1.25 gm of a chalk sample `(CaCO_(3))` is strongly heated, `0.44` gm `CO_(2)^(-)` gas is produced. Determine the percentage `%` yield of the reaction :-

Options

A. `40%`
B. `60%`
C. `80%`
D. `90%`

Answer

To determine the percentage yield of the reaction when 1.25 g of calcium carbonate (CaCO₃) is heated and produces 0.44 g of carbon dioxide (CO₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of calcium carbonate can be represented by the following equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### 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

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

### Step 3: Calculate the number of moles of CaCO₃ used
Using the mass of the chalk sample:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.25 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.0125 \, \text{mol} \]

### Step 4: Calculate the moles of CO₂ produced
The molar mass of CO₂ is:
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol

Total molar mass of CO₂:
\[ 12.01 + 32.00 = 44.01 \, \text{g/mol} \]

Now, calculate the moles of CO₂ produced:
\[ \text{Moles of CO}_2 = \frac{0.44 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.01 \, \text{mol} \]

### Step 5: Determine the theoretical yield of CO₂
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the theoretical yield of CO₂ from 0.0125 moles of CaCO₃ is:
\[ \text{Theoretical moles of CO}_2 = 0.0125 \, \text{mol} \]

### Step 6: Calculate the percentage yield
The percentage yield can be calculated using the formula:
\[ \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage Yield} = \left( \frac{0.01 \, \text{mol}}{0.0125 \, \text{mol}} \right) \times 100 = 80\% \]

### Final Answer
The percentage yield of the reaction is **80%**.

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