[tou]

Hi

I don't know how to solve this:

2Al + 3I₂ => 2AlI₃

1.20g Al and 2.40g I

What are the limiting reagent and the theoretical yield?

I know that I₂ is the limiting reagent, because 2.40/254 = 0.0094mol while 1.20/27 = 0.044mol - thus there is less iodine than aluminum.

Now I thought I have to multiply 0.0094 with 254, that is 2.38g of I₂. But the answer says that the theoretical yield is 2.57g.

I don't get this... 

[agrobert]

First of all you are right in determining that I₂ is the limiting reagent but your calculation is wrong.  You need to account for stoichiometry...(2 mol AlI₃/3 mol I₂)

This will give you the moles of AlI₃ produced theoretically.

What is theoretical yield? What product are you making in your reaction?  I don't think it is I₂...

So from your moles of AlI₃ multiply by the MW of AlI₃ and you should get your answer.  (And the answer is correct)

[tou]

Aha, I see: (2/3) x 0.0094 = 0.0062 x 408 = 2.57

I don't know why I used 254 instead of 408, I guess I was very confused. 

Thank you!