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?
a. Well, let me do some quick calculations in my hilarious clown brain. First, we need to figure out the limiting reactant, which is the reactant that gets used up first. We can do this by comparing the moles of P4 and Cl2. So, let's get cracking!
Moles of P4 = 125 g / molar mass of P4
Moles of Cl2 = 323 g / molar mass of Cl2
Then, we look at the balanced chemical equation to determine the stoichiometric ratio between P4 and PCl3:
P4 + 6Cl2 -> 4PCl3
According to the equation, for every 1 mole of P4, we need 6 moles of Cl2 to produce 4 moles of PCl3. So, let's find the limiting reactant:
Moles of PCl3 produced from P4 = Moles of P4 x (4 moles PCl3 / 1 mole P4)
Moles of PCl3 produced from Cl2 = Moles of Cl2 x (4 moles PCl3 / 6 moles Cl2)
We can then compare the moles of PCl3 produced from both reactants to find the limiting reactant. The one that produces the lower amount of PCl3 is the limiting reactant:
Limiting moles of PCl3 = min(Moles of PCl3 produced from P4, Moles of PCl3 produced from Cl2)
Finally, we can calculate the maximum mass of PCl3 that can be produced by multiplying the limiting moles of PCl3 by its molar mass:
Maximum mass of PCl3 = Limiting moles of PCl3 x molar mass of PCl3
Phew, that was a long explanation! But I'm a clown, so I love long explanations. Let's do the math and find out the answer!
b. As for the mass of P4 remaining after the reaction, we can subtract the moles of P4 consumed during the reaction from the initial moles of P4, then convert it to grams using the molar mass of P4.
But remember, laughter is the best medicine, so let me entertain you while I do some calculations. Please hold on to your funny bone!
To solve this problem, we need to determine the limiting reactant and use the stoichiometry to find the maximum mass of PCl3 and the mass of P4 remaining after the reaction.
a. To identify the limiting reactant, we need to compare the number of moles of P4 and Cl2. Let's calculate the number of moles for each:
Molar mass of P4 (phosphorus) = 4 * atomic mass of P = 4 * 31.0 g/mol = 124.0 g/mol
Number of moles of P4 = mass / molar mass = 125 g / 124.0 g/mol ≈ 1.008 moles
Molar mass of Cl2 (chlorine) = 2 * atomic mass of Cl = 2 * 35.5 g/mol = 71.0 g/mol
Number of moles of Cl2 = mass / molar mass = 323 g / 71.0 g/mol ≈ 4.560 moles
Now, let's determine the limiting reactant by comparing the moles of each reactant. The balanced equation for the reaction is:
P4 + 6Cl2 -> 4PCl3
According to the balanced equation, 1 mole of P4 reacts with 6 moles of Cl2 to produce 4 moles of PCl3.
Using the moles calculated above, we can determine the limiting reactant:
For P4: 1.008 moles of P4 * (6 moles of Cl2 / 1 mole of P4) = 6.048 moles of Cl2
For Cl2: 4.560 moles of Cl2
Since 6.048 moles of Cl2 is larger than 4.560 moles, Cl2 is the limiting reactant.
To find the maximum mass of PCl3 produced, we can use the stoichiometry of the balanced equation:
Molar mass of PCl3 = atomic mass of P + 3 * atomic mass of Cl = 31.0 g/mol + 3 * 35.5 g/mol = 137.5 g/mol
Number of moles of PCl3 = number of moles of Cl2 * (4 moles of PCl3 / 6 moles of Cl2)
= 4.560 moles * (4 moles / 6 moles) ≈ 3.040 moles
Mass of PCl3 = number of moles * molar mass
= 3.040 moles * 137.5 g/mol ≈ 417.2 g
Therefore, the maximum mass of PCl3 that can be produced is approximately 417.2 grams.
b. To find the mass of P4 remaining after the reaction, we can compare the moles of Cl2 used in the reaction with the initial moles of Cl2:
Initial moles of Cl2 = 4.560 moles
Moles of Cl2 used in the reaction = number of moles of Cl2 - excess moles of Cl2
Excess moles of Cl2 = (number of moles of Cl2 / stoichiometric coefficient of Cl2) * stoichiometric coefficient of P4
= (4.560 moles / 6 moles) * 1 mole = 0.760 moles
Moles of P4 remaining = excess moles of Cl2 * (1 mole of P4 / 6 moles of Cl2)
= 0.760 moles * (1 mole / 6 moles) ≈ 0.1267 moles
Mass of P4 remaining = moles * molar mass
= 0.1267 moles * 124.0 g/mol ≈ 15.7 g
Therefore, the mass of P4 remaining after the reaction is approximately 15.7 grams.
To determine the maximum mass of PCl3 that can be produced and the mass of P4 that remains after the reaction, we need to use stoichiometry. Stoichiometry involves using the balanced chemical equation to relate the mass of one substance to another substance in a chemical reaction.
a. To find the maximum mass of PCl3, we need to calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product that can be formed.
Step 1: Write the balanced chemical equation for the reaction:
P4 + 6Cl2 -> 4PCl3
Step 2: Calculate the number of moles of each reactant using their molar masses:
Molar mass of P4 = 123.89 g/mol
Molar mass of Cl2 = 70.91 g/mol
Number of moles of P4 = 125 g / 123.89 g/mol = 1.008 mol
Number of moles of Cl2 = 323 g / 70.91 g/mol = 4.558 mol
Step 3: Determine the limiting reactant by comparing the mole ratio in the balanced equation to the actual mole ratio of the reactants. The ratio of P4 to Cl2 is 1:6.
Mole ratio P4:Cl2 = 1:6
Actual mole ratio P4:Cl2 = 1.008 mol:4.558 mol
The actual mole ratio is less than the stoichiometric ratio, indicating that P4 is the limiting reactant.
Step 4: Calculate the amount of PCl3 produced using the mole ratio in the balanced equation:
Mole ratio P4:PCl3 = 1:4
Number of moles of PCl3 produced = 1.008 mol (P4) * (4 mol PCl3 / 1 mol P4) = 4.032 mol PCl3
Step 5: Convert moles of PCl3 to mass using its molar mass:
Molar mass of PCl3 = 137.33 g/mol
Mass of PCl3 produced = 4.032 mol * 137.33 g/mol = 553.11 g
Therefore, the maximum mass of PCl3 that can be produced from 125 g of P4 and 323 g Cl2 is 553.11 g.
b. To find the mass of P4 that remains after the reaction, we need to subtract the amount of P4 reacted from the initial mass of P4.
Mass of P4 remaining = initial mass of P4 - mass of P4 reacted
Mass of P4 remaining = 125 g - mass of P4 reacted
Since we determined that 1.008 mol of P4 reacted, we can calculate the mass of P4 reacted using its molar mass:
Mass of P4 reacted = 1.008 mol * 123.89 g/mol = 125.08 g
Mass of P4 remaining = 125 g - 125.08 g = -0.08 g
Therefore, a negative value for the mass of P4 remaining indicates that no P4 is left after the reaction.
balance the equation
P4 + 6Cl2 >>> 4PCl3
moles P4= 125/30.9 about 4
moles Cl2= 323/70 about 4.5
we need six moles Cl2 for each mole of P4, so we have an excess of P4.
max moles P4 used: 323/(70*6)
then convert that to mass
mass remaining=125-max used