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The balanced equation for the synthesis of ammonia is 3 H₂(g) + N₂(g) --> 2 NH₃(g).

Calculate:
a. the mass in grams of NH₃ formed from the reaction of 64.0 g of N₂
b. the mass in grams of N₂ required for form 1.00 kg of NH₃

Solution

From the balanced equation, it is known that:

1 mol N₂ ∝ 2 mol NH₃

Use the periodic table to look of the atomic weights of the elements to calculate the weights of the reactants and products:

1 mol of N₂ = 2(14.0 g) = 28.0 g

1 mol of NH₃ is 14.0 g + 3(1.0 g) = 17.0 g

These relations can be combined to give the conversion factors needed to calculate the mass in grams of NH₃ formed from 64.0 g of N₂:

mass NH₃ = 64.0 g N₂ x 1 mol N₂/28.0 g NH₂ x 2 mol NH₃/1mol NH₃ x 17.0 g NH₃/1 mol NH₃

mass NH₃ = 77.7 g NH₃

To obtain the answer to the second part of the problem, the same conversions are used, in a series of three steps:

(1) grams NH₃ --> moles NH₃ (1 mol NH₃ = 17.0 g NH₃)

(2) moles NH₃ --> moles N₂ (1 mol N₂ ∝ 2 mol NH₃)

(3) moles N₂ --> grams N₂ (1 mol N₂ = 28.0 g N₂)

mass N₂ = 1.00 x 10³ g NH₃ x 1 mol NH₃/17.0 g NH₃ x 1 mol N₂/2 mol NH₃ x 28.0 g N₂/1 mol N₂

mass N₂ = 824 g N₂

Answer

a. mass NH₃ = 77.7 g NH₃
b. mass N₂ = 824 g N₂