#29 When 84.8 g of iron (III) oxide reacts with excess of carbon
monoxide, iron is produced.
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
What is the theoretical yield (how much iron do you expect to make?)
of this reaction?
84.8 g Fe2O3 X
1
1 mole Fe2O3 X 2 mole Fe
X
159.7 g Fe2O3
1mole Fe2O3
= 59.3 g Fe
55.85 g Fe
1mole Fe
#30 When 5.00 g copper reacts with excess silver nitrate, silver metal
and copper (II) nitrate are produced. What is the theoretical
yield of silver in this reaction?
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)
5.00 g Cu
1
X
1 mole Cu
63.55 g Cu
X 2 mole Ag
1 mole Cu
X
= 17.0 g Ag
107.9 g Ag
1 mole Ag
#31 If 50.0 g of silicon dioxide is heated with an excess of carbon,
27.9 g of silicon carbide is produced.
SiO2(s) + 3C(s)  SiC(s) + 2CO(g)
What is the % yield in this reaction?
First calculate the theoretical yield:
50.0 g SiO2
1
X
1 mole SiO2
60.09 g SiO2
X 1 mole SiC X
1 mole SiO2
40.09 g SiC
1 mole SiC
= 33.36 g SiC theoretical
27.9 g SiC X 100 = 83.6 % yield
33.36 g SiC
Formula for
% yield:
actual yield X 100
theoretical yield:
#32 If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of
ammonia is produced. What is the percent yield of this reaction?
N2 + 3H2  2NH3
calculate the theoretical yield for both nitrogen and hydrogen
separately The one that produces the least ammonia determines the
theoretical yield, then use that value to determine the % yield:
15.0 g N2 X 1 mole N2
2 mole NH3 X 17 g NH3 = 18.2 g NH
X
3
1
28 g N2
1 mole N2
1 mole NH3
15.0 g H2 X 1 mole H2 X 2 mole NH3 X 17 g NH3 = 85 g NH3
1
2.0 g H2
3 mole H2
1 mole NH3
10.5 g NH3
X 100 = 57.7%
18.2 g NH3
#33 In a chemical reaction, how does an insufficient quantity
of a reactant affect the amount of product formed?
If a reactant runs out, the reaction must
stop. It is like building bicycles, if you
run out of one piece you stop building
bicycles.
#34 How can you gauge the efficiency of a reaction carried out
in the laboratory?
You can compare your results in the lab with the theoretical results
that you can calculate. You can determine the % yield that you
got compared with what might be expected with purely calculated
values.
What you got from the lab
actual yield X 100
theoretical yield:
What you got from the
mathematical calculation
#35 What is the percent yield if 4.65 g of copper is produced
when 1.87 g of aluminum reacts with an excess of copper (II) sulfate?
2 Al(s) + 3 CuSO4(aq) Al2(SO4)3(aq) +
3Cu(s)
1.87 g Al
X 1 mole Al X 3 moles Cu X 63.54 g Cu
1
27.0 g Al
2 moles Al
1 mole Cu
4.65 g Cu X 100 =
6.6 g Cu
70.5%
Formula for
% yield:
= 6.6 g Cu
theoretical
actual yield X 100
theoretical yield: