`X` reacts withy `NaOH` (aqueous solution) to form `Y` and `H_(2)`. The aqueous solution of `Y` is heated to `323-333 K` and on passing `CO_(2)` into it, `Na_(2)CO_(3)` and `Z` were formed. When `Z` is heated to `1200^(@)C, Al_(2)O_(3)` is formed. `X, Y`, and `Z`, respectively, are

Options

A. `Al, AlCl_(3), NaAlO_(2)`
B. `Zn, Na_(2)ZnO_(2)`
C. `Al, Al(OH)_(3), AlCl_(3)`
D. `Al, NaAlO_(2), Al(OH)_(3)`

Answer

D

Explanation

Aluminium is attacked by caustic alkalies with the evolution of hydrogen gas:
`underset((X))(2Al) + 2NaOH+2H_(2)Orarr underset((Y))(2NaAlO_(2))+3H_(2)`
When aqueous solution containing sodium metaaluminate `(Y)` is heated to `323-333 K` and carbon dioxide is circulated through it, aluminium hydroxides `(Y)` separates our as pre-cipitage:
`underset((Y))(2NaAlO_(2))+CO_(2)+3H_(2)Orarr underset((Z))(2Al(OH)_(3))+Na_(2)CO_(3)`
When aluminium hydroxide `(Z)` is heated to `1200^(@)C`, alumina `(Al_(2)O_(3))` is formed:
`underset((Y))(2Al(OH)_(3))overset(1200^(@)C)rarrunderset((Z))(Al_(2)O_(3))+3H_(2)O`
Thus, `X` is `Al, Y` is `NaAlO_(2)`, and `Z` is `Al(OH)_(3)`.