Appearance of milkiness on passing `CO_(2)` in the solution of compound B indicates that compound B is lime water and compound C is `CaCO_(3)`. Since, compound B is obtained by adding `H_(2)O` to compound A, therefore, compound A is quicklime, `CaO`.
The reactions are as follows
(i) `underset((A))underset(“Calcium oxide”)(CaO)+H_(2)O rarr underset((B))underset(“Lime water”)(Ca(OH)_(2))`
(ii) `underset((B))(Ca(OH)_(2))+CO_(2)rarr underset((“Milkiness”))underset(“Calcium carbonate”)(CaCO_(3))+underset((C))(H_(2)O)`
(iii) When excess of `CO_(2)` is passed, milkiness disappears due to the formation of soluble calcium bicarbonate (D).
`underset((C))underset(“Milkiness”)(CaCO_(3))+CO_(2)+H_(2)Orarr underset((D))underset((“SOluble in”H_(2)O))underset(“Calcium bicarbonate”)(Ca(HCO_(3))_(2))`

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)`.