A divalent metal salt X, which contributes to hardness of water, combines with washing soda and forms an insoluble salt Y and common salt. Salt Y is also used for the laboratory preparation of `CO_(2)`. Identify X and Y, respectively.

Options

A. `CaSO_(4),CaCO_(3)`
B. `MgSO_(4),MgCO_(3)`
C. `CaCl_(2),CaCO_(3)`
D. `MgCl_(2),MgCO_(3)`

Explanation

To solve the problem, we need to identify the divalent metal salt (X) that contributes to the hardness of water and the insoluble salt (Y) formed when it reacts with washing soda (sodium carbonate).

### Step-by-Step Solution:

1. **Understanding Hardness of Water**:
- Hardness in water is primarily caused by the presence of divalent metal ions, such as calcium (Ca²⁺) and magnesium (Mg²⁺).

2. **Identifying the Divalent Metal Salt (X)**:
- The problem states that X is a divalent metal salt. The common divalent salts that contribute to water hardness are calcium chloride (CaCl₂) and magnesium chloride (MgCl₂).
- Since we are looking for a salt that reacts with washing soda (Na₂CO₃) to form an insoluble salt and common salt, we can focus on calcium chloride (CaCl₂).

3. **Reaction with Washing Soda**:
- When calcium chloride (CaCl₂) reacts with washing soda (Na₂CO₃), the reaction can be represented as:
\[ CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 \downarrow + 2NaCl \]
- Here, calcium carbonate (CaCO₃) is formed as an insoluble salt (Y), and sodium chloride (NaCl) is formed as common salt.

4. **Identifying the Insoluble Salt (Y)**:
- The insoluble salt formed in the reaction is calcium carbonate (CaCO₃), which is also known to be used in the laboratory preparation of carbon dioxide (CO₂) when heated.

5. **Final Identification**:
- Therefore, the divalent metal salt X is calcium chloride (CaCl₂), and the insoluble salt Y is calcium carbonate (CaCO₃).

### Conclusion:
- **X = CaCl₂ (Calcium Chloride)**
- **Y = CaCO₃ (Calcium Carbonate)**