Question

A). A mixture of methane and argon gases is maintained in a 8.34 L flask at a pressure of 2.24 atm and a temperature of 66 °C. If the gas mixture contains 3.59 grams of methane, the number of grams of argon in the mixture is............ g

. B). A mixture of xenon and hydrogen gases, in a 5.89 L flask at 13 °C, contains 12.5 grams of xenon and 0.288 grams of hydrogen. The partial pressure of hydrogen in the flask is.............. atm and the total pressure in the flask is............. atm.

Answer 1

A)
Given:
P = 2.24 atm
V = 8.34 L
T = 66.0 oC
= (66.0+273) K
= 339 K

find number of moles using:
P * V = n*R*T
2.24 atm * 8.34 L = n * 0.08206 atm.L/mol.K * 339 K
n = 0.6716 mol

Molar mass of CH4,
MM = 1*MM(C) + 4*MM(H)
= 1*12.01 + 4*1.008
= 16.042 g/mol


mass(CH4)= 3.59 g

use:
number of mol of CH4,
n = mass of CH4/molar mass of CH4
=(3.59 g)/(16.04 g/mol)
= 0.2238 mol

Now use:
Mol of Ar = total mol - mol of CH4
= 0.6716 mol - 0.2238 mol
= 0.4478 mol

Molar mass of Ar = 39.95 g/mol

use:
mass of Ar,
m = number of mol * molar mass
= 0.4478 mol * 39.95 g/mol
= 17.89 g
Answer: 17.9 g

B)
I)

Molar mass of H2 = 2.016 g/mol


mass(H2)= 0.288 g

use:
number of mol of H2,
n = mass of H2/molar mass of H2
=(0.288 g)/(2.016 g/mol)
= 0.1429 mol

Given:
V = 5.89 L
n = 0.1429 mol
T = 13.0 oC
= (13.0+273) K
= 286 K

use:
P * V = n*R*T
P * 5.89 L = 0.1429 mol* 0.08206 atm.L/mol.K * 286 K
P = 0.5694 atm
Answer: 0.569 atm

II)

Molar mass of Xe = 131.3 g/mol


mass(Xe)= 12.5 g

use:
number of mol of Xe,
n = mass of Xe/molar mass of Xe
=(12.5 g)/(1.313*10^2 g/mol)
= 9.52*10^-2 mol

Given:
V = 5.89 L
n = 0.0952 mol
T = 13.0 oC
= (13.0+273) K
= 286 K

use:
P * V = n*R*T
P * 5.89 L = 0.0952 mol* 0.08206 atm.L/mol.K * 286 K
P = 0.3793 atm

Now use:
Total pressure = p(H2) + p(Xe)
= 0.5694 atm + 0.3793 atm
= 0.9487 atm
Answer: 0.949 atm