80% pure 200 gram of calcium carbonate thermally decomposes to CO2 and H2O find the volume of the gas released at STP
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Explanation:
CaCO
3
⇌CaO+CO
2
One mole of calcium carbonate on decomposition gives one mole of carbon dioxide.
The molecular weight of calcium carbonate is 100 g/mol.
10 g calcium carbonate corresponds to 0.1 mol. It on decomposition will give 0.1 mol of carbon dioxide. But limestone is only 90% pure. Hence, 0.09 mol of carbon dioxide will be obtained.
1 mole of carbon dioxide occupies 22.4 L at STP.
Hence, 0.09 mol of carbon dioxide will occupy 0.09×22.4=2.016 L.
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