If `1 g` of `HCl` and `1 g` of `MnO_(2)` heated together the maximum weight of `Cl_(2)` gas evolved will be :
`[MnO_(2) + 4HCl rarr MnCl_(2) + Cl_(2) + 2H_(2)O]` :

Options

A. `2 g`
B. `0.975 g`
C. `0.486 g`
D. `0.972 g`

Explanation

To solve the problem, we need to determine the maximum weight of chlorine gas (Cl₂) evolved when 1 g of HCl reacts with 1 g of MnO₂ according to the balanced chemical equation provided:

\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O} \]

### Step 1: Calculate the number of moles of HCl and MnO₂

1. **Molar mass of HCl**:
- H: 1 g/mol
- Cl: 35.5 g/mol
- Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

2. **Moles of HCl**:
\[ \text{Moles of HCl} = \frac{\text{mass of HCl}}{\text{molar mass of HCl}} = \frac{1 \text{ g}}{36.5 \text{ g/mol}} \approx 0.0274 \text{ mol} \]

3. **Molar mass of MnO₂**:
- Mn: 54.9 g/mol
- O: 16 g/mol (2 O atoms)
- Molar mass of MnO₂ = 54.9 + (2 \times 16) = 54.9 + 32 = 86.9 g/mol

4. **Moles of MnO₂**:
\[ \text{Moles of MnO}_2 = \frac{\text{mass of MnO}_2}{\text{molar mass of MnO}_2} = \frac{1 \text{ g}}{86.9 \text{ g/mol}} \approx 0.0115 \text{ mol} \]

### Step 2: Determine the limiting reagent

From the balanced equation, we see that 4 moles of HCl react with 1 mole of MnO₂.

- For 0.0115 moles of MnO₂, the required moles of HCl would be:
\[ \text{Required moles of HCl} = 4 \times \text{moles of MnO}_2 = 4 \times 0.0115 \text{ mol} = 0.046 \text{ mol} \]

Since we only have 0.0274 moles of HCl, HCl is the limiting reagent.

### Step 3: Calculate the moles of Cl₂ produced

According to the balanced equation, 4 moles of HCl produce 1 mole of Cl₂. Therefore, the moles of Cl₂ produced from the available HCl is:
\[ \text{Moles of Cl}_2 = \frac{\text{moles of HCl}}{4} = \frac{0.0274 \text{ mol}}{4} \approx 0.00685 \text{ mol} \]

### Step 4: Calculate the mass of Cl₂ produced

The molar mass of Cl₂ is:
- Cl: 35.5 g/mol (2 Cl atoms)
- Molar mass of Cl₂ = 2 \times 35.5 = 71 g/mol

Now, we can calculate the mass of Cl₂ produced:
\[ \text{Mass of Cl}_2 = \text{moles of Cl}_2 \times \text{molar mass of Cl}_2 = 0.00685 \text{ mol} \times 71 \text{ g/mol} \approx 0.486 \text{ g} \]

### Final Answer
The maximum weight of Cl₂ gas evolved is approximately **0.486 g**.

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