Chlorine gas can be produced by the reaction of `HCl (aq)` with `MnO_(2) (s)`. Only `MnCl_(2)` and `H_(2)O(l)` are the by products. What volume of `Cl_(2) (g) ("in litre")` of density `2.84 g//L` will be produced from the reaction of `400 mL` of `0.1 M HCl(aq)` with an excess of `MnO_(2)` ?

Explanation

To solve the problem of how much chlorine gas (Cl₂) is produced from the reaction of hydrochloric acid (HCl) with manganese dioxide (MnO₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 4 \text{HCl} (aq) + \text{MnO}_2 (s) \rightarrow \text{Cl}_2 (g) + \text{MnCl}_2 (aq) + 2 \text{H}_2O (l) \]

### Step 2: Calculate the number of moles of HCl
We know the volume and molarity of HCl:
- Volume of HCl = 400 mL = 0.400 L (convert mL to L by dividing by 1000)
- Molarity of HCl = 0.1 M

Using the formula for moles:
\[ \text{Number of moles of HCl} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
\[ \text{Number of moles of HCl} = 0.400 \, \text{L} \times 0.1 \, \text{mol/L} = 0.040 \, \text{mol} \]

### Step 3: Determine the moles of Cl₂ produced
From the balanced equation, we see that 4 moles of HCl produce 1 mole of Cl₂. Therefore, the moles of Cl₂ produced from 0.040 moles of HCl can be calculated as follows:
\[ \text{Moles of Cl}_2 = \frac{1}{4} \times \text{Moles of HCl} \]
\[ \text{Moles of Cl}_2 = \frac{1}{4} \times 0.040 \, \text{mol} = 0.010 \, \text{mol} \]

### Step 4: Calculate the mass of Cl₂ produced
The molar mass of Cl₂ (chlorine gas) is calculated as:
- Atomic mass of Cl = 35.5 g/mol
- Therefore, molar mass of Cl₂ = 2 × 35.5 g/mol = 71 g/mol

Now, calculate the mass of Cl₂ produced:
\[ \text{Mass of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar mass of Cl}_2 \]
\[ \text{Mass of Cl}_2 = 0.010 \, \text{mol} \times 71 \, \text{g/mol} = 0.71 \, \text{g} \]

### Step 5: Calculate the volume of Cl₂ produced
Using the density of Cl₂ to find the volume:
- Density of Cl₂ = 2.84 g/L

Using the formula for volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
\[ \text{Volume of Cl}_2 = \frac{0.71 \, \text{g}}{2.84 \, \text{g/L}} \approx 0.25 \, \text{L} \]

### Final Answer
The volume of Cl₂ produced from the reaction is approximately **0.25 liters**.

---