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

Answer

D

Explanation

`C_(X)H_(Y)(g)+((4x+y)/(4))O_(2)(g) to xCO_(2)(g)+(y)/(2)H_(2)O(1)`
Volume of `O_(2) "used"=375 xx (20)/(100)=75 mL`
`:.` Form the reaction of combustion
`1 mL C_(x)H_(y) "requires"=(4x+y)/(4)mL O_(2)`
`15 mL =15((4x+y)/(4))=75`
So 4x + y = 20
x = 3
y = 8
`C_3H_8`