Chapter 9
Limiting Reactants
and Percent Yield
A. Limiting Reactant
1. If you are given one dozen loaves of
bread, a gallon of mustard, and three
pieces of ham, how many ham
sandwiches can you make?
2. The limiting reactant is the reactant
you run out of first.
3. The excess reactant is the one you
have left over.
4. The limiting reactant determines how
much product you can make
5. What is the limiting reactant of your
sandwich supplies?
ham
6. What is the excess reactant of your
sandwich supplies?
gallon of mustard
B. How do you find out?
1. Do two stoichiometry problems.
2. The one that makes the least product
is the limiting reagent.
3. For example
Copper reacts with sulfur to form
copper ( I ) sulfide. If 10.6 g of copper
reacts with 3.83 g S how much
product will be formed?
• If 10.6 g of copper reacts with 3.83
g S. How many grams of product
will be formed?
Cu is
• 2Cu + S  Cu2S Limiting
Reactant
Cu2S 159.16 g Cu2S
1 mol Cu 1 mol
10.6 g Cu
63.55g Cu 2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
C. Clarify
1. Limiting Reactant
–used up in a reaction
–determines the amount of product
2. Excess Reactant
–added to ensure that the other
reactant is completely used up
–cheaper & easier to recycle
3. Example
a. Available Ingredients
–4 slices of bread
–1 jar of peanut butter
–1/2 jar of jelly
b. Limiting Reactant
–bread
c. Excess Reactants
–peanut butter and jelly
E. Calculating Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
–limiting reactant
4. Do Stoichiometry!
7. Still another example
• If 10.3 g of aluminum are reacted with
51.7 g of CuSO4 how much copper
will be produced?
• How much excess reagent will
remain?
• Al + CuSO4 -> Al2(SO4)3 + Cu
• Balanced?
• 2 Al + 3 CuSO4 -> Al2(SO4)3 + 3Cu
2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
1st do Al
3mol
Cu
1mol
Al
63.55g Cu
10.3g Al
26.98g Al 2mol Al
1 mol Cu
Now CuSO4
= 36.4g Cu
1mol
51.7g CuSO4 3mol Cu
63.55g Cu
CuSO4 159.56g 3 mol
1 mol Cu
CuSO4
CuSO4
=20.6g Cu
So, limiting reactant?
Al: 36.4g Cu
CuSO4: 20.6g Cu
Limiting reactant: CuSO4
Excess: Al
A. Percent Yield
1. The amount of product made in a
chemical reaction.
2. There are three types:
a. Actual yield- what you get in the lab
when the chemicals are mixed
b. Theoretical yield- what the
balanced equation tells what should
be made
c. Percent yield =
Actual
X 100%
Theoretical
3. Details
• Percent yield tells us how “efficient” a
reaction is.
• Percent yield can not be bigger than
100 %.
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
B. Example 1
• 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?
2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
Actual yield? 6.78g
Why? “cause it told me already!!!!
Theoretical yield?
Stoichiometry!!!!
1mol Al 3 mol Cu 63.55g Cu
3.92g Al
26.98g Al 2mol Al 1 mol Cu
= 13.85g Cu
So, Theoretical yield is 13.85g Cu
What is percent yield?
Actual yield
X 100%
Theoretical yield
6.87g Cu X
100% = 48.95%
13.85g Cu
Example 2
• When 45.8 g of K2CO3 react with
excess HCl, 46.3 g of KCl are
formed. Calculate the theoretical
and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
45.8 g 1 mol 2 mol 74.55
K2CO3 K2CO3 KCl g KCl
= 49.4
138.21 g 1 mol 1 mol g KCl
K2CO3 K2CO3 KCl
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
% Yield =
46.3 g
49.4 g
 100 = 93.7%
Yet another example
If 36.8g of C6H6 react with excess of Cl2
And the actual yield of C6H5Cl is 38.8g,
what is the percent yield?
1st find theoretical yield of C6H5Cl
1 mol
1mol
112.5g
36.8g C6H6
C6H5Cl
C6H5Cl
C6H6 78.06g
1mol
1mol
C6H6
C6H6
C6H5Cl
= 53.03g C6H5Cl
Percent yield
38.8g
X 100% = 73.17%
53.03g
One more time!
If 1.85g of Al react with excess CuSO4
to produce 3.70g of Cu, what is the
percent yield?
Al + CuSO4 -> Al2(SO4)3 + Cu
Balance
2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
Now find theoretical yield
2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
1mol Al 3mol Cu
1.85g Al
26.98g Al 2mol Al
63.55g Cu
1mol Cu
= 6.54g Cu
Now percent yield
3.70g
6.45g
X 100% = 57.36%