The given metal $X$ gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, $X$ must be aluminium.

The white precipitate (compound $A$) obtained is aluminium hydroxide. The compound $B$ formed when an excess of the base is added is sodium tetrahydroxoaluminate $(III)$.

$\mathop {2Al}\limits_{Alu\min imum\,\,(X)}  + \mathop {3NaOH}\limits_{Sodium\,hydroxide}  \to \mathop {Al(O{H_3}) \downarrow }\limits_{White\,ppt.(A)}  + 3N{a^ + }$

$\underset{(A)}{\mathop{Al{{(OH)}_{3}}}}\,+NaOH\to \underset{\begin{smallmatrix} 
 Sodium\,tetrahydroxoalu\min atr\,(III) \\ 
 (so\operatorname{lub}le\,complex\,B) 
\end{smallmatrix}}{\mathop{N{{a}^{+}}{{[Al{{(OH)}_{4}}]}^{-}}}}\,$

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound $C$ ) is obtained.

$\mathop {Al{{(OH)}_3}}\limits_{(A)}  + 3HCl \to \mathop {AlC{l_3}}\limits_{(C)}  + 3{H_2}O$

Also, when compound $A$ is heated strongly, it gives compound $D$. This compound is used to extract metal $X$. Aluminium metal is extracted from alumina. Hence, compound $D$ must be alumina.

$\mathop {2Al{{(OH)}_3}}\limits_{(A)} \xrightarrow{\Delta }\mathop {A{l_2}{O_3}}\limits_{(D)}  + 3{H_2}O$