Question

± Percent Yield The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation.

1.34 g H2 is allowed to react with 10.1 g N2, producing 1.25 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. Hints SubmitMy AnswersGive Up Part 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. Hints SubmitMy AnswersGive Up

Answer 1

N2(g) + 3H2(g) ------> 2NH3(g)
no or moles of N2 = W/G.M.Wt
                     = 10.1/28 = 0.36 moles
   no of moles of H2 = W/G.M.Wt
                     = 1.34/2 = 0.67 moles
1 mole of N2 react with 3 moles of H2
0.36 moles of N2 react with = 3*0.36/1 = 1.08 moles of H2
     H2 is limiting reactant
3 moles of H2 react with N2 to gives 2 moles of NH3
0.67 moles of H2 react with N2 to gives = 2*0.67/3 = 0.45 moles of NH3
mass of NH3 = no of moles *gram molar mass
              = 0.45*17 = 7.65g
theoritical yield of NH3 = 7.65g
percentage yield = actual yield*100/theoritical yield
                     = 1.25*100/7.65   = 16.3% >>>.answer