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Nitric Acid

4 g of an imp...

Question

4 g of an impure sample of $C a C O_{3}$ on treatment with excess HCI produce 0.88 g $C O_{2}$ . What is the per cent purity of $C a C O_{3}$ sample ?

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Solution

$C a c O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$
$44 g C O_{2} \equiv 100 g C a C O_{3}$
$0.88 g C O_{2} \equiv \frac{100 \times 0.88}{44} = 2.0 g C a c O_{3}$
$∴$ Percentage purity $= \frac{2}{4} \times 100 = 50$ %

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