Combustion of ammonia 4NH3 + 3O2 = 2N2 + 6H2O is accompanied by the release of 1530 kJ of heat

Combustion of ammonia 4NH3 + 3O2 = 2N2 + 6H2O is accompanied by the release of 1530 kJ of heat. How much ammonia was oxidized if the thermal effect was 4590 kJ?

4NH3 + 3O2 = 2N2 + 6H2O (l) + 1530 kJ

Let’s make a proportion and find the amount of ammonia substance:

4 mol NH3 – 1530 kJ;

x mol NH3 – 4590 kJ;

x = 4590 * 4/1530 = 12 mol;

Let’s find the volume of ammonia (at n.a.):

V (NH3) = n (NH3) * VM = 12 * 22.4 = 268.8 liters.

Answer: V (NH3) = 268.8 liters.