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Question

60 g of lime stone on heating produced 22 g of CO2. The percentage of CaCO3 in limestone is:

A
80%
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B
60%
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C
83.3%
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D
87.66%
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Solution

The correct option is B 83.3%
The balanced chemical reaction is CaCO3CaO+CO2.
22 g of CO2 corresponds to 0.5 moles.
This will be produced by 0.5 moles of calcium carbonate i.e. 50 g of CaCO3.
The percentage purity of limestone is 5060×100=83.3%.

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