5.97
You can map out the following strategy to solve for the total volume of gas.
grams nitroglycerin
→
moles nitroglycerin
→
moles products
→
volume of products
2
1 mol nitroglycerin
29 mol product
? mol products
2.6
10
g nitroglycerin
8.3 mol
227.09 g nitroglycerin
4 mol nitroglycerin
=
×
×
×
=
Calculating the volume of products:
product
L atm
(8.3 mol) 0.0821
(298 K)
mol K
(1.2 atm)
×
÷
×
=
=
=
2
product
1.7
10
L
n
RT
P
V
×
The relationship between partial pressure and
P
total
is:
P
i
=
Χ
i
P
T
Calculate the mole fraction of each gaseous product, then calculate its partial pressure using the equation above.
165
CHAPTER 5:
GASES
component
moles component
total moles all components
=
Χ
166
CHAPTER 5:
GASES
2
2
CO
12 mol CO
0.41
29 mol product
=
=
Χ
Similarly,
2
H O
Χ
=
0.34,
2
N
Χ
=
0.21, and
2
O
Χ
=
0.034
2
2
CO
CO
T
=
P
P
Χ
(0.41)(1.2 atm)
=
=
2
CO
0.49 atm
P
Similarly,
,
, and
.
=
=
=
2
2
2
H O
N
O
0.41 atm
0.25 atm
0.041 atm
P
P
P
5.98
We need to determine the molar mass of the gas.
Comparing the molar mass to the empirical mass will allow
us to determine the molecular formula.
3
0.001 L
(0.74 atm) 97.2 mL
1 mL
1.85
10
mol
L atm
0.0821
(200
273)K
mol K
-
×
÷
=
=
=
×
×
+
÷
×
PV
n
RT
3
0.145 g
molar mass
78.4 g/mol
1.85
10
mol
-
=
=
×
The empirical mass of CH
=
13.02 g/mol
Since
78.4 g/mol
6.02
6
13.02 g/mol
=
≈
, the molecular formula is (CH)
6
or
C
6
H
6
.
5.99
(a)
NH
4
NO
2
(
s
)
→
N
2
(
g
)
+
2H
2
O(
l
)
(b)
Map out the following strategy to solve the problem.
volume N
2
→
moles N
2
→
moles NH
4
NO
2
→
grams NH
4
NO
2
First, calculate the moles of N
2
using the ideal gas equation.
T
(K)
=
22
°
+
273
°
=
295 K
1 L
86.2 mL
0.0862 L
1000 mL
=
×
=
V
2
2
N
N
=
P
V
n
RT
2
3
N
(1.20 atm)(0.0862 L)
=
4.27
10
mol
L atm
0.0821
(295 K)
mol K
-
=
×
×
÷
×
n
Next, calculate the mass of NH
4
NO
2
needed to produce 4.27
×
10
-
3
mole of N
2
.
3
4
2
4
2
2
2
4
2
1 mol NH NO
64.05 g NH NO
(4.27
10
mol N )
1 mol N
1 mol NH NO
-
=
×
×
×
=
4
2
? g NH NO
0.273 g
167
CHAPTER 5:
GASES
5.100
The reaction is:
HCO
3
-
(
aq
)
+
H
+
(
aq
)
→
H
2
O(
l
)
+
CO
2
(
g
)
The mass of HCO
3
-
reacted is:
3
3
32.5% HCO
3.29 g tablet
1.07 g HCO
100% tablet
-
-
×
=
3
2
2
3
2
3
3
1 mol HCO
1 mol CO
mol CO
produced
1.07 g HCO
0.0175 mol CO
61.02 g HCO
1 mol HCO
-
-
-
-
=
×
×
=
2
2
CO
L atm
(0.0175 mol CO ) 0.0821
(37
273)K
mol K
0.445 L
(1.00 atm)
×
+
÷
×
=
=
=
=
2
CO
445 mL
n
RT
P
V
5.101
No, because an ideal gas cannot be liquefied, since the assumption is that there are no intermolecular forces in
an ideal gas.
5.102
(a)
The number of moles of Ni(CO)
4
formed is:
4
4
1 mol Ni(CO)
1 mol Ni
86.4 g Ni
1.47 mol Ni(CO)
58.69 g Ni
1 mol Ni
×
×
=
The pressure of Ni(CO)
4
is:
L atm
(1.47 mol) 0.0821
(43 + 273)K
mol K
4.00 L
×
÷
×
=
=
=
9.53 atm
nRT
V
P
(b)
Ni(CO)
4
decomposes to produce more moles of gas (CO), which increases the pressure.
Ni(CO)
4
(
g
)
→
Ni(
s
)
+
4CO(
g
)
5.103
The partial pressure of carbon dioxide is higher in the winter because carbon dioxide is utilized less by
photosynthesis in plants.
5.104
Using the ideal gas equation, we can calculate the moles of gas.
