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(B) The balanced chemical equation for the combustion of methane is:$CH_4 \ (g) + 2O_2 \ (g) \rightarrow CO_2 \ (g) + 2H_2O \ (g)$From the stoichiomet

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$0.5$

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(B) The balanced chemical equation for the combustion of methane is:$CH_4 \ (g) + 2O_2 \ (g) ightarrow CO_2 \ (g) + 2H_2O \ (g)$From the stoichiometry,$1 \ mol$ of $CH_4$ produces $1 \ mol$ of $CO_2$.The molar mass of $CO_2$ is $12 + (2 	imes 16) = 44 \ g/mol$.Number of moles of $CO_2$ produced $= \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$.Since $1 \ mol$ of $CH_4$ produces $1 \ mol$ of $CO_2$,to produce $0.5 \ mol$ of $CO_2$,we require $0.5 \ mol$ of $CH_4$.

How many moles of methane are required to produce $22 \ g$ of $CO_2 \ (g)$ after combustion (in $mol$)?

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