Chemical Calculations Problems Page 3
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1
2
3Chemistry 100
Clark College
7. Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form
chloric acid (HClO
), according to the following balanced equation:
3
O → 5 HClO
6 ClO
+ 3 H
+ HCl
2
2
3
a) If 142.0 g of ClO
are mixed with 38.0 g of H
O, how many grams of chloric acid are
2
2
formed?
1 mol ClO
5 mol HClO
84.46 g HClO
2
3
3
ClO
: 142.0 g ClO
x
x
x
= 148.2 g HClO
2
2
3
67.45 g ClO
6 mol ClO
1 mol HClO
2
2
3
1 mol H
O
5 mol HClO
84.46 g HClO
2
3
3
H
O: 38.0 g H
O x
x
x
= 297 g HClO
2
2
3
18.01 g H
O
3 mol H
O
1 mol HClO
2
2
3
ClO
is the limiting reagent, 148.2 g of HClO
are produced
2
3
b) Which reactant is left over, and how much is left over?
297 g HClO3 - 148.2 g HClO3 = 149 g of excess HClO3 could be produced.
1 mol HClO
3 mol H
O
18.01 g H
O
3
2
2
149 g HClO
x
x
x
= 19.1 g H
O is leftover
3
2
84.46 g HClO
5 mol HClO
1 mol H
O
3
3
2
8. In the previous problem, what it the percent yield for the reaction if 120.2 g of chloric acid is
actually produced?
120.2 g HClO
actual
Percent yield =
x 100 =
3
x 100 = 81.11%
theoretical
148.2 g HClO
3
9. The deep blue Cu(NH
)
SO
is made by the reaction of copper (II) sulfate with ammonia, as
3
4
4
described by the following reaction:
→ Cu(NH
CuSO
+ 4 NH
)
SO
4(aq)
3aq)
3
4
4(aq)
a) If you use 10.0 g of ammonia and an excess of copper (II) sulfate, what is the theoretical
yield for the reaction?
1 mol NH
1 mol Cu(NH
)
SO
227.76 g Cu(NH
)
SO
10.0 g NH
x
3
x
3
4
4
x
3
4
4
= 33.4 g
3
17.03 g NH
4 mol NH
1 mol Cu(NH
)
SO
3
3
3
4
4
33.4 g of Cu(NH
)
SO
can be produced.
3
4
4
b) If 18.6 g of Cu(NH
)
SO
is isolated at the end of the reaction, what is the percent yield
3
4
4
for the reaction?
18.6 g
x 100 = 55.6%
33.4 g
Chapter 8 Homework
Su07
Page 3 of 3
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