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

For the alkane ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$ , the combustion reaction is

${\mathrm{C}}_n{\mathrm{H}}_{2n+2}+\left(\frac{3n+1}{2}\right){\mathrm{O}}_2\to n{\mathrm{CO}}_2+(n+1){\mathrm{H}}_2\mathrm{O}$

For 5 L of , ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$ the volume of oxygen required is (5 L) {(3n + 1 )/2}. Equating this to 25 L, we get

$\left(\frac{3n+1}{2}\right)=25\mathrm{L}$

This gives n = 3. Hence, the alkane is propane.