Bài 1: Fe + 2HCl --> FeCl2 + H2

          FeCl2 + 2NaOH --> 2NaCl + Fe(OH)2

          Fe(OH)2 + 2HCl --> FeCl2 + 2H2O 

          3FeCl2 + 2Al(OH)3 ---> 2AlCl3 + 3Fe(OH)2

Bài 2: nAl = 32,4/27 = 1,2 (mol)

           2Al + 3CuCl2 ---> 2AlCl3 + 3Cu

            1,2        1,8              1,2         1,8    (mol)

1) mCu = 1,8*64 = 115,2 (g)

2) V CuCl2 = 1,8/1,5 = 1,2 (lít)

3) AlCl3 + 3NaOH ---> 3NaCl + Al(OH)3

Em tham khảo nha :

\(\begin{array}{l}
1)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2}\\
Fe{(OH)_2} + 2HCl \to FeC{l_2} + 2{H_2}O\\
2Al + 3FeC{l_2} \to 2AlC{l_3} + 3Fe\\
2)\\
a)\\
2Al + 3CuC{l_2} \to 2AlC{l_3} + 3Cu\\
{n_{Al}} = \dfrac{{32,4}}{{27}} = 1,2mol\\
{n_{Cu}} = \dfrac{3}{2}{n_{Al}} = 1,8mol\\
{m_{Cu}} = 1,8 \times 64 = 115,2g\\
b)\\
{n_{CuC{l_2}}} = {n_{Cu}} = 1,8mol\\
{V_{CuC{l_2}}} = \dfrac{{1,8}}{{1,5}} = 1,2l\\
3)\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O
\end{array}\)