if 100.0 g of nitrogen reacts completely with excess hydrogen and 34.0 g NH3 are obtained, what is the percent yield of ammonia?

Question

 Ammonia is synthesized commercially from nitrogen has and hydrogen gas for the production of fertilizers:

 

N2(g) + 3H2(g) -> 2NH3(g)

Answer 1

Hello Andrea

 

Use stoichiometry to convert 100.0 g of N₂ to g of NH₃.  This calculated mass will be the theoretical yield.  34.0 g of NH₃ obtained is called the actual yield.  %-Yield = (actual yield / theoretical yield) x 100%.

 

For the calculation, molar mass of N₂ is 28 g/mol, molar mass of NH₃ is 17 g/mol, and mole ratio of N₂ to NH₃ from the balanced chemical reaction is 1:2.

 

100.0 g N₂ x (1 mol N₂ / 28 g N₂) x (2 mol NH₃ / 1 mol N₂) x (17 g NH₃ / 1 mol NH₃) = 121.4 g NH₃

 

%-Yield = (Actual yield / Theoretical yield) x 100%

 

            = (34.0 g / 121.4 g) x 100%

 

            = 28.0%