An impure sample of calcium carbonate contains 80%pure calcium carbonate. 25 g of the impure sample reacted with excess of hydrochloric acid. Calculate the volume of carbon dioxide at NTP obtained from this sample.

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Solution

$100 g$ of impure calcium carbonate contain :

$= 80 g$ pure calcium carbonate

$25 g$ of impure calcium carbonate sample contain $= \frac{80}{100} \times 25$

$= 20 g$ pure calcium carbonate

The desired equation is:

$C a C O_{3} 1 m o l 100 g + 2 H C l \to C a C l_{2} + C O_{2} 22.4 l i t r e a t N T P + H_{2} O$

$100 g$ pure $C a C O_{3}$ liberate = 22.4 litre $C O_{2}$

$20 g$ pure $C a C O_{3}$ liberate $= \frac{22.4}{100} \times 20$

$= 4.48 l i t r e C O_{2}$

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