When carbon dioxide gas is passed through clear lime water (aqueous calcium hydroxide solution), what insoluble compound is formed that turns the solution milky?

Introduction / Context:
Passing carbon dioxide through lime water is a classic school experiment used to test for the presence of CO2 gas. The reaction produces a characteristic milky appearance in the lime water, which is due to formation of an insoluble compound. Knowing the name and formula of this compound is important for understanding acid–base reactions and carbon dioxide detection tests. This question asks you to identify the compound that forms.

Given Data / Assumptions:

• Lime water is an aqueous solution of calcium hydroxide, Ca(OH)2.
• Carbon dioxide gas, CO2, is bubbled through this solution.
• A white, insoluble solid forms, making the solution appear milky.
• We are to choose the correct name for this solid from the options provided.

Concept / Approach:
The reaction involved is between an acidic oxide, carbon dioxide, and a basic hydroxide, calcium hydroxide. The net result is the formation of a salt and water. In this case, the salt is calcium carbonate, CaCO3, which is only sparingly soluble in water and appears as a fine white precipitate. The balanced equation is: CO2 (g) + Ca(OH)2 (aq) → CaCO3 (s) + H2O (l) The milky appearance of lime water is therefore due to suspended particles of calcium carbonate, confirming the presence of CO2.

Step-by-Step Solution:
Step 1: Identify lime water as Ca(OH)2 in aqueous solution. Step 2: Recognise that CO2 is an acidic oxide that can react with bases to form carbonates. Step 3: Write the reaction between CO2 and Ca(OH)2 to obtain CaCO3 and H2O. Step 4: Note that CaCO3 is an insoluble salt that forms a white precipitate. Step 5: Conclude that the milky appearance is due to calcium carbonate and select it from the options.

Verification / Alternative check:
If CO2 is bubbled through lime water for a short time, the mixture turns milky due to formation of CaCO3. If CO2 is passed for a longer time, the solution can turn clear again because CaCO3 dissolves to form soluble calcium bicarbonate, Ca(HCO3)2. This well known demonstration appears in school textbooks under the heading "test for carbon dioxide." None of the other listed compounds are formed in this simple reaction, which confirms calcium carbonate as the product that causes the milkiness.

Why Other Options Are Wrong:
Copper sulphate is CuSO4 and does not form in this reaction because there is no copper or sulphate involved. Magnesium oxide, MgO, is another basic oxide and has no role here. Baking soda is sodium bicarbonate, NaHCO3, and would require sodium ions and bicarbonate formation, which are not part of the lime water system. Calcium chloride would require chloride ions, which are not present. Thus, the only plausible product is calcium carbonate.

Common Pitfalls:
Students sometimes confuse calcium carbonate with sodium bicarbonate because both involve carbonates, or they may forget that lime water contains calcium, not sodium or magnesium. Another mistake is to think that the white substance is magnesium oxide simply because it is a white powder in other contexts. To avoid confusion, remember that lime water is specifically calcium hydroxide solution and that CO2 plus Ca(OH)2 gives CaCO3 as the milky precipitate.

Final Answer:
When carbon dioxide is passed through lime water, the milky precipitate that forms is calcium carbonate.