A `5.0g` sample of a natural gas consisting of `CH_(4),C_(2)H_(4)` was burnt in excess of oxygen yielding `14.5g CO_(2)` and some `H_(2)O` as product. What is weight percentage of `CH_(4)` and `C_(2)H_(4)` in mixture?

Explanation

Suppose ethylene `(C_(2)H_(4))` in the mixture = x g
The methane `(CH_(4))` in the mixture = (5 - x) g
Equations for complete combustion of `CH_(4)` and `C_(2)H_(4)` are :
`C_(2)H(4)+3 O_(2) rarr 2CO_(2)+2H_(2)O`
`CH_(4)+2O_(2)rarr CO_(2)+2H_(2)O`
`28 g C_(2)H_(4)` produce `CO_(2)=2xx44 g = 88 g`
`therefore x g C_(2)H_(4)` will produce `CO_(2)=(88)/(28)xx x g = (22 x)/(7)g`
`16 g CH_(4)` product `CO_(2)=44 g`
`(5-x)g CH_(4)` will produce `CO_(2)=(44)/(16)xx(5-x)=(11(5-x))/(4)`
`therefore (22x)/(7)+(11(5-x))/(4)=14.5` or `88x +77(5-x)=28xx14.5`
or `11 x = 406 - 385=21` or `x=(21)/(11)`
`therefore %` of `C_(2)H_(4)=(21//11)/(5)xx100=(21)/(55)xx100=38.18 %` .