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(C) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$Initially: $20 \ mL$ of $CH_4$ and $50 \ mL$ of

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$30$

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(C) The combustion reaction of methane is: $CH_{4(g)} + 2O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$Initially: $20 \ mL$ of $CH_4$ and $50 \ mL$ of $O_2$ are taken.According to the stoichiometry,$1 \ volume$ of $CH_4$ requires $2 \ volumes$ of $O_2$.So,$20 \ mL$ of $CH_4$ will react with $2 	imes 20 = 40 \ mL$ of $O_2$.Remaining $O_2 = 50 \ mL - 40 \ mL = 10 \ mL$.Volume of $CO_2$ produced = $20 \ mL$.Since $H_2O$ is produced as a liquid at room temperature,its volume is negligible.Total volume of gas left = $Volume \ of \ remaining \ O_2 + Volume \ of \ CO_2 = 10 \ mL + 20 \ mL = 30 \ mL$.

$20 \ mL$ of methane is completely burnt using $50 \ mL$ of oxygen. The volume of the gas left after cooling to room temperature is (in $mL$)

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