For the redox reation
`MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)rarrMn^(2+)CO_(2)+H_(2)O`
The correct stoichiometric coefficients of `Mno_(4)^(-),C_(2)O_(4)^(2-)` and `H^(+)` respectively:

Options

A. 2,5,16
B. 16,3,12
C. 15,16,12
D. 2,16,5

Answer

To solve the redox reaction:

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

we need to determine the correct stoichiometric coefficients for \(\text{MnO}_4^{-}\), \(\text{C}_2\text{O}_4^{2-}\), and \(\text{H}^{+}\).

### Step 1: Identify Oxidation States
- **Manganese in \(\text{MnO}_4^{-}\)**: The oxidation state is +7.
- **Manganese in \(\text{Mn}^{2+}\)**: The oxidation state is +2.
- **Carbon in \(\text{C}_2\text{O}_4^{2-}\)**: Let the oxidation state of carbon be \(X\). The equation for the oxidation state is:

\[ 2X + 4(-2) = -2 \implies 2X - 8 = -2 \implies 2X = 6 \implies X = +3 \]

- **Carbon in \(\text{CO}_2\)**: The oxidation state is +4.

### Step 2: Determine the Changes in Oxidation States
- **Manganese**: +7 to +2 (reduction).
- **Carbon**: +3 to +4 (oxidation).

### Step 3: Balance the Half-Reactions
1. **Reduction Half-Reaction**:
\[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} \]
- Balance oxygen by adding water:
\[ \text{MnO}_4^{-} + 4\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} \]
- This adds 8 hydrogen atoms, so we need 8 \(\text{H}^{+}\):
\[ \text{MnO}_4^{-} + 8\text{H}^{+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
- Balance charge by adding 5 electrons:
\[ \text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

2. **Oxidation Half-Reaction**:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2 \]
- Balance by adding 2 \(\text{CO}_2\):
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^{-} \]

### Step 4: Equalize the Number of Electrons
- Multiply the oxidation half-reaction by 5:
\[ 5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10e^{-} \]

### Step 5: Combine the Half-Reactions
Combine both half-reactions:
\[ 2\text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 4\text{H}_2\text{O} \]

### Step 6: Write the Final Balanced Equation
The final balanced equation is:
\[ 2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \]

### Conclusion
The stoichiometric coefficients are:
- For \(\text{MnO}_4^{-}\): 2
- For \(\text{C}_2\text{O}_4^{2-}\): 5
- For \(\text{H}^{+}\): 16

Thus, the answer is **2, 5, and 16**.