Volume of `CO_(2)` obtained at STP by the complete decomposition of 9.85 gm `BaCO_(3)` is (Mol. wt. of `BaCO_(3) = 197`)

Options

A. 2.24 litre
B. 1.12 litre
C. 0.85 litre
D. 0.56 litre

Answer

To solve the problem of finding the volume of \( CO_2 \) obtained at STP from the complete decomposition of 9.85 g of \( BaCO_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the decomposition reaction**:
The decomposition of barium carbonate can be represented by the following equation:
\[ BaCO_3 \rightarrow BaO + CO_2 \]
From the equation, we see that 1 mole of \( BaCO_3 \) produces 1 mole of \( CO_2 \).

2. **Calculate the number of moles of \( BaCO_3 \)**:
We use the formula for calculating moles:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} \]
Given weight of \( BaCO_3 = 9.85 \, \text{g} \) and molecular weight of \( BaCO_3 = 197 \, \text{g/mol} \):
\[ \text{Number of moles of } BaCO_3 = \frac{9.85 \, \text{g}}{197 \, \text{g/mol}} \approx 0.05 \, \text{mol} \]

3. **Determine the moles of \( CO_2 \) produced**:
From the balanced equation, we know that 1 mole of \( BaCO_3 \) produces 1 mole of \( CO_2 \). Therefore, the moles of \( CO_2 \) produced will also be:
\[ \text{Moles of } CO_2 = 0.05 \, \text{mol} \]

4. **Calculate the volume of \( CO_2 \) at STP**:
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of \( CO_2 \) produced can be calculated as:
\[ \text{Volume of } CO_2 = \text{Moles of } CO_2 \times 22.4 \, \text{L/mol} \]
\[ \text{Volume of } CO_2 = 0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

5. **Conclusion**:
The volume of \( CO_2 \) obtained at STP from the complete decomposition of 9.85 g of \( BaCO_3 \) is **1.12 liters**.

### Final Answer:
The volume of \( CO_2 \) obtained at STP is **1.12 liters**.