PHYSICAL SCIENCE SHS

1. Percent Yield and Limiting
Reactants
Chapter 12.3

2. Percent Yield
 The percent yield is the percentage of a
certain product actually produced in a
chemical reaction.
 The theoretical yield is predicted by a
stoichiometry problem.

3. A. Percent Yield
100
yield
l
theoretica
yield
actual
yield
% 

calculated on paper
measured in lab

4. A. Percent Yield
 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

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

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

7. Percent Yield
 The reaction between SO2 and oxygen yields
SO3.. Calculate the percent yield of SO3 if 40.0
grams of SO3is formed, when 32 grams of SO2
react with an excess of oxygen.
 SO2 + O2  SO3
 2SO2 + O2  2SO3
3
3
3
2
3
2
2
2
SO
g
39.9916085
SO
mole
1
SO
g
80.0642
SO
mole
2
SO
mole
2
SO
g
64.0648
SO
mole
1
1
SO
grams
32





3
3
3
SO
100.02%
100
SO
g
39.9916085
SO
g
40.



8. B. Limiting Reactants
 Available Ingredients
 4 slices of bread
 1 jar of peanut butter
 1/2 jar of jelly
 Limiting Reactant
 bread
 Excess Reactants
 peanut butter and jelly

9. B. Limiting Reactants
 Limiting Reactant
 Is used up first
 Stops the reaction
 Determines the amount of product
 Excess Reactant
 added to ensure that the other reactant is
completely used up
 cheaper & easier to recycle

10. B. Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
 limiting reactant
 amount of product

11. Limiting Reactant Steps (version 2)
1 Step one
o Write and balance the equation for the reaction.
2 Step two
o Convert known masses to grams of product.
4 Step three
o Determine limiting reactant and amount that can
be made.
5 Step four
o Determine the grams of excess from limiting
reactant.

12. B. Limiting Reactants
 79.1 g of zinc react with 55.4 g of HCl. Identify
the limiting and excess reactants. How many
liters of hydrogen are formed at STP?
Zn + 2HCl  ZnCl2 + H2
79.1 g ? L
55.4 g

13. B. Limiting Reactants
79.1
g Zn
1 mol
Zn
65.39
g Zn
= 27.1 L
H2
1 mol
H2
1 mol
Zn
22.4 L
H2
1 mol
H2
Zn + 2HCl  ZnCl2 + H2
79.1 g ? L
0.90 L
2.5M

14. B. Limiting Reactants
22.4
L H2
1 mol
H2
0.90
L
2.5 mol
HCl
1 L
= 25 L
H2
1 mol
H2
2 mol
HCl
Zn + 2HCl  ZnCl2 + H2
79.1 g ? L
0.90 L
2.5M

15. B. Limiting Reactants
Zn: 27.1 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
HCl: 25 L H2

16. Limiting Reagent or Reactant 2
(problem #2)
 How many grams of ammonia will be produced when
20.0 grams of potassium hydroxide react with 15.0
grams of ammonium sulfate?
 2KOH + (NH4)2SO4  K2SO4 +2NH3 +2H2O

17. (NH4)2SO4 is the limiting reagent because it will make
the least amount of product. It is the reactant that will run
out first.
produced.
be
will
NH
grams
3.87
Step3
3

3
3
3
4
2
4
3
4
2
4
4
2
4
4
2
4
3
3
3
3
NH
grams
3.87
NH
moles
1
NH
grams
17.03052
SO
)
(NH
mole
1
NH
moles
2
SO
)
(NH
g
132.14052
SO
)
(NH
mole
1
1
SO
)
(NH
g
15.0
NH
grams
6.07
NH
moles
1
NH
grams
17.03052
KOH
moles
2
NH
mole
2
KOH
g
56.10564
KOH
mole
1
1
KOH
g
20.0
2
Step











18. over.
left
be
will
KOH
of
grams
7.26
therefore
so
used
was
grams
12.73772
and
originally
grams
20.00
was
There
used.
be
will
KOH
grams
12.73772
KOH
mole
1
KOH
grams
56.10564
SO
)
(NH
moles
1
KOH
mole
2
SO
)
(NH
g
132.14052
SO
)
(NH
mole
1
1
SO
)
(NH
g
15.0
4
Step
4
2
4
4
2
4
4
2
4
4
2
4






Limiting Reagent or Reactant 2