`CaCO_3 + 2HCl to CaCl_2 + H_2O + CO_2`
The volume of `CO_2` gas formed when 2.5 g calcium carbonate are dissolved in excess hydrochloric acid at `0^@C` and 1 atm pressure is [1 mole of any gas at `0^@C` and 1 atm pressure occupies 22.414 1 volume]

Options

A. 1.12L
B. 56.0 L
C. 0.28L
D. 0.56L

Explanation

To find the volume of CO₂ gas formed when 2.5 g of calcium carbonate (CaCO₃) is dissolved in excess hydrochloric acid (HCl), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate the molar mass of calcium carbonate (CaCO₃)
The molar mass of CaCO₃ can be calculated as follows:
- Calcium (Ca) = 40.08 g/mol
- Carbon (C) = 12.01 g/mol
- Oxygen (O) = 16.00 g/mol × 3 = 48.00 g/mol

Thus, the molar mass of CaCO₃ is:
\[ 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol} \]

### Step 3: Calculate the number of moles of calcium carbonate
Using the mass of CaCO₃ given (2.5 g), we can calculate the number of moles:
\[ \text{Number of moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.5 \text{ g}}{100.09 \text{ g/mol}} \approx 0.0249 \text{ moles} \]

### Step 4: Determine the moles of CO₂ produced
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will be the same as the moles of CaCO₃:
\[ \text{Moles of CO}_2 = 0.0249 \text{ moles} \]

### Step 5: Calculate the volume of CO₂ at 0°C and 1 atm
At 0°C and 1 atm, 1 mole of any gas occupies 22.414 liters. Thus, the volume of CO₂ produced can be calculated as follows:
\[ \text{Volume of CO}_2 = \text{moles of CO}_2 \times 22.414 \text{ L/mol} = 0.0249 \text{ moles} \times 22.414 \text{ L/mol} \approx 0.558 \text{ L} \]

### Final Answer
The volume of CO₂ gas formed is approximately **0.558 liters**.

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