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Question

Temporary hardness is due to bicarbonates of Mg2+andCa2+, it is removed by addition of CaO as follows:
Ca(HCO3)2+CaO2CaCO3+H2O
Mass of CaO required to precipitate 2 g CaCO3 is:

A
2g
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B
0.56g
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C
0.28g
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D
1.12g
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Solution

The correct option is D 0.56g
Ca(HCO3)2+CaO2CaCO3+H2O
Molecular mass of CaO=40+16=56 g
Molecular mass of CaO3=40+12(16×3)=100 g
Accoridng to the equation, 2 moles of CaCO3 i.e 200 g of CaCO3 are formed from 56 g CaO
2 g of CaCO3 will be formed from =2×56200=0.56 grams

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