`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 (H2C2O4) are equivalent to 20 mL of 1.0 M KMnO4 based on the given reaction:

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

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

To find the number of moles of KMnO4, we use the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]

Given:
- Molarity of KMnO4 = 1.0 M
- Volume of KMnO4 = 20 mL = 0.020 L

Calculating the moles:

\[ \text{Number of moles of KMnO4} = 1.0 \, \text{mol/L} \times 0.020 \, \text{L} = 0.020 \, \text{mol} \]

### Step 2: Determine the n-factor for KMnO4

In the reaction, each MnO4^- ion gains 5 electrons to become Mn^2+. Therefore, the n-factor for KMnO4 is 5.

### Step 3: Calculate the equivalents of KMnO4

The number of equivalents is calculated as follows:

\[ \text{Equivalents} = \text{Number of moles} \times \text{n-factor} \]

Calculating the equivalents:

\[ \text{Equivalents of KMnO4} = 0.020 \, \text{mol} \times 5 = 0.10 \, \text{equivalents} \]

### Step 4: Use the stoichiometry of the reaction to find equivalents of H2C2O4

From the balanced equation, we see that 2 equivalents of MnO4^- react with 5 equivalents of C2O4^2-. Therefore, the ratio of equivalents is:

\[ \frac{5 \text{ equivalents of } C2O4^{2-}}{2 \text{ equivalents of } MnO4^{-}} = \frac{5}{2} \]

### Step 5: Calculate the equivalents of H2C2O4

Using the equivalents of KMnO4 calculated:

\[ \text{Equivalents of H2C2O4} = \text{Equivalents of KMnO4} \times \frac{5}{2} \]

Calculating the equivalents of H2C2O4:

\[ \text{Equivalents of H2C2O4} = 0.10 \, \text{equivalents} \times \frac{5}{2} = 0.25 \, \text{equivalents} \]

### Final Answer

20 mL of 1.0 M KMnO4 is equivalent to **0.25 equivalents of H2C2O4**.

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