When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation
`xCu+yHNO_(3)rarrCu(NO_(3))_(2)+NO+NO_(2)+H_(2)O`
The coefficients `x` and `y` are

Options

A. `2` and `3`
B. `2` and `6`
C. `1` and `3`
D. `3` and` 8`

Explanation

To solve the problem of determining the coefficients \( x \) and \( y \) in the reaction of copper with nitric acid, we need to balance the given redox reaction:

\[ x \text{Cu} + y \text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + \text{NO} + \text{NO}_2 + \text{H}_2\text{O} \]

### Step 1: Identify Oxidation States
- Copper (Cu) starts at an oxidation state of 0 and is oxidized to +2 in \(\text{Cu(NO}_3)_2\).
- Nitrogen in \(\text{HNO}_3\) is in the +5 oxidation state. In the products, nitrogen in \(\text{NO}\) is +2 and in \(\text{NO}_2\) is +4.

### Step 2: Write Half-Reactions
1. **Oxidation Half-Reaction**:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
For \( x \) moles of Cu, the total electrons lost will be \( 2x \).

2. **Reduction Half-Reactions**:
- For \(\text{NO}_2\):
\[ \text{HNO}_3 + e^- \rightarrow \text{NO}_2 + \text{H}^+ \]
(1 electron is gained for each \(\text{NO}_2\) produced)

- For \(\text{NO}\):
\[ \text{HNO}_3 + 3e^- \rightarrow \text{NO} + 2\text{H}^+ \]
(3 electrons are gained for each \(\text{NO}\) produced)

### Step 3: Balance Electrons
To balance the electrons, we need to find a common multiple. The least common multiple of 2 (from Cu) and 4 (from NO2) and 3 (from NO) is 12.

- For \( x \) moles of Cu, we need \( 6 \) moles of Cu to produce \( 12 \) electrons.
- For \( 4 \) moles of \(\text{NO}_2\), we need \( 4 \) moles of \(\text{HNO}_3\) (4 electrons).
- For \( 2 \) moles of \(\text{NO}\), we need \( 6 \) moles of \(\text{HNO}_3\) (18 electrons).

### Step 4: Combine Half-Reactions
Combining the half-reactions gives us:
\[ 2 \text{Cu} + 4 \text{HNO}_3 \rightarrow 2 \text{Cu(NO}_3)_2 + 2 \text{NO} + 2 \text{NO}_2 + 4 \text{H}_2\text{O} \]

### Step 5: Final Coefficients
From the balanced equation:
- \( x = 2 \) (for Cu)
- \( y = 6 \) (for HNO3)

### Conclusion
Thus, the coefficients \( x \) and \( y \) are:
\[ x = 2, \quad y = 6 \]