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 HNO₃ in the given reaction, we can follow these steps:

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

### Step 2: Determine the Oxidation States
1. **Oxidation state of Nitrogen in HNO₃**:
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of Nitrogen (N) be \( x \).
- The equation for HNO₃ is:
\[ x + 3(-2) + 1 = 0 \]
\[ x - 6 + 1 = 0 \]
\[ x = +5 \]

2. **Oxidation state of Nitrogen in NO**:
- Let the oxidation state of Nitrogen in NO be \( y \).
- The equation for NO is:
\[ y - 2 = 0 \]
\[ y = +2 \]

### Step 3: Calculate the Change in Oxidation State
- The change in oxidation state for Nitrogen from HNO₃ to NO is:
\[ +5 \text{ (in HNO₃)} \rightarrow +2 \text{ (in NO)} \]
- The change is:
\[ 5 - 2 = 3 \]

### Step 4: Determine the N Factor
- The N factor for HNO₃ in this reaction is equal to the total change in oxidation state per molecule of HNO₃ that is reduced to NO. Since 2 moles of NO are produced from 2 moles of HNO₃, the N factor is 3.

### Step 5: Calculate the Equivalent Weight
- The formula for equivalent weight is:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{N Factor}} \]
- Given that the molecular weight of HNO₃ is 63, we can substitute:
\[ \text{Equivalent Weight} = \frac{63}{3} = 21 \]

### Final Answer
The equivalent weight of HNO₃ in the reaction is **21 g/equiv**.

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