When 6.0 mol al react with 13 mol hcl, what is the limiting reactant, and how many moles of h2 can be formed?

2 al + 6hcl _ 2 alcl3 + 3 h2

al is the limiting reactant, 9.0 mol h2 can be formed

hcl is the limiting reactant, 6.5 mol h2 can be formed

al is the limiting reactant, 6.0 mol h2 can be formed

hcl is the limiting reactant, 4.3 mol h2 can be formed

HCl is the limiting reactant, 6.5 mol H2 can be formed

Explanation:

If you check the estequiometric equation for 2 mol of Al you will need 6 mols of HCl. Having 6 mols of Al it is necessary to have 18 mols of HCl for a complete reaction, you have 13 mols. Therefore, there is less HCl than needed; it is the limiting reactant.

On the other hand, in a hypothetic complete reaction with 6 mol of Al and 18 mols of HCl nine mols of H2 would be produced. Here you can calculate taking the direct proportion as follows:

[tex] 13/18*9=6.5 mols of H2[\tex]

So it is possible to produce 6.5 mol of H2.