# Last Stoich I swear

1. Dec 14, 2007

### UWMpanther

[SOLVED] Last Stoich I swear

1. The problem statement, all variables and given/known data
When FeCl3 is ignited in an atmosphere of pure oxygen, this reaction takes place
4FeCl3 + 3O2 -> 2Fe2O3 + 6Cl2

If 3.00 mol of FeCl3 are ignited in the presence of 2.00 mol of O2 gas, how much of which reagent is present in excess and therefore remains unused?

Now I don't know where to go from here. Obviously solve for the limiting reagent. Then I need some type of ratio to solve for how many mol are used :grumpy:

2. Dec 14, 2007

### rocomath

So which one is the limiting reagent? Or do you need help on that part as well? We can start from there if you'd like.

3. Dec 14, 2007

### UWMpanther

I think the limiting reagent is the O2 but I don't know how to caculate it since it is given in mol not g.

4. Dec 14, 2007

### rocomath

It doesn't matter at all. You can use either moles or grams.

Anyways, you're correct about O2 being the LR. It produces only 4.00 mol of Cl2 (Theoretical Yield), while FeCl3 produces 4.50 mol of Cl2.

5. Dec 14, 2007

### UWMpanther

Ok cool I got that far then when I was just guessing. So I basically went something like this:

since in the equation their relationship is 4FeCl2 to 3O2, I just took the 2O2 x 1.5 so it was equal to the 3.0 mol given. So that put 4FeCl2 to 4.5 mol. But I guess that's all I solved for the LR.

So next step would be, I'm assuming a relationship between the two. Maybe:

4.0 mol O2 x 4mol FeCl3/3mol O2 = 5.33mol FeCl3?

6. Dec 14, 2007

### rocomath

Ok since we only produce 4.00 mol Cl2 with O2, we use this same number and find how much FeCl3 is needed to produce 4.00 mol of Cl2.

So working backwards:

$$\mbox{4.00 mol} Cl_2 \times \frac{\mbox{2.00 mol}}{\mbox{3.00 mol}} \frac{FeCl_{2}}{Cl_{2}} = \frac{8}{3}\mbox{mol} FeCl_{3}$$

We started out with 3.00 mol FeCl3, so it's just the difference.

3.00 mol FeCl3 - 8/3 mol FeCl3 = ?