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

Options

A. (a)`0.84 L`
B. (b)`2.24 L`
C. (c )`4.06 L`
D. (d)`1.12 L`

Explanation

To solve the problem of finding the volume of CO₂ released at STP when 9.85 g of BaCO₃ is heated, we can follow these steps:

### Step 1: Write the decomposition reaction
When barium carbonate (BaCO₃) is heated, it decomposes into barium oxide (BaO) and carbon dioxide (CO₂):
\[ \text{BaCO}_3 (s) \rightarrow \text{BaO} (s) + \text{CO}_2 (g) \]

### Step 2: Calculate the molar mass of BaCO₃
To find the number of moles of BaCO₃, we first need to calculate its molar mass:
- 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)

Thus, the molar mass of BaCO₃ is:
\[ \text{Molar mass of BaCO}_3 = 137 + 12 + (3 \times 16) = 137 + 12 + 48 = 197 \text{ g/mol} \]

### Step 3: Calculate the number of moles of BaCO₃
Now, we can calculate the number of moles of BaCO₃ in 9.85 g:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{9.85 \text{ g}}{197 \text{ g/mol}} \]

Calculating this gives:
\[ \text{Number of moles} = \frac{9.85}{197} \approx 0.05 \text{ moles} \]

### Step 4: Determine the moles of CO₂ produced
From the balanced equation, we see that 1 mole of BaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will also be 0.05 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. Therefore, the volume of CO₂ produced can be calculated as follows:
\[ \text{Volume of CO}_2 = \text{Number of moles} \times 22.4 \text{ L/mol} \]
\[ \text{Volume of CO}_2 = 0.05 \text{ moles} \times 22.4 \text{ L/mol} = 1.12 \text{ L} \]

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

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