Question

This manufacture of urea is an important process, since urea is an important commodity in fertilizer and plastics production. This equation of the reaction is:

CO2(g) + NH3(g) = CO(NH2)2 + H2O

This reaction rums in 94% yield. If your production manager shouts at you “Motorola’s case manufacturer needs 450 tons of urea” how much CO2 and NH3(in kg) should you purchase?

Answer 1

First to all you need to balance the reaction:

CO₂ + 2NH₃ --------> CO(NH₂)₂ + H₂O

We do know that you need to produce 450 tons of urea, which means that if we take that into grams it would be:

450 tons * 1000 kg/Tons * 1000 g/kg = 4.5x10⁸ g

We also know that the reaction runs in 94% yield, so we can assume that this quantity would be obtained in the 94% run, so, in the 100%:

4.5x10⁸ * (100/94) = 4.79x10⁸ g

Now, let's calculate the moles of urea: (Molecular weight reported is 60.06 g/mol)

moles = 4.79x10⁸ / 60.06 = 7.98x10⁶ moles

Now, 1 mol of CO₂ ------> 1 mol Urea; so the moles of urea are the same moles of CO₂. And 2 mol of NH₃ produces 1 mol of urea, so we need the double in NH₃. Now let's calculate the mass of CO₂ and NH₃ minimun:

m CO₂ = 7.98x10⁶ * (44) = 351,120,000 g or simply 351.12 tons or 351120 kg.

m NH₃ = 7.98x10⁶ * (17) = 135660000 g or 135600 kg or 135.66 tons

Hope this helps