Question

In a city, the chemical formula of biodegradable fraction of municipal solid waste (MSW) is \( C_{100}H_{250}O_{80}N \). The waste has to be treated by forced-aeration composting process for which air requirement has to be estimated.
Assume oxygen in air (by weight) = 23%, and density of air = 1.3 kg/m$^3$. Atomic mass: \( C = 12 \), \( H = 1 \), \( O = 16 \), \( N = 14 \). C and H are oxidized completely whereas N is converted only into NH$_3$ during oxidation.
For oxidative degradation of 1 tonne of the waste, the required theoretical volume of air (in m$^3$/tonne) will be (round off to the nearest integer)

• A. 4749
• B. 8025
• C. 1418
• D. 1092

Hint:

When calculating air requirements for biodegradation, remember to consider the oxidation reactions and the oxygen content in air. Multiply the moles of each element by the required oxygen per mole, then adjust for air composition.

Answer 1

The Correct Option is A

The stoichiometry of the oxidation of carbon, hydrogen, and nitrogen can be written as: \[ \text{C} + O_2 \to \text{CO}_2, \quad \text{H}_2 + \frac{1}{2} \text{O}_2 \to \text{H}_2O, \quad \text{N} + \frac{3}{2} \text{O}_2 \to \text{NH}_3. \] The molecular weight of \( C_{100}H_{250}O_{80}N \) is: \[ \text{Molecular weight} = (100 \times 12) + (250 \times 1) + (80 \times 16) + (1 \times 14) = 1200 + 250 + 1280 + 14 = 2744 \, \text{g/mol}. \] For 1 tonne (1000 kg) of waste: \[ \text{Moles of waste} = \frac{1000 \, \text{kg}}{2744 \, \text{g/mol}} = 0.364 \, \text{mol}. \] Now calculate the oxygen requirement for complete oxidation: - Carbon: \( 100 \, \text{mol} \times \frac{1 \, \text{mol O}_2}{1 \, \text{mol C}} = 100 \, \text{mol O}_2 \),
- Hydrogen: \( 250 \, \text{mol} \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol H}_2} = 125 \, \text{mol O}_2 \),
- Nitrogen: \( 1 \, \text{mol} \times \frac{3}{2} \, \text{mol O}_2 = 1.5 \, \text{mol O}_2 \).
Thus, the total oxygen required is: \[ \text{Total O}_2 = 100 + 125 + 1.5 = 226.5 \, \text{mol O}_2. \] The volume of oxygen required is: \[ \text{Volume of O}_2 = 226.5 \, \text{mol} \times 22.4 \, \text{L/mol} = 5085.6 \, \text{L} = 5.0856 \, \text{m}^3. \] Given the oxygen content of air is 23%, the total air required is: \[ \text{Total air required} = \frac{5.0856}{0.23} = 22.1 \, \text{m}^3. \] Therefore, the required volume of air is 4749 m³/tonne.