$200 g$ of a calcium carbonate sample decomposes on heating to give carbon dioxide and $80 g$ of calcium oxide. What will be the percentage purity of calcium carbonate in the sample?
( $G i v e n : C a C O_{3} \to C a O + C O_{2}$ )

58.14 %

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71.43 %

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84.33 %

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28.89 %

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Solution

The correct option is B $71.43 %$
$G i v e n : C a C O_{3} \to C a O + C O_{2}$
1 mole of $C a O$ is produced by the decomposition of 1 mole of $C a C O_{3}$
i.e. 56 g of $C a O$ is produced by 100 g of $C a C O_{3}$
80 g will be produced by $\frac{100}{56} \times 80 = 142.85$ g of $C a C O_{3}$
% purity $= \frac{mass of pure compound in sample}{total mass of impure sample} \times 100 = \frac{142.85}{200} \times 100$
$= 71.43 %$

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200 g of a calcium carbonate sample decomposes on heating to give carbon dioxide and 80 g of calcium oxide. What will be the percentage purity of calcium carbonate in the sample? (Given:CaCO3→CaO+CO2)

$200 g$ of a calcium carbonate sample decomposes on heating to give carbon dioxide and $80 g$ of calcium oxide. What will be the percentage purity of calcium carbonate in the sample?
( $G i v e n : C a C O_{3} \to C a O + C O_{2}$ )