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Calculate the weight of `CaO` required to remove hardness of `10^(6) L` of water containing `1.62 g` of `Ca (HCO_(3))_(2)` in `1.0 L`. `(Mw of Ca (HCO_(3))_(2) = 162, mw of CaO = 56)` |
Answer» The reaction is: `underset(1 mol)(CaO) + underset(1 mol)(Ca (HCO_(3))_(2) rarr underset(2 mol)(2CaCO_(3)) + H_(2) O` Moles of `Ca(HCO_(3))_(2)` in `1.0 L` of sample `= (1.62)/(162) = 0.01 mol` Moles of `CaO` required in `1.0 L` of sample `= 0.1 mol L^(-1)` Mole of `CaO` requried in `10^(6) L` of sample `= 0.01 xx 10^(6) mol//10^(6) L = 10^(4) mol (10^(6) L)^(-1)` Weight of `CaO = 10^(4) xx 56 = 5.6 xx 10^(5) g` |
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