Question

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl3). If it is the only product, what mass of PCl 3 is formed?

I had formula
P + 3Cl --> PCl3

but apparently that's wrong and correct way to write this is

2P + 3 Cl2 - > 2PCl2

I then found out that Cl is a diatomic element and so is N, O, F, Br and I. Does that mean that every time those elements need to be written with the 2 subscript?

Answer 1

Yes. Cl2

Answer 2

but why is the problem telling me "form only phosphorus trichloride (PCl3)" but I end up changing PCl3 to PCl2?

Answer 3

54.93 g of PCl3 is formed in this reaction.

Answer 4

I found a really good explanation I'll post it here for you

Answer 5

First, we need to write a balanced equation for this reaction:

2 P(s) + 3 Cl2(g) → 2 PCl3

Next, we need to determine which reactant is the limiting one. Let us assume that it is phosphorus. We can use dimensional analysis and equation coefficients to convert 12.39 g of phosphorus to moles of phosphorus, to moles of chlorine, to grams of chlorine to see if there is enough chlorine to react completely with 12.39 grams of phosphorus. We shall need the following equalities to construct conversion factors:

1 mol P = 30.974 g P
3 mol Cl2 = 2 mol P
1 mol Cl2 = 70.906 g Cl2

[(12.39 g P)/1][(1 mol P)/(30.974 g P)][(3 mol Cl2)/(2 mol P)][(70.906 g Cl2)/(1 mol Cl2)] = 42.54 g Cl2 needed

Since 42.54 g of chlorine is exactly what is given in the problem, both reactants are limiting reactants, because they will both be used up completely in this reaction.

To find the mass of PCl3 formed, we can use dimensional analysis and equation coefficients to convert 12.39 g of phosphorus to moles of phosphorus, to moles of phosphorus trichloride, to grams of phosphorus trichloride. We shall need the following equalities to construct conversion factors:

1 mol P = 30.974 g P
2 mol PCl3 = 2 mol P
1 mol PCl3 = 137.333 g PCl3

[(12.29 g P)/1][(1 mol P)/(30.974 g P)][(2 mol PCl3)/(2 mol P)][(137.333 g PCl3)/(1 mol PCl3)] = 54.93 g PCl3

Answer: 54.93 g of PCl3 is formed in this reaction.

Answer 6

credit to: eveanne1221

Answer 7

Helter, you are rocking Chem!

Answer 8

Thank you so much! Really cleared things up!