One mole of one of sodium salt listed below, having carbon content close to 14.3% produces 1 mole of carbon dioxide upon heating ( atomic mass Na=23, C=12,O=16 ). The salt is

Options

A. `C_(2)H_(5)COONa`
B. `NaHCO_(3)`
C. HCOONa
D. `CH_(3)COONa`

Explanation

To solve the problem, we need to identify which sodium salt has a carbon content close to 14.3% and produces 1 mole of carbon dioxide upon heating. Let's go through the options step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find a sodium salt that has a carbon content of approximately 14.3% and produces 1 mole of CO₂ upon heating.

2. **Calculating the Carbon Percentage**: The percentage of carbon in a compound can be calculated using the formula:
\[ \text{Percentage of Carbon} = \left(\frac{\text{Mass of Carbon in the formula}}{\text{Molar Mass of the compound}}\right) \times 100 \]

3. **Analyzing Each Option**:
- **Option 1: C₂H₅COONa**
- Molar Mass Calculation:
\[ \text{Molar Mass} = (2 \times 12) + (5 \times 1) + 12 + (16 \times 2) + 23 = 96 \, \text{g/mol} \]
- Carbon Percentage:
\[ \text{Percentage of Carbon} = \left(\frac{12}{96}\right) \times 100 = 12.5\% \]
- This is not close to 14.3%.

- **Option 2: NaHCO₃**
- Molar Mass Calculation:
\[ \text{Molar Mass} = 23 + 1 + 12 + (16 \times 3) = 84 \, \text{g/mol} \]
- Carbon Percentage:
\[ \text{Percentage of Carbon} = \left(\frac{12}{84}\right) \times 100 \approx 14.29\% \]
- This is very close to 14.3%.

- **Option 3: HCOONa**
- Molar Mass Calculation:
\[ \text{Molar Mass} = 1 + 12 + 16 + 23 = 68 \, \text{g/mol} \]
- Carbon Percentage:
\[ \text{Percentage of Carbon} = \left(\frac{12}{68}\right) \times 100 \approx 17.64\% \]
- This is not close to 14.3%.

- **Option 4: CH₃COONa**
- Molar Mass Calculation:
\[ \text{Molar Mass} = 12 + 3 + 12 + 16 + 23 = 82 \, \text{g/mol} \]
- Carbon Percentage:
\[ \text{Percentage of Carbon} = \left(\frac{12}{82}\right) \times 100 \approx 14.63\% \]
- This is also not close to 14.3%.

4. **Conclusion**: The only option that has a carbon content close to 14.3% is **NaHCO₃** (sodium bicarbonate).

5. **Reaction Upon Heating**: When NaHCO₃ is heated, it decomposes as follows:
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \]
This confirms that it produces 1 mole of CO₂.

### Final Answer:
The sodium salt is **NaHCO₃** (sodium bicarbonate).

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