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 $E$.
$(1979,4 M)$
Show Answer
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 $ZnO$.
$$ \begin{aligned} & A \stackrel{\Delta}{\longrightarrow} \underset{{ }^{\prime} C}{ZnO}+B \text { (gas) } \ & ZnO _C+2 HCl \longrightarrow ZnCl _2+H _2 O \ & 2 ZnCl _2+K _4\left[Fe(CN) _6\right] \longrightarrow 4 KCl+Zn _2\left[Fe(CN) _6\right] \downarrow \ & \text { White ppt } \end{aligned} $$
(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 $ZnCO _3$.
$$ \begin{aligned} CO _2+Ca(OH) _2 & \longrightarrow H _2 O+CaCO _3 \downarrow \ CaCO _3+CO _2+H _2 O & \longrightarrow Ca\left(HCO _3\right) _2 \ ZnCO _3 & \stackrel{\Delta}{\longrightarrow} ZnO+\underset{B}{CO _2} \end{aligned} $$
(iii) The solution of $A$ gives white ppt of $ZnS D$ with $NH _4 OH$ and excess of $H _2 S$.
$$ \begin{aligned} & ZnCO _3+HCl \longrightarrow \underset{B}{CO _2 \uparrow}+ZnCl _2 \ & ZnCl _2+H _2 S \xrightarrow{NH _4 OH} \underset{D}{2 HCl+\underset{D}{ZnS \downarrow} \text { (white) }} \end{aligned} $$
(iv) The solution of $A$ also gives initially a white ppt $E$ with $NaOH$, which dissolve in excess of reagent.
$$ \begin{aligned} ZnCl _2+2 NaOH & \underset{E(\text { white })}{Zn(OH) _2 \downarrow+2 NaCl} \ Zn(OH) _2+2 NaOH & \longrightarrow \underset{\text { Soluble }}{Na _2\left[Zn(OH) _4\right]} \end{aligned} $$
