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

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$50$

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(C) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$From the stoichiometry of the reaction,$1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$.The molar mass of $CO_2$ is $12 + 2 	imes 16 = 44 \ g/mol$.Number of moles of $CO_2$ produced $= \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$.Since $1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$,the moles of $CH_4$ required $= 0.5 \ mol$.We are given that the moles of $CH_4$ required is $x 	imes 10^{-2}$.$0.5 = x 	imes 10^{-2} \implies x = 0.5 	imes 10^2 = 50$.Thus,the value of $x$ is $50$.

Number of moles of methane required to produce $22 \ g$ $CO_{2(g)}$ after combustion is $x \times 10^{-2}$ moles. The value of $x$ is

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