Molar mass of calcium hydroxide = 74 g/moleMolar mass of magnesium sulfate = 120 g/moleMolar mass of calcium carbonate = 100 g/moleGiven hardness of magnesium sulfate = 60 ppmThe BALANCED chemical reaction will be,MgSO_4+Ca(OH)_2\rightarrow CaSO_4+Mg(OH)_2MgSO 4 +Ca(OH) 2 →CaSO 4 +Mg(OH) 2 First we have to calculate the hardness of magnesium sulfate in TERMS of calcium carbonate.\text{Hardness of }MgSO_4=\frac{\text{Given hardness of }MgSO_4}{\text{Molar mass of }MgSO_4}\times \text{Molar mass of }CaCO_3Hardness of MgSO 4 = Molar mass of MgSO 4 Given hardness of MgSO 4 ×Molar mass of CaCO 3 Now put all the given values in this expression, we get the hardness of magnesium sulfate in terms of calcium carbonate.\text{Hardness of }MgSO_4=\frac{60ppm}{120g/mole}\times 100g/mole=50ppmHardness of MgSO 4 = 120g/mole60ppm ×100g/mole=50ppmNow we have to calculate the amount of LIME required.\text{Amount of lime}=\frac{74}{100}\times 50mg/L\times 5000L=185000mg=185gAmount of lime= 10074 ×50mg/L×5000L=185000mg=185gTherefore, the amount of lime required is, 185 GRAMS