Question

The amount of lime required to remove 60 ppm MgSo4 hardness of 5000 litres of water​?

Answer 1

Answer : The amount of lime required is, 185 grams

Explanation : Given,

Molar mass of calcium hydroxide = 74 g/mole

Molar mass of magnesium sulfate = 120 g/mole

Molar mass of calcium carbonate = 100 g/mole

Given hardness of magnesium sulfate = 60 ppm

The balanced chemical reaction will be,

$MgSO_4+Ca(OH)_2\rightarrow CaSO_4+Mg(OH)_2$

First we have to calculate the hardness of magnesium sulfate in terms of calcium carbonate.

$\text{Hardness of }MgSO_4=\frac{\text{Given hardness of }MgSO_4}{\text{Molar mass of }MgSO_4}\times \text{Molar mass of }CaCO_3$

Now put all the given values in this expression, we get the hardness of magnesium sulfate in terms of calcium carbonate.

$\text{Hardness of }MgSO_4=\frac{60ppm}{120g/mole}\times 100g/mole=50ppm$

Now we have to calculate the amount of lime required.

$\text{Amount of lime}=\frac{74}{100}\times 50mg/L\times 5000L=185000mg=185g$

Therefore, the amount of lime required is, 185 grams