Sodium iodate is treated with calculated amount of sodium bisulphite to prepare one mole of iodine. How many moles of sodium disulphite are required to prepare one mole of iodine form sodium iodate ?

Explanation

To solve the problem of how many moles of sodium bisulfite (NaHSO3) are required to prepare one mole of iodine (I2) from sodium iodate (NaIO3), we need to follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sodium iodate and sodium bisulfite can be represented as follows:

\[ \text{NaIO}_3 + \text{NaHSO}_3 \rightarrow \text{I}_2 + \text{NaHSO}_4 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \]

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

The balanced equation is:

\[ 2 \text{NaIO}_3 + 5 \text{NaHSO}_3 \rightarrow \text{I}_2 + 3 \text{NaHSO}_4 + 2 \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

### Step 3: Determine the stoichiometry
From the balanced equation, we can see that:

- 2 moles of NaIO3 react with 5 moles of NaHSO3 to produce 1 mole of I2.

### Step 4: Calculate the moles of sodium bisulfite required
To find out how many moles of sodium bisulfite are needed to produce 1 mole of iodine, we can set up a proportion based on the stoichiometry:

- If 2 moles of NaIO3 produce 1 mole of I2, then to produce 1 mole of I2, we need 5 moles of NaHSO3.

Thus, the calculation is:

\[ \text{Moles of NaHSO}_3 = \frac{5 \text{ moles of NaHSO}_3}{1 \text{ mole of I}_2} = 5 \text{ moles of NaHSO}_3 \]

### Conclusion
Therefore, **5 moles of sodium bisulfite are required to prepare 1 mole of iodine from sodium iodate.**

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