Question

$200g$ of a calcium carbonate sample decomposes on heating to give carbon dioxide and $80g$ of calcium oxide. What will be the percentage purity of calcium carbonate in the sample?
($Given:CaC{O}_3\to CaO+C{O}_2$)

• A. $58.14\%$
• B. $71.43\%$
• C. $84.33\%$
• D. $28.89\%$

Answer 1

The correct option is B $71.43\%$

$Given:CaC{O}_3\to CaO+C{O}_2$
1 mole of $CaO$ is produced by the decomposition of 1 mole of $CaC{O}_3$
i.e. 56 g of $CaO$ is produced by 100 g of $CaC{O}_3$
80 g will be produced by $\frac{100}{56}\times 80=142.85$ g of $CaC{O}_3$
% purity $=\frac{\text{mass of pure compound in sample}}{\text{total mass of impure sample}}\times 100=\frac{142.85}{200}\times 100$
$=71.43\%$