Correct Answer - 4440 g
Step I. Mass of calcium bicarbonate in well water
100 L of well water contain calcium bicarbonate = 16.2 g
60000 L of well water contain calcium bicarbonate `= (16.2)/(100) xx 60000 = 9720 g`
Step II. Quantity of slaked lime of `Ca(OH)_(2)` required
`underset(40+2+24+96+(=162g))(Ca(HCO_(3))_(2))+underset(40+34(=74g))(Ca(OH)_(2))rarr2CaCO_(3)+2H_(2)O`
162 g of `Ca(HCO_(3))_(2)` require slaked lime = 74 g
9720 g of `Ca(HCO_(3))_(2)` require slaked lime `= (74xx9720)/(162)g=4440g`.