Question

Urea (H2NCONH2) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide:

2 NH3(g) + CO2(g) -->H2NCONH2(s) + H2O(g)


Ammonia gas at 223°C and 90. atm flows into a reactor at a rate of 460. L/min. Carbon dioxide at 223°C and 46 atm flows into the reactor at a rate of 600. L/min. What mass of urea is produced per minute by this reaction assuming 100% yield?

____________g/min.

Answer 1

30540 g/min

Step-by-step explanation:

First, let's calculate the number of moles of each reactant from the ideal gas equation:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (R = 0.082 atm*L/mol*K), and T is the temperature. Assuming 1 min:

NH₃: P = 90 atm, V = 460 L, T = 223ºC + 273 = 496 K

90*460 = n*0.082*496

40.672n = 41400

n = 1,018 mol

CO₂: P = 46 atm, V = 600 L, T = 223ºC + 273 = 496 K

46*600 = n*0.082*496

40.672n = 27600

n = 678.6 mol

For the reaction given, the stoichiometry is 2 mol of NH₃ to 1 mol of CO₂. So let's find which one is in excess, testing from CO₂:

2 mol of NH₃ --------------- 1 mol of CO₂

1,018 mol of NH₃ ---------- x

By a simple direct three rule

2x = 1,018

x = 509 mol of CO₂

So, there are more moles of CO₂ than it's necessary, so it's in excess, and NH₃ is the limiting reactant. The stoichiometry must be done with it, and it is:

2 mol of NH₃ ----------------- 1 mol of H₂NCONH₂

1,018 mol of NH₃ ------------ x

x = 509 mol of H₂NCONH₂

Knowing the molar masses (H = 1g/mol, N= 14 g/mol, C = 12 g/mol, O = 16 g/mol), the molar mass of urea is

2x1 + 14 + 12 + 16 + 14 + 2x1 = 60 g/mol

The mass is the number of moles multiplied by the molar mass:

m = 509x60 = 30540 g

As we did it for 1 min

mass of urea = 30540 g/min