An queous solution containing 2.14 g `KIO_3` was treated with 100 ml of 0.4 M Kl solution, the weight of `I_2` produced is-

Options

A. 6.096 (g)
B. 7.62 (g)
C. 30.48 (g)
D. 18.288 g

Explanation

To solve the problem of determining the weight of \( I_2 \) produced when an aqueous solution containing \( 2.14 \, \text{g} \) of \( KIO_3 \) is treated with \( 100 \, \text{ml} \) of \( 0.4 \, \text{M} \) \( KI \) solution, we will follow these steps:

### Step 1: Calculate the moles of \( KIO_3 \)

1. **Find the molar mass of \( KIO_3 \)**:
- Potassium (K): \( 39.1 \, \text{g/mol} \)
- Iodine (I): \( 126.9 \, \text{g/mol} \)
- Oxygen (O): \( 16.0 \, \text{g/mol} \times 3 = 48.0 \, \text{g/mol} \)
- Total molar mass of \( KIO_3 = 39.1 + 126.9 + 48.0 = 214.0 \, \text{g/mol} \)

2. **Calculate the moles of \( KIO_3 \)**:
\[ \text{Moles of } KIO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.14 \, \text{g}}{214.0 \, \text{g/mol}} = 0.01 \, \text{moles} \]

### Step 2: Calculate the moles of \( KI \)

1. **Calculate the moles of \( KI \)**:
\[ \text{Moles of } KI = \text{Molarity} \times \text{Volume (in L)} = 0.4 \, \text{mol/L} \times 0.1 \, \text{L} = 0.04 \, \text{moles} \]

### Step 3: Determine the limiting reagent

1. **Write the balanced reaction**:
\[ KIO_3 + 5 KI \rightarrow 3 K_2O + 3 I_2 \]

2. **Determine the stoichiometric ratio**:
- From the reaction, \( 1 \, \text{mole of } KIO_3 \) reacts with \( 5 \, \text{moles of } KI \).
- Therefore, \( 0.04 \, \text{moles of } KI \) would react with:
\[ \text{Moles of } KIO_3 \text{ required} = \frac{0.04}{5} = 0.008 \, \text{moles} \]

3. **Compare available moles**:
- Available moles of \( KIO_3 = 0.01 \, \text{moles} \)
- Required moles of \( KIO_3 = 0.008 \, \text{moles} \)

Since \( KI \) is the limiting reagent (as it will be completely consumed), we will use \( KI \) to calculate the amount of \( I_2 \) produced.

### Step 4: Calculate the moles of \( I_2 \) produced

1. **Use the stoichiometry of the reaction**:
- From the balanced equation, \( 5 \, \text{moles of } KI \) produce \( 3 \, \text{moles of } I_2 \).
- Therefore, \( 0.04 \, \text{moles of } KI \) will produce:
\[ \text{Moles of } I_2 = \frac{3}{5} \times 0.04 = 0.024 \, \text{moles} \]

### Step 5: Calculate the mass of \( I_2 \) produced

1. **Find the molar mass of \( I_2 \)**:
- \( I_2 = 2 \times 126.9 \, \text{g/mol} = 253.8 \, \text{g/mol} \)

2. **Calculate the mass of \( I_2 \)**:
\[ \text{Mass of } I_2 = \text{moles} \times \text{molar mass} = 0.024 \, \text{moles} \times 253.8 \, \text{g/mol} = 6.096 \, \text{g} \]

### Final Answer:
The weight of \( I_2 \) produced is \( 6.096 \, \text{g} \).

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