Temporary hardness is due to hco3 of mg2 and ca2 it is removed by addition of ca(hco3)2+ca oh 2 caco3+h2o mass of cao required to participate 2g caco3 is
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Answer:
Correct option is B)
Ca(HCO
3
)
2
+CaO→2CaCO
3
+H
2
O
Molecular mass of CaO=40+16=56 g
Molecular mass of CaO
3
=40+12(16×3)=100 g
Accoridng to the equation, 2 moles of CaCO
3
i.e 200 g of CaCO
3
are formed from 56 g CaO
2 g of CaCO
3
will be formed from =
200
2×56
=0.56 grams
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