When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A,B,C and D. Explain why the milkiness disppears in the last step.

Explanation

When `CO_(2)` is passed through solution of compound (B) , it turns milky , therefore , solution of compound (B) must be lime water , i.e., `Ca(OH)_(2)` solution and milkiness must be due to the formation of insoluble `CaCO_(3)` (C) . Further since , compound (B) is obtained by dissolving compound (A) in water , therefore , compound (A) must be quick lime , CaO
`underset((A)) underset("Calcium oxide")(CaO) + H_(2)O underset("Lime water (B)")(Ca(OH)_(2)) , CO_(2) to underset("Milkiness)")underset("carbonate (C)")underset("Calcium")(CaCO_(3)) + H_(2)O`
When excess of `CO_(2)` is passed milkiness disappears due to formation of soluble calcium bicarbonate (D)
`underset("Milkiness (C)")(CaCO_(3)) + CO_(2) + H_(2)O to underset("(Soluble in" H_(2)O")") underset("Calcium bicarbonate")(Ca(HCO_(3))_(2))`
Thus , compound (A) = CaO , Compound (B) = `Ca(OH)_(2)` , Compound (C) = `CaCO_(3)` and compound (D) = `Ca(HCO_(3))_(2)`.