Calculate the weight of lime (CaO) obtained by heating 270kg of 95% pure lime stone (`CaCO_3`).
`CaCO_3​→CaO+CO_2` ​

Explanation

To calculate the weight of lime (CaO) obtained by heating 270 kg of 95% pure limestone (CaCO3), we can follow these steps:

### Step 1: Calculate the amount of pure CaCO3
First, we need to find out how much pure CaCO3 is present in the 270 kg of limestone. Since the limestone is 95% pure, we can calculate the mass of pure CaCO3 as follows:

\[ \text{Mass of pure CaCO3} = \frac{95}{100} \times 270 \, \text{kg} \]

Calculating this gives:

\[ \text{Mass of pure CaCO3} = 0.95 \times 270 = 256.5 \, \text{kg} \]

### Step 2: Use the stoichiometry of the reaction
The balanced chemical equation for the decomposition of CaCO3 is:

\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

From the equation, we see that 1 mole of CaCO3 produces 1 mole of CaO.

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

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

### Step 4: Calculate the amount of CaO produced
Using the stoichiometric relationship, we can set up a proportion based on the molar masses:

\[ \text{If } 100 \, \text{g of CaCO3 produces } 56 \, \text{g of CaO, \ then } 256500 \, \text{g of CaCO3 will produce } x \, \text{g of CaO. \]

Setting up the equation:

\[ x = \frac{56}{100} \times 256500 \]

Calculating this gives:

\[ x = 0.56 \times 256500 = 143640 \, \text{g} \]

### Step 5: Convert grams to kilograms
Finally, we convert the mass of CaO from grams to kilograms:

\[ \text{Mass of CaO in kg} = \frac{143640 \, \text{g}}{1000} = 143.64 \, \text{kg} \]

### Final Answer
The weight of lime (CaO) obtained by heating 270 kg of 95% pure limestone is approximately **143.64 kg**.

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