Compound A reacts with NH_4Cl and forms a compound B.… — JEE Main 2023 Chemistry PYQ

JEE Main 2023 · 25 Jan · Shift 1

Compound A reacts with

\text{NH}_4\text{Cl}

and forms a compound B. Compound B reacts with

\text{H}_2\text{O}

and excess of

\text{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.

A.
$\text{Ca(OH)}_2$ , $\text{NH}_4^\oplus$ , $(\text{NH}_4)_2\text{CO}_3$
B.
$\text{Ca(OH)}_2$ , $\text{NH}_3$ , $\text{NH}_4\text{HCO}_3$
C.
$\text{CaCl}_2$ , $\text{NH}_3$ , $\text{NH}_4\text{HCO}_3$
D.
$\text{CaCl}_2$ , $\text{NH}_4^\oplus$ , $(\text{NH}_4)_2\text{CO}_3$

Show answer and solution

Correct option: B

Solution

This is the Solvay process. Ca(OH)₂ (A) reacts with NH₄Cl to produce NH₃ (B): Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O. NH₃ (B) then reacts with H₂O and excess CO₂ to form NH₄HCO₃ (C): NH₃ + H₂O + CO₂ → NH₄HCO₃. Finally, NH₄HCO₃ passed through saturated NaCl solution gives NaHCO₃ (sodium hydrogen carbonate) because NaHCO₃ is less soluble and precipitates out, which is the key step of the Solvay process. Hence A = Ca(OH)₂, B = NH₃, C = NH₄HCO₃.

[JEE Main 2023 · 25 Jan · Shift 1]

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