At `300 K` and `1 atm, 15 mL` of a gaseous hydrocarbon requires `375 mL` air containing `20% O_(2)` by volume for complete combustion. After combustion, the gases occupy `330 mL`. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

Options

A. `C_(3)H_(6)`
B. `C_(3)H_(6)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Answer

B

Explanation

`C_(x)H_(y) +(x+y/4)O_(2) to xCO_(3) + y/2 H_(2)O`
Volume of oxygen used,
`VO_(2) = (20 xx 375)/100 = 75` mL
`15(x+y/4) = 75`
`(x+y/4) = 5`
This correspond to formula `C_(3)H_(8)`.
It may be noted that in this case the further information i.e, 330 mL volume is neglected. If we sue that information then none of the answer is correct.