Question

What mass of AlCl3 can be formed from 5.000 mol NiCl2 and 2.000 mol Al?


Al(s) + NiCl2(aq) → AlCl3(aq) + Ni(s)
I balanced it to
2Al + 2NiCl2 -> 2AlCl3 + 3Ni

I'm not sure how to go about this. I converted the moles into grams, but I feel like the balanced equation falls into play too. Am lost and a way to start would be good.

Answer 1

This is a limiting reagent problem. You know that because it lists amounts for BOTH reactants.
First the equation isn't balanced; you have 2 Ni on the left and 3 on the right.
2Al + 3NiCl2 ==> 2AlCl3 + 3Ni

Do these one at a time and use the coefficients in the balanced equation.
5.000 mols NiCl2 x (2 mols AlCl3/3 mol NiCl2) = 5*2/3 = 3.33 mols AlCl3

2.000 mols Al x (2 mols AlCl3/2 mol Al) = 2*2/2 =2 mol AlCl3.

Obviously both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value.
Therefore, 2.000 mol AlCl3 wll be formed.