1. Theoretical and Percent Yield
Limiting Reactant
Theoretical and Percent Yield

2. 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.

3. 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.

4. 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!

5. Limiting 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:
Now Cl2:
10.0 g Al 1 mol Al mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3

6. 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 g of AlCl3 is the calculated product, so that is the theoretical yield.

7. 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 = 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

8. 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?

9. 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 I mol I mol K g K
254 g I mol I mol K
= 4.62 g K USED!
15.0 g K – 4.62 g K = 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!

10. 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)

11. Percentage Yield
Example:  A student conducts a single displacement reaction that produces grams of copper.  If grams of copper should have been produced what is the student's percentage yield?

12. Percentage Yield
2.755g percentage yield = x 100                            g  
percentage yield = %

13. 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:
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
16 g H2
143 g =
Step 3: Calculate % yield
actual
theoretical
138 g H2O
143 g H2O =
% yield =
x 100%
x 100%
96.7% =

14. 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:
2 mol NH3
3 mol H2 x
17.04 g NH3
1 mol NH3 x
# g NH3=
20.0 mol H2
227 g =
Step 3: Calculate % yield
actual
theoretical
40.5 g NH3
227 g NH3 =
% yield =
x 100%
x 100%
17.8% =