Question

A white substance $(A)$ on heating with excess of dil. $HCl$ gave an offensive smelling gas $(B)$ and a solution $(C)$ . Solution $(C)$ on treatment with aqueous $N{H_3}$ did not give any precipitate but on treatment with $NaOH$ Solution gave a precipitate $(D)$ which dissolves in excess of $NaOH$ solution. $(A)$ on strong heating in air gave a strong smelling gas $(E)$ and a solid $(F)$ . Solid $(F)$ dissolved completely in $HCl$ and the solution gave a precipitate with $BaC{l_2}$ in acid solution. Identify $(A)$ to $(F)$ and write the chemical equations for various reactions involved.

Answer 1

We should know about the property of the chemicals used in it. For instance, consider the $HCl$ solution, we should know that if it is reacted with a white substance it will give an offensive smell and the same with other chemicals.

Complete step-by-step answer: If we see that it is give that, when solution $(C)$ is treated with $NaOH$ will give precipitate which is soluble in excess of $NaOH$ , So, the cation should be of amphoteric metal like $Zn$ or $Al$ .
Also, solid $(F)$ is given that it is completely soluble in $HCl$ and it will give white precipitate when reacted with $BaC{l_2}$ . Therefore, anion must be $S{O_4}^{2 - }$ ion.
Now, if we see $(A)$ , it gives offensive smelling gas. Hence, $(A)$ can be $ZnS$ or $A{l_2}{S_3}$ . But when we heat $A{l_2}{S_3}$ in air it does not form $A{l_2}{(S{O_4})_3}$ .
So, the chemical reactions are given below:
$ZnS + 2HCl \to ZnC{l_2} + {H_2}S\left( \uparrow \right)$
So, here $ZnS$ is a substance which results in a solution $(C)$ that is $ZnC{l_2}$ with an offensive smell $(B)$ that is ${H_2}S$ .
$ZnC{l_2} + 2NaOH \to Zn{\left( {OH} \right)_2}\left( \downarrow \right) + 2NaCl$
Now, when $(C)$ is reacted it gives precipitate $(D)$ that is $Zn{\left( {OH} \right)_2}$ .
Now, when $(D)$ is dissolved in excess of sodium hydroxide following reaction will take place:
$Zn{\left( {OH} \right)_2} + NaOH \to N{a_2}Zn{O_2} + 2{H_2}O$
Now, it’s given that $(A)$ is heated strongly heated in air which gave a smelling gas and solid $(F)$ as:
$ZnS + 3{O_2} \to 2ZnO + 2S{O_2}$
$ZnS + 2{O_2} \to ZnS{O_4}$
Hence, $(E)$ is $ZnO$ and $(F)$ is $ZnS{O_4}$ .
It is given that $(F)$ gave a precipitate with $BaC{l_2}$ as:
$ZnS{O_4} + BaC{l_2} \to BaS{O_4}\left( \downarrow \right) + ZnC{l_2}$
So, the substances from $(A)$ to $(F)$ are as follows:
$(A) = ZnS,\;(B) = {H_2}S,\;(C) = ZnC{l_2},\;(D) = Zn{(OH)_2},\;(E) = ZnO,\;(F) = ZnS{O_4}$

Note: When a compound reacts with another if it gives precipitate, then it means, it is from amphoteric metals which is cation. Also, the substance which is soluble in hydrogen chloride and gave precipitate when reacted with $BaC{l_2}$ then, the anion involved is definitely $S{O_4}^{2 - }$ ion.