30mL of a gaseous hydrocarbon requires 90mL of `O_(2)` for complete oxidation, 60mL of `CO_(2)` gas is formed in the process. The molecular formula of the hydrocarbon will be:

Options

A. `C_(2)H_(4)`
B. `C_(4)H(10)`
C. `C_(3)H_(6)`
D. `C_(2)H_(2)`

Explanation

`C_(x)H_(y)+(x+(x)/(4))"mL oxygen"`
`"30mL of hydrocarbon will require 30 "(x+(x)/(4))"mL oxygen"`
`therefore 30(x+(x)/(4))=90`
`x+(x)/(4)=3…….(i)`
`"Volume of "CO_(2) "produced"=30x=60mL`
`therefore x=2`
From eqn.(i), `2+(y)/(4)=3`
`therefore y=4`
Molecular formula of hydrocarbon= `C_(2)H_(4)`