What is the limiting reactant if 0.5 g al is reacted with 3.5 g cucl2? take into account cucl2 is a dihydrate when calculating the molecular weight?

Answer is: CuCl₂·2H₂O is limiting reactant.
Chemical reaction: 3CuCl₂·2H₂O + 2Al → 3Cu + 2AlCl₃ + 6H₂O.
m(Al) = 0,5 g.
m(CuCl₂·2H₂O) = 3,5 g.
n(CuCl₂·2H₂O) = m(CuCl₂·2H₂O) ÷ M(CuCl₂·2H₂O).
n(CuCl₂·2H₂O) = 3,5 g ÷ 170,5 g/mol.
n(CuCl₂·2H₂O) = 0,0205 mol.
n(Al) = 0,5 g ÷ 27 g/mol.
n(Al) = 0,0185 mol.
From chemical reaction: n(Al) : n(CuCl₂·2H₂O) = 2 : 3.
0,0185 mol : n(CuCl₂·2H₂O) = 2 : 3.
n(CuCl₂·2H₂O) = 0,02755 mol; there is no enough n(CuCl₂·2H₂O).

Show an image of the data (˵¯͒ བ¯͒˵)

13) a 14) a15) b16) i don't know sorry17) a (maybe)