Question

A precipitate of ______ would be obtained on adding $HCl$ to a solution of $(S{b_2}{S_3})$ in yellow ammonium sulphide
(A) $S{b_2}{S_3}$
(B) $S{b_2}{S_5}$
(C) $SbS$
(D) $Sb{S_2}$

Answer 1

Hint: Let’s we talk about equilibrium some parts of equilibrium because our this question is to be a part of equilibrium So, when we do any numerical based question of equilibrium we have to determine the relative charges which help us to reach to the equilibrium after which we have to write the concentrations of equilibrium as in the ways of their charges after all solve for the change takes place and the concentrations of the equilibrium.

Complete Step by Step Solution:
As we know that when we react any substance with $HCl$ then the product will made and the ${H_2}S$
Gas will from as with the product.
So, as we do this question by using the reaction basis so,
We react out given $(S{b_2}{S_3})$ with $HCl$ then we get,
As taking yellow ammonium sulphide as $(S{b_2}{S_3})$with reaction to $HCl$ get the product as,
$2{(N{H_4})_3}Sb{S_4} + 6HCl\xrightarrow{{}}S{b_2}{S_5} + 6N{H_4}Cl + 3{H_2}S$
From the above equation we get that,
$S{b_2}{S_5}$ is a orange precipitate obtained in the product,
And we get the precipitate as $S{b_2}{S_5}$.
Hence, the blank will be filled with $S{b_2}{S_5}$.
Therefore, the correct answer is (B).

Note: In this question a term is used in “ yellow ammonium sulphide” as in the form of $(S{b_2}{S_3})$. The formula to find this reactant is ${(N{H_4})_2}{S_x}$ as putting all the values in this equation we get the first reactant and this is a very useful term and used in many questions of this chapter.