Volume of `CO_(2)` obtained at STP by the complete decomposition of 9.85 g `Na_(2)CO_(3)` is

Options

A. 2.24 litre
B. Zero
C. 0.85 litre
D. 0.56 litre

Answer

To find the volume of carbon dioxide (CO₂) obtained at STP from the complete decomposition of 9.85 g of sodium carbonate (Na₂CO₃), we can follow these steps:

### Step 1: Write the decomposition reaction
The thermal decomposition of sodium carbonate can be represented by the following equation:
\[ \text{Na}_2\text{CO}_3 (s) \rightarrow \text{Na}_2\text{O} (s) + \text{CO}_2 (g) \]
From the equation, we can see that 1 mole of sodium carbonate produces 1 mole of carbon dioxide.

### Step 2: Calculate the molar mass of Na₂CO₃
The molar mass of sodium carbonate (Na₂CO₃) can be calculated as follows:
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Adding these together:
\[ \text{Molar mass of Na}_2\text{CO}_3 = 46 + 12 + 48 = 106 \text{ g/mol} \]

### Step 3: Calculate the number of moles of Na₂CO₃
Using the mass of sodium carbonate provided (9.85 g), we can calculate the number of moles:
\[ \text{Number of moles of Na}_2\text{CO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{9.85 \text{ g}}{106 \text{ g/mol}} \approx 0.0930 \text{ moles} \]

### Step 4: Determine the volume of CO₂ produced
From the decomposition reaction, we know that 1 mole of Na₂CO₃ produces 1 mole of CO₂. Therefore, the number of moles of CO₂ produced will also be approximately 0.0930 moles.

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as follows:
\[ \text{Volume of CO}_2 = \text{Number of moles of CO}_2 \times 22.4 \text{ L/mol} = 0.0930 \text{ moles} \times 22.4 \text{ L/mol} \approx 2.08 \text{ L} \]

### Final Answer
The volume of CO₂ obtained at STP by the complete decomposition of 9.85 g of Na₂CO₃ is approximately **2.08 liters**.

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