Calculate the weight of lime (CaO) obtained by heating 175kg of 95% pure lime stone (`CaCO_3`).
`CaCO_3​→CaO+CO_2` ​

Options

A. 86.9Kg
B. 38.9Kg
C. 93.1Kg
D. None of these

Explanation

To solve the problem of calculating the weight of lime (CaO) obtained by heating 175 kg of 95% pure limestone (CaCO₃), we can follow these steps:

### Step 1: Calculate the amount of pure limestone (CaCO₃)
We start by determining the amount of pure limestone in the 175 kg sample. Since the limestone is 95% pure, we can calculate the mass of pure CaCO₃ as follows:

\[ \text{Mass of pure CaCO}_3 = \frac{95}{100} \times 175 \text{ kg} \]

Calculating this gives:

\[ \text{Mass of pure CaCO}_3 = 0.95 \times 175 = 166.25 \text{ kg} \]

### Step 2: Convert the mass of pure CaCO₃ to grams
Since the molar masses are typically in grams, we convert the mass from kilograms to grams:

\[ 166.25 \text{ kg} = 166250 \text{ g} \]

### 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. Now, we need to find the molar masses:
- 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 to find out how much CaO is produced from the mass of CaCO₃ we have:

\[ \text{Mass of CaO} = \left( \frac{56 \text{ g CaO}}{100 \text{ g CaCO}_3} \right) \times 166250 \text{ g CaCO}_3 \]

Calculating this gives:

\[ \text{Mass of CaO} = \frac{56}{100} \times 166250 = 93100 \text{ g} \]

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

\[ 93100 \text{ g} = 93.1 \text{ kg} \]

### Final Answer
The weight of lime (CaO) obtained by heating 175 kg of 95% pure limestone is **93.1 kg**.

---