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Statement-1 : Carbon reduction for $A l_{2} O_{3}$ , is very difficult.
Statement-2 : $A l$ forms carbide the $A l_{4} C_{3}$ which produces propyne when reacts with $H_{2} O$ .

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Statement- $1$ is true, statement- $2$ is true and statement- $2$ is correct explanation for statement- $1$ .

Statement- $1$ is true, statement- $2$ is true and statement- $2$ is NOT the correct explanation for statement- $1$ .

Statement- $1$ is true, statement- $2$ is false.

Statement- $1$ is false, statement- $2$ is true.

answer is C.

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Detailed Solution

Carbon reduction of $A l_{2} O_{3}$ occurs above $2000^{\circ} C$

Melting point of $A l$ $: 660^{\circ} C$

Boiling point of $A l : 2467^{\circ} C$

hence some $A l$ goes into vapourphase also a part of liberated $A l$ combines with liberated carbon and form $A l_{4} C_{3}$ , therefore carbon reduction of $A l_{2} O_{3}$ is difficult

$A l_{4} C_{3} + 12 H_{2} O \to 4 A l {(O H)}_{3} + 3 C H_{4}$

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