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Question

(C) The balanced chemical equation for the combustion of methane is:$CH_{4}(g) + 2O_{2}(g) \longrightarrow CO_{2}(g) + 2H_{2}O(l)$From the stoichiomet

Vedclass · Accepted Answer

$11.2$

Vedclass · Answer

(C) The balanced chemical equation for the combustion of methane is:$CH_{4}(g) + 2O_{2}(g) \longrightarrow CO_{2}(g) + 2H_{2}O(l)$From the stoichiometry of the reaction,$1 \ mol$ of $CH_{4}$ requires $2 \ mol$ of $O_{2}$ for complete combustion.At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ dm^{3}$.Therefore,$2 \ mol$ of $O_{2}$ occupies $2 	imes 22.4 \ dm^{3} = 44.8 \ dm^{3}$.For $0.25 \ mol$ of $CH_{4}$,the volume of $O_{2}$ required is:$0.25 \ mol 	imes (2 \ mol \ O_{2} / 1 \ mol \ CH_{4}) 	imes 22.4 \ dm^{3}/mol = 0.5 \ mol 	imes 22.4 \ dm^{3}/mol = 11.2 \ dm^{3}$.Thus,the correct option is $C$.

The volume of oxygen required for complete combustion of $0.25 \ mol$ of methane at $S.T.P.$ is (in $dm^{3}$)

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