A mixture of ethyne gas (C2H2) and methane gas (CH4) occupied a certain volume at a total pressure of 16.8 kPa

A mixture of ethyne gas (C2H2) and methane gas (CH4) occupied a certain volume at a total pressure of 16.8 kPa. When the sample was burned, the products collected were CO2 gas and H2O vapor. The CO2 was collected and its pressure was found to be 25.2 kPa in the same volume and the same temperature as the original mixture. What was the percentage of methane in the original mixture?


I really need help on how to solve this. Can someone write out the steps? We are getting a ton of questions like this and I HAVE to know how to solve.

Thanks!

since T and V are held constant, moles are directly proportional to partial pressure of the gases
so we can use the coefficients in a balanced equation to find the relative amounts of each
actually
we only need a partially balanced equation - balanced for the gases in question

___CH4 + ___C2H2 + ___O2 ---> ___CO2 + ___H2O
let's use x and y
x CH4 + y C2H2 + ___O2 ---> (x+2y) CO2 + ___H2O
the C's are balanced and that's all that's necessary

now taking the x and y as partial pressures
x + y = 16.8 (before reaction)
x + 2y = 25.2 (after reaction)
now subtract the top from the bottom
and you get
y = 8.4
and x = 16.8-8.4 = 8.4
so
%/100 = x/16.8 = 8.4/16.8
50% methane in the mix