Question

How many liters of methanol would be required daily to remove the nitrogen from a 200,000-L/day sewage treatment plant producing an effluent containing 50 mg/L of nitrogen? Assume that the nitrogen has been converted to NO3- in the plant. The density of CH3OH is 0.791 kg/L.

The denitrifying reaction is: 6 NO3- + 5 CH3OH + 6 H+ (Denitrifying bacteria) → 3 N2(g) + 5 CO2 + 13 H2O

Answer 1

Total volume of sewage per day = 200000 L

Concentration of Nitrogen = 50 mg/L

1 L sewagee = 50 mg of N

Actual concentration of N in 200000 L of sewage = 200000 x 50 = 10000000 = 10⁷ mg = 10 x 10⁶ mg = 10 Kg

(note 1 Kg = 10³ g = 10³ x 10³ mg = 10⁶ mg)

The denitrifying reaction is: 6 NO3- + 5 CH3OH + 6 H+ (Denitrifying bacteria) → 3 N2(g) + 5 CO2 + 13 H2O

It's clear that,

6 moles of NO₃^- 5 moles of H3COH

1 mole of NO₃^- 1 mole of N = 14 g.............(Atomic mass of N= 14 g/mole)

Molar mass of Methanol (H3COH) = 32.04 g/mole

Hence mass relation is writen as,

6 x 14 g of N in NO₃^- 5 x 32.04 g of H3COH

84 g of N in NO₃^- 160.20 g of H3COH

i.e. 84 g of N in NO₃^- 160.20 g of H3COH.........(10^-3 factor cancelled out).

Let us calculate mass of methanol corresponds to 10 Kg of N in sewage.

If, 84 Kg of N in NO₃^- 160.20 Kg of H3COH

Then, 10 Kg of N in NO₃^- say 'A' Kg of H3COH

On cross multiplication,

A x 84 = 10 x 160.20

A x 84 = 1602.0

A = 1602.0 / 84

A = 19.071 Kg

Mass of Methanol required is 19.071 Kg.

Now density = Mass / Volume

Volume = Mass / Density

Mass of Methanol = 19.071 Kg

Density of methanol = 0.791 Kg/L

Then,

Volume of methanol = 19.071 / 0.791 = 24.11 L

24.11 L of Methanol will be required to remove N from 200000 L of sewage per day.

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Note : Here I considered N as atomic species and used atomic mass 14 in calculation. If it is N2 diatomic state just 28 g/mole (instead 14) have to be used in calculation.

84 g of N in NO₃^- 160.20 g of H3COH