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 find the volume of CO₂ obtained from the complete decomposition of 9.85 g of BaCO₃ at STP, we can follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of barium carbonate (BaCO₃) can be represented by the following balanced chemical equation:
\[ \text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2 \]
From this equation, we can see that 1 mole of BaCO₃ produces 1 mole of CO₂.

### Step 2: Calculate the number of moles of BaCO₃
To calculate the number of moles of BaCO₃, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass of BaCO₃ = 9.85 g
- Molar mass of BaCO₃ = 197 g/mol

Now substituting the values:
\[ \text{Number of moles of BaCO}_3 = \frac{9.85 \, \text{g}}{197 \, \text{g/mol}} \approx 0.05 \, \text{moles} \]

### Step 3: Determine the number of moles of CO₂ produced
From the balanced equation, we know that 1 mole of BaCO₃ produces 1 mole of CO₂. Therefore, the number of moles of CO₂ produced will also be:
\[ \text{Number of moles of CO}_2 = 0.05 \, \text{moles} \]

### Step 4: Calculate the volume of CO₂ at STP
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Thus, we can calculate the volume of CO₂ produced using the formula:
\[ \text{Volume (L)} = \text{Number of moles} \times 22.4 \, \text{L/mol} \]
Substituting the values:
\[ \text{Volume of CO}_2 = 0.05 \, \text{moles} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

### Final Answer
The volume of CO₂ obtained at STP by the complete decomposition of 9.85 g of BaCO₃ is **1.12 liters**.

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