What volume and what mass of ammonia is required to obtain 5 tons of ammonium nitrate?

Find the chemical amount of ammonium nitrate. For this purpose, we divide its weight by its molar mass.
M NH4NO3 = 14 + 4 + 14 + 16 x 3 = 80 grams / mol;
N NH4NO3 = 5,000,000 / 80 = 62,500 mol;
This amount of ammonia will be required. Let’s define its mass and volume.
Let’s calculate its volume.
To do this, multiply the amount of substance by the standard volume of 1 mole of gas (which is 22.40 liters)
V NH3 = 62,500 x 22.4 = 1,400,000 liters = 1,400 m3;
Let’s calculate its weight.
M NH3 = 14 + 3 = 17 grams / mol;
m NH3 = 62,500 x 17 = 1,062,500 grams = 1.062 tons;