Calculate the mass of `MnO_(2)`which will be completely oxidises 30 mL of `0. 1M H_(2)C_(2)O_(4)` in acidic medium?

Explanation

To solve the problem of calculating the mass of `MnO2` that will completely oxidize 30 mL of `0.1M H2C2O4` in acidic medium, we will follow these steps:

### Step 1: Calculate the number of equivalents of `H2C2O4`
The first step is to determine the number of equivalents of oxalic acid (`H2C2O4`). The n-factor for `H2C2O4` is 2 because it undergoes oxidation from +3 to +4 for each carbon atom, and since there are 2 carbon atoms, the total change is 2 electrons.

\[ \text{Number of equivalents} = n \times \text{moles} \]

Where:
- n = n-factor of `H2C2O4` = 2
- Molarity (M) = 0.1 M
- Volume (V) = 30 mL = 0.030 L

Calculating the number of moles of `H2C2O4`:

\[ \text{Moles of } H2C2O4 = M \times V = 0.1 \, \text{mol/L} \times 0.030 \, \text{L} = 0.003 \, \text{mol} \]

Now, calculating the number of equivalents:

\[ \text{Equivalents of } H2C2O4 = n \times \text{moles} = 2 \times 0.003 = 0.006 \, \text{equivalents} \]

### Step 2: Determine the equivalents of `MnO2`
Since `MnO2` reacts with `H2C2O4` in a 1:1 ratio in terms of equivalents, the equivalents of `MnO2` will also be 0.006 equivalents.

### Step 3: Calculate the number of moles of `MnO2`
The n-factor for `MnO2` is also 2, as it goes from +4 to +2 during the reaction. Therefore, we can calculate the number of moles of `MnO2`:

\[ \text{Equivalents of } MnO2 = n \times \text{moles} \]

Rearranging gives:

\[ \text{Moles of } MnO2 = \frac{\text{Equivalents}}{n} = \frac{0.006}{2} = 0.003 \, \text{mol} \]

### Step 4: Calculate the mass of `MnO2`
Now, we can find the mass of `MnO2` using its molar mass. The molar mass of `MnO2` is approximately 87 g/mol.

Using the formula:

\[ \text{Mass} = \text{moles} \times \text{molar mass} \]

\[ \text{Mass of } MnO2 = 0.003 \, \text{mol} \times 87 \, \text{g/mol} = 0.261 \, \text{g} \]

### Final Answer
The mass of `MnO2` required to completely oxidize 30 mL of `0.1M H2C2O4` in acidic medium is **0.261 grams**.

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