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Minerals, Ores and their Formation

Calcium carbo...

Question

Calcium carbonate decomposes completely, on heating, into lime $(C a O)$ and carbon dioxide $(C O_{2})$ . 1 kg of calcium carbonate is completely decomposed by heat, when 560 g of lime are obtained. How much quantity of carbon dioxide in grams, moles and litres at NTP is produced in the process ?

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Solution

According to law of conservation of mass, Mass of lime + Mass of carbon dioxide = Mass of calcium carbonate
560 g + Mass of $C O_{2}$ = 1000 g
Mass of $C O_{2}$ =1000 - 560 = 440 g
Molecular mass of $C O_{2}$ = 12 + 32 = 44 g (1 mole)
No. of moles in 440 g of $C O_{2} = \frac{440}{44} = 10$
1 mole of $C O_{2}$ occupies volume at NTP = 22.4 litre
10 moles of $C O_{2}$ will occupy volume at NTP $= 22.4 \times 10 = 224$ litre

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