The mass of urea produced per minute is 56.28 g assuming 100% yield.
Urea is produced from the reaction of ammonia and carbon dioxide:
2NH₃ + CO₂ → CO(NH₂)₂ + H₂O
Ammonia gas at 223.8°C and 90 atm flows into a reactor at a rate of 500 L/min.
Carbon dioxide at 223.8°C and 45 atm flows into the reactor at a rate of 600 L/min.
Find:
The mass of urea produced per minute assuming 100% yield.
Steps to solve:
Calculate the molar flow rates of ammonia and carbon dioxide.
n_NH₃ = (500 L/min) * (1 mol/22.4 L) * (273.15 K / (273.15 K + 223.8 K)) * (90 atm) = 1.875 mol/min
n_CO₂ = (600 L/min) * (1 mol/22.4 L) * (273.15 K / (273.15 K + 223.8 K)) * (45 atm) = 1.425 mol/min
Determine the limiting reactant.
The stoichiometry of the reaction is 2 moles of NH₃ to 1 mole of CO₂. Therefore, we need to compare the molar flow rates of NH₃ and CO₂ to see which one is the limiting reactant.
n_NH₃ / 2 = 0.938 mol/min
Since n_NH₃ / 2 < n_CO₂, NH₃ is the limiting reactant.
Calculate the theoretical yield of urea.
The theoretical yield of urea is equal to the moles of the limiting reactant (NH₃) multiplied by the stoichiometric coefficient of urea in the balanced chemical equation (1 mol of urea per 2 mol of NH₃).
n_urea = n_NH₃ / 2 = 0.938 mol/min
Calculate the mass of urea produced per minute.
The mass of urea produced per minute is equal to the number of moles of urea multiplied by the molar mass of urea (60 g/mol).
m_urea = n_urea * 60 g/mol = 56.28 g/min
Therefore, the mass of urea produced per minute is 56.28 g assuming 100% yield.