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(A) $1$. Calculate the volume of $O_2$ in $150 \ mL$ of air: $V_{O_2} = 150 	imes \frac{21}{100} = 31.5 \ mL$. $2$. The reaction is $CH_{4(g)} + 2O_{

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$200$

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(A) $1$. Calculate the volume of $O_2$ in $150 \ mL$ of air: $V_{O_2} = 150 	imes \frac{21}{100} = 31.5 \ mL$. $2$. The reaction is $CH_{4(g)} + 2O_{2(g)} 	o CO_{2(g)} + 2H_2O_{(g)}$. $3$. According to the stoichiometry,$1 \ mL$ of $CH_4$ requires $2 \ mL$ of $O_2$. Thus,$50 \ mL$ of $CH_4$ would require $100 \ mL$ of $O_2$. $4$. Since we only have $31.5 \ mL$ of $O_2$,$O_2$ is the limiting reagent. $5$. $31.5 \ mL$ of $O_2$ will react with $15.75 \ mL$ of $CH_4$ to produce $15.75 \ mL$ of $CO_2$ and $31.5 \ mL$ of $H_2O_{(g)}$. $6$. Remaining $CH_4 = 50 - 15.75 = 34.25 \ mL$. $7$. Volume of $N_2$ (inert) $= 150 	imes \frac{79}{100} = 118.5 \ mL$. $8$. Total volume $= V_{CH_4(rem)} + V_{CO_2} + V_{H_2O} + V_{N_2} = 34.25 + 15.75 + 31.5 + 118.5 = 200 \ mL$.

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