Iodine can be prepared by the following reactions.
`2NaIO_(3) + 5NaSO_(3) rarr 2NaSO_(4) + 2Na_(2) SO_(4) + H_(2)O + I_(2)`
How much `NaIO_(3)` is reuired to produce `127 g` is `I_(2)`?

Options

A. `1.98 kg`
B. `3.96 kg`
C. `5.94 kg`
D. `0.99 kg`

Explanation

To determine how much \( \text{NaIO}_3 \) is required to produce \( 127 \, \text{g} \) of \( \text{I}_2 \), we can follow these steps:

### Step 1: Calculate the moles of \( \text{I}_2 \)
First, we need to find the number of moles of \( \text{I}_2 \) produced from the given mass.

\[ \text{Molar mass of } \text{I}_2 = 2 \times 127 \, \text{g/mol} = 254 \, \text{g/mol} \]

Now, we can calculate the moles of \( \text{I}_2 \):

\[ \text{Moles of } \text{I}_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{127 \, \text{g}}{254 \, \text{g/mol}} = 0.5 \, \text{moles} \]

### Step 2: Determine the moles of \( \text{NaIO}_3 \) required
From the balanced chemical equation:

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

We can see that 2 moles of \( \text{NaIO}_3 \) produce 1 mole of \( \text{I}_2 \). Therefore, to find the moles of \( \text{NaIO}_3 \) required for 0.5 moles of \( \text{I}_2 \):

\[ \text{Moles of } \text{NaIO}_3 = 0.5 \, \text{moles of } \text{I}_2 \times \frac{2 \, \text{moles of } \text{NaIO}_3}{1 \, \text{mole of } \text{I}_2} = 1 \, \text{mole of } \text{NaIO}_3 \]

### Step 3: Calculate the mass of \( \text{NaIO}_3 \)
Now, we need to find the mass of \( \text{NaIO}_3 \) using its molar mass.

The molar mass of \( \text{NaIO}_3 \) can be calculated as follows:

\[ \text{Molar mass of } \text{NaIO}_3 = \text{Atomic mass of Na} + \text{Atomic mass of I} + 3 \times \text{Atomic mass of O} \]
\[ = 23 \, \text{g/mol} + 127 \, \text{g/mol} + 3 \times 16 \, \text{g/mol} = 23 + 127 + 48 = 198 \, \text{g/mol} \]

Now, we can calculate the mass of \( \text{NaIO}_3 \):

\[ \text{Mass of } \text{NaIO}_3 = \text{Moles} \times \text{Molar mass} = 1 \, \text{mole} \times 198 \, \text{g/mol} = 198 \, \text{g} \]

### Step 4: Convert the mass to kilograms
To express the mass in kilograms:

\[ 198 \, \text{g} = 0.198 \, \text{kg} \]

### Final Answer
The amount of \( \text{NaIO}_3 \) required to produce \( 127 \, \text{g} \) of \( \text{I}_2 \) is \( 0.198 \, \text{kg} \).

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