When water is added to compound $(A)$ of calcium,a 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,the milkiness disappears due to the formation of compound $(D)$. Identify the compounds $A$,$B$,$C$,and $D$. Explain why the milkiness disappears in the last step.

(A) Compound $(A)$ is $CaO$ (Calcium oxide). When water is added to it,it forms compound $(B)$,which is $Ca(OH)_2$ (Calcium hydroxide or lime water).
$(i)$ $CaO + H_2O \rightarrow Ca(OH)_2$
When $CO_2$ is passed through $(B)$,it forms $CaCO_3$ (Calcium carbonate),which is compound $(C)$ and is insoluble in water,causing the solution to turn milky.
$(ii)$ $Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$
When excess $CO_2$ is passed,$CaCO_3$ reacts with $CO_2$ and water to form $Ca(HCO_3)_2$ (Calcium bicarbonate),which is compound $(D)$. Since $Ca(HCO_3)_2$ is soluble in water,the milkiness disappears.
$(iii)$ $CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2$