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 $K _4 Fe(CN) _6$ solution. $A$ dissolves in dil. $HCl$ 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 $NaOH$ 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 $ZnO$. $C$ with dil. $HCl$ and $K _4\left[Fe(CN) _6\right]$ gives white ppt. This confirms that the compound $C$ must be $ZnO$.
$ A \xrightarrow{\Delta} \underset{‘C’}{\mathrm{ZnO}} +B \text { (gas) } $
$ \underset{C}{\mathrm{ZnO}}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{O} $
$ 2 \mathrm{ZnCl}_2+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow 4 \mathrm{KCl}+ \underset{\text { White ppt}}{\mathrm{Zn}_2[\mathrm{Fe}(\mathrm{CN})_6]} \downarrow $
(ii) The gas $B$ turns lime water milky and milkiness disappear with continuous passage of gas. Hence, the gas is $CO _2$ and compound $A$ in $ZnCO _3$.
$\underset{B}{\mathrm{CO}_2}+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CaCO}_3 \downarrow $
$\mathrm{CaCO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 $
$\underset{A}{\mathrm{ZnCO}_3} \xrightarrow{\Delta} \underset{C} {\mathrm{ZnO}}+\underset{B}{\mathrm{CO}_2}$
(iii) The solution of $A$ gives white ppt of $ZnS D$ with $NH _4 OH$ and excess of $H _2 S$.
$ \mathrm{ZnCO}_3+\mathrm{HCl} \longrightarrow \underset{B}{\mathrm{CO}_2} \uparrow+\mathrm{ZnCl}_2 $
$ \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{~S} \xrightarrow{\mathrm{NH}_4 \mathrm{OH}} 2 \mathrm{HCl}+\underset{D}{\mathrm{ZnS}} \downarrow \text { (white) }$
(iv) The solution of $A$ also gives initially a white ppt $E$ with $NaOH$, which dissolve in excess of reagent.
$ \mathrm{ZnCl}_2+2 \mathrm{NaOH} \longrightarrow \underset{E \text { (white) }}{\mathrm{Zn}(\mathrm{OH})_2} \downarrow+2 \mathrm{NaCl} $
$ \mathrm{Zn}(\mathrm{OH})_2+2 \mathrm{NaOH} \longrightarrow \underset{\text{Soluble}}{ \mathrm{Na}_2[\mathrm{Zn}(\mathrm{OH})_4]} $
BETA
