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.

A mixture of C_(2)H_(6) and C_(4)H_(4) occupies 40 litres at 1 atm and

1.	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.
Answer» Solution :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`.