`5H_(2)C_(2)O_(4)(aq)+2MnO_(4)(aq)+6H^(+)(aq) to 2Mn^(2+)(aq)+10CO_(2)(g)+8H_(2)O(l)` Oxalic acid, `H_(2)C_(2)O_(2)` , reacts with permanganate ion accroding to the balanced equation above. How many mL of 0.0154 M `KMnO_(4)` solution are required to react with 25.0mL of 0.0208 M `H_(2)C_(2)O_(4)` solution?

Options

A. 13.5 mL
B. 18.5 mL
C. 33.8 mL
D. 84.4 mL

Answer

To solve the problem, we need to determine how many mL of a 0.0154 M KMnO4 solution are required to react with 25.0 mL of a 0.0208 M H2C2O4 solution based on the balanced chemical equation provided.

### Step-by-Step Solution:

1. **Identify the Reaction and Stoichiometry**:
The balanced chemical equation is:
\[ 5 \text{H}_2\text{C}_2\text{O}_4 + 2 \text{MnO}_4^- + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]
From the equation, we can see that 5 moles of oxalic acid (H2C2O4) react with 2 moles of permanganate ion (MnO4^-).

2. **Calculate the Number of Moles of H2C2O4**:
We need to find the number of moles of H2C2O4 in 25.0 mL of a 0.0208 M solution.
\[ \text{Moles of H2C2O4} = \text{Molarity} \times \text{Volume (in L)} = 0.0208 \, \text{mol/L} \times 0.025 \, \text{L} = 0.00052 \, \text{mol} \]

3. **Determine the Equivalent of H2C2O4**:
The n-factor for H2C2O4 can be determined from the reaction. Since 5 moles of H2C2O4 yield 10 moles of CO2, the n-factor is 2 (as each mole of H2C2O4 provides 2 moles of CO2).
\[ \text{Equivalents of H2C2O4} = \text{Moles} \times \text{n-factor} = 0.00052 \, \text{mol} \times 2 = 0.00104 \, \text{equivalents} \]

4. **Calculate the Equivalents of KMnO4 Needed**:
From the stoichiometry of the reaction, 2 moles of MnO4^- react with 5 moles of H2C2O4, which means the n-factor for KMnO4 is 5.
\[ \text{Equivalents of KMnO4} = \text{Equivalents of H2C2O4} = 0.00104 \, \text{equivalents} \]

5. **Find the Volume of KMnO4 Required**:
Using the molarity of KMnO4 (0.0154 M), we can find the volume required to provide 0.00104 equivalents.
\[ \text{Volume (L)} = \frac{\text{Equivalents}}{\text{Molarity}} = \frac{0.00104 \, \text{equivalents}}{0.0154 \, \text{mol/L}} = 0.0675 \, \text{L} \]
Converting to mL:
\[ \text{Volume (mL)} = 0.0675 \, \text{L} \times 1000 \, \text{mL/L} = 67.5 \, \text{mL} \]

6. **Final Calculation**:
Since we need to find the volume of KMnO4 that corresponds to the equivalents of H2C2O4, we need to adjust based on the stoichiometric ratio from the balanced equation:
\[ \text{Volume of KMnO4} = \frac{0.00104 \, \text{equivalents}}{5/2} = 0.000416 \, \text{equivalents} \]
\[ \text{Volume (mL)} = \frac{0.000416 \, \text{equivalents}}{0.0154 \, \text{mol/L}} \times 1000 \, \text{mL/L} = 27.0 \, \text{mL} \]

### Final Answer:
The volume of 0.0154 M KMnO4 solution required to react with 25.0 mL of 0.0208 M H2C2O4 solution is approximately **27.0 mL**.