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, we need to balance the chemical reaction of copper (Cu) with nitric acid (HNO₃) to produce copper nitrate (Cu(NO₃)₂), nitric oxide (NO), nitrogen dioxide (NO₂), and water (H₂O). The goal is to find the coefficients \(x\) and \(y\) for the reaction:

\[ xCu + yHNO_3 \rightarrow Cu(NO_3)_2 + NO + NO_2 + H_2O \]

### Step 1: Write the unbalanced equation
The unbalanced equation based on the information provided is:

\[ Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO + NO_2 + H_2O \]

### Step 2: Identify the products and reactants
From the problem, we know that when copper reacts with nitric acid, it produces:
- Copper nitrate (Cu(NO₃)₂)
- Nitric oxide (NO)
- Nitrogen dioxide (NO₂)
- Water (H₂O)

### Step 3: Write two separate balanced equations
Based on the information given, we can write two separate reactions for the formation of NO and NO₂.

1. For the formation of NO:
\[ 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O \]

2. For the formation of NO₂:
\[ Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O \]

### Step 4: Combine the two equations
Now, we combine these two equations to account for both gases produced in equal volumes:

Adding the two equations gives us:
\[ 4Cu + 12HNO_3 \rightarrow 4Cu(NO_3)_2 + 2NO + 2NO_2 + 6H_2O \]

### Step 5: Simplify the equation
We can simplify this equation by dividing all coefficients by 2:

\[ 2Cu + 6HNO_3 \rightarrow 2Cu(NO_3)_2 + NO + NO_2 + 3H_2O \]

### Step 6: Identify coefficients \(x\) and \(y\)
From the simplified balanced equation:
- The coefficient \(x\) (for Cu) is 2.
- The coefficient \(y\) (for HNO₃) is 6.

### Final Answer
Thus, the coefficients \(x\) and \(y\) are:
- \(x = 2\)
- \(y = 6\)