Compound A reacts with NH_4Cl and forms a com | vedclass

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(C) The reaction sequence is based on the Solvay process and the preparation of ammonia.$1$. $Ca(OH)_2 + 2NH_4Cl \longrightarrow CaCl_2 + 2NH_3 + 2H_2

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$Ca(OH)_2, NH_3, NH_4HCO_3$

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(C) The reaction sequence is based on the Solvay process and the preparation of ammonia.$1$. $Ca(OH)_2 + 2NH_4Cl \longrightarrow CaCl_2 + 2NH_3 + 2H_2O$Here,$A = Ca(OH)_2$ and $B = NH_3$.$2$. $NH_3 + H_2O + CO_2 	ext{ (excess)} \longrightarrow NH_4HCO_3$Here,$C = NH_4HCO_3$.$3$. $NH_4HCO_3 + NaCl \longrightarrow NaHCO_3 + NH_4Cl$Thus,the compounds are $A = Ca(OH)_2, B = NH_3, C = NH_4HCO_3$.

Compound $A$ reacts with $NH_4Cl$ and forms a compound $B$. Compound $B$ reacts with $H_2O$ and excess of $CO_2$ to form compound $C$,which on reaction with saturated $NaCl$ solution forms sodium hydrogen carbonate. Compounds $A, B$ and $C$ are respectively:

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