For the redox recation
`MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)toMn^(2+)+CO_(2)+H_(2)O`
the correct coefficients of the reactants for the blalanced equation are

Options

A. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,16,5):}`
B. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}`
C. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(16,5,2):}`
D. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(5,16,2):}`

Answer

To balance the redox reaction given by:

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

we will use the oxidation number method. Here are the steps to find the correct coefficients of the reactants for the balanced equation:

### Step 1: Assign Oxidation States
- For Mn in \(\text{MnO}_4^{-}\): Let the oxidation state be \(x\). The equation is \(x - 8 = -1\), so \(x = +7\).
- For C in \(\text{C}_2\text{O}_4^{2-}\): Let the oxidation state be \(y\). The equation is \(2y - 8 = -2\), so \(2y = +6\) and \(y = +3\).
- For Mn in \(\text{Mn}^{2+}\): The oxidation state is \(+2\).
- For C in \(\text{CO}_2\): The oxidation state is \(z\). The equation is \(z - 4 = 0\), so \(z = +4\).

### Step 2: Identify Changes in Oxidation States
- Mn changes from \(+7\) to \(+2\) (a decrease of \(5\)).
- Each C changes from \(+3\) to \(+4\) (an increase of \(1\)). Since there are 2 C atoms in \(\text{C}_2\text{O}_4^{2-}\), the total increase is \(2\).

### Step 3: Balance the Changes in Oxidation States
To balance the changes in oxidation states, we need to equalize the total decrease and increase:
- Decrease: \(5\) (from Mn)
- Increase: \(2\) (from C)

To equalize, we multiply the oxidation half-reaction by \(2.5\) for the decrease and \(5\) for the increase. However, since we cannot have fractional coefficients, we will multiply the entire reaction by \(2\) to avoid fractions.

### Step 4: Write the Half-Reactions
1. **Reduction half-reaction**:
\[ 2 \text{MnO}_4^{-} \rightarrow 2 \text{Mn}^{2+} \]
2. **Oxidation half-reaction**:
\[ 5 \text{C}_2\text{O}_4^{2-} \rightarrow 10 \text{CO}_2 \]

### Step 5: Balance Oxygen and Hydrogen
Now, we need to balance the oxygen and hydrogen atoms:
- Total oxygen on the left: \(2 \times 4 + 5 \times 4 = 28\)
- Total oxygen on the right: \(10 \times 2 + 8 \text{(from water)} = 20 + 8 = 28\)

To balance the hydrogen, we add \(16 \text{H}^{+}\) to the left side:
\[ 2 \text{MnO}_4^{-} + 5 \text{C}_2\text{O}_4^{2-} + 16 \text{H}^{+} \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]

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

### Step 7: Identify Coefficients
The coefficients of the reactants are:
- For \(\text{MnO}_4^{-}\): \(2\)
- For \(\text{C}_2\text{O}_4^{2-}\): \(5\)
- For \(\text{H}^{+}\): \(16\)

### Final Answer
Thus, the correct coefficients of the reactants for the balanced equation are \(2, 5, 16\).