Question

A solid waste of composition \( C_{60}H_{135}O_{50}N_5 \) is to be composted aerobically in a closed vessel mechanical composting facility. Given: all ammonia generated escapes the facility; air contains 23% of Oxygen by weight; 100% excess air requirement for the closed vessel composting facility. The atomic weights: C – 12, H – 1, O – 16, N – 14. The actual air required for composting is:

Hint:

- The oxygen requirement depends on the waste composition. Each element (C, H, O, N) contributes differently to the total oxygen needed for the aerobic process.
- Excess air is often provided to ensure that the composting process is efficient, particularly in closed vessel systems.

Answer 1

Step 1: Calculate Molecular Weight of Waste

  
  C60H135O50N5 = 60 × 12 + 135 × 1 + 50 × 16 + 5 × 14
              = 720 + 135 + 800 + 70
              = 1725 g/mol = 1.725 kg/mol
  
  

Step 2: Write and Balance the Composting Reaction

  
  C60H135O50N5 + a O2 → b CO2 + c H2O + d NH3
  
  

Balancing: Carbon: 60 → b = 60
Hydrogen: 135 = 2c + 3d
Nitrogen: 5 → d = 5
Oxygen: 50 + 2a = 2b + c
Solving:

  
  d = 5
  2c + 3(5) = 135 ⟶ c = 60
  50 + 2a = 2(60) + 60 ⟶ a = 65
  
  

Balanced equation:

  
  C60H135O50N5 + 65 O2 → 60 CO2 + 60 H2O + 5 NH3
  
  

Step 3: Calculate Theoretical Oxygen Requirement

  
  O2 required per mole waste = 65 moles × 32 g/mol = 2080 g = 2.08 kg
  O2 per kg waste = 2.08 kg / 1.725 kg ≈ 1.206 kg O2 / kg waste
  
  

Step 4: Convert Oxygen to Air Requirement

  
  Air = O2 required / % O2 in air = 1.206 / 0.23 ≈ 5.243 kg air / kg waste
  
  

Step 5: Apply 100% Excess Air

  
  Actual air = Theoretical air × 2 = 5.243 × 2 ≈ 10.486 kg air / kg waste
  
  

Final Answer

  
  10.5 kg air per kg waste