Temporary hardness is due to bicarbonates of Mg2+ and Ca2+. It is removed by addition of CaO as follows: Ca(HCO3)2+CaO→2CaCO3+H2O Mass of CaO required to precipitate 2gCaCO3
A
2 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 B 0.56 g
mole of CaCO3=2100=0.02
From stoichiometry,
1 mole of CaO is required to precipitate 2 moles of CaCO3 and hence moles of CaO required to precipitate 0.02 mole of CaCO3
= 12×0.02=0.01
so mass of CaO =0.01×56=0.56g