$A$ certain salt $X$ gives the following results:
$(i)$ Its aqueous solution is alkaline to litmus.
$(ii)$ It swells up to a glassy material $Y$ on strong heating.
$(iii)$ When conc. $H_{2}SO_{4}$ is added to a hot solution of $X,$ white crystals of an acid $Z$ separate out.
Write equations for all the above reactions and identify $X, Y,$ and $Z$.

(N/A) The given salt is alkaline to litmus. Therefore,$X$ is a salt of a strong base and a weak acid. Also,when $X$ is strongly heated,it swells to form substance $Y$. Therefore,$X$ must be borax $(Na_{2}B_{4}O_{7} \cdot 10H_{2}O)$.
$(i)$ Aqueous solution of borax is alkaline due to hydrolysis:
$Na_{2}B_{4}O_{7} + 7H_{2}O \rightarrow 2NaOH + 4H_{3}BO_{3}$
$(ii)$ On heating,borax loses water and swells to form a glassy bead $Y$ consisting of sodium metaborate $(NaBO_{2})$ and boric anhydride $(B_{2}O_{3})$:
$Na_{2}B_{4}O_{7} \cdot 10H_{2}O$ $\xrightarrow{\Delta} Na_{2}B_{4}O_{7}$ $\xrightarrow{\Delta} 2NaBO_{2} + B_{2}O_{3} (Y)$
$(iii)$ When concentrated $H_{2}SO_{4}$ is added to a hot solution of borax,white crystals of orthoboric acid $(Z)$ are formed:
$Na_{2}B_{4}O_{7} + H_{2}SO_{4} + 5H_{2}O \rightarrow Na_{2}SO_{4} + 4H_{3}BO_{3} (Z)$
Thus,$X = Na_{2}B_{4}O_{7} \cdot 10H_{2}O$,$Y = NaBO_{2} + B_{2}O_{3}$,and $Z = H_{3}BO_{3}$.