How much air is required to burn 77.5 kg of phosphorus containing 20% impurities?

1. Let’s compose the equation of phosphorus combustion:

4P + 5O2 = 2P2O5;

2. find the mass of phosphorus:

m (P) = (1 – w (impurities) * m (sample) = 0.8 * 77.5 = 62 kg;

3.Calculate the chemical amount of phosphorus:

n (P) = m (P): M (P) = 62: 31 = 2 kmol;

4.determine the amount of oxygen:

n (O2) = n (P) * 5: 4 = 2 * 5: 4 = 2.5 kmol;

5.Calculate the volume of oxygen:

V (O2) = n (O2) * Vm = 2.5 * 22.4 = 56 m ^ 3;

6. air contains 21% oxygen, we find the volume of air that went into the reaction:

V (air) = V (O2): ϕ (O2) = 70: 0.21 = 266.67 m ^ 3.

Answer: 266.67 m ^ 3.