$Al\rightarrow AlCl_3\rightarrow Al(OH)_3\rightarrow Al_2O_3\rightarrow Al\rightarrow Al_2O_3\rightarrow Al(NO_3)_3\rightarrow Al(OH)_3$ $PTHH:\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ 2Al_2O_3\xrightarrow[criolit]{đpnc}4Al+3O_2\\ 4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ Al_2O_3+6HNO_3\to 2Al(NO_3)_3+3H_2O\\ Al(NO_3)_3+3NaOH\to Al(OH)_3\downarrow+3NaNO_3$

Đáp án:

Giải thích các bước giải:

 Al → `Al_2O_3` → `AlCl_3` → `Al(OH)_3` → `Al_2(SO_4)_3` → `Al(OH)_3`→ `Al_2O_3`  → Al

Phương trình phản ứng:

4Al + 3`O_2` $\xrightarrow[]{t°}$ 2`Al_2O_3`

`Al_2O_3` + 6HCl → 2`AlCl_3` + 3`H_2O`

`AlCl_3` + 3NaOH → `Al(OH)_3` + 3NaCl

`2Al(OH)_3` + `3H_2SO_4` → `Al_2(SO_4)_3` + 6`H_2O`

`Al_2(SO_4)_3` + 6NaOH → 3`Na_2SO_4` + 2`Al(OH)_3`

`2Al(OH)_3` $\xrightarrow[]{t°}$ `Al_2O_3` + 3`H_2O`

2`Al_2O_3` $\xrightarrow[đpnc]{criolit}$4Al + 3`O_2`

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