Question 1
A town's waste water following primary treatment contains 450mg per litre of organic matter (CH2O). It is treated in a secondary treatment plant that can process 200,000 litres per day. During its passage through the plant, 50% of the organic matter is oxidised:CH2O + O2 ----------> CO2 + H2O
If the percentage of oxygen in the air is 20% by volume, what volume of air is required to oxidise the organic matter in the waste water?
Answer
If you answered 180 litres x 103, you were correct: and this is how to work it out:
450mg x 200,000 litres = 90kg organic matter;
Mr(relative molecular mass) CH2O = 30g per mol, so 90kg = 90,000/30 = 3000 mol;
50% oxidised = 1500 mol, so (acc to eqn) 1500 mol oxygen needed;
If 20% air is oxygen, mass of air = 1500 x 100/20 = 7500 mol;
Volume of air with this mass = 7500 x 24 litres = 180,000 litres (or 180 litres x 103)
