When metal $X$ is treated with sodium hydroxide,a white precipitate $(A)$ is obtained,which is soluble in excess of $NaOH$ to give soluble complex $(B)$. Compound $(A)$ is soluble in dilute $HCl$ to form compound $(C)$. The compound $(A)$ when heated strongly gives $(D)$,which is used to extract metal. Identify $(X), (A), (B), (C)$ and $(D)$. Write suitable equations to support their identities.

(A) The given metal $X$ reacts with sodium hydroxide to form a white precipitate that dissolves in excess of $NaOH$. This behavior is characteristic of aluminium.
$X = Al$ (Aluminium)
$A = Al(OH)_3$ (Aluminium hydroxide)
$B = Na[Al(OH)_4]$ (Sodium tetrahydroxoaluminate$(III)$)
$C = AlCl_3$ (Aluminium chloride)
$D = Al_2O_3$ (Alumina)
Equations:
$1$. $2Al + 2NaOH + 6H_2O \to 2Na[Al(OH)_4] + 3H_2$
$2$. $Al^{3+} + 3OH^- \to Al(OH)_3 \downarrow$ (White precipitate $A$)
$3$. $Al(OH)_3 + NaOH \to Na[Al(OH)_4]$ (Soluble complex $B$)
$4$. $Al(OH)_3 + 3HCl \to AlCl_3 + 3H_2O$ (Compound $C$)
$5$. $2Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3H_2O$ (Compound $D$ used for extraction)