Aluminum reacts with Chlorine gas to form AlCl3. [2Al(s)+3Cl2(g)=2AlCl3(s)]

Question

Given 11.0g of Al and 16.0g Cl2, if you had excess Al, how many moles of AlCl3 could be produced from 16.0g of Cl2?

Answer 1

You post is confusing. I can't tell if it's a limiting reagent problem or not. I'll work both.
If you had 11.0 g Al and all the Cl2 needed, you could produce how many mols of AlCl3? That's
mols Al = 11.0/27 = 0.407
mols AlCl3 = 0.407 x (2 mol AlCl3/2 mol Al) = 0.407 x 2/2 = 0.407 mols AlCl3. g = mols x molar mass.

If you had 16.0 g Cl2 and all the Al needed, how much AlCl3 could be formed? That's
mols Cl2 = 16.0/71 = 0.225 mols.
mols AlCl3 = 0.225 x (2 mols AlCl3/3 mols Cl2) = 0.225 x 2/3 = 0.15 mols AlCl3. g = mols x molar mass

The answers are above for each (11 or 16) with excess other reagent. If they are together you will be able to produce the SMALLER of the two values.