Question

A city has 450,000 people. Suppose each person produces around 2 kg of waste daily that goes into landfills. (a) How much methane, in moles, would be produced if all of this waste was cellulose (-CH₂O-) based and could react anaerobically into methane? (b) How many homes could be heated from this is each person used 10⁸ kJ of energy per year if this methane was saved and combusted (heat of combustion for methane is -890 kJ/mol).

Answer 1

(a) Anaerobic production of CH₄ from C₆H₁₂O₆ is given by the following equation,

From question, it is given that one person produces 2 Kg of waste then 450,000 persons will produce 2 x 450,000 Kg = 900,000 Kg of wastes.

One mole of C₆H₁₂O₆ produces four moles of methane. 1 mole of C₆H₁₂O₆ = 180 g of C₆H₁₂O₆ and 1 mole of CH₄ = 16 g of CH₄ , therefore, 180 g of C₆H₁₂O₆ produces 64 g CH₄.

Since, 180 g of C₆H₁₂O₆ = 0.180 Kg of C₆H₁₂O₆ produces 64 g of CH₄ = 0.064 Kg of CH₄

Therefore, 900,000 Kg of C₆H₁₂O₆ will produce = ? Kg of CH₄ , calculated as follows:

Hence, 2,00,00,000 moles of methane will be produced.

(b) Heat of combustion for methane = - 890 kJ/mol. Since, 1 mole of CH₄ produces -890 kJ of heat therefore, 2,00,00,000 moles of CH₄ will produce, 2,00,00,000 x -890 kJ of heat = -17,80,00,00,000 kJ of heat.

Since, one person uses 108 kJ of energy per year therefore, the number of persons that can use -17,80,00,00,000 kJ of heat per year will be,

Hence, 1,48,14,815 homes can be heated.