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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_{2}O_{(g)}$$(i)$ $16 \ g$ of $CH

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) The balanced chemical equation for the combustion of methane is:$CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_{2}O_{(g)}$$(i)$ $16 \ g$ of $CH_{4}$ is equal to $1 \ 	ext{mole}$.$(ii)$ From the balanced equation,$1 \ 	ext{mole}$ of $CH_{4(g)}$ produces $2 \ 	ext{moles}$ of $H_{2}O_{(g)}$.$(iii)$ The molar mass of $H_{2}O$ is $(2 	imes 1 + 16) = 18 \ g/	ext{mol}$.$(iv)$ Therefore,$2 \ 	ext{moles}$ of $H_{2}O_{(g)} = 2 	imes 18 \ g = 36 \ g$ of $H_{2}O$.

Calculate the amount of water $(g)$ produced by the combustion of $16 \ g$ of methane. (in $g$)

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