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_(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) to xCO_(2)(g) + (y)/(2)H_(2)O(l)`
15 mL hydrocabon requires `15(x+(y)/(4))mL` oxygen
`therefore " " 15(x+(y)/(4)) = 75 `
` x + (y)/(4) = 5`
`therefore` Hydrocarbon will be `C_(3)H_(8)`.