1 " mol of "`IO_(3)^(ɵ)` ions is heated with excess of `I^(ɵ)` ions in the presence of acidic conditions as per the following equation
`IO_(3)^(ɵ)+I^(ɵ)toI_(2)`
How many moles of acidified hypo solution will be required to react completely with `I_(2)` thus produced?

Options

A. 1
B. 3
C. 5
D. 6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equations
The first reaction involves the reduction of iodate ions (`IO3^-`) by iodide ions (`I^-`) to form iodine (`I2`):
\[ \text{IO}_3^- + \text{I}^- \rightarrow \text{I}_2 \]

The second reaction involves the reaction of iodine with sodium thiosulfate (hypo) to form iodide ions and tetrathionate ions:
\[ \text{I}_2 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2\text{I}^- + \text{Na}_2\text{S}_4\text{O}_6 \]

### Step 2: Determine the number of moles of `I2` produced
From the first reaction, 1 mole of `IO3^-` reacts with excess `I^-` to produce 1 mole of `I2`. Therefore, the amount of `I2` produced is:
\[ \text{Moles of } I_2 = 1 \text{ mole} \]

### Step 3: Calculate the equivalence of `IO3^-`
The n-factor for `IO3^-` can be calculated based on the change in oxidation state of iodine:
- In `IO3^-`, the oxidation state of iodine is +5.
- In `I2`, the oxidation state of iodine is 0.

The change in oxidation state is:
\[ 5 \text{ (from IO3^-)} \rightarrow 0 \text{ (in I2)} \]

Thus, the n-factor for `IO3^-` is 5 (as 5 electrons are gained). Therefore, the equivalence of `IO3^-` is:
\[ \text{Equivalence of } IO3^- = \text{n-factor} \times \text{moles} = 5 \times 1 = 5 \]

### Step 4: Determine the equivalence of `I2`
From the second reaction, 1 mole of `I2` will react with 1 mole of sodium thiosulfate (`Na2S2O3`) to produce 2 moles of `I^-`. The equivalence of `I2` is also 1 (since 1 mole of `I2` produces 2 moles of `I^-`, which means it consumes 2 equivalents of thiosulfate). Therefore, the equivalence of `I2` produced is:
\[ \text{Equivalence of } I2 = 1 \]

### Step 5: Relate the equivalences
From the reactions, we can relate the equivalences:
\[ \text{Equivalence of } IO3^- = \text{Equivalence of } I2 = \text{Equivalence of } Na2S2O3 \]

### Step 6: Calculate the moles of `Na2S2O3` required
Since the equivalence of `IO3^-` is 5, the moles of `Na2S2O3` required to completely react with the `I2` produced is:
\[ \text{Moles of } Na2S2O3 = \frac{\text{Equivalence of } IO3^-}{\text{n-factor of } Na2S2O3} = \frac{5}{1} = 5 \]

### Final Answer
Thus, the number of moles of acidified hypo solution required to react completely with `I2` produced is:
**5 moles**

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