What volume of `CO_2` at STP is obtained by thermal decomposition of 20g `KHCO_3` ? [Atomic weight of `K=39 g mol ^(−1)`]
2`KHCO_3(s)⟶K_2O(s)+2CO_2 (g)+H_2O(g)`

Answer

To solve the problem, we need to calculate the volume of \( CO_2 \) produced from the thermal decomposition of 20 g of \( KHCO_3 \) at STP (Standard Temperature and Pressure).

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The thermal decomposition of potassium bicarbonate (\( KHCO_3 \)) is represented by the equation:
\[ 2 KHCO_3(s) \rightarrow K_2O(s) + 2 CO_2(g) + H_2O(g) \]

2. **Calculate the molar mass of \( KHCO_3 \):**
The molar mass can be calculated using the atomic weights:
- K = 39 g/mol
- H = 1 g/mol
- C = 12 g/mol
- O = 16 g/mol (3 oxygen atoms)

Therefore, the molar mass of \( KHCO_3 \) is:
\[ 39 + 1 + 12 + (16 \times 3) = 39 + 1 + 12 + 48 = 100 \text{ g/mol} \]

3. **Calculate the number of moles of \( KHCO_3 \):**
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For 20 g of \( KHCO_3 \):
\[ \text{Number of moles} = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ moles} \]

4. **Determine the moles of \( CO_2 \) produced:**
From the balanced equation, 2 moles of \( KHCO_3 \) produce 2 moles of \( CO_2 \). Therefore, 0.2 moles of \( KHCO_3 \) will produce:
\[ \text{Moles of } CO_2 = 0.2 \text{ moles } KHCO_3 \times \frac{2 \text{ moles } CO_2}{2 \text{ moles } KHCO_3} = 0.2 \text{ moles } CO_2 \]

5. **Calculate the volume of \( CO_2 \) at STP:**
At STP, 1 mole of any gas occupies 22.4 liters. Thus, the volume of \( CO_2 \) produced is:
\[ \text{Volume of } CO_2 = \text{moles of } CO_2 \times 22.4 \text{ L/mol} = 0.2 \text{ moles} \times 22.4 \text{ L/mol} = 4.48 \text{ L} \]

### Final Answer:
The volume of \( CO_2 \) obtained by the thermal decomposition of 20 g of \( KHCO_3 \) at STP is **4.48 liters**.

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