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Gay Lussac's Law, Avagadro's Law

1 mole of C...

Question

$1$ mole of $C_{2} H_{x}$ is oxidised to $C O_{2}$ and $H_{2} O$ by $3$ moles $O_{2}$ . Thus, $140$ g $C_{2} H_{x}$ has ______ moles.

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Solution

The balanced chemical equation for the combustion reaction is as given below:
$C_{2} H x + (\frac{8 + x}{4}) O_{2} \to 2 C O_{2} + \frac{x}{2} H_{2} O$ .
But $1$ mole of $C_{2} H_{x}$ requires $3$ moles of oxygen for complete combustion.
Hence, $3 = \frac{8 + x}{4}$
or, $x = 4$ .
The molecular weight of $C_{2} H_{4}$ is $28$ g/mol.
Hence, $140$ g of $C_{2} H_{4}$ corresponds to $\frac{140}{28} = 5$ moles.

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