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(A) The combustion reaction is: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$Moles of $CH_4 = \frac{100 \, g}{16 \, g/mol} = 6.25 \, mol$Moles of $O_2 = \fra

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$143$

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(A) The combustion reaction is: $CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$Moles of $CH_4 = \frac{100 \, g}{16 \, g/mol} = 6.25 \, mol$Moles of $O_2 = \frac{208 \, g}{32 \, g/mol} = 6.5 \, mol$To find the limiting reagent,calculate the ratio of moles to stoichiometric coefficient:For $CH_4: \frac{6.25}{1} = 6.25$For $O_2: \frac{6.5}{2} = 3.25$Since the ratio for $O_2$ is smaller,$O_2$ is the limiting reagent.From the stoichiometry,$2 \, mol$ of $O_2$ produces $1 \, mol$ of $CO_2$.Therefore,$6.5 \, mol$ of $O_2$ will produce $\frac{6.5}{2} = 3.25 \, mol$ of $CO_2$.Mass of $CO_2 = 3.25 \, mol 	imes 44 \, g/mol = 143 \, g$.

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