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Q.

2 mol of 50% pure Ca(HCO3)2 on heating forms 1 mol of CO2. The % yield of CO2 is

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a

50%

b

100%

c

75%

d

90%

answer is B.

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Detailed Solution

2CaHCO32⟶CaCO3+CO2+H2O2mol                                   1molThe percent yield of CO2 is 100% whether Ca(HCO3)2, is 100% pure or 50% pure.
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