Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$ 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \underset{\text { Pressure }}{\text { Heat }}{\mathrm{H}}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ Ammonia gas at \(223^{\circ} \mathrm{C}\) and \(90 .\) atm flows into a reactor at a rate of \(500 . \mathrm{L} / \mathrm{min}\). Carbon dioxide at \(223^{\circ} \mathrm{C}\) and 45 atm flows into the reactor at a rate of \(600 . \mathrm{L} / \mathrm{min} .\) What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Step by step solution

01

Calculate moles of ammonia and carbon dioxide entering the reactor

First, we need to find the number of moles for both ammonia (NH₃) and carbon dioxide (CO₂) entering the reactor every minute. We can do that by using the Ideal Gas Law: \(PV = nRT\) Where: P = pressure (atm) V = volume (L) n = moles R = universal gas constant (0.0821 L atm / mol K) T = temperature (K) For ammonia: P = 90 atm V = 500 L/min T = 223 °C + 273.15 = 496.15 K For carbon dioxide: P = 45 atm V = 600 L/min T = 223 °C + 273.15 = 496.15 K Now let's calculate moles for both gases.

02

Calculate moles of NH₃ and CO₂

Solving the Ideal Gas Law for moles (n): \(n = PV / RT\) For ammonia (NH₃): \(n_{NH_3} = \frac{90 \,\text{atm} × 500 \,\text{L/min}}{0.0821 \frac{\text{L atm}}{\text{mol K}} × 496.15 \,\text{K}}\) \(n_{NH_3} = 553.25 \,\text{mol/min}\) For carbon dioxide (CO₂): \(n_{CO_2} = \frac{45 \,\text{atm} × 600 \,\text{L/min}}{0.0821 \frac{\text{L atm}}{\text{mol K}} × 496.15 \,\text{K}}\) \(n_{CO_2} = 661.19 \,\text{mol/min}\)

03

Determine the limiting reactant

Next, we need to determine the limiting reactant, which is the reactant that runs out first and limits the production of urea, based on the stoichiometry and the number of moles calculated in the previous step. From the balanced chemical equation: \(2 \,\text{mol}\, NH_3 + 1\, \text{mol}\, CO_2 \rightarrow 1\, \text{mol}\, H_2NCONH_2 + 1\, \text{mol}\, H_2O\) Calculate moles ratio of NH₃ to CO₂ Moles ratio = moles of NH₃ / moles of CO₂ Moles ratio = \(553.25\, \text{mol/min} / 661.19\, \text{mol/min}\) Moles ratio = 0.83 Since the moles ratio of 0.83 is less than the stoichiometric ratio of 2:1 (NH₃ to CO₂), NH₃ (ammonia) is the limiting reactant.

04

Calculate moles of urea produced

Using stoichiometry, we can now find the moles of urea produced: For every 2 moles of NH₃, 1 mol of urea (H₂NCONH₂) is produced. Moles of urea = \(553.25\,\text{mol/min NH}_3 × \frac{1\,\text{mol H}_2\text{NCONH}_2}{2\,\text{mol NH}_3}\) Moles of urea = 276.63 mol/min

05

Calculate the mass of urea produced per minute

Now, let's find the mass of urea produced per minute using the molar mass of urea, which is: Molar mass of urea = 2(14.01) + 4(1.01) + 1(12.01) + 1(16.00) = 60.06 g/mol Using the moles of urea and its molar mass, calculate the mass of urea produced per minute: Mass of urea = moles of urea × molar mass of urea Mass of urea = \(276.63\, \text{mol/min} × 60.06 \,\text{g/mol}\) Mass of urea = 16609.68 g/min So, the mass of urea produced per minute is approximately 16609.68 grams, assuming 100% yield.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!