Percentage yield = (actual yield / theoretical yield) x 100%
Hence,
actual yield = percentage yield x theoretical yield / 100
The balanced reaction equation for the production of ammonia is
N₂(g) + 3H₂(g) → 2NH₃(g)
The mass of N₂(g) used = 2.00 kg = 2 x 10³ g
Molar mass of N₂(g) = 28 g mol⁻¹
moles (mol) = mass (g) / molar mass (g/mol)
Hence, moles of N₂(g) = 2 x 10³ g / 28 g mol⁻¹
= 71.428 mol
Stoichiometric ratio between N₂(g) and NH₃(g) is 1 : 2
Hence, produced NH₃(g) moles = 71.428 mol x 2
= 142.856 mol
Molar mass of NH₃(g) = 17 g mol⁻¹
mass of NH₃(g) = 142.856 mol x 17 g mol⁻¹ = 2428.552 g
Hence, the theoretical yield = 2428.552 g
Then the actual yield of NH₃(g) produced = (68.2 x 2428.552 g) / 100
= 1656.27 g
