What will be the volume of CO2 at STP (atm) obtained on heating 10 grams of 90% pure limestone?
A
22.4L
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B
2.02L
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C
2.24L
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D
20.2L
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Solution
The correct option is B2.02L CaCO3→CaO+CO2
Amount of pure CaCO3=10×0.90=9g
1 mol of CaCO3 decomposes to form 1 mol of CO2 = 22.4L of CO2 gas, i.e. 100g of CaCO3 decomposes to form 22.4L of CO2 gas. 9g of CaCO3 produces = 22.4100×9=2.02L of CO2 gas