"A" obtained by Ostwald's method involving air oxidation of NH$_3$, upon further air oxidation produces "B", "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test.
Identify A and B, respectively :
Solution
Ostwald's process is :
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$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow \underset{(A)}{4 \mathrm{NO}}+6 \mathrm{H}_{2} \mathrm{O}$
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$\underset{\text { (A) }}{2 \mathrm{NO}}+\mathrm{O}_{2} \longrightarrow \underset{\text { (B) }}{2 \mathrm{NO}_{2}}$
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$4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \longrightarrow 4 \mathrm{HNO}_{3}$
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$\therefore \mathrm{A}$ and $\mathrm{B}$ are $\mathrm{NO}$ and $\mathrm{NO}_{2}$ respectively
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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