How many grams of Cu would be created if 9.1g of Al was reacted via the following reaction?
2Al + 3CuCl2 → 3Cu + 2AlCl3
3 mols of Cu for every 2 mols of Al
mols of Al = 9.1 g * ( 1 mol / 27 g)
mols of Cu = (3/2) * ( 9.1/27)
grams of Cu = mols of Cu * ( 63.5 grams/1mol)