`H_(2)C_(2)O_(4).2H_(2)O` (Mol wt =126) can be oxidised into`CO_(2)` by acidified `KMnO_(4)`. 6.3 gms of oxalic acid can not be oxidised

Options

A. 3.16 gms of `KMnO_(4)`
B. 200 mlof 0.1 M `KMnO_(4)`
C. 0.1 mole of `KMnO_(4)`
D. 0.02 moles of `KMnO_(4)`

Answer

To solve the problem of determining whether 6.3 grams of oxalic acid can be oxidized by acidified KMnO4, we need to calculate the equivalents of oxalic acid and compare it with the equivalents of KMnO4 provided in the options. Here’s a step-by-step solution:

### Step 1: Calculate the Gram Equivalent of Oxalic Acid
1. **Molecular Weight of Oxalic Acid**: Given as 126 g/mol.
2. **Gram Equivalent Weight**: Since oxalic acid (H2C2O4) is a dibasic acid, its gram equivalent weight is calculated as:
\[ \text{Gram Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Number of H+ ions}} = \frac{126}{2} = 63 \text{ g/equiv} \]

### Step 2: Calculate the Number of Equivalents of Oxalic Acid
1. **Weight of Oxalic Acid**: Given as 6.3 g.
2. **Number of Equivalents**:
\[ \text{Number of Equivalents} = \frac{\text{Weight of Oxalic Acid}}{\text{Gram Equivalent Weight}} = \frac{6.3}{63} = 0.1 \text{ equivalents} \]

### Step 3: Analyze the Options for KMnO4
Now we need to check each option to see if they provide 0.1 equivalents of KMnO4.

#### Option 1: 3.16 g of KMnO4
1. **Molecular Weight of KMnO4**: 158 g/mol.
2. **Number of Equivalents**:
\[ \text{Number of Equivalents} = \frac{3.16 \text{ g}}{158 \text{ g/mol}} \times 5 = \frac{3.16 \times 5}{158} = 0.1 \text{ equivalents} \]

#### Option 2: 200 mL of 0.1 M KMnO4
1. **Normality Calculation**:
\[ \text{Number of Equivalents} = 0.1 \text{ M} \times 200 \text{ mL} = 0.1 \text{ equivalents} \]

#### Option 3: 0.1 moles of KMnO4
1. **Number of Equivalents**:
\[ \text{Number of Equivalents} = 0.1 \text{ moles} \times 5 = 0.5 \text{ equivalents} \]

#### Option 4: 0.02 moles of KMnO4
1. **Number of Equivalents**:
\[ \text{Number of Equivalents} = 0.02 \text{ moles} \times 5 = 0.1 \text{ equivalents} \]

### Conclusion
The only option that does not provide 0.1 equivalents of KMnO4 is **Option 3**, which provides 0.5 equivalents. Therefore, the answer is:

**Option 3: 0.1 moles of KMnO4 cannot oxidize 6.3 g of oxalic acid.**

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