`5L` of an alkane requires `25L` of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :

Options

A. Isobutane
B. Ethane
C. Butane
D. Propane

Answer

D

Explanation

Since the compound undergoing combustion is an alkane. Hence the combustion reaction can be written as
`underset(5L)(C_(n)H_(2n+2))+underset(25L)(((3n+1)/(2)))O_(2) to nCO_(2)+(n+1)H_(2)O`
Since volumes are measured at constant T & P. Hence according to Avogadro.s law
`"Volume" prop "mole"`
`:. n_("alkane")=((2)/(3n+1)) xx n_(O_(2))`
`5=(2)/(3n+1) xx 25`
`:. n =3`
Hence alkane is propane (`C_(3)H_(8)`).