1. No category

Chem 2 AP Homework #5-3: Problems pg. 206 #63, 64, 68, 70, 102
63.
A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure
is 1.50 atm. Calculate the partial pressures of the gases.
The number of moles of the combined gases is:
n = n CH 4 + n C 2H 6 + n C3H8 = 0.31 mol + 0.25 mol + 0.29 mol = 0.85 mol
The mole fractions of each gas are:
Χ CH 4 =
0.31 mol
= 0.36
0.85 mol
Χ C2H6 =
0.25 mol
= 0.29
0.85 mol
Χ C 3H8 =
0.29 mol
= 0.34
0.85 mol
The partial pressures are:
PCH4 = Χ CH 4 × Ptotal = 0.36 × 1.50 atm = 0.54 atm
PC2H6 = Χ C 2H 6 × Ptotal = 0.29 × 1.50 atm = 0.44 atm
PC3H8 = Χ C3H8 × Ptotal = 0.34 × 1.50 atm = 0.51 atm
64.
A 2.5-L flask at 15°C contains a mixture of N2, He, and Ne at parial pressures of 0.32 atm for N2,
0.15 atm for He, and 0.42 atm for Ne.
(a) Calculate the total pressure of the mixture.
P total = 0.32 atm + 0.15 atm + 0.42 atm = 0.89 atm
(b) Calculate the volume in liters at STP occupied by He and Ne if the N2 is removed
selectively.
Initial conditions
P1 = (0.15 + 0.42)atm = 0.57 atm
T1 = (15 + 273)K = 288 K
V1 = 2.5 L
Final Conditions
P2 = 1.0 atm
T2 = 273 K
V2 = ?
P1V1
PV
PV T
(0.57 atm)(2.5 L)(273 K)
= 2 2 , so V2 = 1 1 2 =
= 1.4 L at STP
T1
T2
P2T1
(1.0 atm)(288 K)
68.
A sample of zinc metal reacts completely with excess hydrochloric acid:
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)
The hydrogen gas produced is collected over water at 25.0°C using an arrangement similar to
that shown in Figure 5.15. The volume of the gas is 7.80 L, and the pressure is 0.980 atm.
Calculate the mass of zinc consumed in the reaction (Pvap of H2O at 25.0°C is 23.8 mmHg).
PTotal = PH2 + PH2O and PH2 = PTotal − PH2O
" 1 atm %
PH 2 = 0.980 atm − (23.8 mmHg )$
' = 0.949 atm
# 760 mmHg &
nH2 =
PH2V
(0.949 atm)(7.80 L)
=
= 0.303 mol H 2
L⋅atm
RT
(0.0821 mol⋅K
)(25 + 273)K
? g Zn = 0.303 mol H 2 ×
1 mol Zn 65.39 g Zn
×
= 19.8 g Zn
1 mol H 2
1 mol Zn
2
HOMEWORK #5-3 ANSWER KEY
70.
A sample of NH3 gas is completely decomposed to N2 and H2 gases over heated iron wool. If
the total pressure is 866mmHg, calculate the partial pressures of N2 and H2.
→ N2(g) + 3H2(g)
2NH3(g) !!
Χ H2 =
3 mol
1 mol
= 0.750 ; Χ N2 =
= 0.250
3 mol + 1 mol
3 mol + 1 mol
PH2 = Χ H 2 PT = (0.750)(866 mmHg ) = 650 mmHg
PN2 = Χ N 2 PT = (0.250)(866 mmHg ) = 217 mmHg
102.
Consider the following apparatus. Calculate the parial pressures
of He and Ne after the stopcock is open. The temperature
remains constant at 16°C
First calculate the number of moles of each gas in the mixture.
n He =
PV
(0.63 atm )(1.2 L )
=
= 0.032 mol He
L⋅atm
RT
(0.0821 mol⋅K
)(16 + 273)K
n Ne =
PV
(2.8 atm )(3.4 L )
=
= 0.40 mol Ne
L⋅atm
RT
(0.0821 mol⋅K
)(16 + 273)K
The total pressure is:
PTotal =
L⋅atm
(0.032 + 0.40)mol (0.0821 mol⋅K
)(16 + 273)K = 2.2 atm
(nHe + nNe )RT
=
VTotal
(1.2 + 3.4)L
Pi = ΧiPT. The partial pressures of He and Ne are:
PHe =
0.032 mol
0.40 mol
× 2.2 atm = 0.16 atm ; PNe =
× 2.2 atm = 2.0 atm
(0.032 + 0.40)mol
(0.032 + 0.40)mol
Another way to calculate:
After mixing, the total volume is now 4.6 L.
L⋅atm
(0.032 mol He) (0.0821 mol⋅K
)(289K ) = 0.17 atm
n RT
PHe = He
=
V
4.6L
PNe =
L⋅atm
(0.40 mol Ne) (0.0821 mol⋅K
)(289K ) = 2.1 atm
nNe RT
=
V
4.6L
One final way:
Alternately, one could calculate the partial pressure of each gas by using Boyle’s Law and allowing each
gas to expand from its bulb to the entire container (V2:
PHe =
PiV i (0.63 atm)(1.2 L)
PV (2.8 atm)(3.4 L)
=
= 0.16 atm ; PNe = i i =
= 2.1 atm
VT
(1.2 + 3.4)L
VT
(1.2 + 3.4)L