SATHEE: Qualitative Analysis - Result Question 79

Qualitative Analysis - Result Question 79

####82. A white amorphous powder $A$ on heating yields a colourless, non-combustible gas $B$ and a solid $C$ . The later compound assumes a yellow colour on heating and changes to white on cooling. $C$ dissolves in dilute hydrochloric acid and the resulting solution gives a white precipitate with $K_{4} F e (C N)_{6}$ solution. $A$ dissolves in dil. $H C l$ with the evolution of gas, which is identical in all respect with $B$ .

The gas $B$ turns lime water milky, but the milkiness disappears with the continuous passage of gas. The solution of $A$ as obtained above, gives a white ppt $E$ on addition of $N a O H$ solution, which dissolves on further addition of base. Identify the compounds $A, B, C, D$ and $E$ .

$(1979, 4 M)$

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Solution:

(i) The compound $C$ produced by heating $A$ is white in colour and changes to yellow on heating, thus compound $C$ may be $Z n O$ . $C$ with dil. $H C l$ and $K_{4} [F e (C N)_{6}]$ gives white ppt. This confirms that the compound $C$ must be $Z n O$ .

$\begin{aligned} A \overset{Δ}{⟶} \underset{^{'} C}{Z n O} + B (gas) & Z n O_{C} + 2 H C l ⟶ Z n C l_{2} + H_{2} O & 2 Z n C l_{2} + K_{4} [F e (C N)_{6}] ⟶ 4 K C l + Z n_{2} [F e (C N)_{6}] ↓ & White ppt \end{aligned}$

(ii) The gas $B$ turns lime water milky and milkiness disappear with continuous passage of gas. Hence, the gas is $C O_{2}$ and compound $A$ in $Z n C O_{3}$ .

$\begin{aligned} C O_{2} + C a (O H)_{2} & ⟶ H_{2} O + C a C O_{3} ↓ C a C O_{3} + C O_{2} + H_{2} O & ⟶ C a {(H C O_{3})}_{2} Z n C O_{3} & \overset{Δ}{⟶} Z n O + \underset{B}{C O_{2}} \end{aligned}$

(iii) The solution of $A$ gives white ppt of $Z n S D$ with $N H_{4} O H$ and excess of $H_{2} S$ .

$\begin{aligned} Z n C O_{3} + H C l ⟶ \underset{B}{C O_{2} ↑} + Z n C l_{2} & Z n C l_{2} + H_{2} S \overset{N H_{4} O H}{\to} \underset{D}{2 H C l + \underset{D}{Z n S ↓} (white)} \end{aligned}$

(iv) The solution of $A$ also gives initially a white ppt $E$ with $N a O H$ , which dissolve in excess of reagent.

$\begin{aligned} Z n C l_{2} + 2 N a O H & \underset{E (white)}{Z n (O H)_{2} ↓ + 2 N a C l} Z n (O H)_{2} + 2 N a O H & ⟶ \underset{Soluble}{N a_{2} [Z n (O H)_{4}]} \end{aligned}$