`0.56 g` of lime stone was treated with oxalic acid to give `CaC_(2)O_(4)`. The precipitate decolorized `45 ml` of `0.2N KMnO_(4)` in acid medium. Calculate `%` of `CaO` in lime stone.

Answer

To solve the problem step by step, we will follow the chemical reactions and calculations involved in determining the percentage of CaO in limestone after it reacts with oxalic acid and KMnO4.

### Step 1: Understanding the reaction
Limestone primarily consists of calcium carbonate (CaCO3). When treated with oxalic acid (H2C2O4), it forms calcium oxalate (CaC2O4) as a precipitate. The reaction can be represented as:
\[ \text{CaCO}_3 + \text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Determine the milliequivalents of KMnO4
The given volume of KMnO4 solution is 45 ml with a normality of 0.2 N. The milliequivalents of KMnO4 can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (L)} \]
Convert 45 ml to liters:
\[ 45 \, \text{ml} = 0.045 \, \text{L} \]
Now calculate the milliequivalents:
\[ \text{Milliequivalents of KMnO}_4 = 0.2 \, \text{N} \times 0.045 \, \text{L} = 0.009 \, \text{equivalents} = 9 \, \text{milliequivalents} \]

### Step 3: Relate the milliequivalents of CaC2O4 to KMnO4
The reaction between CaC2O4 and KMnO4 shows that 1 mole of CaC2O4 will react with 5 moles of KMnO4. Therefore, the valency factor for CaC2O4 is 2 (as it can donate 2 electrons). The milliequivalents of CaC2O4 will also equal the milliequivalents of KMnO4:
\[ \text{Milliequivalents of CaC}_2\text{O}_4 = \text{Milliequivalents of KMnO}_4 = 9 \, \text{milliequivalents} \]

### Step 4: Calculate the milliequivalents of CaO
Since the milliequivalents of CaO will be equal to the milliequivalents of CaC2O4, we have:
\[ \text{Milliequivalents of CaO} = 9 \, \text{milliequivalents} \]

### Step 5: Calculate the weight of CaO
The equivalent weight of CaO can be calculated as follows:
- Molar mass of CaO = 40 (Ca) + 16 (O) = 56 g/mol
- The N factor for CaO is 2 (as it can donate 2 moles of electrons).

Using the formula:
\[ \text{Weight} = \text{Milliequivalents} \times \text{Equivalent weight} \]
\[ \text{Weight of CaO} = 9 \, \text{milliequivalents} \times \frac{56 \, \text{g/mol}}{1000} = 0.504 \, \text{g} \]

### Step 6: Calculate the percentage of CaO in limestone
Now, we can calculate the percentage of CaO in the original limestone sample:
\[ \text{Percentage of CaO} = \left( \frac{\text{Weight of CaO}}{\text{Weight of limestone}} \right) \times 100 \]
\[ \text{Percentage of CaO} = \left( \frac{0.504 \, \text{g}}{0.56 \, \text{g}} \right) \times 100 \approx 90\% \]

### Final Result
The percentage of CaO in limestone is approximately **90%**.