The moles of methane required to produce 81 | vedclass

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(B) The balanced chemical equation for the combustion of methane is:$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$From the stoichiometry,$1 \ mol$ of $CH_4$ p

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(B) The balanced chemical equation for the combustion of methane is:$CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$From the stoichiometry,$1 \ mol$ of $CH_4$ produces $2 \ mol$ of $H_2O$.Given mass of water = $81 \ g$.Molar mass of $H_2O = 18 \ g/mol$.Moles of $H_2O$ produced = $\frac{81 \ g}{18 \ g/mol} = 4.5 \ mol$.Since $2 \ mol$ of $H_2O$ are produced from $1 \ mol$ of $CH_4$,the moles of $CH_4$ required = $\frac{4.5}{2} = 2.25 \ mol$.Expressing $2.25 \ mol$ in terms of $10^{-2} \ mol$:$2.25 = 225 	imes 10^{-2} \ mol$.Thus,the nearest integer is $225$.

The moles of methane required to produce $81 \ g$ of water after complete combustion is $.... \times 10^{-2} \ mol.$ [nearest integer]

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