Question

A white substance $A$ reacts with dilute $H_2 SO_4$ to produce a colourless gas $B$ and a colourless solution $C$. The reaction between $B$ and acidified $K_2 Cr_2 O_7$ solution produces a green solution and a slightly coloured precipitate $D$. The substance $D$ burns in air to produce a gas $E$ which reacts with $B$ to yield $D$ and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous $NH_3$ or $NaOH$ to $C$ produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify $A, B, C, D$ and $E$. Write the equations of the reactions involved

Answer 1

Since, the white substance $A$ gives a colourless gas $B$ with dil, $H_2 SO_4$, such gas may be $H_2 S$ So, the substance $A$ may be a metal sulphide ($Na/K/Zn$, etc.) When $H_2 S$ gas reacts with acidified $K_2 Cr_2 O_7$, it gives green coloured solution of $Cr_2(SO_4)_3$ along with slightly yellow ppt of $D$ as sulphur. $K_2 Cr_2 O_7 + 4H_2 SO_4 + 3H_2 S \longrightarrow \, K_2 SO_4 + \underset{\text{Green}}{Cr_2 (SO_4)_3}$ $ + \underset{D}{3S} + 7H_2$ $S$ on burning in air gives $SO_2 (E)$. Substance $E$ on reaction with $B (H_2S)$ produces $D (s)$ : $ \underset {B}{2H_2 S} + \underset {E}{SO_2} \longrightarrow \, 2H_2 O (l) + \underset {D}{3S \, \downarrow}$ Anhydrous $CuSO_4$ produces blue colour in water. Solution $C$ produces ppt first with $NH_3 / NaOH$ which dissolve in excess $NH_3 / NaOH.$ Hence, A must be $ZnS$. $ \underset {A}{ZnS} + dil. H_2 SO_4 \longrightarrow \, \underset{C}{ZnSO_4 \, (aq)} + H_2 S (g)$ $\underset{C}{ZnSO_4} + 2NaOH \longrightarrow \underset{\text{White}}{Zn(OH)_2 \, \downarrow} + Na_2 SO_4$ $ Zn(OH)_2 + 2NaOH \longrightarrow \, [Zn(OH)_4]^{2-} + 2Na^+$