When 4.5 mol Al reacts with 11.5 mol HCl,what is the limiting reactant, and how many moles of H2 can be formed?

Answer 1

LR = limiting reagent.
2Al + 6HCl ==> 3H2 + 2AlCl3

How many mols H2 produced if we had 4.5 mol Al and all of the HCl we needed. That would be
4.5 x (3 mols H2/2 mol Al) = 4.5 x 3/2 = 6.75.

How many mols H2 produced if we had 11.5 mols HCl and all of the Al we needed. That's
11.5 x (3 mols H2/6) = 11.5 x 3/6 = 4.75 mols H2.
Obviously, both answers can't be right; in LR problems the correct answer is ALWAYS the SMALLER value AND the reagent providing that smaller value is the LR. So you will have 5.75 mols H2 produced.