

InterviewSolution
Saved Bookmarks
1. |
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` |
Answer» (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` |
|