A white substance (A) on heating with excess of dil HCI gave an offensive smelling gas (B) ans 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 HCI and the solution gave a precipitate with ${B a C I}_{2}$ in acid solution .Identify A to F and write the chemical equations for various reactions involved.

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Solution

Solution:-
Solution $C$ gives precipitate with $N a O H$ solution which is soluble in excess of $N a O H$ solution, hence the cation should be of amphoteric metal like $Z n$ or $A l$ . Again solid $F$ is soluble in $H C l$ and gives white precipitate with $B a {C l}_{2}$ . Therefore anion must be ${S O_{4}}^{2 -}$ ion.
Now $A$ gives offensive smelling gas hence the $A$ may be $Z n S$ or ${A l}_{2} S_{3}$ . But ${A l}_{2} S_{3}$ on heating in air does not form ${A l}_{2} {(S O_{4})}_{3}$ .
$Z n S (A) + 2 H C l ⟶ Z n {C l}_{2} (C) + H_{2} S (B) ↑$
$Z n {C l}_{2} (C) + 2 N a O H ⟶ Z n {(O H)}_{2} (D) ↓ + 2 N a C l$
$D$ dissolves in excess of $N a O H$ .
$Z n {(O H)}_{2} (D) + N a O H (excess) ⟶ {N a}_{2} Z n O_{2} + 2 H_{2} O$
$A$ on strong heating in air gave a strong smelling gas $E$ and a solid $F$ .
$Z n S (A) + 3 O_{2} ⟶ 2 Z n O (E) + 2 S O_{2}$
$Z n S (A) + 2 O_{2} ⟶ Z n S O_{4} (F)$
$F$ gave a precipitate with $B a {C l}_{2}$ .
$Z n S O_{4} (F) + B a {C l}_{2} ⟶ B a S O_{4} ↓ (ppt.) + Z n {C l}_{2}$
Hence
$A \to Z n S$
$B \to H_{2} S$
$C \to Z n {C l}_{2}$
$D \to Z n {(O H)}_{2}$
$E \to Z n O$
$F \to Z n S O_{4}$

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