Calculate the weight of lime (CaO) obtained by heating 250kg of 90% pure lime stone (`CaCO_3`).
`CaCO_3​→CaO+CO_2` ​

Options

A. 126Kg
B. 100Kg
C. 52Kg
D. 98Kg

Explanation

To solve the problem of calculating the weight of lime (CaO) obtained by heating 250 kg of 90% pure limestone (CaCO3), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Amount of Pure Limestone**:
- Given that the limestone is 90% pure, we first need to find out how much of the 250 kg is pure CaCO3.
\[ \text{Amount of pure CaCO}_3 = \frac{90}{100} \times 250 \text{ kg} = 225 \text{ kg} \]

2. **Convert the Mass of Pure Limestone to Grams**:
- Since we will be using molar masses in grams, we convert the mass of pure CaCO3 from kilograms to grams.
\[ 225 \text{ kg} = 225 \times 1000 \text{ g} = 225000 \text{ g} \]

3. **Use the Decomposition Reaction**:
- The decomposition reaction of calcium carbonate is:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
- From the reaction, we see that 1 mole of CaCO3 produces 1 mole of CaO.

4. **Calculate Molar Masses**:
- The molar mass of CaCO3 (calcium carbonate) is approximately 100 g/mol.
- The molar mass of CaO (calcium oxide) is approximately 56 g/mol.

5. **Calculate the Amount of CaO Produced**:
- We know that 100 g of CaCO3 produces 56 g of CaO. We can set up a proportion to find out how much CaO is produced from 225000 g of CaCO3.
\[ \text{Amount of CaO} = \left(\frac{56 \text{ g CaO}}{100 \text{ g CaCO}_3}\right) \times 225000 \text{ g CaCO}_3 \]
\[ \text{Amount of CaO} = \frac{56}{100} \times 225000 = 126000 \text{ g} \]

6. **Convert the Mass of CaO to Kilograms**:
- Finally, we convert the mass of CaO from grams back to kilograms.
\[ 126000 \text{ g} = \frac{126000}{1000} \text{ kg} = 126 \text{ kg} \]

### Final Answer:
The weight of lime (CaO) obtained by heating 250 kg of 90% pure limestone is **126 kg**.

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