If 1 gm of HCl and 1 gm of `MnO_2` are heated together, the maximum weight of` Cl_2` gas evolved will be:

Options

A. 2g
B. 12g
C. 0.985g
D. 0.486g

Explanation

To solve the problem of determining the maximum weight of Cl₂ gas evolved when 1 gram of HCl and 1 gram of MnO₂ are heated together, we will follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between hydrochloric acid (HCl) and manganese dioxide (MnO₂) can be represented as:
\[ \text{MnO}_2 + 4 \text{HCl} \rightarrow \text{Cl}_2 + \text{MnCl}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Determine the molar masses.
- Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol
- Molar mass of MnO₂ = 54.9 (Mn) + 2 × 16 (O) = 86.9 g/mol
- Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol

### Step 3: Calculate the moles of HCl and MnO₂.
- Moles of HCl:
\[ \text{Moles of HCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{36.5 \text{ g/mol}} \approx 0.0274 \text{ mol} \]

- Moles of MnO₂:
\[ \text{Moles of MnO}_2 = \frac{1 \text{ g}}{86.9 \text{ g/mol}} \approx 0.0115 \text{ mol} \]

### Step 4: Identify the limiting reagent.
From the balanced equation, we see that 4 moles of HCl are required for every 1 mole of MnO₂. Therefore, we need:
\[ 4 \times 0.0115 \text{ mol} = 0.046 \text{ mol of HCl} \]
Since we only have 0.0274 mol of HCl, HCl is the limiting reagent.

### Step 5: 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:
\[ \text{Moles of Cl}_2 = \frac{0.0274 \text{ mol HCl}}{4} \approx 0.00685 \text{ mol Cl}_2 \]

### Step 6: Calculate the mass of Cl₂ produced.
Now, we can calculate the mass of Cl₂ produced using its molar mass:
\[ \text{Mass of Cl}_2 = \text{moles} \times \text{molar mass} = 0.00685 \text{ mol} \times 71 \text{ g/mol} \approx 0.486 \text{ g} \]

### Conclusion:
The maximum weight of Cl₂ gas evolved is approximately **0.486 grams**.

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