`NalO_(3)` reacts with `NaHSO_(3)` according to eaquation `Io_(3)^(-)+3HSO_(3)^(-)to I^(-)+3H^(+)3SO_(4)^(2-)`The weight of `NaHSO_(3)` required to react with 100 mL of solution containing 0.58 gm of is :

Options

A. 5.2gm
B. 4.57gm
C. 2.3gm
D. None of the above

Explanation

To solve the problem, we need to determine the weight of `NaHSO3` required to react with `NaIO3` based on the provided chemical equation.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Weight of `NaIO3` = 0.58 g
- Volume of solution = 100 mL
- The balanced reaction is:
\[ IO_3^{-} + 3 HSO_3^{-} \rightarrow I^{-} + 3 H^{+} + 3 SO_4^{2-} \]

2. **Calculate the Molar Mass of `NaIO3`:**
- Molar mass of `NaIO3` = Na (23 g/mol) + I (127 g/mol) + 3 × O (16 g/mol)
- Molar mass = 23 + 127 + 48 = 198 g/mol

3. **Calculate the Number of Moles of `NaIO3`:**
\[ \text{Moles of NaIO3} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.58 \, \text{g}}{198 \, \text{g/mol}} \approx 0.00293 \, \text{mol} \]

4. **Determine the Equivalent of `NaIO3`:**
- From the balanced equation, `NaIO3` has a valence factor of 6 (since it changes from +5 to -1).
\[ \text{Equivalent of NaIO3} = \text{moles} \times \text{valence factor} = 0.00293 \, \text{mol} \times 6 \approx 0.01758 \, \text{equivalents} \]

5. **Determine the Equivalent of `NaHSO3`:**
- From the balanced equation, `NaHSO3` has a valence factor of 2 (since it changes from +4 to +6).
\[ \text{Equivalent of NaHSO3} = \text{equivalents of NaIO3} = 0.01758 \, \text{equivalents} \]

6. **Calculate the Number of Moles of `NaHSO3`:**
\[ \text{Moles of NaHSO3} = \frac{\text{equivalents}}{\text{valence factor}} = \frac{0.01758}{2} \approx 0.00879 \, \text{mol} \]

7. **Calculate the Molar Mass of `NaHSO3`:**
- Molar mass of `NaHSO3` = Na (23 g/mol) + H (1 g/mol) + S (32 g/mol) + 3 × O (16 g/mol)
- Molar mass = 23 + 1 + 32 + 48 = 104 g/mol

8. **Calculate the Weight of `NaHSO3`:**
\[ \text{Weight of NaHSO3} = \text{moles} \times \text{molar mass} = 0.00879 \, \text{mol} \times 104 \, \text{g/mol} \approx 0.914 \, \text{g} \]

### Final Answer:
The weight of `NaHSO3` required to react with 100 mL of solution containing 0.58 g of `NaIO3` is approximately **0.914 g**.