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 3H_2 (g) + N_2 (g) rightarrow 2NH_3 (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. 1.29 g H_2 is allowed to react with 9.56 g N_2, producing 1.88 g NH_3.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.

Explanation / Answer

number of moles of H2 = 1.29 g / 2.01 g.mol^-1 = 0.642 mole

number of moles of N2 = 9.56g / 28.013 g.mol^-1 = 0.341 mole

from the balanced equation we can say that

3 mole of H2 requires 1 mole of N2 so

0.642 mole of H2 will require

= 0.642 mole of H2*(1 mole of N2/3 mole of H2)

= 0.214 mole of N2

but we have 0.341 mole of N2 which is in excess so H2 is limiting reactant

from the balanced equation we can say that

3 mole of H2 produces 2 mole of NH3 so

0.642 mole of H2 will produce

= 0.642 mole of H2*(2 mole of NH3 /3 mole of H2)

= 0.428 mole of NH3

1 mole of NH3 = 17.031 g

so 0.428 mole of NH3 = 17.031*0.428 = 7.29 g

Therefore, theoretical yield of NH3 = 7.29 g

percent yield = (actual yield/theoretical yield)*100

percent yield = (1.88/7.29)*100 = 25.8

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