our
baking can be described as follows:
Theoretical yield 60 cookies possible
Actual yield
48 cookies to eat
Percent yield
48 cookies x 100 = 80% yield
60 cookies
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Percent Yield



The theoretical yield is the maximum amount
of product calculated using the balanced
equation.
The actual yield is the amount of product
obtained when the reaction is run.
Percent yield is the ratio of actual yield
compared to the theoretical yield.
Percent Yield =
Actual Yield (g)
x 100
Theoretical Yield (g)
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Sample Exercise % Yield
Without proper ventilation and limited
oxygen, the reaction of carbon and oxygen
produces carbon monoxide.
2C + O2
2CO
What is the percent yield if 40.0 g of CO are
produced from the reaction of 30.0 g O2?
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Sample Exercise % Yield (cont.)
1.
Calculate theoretical yield of CO.
30.0 g O2 x 1 mole O2 x 2 mole CO x 28.0 g CO
32.0 g O2 1 mole O2
1 mole CO
= 52.5 g CO (theoretical)
2.
Calculate the percent yield.
40.0 g CO (actual)
x 100 = 76.2 % yield
52.5 g CO(theoretical)
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Learning Check
In the lab, N2 and 5.0 g of H2 are reacted and
produce 16.0 g of NH3. What is the
percent yield for the reaction?
N2(g) + 3H2(g)
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
2NH3(g)
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Solution
N2(g) + 3H2(g)
2NH3(g)
5.0 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.0 g H2
3 moles H2
1 mole NH3
= 28.3 g NH3 (theoretical)
Percent yield = 16.0 g NH3 x 100 = 56.5 %
28.3 g NH3
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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7.8 Systems at Equilibrium

The forward and reverse reactions in a reversible
reaction are indicated by double arrows. The
reversible reaction of substances A and B is
written as
forward reaction
aA
bB
reverse reaction

At equilibrium, the rate of the forward reaction is
equal to the rate of the reverse reaction.
Rate of forward reaction = Rate of reverse reaction
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Equilibrium Constants



aA
bB
An equilibrium expression is written to give the
equilibrium constant (Keq).
Keq =
[B]b =
[Products]
[A]a
[Reactants]
The square brackets indicate the moles/liter of each
substance (concentration).
The coefficients are written as superscripts that
raise the moles/liter to a specific power.
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Writing a Keq Expression

Consider the following reaction:
2CO(g) + O2(g)
2CO2(g)

For the Keq expression, place the products in the
numerator and the reactants in the denominator.
Keq =
[CO2]
[CO] [O2]

Write the coefficients as superscripts.
Keq =
[CO2]2
[CO]2 [O2]
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Learning Check
The Keq expression for the following
reaction is:
N2(g) +3Cl2(g)
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
2NCl3(g)
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Solution
The Keq expression for the following reaction is:
N2(g) + 3Cl2(g)
2NCl3(g)
[NCl3]2
[N2][Cl2]3
products
reactants
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Equilibrium Constants
When equilibrium is reached, if:
Keq > 1 favors products
Keq < 1 favors reactants
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Learning Check
For each Keq, indicate whether there are
mostly (1) reactants or (2) products at equilibrium.
__A. H2(g) + F2(g)
Keq = 1 x 1095
2HF(g)
__B. 3O2(g)
2O3(g)
Keq = 1.8 x 10-7
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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Solution
A. H2(g) + F2(g)
2HF(g)
Keq = 1 x 1095
Keq > 1 favors products
B. 3O2(g)
2O3(g)
Keq = 1.8 x 10-7
Keq < 1 favors reactants
Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
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