Calculate the weight of lime (CaO) obtained by heating 150kg of 95% pure lime stone (`CaCO_3`).
`CaCO_3​→CaO+CO_2` ​

Options

A. 89.7Kg
B. 79.8Kg
C. 97.8Kg
D. None of these

Explanation

To calculate the weight of lime (CaO) obtained by heating 150 kg of 95% pure limestone (CaCO3), we can follow these steps:

### Step 1: Calculate the amount of pure limestone (CaCO3)
First, we need to find out how much of the 150 kg of limestone is pure CaCO3. Since the limestone is 95% pure, we can calculate the mass of pure CaCO3 as follows:

\[ \text{Mass of pure CaCO}_3 = \frac{95}{100} \times 150 \text{ kg} \]

Calculating this gives:

\[ \text{Mass of pure CaCO}_3 = 0.95 \times 150 = 142.5 \text{ kg} \]

### Step 2: Convert the mass of CaCO3 to grams
To make calculations easier, we convert the mass of CaCO3 from kilograms to grams:

\[ 142.5 \text{ kg} = 142500 \text{ g} \]

### Step 3: Use the stoichiometry of the reaction
The reaction given 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.

### Step 4: Calculate the molar masses
We need the molar masses to relate the grams of CaCO3 to grams of CaO:

- Molar mass of CaCO3 = 100 g/mol
- Molar mass of CaO = 56 g/mol

### Step 5: Set up the conversion from CaCO3 to CaO
Using the molar mass ratio, we can find out how much CaO is produced from 142500 g of CaCO3:

\[ \text{Mass of CaO} = \left( \frac{56 \text{ g CaO}}{100 \text{ g CaCO}_3} \right) \times 142500 \text{ g CaCO}_3 \]

Calculating this gives:

\[ \text{Mass of CaO} = \frac{56}{100} \times 142500 = 79800 \text{ g} \]

### Step 6: Convert the mass of CaO back to kilograms
Finally, we convert the mass of CaO from grams back to kilograms:

\[ 79800 \text{ g} = 79.8 \text{ kg} \]

### Final Answer
The weight of lime (CaO) obtained by heating 150 kg of 95% pure limestone is **79.8 kg**.

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