A mixture of `Na_2C_2O_4` and `KHC_2O_4`. `H_2C_2O_4` required equal volume of 0.1 M `KMnO_4` and 0.1 M NaOH separately. What is the molar ratio of `Na_2C_2O_4` and `KHC_2O_4`. `H_2C_2O_4` in the mixture?

Answer

To determine the molar ratio of `Na2C2O4` and `KHC2O4.H2C2O4` in the mixture, we will analyze the reactions with `KMnO4` and `NaOH` separately.

### Step-by-Step Solution:

1. **Define Variables**:
- Let the number of moles of `Na2C2O4` be \( x \).
- Let the number of moles of `KHC2O4.H2C2O4` be \( y \).

2. **Determine Oxalate Ions**:
- `Na2C2O4` contributes \( 1 \) mole of `C2O4^2-` per mole, so from \( x \) moles, it contributes \( x \) moles of `C2O4^2-`.
- `KHC2O4.H2C2O4` contributes \( 2 \) moles of `C2O4^2-` per mole (1 from `KHC2O4` and 1 from `H2C2O4`), so from \( y \) moles, it contributes \( 2y \) moles of `C2O4^2-`.

3. **Total Oxalate Ions**:
- The total moles of `C2O4^2-` in the mixture is \( x + 2y \).

4. **Reaction with KMnO4**:
- The balanced reaction is:
\[ 5 \text{C2O4}^{2-} + 2 \text{MnO4}^- \rightarrow 10 \text{CO2} + 2 \text{Mn}^{2+} \]
- From the stoichiometry, \( 5 \) moles of `C2O4^2-` react with \( 2 \) moles of `MnO4^-`.
- Therefore, the moles of `MnO4^-` required for the total oxalate is:
\[ \text{Moles of } \text{MnO4}^- = \frac{2}{5}(x + 2y) \]

5. **Using Molarity and Volume**:
- The volume of `KMnO4` used is \( V \) mL, and the concentration is \( 0.1 \, M \).
- Thus, the moles of `KMnO4` used can be expressed as:
\[ \text{Moles of } \text{MnO4}^- = \frac{0.1 \times V}{1000} \]

6. **Setting Equations Equal**:
- Since the volume of `KMnO4` and `NaOH` used is the same, we can set the two expressions for moles of `MnO4^-` equal:
\[ \frac{2}{5}(x + 2y) = \frac{0.1 \times V}{1000} \]

7. **Reaction with NaOH**:
- The reaction of `KHC2O4` with `NaOH` produces `H+` ions, which can be represented as:
\[ KHC2O4 + NaOH \rightarrow KNaC2O4 + H2O \]
- The moles of `H+` produced from `KHC2O4` is \( 3y \) (since each `KHC2O4` releases \( 3 \) moles of `H+`).

8. **Setting Up the NaOH Reaction**:
- The moles of `NaOH` used is equal to the moles of `H+` produced:
\[ 3y = \frac{0.1 \times V}{1000} \]

9. **Solving the Equations**:
- From the equations:
\[ \frac{2}{5}(x + 2y) = 3y \]
- Rearranging gives:
\[ 2x + 4y = 15y \]
\[ 2x = 11y \quad \Rightarrow \quad x = \frac{11}{2}y \]

10. **Finding the Molar Ratio**:
- The molar ratio \( \frac{x}{y} \) is:
\[ \frac{x}{y} = \frac{11/2}{1} = 5.5 \]

### Final Answer:
The molar ratio of `Na2C2O4` to `KHC2O4.H2C2O4` is \( 5.5:1 \).