Two part question: Part1. What is the theoretical yield for this reaction under the given conditions? 1.56g H2 is allowed to react with 9.90g N2, producing 2.34g NH3. Part 2: What is the percent yield for this reaction under the given conditions?
AnswerPart 1: First find the limiting reagent: 3H2 + N2 —> 2NH3 molH2= 1.56g/2g/mol =0.78 mol molN2= 9.90g/28g/mol= 0.354 mol 3 mol H2 —– 1 mol N2 0.78 mol H2 —- x=0.26 mol N2 —-> we have 0.354 mol N2, but we only need 0.26mol, so, N2 is in excess. Now we find the theoretical yield: 3 mol H2 —- 2 mol NH3 0.78 mol H2 — x= 0.52 mol NH3 —-> mass= 0.52mol x 17g/mol= 8.84g (Theoretical yield) Part 2:
%yield= produced amount/theoretical yield x100 =2.34g/8.84g x100= 26.5%