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

Correct Answer: 10.4

Step 1: Calculate Molecular Weight of Waste

C₆₀H₁₃₅O₅₀N₅

\[ \text{Molecular weight} = (60 \times 12) + (135 \times 1) + (50 \times 16) + (5 \times 14) \]
\[ = 720 + 135 + 800 + 70 = 1725~\text{g/mol} = 1.725~\text{kg/mol} \]

Step 2: Write and balance the composting reaction

General reaction:
C₆₀H₁₃₅O₅₀N₅ + a O₂ → b CO₂ + c H₂O + d NH₃

Balancing atoms:
Carbon: b = 60
Nitrogen: d = 5
Hydrogen: 135 = 2c + 3d → c = 60
Oxygen: 50 + 2a = 2b + c → a = 65

Balanced equation:

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

Step 3: Calculate theoretical oxygen requirement

O₂ required per mole of waste:
65 × 32 = 2080 g = 2.08 kg O₂

O₂ required per kg of waste:
\[ \frac{2.08}{1.725} = 1.206~\text{kg O}_2/\text{kg waste} \]

Step 4: Convert oxygen requirement to air requirement

Assuming oxygen content in air = 23% by mass:
\[ \text{Theoretical air} = \frac{1.206}{0.23} = 5.243~\text{kg air/kg waste} \]

Step 5: Apply 100% excess air

\[ \text{Actual air required} = 2 \times 5.243 = 10.486~\text{kg air/kg waste} \]

Final Answer:
10.5 kg air per kg waste