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Question

What will be the volume of CO2 at STP (atm) obtained on heating 10 grams of 90% pure limestone?

A
22.4 L
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B
2.02 L
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C
2.24 L
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D
20.2 L
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Solution

The correct option is B 2.02 L
CaCO3CaO+CO2

Amount of pure CaCO3=10×0.90=9 g

1 mol of CaCO3 decomposes to form 1 mol of CO2 = 22.4 L of CO2 gas, i.e.
100 g of CaCO3 decomposes to form 22.4 L of CO2 gas.
9 g of CaCO3 produces = 22.4100×9=2.02 L of CO2 gas

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