`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 will follow a systematic approach to determine the weight of NaHSO₃ required to react with 100 mL of a solution containing 0.58 g of NaIO₃.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ IO_3^{-} + 3 HSO_3^{-} \rightarrow I^{-} + 3 H^{+} + 3 SO_4^{2-} \]
From this equation, we can see that 1 mole of IO₃⁻ reacts with 3 moles of HSO₃⁻.

2. **Calculate the Molar Mass of NaIO₃**:
The molar mass of NaIO₃ is calculated as follows:
- Sodium (Na) = 23 g/mol
- Iodine (I) = 127 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of NaIO₃} = 23 + 127 + 48 = 198 \text{ g/mol} \]

3. **Calculate the Moles of NaIO₃**:
Given the mass of NaIO₃ is 0.58 g, we can calculate the number of moles:
\[ \text{Moles of NaIO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.58 \text{ g}}{198 \text{ g/mol}} \approx 0.00293 \text{ mol} \]

4. **Determine the Moles of NaHSO₃ Required**:
From the balanced equation, 1 mole of IO₃⁻ reacts with 3 moles of HSO₃⁻. Therefore, the moles of NaHSO₃ required are:
\[ \text{Moles of NaHSO₃} = 3 \times \text{Moles of NaIO₃} = 3 \times 0.00293 \text{ mol} \approx 0.00879 \text{ mol} \]

5. **Calculate the Molar Mass of NaHSO₃**:
The molar mass of NaHSO₃ is calculated as follows:
- Sodium (Na) = 23 g/mol
- Hydrogen (H) = 1 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of NaHSO₃} = 23 + 1 + 32 + 48 = 104 \text{ g/mol} \]

6. **Calculate the Mass of NaHSO₃ Required**:
Now, we can find the mass of NaHSO₃ required using the number of moles calculated:
\[ \text{Mass of NaHSO₃} = \text{Moles of NaHSO₃} \times \text{Molar mass of NaHSO₃} \]
\[ \text{Mass of NaHSO₃} = 0.00879 \text{ mol} \times 104 \text{ g/mol} \approx 0.914 \text{ g} \]

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