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(C) The balanced chemical equation for the combustion of methane is:$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$According to the stoichiomet

Vedclass · Accepted Answer

$11.2$

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(C) The balanced chemical equation for the combustion of methane is:$CH_4(g) + 2O_2(g) \longrightarrow CO_2(g) + 2H_2O(l)$According to the stoichiometry of the reaction,$1 \ mole$ of $CH_4$ requires $2 \ moles$ of $O_2$.At $	ext{S.T.P.}$,$1 \ mole$ of any gas occupies $22.4 \ dm^3$.Therefore,$2 \ moles$ of $O_2$ occupy $2 	imes 22.4 \ dm^3 = 44.8 \ dm^3$.For $0.25 \ mole$ of $CH_4$,the volume of $O_2$ required is:$V = 0.25 	imes 44.8 \ dm^3 = 11.2 \ dm^3$.

What is the volume of oxygen required for complete combustion of $0.25 \ mole$ of methane at $\text{S.T.P.}$ (in $dm^3$)?

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