Theoretical yield and percent yield of 1.92 g H2 is allowed to react with 9.73 g N2, producing 2.04 g NH3

Answer 1

First, write a correctly balanced equation for the reaction taking place:

N₂ + 3H₂ ==> 2NH₃

In order to find the percent yield, we must first find the theoretical yield, and to do that, we must first find the limiting reactant:

H₂: 1.92 g H₂ x 1 mol H₂/2.0 g H₂ = 0.96 mol H₂ x 2 mol NH₃/3 mol H₂ = 0.64 moles NH₃

N₂: 9.73 g N₂ x 1 mol N₂/28.0 g N₂ = 0.335 mol N₂ x 2 mol NH₃/1 mol N₂ = 0.67 moles NH₃

Thus, H₂ is limiting.

Theoretical yield:

Using the limiting reactant, we see that only 0.64 moles of NH₃ can be produced. Converting this to grams, ..

0.64 moles NH₃ x 17.0 g/mol = 10.88 g NH₃ = Theoretical yield

Percent yield:

Actual yield / theoretical yield (x100) = percent yield

2.04 g / 10.88 g (x100) = 18.8 percent