`A overset("Calcination")rarr CaO + MgO + underset("Colourless gas")((B))`
When the colourless gas `(B)` is passed through lime water, initially a milky solution is obtained and on excess passage of gas `(B)` through the lime water, milkiness disappears.
Name of the product `(A)` is :

Options

A. Dolomite
B. Limenstone
C. Magnesite
D. Malachite

Explanation

To solve the question, we need to identify the compound \( A \) that, upon calcination, produces \( CaO \), \( MgO \), and a colorless gas \( B \). The behavior of gas \( B \) when passed through lime water will help us identify it.

### Step-by-step Solution:

1. **Understanding Calcination**:
- Calcination is a thermal treatment process in the absence of air, often used to extract metals from their ores. In this case, we are looking for a compound that, when calcined, yields \( CaO \) (calcium oxide) and \( MgO \) (magnesium oxide).

2. **Identifying the Products**:
- The products of calcination are \( CaO \) and \( MgO \). This suggests that the original compound \( A \) must contain both calcium and magnesium in its formula.

3. **Considering Possible Compounds**:
- A common compound that contains both calcium and magnesium is dolomite, which has the formula \( CaCO_3 \cdot MgCO_3 \).

4. **Writing the Calcination Reaction**:
- The reaction for the calcination of dolomite can be written as:
\[ CaCO_3 \cdot MgCO_3 \xrightarrow{\text{calcination}} CaO + MgO + CO_2 \]
- Here, \( CO_2 \) is the colorless gas \( B \).

5. **Analyzing the Behavior of Gas \( B \)**:
- When \( CO_2 \) is passed through lime water (which is a solution of \( Ca(OH)_2 \)), it reacts to form calcium carbonate \( CaCO_3 \), which is insoluble and causes the solution to turn milky:
\[ Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O \]
- The formation of a white precipitate indicates the presence of \( CO_2 \).

6. **Excess \( CO_2 \) and Disappearance of Milkiness**:
- When excess \( CO_2 \) is passed through the lime water, the milkiness disappears because the calcium carbonate reacts with more \( CO_2 \) to form soluble calcium bicarbonate:
\[ CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2 \]
- This reaction explains why the milkiness disappears upon the addition of excess gas \( B \).

7. **Conclusion**:
- Based on the above analysis, we can conclude that the name of the product \( A \) is **dolomite**.

### Final Answer:
The name of the product \( A \) is **dolomite**.