Chemistry Forum

Hello everyone, would you help me understand how to solve the following exercise?
"For complete combustion of 5.00 dm ^ 3 of a mixture of methane, CH₄, and ethane, C₂H₆, 13.50 dm ^ 3 of oxygen are needed, measured at the same conditions of temperature and pressure:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂O
2C₂H₆ (g) + 7O₂ (g) -> 4CO₂ (g) + 6H₂O
calculate the molar fractions of the two hydrocarbons. "
thanks so much

pV = nRT

P, R and T are constant so n~V

V(CH₄) + V(C₂H₆) = 5 dm^3

2V(O₂(CH₄)) + 3,5V(O₂(C₂H₆)) = 13,5 dm^2

Substitute Ethane by 5dm^3 - V(CH₄) = V(C₂H₆)

x + y = 5
2 x + 3,5 y =13,5

thank you very much, really. I had guessed the proportions but I hadn't put them into a system.

What is your result?

here is my development :
x+y=5 y=2,33
2x+7/2y=13,5 x=2,66
n(tot)=V/Vm= 5 L/22,414 L= 0,223
n(CH₄)=2,66/22,414=0,1189 mol; X(CH₄)=n(CH₄)/n(tot)=0,1189/0,223=0,533
n(C₂H₆)=2,33/22,414=0,1041 mol; X(C₂H₆)=n(C₂H₆)/n(tot)=0,1041/0,223=0,467
it's correct?

The candidate has 99 points. If you reach 100 points you get a light cookie.

But to achieve this calculation it goes faster 2,666/5 = 0,5332
And 2,333/5 = 0,4666

Because V~ n.

thanks for the help, I asked for help on an Italian forum but I have not received an answer. It is the first time that I have encountered an exercise of this kind, and among the exercises carried out by my professor there was none of the kind.