An alkali metal $A$ gives a compound $B$ (molecular mass $= 40$) on reacting with water. The compound $B$ gives a soluble compound $C$ on treatment with aluminium oxide. Identify $A$,$B$ and $C$ and give the reaction involved.
(A) The alkali metal $A$ is Sodium $(Na)$.
When $Na$ reacts with water,it forms Sodium hydroxide $(NaOH)$ as compound $B$.
The molecular mass of $NaOH = 23 + 16 + 1 = 40$.
When $NaOH$ $(B)$ reacts with aluminium oxide $(Al_{2}O_{3})$,it forms Sodium aluminate $(NaAlO_{2})$ as compound $C$,which is soluble in water.
The reactions are:
$2Na + 2H_{2}O \rightarrow 2NaOH + H_{2}$
$Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O$
When $Na$ reacts with water,it forms Sodium hydroxide $(NaOH)$ as compound $B$.
The molecular mass of $NaOH = 23 + 16 + 1 = 40$.
When $NaOH$ $(B)$ reacts with aluminium oxide $(Al_{2}O_{3})$,it forms Sodium aluminate $(NaAlO_{2})$ as compound $C$,which is soluble in water.
The reactions are:
$2Na + 2H_{2}O \rightarrow 2NaOH + H_{2}$
$Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O$
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