Question

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.

1.10 g H2 is allowed to react with 9.72 g N2, producing 1.68 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.

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.

Answer 1

moles of H2 burnt = mass/molar mass
                                      = 1.10/1
                                       = 1.10 mol

moles of N2 burnt = mass/molar mass
                                      = 9.72/28
                                       = 0.347 mol

Since 1 mol of N2 reacts with 3 mol of H2
0.347 mol of N2 will require = 0.347*3 = 1.041 mol of H2
we have more H2 than this

so N2 is limiting reagent
mol of NH3 formed = (2)*moles of N2
                                         = 2*0.347
                                         =0.694 mol
mass of NH3 = moles of NH3*molar mass
                           = 0.694*17
                            = 11.8 g
Answer: 11.8 g

B)
percent yield = actual yield *100/ theoretical yield
                             = 11.8*100/12.5
                              = 94.4 %
Answer: 94.4 %