Answer it with respect to gay lussacs law

Dear Student
$$\begin{gathered} 22.4 d{m}^3= 22.4 litre = 1mole \\ 11. 2d{m}^3 = 11.2 litre = 0.5 moles \\ Now C{H}_4 + 2O_2 \to C{O}_2 + 2H_{2}O \\ so 1mol C{H}_4 requires 1 mol O_2 to burn completely. \\ also, 2H_2 + O_2 \to 2H_{2}O \\ 2 mol H_2 requires 1 mol O_2 to burn completely. \\ So O_2 required by 0.5 mol H_2= 0.25 moles \\ Hence total moles of oxygen required=1+0.25=1.25 moles \end{gathered}$$
Hence volume required = 1.25 x 22.4 litres=28 litres
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