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

Let `x=` moles of `C_(2)H_(6)` and `y=` mols of `C_(2)H_(4)`
`C_(2)H_(6)+(7)/(2)O_(2)to2CO_(2)+3H_(2)O`
`C_(2)H_(4)+3O_(2)to2CO_(2)+2H_(2)O`
Moles of `O_(2)` used `=((7x)/(2)+3y)`
Moles of `O_(2)` available `=(130)/(32)`
`:. (7x)/(2)+3y=(130)/(32)`
`implies 3.5x+3y=4.06`
Also, `x+y=(PV)/(RT)=(1xx40)/(0.0821xx400)=1.218`
Solving for `x` and `y`, we get
`x=0.812` and `y=0.406`
`x_(C_(2)H_(6))=(0.812)/(1.218)=0.67` and `y_(C_(2)H_(4))=0.33`