\(n_{Al}=\dfrac{32,4}{27}=1,2\left(mol\right)\)
PTHH: 2Al + 3CuCl2--> 2AlCl3 + 3Cu
_____1,2-->1,8-------->1,2------->1,8
1) mCu = 1,8.64 = 115,2 (g)
2) \(V_{ddCuCl_2}=\dfrac{1,8}{1,5}=1,2\left(l\right)\)
3)
\(AlCl_3+3NaOH\rightarrow3NaCl+Al\left(OH\right)_3\downarrow\)
\(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)