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.

(N/A) The given metal $X$ is $Al$. The reactions are as follows:
$1$. $2 Al + 2 NaOH + 6 H_2O \rightarrow 2 Na[Al(OH)_4] + 3 H_2$ (Note: $Al(OH)_3$ is the intermediate white precipitate $(A)$).
$2$. $Al(OH)_3 + NaOH \rightarrow Na[Al(OH)_4]$ (Soluble complex $(B)$).
$3$. $Al(OH)_3 + 3 HCl \rightarrow AlCl_3 + 3 H_2O$ (Compound $(C)$ is $AlCl_3$).
$4$. $2 Al(OH)_3 \xrightarrow{\Delta} Al_2O_3 + 3 H_2O$ (Compound $(D)$ is $Al_2O_3$,which is used in the extraction of $Al$).