Question

A mixture of CH4 (g) and C2H6 (g) has a total pressure of 0.53 atm. Just enough O2 was added to the mixture to bring about it's complete combustion to CO2 (g) and H2O (g). The total pressure of the two product gases is found to be 2.2 atm. Assuming constant volume and temperature, find the mole fraction of CH4 in the original mixture.

Answer 1

In each condition, the pressure will be directly proportional to the moles of gas in the container.
Looking at the balanced equation for the combustion of each compound,

CH4 + 2 O2 --> CO2 + 2 H2O
C2H6 + 7/2 O2 --> 2 CO2 + 3 H2O

If you let x = pressure of CH4 in the original mixture, then its combustion would produce x atm of CO2 and 2x atm of H2O

Likewise if y = pressure of C2H6, then its combustion would produce 2y atm of CO2 and 3y atm of H2O.

So, x + y = 0.53 atm and 3x + 5y = 2.2 atm
Solving these two equations:
x+y = .53--> x = .53 - y

Substituting this into the second equation and solving for y gives:

3(.53-y)+5y = 2.2
2y = 0.61
y = .305 atm = pressure of C2H6
x = .53 - .305 = .225 atm = pressure of CH4

The mole fraction of CH4 is equal to the ratio of the partial pressure of CH4 to the total pressure:

.225 / .53 = 0.425

And the mole fraction of C2H6 , calculated the same way is : .305/.53 = 0.575