Ammonia is used as the feedstock to fertilizers production. In the Haber-Bosch p

Question

Ammonia is used as the feedstock to fertilizers production. In the Haber-Bosch process, the ammonia (NH3) is produced by reacting nitrogen (N2) with hydrogen (H2) via metal catalyst under high temperatures and pressures. The product stream is analyzed and found to contain 70.5 mole% NH3, 20.5 mole% N2 and 9.0 mole% H2. The feed to the reactor contains only N2 and H2. By performing the element balance for the process, calculate the fractional conversion on the limiting reactant and the theoretical percentage by which the other reactant is in excess.

Explanation / Answer

The reaction is as follows;

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

Finally we have 70.5 mol% NH3 , 20.5% N2 and 9.0 mol% H2

Nitrogen Balance:

Final Nitrogen moles = 70.5 moles + 2 x 20.5 moles = 111.5 moles Nitrogen element

So , initial Moles of Nitrogen element = 111.5 moles

Initial moles of N2 gas = 0.5 x 111.5 = 55.75 moles

Final Hydrogen moles = 3 x 70.5 moles + 2 x 9 moles = 229.5 moles

Initial moles of Hydrogen element =229.5 moles

Initial moles of Hydrogen gas = 114.75 moles

From equation 1 mol of N2 reacts with 3 moles of H2.

Here we have 55.75 moles of N2, so we required 55.75 x 3 = 167.25 moles of H2

but we have only 114.75 moles.

So, the limiting reactant is H2 and excess raecatant is N2

Theoritical moles of Ammonia formed = 2/3 x 114.75 = 76.5 moles

Frcational conversion of Ammonia = actual moles/theoritical moles

= 70.5 / 76.5

= 0.922

Amount of N2 required for Given hydrogen = 1/3 x 114.75 = 38.25 moles

Actual moles present = 55.75 moles

Percentage excess reactant = ((55.75 - 38.25)/55.75 ) x 100 = 31.4%

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