At 300 K and 1 atm, 15mL of a gaseous hydrocarbon requires 375mL air containing 20% `O_(2)` by volume for complete combustion. After combination the gases occupy 330 mL. Assuming that the water formed isin 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_(8)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Answer

B

Explanation

Combustion of hydrocarbon `C_(x)H_(y)` is given as,
`C_(x)H_(y)(g)+(x+(y)/(4))O_(2)(g)rarrxCO_(2)(g)+(y)/(2)H_(2)O(l)`
15mL hydrocarbon requires `15(x+(y)/(4))`mL oxygen.
`therefore 15(x+(y)/(4))=75` or `x+(y)/(4)=5`
`therefore` Hydrocarbon will be `C_(3)H_(8)`.