Question

The volume in litres of $C{O}_2$ liberated at STP, when $10$ g of $90\%$ pure limestone is heated completely, is:

• A. $2.016$
• B. $20.16$
• C. $2.24$
• D. $22.4$

Answer 1

The correct option is A $2.016$

$CaC{O}_3\rightleftharpoons CaO+C{O}_2$

One mole of calcium carbonate on decomposition gives one mole of carbon dioxide.

The molecular weight of calcium carbonate is $100$ g/mol.
$10$ g calcium carbonate corresponds to $0.1$ mol. It on decomposition will give $0.1$ mol of carbon dioxide. But limestone is only $90\%$ pure. Hence, $0.09$ mol of carbon dioxide will be obtained.

$1$ mole of carbon dioxide occupies $22.4$ L at STP.
Hence, $0.09$ mol of carbon dioxide will occupy $0.09\times 22.4=2.016$ L.