$20$ g of sulphur on burning in air produced $11.2$ litres of $S O_{2}$ at STP. The percentage of unreacted sulphur is:

80 %

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20 %

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60 %

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40 %

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Solution

The correct option is B $20 %$
The balanced chemical reaction is $S + O_{2} \to S O_{2}$ .

At STP, $11.2$ L $S O_{2}$ corresponds to $0.5$ moles.

$0.5$ moles of $S O_{2}$ is produced by $0.5$ moles of $S$ .

Hence, $0.5$ moles of $S$ = 16 g.

The percentage of unreacted sulfur is $\frac{20 - 16}{20} \times 100 = 20 %$ .

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Illustrate the law of definite proportions from the following data:-1)0.32g of sulphur on burning in air produced 224ml of SO2 at N.T.P

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\frac{1}{8} S_{8} (s) + O_{2} (g) \to S O_{2} (g)

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