Temporary hardness is due to HCO−3 of Mg2+ and Ca2+. It is removed by addition of CaO. Ca(HCO3)2+CaO⟶2CaCO3+H2O Mass of CaO required to precipitate 2 g CaCO3 is :
A
2.00 g
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B
0.56 g
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C
0.28 g
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D
1.12 g
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Solution
The correct option is D0.56 g Ca(HCO3)2+CaO⟶2CaCO3+H2O According to this, for 2 moles of CaCO3, CaO required is 1 mole. So, for 2 g CaCO3 (0.02 mol), CaO required is 0.01 mol =0.56 g.