Question

A gas mixture with a total pressure of 760 mmHg contains each of the following gases at the indicated partial pressures: 118 mmHg CO2, 204 mmHg Ar, and 177 mmHg O2. The mixture also contains helium gas. What mass of helium gas is present in a 12.6-L sample of this mixture at 278 K ?

Answer 1

1st find the pressure of Helium gas

pTotal = sum of all partial pressures

760 = 118 + 204 + 177 + p(He)

p(He) = 261 mmHg

we have:

P = 261.0 mm Hg

= (261.0/760) atm

= 0.3434 atm

V = 12.6 L

T = 278.0 K

find number of moles using:

P * V = n*R*T

0.3434 atm * 12.6 L = n * 0.08205 atm.L/mol.K * 278 K

n = 0.1897 mol

Molar mass of He = 4.003 g/mol

we have below equation to be used:

mass of He,

m = number of mol * molar mass

= 0.1897 mol * 4.003 g/mol

= 0.759 g

Answer: 0.759 g