How many gram of `I_(2)` are present in a solution which requires `40 mL` of `0.11N Na_(2)S_(2)O_(3)` to react with it, `S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I`

Options

A. `12.7 g`
B. `0.558 g`
C. `25.4 g`
D. `11.4 g`

Explanation

To solve the problem of how many grams of \( I_2 \) are present in a solution that requires \( 40 \, \text{mL} \) of \( 0.11 \, N \, Na_2S_2O_3 \) to react with it, we can follow these steps:

### Step 1: Calculate the number of equivalents of \( Na_2S_2O_3 \)

The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (L)} \]
Given:
- Normality \( N = 0.11 \, N \)
- Volume \( V = 40 \, \text{mL} = 0.040 \, \text{L} \)

Calculating the equivalents:
\[ \text{Number of equivalents of } Na_2S_2O_3 = 0.11 \times 0.040 = 0.0044 \, \text{equivalents} \]

### Step 2: Relate the equivalents of \( I_2 \) to \( Na_2S_2O_3 \)

From the balanced reaction:
\[ S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^- \]
We see that 1 equivalent of \( S_2O_3^{2-} \) reacts with 1 equivalent of \( I_2 \). Therefore, the number of equivalents of \( I_2 \) is equal to the number of equivalents of \( Na_2S_2O_3 \):
\[ \text{Number of equivalents of } I_2 = 0.0044 \, \text{equivalents} \]

### Step 3: Calculate the mass of \( I_2 \)

Using the formula for equivalents:
\[ \text{Number of equivalents} = \frac{\text{Weight (g)}}{\text{Molar mass (g/mol)} \times \text{Valency factor}} \]
For \( I_2 \):
- Molar mass = \( 254 \, \text{g/mol} \)
- Valency factor = 2 (since \( I_2 \) gives 2 moles of \( I^- \))

Rearranging the formula to find the weight:
\[ \text{Weight of } I_2 = \text{Number of equivalents} \times \text{Molar mass} \times \text{Valency factor} \]
Substituting the values:
\[ \text{Weight of } I_2 = 0.0044 \times 254 \times 2 \]
Calculating:
\[ \text{Weight of } I_2 = 0.0044 \times 254 \times 2 = 0.558 \, \text{g} \]

### Final Answer
The mass of \( I_2 \) present in the solution is \( 0.558 \, \text{g} \).

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