Calculate the weight of lime (CaO) obtained by heating 250kg of 95% pure lime stone (`CaCO_3`).
`CaCO_3​→CaO+CO_2` ​

Options

A. 133Kg
B. 106.4Kg
C. 152.4Kg
D. 100Kg

Answer

To solve the problem of calculating the weight of lime (CaO) obtained by heating 250 kg of 95% pure limestone (CaCO3), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the amount of pure CaCO3:**
- Given that the limestone is 95% pure, we can calculate the amount of pure CaCO3 in 250 kg of limestone.
- \[ \text{Amount of pure } CaCO_3 = \frac{95}{100} \times 250 \text{ kg} \]
- \[ = 237.5 \text{ kg} \]

2. **Write the balanced chemical equation:**
- The decomposition of calcium carbonate (CaCO3) can be represented as:
\[ CaCO_3 \rightarrow CaO + CO_2 \]

3. **Determine the molar masses:**
- Molar mass of CaCO3 = 100 g/mol
- Molar mass of CaO = 56 g/mol

4. **Calculate the amount of CaO produced:**
- From the balanced equation, 1 mole of CaCO3 produces 1 mole of CaO. Therefore, we can set up a proportion:
- \[ \text{If } 100 \text{ g of } CaCO_3 \text{ produces } 56 \text{ g of } CaO, \]
- Then, for 237,500 g of CaCO3:
\[ \text{Amount of } CaO = \left(\frac{56 \text{ g CaO}}{100 \text{ g CaCO}_3}\right) \times 237,500 \text{ g CaCO}_3 \]
- \[ = 133,200 \text{ g CaO} \]

5. **Convert grams to kilograms:**
- To convert grams to kilograms, divide by 1000:
\[ \text{Weight of } CaO = \frac{133,200 \text{ g}}{1000} = 133.2 \text{ kg} \]

### Final Answer:
The weight of lime (CaO) obtained by heating 250 kg of 95% pure limestone is **133.2 kg**.

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