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) ---Heat pressure---> H2NCONH2(s) + H2O(g)
Ammonia gas at 223°C and 90. atm flows into a reactor at a rate of 610. L/min. Carbon dioxide at 223°C and 48 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?
i do not know if i got it right, please help!! i got 42000 g/min
Well, let's break it down step by step and see if your answer is correct.
First, we need to calculate the number of moles of ammonia and carbon dioxide flowing into the reactor per minute.
For ammonia:
Using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can rearrange the equation to solve for n:
n = PV/RT
Given:
P = 90. atm
V = 610. L/min
R = 0.0821 L.atm/(mol.K)
T = 273°C + 223°C = 496 K
Calculating the number of moles of ammonia:
n(NH3) = (90. atm)(610. L/min)/(0.0821 L.atm/(mol.K))(496 K)
n(NH3) = 836.5 mol/min
Now, let's do the same for carbon dioxide:
Given:
P = 48. atm
V = 600. L/min
R = 0.0821 L.atm/(mol.K)
T = 273°C + 223°C = 496 K
n(CO2) = (48. atm)(600. L/min)/(0.0821 L.atm/(mol.K))(496 K)
n(CO2) = 878.6 mol/min
Now, let's look at the balanced chemical equation:
2 NH3(g) + CO2(g) ---> H2NCONH2(s) + H2O(g)
According to the stoichiometry, we can see that 2 moles of ammonia react with 1 mole of carbon dioxide to produce 1 mole of urea.
Since we have the number of moles of ammonia and carbon dioxide flowing into the reactor per minute, the limiting reactant in this case is carbon dioxide because 878.6 mol/min is more than double 836.5 mol/min.
Therefore, we can calculate the theoretical yield of urea based on the number of moles of carbon dioxide:
Using the molar mass of carbon dioxide (44.01 g/mol) and urea (60.06 g/mol):
m(Urea) = n(CO2) × M(Urea)
m(Urea) = 878.6 mol/min × 60.06 g/mol
m(Urea) = 52719.4 g/min
So, with 100% yield, the mass of urea produced per minute would be approximately 52719.4 g/min.
Your answer of 42000 g/min is incorrect. The correct answer, assuming 100% yield, is approximately 52719.4 g/min.
To solve this problem, we need to use the stoichiometry of the balanced chemical equation and the given conditions to calculate the mass of urea produced per minute.
Step 1: Calculate the number of moles of ammonia (NH3) and carbon dioxide (CO2) entering the reactor per minute.
Using the ideal gas law, we can calculate the number of moles from the given volumes, temperatures, and pressures.
For ammonia:
PV = nRT
n = (PV) / (RT)
n = (90 atm * 610 L) / [(0.0821 L·atm/(mol·K)) * (223 + 273) K]
n = 18.3 mol/min
For carbon dioxide:
PV = nRT
n = (PV) / (RT)
n = (48 atm * 600 L) / [(0.0821 L·atm/(mol·K)) * (223 + 273) K]
n = 11.7 mol/min
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we compare the mole ratios between the reactants and the product in the balanced equation. The reactant that produces the least amount of product is the limiting reactant.
From the balanced chemical equation:
2 NH3(g) + CO2(g) -> H2NCONH2(s) + H2O(g)
The mole ratio between NH3 and H2NCONH2 is 2:1.
The mole ratio between CO2 and H2NCONH2 is 1:1.
Since the mole ratio of CO2 to H2NCONH2 is 1:1, it means that 11.7 mol/min of CO2 will react with 11.7 mol/min of NH3. Therefore, NH3 is the limiting reactant.
Step 3: Calculate the mass of urea produced.
From the balanced equation, we know that 2 moles of NH3 react to produce 1 mole of urea (H2NCONH2).
Molar mass of NH3 = 17.03 g/mol
Molar mass of H2NCONH2 = 60.06 g/mol
Using the molar ratio:
2 mol NH3 : 1 mol H2NCONH2
Mass of urea produced per minute = (2 mol NH3 * 60.06 g/mol) / (1 mol H2NCONH2) = 120.12 g/min
Therefore, the mass of urea produced per minute by this reaction, assuming 100% yield, is 120.12 g/min.