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A mixture of ethane (C(2)H(6)) and ethen...

A mixture of ethane `(C_(2)H_(6))` and ethene `(C_(2)H_(4))` occupies `40 L` at `1.00 atm` and at `400 K`. The mixture reacts completely with `130 g` of `O_(2)` to produce `CO_(2)` and `H_(2)O`. Assuming ideal gas behaviour, calculate the mole fractions of `C_(2)H_(4)` and `C_(2)H_(6)` in the mixture.

Text Solution

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The correct Answer is:
(0.34 , 0.66)
The total moles of gaseous mixture `= (pV)/(RT) = (1 xx 40)/(0.082 xx 400) = 1.22`
Let the mixture contain x mole of ethane.
Therefore, `C_(2)H_(6) + (7)/(2) O_(2) rArr 2CO_(2) + 3H_(2)O, C_(2)H_(4) + 3O_(2) rarr 2CO_(2) + 2H_(2)O`
Total moles of `O_(2)` required `= (7)/(2) X + 3 (1.22 -X) = (X)/(2) + 3.66`
`rArr (130)/(32) = (X)/(2) + 3.66 rArr X = 0.805` mole ethane and 0.415 mole ethane.
`rArr` Moles fraction of ethane `=(0.805)/(1.22) = 0.66`
Moles fraction of ethane `= 1 - 0.66 = 0.34 , C_(2) H_(6) = 0.66, C_(2)H_(4) = 0.34`
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