Question
A metal ' $X$' for crystalline ionic solid with dihydrogen. The solid produces hydrogen with $\mathrm{H}_2 \mathrm{O}$ and produces $\mathrm{H}_2$ gas and compound ' $Y$ ' is formed. ' $Y$ ' reacts with Aluminium metal and liberates $\mathrm{H}_2$ gas. ' $Y$ ' is also manufactured by electrolysis of brine solution.
Identify ' $X$ ' and ' $Y$ ' and write reaction involved.

Answer

$2\text{Na}+\text{H}_2\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }2\text{NaH}\\\ \ \ \text{'X'}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{sodium hydride}\\^\text{sodium}$
$2\text{NaH}+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ }2\text{NaOH}+3\text{H}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{'Y'}$
$2\text{Al}+2\text{NaOH}+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }2\text{NaAlO}_2+3\text{H}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{'Y'}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Sodium meta}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{aluminate}$
$2\text{H}_2\text{O}+2\text{NaCl}(\text{aq)}\xrightarrow[]{\text{electrolysis}}2\text{NaOH}+\text{H}_2+\text{Cl}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{'Y'}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Sodium hydroxide}$

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