`HNO_(3)` oxidies `NH_(4)^(+)` ions to nitrogen and itself gets reduced to `NO_(2)`. The moles of `HNO_(3)` required by 1 mole of `(NH_(4))_(2)SO_(4)` is:

Options

A. 4
B. 5
C. 6
D. 2

Explanation

To determine the moles of \( HNO_3 \) required by 1 mole of \( (NH_4)_2SO_4 \), we will follow these steps:

### Step 1: Identify the oxidation states
- In \( NH_4^+ \), nitrogen has an oxidation state of -3.
- In \( N_2 \), nitrogen has an oxidation state of 0.
- In \( NO_2 \), nitrogen has an oxidation state of +4.
- In \( HNO_3 \), nitrogen has an oxidation state of +5.

### Step 2: Determine the change in oxidation states
- For \( NH_4^+ \) to \( N_2 \):
\[ \text{Change in oxidation state} = 0 - (-3) = 3 \]
- For \( HNO_3 \) to \( NO_2 \):
\[ \text{Change in oxidation state} = 5 - 4 = 1 \]

### Step 3: Calculate the valency factors
- The valency factor for \( NH_4^+ \) is 3 (since it goes from -3 to 0).
- The valency factor for \( HNO_3 \) is 1 (since it goes from +5 to +4).

### Step 4: Relate the moles of \( HNO_3 \) to \( NH_4^+ \)
- From the stoichiometry of the reaction:
\[ 1 \text{ mole of } NH_4^+ \text{ requires } 3 \text{ moles of } HNO_3 \]

### Step 5: Consider the moles of \( NH_4^+ \) in \( (NH_4)_2SO_4 \)
- \( (NH_4)_2SO_4 \) dissociates to give 2 moles of \( NH_4^+ \):
\[ 1 \text{ mole of } (NH_4)_2SO_4 \rightarrow 2 \text{ moles of } NH_4^+ \]

### Step 6: Calculate the total moles of \( HNO_3 \) required
- Since 1 mole of \( NH_4^+ \) requires 3 moles of \( HNO_3 \):
\[ 2 \text{ moles of } NH_4^+ \text{ will require } 2 \times 3 = 6 \text{ moles of } HNO_3 \]

### Final Answer
Thus, 1 mole of \( (NH_4)_2SO_4 \) requires **6 moles of \( HNO_3 \)**.

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