Respuesta :

3CuCl₂ + 2Al = 3Cu + 2AlCl₃

m(CuCl₂)=10.5 g
M(CuCl₂)=134.45 g/mol
n(CuCl₂)=10.5/134.45=0.0781 mol

m(Al)=12.4 g
M(Al)=26.98 g/mol
n(Al)=12.4/26.98=0.4596

CuCl₂:Al = 3:2
0.0781:0.4596 ⇒ 3:17.7

the limiting reactant is CuCl₂

k0kichi0ma

Answer:

CuCI2 is the limiting reactant.