8.7 gm of pure `MnO_(2)` is heated with an excess of HCl and the gas evolved is passed into a solution of KI. The amount of `I_(2)` liberated is [Atomic mass O = 16, Mn = 55]

Options

A. 0.2 mole
B. 25.4 gm
C. 15.4 gm
D. 7.7 gm

Explanation

To solve the problem step by step, we will analyze the reactions involved and perform the necessary calculations.

### Step 1: Determine the number of moles of MnO₂
Given:
- Mass of MnO₂ = 8.7 g
- Molar mass of MnO₂ = 55 (Mn) + 16×2 (O) = 55 + 32 = 87 g/mol

Now, we can calculate the number of moles of MnO₂:
\[ \text{Number of moles of MnO₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{8.7 \, \text{g}}{87 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 2: Determine the amount of Cl₂ produced
From the balanced chemical equation:
\[ \text{MnO₂} + 4 \text{HCl} \rightarrow \text{Cl₂} + \text{MnCl₂} + 2 \text{H₂O} \]
We see that 1 mole of MnO₂ produces 1 mole of Cl₂. Therefore, 0.1 moles of MnO₂ will produce:
\[ 0.1 \, \text{moles of Cl₂} \]

### Step 3: Determine the amount of I₂ produced from Cl₂
The reaction between Cl₂ and KI is:
\[ \text{Cl₂} + 2 \text{KI} \rightarrow 2 \text{KCl} + \text{I₂} \]
From the equation, we see that 1 mole of Cl₂ produces 1 mole of I₂. Therefore, 0.1 moles of Cl₂ will produce:
\[ 0.1 \, \text{moles of I₂} \]

### Step 4: Calculate the mass of I₂ produced
The molar mass of I₂ (Iodine) is:
\[ \text{Molar mass of I₂} = 2 \times 127 = 254 \, \text{g/mol} \]
Now, we can calculate the mass of I₂ produced:
\[ \text{Mass of I₂} = \text{number of moles} \times \text{molar mass} = 0.1 \, \text{moles} \times 254 \, \text{g/mol} = 25.4 \, \text{g} \]

### Final Answer
The amount of I₂ liberated is **25.4 g**.

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