\(\begin{array}{l}
\rm{1.1\quad \rm HCl \xrightarrow{(1)} Cl_2 \xrightarrow{(2)} HCl \xrightarrow{(3)} FeCl_2 \xrightarrow{(4)} Fe(OH)_2 \xrightarrow{(5)} FeCl_2\\
(1)\quad 2HCl \xrightarrow{\quad đpdd\quad} H_2\uparrow + Cl_2\uparrow\\
(2)\quad Cl_2 + H_2 \xrightarrow{\quad t^\circ\quad} 2HCl\uparrow \\
(3)\quad 2HCl + Fe \longrightarrow FeCl_2 + H_2\uparrow \\
(4)\quad FeCl_2 + 2NaOH \longrightarrow Fe(OH)_2\downarrow + 2NaCl\\
(5)\quad Fe(OH)_2 + HCl \longrightarrow FeCl_2 + H_2O\\
1.2\quad Mg \xrightarrow{(1)} MgCl_2 \xrightarrow{(2)} Mg(OH)_2 \xrightarrow{(3)} MgCl_2\\
(1)\quad Mg + 2HCl \longrightarrow MgCl_2 + H_2\uparrow\\
(2)\quad MgCl_2 + 2NaOH \longrightarrow Mg(OH)_2\downarrow + 2NaCl\\
(3)\quad Mg(OH)_2 + 2HCl \longrightarrow MgCl_2 + H_2O\\
1.3 \quad MnO_2 \xrightarrow{(1)} Cl_2 \xrightarrow{(2)} HCl \xrightarrow{(3)} Cl_2 \xrightarrow{(4)} FeCl_3 \xrightarrow{(5)} Fe(OH)_3 \xrightarrow{(6)} Fe_2O_3 \xrightarrow{(7)} FeCl_3\\
(1)\quad MnO_2 + 4HCl \longrightarrow MnCl_2 + Cl_2\uparrow + 2H_2O\\
(2)\quad Cl_2 + H_2 \xrightarrow{\quad t^\circ\quad} 2HCl\uparrow\\
(3)\quad 2HCl \xrightarrow{\quad đpdd\quad} H_2\uparrow + Cl_2\uparrow\\
(4)\quad 3Cl_2 + 2Fe \xrightarrow{\quad t^\circ\quad} 2FeCl_3\\
(5)\quad FeCl_3 + 3NaOH \longrightarrow Fe(OH)_3\downarrow + 3NaCl\\
(6)\quad 2Fe(OH)_3 \xrightarrow{\quad t^\circ\quad} Fe_2O_3 + 3H_2O\\
(7)\quad Fe_2O_3 + 6HCl \longrightarrow 2FeCl_3 + 3H_2O}\\
2)\\
\rm 2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2\uparrow\\
n_{Al} = \dfrac{m}{M} = \dfrac{5,4}{27} = 0,2\ mol\\
\Rightarrow \begin{cases}n_{AlCl_3} = n_{Al} = 0,2\ mol\\n_{H_2} = \dfrac32n_{Al} =0,3\ mol\end{cases}\\
\Rightarrow \begin{cases}m_{AlCl_3} = n\times M = 0,2 \times 133,5 =26,7\ g\\V_{H_2} = n\times 22,4 = 0,3 \times 22,4 = 6,72\ l\end{cases}\\
3)\\
\rm Zn + 2HCl \longrightarrow ZnCl_2 + H_2\uparrow\\
n_{H_2} = \dfrac{V}{22,4} = \dfrac{6,72}{22,4} = 0,3\ mol\\
\Rightarrow n_{Zn} = n_{ZnCl_2} = n_{H_2} = 0,3 \ mol\\
\Rightarrow \begin{cases}m_{Zn} = n\times M = 0,3\times 65 = 19,5\ g\\m_{ZnCl_2} = n\times M = 0,3 \times 136 = 40,8\ g\end{cases}\\
4)\\
\rm FeO + 2HCl \longrightarrow FeCl_2 + H_2O\\
n_{FeO} = \dfrac{m}{M} = \dfrac{14,4}{72} = 0,2\ mol\\
\Rightarrow n_{FeCl_2} = n_{FeO} = 0,2\ mol\\
\Rightarrow m_{FeCl_2} = n\times M = 0,2 \times 127 = 25,4\ g
\end{array}\)
