Limiting reactant, theoretical, percentage yield problem

Question

I'm new to this concept, so it's a little bit confusing.. I need some help. Thanks.

A chemist synthesised phosphorous trichloride by mixing 12.0 g of P with 35.0 g Cl2. The limiting reactant is Cl2. The equation for the reaction is:

2P (s) + 3Cl2 (g) -> 2PCl3 (l)

(i) Calculate the theoretical moles of phosphorous trichloride (PCl3) that should be formed according to above balanced chemical reaction.

(ii) Using the moles of phosphorous trichloride (PCl3) in part (i), calculate the theoretical yield of PCl3 in grams.

(iii) Calculate the percentage yield of PCl3, if the mass of PCl3 obtained experimentally is 42.4 g.

Answer 1

The limiting reactant is that reactant that will run out first, leaving the other reactant(s) in excess. It's sort of like having 6 hot dogs and 8 buns. The hot dogs run out before the buns and so you can only make 6 products, not 8.

In this problem, they tell us that Cl₂ is limiting, so the moles of Cl₂ will determine how much product (PCl₃) can be made.

2P(s) + 3Cl₂(g) ==> 2PCl₃(l) .. balanced equation

First, we find the MOLES of Cl₂: 35.0 g Cl₂ x 1 mol Cl₂ / 71 g = 0.493 mols Cl₂

(i) Theoretical moles PCl₃

0.493 mols Cl₂ x 2 mol PCl₃ / 3 mol Cl₂ = 0.329 mols PCl₃ is the theoretical yield

(ii) Converting to grams:

molar mass PCl₃ = 137 g / mol

0.329 mols PCl₃ x 137 g / mol = 45.1 g PCl₃

(iii) Percent yield = actual yield / theoretical yield (x100%)

42.4 g / 45.1 g (x100) = 94.0% yield