In following reaction
`yMnO_(4)^(-)+xH^(+)+C_(2)O_(4)^(2-)rarryMn^(++)+2CO_(2)+(x)/(2)H_(2)O,` x and y are

Options

A. 2 and 16
B. 16 and 2
C. 8 and 16
D. 5 and 2

Explanation

To solve the reaction given in the question, we need to balance the redox reaction involving permanganate ion (MnO4^-) and oxalate ion (C2O4^2-).

The reaction can be summarized as follows:

1. **Identify the reactants and products**:
- Reactants: MnO4^-, H^+, C2O4^2-
- Products: Mn^2+, CO2, H2O

2. **Write the half-reactions**:
- The reduction half-reaction for MnO4^- to Mn^2+:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
- The oxidation half-reaction for C2O4^2- to CO2:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 \text{e}^- \]

3. **Balance the electrons in the half-reactions**:
- To balance the electrons, we need to multiply the oxidation half-reaction by 5:
\[ 5 \text{C}_2\text{O}_4^{2-} \rightarrow 10 \text{CO}_2 + 10 \text{e}^- \]

4. **Combine the half-reactions**:
- Now we can add the balanced half-reactions:
\[ 2 \text{MnO}_4^- + 5 \text{C}_2\text{O}_4^{2-} + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]

5. **Identify the values of x and y**:
- From the balanced equation, we see that:
- y (the coefficient of MnO4^-) = 2
- x (the coefficient of H^+) = 16

Thus, the values of x and y are:
- \( x = 16 \)
- \( y = 2 \)

### Final Answer:
- \( x = 16 \)
- \( y = 2 \)