wilson (tw8656) – H03: Mixtures and KMT – mccord – (50960)
1
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before answering.
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McCord’s MWF
11am class (50960) must report to the correct
room for the exams.
There are 2 different
rooms and the class is split according to the
first letter of your last name.
Go to the
right room or you WILL receive a zero on the
exams.
AL go to UTC 2.112A
MZ go to UTC 2.102A
001(part1of4)10.0points
Iron pyrite (FeS
2
) is the form in which much of
the sulfur exists in coal. In the combustion of
coal, oxygen reacts with iron pyrite to produce
iron(III) oxide and sulfur dioxide, which is a
major source of air pollution and a substantial
contributor to acid rain. What mass of Fe
2
O
3
is produced from the reaction is 77 L of oxygen
at 3
.
58 atm and 149
◦
C with an excess of iron
pyrite?
Correct answer: 231
.
121 g.
Explanation:
P
= 3
.
58 atm
T
= 149
◦
C + 273 = 422 K
R
= 0
.
08206
L
·
atm
K
·
mol
V
= 77 L
MW
Fe
2
O
3
= 2(55
.
845 g
/
mol)
+ 3(15
.
9994 g
/
mol)
= 159
.
688 g
/
mol
The balanced equation is
4 FeS
2
(s) + 11 O
2
(g)
→
2 Fe
2
O
3
(s) + 8 SO
2
(g)
Applying the ideal gas law to the O
2
,
P V
=
n R T
n
=
P V
R T
=
(3
.
58 atm) (77 L)
(
0
.
08206
L
·
atm
K
·
mol
)
(422 K)
= 7
.
96031 mol
.
From stoichiometry and the molar mass of
Fe
2
O
3
,
m
Fe
2
O
3
= (159
.
688 g
/
mol Fe
2
O
3
)
×
2 mol Fe
2
O
3
11 mol O
2
(7
.
96031 mol O
2
)
= 231
.
121 g Fe
2
O
3
.
002(part2of4)10.0points
If the sulfur dioxide that is generated above
is dissolved to form 5
.
5 L of aqueous solu
tion, what is the molar concentration of the
resulting sulfurous acid (H
2
SO
3
) solution?
Correct answer: 1
.
0526 M.
Explanation:
V
= 5
.
5 L
SO
2
(g) + H
2
O(
ℓ
)
→
H
2
SO
3
(aq)
.
From the stoichiometry,
n
SO
2
= (7
.
96031 mol)
parenleftbigg
8
n
SO
2
11
n
O
2
parenrightbigg
= 5
.
78931 mol
.
5
.
78931 mol of SO
2
will dissolve in 5
.
5 L of
water to form a solution that is
5
.
78931 mol
5
.
5 L
= 1
.
0526 M in H
2
SO
4
.
003(part3of4)10.0points
What mass of SO
2
is produced in the burning
of 1 tonne (1 t = 1000 kg) of highsulfur coal,
if the coal is 4% pyrite by mass?
Correct answer: 42
.
7181 kg.
Explanation:
m
coal
= 1000 kg
m
FeS
2
= 4%(1000 kg) = 40 kg = 40000 g
MW
FeS
2
= 55
.
845 g
/
mol + 2(32
.
065 g
/
mol)
= 119
.
975 g
MW
SO
2
= 32
.
065 g
/
mol + 2(15
.
9994 g
/
mol)
= 64
.
0638 g
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wilson (tw8656) – H03: Mixtures and KMT – mccord – (50960)
2
m
SO
2
= 1000 kg coal
parenleftbigg
40000 g FeS
2
1000 kg coal
parenrightbigg
×
parenleftbigg
1 mol FeS
2
119
.
975 g FeS
2
parenrightbigg
×
parenleftbigg
8 mol SO
2
4 mol FeS
2
parenrightbigg parenleftbigg
64
.
0638 g SO
2
1 mol SO
2
parenrightbigg
= 42718
.
1 g = 42
.
7181 kg SO
2
.
004(part4of4)10.0points
From the previous problem, what is the vol
ume of the SO
2
gas at 0
.
8 atm and 34
◦
C?
Correct answer: 20998
.
1 L.
Explanation:
R
= 0
.
08206
L
·
atm
K
·
mol
T
= 34
◦
C + 273 = 307 K
P
= 0
.
8 atm
The number of moles of SO
2
is
n
SO
4
= (42718
.
1 g SO
2
)
1 mol SO
2
64
.
0638 g SO
2
= 666
.
806 mol SO
2
.
From the ideal gas law,
V
=
n T R
P
=
(666
.
806 mol) (307 K)
0
.
8 atm
×
parenleftbigg
0
.
08206
L
·
atm
K
·
mol
parenrightbigg
= 20998
.
1 L
.
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 Fall '07
 Fakhreddine/Lyon
 Chemistry

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