F4 calculation

1. NH3 + 2O2 → HNO3 + H2O
2. HNO3 + NH3 → NH4NO3

According to the equations, we should use 8.5 tonnes of ammonia in state one and the other 8.5 tonnes in state 2. (By mole ratio, NH3 : HNO3 = 1:1)

Molar mass of ammonia, NH3
= 14.0 + 1.0 X 3
= 17.0 g/mol

No. of mole of NH3
= mass / molar mass
= 1.7 X 10^7 / 17.0
= 1.0 X 10^6 mol

From equation 2, mole ratio of NH3 : NH4NO3 = 1:1
So, no. of mole of NH4NO3 = 1.0 X 10^6 mol

Molar mass of NH4NO3
= 14.0 X 2 + 1.0 X 4 + 16.0 X 3
= 80.0 g/mol

By mass = molar mass X no. of mole
Mass of NH4NO3 = 80.0 X 1.0 X 10^6
= 8.0 X 10^7 g
= 80 tonnes

So, the maximum mass of fertilizer that can be made is 80 tonnes.

(P.S. relative atomic mass of H = 1.0 , N = 14.0 , O = 16.0)

Overall reaction:
2 NH3 + 2 O2 --> NH4NO3 + H2O
i.e. 2 mole of NH3 (mw = 17) yield 1 mole of NH4NO3 (mw = 80)
Thus, assuming 100 % conversion,
max. mass of NH4NO3 = 17 tonnes / 17 / 2 * 80 = 40 tonnes