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)

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.

1.61g H2 is allowed to react with 9.91g N2, producing 1.64g NH3.

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

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

Explanation / Answer

1. Before we can begin part a, we need to find which reactant is the limiting reagent.

85 g H2 * 1 mol/2.00 g = 42.5 mol H2 [have]

9.64 g N2 * 1 mol/28.0 g = 0.344 mol N2 [have]

42.5 mol H2 * 1 mol N2/3 mol H2 = 14.2 mol N2 [needed - we don't have]

0.344 mol N2 * 3 mol H2/1 mol N2 = 1.03 mol H2 [needed - we have]

From this, we can see that N2 is the limiting reagent and that H2 is in excess.

Now, moving on to part a.

a. 0.344 mol N2 * 2 mol NH3/1 mol N2 = 0.688 mol NH3 [theoretical]

0.688 mol NH3 * 17.0 g/1 mol = 11.7 g NH3 [theoretical yield]

b. percent yield = actual/theoretical x 100% = 1.31 g/11.7 g x 100% = 11.2%

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