Oxalic acid, `H_(2)C_(2)O_(4)`, reacts with paramagnet ion according to the balanced equation `5H_(2)C_(2)O_(4)(aq)+2MnO_(4)^(-)(aq)+6H^(+)(aq)hArr2Mn^(2+)(aq)+10CO_(2)(g)+8H_(2)O(l)`. The volume in mL of 0.0162 M `KMnO_(4)` solution required to react with 25.0 mL of 0.022 M `H_(2)C_(2)O_(4)` solution is

Options

A. 13.6
B. 18.5
C. 33.8
D. 84.4

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation given is:
\[ 5H_2C_2O_4(aq) + 2MnO_4^-(aq) + 6H^+(aq) \rightarrow 2Mn^{2+}(aq) + 10CO_2(g) + 8H_2O(l) \]

### Step 2: Determine the n-factor for each reactant
- For \( KMnO_4 \):
- Manganese changes from +7 in \( MnO_4^- \) to +2 in \( Mn^{2+} \).
- The change in oxidation state is \( 7 - 2 = 5 \).
- Therefore, the n-factor for \( KMnO_4 \) is 5.

- For \( H_2C_2O_4 \):
- Each carbon in \( C_2O_4^{2-} \) changes from +3 to +4 (as determined from oxidation states).
- Since there are 2 carbons, the total change is \( 1 \times 2 = 2 \).
- Therefore, the n-factor for \( H_2C_2O_4 \) is 2.

### Step 3: Calculate the equivalents of each reactant
- For \( H_2C_2O_4 \):
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
\[ \text{Normality of } H_2C_2O_4 = 0.022 \, \text{mol/L} \times 2 = 0.044 \, \text{N} \]
\[ \text{Equivalents of } H_2C_2O_4 = \text{Normality} \times \text{Volume (L)} = 0.044 \, \text{N} \times 0.025 \, \text{L} = 0.0011 \, \text{equivalents} \]

- For \( KMnO_4 \):
\[ \text{Equivalents of } KMnO_4 = \text{Normality} \times \text{Volume (L)} = \text{Normality} \times V \]
\[ \text{Normality of } KMnO_4 = 0.0162 \, \text{mol/L} \times 5 = 0.081 \, \text{N} \]
\[ \text{Equivalents of } KMnO_4 = 0.081 \, \text{N} \times V \]

### Step 4: Set the equivalents equal
Since the equivalents of \( H_2C_2O_4 \) and \( KMnO_4 \) are equal:
\[ 0.0011 = 0.081 \times V \]

### Step 5: Solve for \( V \)
\[ V = \frac{0.0011}{0.081} \approx 0.01358 \, \text{L} = 13.58 \, \text{mL} \]

### Step 6: Round the answer
The volume of \( KMnO_4 \) solution required is approximately 13.6 mL.

### Final Answer:
The volume in mL of 0.0162 M \( KMnO_4 \) solution required to react with 25.0 mL of 0.022 M \( H_2C_2O_4 \) solution is **13.6 mL**.

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