Qualitative Analysis - Result Question 62
65. An orange solid $A$ on heating gave a green residue $B$, a colourless gas $C$ and water vapour. The dry gas $C$ on passing over heated $Mg$ gave a white solid $D . D$ on reaction with water gave a gas $E$ which formed dense white fumes with $HCl$. Identify $A$ to $E$ and give the reaction involved.
(1993, 3M)
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Solution:
$\underset{\text { Orange solid }}{A} \stackrel{\Delta}{\longrightarrow} \underset{\text { Green }}{B \downarrow}+\underset{\text { Colourless }}{C \uparrow}+H _2 O(v) \uparrow$
$C+Mg \longrightarrow D \text { (white solid) }$
$D+H _2 O \longrightarrow E(g) \xrightarrow{HCl}$ White fumes.
Hence, $E$ must be ammonia gas so $D$ must be $Mg _3 N _2$ and $C$ is $N _2(g)$. This $N _2$ is obtained on strong heating of $\left(NH _4\right) _2 Cr _2 O _7$ because $\left(NH _4\right) _2 Cr _2 O _7$ is orange solid and produces green $Cr _2 O _3$ residue on heating.
$ \underset{\text { Orange (A)}}{(NH _4) _2 Cr _2 O _7 }\stackrel{\Delta}{\longrightarrow} \underset{\text { Green (B)}}{Cr _2 O _3}+\underset{C}{N _2}+4 H _2 O $
$ N _2+Mg \stackrel{\Delta}{\longrightarrow} \underset{D}{Mg _3 N _2} \xrightarrow{H _2 O} \underset{E}{NH_3} + {Mg (OH) _2} $
$ \underset{E}{NH _3}+HCl \longrightarrow NH _4 Cl \text { (white fumes) } $
BETA
