How many moles of methane are required in the | 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 balanced equation,$1 \ mol$ of $CH_4$ p

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$0.5$

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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 balanced equation,$1 \ mol$ of $CH_4$ produces $1 \ mol$ of $CO_2$.The molar mass of $CO_2$ is $12 + 2 	imes 16 = 44 \ g/mol$.The number of moles of $CO_2$ produced is $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$.Since $1 \ mol$ of $CH_4$ produces $1 \ mol$ of $CO_2$,to produce $0.5 \ mol$ of $CO_2$,$0.5 \ mol$ of $CH_4$ is required.

How many moles of methane are required in the combustion reaction of methane to produce $22 \ g$ of $CO_{2(g)}$ (in $mol$)?

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