`KMnO_(4)` reacts with oxalic acid according to the equation
`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) to 2Mn^(2+)+10CO_(2)+8H_(2)O`
Here, 20mL of 1.0M `KMnO_(4)` is equivalent to:

Options

A. `120mL "of"0.25M H_(2)C_(2)O_(4)`
B. `150mL "of" 0.10M H_(2)C_(2)O_(4)`
C. `25mL "of" 0.20M H_(2)C_(2)O_(4)`
D. `50mL "of" 0.20M H_(2)C_(2)O_(4)`

Answer

To solve the problem, we need to determine how many equivalents of oxalic acid (C₂O₄²⁻) are equivalent to 20 mL of 1.0 M KMnO₄ based on the given balanced chemical equation:

\[ 2 \text{MnO}_4^{-} + 5 \text{C}_2\text{O}_4^{2-} + 16 \text{H}^{+} \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]

### Step 1: Calculate the number of equivalents of KMnO₄

1. **Identify the n-factor of KMnO₄**:
- In the reaction, KMnO₄ reduces from Mn in the +7 oxidation state to Mn in the +2 oxidation state.
- The change in oxidation state is \( +7 \) to \( +2 \), which is a change of \( 5 \) (i.e., \( 7 - 2 = 5 \)).
- Therefore, the n-factor of KMnO₄ is \( 5 \).

2. **Calculate the number of equivalents**:
- The formula for equivalents is given by:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \]
- Convert volume from mL to L:
\[ 20 \, \text{mL} = 0.020 \, \text{L} \]
- Now substitute the values:
\[ \text{Number of equivalents} = 1.0 \, \text{M} \times 0.020 \, \text{L} \times 5 = 0.1 \, \text{equivalents} \]

### Step 2: Relate the equivalents of KMnO₄ to oxalic acid

From the balanced equation, we see that:
- 2 equivalents of KMnO₄ react with 5 equivalents of oxalic acid.

### Step 3: Calculate the equivalents of oxalic acid

Using the ratio from the balanced equation:
\[ \frac{5 \, \text{equivalents of C}_2\text{O}_4^{2-}}{2 \, \text{equivalents of KMnO}_4} = \frac{x}{0.1} \]
Where \( x \) is the number of equivalents of oxalic acid.

Cross-multiplying gives:
\[ 2x = 0.1 \times 5 \]
\[ 2x = 0.5 \]
\[ x = 0.25 \, \text{equivalents of C}_2\text{O}_4^{2-} \]

### Conclusion

20 mL of 1.0 M KMnO₄ is equivalent to 0.25 equivalents of oxalic acid.

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