Calculate the volume and mass of ammonia required to obtain 5 tons of ammonium nitrate.

Given:
m (NH4NO3) = 5 t = 5,000,000 g

To find:
V (NH3) -?
m (NH3) -?

Solution:
1) NH3 + HNO3 => NH4NO3;
2) Mr (NH4NO3) = Ar (N) * 2 + Ar (H) * 4 + Ar (O) * 3 = 14 * 2 + 1 * 4 + 16 * 3 = 80 g / mol;
Mr (NH3) = Ar (N) + Ar (H) * 3 = 14 + 1 * 3 = 17 g / mol;
3) n (NH4NO3) = m (NH4NO3) / Mr (NH4NO3) = 5,000,000 / 80 = 62500 mol;
4) n (NH3) = n (NH4NO3) = 62500 mol;
5) V (NH3) = n (NH3) * Vm = 62500 * 22.4 = 1400000 l;
6) m (NH3) = n (NH3) * Mr (NH3) = 62500 * 17 = 1062500 g.

Answer: The volume of NH3 is 1,400,000 liters; weight – 1062500 g.