A mixture of ethane and ethene occupies `40 mL` at `1.0 atm`, and `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)+2H_(2)O(l)`
Let the moles of ethene be `a`.
Volume of `O_(2)` required by ethene `=(7)/(2)a`
Volume of ethene `=1.218-a`
Volume of `O_(2)` required by ethene `=3(1.218-a)`
Given
`(7)/(2)a+3(1.218-a)=(130)/(32)` (moles of oxygen)
or `(7)/(2)a+3.654-3a=4.0625`
`(1.218-a)=(1.218-0.817)=0.401` (moles of ethene)
Mole fraction of ethane `=(0.817)/(1.218)=0.6707`
Mole fraction of ethene `=(1-0.6707)=0.3293`