Question

$60$ g of lime stone on heating produced $22$ g of $C{O}_2$. The percentage of $CaC{O}_3$ in limestone is:

• A. $80\%$
• B. $60\%$
• C. $83.3\%$
• D. $87.66\%$

Answer 1

Answer: (C)

$83.3\%$

The balanced chemical reaction is $CaC{O}_3\to CaO+C{O}_2$.
$22$ g of $C{O}_2$ corresponds to $0.5$ moles.
This will be produced by 0.5 moles of calcium carbonate i.e. $50$ g of $CaC{O}_3$.
The percentage purity of limestone is $\frac{50}{60}\times 100=83.3\%$.