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± Percent Yield The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process,...

± 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.

1.26 g H2 is allowed to react with 9.69 g N2, producing 1.93 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.

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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.

Solutions

Expert Solution

3H2(g)+N2(g)→2NH3(g)

no of moles of H2 = W/G.M.Wt   = 1.26/2 = 0.63moles

no of moles of N2 = W/G.M.Wt    = 9.69/28 = 0.346moles

3 moles of H2 react with 1 mole of N2

0.63 moles of H2 react with = 1*0.63/3   = 0.21 moles of N2

   H2 is limiting reactant

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

0.63 moles of H2 react with N2 to gives = 2*0.63/3   = 0.42 moles of NH3

mass of NH3 = no of moles * gram molar mass

                        = 0.42*17   = 7.14g

Theoritical yield   = 7.14g

Percentage yield = actual yiled*100/theoritical yield

                              = 1.93*100/7.14   = 27% >>>>>answer


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