What volume of `CO_(2)` at STP will evolve when 1 g of `CaCO_(3)` reacts with excess of dill . HCI ?

Options

A. 224 ml
B. 112 ml
C. 56 ml
D. 448 ml

Answer

To solve the problem of determining the volume of CO₂ that evolves when 1 g of CaCO₃ reacts with excess dilute HCl at STP, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) can be represented as:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2 O (l) \]

### Step 2: Calculate the molar mass of CaCO₃
The molar mass of CaCO₃ is 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

Adding these together:
\[ \text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 48.00 = 100.09 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of CaCO₃
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles of CaCO}_3 = \frac{1 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.00997 \, \text{mol} \]

### 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 number of moles of CO₂ produced is also approximately:
\[ \text{Number of moles of CO}_2 \approx 0.00997 \, \text{mol} \]

### Step 5: Calculate the volume of CO₂ at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as:
\[ \text{Volume of CO}_2 = \text{Number of moles} \times 22.4 \, \text{L/mol} \]
Substituting the values:
\[ \text{Volume of CO}_2 = 0.00997 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.223 \, \text{L} \]

### Step 6: Convert the volume to milliliters
To convert liters to milliliters:
\[ 0.223 \, \text{L} = 0.223 \times 1000 \, \text{mL} = 223 \, \text{mL} \]

### Final Answer
The volume of CO₂ evolved when 1 g of CaCO₃ reacts with excess dilute HCl at STP is approximately **223 mL**.

---