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 1.86 gH_2 is allowed to react worth 104 gN_2. producing 1.94 g N_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

1.86 g H2 is allowed to recat with 10.4 g N2 producing 1.94 g NH3

3 H2 + N2 >> 2 NH3

Moles H2 = 1.86 g / 2.016 g/mol = 0.923
Moles N2 = 10.4 g / 28.0134 g/mol = 0.371

Moles H2 needed = 3 x 0.371 = 1.13 moles

we have only 0.923 moles of H2 so it is the limiting reactant

H2 is limiting agent , due to following reasons:

now calculate the moles of NH3 as follows:

0.923 mole H2 * 2 mole NH3/3 mole H2

= 0.615 moleNH3

Mass NH3 = 0.615 x 17.0307 g/mol =10.5 g Theoretical yield

% yield = obsereved yield / Theoretical yield *100

1.94 x100 / 10.5 = 18.5%

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