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Laws of Definite Proportions, Multiple Proportions

150 mL of N...

Question

$150 m L$ of $N / 10 H C l$ is required to react completely with $1.0 g$ of a sample of limestone. Calculate the percentage purity of calcium carbonate.

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Solution

$150 m L \frac{N}{10} H C l \equiv 150 m L \frac{N}{10} C a C O_{3}$
$C a C O_{3} M o l . m a s s + 2 H C l 2 g e q . \to C a C l_{2} + H_{2} O + C O_{2}$
Eq. mass of $C a C O_{3} = \frac{40 + 12 + 48}{2} = \frac{100}{2} = 50$
Mass of $C a C O_{3}$ present in $150 m L N / 10$ solution,
$[N \times E \times \frac{V}{1000}] = 50 \times \frac{1}{10} \times \frac{150}{1000} = 0.75 g$
$P u r i t y = \frac{0.75}{1} \times 100 = 75$ %.

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