Calculate the weight of lime (CaO) obtained by heating 240kg of 95% pure lime stone (`CaCO_3`).
`CaCO_3​→CaO+CO_2` ​

Explanation

To calculate the weight of lime (CaO) obtained by heating 240 kg of 95% pure limestone (CaCO3), we can follow these steps:

### Step 1: Calculate the amount of pure limestone
First, we need to determine the amount of pure CaCO3 in the 240 kg of limestone. Since the limestone is 95% pure, we calculate:

\[ \text{Amount of pure } CaCO_3 = \frac{95}{100} \times 240 \text{ kg} \]

Calculating this gives:

\[ \text{Amount of pure } CaCO_3 = 0.95 \times 240 = 228 \text{ kg} \]

### Step 2: Convert kg to grams
Next, we convert the amount of pure CaCO3 from kilograms to grams since the molar masses are typically expressed in grams.

\[ 228 \text{ kg} = 228 \times 1000 \text{ g} = 228000 \text{ g} \]

### Step 3: Use the balanced chemical equation
The balanced chemical equation for the decomposition of calcium carbonate is:

\[ CaCO_3 \rightarrow CaO + CO_2 \]

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

### Step 4: Calculate the molar masses
The molar mass of CaCO3 is 100 g/mol, and the molar mass of CaO is 56 g/mol.

### Step 5: Calculate the amount of CaO produced
Using the stoichiometry from the balanced equation, we can find the amount of CaO produced from the amount of CaCO3:

\[ \text{Weight of } CaO = \left( \frac{\text{Molar mass of } CaO}{\text{Molar mass of } CaCO_3} \right) \times \text{Weight of pure } CaCO_3 \]

Substituting the values:

\[ \text{Weight of } CaO = \left( \frac{56 \text{ g}}{100 \text{ g}} \right) \times 228000 \text{ g} \]

Calculating this gives:

\[ \text{Weight of } CaO = 0.56 \times 228000 = 127680 \text{ g} \]

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

\[ 127680 \text{ g} = \frac{127680}{1000} \text{ kg} = 127.68 \text{ kg} \]

### Final Answer
The weight of lime (CaO) obtained by heating 240 kg of 95% pure limestone is **127.68 kg**.

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