Em tham khảo nha:

\(\begin{array}{l}
1)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{1,68}}{{22,4}} = 0,075\,mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 2,3\\
a + 1,5b = 0,075
\end{array} \right.\\
 \Rightarrow a = 0,025;b = \dfrac{1}{{30}}\\
\% {m_{Fe}} = \dfrac{{0,025 \times 56}}{{2,3}} \times 100\%  = 60,87\% \\
2)\\
{n_{NaOH}} = 0,4 \times 1,5 = 0,6\,mol\\
{n_{AlC{l_3}}} = 0,3 \times 1 = 0,3\,mol\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
\text{  Lập tỉ lệ ta có : } \dfrac{{{n_{AlC{l_3}}}}}{1} > \dfrac{{{n_{NaOH}}}}{3}(0,3 > 0,2)\\
 \Rightarrow \text{ $AlCl_3$ dư tính theo NaOH } \\
{n_{Al{{(OH)}_3}}} = \dfrac{{0,6}}{3} = 0,2\,mol\\
{m_{Al{{(OH)}_3}}} = 0,2 \times 78 = 15,6g \Rightarrow m = 15,6\\
3)\\
{n_{NaOH}} = 0,5 \times 1,4 = 0,7\,mol\\
{n_{AlC{l_3}}} = 0,2 \times 1 = 0,2mol\\
Al(N{O_3)_3} + 3NaOH \to Al{(OH)_3} + 3NaN{O_3}\\
\text{  Lập tỉ lệ ta có : }\dfrac{{{n_{Al(N{O_3)_3}}}}}{1} < \dfrac{{{n_{NaOH}}}}{3}(0,2 < \dfrac{{0,7}}{3})\\
 \Rightarrow \text{ NaOH dư tính theo $Al(NO_3)_3$ }\\
{n_{NaOH}} \text{ dư }= 0,7 - 0,2 \times 3 = 0,1\,mol\\
{n_{Al{{(OH)}_3}}} = {n_{Al(N{O_3)_3}}} = 0,2\,mol\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O\\
{n_{Al{{(OH)}_3}}} = {n_{NaOH}}  \text{ dư } = 0,1\,mol\\
{n_{Al{{(OH)}_3}}}  \text{ dư }= 0,2 - 0,1 = 0,1\,mol\\
{m_{Al{{(OH)}_3}}} = 0,1 \times 78 = 7,8g \Rightarrow m = 7,8
\end{array}\)