Question:
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
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
Updated On: Feb 4, 2026
- $CaCl _2, NH _4^{+},\left( NH _4\right)_2 CO _3$
- $CaCl _2, NH _3, NH _4 HCO _3$
- $Ca ( OH )_2, NH _3, NH _4 HCO _3$
- $Ca ( OH )_2, NH _4^{+},\left( NH _4\right)_2 CO _3$
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The Correct Option is C
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1. Reaction 1: Ca(OH)$_2$ reacts with NH$_3$Cl to form NH$_3$, CaCl$_2$, and H$_2$O.
2. Reaction 2: NH$_3$ reacts with H$_2$O and CO$_2$ to form NH$_4$HCO$_3$.
Thus, A = Ca(OH)$_2$, B = NH$_3$, and C = NH$_4$HCO$_3$.
Ammonium bicarbonate (NH$_4$HCO$_3$) is formed when ammonia reacts with water and carbon dioxide, a key step in the Solvay process.
2. Reaction 2: NH$_3$ reacts with H$_2$O and CO$_2$ to form NH$_4$HCO$_3$.
Thus, A = Ca(OH)$_2$, B = NH$_3$, and C = NH$_4$HCO$_3$.
Ammonium bicarbonate (NH$_4$HCO$_3$) is formed when ammonia reacts with water and carbon dioxide, a key step in the Solvay process.
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Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
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- Group 13 Elements: Boron family
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