1.87 g H2 is allowed to react with 9.98 g N2 , producing 2.28 g NH3 . What is the theoretical yield in grams for this reaction under the given conditions?

Question

b) What is the percent yield for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

The first step to solve this problem is to write out a balanced chemical equation. With that, we get 3H₂ + N₂ --> 2NH₃. Now, we have to determine if the hydrogen or nitrogen is the limiting reactant in this example. In order to do that, we take the amount of H₂ we have and determine the amount of NH₃ we would expect if we had an abundance of N₂. This would look something like this: (1.87g H₂)*(1 mol/2.02 g)*(2 mol NH₃/3 mol H₂)(17.04 g NH₃/1 mol NH₃). The amount of NH₃ we would expect from 1.87 g of H₂ is 10.5 g. We do the same equation but with N₂ and we would get 12.1 g of expected NH₃. During the chemical reaction, all of the H₂ would be used up when we have 10.5 g of NH₃ which makes H₂ the limiting reactant and 10.5 g NH₃ the theoretical yield. If the theoretical yield was 10.5 g and the experimental yield was 2.28 g, we divide the experimental yield by the theoretical yield to get the percent yield for part B. The percent yield is .217 = 21.7%.