Question

Find the value of x and y.
$AgN{O_3} + xHCl \to AgCl + HN{O_3}$
$AgCl + N{H_y}OH \to [Ag{(N{H_3})_2}Cl] + N{H_4}Cl$
A. $x = 2,y = 3$
B. $x = 1,y = 4$
C. $x = 1,y = 3$
D. $x = 2,y = 4$

Answer 1

According to the law of conservation of mass, mass can neither be created nor be destroyed, that is why we balance every chemical equation. Precipitate of silver chloride is soluble in the solution of ammonium hydroxide.

Complete step by step answer:
When hydrochloric acid is added to the solution of a nitrate of silver(I), the corresponding silver chloride is formed as a curdy white precipitate. It is a recommended test for hydrochloric acid in laboratories. It is a balanced reaction such that the stoichiometric coefficients of each compound is one.
$AgN{O_3} + xHCl \to AgCl + HN{O_3}$ where $x = 1$ as it is a balanced equation in which right hand side number of atoms are equal to left hand side number of atoms of each species.
The precipitate of silver chloride when dissolved in ammonium hydroxide results in formation of a complex compound called diamine silver(I) chloride. The white precipitate can be recovered by adding nitric acid.
$AgCl + N{H_y}OH \to [Ag{(N{H_3})_2}Cl] + N{H_4}Cl$ where $y = 4$ as it will be ammonium hydroxide in which the curdy white precipitate of silver chloride is soluble. Thus, it is a precipitation reaction.
Both the equations can be rewritten as:
$AgN{O_3} + HCl \to AgCl + HN{O_3}$
$AgCl + N{H_4}OH \to [Ag{(N{H_3})_2}Cl] + N{H_4}Cl$

Hence, the correct option is (B).

Note:
Ammonium hydroxide is a weak base prepared by dissolving anhydrous ammonia into water. On adding excess ammonium hydroxide to the silver chloride precipitate, it dissociates into cation ${[Ag{(N{H_3})_2}]^ + }$ and anion $C{l^ - }$ as by-product. These are not stable products and thus revert back to silver chloride and ammonia will be given off as gas.