In the reaction
`xHI+yHNO_(3) rarr NO+I_(2)+H_(2)O`

Options

A. `x=3`, `y=2`
B. `x=2`, `y=3`
C. `x=6`, `y=2`
D. `x=6`, `y=1`

Explanation

To solve the reaction \( xHI + yHNO_3 \rightarrow NO + I_2 + H_2O \) and find the values of \( x \) and \( y \), we will follow these steps:

### Step 1: Identify Oxidation and Reduction
In this reaction, we need to identify which species are oxidized and which are reduced.

- **Oxidation**: The iodide ion \( I^- \) in \( HI \) is oxidized to \( I_2 \). The oxidation state of iodine changes from -1 in \( HI \) to 0 in \( I_2 \).

- **Reduction**: The nitrogen in \( HNO_3 \) is reduced to \( NO \). The oxidation state of nitrogen changes from +5 in \( HNO_3 \) to +2 in \( NO \).

### Step 2: Write Half-Reactions
Now, we can write the half-reactions for oxidation and reduction.

- **Oxidation Half-Reaction**:
\[ 2I^- \rightarrow I_2 + 2e^- \]

- **Reduction Half-Reaction**:
\[ 2HNO_3 + 6e^- \rightarrow 2NO + 4H_2O \]

### Step 3: Balance Electrons
To balance the electrons lost in oxidation with those gained in reduction, we need to multiply the oxidation half-reaction by 3 (to get 6 electrons):

- **Balanced Oxidation Half-Reaction**:
\[ 6I^- \rightarrow 3I_2 + 6e^- \]

### Step 4: Combine Half-Reactions
Now we can combine the balanced half-reactions:

\[ 6I^- + 2HNO_3 \rightarrow 3I_2 + 2NO + 4H_2O \]

### Step 5: Determine Coefficients
From the combined balanced equation, we can identify the coefficients:

- For \( HI \): Since \( 6I^- \) comes from \( 6HI \), we have \( x = 6 \).
- For \( HNO_3 \): We have \( 2HNO_3 \), so \( y = 2 \).

### Final Balanced Equation
The final balanced equation is:
\[ 6HI + 2HNO_3 \rightarrow 3I_2 + 2NO + 4H_2O \]

### Conclusion
Thus, the values of \( x \) and \( y \) are:
- \( x = 6 \)
- \( y = 2 \)