± 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.

1.34 g H2 is allowed to react with 10.1 g N2, producing 1.25 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.

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± 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.

1.34 g H2 is allowed to react with 10.1 g N2, producing 1.25 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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SubmitMy AnswersGive Up

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.

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Explanation / Answer

N2(g) + 3H2(g) ------> 2NH3(g)
no or moles of N2 = W/G.M.Wt
                     = 10.1/28 = 0.36 moles
   no of moles of H2 = W/G.M.Wt
                     = 1.34/2 = 0.67 moles
1 mole of N2 react with 3 moles of H2
0.36 moles of N2 react with = 3*0.36/1 = 1.08 moles of H2
     H2 is limiting reactant
3 moles of H2 react with N2 to gives 2 moles of NH3
0.67 moles of H2 react with N2 to gives = 2*0.67/3 = 0.45 moles of NH3
mass of NH3 = no of moles *gram molar mass
              = 0.45*17 = 7.65g
theoritical yield of NH3 = 7.65g
percentage yield = actual yield*100/theoritical yield
                     = 1.25*100/7.65   = 16.3% >>>.answer
     

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