Chapter 8 Lecture
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Transcript Chapter 8 Lecture
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
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
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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
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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
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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
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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
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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
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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.
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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
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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+BC
A is LR
Calculate
amount of C
B is LR
Calculate
amount of C
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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
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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
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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
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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
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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
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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
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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 %
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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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