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

Question

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) rightarrow 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. 1.62g H2 is allowed to react with 9.92g N2, producing 2.85g NH3. What is the theoretical yield for this reaction under the given conditions? Express your answer numerically in grams. What is the percent yield for this reaction under the given conditions? Express the percentage numerically.

Explanation / Answer

A)N2+3H2---->2NH3

From the equation it is clear that 1 mole N2 reacts with 3 moles H2 to produce 2 moles NH3

1.62 g H2=1.62/2

=0.81 moles

9.92 g N2=9.92/14

=0.7085 moles

So 0.81 moles H2 react with (0.81)(1/3)=0.27 moles N2 thereby producing (0.27)(2)=0.54 moles NH3

So theoretical yield of NH3=(0.54)(17)

=9.18 g

B)Given actual yield=2.85 g

So % yield of NH3=Actial yield /theoretical yield x100

=(2.85/9.18)(100)

=31.04%

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