What maximum amount of ammonia in kilograms can be synthesized from 5.22 kg of h2 and 31.5 kg of n2?

Consider the unbalanced reaction,

H₂ + N₂   ===>   NH₃

Balance it to get

3 H₂ + N₂   ===>   2 NH₃

This tells you that for every 3 moles of H₂ and 1 mole of N₂ among the reactants, 2 moles of NH₃ are produced.

Look up the molar masses of each component element:

• H : 1.008 g/mol     ===>   H₂ : 2.016 g/mol

• N : 14.007 g/mol     ===>   N₂ : 28.014 g/mol

• NH₃ : 17.031 g/mol

Convert the given mass to moles for each reactant:

• H₂ :

(5.22 kg) × (1000 g/kg) × (1/2.016 mol/g) ≈ 2589.29 mol

• N₂ :

(31.5 kg) × (1000 g/kg) × (1/28.014 mol/g) ≈ 1124.44 mol

Now,

(2589.29 mol H₂) : (1124.44 mol N₂) ≈ (2.30 mol H₂) : (1 mol N₂)

so that the amount of N₂ consumed is

(2589.29 mol H₂) × (1 mol N₂) / (3 mol H₂) ≈ 863.095 mol N₂

leaving an excess of about 261.343 mol N₂, and producing

(2589.29 mol H₂) × (2 mol NH₃) / (3 mol H₂) ≈ 1726.19 mol NH₃

Convert this to a mass :

(1726.19 mol) × (17.031 g/mol) × (1/1000 kg/g) ≈ 29.3987 kg

So, up to 29.4 kg of NH₃ can be synthesized.

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