± Percent Yield 3H2(g)+N2(g)2NH3(g) The ammonia produced in the Haber-Bosch proc

Question

± Percent Yield

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.

± 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.

SubmitHintsMy AnswersGive UpReview Part

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.

Explanation / Answer

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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