A neutral white sodium salt (A) on heating liberates a gas (B), leaving a highly alkaline reside (C ). Thegas (B) is colourless, odourless and turns lime water milky. (A) is

Options

A. `NaNO_(3)`
B. `NaHCO_(3)`
C. `Na_(2)CO_(3)`
D. `NaCI`

Explanation

To solve the problem step by step, we will analyze the information provided and deduce the identities of the substances involved.

### Step 1: Identify the Gas (B)
The problem states that the gas (B) is colorless, odorless, and turns lime water milky. This is a characteristic of carbon dioxide (CO2). Therefore, we can conclude:
- **Gas (B) = CO2**

**Hint for Step 1:** Remember that CO2 is known for turning lime water milky due to the formation of calcium carbonate.

### Step 2: Identify the Alkaline Residue (C)
The problem mentions that the heating of the neutral white sodium salt (A) leaves a highly alkaline residue (C). Since we are dealing with sodium compounds, we need to consider which sodium salt can produce an alkaline residue upon heating.

**Hint for Step 2:** Think about sodium salts that release CO2 when heated and leave behind a strong base.

### Step 3: Consider Possible Sodium Salts (A)
We need to identify a neutral white sodium salt (A) that, when heated, produces CO2 and leaves behind an alkaline residue (C). The options for sodium salts include:
- NaCl (sodium chloride)
- NaNO3 (sodium nitrate)
- NaHCO3 (sodium bicarbonate)
- Na2CO3 (sodium carbonate)

**Hint for Step 3:** Focus on the salts that contain carbonate or bicarbonate, as they are more likely to release CO2 upon heating.

### Step 4: Analyze Each Salt
1. **NaCl**: Does not contain carbon, so it cannot produce CO2.
2. **NaNO3**: Does not produce CO2 upon heating.
3. **NaHCO3**: When heated, it decomposes to produce Na2CO3, CO2, and H2O.
- Reaction: 2 NaHCO3 → Na2CO3 + CO2 + H2O
4. **Na2CO3**: Does not produce CO2 upon heating.

**Hint for Step 4:** Only sodium bicarbonate (NaHCO3) produces CO2 when heated.

### Step 5: Identify the Residue (C)
From the reaction of NaHCO3, we see that the residue left after heating is sodium carbonate (Na2CO3), which is highly alkaline.

**Hint for Step 5:** Remember that sodium carbonate is a strong base and will contribute to the alkalinity of the residue.

### Conclusion
Putting all the pieces together, we identify:
- **(A) = NaHCO3 (sodium bicarbonate)**
- **(B) = CO2 (carbon dioxide)**
- **(C) = Na2CO3 (sodium carbonate)**

Thus, the neutral white sodium salt (A) is sodium bicarbonate (NaHCO3).

**Final Answer:** A = NaHCO3