Question

. Carbamide (CH4N2O), can be synthesized by the reaction of ammonia (NH3) with carbon

dioxide (CO2):

2 NH3 (aq) + CO2 (aq) CH4N2O (aq) + H2O (l)

An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia

with 105 kg of carbon dioxide. Determine the limiting reactant, theoretical yield of urea and percent yeild for the reaction​

Answer 1

• NH₃ is the limiting reactant.

• Theoretical yield = 120 kg

• % yield = 72.9 %

Step-by-step explanation:

• 2NH₃ (aq) + CO₂ (aq) → CH₄N₂O (aq) + H₂O (l)

First we convert the given masses of reactants to moles, using their respective molar masses:

• 68.2 kg NH₃ ÷ 17 kg/kmol = 4.01 kmol NH₃

• 105 kg CO₂ ÷ 44 kg/kmol = 2.39 kmol CO₂

2.39 kmol of CO₂ would react completely with (2.39 * 2) 4.78 kmol of NH₃. There are not as many NH₃ kmoles so NH₃ is the limiting reactant.

We calculate how much urea would form with a 100% yield, using the moles of limiting reactant:

• 4.01 kmol NH₃ *
  `(1kmolCH_4N_2O)/(2kmolNH_3)` = 2.00 kmol CH₄N₂O

We convert that amount to kg:

• 2.00 kmol CH₄N₂O * 60 kg/kmol = 120 kg CH₄N₂O

Finally we calculate the percent yield:

• 87.5 kg / 120 kg * 100% = 72.9 %