Determine the theoretical yield of urea Express your answer using four significant figures.

Question

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

In an industrial synthesis of urea, a chemist combines 125.0 kg of ammonia with 211.4 kg of carbon dioxide and obtains 183.8 kg of urea.

Answer 1

2NH₃(aq) + CO₂(aq) ==> CH₄N₂O(aq) + H₂O(l) ... balanced equation

Whenever you area given the amounts of BOTH reactants, you must determine which reactant (if any) is limiting. We can do that by dividing the moles of each by their respective coefficient in the balanced equation, as follows:

mol NH₃ = 125.0 kg NH₃ x 1 kmol/17 kg = 7.353 kmoles NH₃ (÷2=3.68)

mol CO₂ = 211.4 kg CO₂ x 1 kmol/44 kg = 4.80 kmoles CO₂ (÷1=4.80)

Therefore the 7.35 kmoles of NH₃ is limiting (3.68 < 4.80). So now we use the 7.35 kmoles NH₃ to determine the theoretical yield of urea.

7.353 kmol NH₃ x 1 kmol CH₄N₂O / 2 kmol NH₃ x 60.06 kg urea/mol = 220.8 kg CH₄N₂O