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20 gms of sulphur on burning in air produced 11.2 its of ${SO}_{2}$ at STP. The percentage of unreacted sulphur

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answer is B.

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Detailed Solution

$S + O_{2} \to {SO}_{2}$

1mol sulphate = 1 mole ${SO}_{2} = 22.4 Lt {SO}_{2}$

$11.2 Lt {SO}_{2} = \frac{1}{2} mole sulphure = 16 grams sulphur$

$% unreacted = \frac{4}{20} \times 100 = 20 %$

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