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!