JEE Main 2023 — S Block Elements Question with Solution

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

CaO (s) + H${}_2$O (l) $\to$ 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) $\to$ 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) $\to$ Ca(HCO${}_3$)${}_2$ (aq)

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

Option D : Slaked lime