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Calculate the quantity of lime required ...

Calculate the quantity of lime required to soften `10^(3)L` of `H_2O` which contains `7.5 g L^(-1)` of `Ca(HCO_3)_2` and `5.0 g L^(-1) of Mg(HCO_3)_2`

Text Solution

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(a). `Ca(HCO_(3))_(3)+Ca(OH)_(2)to2CaCO_(3)+2H_(2)O`
(b). `Mg(HCO_(3))_(2)+Ca(OH)_(2)toCaCO_(3)+cancel(MgCO_(3))+2H_(2)O`
`underline(cancel(MgCO_(3))+Ca(OH)_(2)toMg(OH)_(2)+CaCO_(3))`
`underline(Mg(HCO_(3))_(2)+2Ca(OH)_(2)to2CaCO_(3)+Mg(OH)_(2)+2H_(2)O)`
`[mW of Ca(HCO_(3))_(3),Mg(HCO_(3))_(2), and Ca(OH)_(2)`,
Respectively, are : 162,146 and 74 `mol^(-1)]`
From reaction (a),
Weight of `Ca(OH)_(2)=(74xx7.5)/(162)=3.42g`
From reaction (b),
Weight of `Ca(OH)_(2)=(2xx74xx5.0)/(146)=5.06g`
Total weight of `Ca(OH)_(2)` in `1 L" of "H_(2)O=3.42+5.06=8.48L^(-1)`
Weight of `Ca(OH)_(2)` in `10^(3)L=8.48xx10^(3)g`
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