The mass of CaO obtained by heating 100 kg of 95% pure limestone `(CaCO_3)` is-

Options

A. 56 kg
B. 28 kg
C. 53.2 kg
D. 50 kg

Answer

To find the mass of CaO obtained by heating 100 kg of 95% pure limestone (CaCO₃), we can follow these steps:

### Step 1: Calculate the mass of pure CaCO₃ in the limestone
Since the limestone is 95% pure, we first calculate the mass of pure CaCO₃ in 100 kg of limestone.

\[ \text{Mass of pure CaCO}_3 = \text{Total mass of limestone} \times \text{Purity} \]
\[ \text{Mass of pure CaCO}_3 = 100 \, \text{kg} \times 0.95 = 95 \, \text{kg} \]

### Step 2: Write the decomposition reaction of CaCO₃
When calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂) according to the following reaction:

\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 3: Calculate the molar masses
Next, we need to find the molar masses of CaCO₃ and CaO.

- Molar mass of CaCO₃:
- Ca: 40 g/mol
- C: 12 g/mol
- O₃: 16 g/mol × 3 = 48 g/mol
- Total: 40 + 12 + 48 = 100 g/mol

- Molar mass of CaO:
- Ca: 40 g/mol
- O: 16 g/mol
- Total: 40 + 16 = 56 g/mol

### Step 4: Use stoichiometry to find the mass of CaO produced
From the reaction, 1 mole of CaCO₃ produces 1 mole of CaO. Therefore, we can set up a ratio based on the molar masses:

\[ \text{Mass of CaO produced} = \left( \frac{\text{Molar mass of CaO}}{\text{Molar mass of CaCO}_3} \right) \times \text{Mass of pure CaCO}_3 \]
\[ \text{Mass of CaO produced} = \left( \frac{56 \, \text{g/mol}}{100 \, \text{g/mol}} \right) \times 95 \, \text{kg} \]
\[ \text{Mass of CaO produced} = 0.56 \times 95 \, \text{kg} = 53.2 \, \text{kg} \]

### Final Answer
The mass of CaO obtained by heating 100 kg of 95% pure limestone is **53.2 kg**.

---