What volume of oxygen gas (${\text{O}}_2$) measured at $0^{\circ }\text{C}$ and $1\text{atm}$, is needed to burn completely $1\text{L}$ of propane gas (${\text{C}}_3{\text{H}}_8$) measured under the same conditions?

The balanced chemical equation for the combustion of propane is:
$${\text{C}}_3{\text{H}}_8(g)+5{\text{O}}_2(g)\to 3{\text{CO}}_2(g)+4{\text{H}}_2\text{O}(l)$$
According to Gay-Lussac's law of combining volumes, the ratio of volumes of gases is the same as their stoichiometric coefficients at the same $T$ and $P$.$1{\text{volume of C}}_3{\text{H}}_8$ requires $5{\text{volumes of O}}_2$.So, $1{\text{L of C}}_3{\text{H}}_8$ requires $5{\text{L of O}}_2$.