QJEEM23Ch033b: JEE Main 2023 Chemistry - Class 11 JEEM23Ch033b Solution: Lime reacts exothermally with water to give 'A' which has low…

Lime (Calcium oxide, CaO) reacts exothermally with water to form slaked lime (Calcium hydroxide, Ca(OH) $_2$ ):

CaO (s) + H $_2$ O (l) $\rightarrow$ Ca(OH) $_2$ (s)

This slaked lime has low solubility in water and is used for the test of carbon dioxide (CO $_2$ ). In this test, the carbon dioxide reacts with the calcium hydroxide to form calcium carbonate (CaCO $_3$ ), an insoluble white precipitate:

Ca(OH) $_2$ (aq) + CO $_2$ (g) $\rightarrow$ CaCO $_3$ (s) + H $_2$ O (l)

Further reaction of calcium carbonate with carbon dioxide forms calcium bicarbonate [Ca(HCO $_3$ ) $_2$ ], a soluble compound:

CaCO $_3$ (s) + CO $_2$ (g) + H $_2$ O (l) $\rightarrow$ Ca(HCO $_3$ ) $_2$ (aq)

Thus, 'A' in this case is slaked lime. So, the correct option is:

Option D : Slaked lime