The balanced chemical reaction:
3CuCl2+2Al->3Cu+2AlCl3
We
are given the amount of the reactant to be used for this reaction. These values
will be the starting point of the calculations.
2.50 g CuCl2 ( 1 mol CuCl2 / 134.45 g CuCl2) = 0.019 mol CuCl2
5 g Al ( 1 mol Al / 26.98 g Al ) = 0.185 mol Al
The mole ratio of the reactants is 3:2. In every 3 moles of CuCl2, 2 mol of Al is required. Therefore, the limiting reactant is CuCl2.
