N2 + 3 H2 → 2 NH3 What mass of ammonia should he be able to produce if he reacts 70.0 g of H2 with excess N2? f he recovers 250.0 g of ammonia, what is his percent yield?

Answer 1

N₂ + 3H₂ → 2NH₃ ... balanced equation

Theoretical yield of NH_3:

70.0 g H₂ x 1 mol H₂/2 g x 2 mol NH₃/3 mol H₂ x 17 g NH₃/mol NH₃ = 397 g NH₃

Percent yield = actual yield/theoretical yield (x100%):

250 g/397 g (x100%) = 63.0% yield