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A mixture of C(2)H(6) and C(4)H(4) occup...

A mixture of `C_(2)H_(6)` and `C_(4)H_(4)` occupies 40 litres at 1 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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Let the moles of `C_(2)H_(6)` and `C_(2)H_(4)` be `n_(1)` and `n_(2)` respectively. Applying ideal gas equation, `pV = n RT`,
`1 xx 40 = (n_(1) + n_(2)) xx 0.0821 xx 400`
or, `(n_(1) + n_(2)) = (40)/(0.0821 xx 400)` ……(1)
`C_(2)H_(6) + C_(2)H_(4) + O_(2) rarr CO_(2) + H_(2)O`
Applying POAC for C,
`2n_(1) + 2n_(2)` = moles of `CO_(2)` ......(2)
Applying POAC for H,
`6 n_(1) + 4n_(2) = 2 xx "moles of " H_(2)O` ........(3) Applying POAC for o,
`2 xx (130)/(32) = 2 xx " moles of " CO_(2) + " moles of " H_(2)O` ........(4)
From equations (2), (3) and (4) we get,
`7 n_(1) + 6n_(2) = (260)/(32)` .......(5)
Solving equations (1) and (5) we get,
`n_(1) = 0.8168`
`n_(2) = 0.4012`
`therefore` mole fraction of `C_(2)H_(6) = (0.8168)/(0.8168 + 0.4012) = 0.67`
Mole fraction of `C_(2)H_(4) = 1-0.67 = 0.33`.
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