a mixture of methane CH4 and ethane C2H6 is stored in a container at 294 mmHg. The gases are burned in air to form CO2 and H2O. If the pressure of CO2 is 346mmHg at the same temp and volume as the original mixture, calculate the ratio of moles of CO2 produced to moles of the original gas mixture.

Answer 1

n = PV/RT so the ratio is
(PV/RT)of CO2 = (PV/RT)gases
Since V, R, and T are constant, then the ratio is simply PCO2/Pgases = 346/294 = ??
The problem doesn't ask for it but you can calculate the PCH4 and PC2H6 if you wish as below. I wouldn't have done this but I misread the problem and calculated the actual pressures first before I did the part above. I hated for the below to go to waste.

You must note that the CH4 produces 1 mole CO2 for every 1 mole CH4 burned while C2H6 produces 2 moles CO2 for every mole of C2H6 burned.
PCH4 + PC2H6 = 294
PCO2fromCH4 + 2PCO2fromC2H6 = 346
but 2PC2H6 = PCO2fromC2H6.
Rewrite
PCO2from CH4 + 2PCO2fromC2H6 = 346. Substitute
PCH4 + 2PC2H6 = 294
Subtract
PC2H6 = 346-394=52
Therefore, PC2H6=52 mm Hg.
PCH4 = 294-52 = 242 mm Hg.

Answer 2

Of course moles of the reactants are NOT equal to the moles of products. The ratio is (PV/RT)_CO2/(PV/RT)_{gases/sub> so ratio of moles is ratio of PCO2/Pgases.}