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(B) The combustion reaction of methane is: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. From the stoichiometry,$1 \ mole$ of $CH_4$ produces $1 \ mole$ of

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$80$

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(B) The combustion reaction of methane is: $CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$. From the stoichiometry,$1 \ mole$ of $CH_4$ produces $1 \ mole$ of $CO_2$. The reaction of $CO_2$ with calcium hydroxide is: $Ca(OH)_2 + CO_2 ightarrow CaCO_3 \downarrow + H_2O$. The molar mass of $Ca(OH)_2 = 40 + 2 	imes (16 + 1) = 74 \ g/mol$. The molar mass of $CaCO_3 = 40 + 12 + 3 	imes 16 = 100 \ g/mol$. Given $500 \ g$ of $CaCO_3$ precipitate is formed,which corresponds to $500 / 100 = 5 \ moles$ of $CaCO_3$. Since $1 \ mole$ of $CaCO_3$ is produced from $1 \ mole$ of $CO_2$,$5 \ moles$ of $CO_2$ must have been produced. From the combustion reaction,$5 \ moles$ of $CO_2$ are produced from $5 \ moles$ of $CH_4$. The molar mass of $CH_4 = 12 + 4 	imes 1 = 16 \ g/mol$. Therefore,the mass of $CH_4$ $(x)$ = $5 \ moles 	imes 16 \ g/mol = 80 \ g$.

$x \ g$ of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing $370 \ g$ of $Ca(OH)_2$. The weight of white precipitate obtained was $500 \ g$. What is the value of $x$ (in $g$)? (Given: $C=12, H=1, Ca=40, O=16 \ u$)

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