Question

A mixture of CH4(g) and C2H6(g) has a total pressure of 0.55 atm . Just enough O2(g) is added to the mixture to bring about its complete combustion to CO2(g) and H2O(g). The total pressure of the two product gases is found to be 2.3 atm . Assuming constant volume and temperature, find the mole fraction of CH4 in the mixture. Please type out answer step by step. Thank you!

Answer 1

P( CH4(g) + C2H6(g) ) = 0.55 atm

Since Pressure is directly proprtional to moles

n(CH4) + n(C2H6) = 0.55

n(CH4) = 0.55 - n(C2H6)

CH4(g) + 2 O2(g) -------> CO2(g) + 2 H2O (g)

x                                       x               2x

C2H6(g) + 7/2 O2(g) -------> 2 CO2(g) + 3 H2O (g)

0.55 - x                                  1.1 - 2x        1.65 - 3x

Now,

Total pressure of two product gases = 2.3 atm

3x + ( 1.1 - 2x) + ( 1.65 - 3x) = 2.3

2.75 - 2.3 = 2x

x = 0.225

Moles of CH4 = 0.225

Moles of C2H6 = 0.55 - 0.225 = 0.325

Mole fraction of CH4 = 0.225 / 0.55 = 0.409