Compound A reacts with NH$_4$Cl and forms a compound B. Compound B reacts with H$_2$O and excess of CO$_2$ to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound A, B and C, are respectively :
Solution
$\underset{(B)}{\mathrm{Ca}(\mathrm{OH})_{2}}+\mathrm{NH}_{4} \mathrm{Cl} \longrightarrow \mathrm{CaCl}_{2}+\underset{(B)}{\mathrm{NH}_{3}}+\mathrm{H}_{2} \mathrm{O}$
<br/><br/>
$$
\begin{aligned}
& \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O}+\underset{\text { Excess }}{\mathrm{CO}_{2}} \longrightarrow \underset{(C)}{\mathrm{NH}_{4} \mathrm{HCO}_{3}} \\\\
& \mathrm{NH}_{4} \mathrm{HCO}_{3}+\mathrm{NaCl} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}+\mathrm{NaHCO}_{3}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
This question is part of PrepWiser's free JEE Main question bank. 300 more solved questions on p-Block Elements are available — start with the harder ones if your accuracy is >70%.