Convert 74.2 g N₂ to moles. (N₂ molar mass ≈ 28.014 g/mol)
(74.2 g) (1/28.014 mol/g) ≈ 2.65 mol N₂
According to the balanced reaction, 1 mol N₂ and 3 mol H₂ get consumed together to produce 2 mol NH₃, which means we consume up to
3 (2.65 mol N₂) ≈ 7.95 mol H₂
That is, all the N₂ is consumed and we have an excess of about 6.05 mol H₂. We end up with
2 (2.65 mol N₂) ≈ 5.30 mol NH₃
or (with NH₃ molar mass ≈ 17.031 g/mol)
(5.30 mol) (17.031 g/mol) ≈ 90.3 g NH₃
