The volume of $C O_{2}$ , liberated at STP when 10 g of 90% pure lime is heated completely is:

2.016 l

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20.16 l

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2.24 l

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22.4 l

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Solution

The correct option is D $2.016 l$
On heating;
$C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)$

100g of $C a C O_{3}$ liberates = 22.4 l of $C O_{2}$
10g of 90%pure $C a C O_{3}$ = $10 \times \frac{90}{100}$ = 9g
$∴$ 9g of $C a C O_{3}$ liberates = $\frac{22.4}{100} \times 9$ = $2.016 l$

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