The number of modes of `As_(2)S_(3)` oxidised to `H_(3)AsO_(4) " and " H_(2)SO_(4) ` by one mode of `HNO_(3)` (reduced to NO) is .

Options

A. `3/2`
B. `5/2`
C. `2/3`
D. `3/28`

Explanation

To solve the problem of determining the number of moles of \( As_2S_3 \) oxidized to \( H_3AsO_4 \) and \( H_2SO_4 \) by one mole of \( HNO_3 \) (which is reduced to \( NO \)), we will follow these steps:

### Step 1: Write the balanced chemical equation
We start by writing the unbalanced reaction:
\[ As_2S_3 + HNO_3 \rightarrow H_3AsO_4 + H_2SO_4 + NO \]

### Step 2: Determine oxidation states
Next, we need to assign oxidation states to each element in the reactants and products:
- In \( As_2S_3 \):
- \( As \): +3
- \( S \): -2
- In \( HNO_3 \):
- \( N \): +5
- In \( H_3AsO_4 \):
- \( As \): +5
- \( O \): -2
- In \( H_2SO_4 \):
- \( S \): +6
- In \( NO \):
- \( N \): +2

### Step 3: Identify changes in oxidation states
Now we calculate the changes in oxidation states:
- \( As \) changes from +3 to +5 (increase of 2 per atom, total for 2 atoms = 4)
- \( S \) changes from -2 to +6 (increase of 8 for 3 sulfur atoms = 24)
- \( N \) changes from +5 to +2 (decrease of 3 for 1 nitrogen atom)

### Step 4: Balance the changes in oxidation states
We need to balance the total increase and decrease in oxidation states:
- Total increase in oxidation states from \( As \) and \( S \): \( 4 + 24 = 28 \)
- Total decrease in oxidation states from \( N \): \( 3 \)

To balance the equation, we can set up a ratio:
\[ 3 \text{ moles of } HNO_3 \text{ will balance with } 28 \text{ moles of } As_2S_3 \]

### Step 5: Set up the proportion for the reaction
Since we are given 1 mole of \( HNO_3 \), we can set up the proportion:
\[ \frac{3 \text{ moles of } As_2S_3}{28 \text{ moles of } HNO_3} = \frac{x \text{ moles of } As_2S_3}{1 \text{ mole of } HNO_3} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives us:
\[ 3 = 28x \]
\[ x = \frac{3}{28} \]

### Conclusion
Thus, the number of moles of \( As_2S_3 \) oxidized by one mole of \( HNO_3 \) is:
\[ \frac{3}{28} \]

### Final Answer
The number of moles of \( As_2S_3 \) oxidized to \( H_3AsO_4 \) and \( H_2SO_4 \) by one mole of \( HNO_3 \) is \( \frac{3}{28} \).

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