Question

Constants | Periodic Table 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.

1.93 g H2 is allowed to react with 10.2 g N2, producing 1.09 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. View Available Hint(s) nothingnothing Submit 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. View Available Hint(s) nothingnothing Submit Return to Assignment

Answer 1

3H2(g)+N2(g)→2NH3(g)

no of moles of H2 = W/G.M.Wt

                              = 1.93/2   = 0.965 moles

no of moles of N2   = W/G.M.Wt

                              = 10.2/28    = 0.364moles

1 mole of N2 react with 3 moles of H2

0.364 moles of N2 react with = 3*0.364/1   = 1.092 moles of H2 is required

H2 is limiting reactant

3H2(g)+N2(g)→2NH3(g)

3 moles of H2 react with excess of N2 to gives 2 moles of NH3

0.965 moles of H2 react with excess of N2 to gives = 2*0.965/3   = 0.643 moles of NH3

mass of NH3 = no of moles * gram molar mass

                      = 0.643*17   = 10.931g

Theoretical yield of NH3 = 10.931g

percent yield   = actual yield *100/theoretical yield

                       = 1.09*100/10.931   = 9.97% >>>>answer