Question

A mixture of methane (CH4) and ethane (C2H6) is stored in a container at 294 mm Hg. The gases are burned in air to form CO2 and H2O. If the pressure of CO2 is 351 mm Hg measured at the same temperature and volume as the original mixture, calculate the mole fraction of the gases.

Answer 1

Explanation:

It is given that total pressure is 294 mm Hg. This also means that

     $P_{CH_{4}} + P_{C_{2}H_{6}}$ = 294 mm Hg

When we balance the carbon atoms then, 1 mole of $CH_{4}$ will yield 1 mole of $CO_{2}$ whereas 1 mole of $C_{2}H_{6}$ will yield 2 moles of $CO_{2}$.

So, $P_{CH_{4}}(\frac{1 mol CO_{2}}{1 mol CH_{4}}) + P_{C_{2}H_{6}}(\frac{2 mol CO_{2}}{1 mol C_{2}H_{6}}) = P_{CO_{2}}$

     $P_{CH_{4}}(1) + (294 - P_{CH_{4}})(2)$ = 351 mm Hg

     $P_{CH_{4}}$ = 2(294) - 351

                      = 237 mm Hg

We know that for ideal gas, mole fraction is the ratio of partial pressure and total pressure. And, the sum of moles fraction is always equal to 1.

Therefore, $\chi_{CH_{4}} = \frac{237}{294}$

                              = 0.806

     $\chi_{C_{2}H_{6}}$ = 1 - 0.806

                   = 0.194

Thus, we can conclude that the mole fraction of the gases are 0.806 and 0.194.