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`

Explanation

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

### Step 1: Calculate the milliequivalents of \( Na_2S_2O_3 \)

First, we need to calculate the milliequivalents (mEq) of \( Na_2S_2O_3 \) used in the reaction. The formula for milliequivalents is:

\[ \text{mEq} = \text{Normality} \times \text{Volume (L)} \]

Given:
- Normality of \( Na_2S_2O_3 = 0.11 \, N \)
- Volume = 40 mL = 0.040 L

Calculating:

\[ \text{mEq of } Na_2S_2O_3 = 0.11 \, N \times 0.040 \, L = 0.0044 \, \text{equivalents} = 4.4 \, \text{mEq} \]

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

From the balanced chemical equation:

\[ S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I \]

We see that 1 mole of \( I_2 \) reacts with 1 mole of \( S_2O_3^{2-} \). Therefore, the mEq of \( I_2 \) will also be equal to the mEq of \( Na_2S_2O_3 \):

\[ \text{mEq of } I_2 = 4.4 \, \text{mEq} \]

### Step 3: Determine the n-factor for \( I_2 \)

The n-factor is determined by the change in oxidation state. In this reaction:
- The oxidation state of \( I_2 \) changes from 0 to -1 for each iodine atom.
- Since there are 2 iodine atoms in \( I_2 \), the total change is:

\[ \text{n-factor} = 2 \, \text{(for 2 iodine atoms)} \]

### Step 4: Calculate the millimoles of \( I_2 \)

Using the relationship between milliequivalents, millimoles, and n-factor:

\[ \text{mEq} = \text{millimoles} \times \text{n-factor} \]

Rearranging gives us:

\[ \text{millimoles of } I_2 = \frac{\text{mEq of } I_2}{\text{n-factor}} = \frac{4.4 \, \text{mEq}}{2} = 2.2 \, \text{mmol} \]

### Step 5: Calculate the mass of \( I_2 \)

Now we can calculate the mass of \( I_2 \) using its molar mass. The molar mass of \( I_2 \) is approximately 254 g/mol.

Using the formula:

\[ \text{mass} = \text{millimoles} \times \text{molar mass} \times \frac{1 \, \text{g}}{1000 \, \text{mg}} \]

Calculating:

\[ \text{mass of } I_2 = 2.2 \, \text{mmol} \times 254 \, \text{g/mol} = 558.8 \, \text{mg} = 0.558 \, \text{g} \]

### Final Answer

The mass of \( I_2 \) present in the solution is **0.558 grams**.

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