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 `O_(2)`
`PV=nRT`
or `1xx40=nxx0.0821xx400`
or `n=1.218`
`C_(2)H_(6)(g)+(7)/(2)O_(2)(g)to2CO_(2)(g)+3H_(2)O(l)`
`C_(2)H_(4)(g)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(l)`
Let the moles of ethane be a.
Volume of `O_(2)` requried by ethane `=(7)/(2a)`
Volume of ethene`=(1.218-a)`
Volume of `O_(2)` required by ethene`=3(1.218-a)`
Given
`(7)/(2)a3(1.218-a)=(130)/(32)` (moles of oxygen)
or `(7)/(2)a+3.654-3a=4.0625`
or `a=0.817` (moles of ethane)
`therefore1.218-a=1.218-0.817`
`=0.401` (miles of ethane)
Mole fraction ethane `=(0.817)/(1.218)=0.6707`
Mole fraction of ethene`=1-0.6707=0.3293`