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.

1.71 g H2 is allowed to react with 10.1 g N2, producing 1.36 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.

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

Express your answer to three significant figures and include the appropriate units.

Answer 1

3H2(g)+N2(g)→2NH3(g)

Number of moles of H2 = 1.71/2 = 0.855 moles

Number of moles of N2 = 10.1/28 = 0.3607 moles

1 mole of N2 requires 3 moles of H2, hence H2 is the limiting reagent in the reaction

Number of moles of NH3 = 2/3 * 0.855 = 0.57 moles

Theoritical Yield of NH3 = number of moles * molar mass of NH3

=> 0.57 moles * 17 gm/mol

=> 9.69 grams

Percent Yield of the reaction = actual yield/theoritical yield * 100

=> 1.36/9.69 * 100

=> 14.035%