nanni (arn437) – HW 03 Mixtures and Kinetic Theory – vanden bout – (50985)
1
This printout should have 19 questions.
Multiplechoice questions may continue on
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beFore answering.
001 (part 1 of 4) 10.0 points
Iron pyrite (±eS
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 ±e
2
O
3
is produced From the reaction is 62 L oF oxygen
at 3
.
27 atm and 165
◦
C with an excess oF iron
pyrite?
Correct answer: 163
.
774 g.
Explanation:
P
= 3
.
27 atm
T
= 165
◦
C + 273 = 438 K
R
= 0
.
08206
L
·
atm
K
·
mol
V
= 62 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 ±eS
2
(s) + 11 O
2
(g)
→
2 ±e
2
O
3
(s) + 8 SO
2
(g)
Applying the ideal gas law to the O
2
,
P V
=
nRT
n
=
P V
RT
=
(3
.
27 atm) (62 L)
(
0
.
08206
L
·
atm
K
·
mol
)
(438 K)
= 5
.
64071 mol
.
±rom stoichiometry and the molar mass oF
±e
2
O
3
,
m
2
O
3
= (159
.
688 g
/
mol ±e
2
O
3
)
×
2 mol ±e
2
O
3
11 mol O
2
(5
.
64071 mol O
2
)
= 163
.
774 g ±e
2
O
3
.
002 (part 2 of 4) 10.0 points
IF the sulFur dioxide that is generated above
is dissolved to Form 6
.
6 L oF aqueous solu
tion, what is the molar concentration oF the
resulting sulFurous acid (H
2
SO
3
) solution?
Correct answer: 0
.
621566 M.
Explanation:
V
= 6
.
6 L
SO
2
(g) + H
2
O(
ℓ
)
H
2
SO
3
(aq)
.
±rom the stoichiometry,
n
SO
2
= (5
.
64071 mol)
p
8
n
SO
2
11
n
O
2
P
= 4
.
10233 mol
.
4
.
10233 mol oF SO
2
will dissolve in 6
.
6 L oF
water to Form a solution that is
4
.
10233 mol
6
.
6 L
= 0
.
621566 M in H
2
SO
4
.
003 (part 3 of 4) 10.0 points
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
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
m
SO
2
= 1000 kg coal
p
40000 g ±eS
2
1000 kg coal
P
×
p
1 mol ±eS
2
119
.
975 g ±eS
2
P
×
p
8 mol SO
2
4 mol ±eS
2
Pp
64
.
0638 g SO
2
1 mol SO
2
P
= 42718
.
1 g = 42
.
7181 kg SO
2
.
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View Full Documentnanni (arn437) – HW 03 Mixtures and Kinetic Theory – vanden bout – (50985)
2
004 (part 4 of 4) 10.0 points
From the previous problem, what is the vol
ume of the SO
2
gas at 1 atm and 44
◦
C?
Correct answer: 17345
.
6 L.
Explanation:
R
= 0
.
08206
L
·
atm
K
·
mol
T
= 44
◦
C + 273 = 317 K
P
= 1 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
=
nT R
P
=
(666
.
806 mol) (317 K)
1 atm
×
p
0
.
08206
L
·
atm
K
·
mol
P
= 17345
.
6 L
.
005
10.0 points
Two gases are contained in gas bulbs con
nected by a valve. Gas
A
is present in a 1 liter
bulb at a pressure of 835 torr. Gas
B
ex
erts a pressure of 202 torr in a 1 liter bulb.
The valve is opened and the two gases equili
brate. What is the partial pressure of gas A
expressed after equilibration?
Correct answer: 417
.
5 torr.
Explanation:
V
A
= 1 L
V
B
= 1 L
P
A
= 835 torr
P
B
= 202 torr
V
total
= 2 L
P
A
V
A
=
P
t
V
t
P
t
=
P
A
V
A
V
t
=
(835 torr) (1 L)
2 L
= 417
.
5 torr
006
10.0 points
A reminder.
.. partial pressures are calculated
from the total pressure via mole fractions,
not percent by mass. So you WILL have to
convert the percent by mass to mole fraction
BEFORE you answer this problem.
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 Fall '07
 Fakhreddine/Lyon
 Kinetic theory

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