A 10 mg effervescent tablet containing sodium bicarbonate and
oxalic acid releases 0.25 ml of `CO_(2)` at T = 298.15 K and p = 1 bar. If molar volume of `CO_(2)` is 25.9 L under such condition, what is the
percentage of sodium bicarbonate in each tablet ? [Molar mass of `NaHCO_(3) = 84 g mol^(-1)`]

Options

A. `16.8`
B. `8.4`
C. `0.84`
D. `33.6`

Answer

To find the percentage of sodium bicarbonate (NaHCO₃) in a 10 mg effervescent tablet that releases 0.25 ml of CO₂, we can follow these steps:

### Step 1: Write the Reaction
The reaction between sodium bicarbonate and oxalic acid can be represented as:
\[ 2 \text{NaHCO}_3 + \text{H}_2\text{C}_2\text{O}_4 \rightarrow 2 \text{CO}_2 + \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{O} \]

### Step 2: Calculate the Density of CO₂
Given the molar volume of CO₂ is 25.9 L, we can find the density (d) using the molar mass of CO₂ (44 g/mol):
\[ d = \frac{\text{molar mass}}{\text{molar volume}} = \frac{44 \text{ g/mol}}{25.9 \text{ L/mol}} \approx 1.70 \text{ g/L} \]

### Step 3: Calculate the Mass of CO₂ Released
Convert the volume of CO₂ released from ml to liters:
\[ 0.25 \text{ ml} = 0.25 \times 10^{-3} \text{ L} \]
Now, calculate the mass of CO₂ using the density:
\[ \text{mass of CO}_2 = \text{density} \times \text{volume} = 1.70 \text{ g/L} \times 0.25 \times 10^{-3} \text{ L} = 0.425 \times 10^{-3} \text{ g} = 0.000425 \text{ g} \]

### Step 4: Calculate the Moles of CO₂
Now, convert the mass of CO₂ to moles:
\[ \text{moles of CO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.000425 \text{ g}}{44 \text{ g/mol}} \approx 9.66 \times 10^{-6} \text{ mol} \]

### Step 5: Relate Moles of CO₂ to Moles of NaHCO₃
From the balanced reaction, we see that 2 moles of NaHCO₃ produce 2 moles of CO₂. Thus, the moles of NaHCO₃ is also:
\[ \text{moles of NaHCO}_3 = 9.66 \times 10^{-6} \text{ mol} \]

### Step 6: Calculate the Mass of NaHCO₃
Now, convert the moles of NaHCO₃ to mass:
\[ \text{mass of NaHCO}_3 = \text{moles} \times \text{molar mass} = 9.66 \times 10^{-6} \text{ mol} \times 84 \text{ g/mol} \approx 0.00081 \text{ g} = 0.81 \text{ mg} \]

### Step 7: Calculate the Percentage of NaHCO₃ in the Tablet
Now, we can find the percentage of NaHCO₃ in the tablet:
\[ \text{percentage of NaHCO}_3 = \left( \frac{\text{mass of NaHCO}_3}{\text{mass of tablet}} \right) \times 100 = \left( \frac{0.81 \text{ mg}}{10 \text{ mg}} \right) \times 100 = 8.1\% \]

### Final Answer
The percentage of sodium bicarbonate in each tablet is approximately **8.1%**.