What are the products formed in sequence when excess of CO$$_2$$ is passed in slaked lime?

First, recall that slaked lime is chemically represented as $$\text{Ca(OH)}_2$$.

When carbon dioxide gas $$\text{CO}_2$$ is bubbled through slaked lime, the very first reaction that takes place is the formation of calcium carbonate. We state the reaction:

$$\text{Ca(OH)}_2 \;+\; \text{CO}_2 \;\longrightarrow\; \text{CaCO}_3 \;+\; \text{H}_2\text{O}$$

Here, the hydroxide ion $$\text{OH}^-$$ of calcium hydroxide combines with the acidic oxide $$\text{CO}_2$$ to give the neutral salt calcium carbonate $$\text{CaCO}_3$$ along with water $$\text{H}_2\text{O}$$. So the first solid product deposited in the medium is calcium carbonate.

Now we are told in the problem that an excess of carbon dioxide is passed. Once all the available $$\text{Ca(OH)}_2$$ has changed to $$\text{CaCO}_3$$, the additional $$\text{CO}_2$$ does not remain idle; it reacts further with the freshly formed calcium carbonate in the presence of water. The well-known carbonation reaction is:

$$\text{CaCO}_3 \;+\; \text{CO}_2 \;+\; \text{H}_2\text{O} \;\longrightarrow\; \text{Ca(HCO}_3)_2$$

This equation shows that calcium carbonate absorbs another molecule of carbon dioxide and one molecule of water to produce calcium hydrogen carbonate, written as $$\text{Ca(HCO}_3)_2$$. Hydrogen carbonate is soluble in water, so the white precipitate of $$\text{CaCO}_3$$ gradually disappears as it converts into the soluble bicarbonate.

Thus, in chronological order, the sequence of products obtained is

$$\boxed{\text{First: } \text{CaCO}_3 \quad\longrightarrow\quad \text{Then: } \text{Ca(HCO}_3)_2}$$

Comparing this sequence with the options provided, we find that Option C lists exactly this order: “CaCO$$_3$$, Ca(HCO$$_3$$)$$_2$$”.

Hence, the correct answer is Option C.