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Avogadro's Law & Gay Lussac's Law

A 10 g samp...

Question

A $10 g$ sample of natural gas containing $C H_{4}$ and $C_{2} H_{4}$ was burnt in excess of oxygen to give $29.0$ grams of $C O_{2}$ and some water. How many grams of water are formed?

9.42 g

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18.81 g

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11.42 g

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15.31 g

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Solution

The correct option is B $15.31 g$
The combustion reactions are as follows:
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ ...... $(1)$
$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$ ...... $(2)$
Add equations $(1)$ and $(2)$ , we get
$C H_{4} + C_{2} H_{4} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$
Thus, $3$ moles of $C O_{2}$ corresponds to $4$ moles of $H_{2} O$ .
Thus, $3 \times 44$ g of $C O_{2}$ corresponds to $4 \times 18$ g of $H_{2} O$ .
Thus, $29.0$ g of $C O_{2}$ corresponds to $\frac{29.0}{3 \times 44} \times 4 \times 18 = 15.82$ g of $H_{2} O$ .
Hence, $15.82$ g of water are formed.

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