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

### Step 1: Write the decomposition reaction
When sodium bicarbonate is heated, it decomposes into sodium carbonate (Na₂CO₃), water (H₂O), and carbon dioxide (CO₂). The balanced equation for the reaction is:
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate the moles of CO₂ produced
We know that 4.4 g of CO₂ is produced. To find the number of moles of CO₂, we use the formula:
\[ \text{Moles of CO}_2 = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of CO₂ is calculated as follows:
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Total molar mass of CO₂ = 12 + 32 = 44 g/mol

Now, we can calculate the moles of CO₂:
\[ \text{Moles of CO}_2 = \frac{4.4 \text{ g}}{44 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Determine the moles of NaHCO₃ required
From the balanced equation, we see that 2 moles of NaHCO₃ produce 1 mole of CO₂. Therefore, the moles of NaHCO₃ required can be calculated as:
\[ \text{Moles of NaHCO}_3 = 2 \times \text{Moles of CO}_2 = 2 \times 0.1 = 0.2 \text{ moles} \]

### Step 4: Calculate the mass of NaHCO₃
Now, we need to find the mass of 0.2 moles of NaHCO₃. The molar mass of NaHCO₃ is calculated as follows:
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
- Total molar mass of NaHCO₃ = 23 + 1 + 12 + 48 = 84 g/mol

Now we can calculate the mass of NaHCO₃:
\[ \text{Mass of NaHCO}_3 = \text{Moles} \times \text{Molar mass} = 0.2 \text{ moles} \times 84 \text{ g/mol} = 16.8 \text{ g} \]

### Step 5: Calculate the percentage purity of NaHCO₃
Finally, we can find the percentage purity of NaHCO₃ in the original sample using the formula:
\[ \text{Percentage purity} = \left( \frac{\text{mass of pure NaHCO}_3}{\text{mass of impure sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage purity} = \left( \frac{16.8 \text{ g}}{33.6 \text{ g}} \right) \times 100 = 50\% \]

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

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