What is the equivalent weight of `HNO_(3)` in the given reaction?
`4Zn+10HNO_(3) rarr 4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`

Options

A. `(63)/(10)`
B. `(63)/(9)`
C. `(63)/(8)xx10`
D. `(63)/(4)xx14`

Explanation

To determine the equivalent weight of \( HNO_3 \) in the given reaction:

\[ 4Zn + 10HNO_3 \rightarrow 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O \]

we will follow these steps:

### Step 1: Identify the change in oxidation states of nitrogen in \( HNO_3 \) and \( NH_4NO_3 \)

In the reaction, nitrogen in \( HNO_3 \) is in the +5 oxidation state, while in \( NH_4NO_3 \), nitrogen in the ammonium ion \( NH_4^+ \) is in the -3 oxidation state.

### Step 2: Calculate the change in oxidation state

The change in oxidation state for nitrogen from \( +5 \) (in \( HNO_3 \)) to \( -3 \) (in \( NH_4^+ \)) is calculated as follows:

\[ \text{Change} = +5 - (-3) = +5 + 3 = 8 \]

This indicates that 8 electrons are involved in the reduction of nitrogen from \( HNO_3 \) to \( NH_4^+ \).

### Step 3: Determine the n-factor for \( HNO_3 \)

The n-factor of a substance in a redox reaction is defined as the number of moles of electrons exchanged per mole of the substance. Since 10 moles of \( HNO_3 \) are reacting to produce 1 mole of \( NH_4NO_3 \), we can set up the following relationship:

\[ \text{n-factor of } HNO_3 = \frac{\text{Total change in electrons}}{\text{Moles of } HNO_3} = \frac{8 \text{ electrons}}{10 \text{ moles of } HNO_3} = 0.8 \]

### Step 4: Calculate the equivalent weight of \( HNO_3 \)

The equivalent weight is calculated using the formula:

\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} \]

The molar mass of \( HNO_3 \) is approximately 63 g/mol. Thus, we can substitute the values:

\[ \text{Equivalent weight of } HNO_3 = \frac{63 \text{ g/mol}}{0.8} = 78.75 \text{ g/equiv} \]

### Final Answer

The equivalent weight of \( HNO_3 \) in the given reaction is \( 78.75 \text{ g/equiv} \).

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