Calculate the mass and volume of ammonia required to obtain 224 g of ammonium nitrate

| May 9, 2021 | education

Calculate the mass and volume of ammonia required to obtain 224 g of ammonium nitrate if the yield of the reaction product is 85% of the theoretically possible.

1. Let’s compose the reaction equation:

NH3 + HNO3 = NH4NO3;

2.Calculate the practical chemical amount of ammonium nitrate:

ntrack (NH4NO3) = mtrack (NH4NO3): M (NH4NO3);

M (NH4NO3) = 14 + 4 + 14 + 3 * 16 = 80 g / mol;

ntract (NH4NO3) = 224: 80 = 2.8 mol;

3.Calculate the theoretical amount of nitrate:

ntheor (NH4NO3) = npract (NH4NO3): ν = 2.8: 0.85 = 3.29 mol;

4.determine the amount and mass of ammonia:

n (NH3) = ntheor (NH4NO3) = 3.29 mol;

m (NH3) = n (NH3) * M (NH3) = 3.29 * 17 = 55.93 g;

5. find the volume of ammonia:

V (NH3) = n (NH3) * Vm = 3.29 * 22.4 = 73.7 liters.

Answer: 55.93 g, 73.7 liters.



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