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S Block Elements

A metal X rea...

Question

A metal X reacts with aqueous NaOH solution to form Y and a highly inflammable gas. Solution Y is heated and $C O_{2}$ IS poured through it. Z precipitates out and $N a_{2} C O_{3}$ is formed. Z on heating gives $A l_{2} O_{3}$ . Identify X, Y and Z.

A

A l - X

,

N a A l O_{2} - Y

,

A l (O H)_{3} - Z

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B

A l_{2} O_{3} - X

,

N a A l O_{2} - Y

,

A l_{2} C O_{3} - Z

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C

A l_{2} O_{3} - X

,

[N a_{2} A l O_{2}]^{+} O h^{-} - Y

,

A l (O H)_{3} - Z

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D

Al-X,

A l (O H)_{3} - Y

,

A l_{2} O_{3} - Z

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Solution

The correct option is C $A l - X$ , $N a A l O_{2} - Y$ , $A l (O H)_{3} - Z$
Solution:-
$X + N a O H (a q) ⟶ Y + highly inflammable gas$
$Y + C O_{2} + H_{2} O ⟶ Z + {N a}_{2} C O_{3}$
$Z - ------ \to O n h e a t i n g {A l}_{2} O_{3}$
${A l}_{2} O_{3}$ is formed when $A l {(O H)}_{3}$ is heated.
$2 A l {(O H)}_{3} - ------ \to O n h e a t i n g {A l}_{2} O_{3} + 3 H_{2} O$
Thus Z is $A l {(O H)}_{3} .$
When $N a A l O_{2}$ is heated and $C O_{2}$ is poured through it, $A l {(O H)}_{3}$ precipitates out and $N a_{2} C O_{3}$ is formed.
$2 N a A l O_{2} + C O_{2} + 3 H_{2} O ⟶ 2 A l {(O H)}_{3} + {N a}_{2} C O_{3}$
Thus Y is $N a A l O_{2}$
When $A l$ reacts with aqueous $N a O H$ solution, it forms $N a A l O_{2} a n d H_{2} (Highly inflammable gas)$
$A l + N a O H (a q) ⟶ N a A l O_{2} + H_{2} (highly inflammable gas)$
Thus X is $A l$ .
Hence, (A) is correct answer.

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