1. Liquid phosphorus trichloride, PCl3 is made by the direct combination of phosphorus and chlorine
a. What is the maximum mass of PCl3 that can be produced from 125 g of P4 and 323 g Cl2?
b. What mass of P4 remains after the reaction?
2.2. Billions of kilograms of urea, are produced annually for use as a fertilizer. The reaction used is given below.
NH3(g) + CO2(g) → CO(NH2)2(s) + H2O(l)
The typical starting reaction mixture has a 3:1 mole ratio of NH3 to CO2. If 47.7 g urea forms per mole of that reacts, what is the
2.2.1. Theoretical yield; [5]
2.2.2. Actual yield; [2]
2.2.3. Percent yield? [2]
To calculate the maximum mass of PCl3 that can be produced from the given mass of P4 and Cl2, we need to determine the limiting reactant first. The limiting reactant is the reactant that gets completely consumed and determines the amount of product that can be formed.
a. To find the limiting reactant, we need to compare the moles of P4 and Cl2. We'll use the balanced chemical equation for the reaction:
P4 + 6Cl2 -> 4PCl3
First, we convert the given masses of P4 and Cl2 to moles using their respective molar masses:
Molar mass of P4 = 123.9 g/mol
Molar mass of Cl2 = 70.9 g/mol
Moles of P4 = 125 g / 123.9 g/mol = 1.008 mol
Moles of Cl2 = 323 g / 70.9 g/mol = 4.558 mol
Now we look at the stoichiometric ratio between P4 and PCl3. From the balanced equation, we see that 1 mole of P4 produces 4 moles of PCl3.
Since the stoichiometry ratio between P4 and PCl3 is 1:4, we can calculate the maximum moles of PCl3 that can be produced from P4:
Maximum moles of PCl3 = (1.008 mol P4) * (4 mol PCl3 / 1 mol P4) = 4.032 mol PCl3
Next, we need to determine the mass of PCl3 that can be produced from the maximum moles:
Mass of PCl3 = (4.032 mol PCl3) * (137.3 g/mol PCl3) = 553.33 g
Therefore, the maximum mass of PCl3 that can be produced from 125 g of P4 and 323 g of Cl2 is 553.33 g.
b. To calculate the mass of P4 that remains after the reaction, we can subtract the mass of P4 that was reacted from the initial mass of P4:
Mass of P4 remaining = Initial mass of P4 - Mass of P4 reacted
We know that the initial mass of P4 is 125 g. From the stoichiometry of the reaction, we found that 1.008 mol of P4 reacts to produce 4 moles of PCl3.
So, the moles of P4 that reacted = (1.008 mol P4) * (4 mol PCl3 / 1 mol P4) = 4.032 mol PCl3
Now we can calculate the mass of P4 that reacted:
Mass of P4 reacted = (4.032 mol PCl3) * (123.9 g/mol P4) = 500 g
Therefore, the mass of P4 remaining after the reaction is:
Mass of P4 remaining = 125 g - 500 g = -375 g
However, this result is not physically meaningful since it implies a negative mass. Therefore, we can conclude that all the P4 has been completely consumed in the reaction, and no P4 remains.