Giải:\(\eqalign{(1)\;4Al + 3{O_2}\buildrel {{t^0}} \over& (2)\;A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O. \cr & (3)\;AlC{l_3} + 3NaO{H}_{đủ} \to Al{(OH)_3} \downarrow + 3NaCl. \cr & (4)\;Al{(OH)_3} + 3HCl \to AlC{l_3} + 3{H_2}O. \cr & (5)\;Al{(OH)_3} + KOH \to K\left[ {Al{{(OH)}_4}} \right]. \cr} \)\(\eqalign{& (6)\;2Al{(OH)_3}\buildrel {{t^0}} \over\longrightarrow A{l_2}{O_3} + 3{H_2}O. \cr & (7)\;A{l_2}{O_3} + 2NaOH + 3{H_2}O \to 2Na\left[ {Al{{(OH)}_4}} \right]. \cr} \)