Less than 6.0 grams of copper is formed, and some aluminum is left in the reaction mixture
From the equation of the reaction:
3CuCl2 + 2Al → 2AlCl3 + 3Cu
Mole ratio of CuCl2 and Al = 3:2
Mole of 1.5 g aluminum foil = 1.5/26.98
= 0.06 moles
Mole of 14 g CuCl2 = 14/134.45
= 0.10 moles
For every 3 moles of CuCl2, 2 mole of Al is needed.
Thus, for 0.10 moles of CuCl2, the mole of Al needed would be:
0.10 x 2/3 = 0.07 moles
CuCl2 is, thus, slightly in excess while Al is the limiting reagent.
Mole ratio of Al and Cu = 2:3
For 0.06 mole Al, the mole of Cu will be: 0.06x3/2 = 0.09 moles
Mass of 0.09 moles Cu = 0.09 x 63.55
= 5.71 g
Thus, less than 6.0 grams Cu is formed while a little CuCl2 would be left in the reaction mixture.
More on stoichiometric calculations can be found here: https://brainly.com/question/8062886
