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.

Explanation

For a gaseous mixture of `C_(2)H_(6)` and `C_(2)H_(14)`
`PV=nRT`
`:. 1xx40=nxx0.082xx400`
`:.` Total mole of `(C_(2)H_(6)+C_(2)H_(4))=1.2195`
Let mole `C_(2)H_(6)` and `C_(2)H_(4)` be `a` and `b` respectively.
`a+b=1.2195`….(1)
`C_(2)H_(6)+(7//2)O_(2)rarr2CO_(2)+3H_(2)O`
`C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O`
`:.` Mole of `O_(2)` needed for complete reaction of mixture
`=7a//2+3b`
`:. (7a)/(2)+3b=(130)/(32)`....(2)
By eqs.(1) and (2), `a=0.808, b=0.4115`
`:.` Mole fraction of `C_(2)H_(6)-0.808//1.2195=0.066`
and Mole fraction of `C_(2)H_(4)=0.34`