A gaseous mixture of propane and butane of volume 3litre on complete combustion produces 11.0 litre CO₂ under standard conditions of temperature and pressure. The ratio of volume of butane to propane is:

Suppose, the volume of propane = V L
$\therefore \underset{\mathrm{V}}{{\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)}+\underset{5\mathrm{V}}{5 {\mathrm{O}}_2}\left(\mathrm{g}\right)⟶\underset{3\mathrm{V}}{3}{\mathrm{CO}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
$\underset{(3-\mathrm{v})}{{\mathrm{C}}_4{\mathrm{H}}_{10}\left(\mathrm{g}\right)}+\underset{\frac{13}{2}(3-\mathrm{V})}{\frac{13}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)}⟶\underset{4(3-\mathrm{v})}{4{\mathrm{CO}}_2\left(\mathrm{g}\right)}+5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
$\because$Total volume of CO₂ produced = 11 L;
$\because$ 3V + 4(3 -V) =11; $\therefore$V=1
$\therefore$ Volume of butane = (3-1) = 2 L
Thus, the ratio of volume of butane to Propane = 2 :1