Limiting Reactant
Theoretical and Percent Yield
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
Limiting Reactant - The reactant in a chemical reaction that limits the
amount of product that can be formed. The reaction will stop when all of the
limiting reactant is consumed.
Excess Reactant - The reactant in a chemical reaction that remains when a
reaction stops when the limiting reactant is completely consumed. The
excess reactant remains because there is nothing with which it can react.
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.
• Be sure to pick the same product!
Limiting
Limiting
Reactant
Reactant: 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:
10.0 g Al
1 mol Al
27.0 g Al
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
• Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
Theoretical Yield
• We get 49.4g of aluminum chloride from the given
amount of aluminum, but only 43.9g of aluminum
chloride from the given amount of chlorine.
Therefore, chlorine is the limiting reactant.
• Theoretical Yield – The predicted amount of
product is the theoretical yield. 43.9 g of AlCl3 is
the calculated product, so that is the theoretical
yield.
Limiting Reactant Practice
• 15.0 g of potassium reacts with 15.0 g of
iodine. Calculate which reactant is limiting
and how much product is made.
2 K + I2  2 KI
15.0 g K x 1mole x 2 KI x 166 g = 63.7 g KI
40 g
2 K 1 mole
15. 0 g I2 x 1 mole x 2 KI x 166g = 19.8 g KI
252 g I2
1 mole
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?
Finding Excess Practice
• 15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2  2 KI
• We found that Iodine is the limiting reactant, and
19.6 g of potassium iodide are produced.
15.0 g I2
1 mol I2
2 mol K
39.1 g K
254 g I2
1 mol I2
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given amount
of excess
reactant
Amount of
excess
reactant
actually
used
Note that we started with
the limiting reactant! Once
you determine the LR, you
should only start with it!
Percentage Yield
Percentage yield is the percent of the
theoretical yield that was made by our
experiment.
actual amount of product
percentage yield = ---------------------------------------- x 100
theoretical yield
(the amount you expect to get from the reaction)
Percentage Yield
Example: A student conducts a single
displacement reaction that produces
2.755 grams of copper. If 3.150 grams
of copper should have been produced
what is the student's percentage yield?
Percentage Yield
2.755g
percentage yield = --------------- x 100
3.150g
percentage yield = 87.46 %
Sample problem
Q - What is the % yield of H2O if 138 g H2O is
produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield.
Actual is given, theoretical is calculated:
# g H2O= 16 g H2 x1 mol H2 x2 mol H2Ox 18.02 g H2O= 143 g
2.02 g H2 2 mol H2 1 mol H2O
Step 3: Calculate % yield
% yield = actual x 100% = 138 g H2Ox 100% = 96.7%
theoretical
143 g H2O
Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is
produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield.
Actual is given, theoretical is calculated:
# g NH3= 20.0 mol H2 x 2 mol NH3 x 17.04 g NH3 = 227 g
3 mol H2
1 mol NH3
Step 3: Calculate % yield
% yield = actual x 100% = 40.5 g NH3 x 100% = 17.8%
theoretical
227 g NH3