The volume of `CO_(2)` obtained by the complete decomposition of one mole of `NaHCO_(3)` at STP is

Options

A. 2.4 L
B. 11.2 L
C. 44.8 L
D. 4.48 L

Answer

To find the volume of carbon dioxide (CO₂) produced by the complete decomposition of one mole of sodium bicarbonate (NaHCO₃) at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of sodium bicarbonate can be represented by the following chemical equation:
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]
This equation shows that 2 moles of sodium bicarbonate decompose to produce 1 mole of carbon dioxide.

### Step 2: Determine the moles of CO₂ produced from 1 mole of NaHCO₃
From the balanced equation, we see that:
- 2 moles of NaHCO₃ produce 1 mole of CO₂.
- Therefore, 1 mole of NaHCO₃ will produce \(\frac{1}{2}\) moles of CO₂.

### Step 3: Calculate the volume of CO₂ at STP
At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters. Thus, the volume of \(\frac{1}{2}\) mole of CO₂ can be calculated as follows:
\[ \text{Volume of } \frac{1}{2} \text{ mole of CO}_2 = \frac{1}{2} \times 22.4 \text{ L} = 11.2 \text{ L} \]

### Conclusion
The volume of carbon dioxide obtained by the complete decomposition of one mole of sodium bicarbonate at STP is **11.2 liters**.

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