1. Chapter 6 Chemical Reactions and Quantities
Percent Yield and Limiting Reactants

2. Vocabulary Theoretical Yield Actual Yield Percent Yield
Limiting Reactants
Excess

3. Theoretical, Actual, and Percent Yield
Theoretical yield:
the maximum amount of product, which is calculated using the balanced equation.
Actual yield:
the amount of product obtained when the reaction takes place
Percent yield:
the ratio of actual yield to theoretical yield
Percent yield = actual yield (g) x theoretical yield (g)

4. Guide to Calculations for Percent Yield

5. Calculating Percent Yield
Suppose you have prepared cookie dough to make 5
dozen cookies. The phone rings and you answer. While
you talk, a sheet of 12 cookies burns, and you have to
throw them out. The rest of the cookies you make are
okay. What is the percent yield of edible cookies?
Theoretical yield: 60 cookies possible
Actual yield: cookies to eat
Percent yield: 48 cookies x 100% = 80.% yield cookies

6. Learning Check
With a limited amount of oxygen, the reaction of carbon and oxygen produces carbon monoxide.
2C(g) + O2(g) CO(g)
What is the percent yield if 40.0 g of CO are produced when 30.0 g of O2 are used?
1) 25.0%
2) 75.0%
3) 76.2%

7. Solution 3) 76.2% yield STEP 1 Given: 40.0 g of CO produced (actual)
30.0 g of O2 used
Need: percent yield of CO
STEP 2 Write a plan to calculate % yield of CO:
g of O moles of moles of g of CO
O CO (theoretical)
Percent yield of CO = g of CO (actual) x 100%
g of CO (theoretical)

8. Solution (continued) STEP 3 Write conversion factors:
1 mole of O2 = 32.0 g of O2
1 mole O2 and g O2
32.0 g O mole O2
1 mole of O2 = 2 moles of CO
1 mole O2 and 2 moles CO
2 moles CO mole O2
1 mole of CO = g of CO
1 mole CO and g CO
28.0 g CO mole CO

9. Solution (continued)
STEP 4 Setup to calculate theoretical yield in g of O2:
30.0 g O2 x 1 mole O2 x 2 moles CO x g CO
32.0 g O mole O mole CO
= g of CO (theoretical)
Setup to calculate percent yield:
40.0 g CO (actual) x = 76.2% yield (3)
52.5 g CO (theoretical)

10. Learning Check When N2 and 5.00 g of H2 are mixed, the reaction
produces 16.0 g of NH3. What is the percent yield
for the reaction?
N2(g) + 3H2(g) NH3(g)
1) 31.3% of NH3
2) 56.9% of NH3
3) 80.0% of NH3

11. Solution 2) 56.9% STEP 1 Given: 16.0 g of NH3 produced (actual)
2) 56.9%
STEP 1 Given: g of NH3 produced (actual)
5.00 g of H2 used
Need: percent yield of NH3
STEP 2 Write a plan to calculate % yield of NH3:
g of H moles of moles of g of NH H NH (theoretical)
Percent yield of NH3 = g of NH3 (actual) x 100%
g of NH3 (theoretical)

12. Solution (continued) STEP 3 Write conversion factors:
1 mole of H2 = 2.02 g of H2
1 mole H2 and g H2
2.02 g H mole H2
1 mole of H2 = 2 moles of NH3
1 mole H and 2 moles NH3
2 moles NH mole H2
1 mole of NH3 = g of NH3
1 mole NH3 and g NH3
17.0 g NH mole NH3

13. Solution (continued)
STEP 4 Setup to calculate theoretical yield of g of NH3:
5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.02 g H moles H mole NH3
= g of NH3 (theoretical)
Setup to calculate percent yield:
Percent yield = g NH3 x 100 = 56.9% yield (2)
28.1 g NH3

14. Limiting Reactant A limiting reactant in a chemical reaction is the
substance that
is used up
limits the amount of product that can form and stops the reaction

15. Reacting Amounts
In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon.
How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives?
What is the limiting item?

16. Reacting Amounts (continued)
Only 4 place settings are possible.
Initially Used Left over
Plates
Forks
Spoons
Knives
The limiting item is the spoon.

17. Example 1 of an Everyday Limiting Reactant
How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches could be made.
The bread is the limiting item.

18. Example 2 of an Everyday Limiting Reactant
How many peanut butter sandwiches could be made
from 8 slices bread and 1 tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1 sandwich
could be made. The peanut butter is the limiting item.

19. Guide to Calculating Product from a Limiting Reactant

20. Limiting Reactant When 4.00 moles of H2 is mixed with 2.00 moles of
Cl2, how many moles of HCl can form?
H2(g) Cl2(g) 2HCl(g)
4.00 moles moles ??? Moles
Calculate the moles of product that each reactant, H2 and Cl2, could produce.
The limiting reactant is the one that produces the smaller number of moles of product.

21. Limiting Reactant (continued)
HCl from H2
4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl
1 moles H2
HCl from Cl2
2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl
1 mole Cl2
4.00 moles of HCl is the smaller number of moles
produced. Thus, Cl2 will be used up.
The limiting reactant is Cl2.

22. Check Calculations Equation Initially H2 4.00 moles Cl2 2.00 moles
2HCl
0 mole
Reacted/
Formed
–2.00 moles
+4.00 moles
Left after reaction
(4.00 – 2.00)
Excess
0 moles
(2.00 – 2.00)
Limiting
( )
Product possible

23. Limiting Reactants
If 4.80 moles Ca are mixed with 2.00 moles N2, which is
the limiting reactant? 3Ca(s) + N2(g) Ca3N2(s)
Moles of Ca3N2 from Ca
4.80 moles Ca x 1 mole Ca3N2 = moles of Ca3N moles Ca (Ca is used up)
Moles of Ca3N2 from N2
2.00 moles N2 x 1 mole Ca3N2 = moles of Ca3N2
1 mole N2
Ca is used up. Thus, Ca is the limiting reactant.

24. Learning Check What is the mass of water that can be produced when
8.00 g of H2 and 24.0 g of O2 react?
2H2(g) + O2(g) H2O(l)
1) 8.0 g of H2O
2) g of H2O
3) 72 g of H2O

25. Solution Moles of H2O from H2:
3) 72 g of H2O
Moles of H2O from H2:
8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O
2.0 g H moles H2
Moles of H2O from O2:
24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O
32.0 g O mole O Smaller number
of moles of H2O
1.50 moles H2O x g H2O = g of H2O
1 mole H2O