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(B) 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

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$0.25$

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(B) 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 $CH_4$ produces $1 \ mole$ of $CO_2$.The molar mass of $CO_2$ is $12 + (2 	imes 16) = 44 \ g \ mol^{-1}$.Moles of $CO_2$ produced = $\frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{11 \ g}{44 \ g \ mol^{-1}} = 0.25 \ mol$.Since $1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$,the number of moles of $CH_4$ required to produce $0.25 \ mol$ of $CO_2$ is $0.25 \ mol$.

The number of moles of methane required to produce $11 \ g$ of $CO_{2(g)}$ after complete combustion is:
(Given molar mass of methane in $g \ mol^{-1}: 16$)

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The number of moles of methane required to produce $11 \ g$ of $CO_{2(g)}$ after complete combustion is:(Given molar mass of methane in $g \ mol^{-1}: 16$)