4g of mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` on heating liberates 448 ml of `CO_(2)` at STP. The percentage of `Na_(2)CO_(3)` in the mixture is

Options

A. 84
B. 16
C. 54
D. 80

Answer

To solve the problem, we need to determine the percentage of Na₂CO₃ in a 4g mixture of Na₂CO₃ and NaHCO₃ that liberates 448 ml of CO₂ upon heating. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
When sodium bicarbonate (NaHCO₃) is heated, it decomposes to form sodium carbonate (Na₂CO₃), carbon dioxide (CO₂), and water (H₂O):
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]
Na₂CO₃ does not decompose upon heating.

### Step 2: Calculate Moles of CO₂
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters (or 22400 ml). Therefore, we can calculate the moles of CO₂ produced:
\[ \text{Moles of } CO_2 = \frac{\text{Volume of } CO_2 \text{ (in ml)}}{22400 \text{ ml/mol}} = \frac{448 \text{ ml}}{22400 \text{ ml/mol}} = 0.02 \text{ moles} \]

### Step 3: Relate Moles of CO₂ to NaHCO₃
From the balanced equation, 2 moles of NaHCO₃ produce 1 mole of CO₂. Therefore, the moles of NaHCO₃ that decomposed can be calculated as:
\[ \text{Moles of NaHCO}_3 = 2 \times \text{Moles of } CO_2 = 2 \times 0.02 = 0.04 \text{ moles} \]

### Step 4: Calculate Mass of NaHCO₃
The molar mass of NaHCO₃ is approximately 84 g/mol. Thus, the mass of NaHCO₃ that decomposed is:
\[ \text{Mass of NaHCO}_3 = \text{Moles} \times \text{Molar Mass} = 0.04 \text{ moles} \times 84 \text{ g/mol} = 3.36 \text{ g} \]

### Step 5: Calculate Mass of Na₂CO₃
Since the total mass of the mixture is 4 g, we can find the mass of Na₂CO₃ in the mixture:
\[ \text{Mass of Na}_2\text{CO}_3 = \text{Total mass} - \text{Mass of NaHCO}_3 = 4 \text{ g} - 3.36 \text{ g} = 0.64 \text{ g} \]

### Step 6: Calculate Percentage of Na₂CO₃
To find the percentage of Na₂CO₃ in the mixture, we use the formula:
\[ \text{Percentage of Na}_2\text{CO}_3 = \left( \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Total mass}} \right) \times 100 = \left( \frac{0.64 \text{ g}}{4 \text{ g}} \right) \times 100 = 16\% \]

### Final Answer
The percentage of Na₂CO₃ in the mixture is **16%**.

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