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)

1.40g H2 is allowed to react with 10.3g N2, producing 2.28g NH3.

Part A

What is the theoretical yield for this reaction under the given conditions?

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

Moles of H2= Mass/mol wt=1.40/2=0.70mol

Moles of N2=10.3/28=0.368 mol

A. 1 mol N2 requires 3mol H2

Thus moles of H2 required=0.368*3=1.104 mol

As mol of H2 req > moles of H2 present

H2 is limiting reagent.

Thus theoritical yeild of NH3=(2/3)*moles of H2

=(2/3)*0.70=0.467 mol

Theoritical yeild= 0.467*17

=7.93 g

b. Percent yeild= (Actual yeild/theoritical yeild)*100

=(2.28/7.93)*100

=28.7 %

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