gaseous mixtures

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anto007
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gaseous mixtures

Post by anto007 »

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
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ChenBeier
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Re: gaseous mixtures

Post by ChenBeier »

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
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anto007
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Re: gaseous mixtures

Post by anto007 »

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

Post by ChenBeier »

What is your result?
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anto007
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Re: gaseous mixtures

Post by anto007 »

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?
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ChenBeier
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Re: gaseous mixtures

Post by ChenBeier »

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.
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anto007
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Re: gaseous mixtures

Post by anto007 »

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.
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