Question

A mixture of methane gas, CH4(g) , and pentane gas, C5H12(g) , has a pressure of 0.5799 atm when placed in a sealed container. The complete combustion of the mixture to carbon dioxide gas, CO2(g) , and water vapor, H2O(g) , was achieved by adding exactly enough oxygen gas, O2(g) , to the container. The pressure of the product mixture in the sealed container is 2.378 atm . Calculate the mole fraction of methane in the initial mixture, assuming the temperature and volume remain constant.

Answer 1

Mole fraction of methane is 0,8624

Step-by-step explanation:

The complete combustion of methane gas and pentane gas is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

C₅H₁₂(g) + 8O₂(g) → 5CO₂(g) + 6H₂O(g)

The initial pressure is 0,5799atm, that is the sum of partial pressures of methane and pentane gases:

`P_(initial) = P_(methane) + P_(pentane)`

0,5799atm = x + y (1)

Where x is partial pressure of methane and y is partial pressure of pentane.

As x moles of methane produce x moles of CO₂ and 2x moles of H₂O and y moles of pentane produce 5y moles of CO₂ and 6y moles of H₂O, it is possible to write:

2,378atm = x + 2x + 5y + 6y

2,378atm = 3x + 11y (2)

Now, replacing (1) in (2):

2,378atm = 1,7397atm - 3y + 11y

0,6383atm = 8y

y = 0,0798 atm

x = 0,5001 atm

The partial pressure of methane over total initial pressure will be the mole fraction of methane:

`(0,5001)/(0,5799)` = 0,8624

I hope it helps!