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.
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.
1 answer
1 answer