1. Limiting Reactants and Percent Yield
Chapter 9
Limiting Reactants
and Percent Yield

2. 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.

3. 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

4. 2. The one that makes the least product is the limiting reagent.
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?

5. Cu is Limiting Reactant
If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
2Cu + S ® Cu2S
Cu is Limiting Reactant
1 mol Cu
1 mol Cu2S
g Cu2S
10.6 g Cu
63.55g Cu
2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
= 13.3 g Cu2S
1 mol S
1 mol Cu2S
g Cu2S
3.83 g S
32.06g S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S

6. 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

7. 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

8. 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!

9. 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 CuSO4 -> Al2(SO4)3 + 3Cu

10. 2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
1st do Al
Now CuSO4
3mol Cu
2mol Al
1mol Al
63.55g Cu
10.3g Al
26.98g Al
1 mol Cu
= 36.4g Cu
1mol CuSO4
51.7g CuSO4
3mol Cu
63.55g Cu
3 mol CuSO4
159.56g CuSO4
1 mol Cu
=20.6g Cu

11. So, limiting reactant?
Al: 36.4g Cu
CuSO4: 20.6g Cu
Limiting reactant: CuSO4
Excess: Al

13. 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

14. c. Percent yield =
Actual Theoretical
3. Details
Percent yield tells us how “efficient” a reaction is.
Percent yield can not be bigger than 100 %.
X 100%

15. measured in lab
calculated on paper

16. 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?

17. 2Al + 3 CuSO4 ® Al2(SO4)3 + 3Cu
Actual yield? 6.78g
Why? “cause it told me already!!!!
Theoretical yield?
Stoichiometry!!!!
3.92g Al
1mol Al
3 mol Cu
63.55g Cu
26.98g Al
1 mol Cu
2mol Al
= 13.85g Cu

18. So, Theoretical yield is 13.85g Cu
What is percent yield?
Actual yield
Theoretical yield
6.87g Cu
13.85g Cu X
100% X
100%
= 48.95%

19. 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

20. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g
Theoretical Yield:
1 mol
K2CO3 g
45.8 g
K2CO3
2 mol
KCl
1 mol
K2CO3
74.55
g KCl
1 mol
KCl
= 49.4
g KCl

21. K2CO3 + 2HCl  2KCl + H2O + CO2
49.4 g
45.8 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
46.3 g
49.4 g
% Yield =
 100 =
93.7%

22. 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 C6H6
1mol C6H5Cl
112.5g C6H5Cl
36.8g C6H6
1mol C6H6
78.06g C6H6
1mol C6H5Cl
= 53.03g C6H5Cl

23. Percent yield
38.8g
53.03g X
100%
= 73.17%

24. 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) Cu
Balance
2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
Now find theoretical yield

25. 2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu
1mol Al
63.55g Cu
3mol Cu
1.85g Al
26.98g Al
2mol Al
1mol Cu
= 6.54g Cu
Now percent yield
3.70g
6.45g
X 100%
= %