At 300 K and 1 atm, 15 mL of 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_(10)`
B. `C_(3)H_(8)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Answer

B

Explanation

Relevant reaction for the combustion of hydrocarbon,
`{:(,C_(x)H_(y)(g),+,(x+(y)/(4))O_(2)(g),to,xCO_(2)(g),+,(y)/(2)H_(2)O(l)),("Initially:",15mL,,15(x+(y)/(4))mL,,0,,),("finally:",0,,0,,15xmL,,):}`
Amount of `O_(2)` in 375 mL of air `=(20)/(100)xx375=75mL`
`therefore 15(x+(y)/(4))=75 or, x+(y)/(4)=5`
only `C_(3)H_(8)` follow the above equation.