Chapter 12
Stoichiometry
Section 12.2
Chemical Calculations
1
Mole Ratios
These mole ratios can be used to calculate the moles of one
chemical from the given amount of a different chemical
Example: How many moles of chlorine are needed to react
with 5 moles of sodium (without any sodium left over)?
Step 1: Write the chemical equation.
Step 2: Balance the equation.
2 Na + Cl2  2 NaCl
Step 3: 2 mol
Step 4: 5 mol
Step 5:
1 mol (Write the mole ratio)
(Write the given)
x
5 1
x
 2.5mol Cl2
2
2
Mole - Mole Conversions
Example:

How many moles of sodium chloride will
be produced if you react 2.6 moles of
chlorine gas with an excess (more than
you need) of sodium metal?
2 Na + Cl2
1 mol
2.6 mol
2 NaCl
2 mol
x
2.6  2
x
 5.2mol NaCl
1
3
Mole - Mole Conversions
Example:
 How
many moles of O2 are
produced when 3.34 moles of
Al2O3 decompose?
2 Al2O3 Al
4 + 3 O2
2 mol
3.34 mol
3 mol
x
3.34  3
x
 5.01mol O2
2
4
Mole - Mass Conversions
Most of the time in chemistry, the amounts
are given in grams instead of moles
We still go through moles and use the
mole ratio, but now we also use molar
mass to get to grams
5

Example: How many grams of chlorine
are required to react completely with
5.00 moles of sodium to produce sodium
chloride?
2 Na + Cl2  2 NaCl
2 mol
5 mol
1 mol
x
5 1
x
 2.5mol Cl2
2
Change moles to grams: No. of moles X molar mass
2.5 X 71 = 178 g Cl2
6
Example:

Calculate the mass in grams of Iodine
required to react completely with 0.50
moles of aluminum.
3 I2 + 2 Al AlI
2 3
3 mol
x
2 mol
0.5 mol
3  0.5
x
 0.75mol I 2
2
Change moles to grams: No. of moles X molar mass
0.75 X 254 = 191 g I2
7
Mass – Mole Conversions
We can also start with mass and convert to
moles of product or another reactant
We use molar mass and the mole ratio to get
to moles of the compound of interest
8
Example:

Calculate the number of moles of ethane (C2H6)
needed to produce 10.0 g of water
2 C2H6 + 7 O2  4 CO2 + 6 H2O
2 mol
x
6 mol
10 g
But before making cross multiplication, change 10 grams to moles
10 / 18.02 = 0.555 mol
0.555  2
x
 0.185mol C 2 H 6
6
9
Mass – Mass Conversion
Example:

Calculate how many grams of ammonia are produced
when you react 2.00g of nitrogen with excess hydrogen.
N2 + 3 H2  2 NH3
1 mol
2g
2 mol
x
But before making cross multiplication, change 2 grams to moles
2 / 28 = 0.0714 mol
0.0714  2
x
 0.143mol NH3
1
Change moles to grams: No. of moles X molar mass
0.143 X 17.03 = 2.44 g NH3 10
Example:
 Calculate how many grams of oxygen are
required to make 10.0 g of aluminum oxide
4
Al + 3 O2  2 Al2O3
3 mol
x
2 mol
10 g
But before making cross multiplication, change 10 grams to moles
10 / 102 = 0.098 mol
0.098  3
x
 0.147 mol O 2
2
Change moles to grams: No. of moles X molar mass
0.147 X 32 = 4.7 g O2
11
Example:
• If 3.84 moles of C2H2 are burned, how many
moles of O2 are needed? (9.6 mol)
2 C2H2 + 5 O2  4 CO2 + 2 H2O
2 mol
3.84 mol
5 mol
x
3.84  5
x
 9.6mol O 2
2
12
Example:
• How many moles of C2H2 are needed to produce
8.95 mole of H2O? (8.95 mol)
2 C2H2 + 5 O2  4 CO2 + 2 H2O
2 mol
x
2 mol
8.95 mol
8.95  2
x
 8.95mol C 2 H 2
2
13
Example:
• If 2.47 moles of C2H2 are burned, how many
moles of CO2 are formed? (4.94 mol)
2 C2H2 + 5 O2  4 CO2 + 2 H2O
2 mol
2.47 mol
4 mol
x
2.47  4
x
 4.94mol CO2
2
14
Mass – Mass Problem:
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al + 3 O2  2Al2O3
4 mol
6.5 g
2 mol
x
But before making cross multiplication, change 6.5 grams to moles
6.5 / 27 = 0.241 mol Al
0.241 2
x
 0.121mol Al 2 O 3
4
Change moles to grams: No. of moles X molar mass
0.121 X 102 = 12.3 g Al2O3 15
Another example:
If 10.1 g of Fe are added to a solution of Copper (II)
Sulfate, how many grams of solid copper would
form?
2 Fe + 3 CuSO4  Fe2(SO4)3 + 3 Cu
2 mol
10.1 g
3 mol
x
before making cross multiplication, change 10.1 grams to moles
10.1 / 55.8 = 0.181 mol Fe
0.181 3
x
 0.272mol Cu
2
Change moles to grams: No. of moles X molar mass
0.272 X 63.5 = 17.3 g Cu 16
Volume – Volume Calculations:

How many liters of CH4 at STP are required to
completely react with 17.5 L of O2 ?
CH4 + 2 O2  CO2 + 2 H2O
1 mol
x
2 mol
17.5 L
before making cross multiplication, change 17.5 L to moles
17.5 / 22.4 = 0.781 mol O2
0.781 1
x
 0.391mol CH 4
2
Change moles to liters: No. of moles X 22.4
0.391 X 22.4 = 8.76 L CH417
Section 12.3
18
Limiting Reactant: Cookies
1 cup butter, 1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4
dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9
dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?
Limiting Reactant



Most of the time in chemistry we have more
of one reactant than we need to completely
use up other reactant.
That reactant is said to be in excess
(there is too much).
The other reactant limits how much product
we get. Once it runs out, the reaction
s.
This is called the limiting reactant.
Limiting Reactant




To find the correct answer, we have to try all of the
reactants. We have to calculate how much of a
product we can get from each of the reactants to
determine which reactant is the limiting one.
The lower amount of a product is the correct
answer.
The reactant that makes the least amount of product
is the limiting reactant. Once you determine the
limiting reactant, you should ALWAYS start with it!
Be sure to pick a product! You can’t compare to see
which is greater and which is lower unless the
product is the same!
Example

10.0g of aluminum reacts with 35.0 grams of chlorine gas to
produce aluminum chloride. Which reactant is limiting, which
is in excess, and how much product is produced?
2 Al + 3 Cl2

2 AlCl3
Start with Al:
2 mol
10 g
2 mol
x
But before making cross multiplication, change 10 grams to moles
10 / 27 = 0.370 mol
0.370  2
x
 0.370mol AlCl 3 0.370 x 133.5 = 49.4 g
2

Now Cl2:
3 mol
35 g
2 mol
x
But before making cross multiplication, change 35 grams to moles
35 / 71 = 0.493 mol
0.493  2
x
 0.329mol AlCl 3 0.329 x 133.5 = 43.9 g
3
LR Example Continued

We get 49.4 g of aluminum chloride from
the given amount of aluminum, but only
43.9 g of aluminum chloride from the
given amount of chlorine. Therefore,
chlorine is the limiting reactant. Once the
33.6 g of chlorine is used up, the reaction
comes to a complete
.
Practice
 If 10.6 g of copper reacts with 3.83 g
sulfur, how many grams of product (cuprous
sulfide) will be formed?
24

15.0 g of potassium reacts with 15.0 g of
iodine. Calculate which reactant is limiting
and how much product is made.
25
Finding the Amount of Excess


By calculating the amount of the excess
reactant needed to completely react with
the limiting reactant, we can subtract that
amount from the given amount to find the
amount of excess.
Can we find the amount of excess
potassium in the previous problem?
Practice
15.0 g of potassium reacts with 15.0 g of iodine.
If 19.6 g of potassium iodide is produced, find the
limiting and excess reactants then how much of
the excess amount is left over?
27
Potassium superoxide, KO2, is used in rebreathing gas masks to
generate oxygen.
a. How many moles of O2 can be produced from 0.15 mol KO2
and 0.10 mol H2O?
b. Determine the limiting and excess reactants.
c. How much of the excess is left over?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
Limiting Reactant: Recap
1.
2.
3.
4.
5.
6.
7.
You can recognize a limiting reactant problem
because there is MORE THAN ONE GIVEN
AMOUNT.
Convert ALL of the reactants to the SAME product
(pick any product you choose.)
The lowest answer is the correct answer.
The reactant that gave you the lowest answer is
the LIMITING REACTANT.
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the
amount used from the given amount.
If you have to find more than one product, be
sure to start with the limiting reactant. You don’t
have to determine which is the LR over and over
again!
Yield





The amount of product made in a
chemical reaction.
There are three types
Actual yield- what you get in the lab when
the chemicals are mixed
Theoretical yield- what the balanced
equation tells you should make.
Percent yield =
Actual
x 100 %
Theoretical
Percent Yield

To determine percentage yield, you need
two pieces of information:
1) Theoretical yield =
Comes from the stoichiometry
2) Actual yield =
Comes from experimental data
Percent Yield =
Actual Yield
Theoretical Yield
100%
Example






6.78 g of copper is produced when 3.92 g
of Al are reacted with excess copper (II)
sulfate.
2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
What is the actual yield?
What is the theoretical yield?
What is the percent yield?
If you had started with 9.73 g of Al, how
much copper would you expect?
Details


Percent yield tells us how “efficient” a
reaction is.
Percent yield can not be bigger than 100 %.
12.75 moles of NaClO3 will produce how many grams of
O2? Calculate the percent yield of oxygen if 20 grams is
produced.
NaClO3
NaCl + O2
35