Chemistry - s Block Elements Question with Solution | TestHub

Here$\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{l}}_2$is formed as a by-product
$\mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2\to \mathrm{N}{\mathrm{H}}_4\mathrm{H}\mathrm{C}{\mathrm{O}}_3$…(i)
$\mathrm{N}{\mathrm{a}}^{+}+\mathrm{C}{\mathrm{l}}^{-}+\mathrm{N}{\mathrm{H}}_4^{+}+\mathrm{H}\mathrm{C}{\mathrm{O}}_3^{-}\to \underset{(\mathrm{p}\mathrm{p}\mathrm{t}.)}{\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{C}{\mathrm{O}}_3}+\mathrm{N}{\mathrm{H}}_4^{+}+\mathrm{C}{\mathrm{l}}^{-}$…(ii)
For step (i),$\mathrm{C}{\mathrm{O}}_2$is produced by heating limestone.
$\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{O}}_3\to \mathrm{C}\mathrm{a}\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$
CaO so-obtained is slaked with${\mathrm{H}}_2\mathrm{O}$and$\mathrm{N}{\mathrm{H}}_4\mathrm{C}\mathrm{l}$produced in step (ii) reacts to form$\mathrm{C}\mathrm{a}\mathrm{C}{\mathrm{l}}_2$as a by-product.
$\text{CaO}\text{+}{\text{H}}_{\text{2}}\text{O}\to {\text{Ca(OH)}}_{\text{2}}\stackrel{{\text{NH}}_{\text{4}}\text{Cl}}{\to }{\text{CaCl}}_{\text{2}}\text{+}{\text{2NH}}_{\text{3}}\text{+}{\text{2H}}_{\text{2}}\text{O}$