Question

The amount of lime required to remove 60 PPM magnesium sulphide hardness of 5000 litres of water is

Answer 1

Answer:

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)_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/mole

60ppm

×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=185gAmount of lime=

100

74

×50mg/L×5000L=185000mg=185g

Therefore, the amount of lime required is, 185 grams