A mixture of `n_(1)` moles of `Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` is titrated separately with `H_(2)O_(2)` and `KOH`, to reach at equivalence point. Which of the following statement is/are correct?

Options

A. Moles of `H_(2)O_(2)` and `KOH` are `n_(1)+n_(2)` and `n_(2)`
B. Moles of `H_(2)O_(2)` and `KOH` is `n_(1)+(n_(2))/(2)` and `n_(1)`
C. n-factors of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)`, respectively, are 1 and 2.
D. n-factors of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and KOH, respectively, are 2 and 1.

Explanation

To solve the problem, we need to analyze the titration of a mixture of sodium oxalate (Na₂C₂O₄) and sodium hydrogen oxalate (NaHC₂O₄) with hydrogen peroxide (H₂O₂) and potassium hydroxide (KOH) separately, and determine the correct statements based on the stoichiometry of the reactions.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Moles**:
- Let \( n_1 \) be the moles of Na₂C₂O₄.
- Let \( n_2 \) be the moles of NaHC₂O₄.

2. **Titration with H₂O₂**:
- Both Na₂C₂O₄ and NaHC₂O₄ react with H₂O₂.
- The n-factor for both Na₂C₂O₄ and NaHC₂O₄ when reacting with H₂O₂ is 2.
- Therefore, the equivalents of H₂O₂ can be expressed as:
\[ \text{Equivalents of H₂O₂} = \text{Equivalents of Na₂C₂O₄} + \text{Equivalents of NaHC₂O₄} \]
- This gives us:
\[ \text{Equivalents of H₂O₂} = 2n_1 + 2n_2 \]
- Simplifying, we find:
\[ \text{Moles of H₂O₂} = n_1 + n_2 \]

3. **Titration with KOH**:
- Only NaHC₂O₄ reacts with KOH in an acid-base reaction.
- The n-factor for NaHC₂O₄ when reacting with KOH is 1.
- Therefore, we have:
\[ \text{Equivalents of KOH} = \text{Equivalents of NaHC₂O₄} \]
- This gives us:
\[ \text{Moles of KOH} = n_2 \]

4. **Analyzing the Statements**:
- **Statement A**: Moles of H₂O₂ = \( n_1 + n_2 \) and Moles of KOH = \( n_2 \) (Correct)
- **Statement B**: Moles of H₂O₂ = \( n_1 + n_2 \) and Moles of KOH = \( n_1 + n_2 \) (Incorrect)
- **Statement C**: The n-factor of NaHC₂O₄ with KOH is 1 and with H₂O₂ is 2 (Correct)
- **Statement D**: The n-factor of Na₂C₂O₄ with H₂O₂ is 2 and with KOH is 1 (Incorrect, as Na₂C₂O₄ does not react with KOH)

5. **Conclusion**:
- The correct statements are A and C.