The Haber-Bosch process is a very important industrial process. In the Haber-Bos

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.

1.95 g H2 is allowed to react with 9.90 g N2, producing 1.84 g NH3.

A) What is the theoretical yield in grams for this reaction under the given conditions?

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

Explanation / Answer

3H2(g)+N2(g)2NH3(g)

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

                               = 1.95/2 = 0.975 moles

no of moles of N2 = W/G.M.Wt

                             = 9.9/28   = 0.354moles

1 mole of N2 react with 3 moles of H2

0.354 moles of N2 react with = 3*0.354/1   = 1.062moles of H2

H2 is limiting reactant

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

0.975 moles of H2 react with N2 to gives = 2*0.975/3   = 0.65 moles of NH3

mass of NH3 = no of moles * gram molar mass

                        = 0.65*17   = 11.05g

Theoritical yield of NH3 = 11.05g

actual yield of NH3              = 1.84g

percentage yiled    = actual yield*100/theoritical yield

                              = 1.84*100/11.05   = 16.65% >>>>answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.