In the following reaction (unbalanced), equivalent weight of `As_(2)S_(3)` is related to molecular weight `M` by
`As_(2)S_(3)+H+NO_(3)^(-) rarr NO+H_(2)O+AsO_(4)^(3-)+SO_(4)^(2-)`

Options

A. `(M)/(2)`
B. `(M)/(4)`
C. `(M)/(24)`
D. `(M)/(28)`

Answer

To find the equivalent weight of \( As_2S_3 \) in the given reaction, we will follow these steps:

### Step 1: Write the unbalanced reaction
The unbalanced reaction is:
\[ As_2S_3 + H^+ + NO_3^- \rightarrow NO + H_2O + AsO_4^{3-} + SO_4^{2-} \]

### Step 2: Balance the reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides.

- **Arsenic (As)**: There are 2 As atoms in \( As_2S_3 \) and 2 As atoms in \( AsO_4^{3-} \).
- **Sulfur (S)**: There are 3 S atoms in \( As_2S_3 \) and 3 S atoms in \( SO_4^{2-} \).

Thus, the balanced reaction is:
\[ As_2S_3 + 6 H^+ + 6 NO_3^- \rightarrow 6 NO + 3 H_2O + 2 AsO_4^{3-} + 3 SO_4^{2-} \]

### Step 3: Determine oxidation states
Next, we need to determine the oxidation states of arsenic and sulfur in the reactants and products.

- In \( As_2S_3 \):
- Let the oxidation state of As be \( +3 \) (since \( 2x + 3(-2) = 0 \) gives \( x = +3 \)).
- The oxidation state of S is \( -2 \).

- In \( AsO_4^{3-} \):
- The oxidation state of As is \( +5 \) (since \( x + 4(-2) = -3 \) gives \( x = +5 \)).

- In \( SO_4^{2-} \):
- The oxidation state of S is \( +6 \) (since \( x + 4(-2) = -2 \) gives \( x = +6 \)).

### Step 4: Calculate the change in oxidation states
Now, we calculate the change in oxidation states for both arsenic and sulfur:

- For arsenic:
- Change from \( +3 \) to \( +5 \): \( 5 - 3 = +2 \) (for 2 As atoms, total change = \( 2 \times 2 = 4 \)).

- For sulfur:
- Change from \( -2 \) to \( +6 \): \( 6 - (-2) = 8 \) (for 3 S atoms, total change = \( 3 \times 8 = 24 \)).

### Step 5: Calculate the total number of electrons transferred
The total number of electrons transferred in the reaction is:
\[ 4 \text{ (from As)} + 24 \text{ (from S)} = 28 \]

### Step 6: Determine the equivalent weight
The equivalent weight (or equivalent mass) of \( As_2S_3 \) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight (M)}}{\text{Valency Factor}} \]

From our calculations, the valency factor is 28. Therefore, the equivalent weight of \( As_2S_3 \) is:
\[ \text{Equivalent Weight} = \frac{M}{28} \]

### Final Answer
The equivalent weight of \( As_2S_3 \) in this reaction is:
\[ \frac{M}{28} \]

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