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Gay Lussac's Law, Avagadro's Law

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Question

What mass (in grams) of $C H_{4}$ produces $3.01 \times 10^{23} H_{2} O$ molecules when burnt to excess $O_{2}$ ?

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Solution

The balanced chemical equation for the complete consumption of methane is as given below:
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$3.01 \times 10^{23}$ water molecules corresponds to $\frac{3.01 \times 10^{23}}{6.02 \times 10^{23}} = 0.5$ mol.
They will be produced from the combustion of $0.5 \times \frac{1}{2} = 0.25$ mol of methane.
This corresponds to $0.25 \times 16 = 4$ g of methane.

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