Iodine can be prepared by the following reactions.
`2NaIO_(3) + 5NaHSO_(3) rarr 2NaSO_(4) + 2Na_(2) SO_(4) + H_(2)O + I_(2)`
How much `NaHSO_(3)` is required to produce `381 g` of `I_(2)`?

Options

A. `156.0 g`
B. `390.0 g`
C. `520.0 g`
D. `780.0 g

Explanation

To solve the problem of how much sodium bisulfite (NaHSO₃) is required to produce 381 g of iodine (I₂), we can follow these steps:

### Step 1: Write down the balanced chemical equation.
The balanced chemical equation is:
\[ 2 \text{NaIO}_3 + 5 \text{NaHSO}_3 \rightarrow 2 \text{Na}_2\text{SO}_4 + 2 \text{NaSO}_4 + \text{H}_2\text{O} + \text{I}_2 \]

### Step 2: Determine the molar mass of iodine (I₂).
The molar mass of iodine (I₂) is calculated as follows:
- Atomic mass of iodine (I) = 126.90 g/mol
- Therefore, molar mass of I₂ = \( 2 \times 126.90 \, \text{g/mol} = 253.80 \, \text{g/mol} \)

### Step 3: Calculate the number of moles of iodine (I₂) produced.
Using the mass of iodine given (381 g), we can calculate the number of moles:
\[ \text{Number of moles of I}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{381 \, \text{g}}{253.80 \, \text{g/mol}} \approx 1.5 \, \text{moles} \]

### Step 4: Use stoichiometry to find the moles of NaHSO₃ required.
From the balanced equation, we see that 5 moles of NaHSO₃ are required to produce 1 mole of I₂. Therefore, for 1.5 moles of I₂:
\[ \text{Moles of NaHSO}_3 = 5 \times 1.5 = 7.5 \, \text{moles} \]

### Step 5: Calculate the molar mass of NaHSO₃.
The molar mass of sodium bisulfite (NaHSO₃) is calculated as follows:
- Atomic mass of Na = 22.99 g/mol
- Atomic mass of H = 1.01 g/mol
- Atomic mass of S = 32.07 g/mol
- Atomic mass of O = 16.00 g/mol (3 oxygen atoms)

Thus,
\[ \text{Molar mass of NaHSO}_3 = 22.99 + 1.01 + 32.07 + (3 \times 16.00) = 104.06 \, \text{g/mol} \]

### Step 6: Calculate the mass of NaHSO₃ required.
Now, we can find the mass of NaHSO₃ needed:
\[ \text{Mass of NaHSO}_3 = \text{moles} \times \text{molar mass} = 7.5 \, \text{moles} \times 104.06 \, \text{g/mol} \approx 780.45 \, \text{g} \]

### Conclusion
The mass of sodium bisulfite (NaHSO₃) required to produce 381 g of iodine (I₂) is approximately 780 g.

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