Compound X is responsible for temporary hardness. On boiling X gives a precipitate of Y then X & Y are

Options

A. `Mg(HCO_(3))_(2),MgCO_(3)`
B. `Mg(HCO_(3))_(2),Mg(OH)_(2)`
C. `Ca(HCO_(3))_(2),Ca(OH)_(2)`
D. `Ca(HCO_(3))_(2),CaO`

Answer

To solve the problem, we need to identify the compounds X and Y based on the information provided about temporary hardness and the reaction that occurs upon boiling.

### Step-by-Step Solution:

1. **Understanding Temporary Hardness**:
- Temporary hardness in water is primarily caused by the presence of bicarbonates of calcium and magnesium. These bicarbonates dissolve in water and can be represented as:
- Calcium bicarbonate: \( \text{Ca(HCO}_3\text{)}_2 \)
- Magnesium bicarbonate: \( \text{Mg(HCO}_3\text{)}_2 \)

2. **Identifying Compound X**:
- Since the question states that compound X is responsible for temporary hardness, we can conclude that X must be one of the bicarbonates of calcium or magnesium. Therefore, X can be either:
- \( \text{Ca(HCO}_3\text{)}_2 \) (Calcium bicarbonate)
- \( \text{Mg(HCO}_3\text{)}_2 \) (Magnesium bicarbonate)

3. **Boiling Compound X**:
- When we boil bicarbonates, they decompose to form carbonates, water, and carbon dioxide gas. The reactions are as follows:
- For calcium bicarbonate:
\[ \text{Ca(HCO}_3\text{)}_2 \, (aq) \xrightarrow{\text{heat}} \text{CaCO}_3 \, (s) + \text{H}_2\text{O} \, (l) + \text{CO}_2 \, (g) \]
- For magnesium bicarbonate:
\[ \text{Mg(HCO}_3\text{)}_2 \, (aq) \xrightarrow{\text{heat}} \text{MgCO}_3 \, (s) + \text{H}_2\text{O} \, (l) + \text{CO}_2 \, (g) \]

4. **Identifying Compound Y**:
- The precipitate formed upon boiling will be either calcium carbonate (\( \text{CaCO}_3 \)) or magnesium carbonate (\( \text{MgCO}_3 \)). Since both yield a white precipitate, we can conclude that Y is the carbonate formed from the respective bicarbonate.

5. **Final Identification**:
- If we consider the options given in the question, we can conclude:
- If X is \( \text{Ca(HCO}_3\text{)}_2 \), then Y is \( \text{CaCO}_3 \).
- If X is \( \text{Mg(HCO}_3\text{)}_2 \), then Y is \( \text{MgCO}_3 \).
- However, since the question emphasizes a single pair, we can state:
- **X = Calcium bicarbonate (\( \text{Ca(HCO}_3\text{)}_2 \))**
- **Y = Calcium carbonate (\( \text{CaCO}_3 \))**

### Conclusion:
- Therefore, the answer is:
- **X = Calcium bicarbonate (\( \text{Ca(HCO}_3\text{)}_2 \))**
- **Y = Calcium carbonate (\( \text{CaCO}_3 \))**