The volume of $O_2$ required for the combustion of 10 lit of methane under the same condition is

Methane in the presence of oxygen forms carbon dioxide and water.

The balanced chemical equation is:

${\mathrm{CH}}_4+2{\mathrm{O}}_2\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

Here, 1 mole of methane combines with 2 moles of $O_2$.

$22.4 \mathrm{L}$ of ${CH}_4$ (at STP) requires $44.8 \mathrm{L}{\mathrm{ofO}}_2$.

$10 \mathrm{L}$ of ${CH}_4$ requires $=\frac{44.8}{22.4}\times 10=20 \mathrm{Lof}{\mathrm{O}}_2$.