Question

PCl3 can be produced from the reaction of P4 with Cl2. P4(s) + 6 Cl2(g) --> 4 PCl3(g). If 1.00 g P4 reacts with 1.00 g Cl2, which reactant is the limiting reactant? What mass of product may be produced?

Answer 1

Convert both of the masses of reactants to the amount of moles of their respective reactant. Then, use stoichiometry for both reactants to figure out how many moles of product will be created. Whichever reactant produces the least amount of moles of product will be the limiting reactant.

Answer 2

How do I use the stoichiometry to tell how much product they can produce? I know 1g of P =0.032moles and 1g of Cl = 0.028moles. That is about as far as I can get.

Answer 3

Use mole ratios. Take the amount of product you have a multiply it by the mole ratio of PCl3 to product.

Answer 4

So, (x mol P4)(4 mol PCl3/1 mol P4) = y mol of PCl3 from x mol P4
You get my drift?

Answer 5

And I'll repeat, whichever reactant that will potentially create the lowest amount of moles of product will be your limiting reactant.

Answer 6

Okay, so I get 4 moles of PCL3 for every 1 mole of P4?

Answer 7

and 20 moles of PCl3 for every one mole of Cl2?

Answer 8

No, you're supposed to use the amount of moles when doing stoichiometry.
For P4: (0.032mol/g)*(4 mol of P)*(4 mol PCl3 / 1 mol P4)

Answer 9

Sorry if I'm not being that active, I can do better, but I'm getting sidetracked by basketball and I don't have a calculator right now.

Answer 10

So I get .512moles of PCl3 for every 0.032 moles of P4?

Answer 11

For Cl2, there are 4 moles of PCl3 for every 6 moles of Cl2, potentially.

Answer 12

No, you sorta messed up at the beginning. You were supposed to find the amount of moles of P4, not P, in 1 gram.

Answer 13

(amount of moles of P4)(4 mole PCl3 / 1 mol P4)

Answer 14

I'm getting the mole ratio from the balanced equation, using the amount of product that can be made to the amount of each reactant we have.

Answer 15

Oh so I'm doing math for no reason. because 1mol P4 yields 4mol PCl3 because of the equation.

Answer 16

Exactly. Now we need to know the actual amount of moles of P4 we need and apply the ratio. So, it'll be 1 g / (4*molar mass of P), since there are 4 P atoms in that reactant.

Answer 17

Which is 0.008mol?

Answer 18

Idk lol. Don't have a calc or periodic table, but I'll trust ya on that. So, what's the potential amount of PCl3 that is yielded by 0.008 mol of P4?
(0.008 mol P4)(4 mol PCl3 / 1 mol P4) = 0.032 mol PCl3, I believe lol. Check that work.

Do the exact same process for the Cl2 reactant.

Answer 19

For the Cl2, the mole ratio for the stoichiometry is (4 moles Cl3 / 6 mol Cl2)

Answer 20

That work is right, but I'm not sure how you came up with that equation. How do you know what to multiply and divide by?

Answer 21

I use the units, which are the mols in this case. That's why I'm including them. I want the moles of the reactant to cancel out so I'm left with the mols of the product.

Answer 22

This is lost in schools, and it sucks. Using units is a great way to figure out a problem.

Answer 23

Thank you very much for this lol. All my teacher does is read from a power point and won't go over the math involved. So let me walk myself through this math.

Answer 24

Ahh, I see.

Answer 25

1g of Cl2 converted to moles is 1mol/70.9g= 0.014mol

Answer 26

Yep, now do the stoichiometry, and compare it to the mole of PCl3 we would get from the P4.

Answer 27

Now I do what you did before which is. (0.014mol Cl2)(4 mol PCl3/6mol Cl2)

Answer 28

Yup yup,

Answer 29

okay I get 0.009mol PCl3

Answer 30

So Cl2 is the limiter

Answer 31

Absoutely. So, when this reaction plays out, ideally, we'll get only 0.009 mol of PCl3. All of the 1g of Cl2 will be used, but not all of the 1g of P4 will be used. However, the question just asks us what the mass of the product will be. So, multiply the amount of PCl3 by its molar mass and voila.

Answer 32

okay I got 1.24g PCl3. that is not one of my answers but I'm just going to pick the closest one lol

Answer 33

Lol ok. Probably did some wrong math somewhere, but it's k. The concept stays the same.

Answer 34

could you help me with Ph problems I have or do I need to post a new thing?

Answer 35

Sorry, I don't know about pH stuff. >.<

Answer 36

Might as well post a new question.

Answer 37

yeah. its only off by 0.03g

Answer 38

okay. thanks again man you are a lifesaver. lol