When 9.0g of Al were treated with an excess of chlorine, 20.0g of Al2Cl6 were collected.

Question

2AL + 3Cl6 -------> Al2Cl6

What was the percentage yield?

This is what i did so far

# moles in Al = 0.333

# moles in Cl2 = 0.282

The limiting reactant = Cl2

27g of Al forms 71g of Cl2
9g of Al should form 9/27 * 71 = 23.7g

The ratio between Cl2 & Al2Cl6 is 3 : 1

I'm stuck There

need Help here tnx

When 9.0g of Al were treated with an excess of chlorine, 20.0g of Al2Cl6 were collected.

2AL + 3Cl6 -------> Al2Cl6

What was the percentage yield?

This is what i did so far

# moles in Al = 0.333

# moles in Cl2 = 0.282

The limiting reactant = Cl2 <b< absolutely not. There was an excess of Cl2, according to the problem. There was no limiting reactant.

27g of Al forms 71g of Cl2 Huh? Aluminum does not form Cl2
9g of Al should form 9/27 * 71 = 23.7g

The ratio between Cl2 & Al2Cl6 is 3 : 1

I'm stuck There
You are going the wrong way, not stuck. For every mole of aluminum, you should get 1/2 mole of Al2Cl6. You used .333mole Al (I did not check that calc), so you should get .333/2 moles of Al2Cl6.

Precentage yield= grams got of product/grams should have got * 100>/b>

You started just great but faltered after the first step. The # mols Al is correct at 0.333.
BUT, the problem says you have an EXCESS of Cl2 so you have all you need and there is no limiting reagent (well, of course the Al is the limiting reagent but this is a regular stoichiometry problem).
Step 2. Convert mols Al to mols Al2Cl6 (is that the problem or is that what you think is formed?).
Step 3. Convert mols Al2Cl6 to grams Al2Cl6. This is the theoretical yield and what you would get at 100%.
Step 4.
% yield = [20.0/theoretical]x 100 = ??

I think you can work it from here but post your work if you get stuck.

Step 2. Convert mols Al to mols Al2Cl6 (is that the problem or is that what you think is formed?).

Lost there

OO

i got it

the yield is 45%.. the ans. is in the book