Calculate the actual yield for the production of ammonia gas (nh3) from hydrogen and nitrogen gases if the percent yield is 68.2% and you begin with 2.00 kg of nitrogen gas

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

And a) oxygen i believe