`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt to

Options

A. (a)`20 "ml of" 0.5 M H_(2)C_(2)O_(4)`
B. (b)`50 "ml of" 0.1 M H_(2)C_(2)O_(4)`
C. (c )`50 " ml of" 0.5 M H_(2)C_(2)O_(4)`
D. (d)`20 "ml of" 0.1 M H_(2)C_(2)O_(4)`

Explanation

To determine how many equivalents of oxalic acid are present when 20 mL of 0.1 M KMnO4 reacts, we will follow these steps:

### Step 1: Calculate the moles of KMnO4
The first step is to calculate the number of moles of KMnO4 in the given volume and concentration.

\[ \text{Moles of KMnO}_4 = \text{Volume (L)} \times \text{Molarity (mol/L)} \]

Given:
- Volume = 20 mL = 0.020 L
- Molarity = 0.1 M

\[ \text{Moles of KMnO}_4 = 0.020 \, \text{L} \times 0.1 \, \text{mol/L} = 0.002 \, \text{mol} \]

### Step 2: Determine the stoichiometry of the reaction
From the balanced chemical equation:

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

We see that 2 moles of KMnO4 react with 5 moles of oxalic acid (C2O4^2-).

### Step 3: Calculate the equivalents of oxalic acid
Using the stoichiometric ratio from the balanced equation, we can find how many moles of oxalic acid correspond to the moles of KMnO4 we calculated.

From the equation:
- 2 moles of KMnO4 react with 5 moles of C2O4^2-.

Thus, the ratio of KMnO4 to C2O4^2- is:

\[ \frac{5 \, \text{moles of C}_2\text{O}_4^{2-}}{2 \, \text{moles of KMnO}_4} \]

Now, we can find the moles of C2O4^2- that react with 0.002 moles of KMnO4:

\[ \text{Moles of C}_2\text{O}_4^{2-} = 0.002 \, \text{mol KMnO}_4 \times \frac{5 \, \text{mol C}_2\text{O}_4^{2-}}{2 \, \text{mol KMnO}_4} = 0.002 \times \frac{5}{2} = 0.005 \, \text{mol C}_2\text{O}_4^{2-} \]

### Step 4: Conclusion
Thus, 20 mL of 0.1 M KMnO4 is equivalent to 0.005 moles of oxalic acid.