33.6 g of an impure sample of sodium bicarbonate was heated strongly and it gave 4.4 g `CO_2`. The percentage purity of `NaHCO_3` will be:

Options

A. 0.25
B. 0.5
C. 0.75
D. 1

Answer

To determine the percentage purity of sodium bicarbonate (NaHCO₃) in the given impure sample, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the mass of CO₂ produced:**
- Given that 4.4 g of CO₂ is produced from the decomposition of the impure sample.

2. **Write the balanced chemical reaction:**
- The decomposition of sodium bicarbonate can be represented as:
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]
- From the reaction, we see that 2 moles of NaHCO₃ produce 1 mole of CO₂.

3. **Calculate the molar masses:**
- Molar mass of NaHCO₃ = 23 (Na) + 1 (H) + 12 (C) + 3*16 (O) = 84 g/mol
- Molar mass of CO₂ = 12 (C) + 2*16 (O) = 44 g/mol

4. **Determine the moles of CO₂ produced:**
- Moles of CO₂ produced = mass / molar mass
\[ \text{Moles of CO}_2 = \frac{4.4 \, \text{g}}{44 \, \text{g/mol}} = 0.1 \, \text{mol} \]

5. **Calculate the moles of NaHCO₃ that produced the CO₂:**
- From the balanced equation, 2 moles of NaHCO₃ produce 1 mole of CO₂.
- Therefore, moles of NaHCO₃ = 2 * moles of CO₂
\[ \text{Moles of NaHCO}_3 = 2 \times 0.1 \, \text{mol} = 0.2 \, \text{mol} \]

6. **Calculate the mass of NaHCO₃ that decomposed:**
- Mass of NaHCO₃ = moles × molar mass
\[ \text{Mass of NaHCO}_3 = 0.2 \, \text{mol} \times 84 \, \text{g/mol} = 16.8 \, \text{g} \]

7. **Calculate the percentage purity of NaHCO₃:**
- The percentage purity formula is:
\[ \text{Percentage Purity} = \left( \frac{\text{Mass of pure NaHCO}_3}{\text{Mass of impure sample}} \right) \times 100 \]
- Plugging in the values:
\[ \text{Percentage Purity} = \left( \frac{16.8 \, \text{g}}{33.6 \, \text{g}} \right) \times 100 \approx 50\% \]

### Final Answer:
The percentage purity of NaHCO₃ in the sample is approximately **50%**.