When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown…

Step 1: Identify gas X

When an ammonium salt (containing $\ce{NH4+}$) is treated with a strong base like sodium hydroxide ($\ce{NaOH}$) and heated, ammonia gas ($\ce{NH3}$) is evolved.

$\ce{NH4+ + OH- -> NH3 ^ + H2O}$

So, the gas X is $\ce{NH3}$.

Step 2: Identify reagent Y and the brown precipitate

The standard test for ammonia gas is its reaction with Nessler's reagent, which is an alkaline solution of potassium tetraiodomercurate(II) ($\ce{K2HgI4}$). When ammonia gas is passed through Nessler's reagent, a brown precipitate of iodide of Millon's base is formed.

Reagent Y = $\ce{K2HgI4} + \ce{KOH}$

$\ce{2K2HgI4 + NH3 + 3KOH -> HgO.Hg(NH2)I v + 7KI + 2H2O}$ The compound $\ce{HgO.Hg(NH2)I}$ is the brown precipitate.

Step 3: Conclusion

Therefore, X is $\ce{NH3}$ and Y is $\ce{K2HgI4} + \ce{KOH}$.

Key Points to Remember:

• Any ammonium salt releases ammonia gas when reacted with a strong base.
• Nessler's reagent is alkaline $\ce{K2HgI4}$ ($\ce{K2HgI4} + \ce{KOH}$ or $\ce{NaOH}$).
• The brown precipitate with Nessler's reagent is called iodide of Millon's base.