Lime reacts exothermally with water to give 'A' which has low solubility in water. Aqueous solution of 'A' is often used for the test of CO$_2$, a test in which insoluble B is formed. If B is further reacted with CO$_2$ then soluble compound is formed, 'A' is :

<p>Lime (Calcium oxide, CaO) reacts exothermally with water to form slaked lime (Calcium hydroxide, Ca(OH)$_2$):</p> <p>CaO (s) + H$_2$O (l) $\rightarrow$ Ca(OH)$_2$ (s)</p> <p>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:</p> <p>Ca(OH)$_2$ (aq) + CO$_2$ (g) $\rightarrow$ CaCO$_3$ (s) + H$_2$O (l)</p> <p>Further reaction of calcium carbonate with carbon dioxide forms calcium bicarbonate [Ca(HCO$_3$)$_2$], a soluble compound:</p> <p>CaCO$_3$ (s) + CO$_2$ (g) + H$_2$O (l) $\rightarrow$ Ca(HCO$_3$)$_2$ (aq)</p> <p>Thus, &#39;A&#39; in this case is slaked lime. So, the correct option is:</p> <p>Option D : Slaked lime</p>