The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the following reaction is
`3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`

Options

A. `(4xx63)/(3)`
B. `(63)/(5)`
C. `(63)/(3)`
D. `(63)/(8)`

Explanation

To find the equivalent weight of \( HNO_3 \) in the given reaction, we will follow these steps:

### Step 1: Write down the reaction
The reaction provided is:
\[ 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O \]

### Step 2: Identify the change in oxidation states
In the reaction, we need to determine the change in oxidation states of nitrogen in \( HNO_3 \) and \( NO \).

- In \( HNO_3 \), the oxidation state of nitrogen (N) is +5.
- In \( NO \), the oxidation state of nitrogen is +2.

### Step 3: Calculate the change in oxidation state
The change in oxidation state for nitrogen can be calculated as follows:
- Change from +5 (in \( HNO_3 \)) to +2 (in \( NO \)) is a reduction of 3 units.

### Step 4: Determine the number of moles of \( HNO_3 \) reacting
From the balanced equation, we see that 8 moles of \( HNO_3 \) are used in the reaction.

### Step 5: Calculate the number of equivalents
The number of equivalents (n-factor) of \( HNO_3 \) can be calculated based on the number of moles reacting and the change in oxidation state:
- Since 3 moles of electrons are gained per mole of \( HNO_3 \) (because each \( HNO_3 \) contributes to the reduction of nitrogen), the total number of equivalents for 8 moles of \( HNO_3 \) is:
\[ \text{n-factor} = \text{change in oxidation state} \times \text{number of moles} = 3 \times 8 = 24 \]

### Step 6: Calculate the equivalent weight
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]

Given that the molecular weight of \( HNO_3 \) is 63:
\[ \text{Equivalent Weight of } HNO_3 = \frac{63}{3} = 21 \]

### Final Answer
The equivalent weight of \( HNO_3 \) in this reaction is 21.

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