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. 0.84L
B. 2.24L
C. 4.06L
D. 1.12L

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 BaCO₃, we need to add 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 (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 provided (9.85 g), we can calculate the number of moles:
\[ \text{Number of moles of BaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{9.85 \text{ g}}{197 \text{ g/mol}} \approx 0.0500 \text{ moles} \]

### Step 4: Determine the volume 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 approximately 0.0500 moles.

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as follows:
\[ \text{Volume of CO}_2 = \text{Number of moles of CO}_2 \times 22.4 \text{ L/mol} = 0.0500 \text{ moles} \times 22.4 \text{ L/mol} \approx 1.12 \text{ L} \]

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

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