Assuming full decomposition, the volume of `CO_(2)` released at STP on heating 9.85 g of `BaCO_(3)` (At mass Ba = 137) will be

Options

A. 2.24 lit.
B. 1.12 lit
C. 0.84 lit
D. 0.56 lit

Answer

To solve the problem of calculating the volume of CO₂ released at STP from the decomposition of 9.85 g of BaCO₃, we will follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of barium carbonate (BaCO₃) can be represented by the following chemical equation:
\[ \text{BaCO}_3 (s) \rightarrow \text{BaO} (s) + \text{CO}_2 (g) \]

### Step 2: Calculate the molar mass of BaCO₃
To find the molar mass of barium carbonate (BaCO₃), we need to sum the atomic masses of its constituent elements:
- Atomic mass of Ba = 137 g/mol
- Atomic mass of C = 12 g/mol
- Atomic mass of O = 16 g/mol (and there are 3 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of BaCO}_3 = 137 + 12 + (16 \times 3) = 137 + 12 + 48 = 197 \text{ g/mol} \]

### Step 3: Calculate the number of moles of BaCO₃
Using the mass of BaCO₃ given (9.85 g), we can calculate the number of moles using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{9.85 \text{ g}}{197 \text{ g/mol}} \approx 0.05 \text{ moles} \]

### Step 4: Determine the number of moles of CO₂ produced
From the balanced equation, we see 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 5: 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:
\[ \text{Volume of CO}_2 = \text{Number of moles} \times 22.4 \text{ L/mol} = 0.05 \text{ moles} \times 22.4 \text{ L/mol} = 1.12 \text{ L} \]

### Final Answer
The volume of CO₂ released at STP on heating 9.85 g of BaCO₃ is **1.12 liters**.

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