What is the volume of O_2 gas required at STP | vedclass

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(B) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$From the stoichiometry,$1 \ mol$ of $CH_4$ requi

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$11.2$

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(B) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$From the stoichiometry,$1 \ mol$ of $CH_4$ requires $2 \ mol$ of $O_2$.Moles of $CH_4 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$.Since $1 \ mol$ $CH_4$ requires $2 \ mol$ $O_2$,then $0.25 \ mol$ $CH_4$ requires $0.25 	imes 2 = 0.50 \ mol$ of $O_2$.At $STP$,the volume of $1 \ mol$ of gas is $22.4 \ L$.Therefore,the volume of $0.50 \ mol$ of $O_2 = 0.50 	imes 22.4 \ L = 11.2 \ L$.

What is the volume of $O_2$ gas required at $STP$ for the combustion of $4 \ g$ of methane gas (in $L$)?

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