Calcualte the volume of `Cl_(2)` gas (in ml) liberated at 1 atm 273 K when 1.74 gm `MnO_(2)` reacts with 2.19 gm HCl according to the following with % yeild 40.
`MnO_(2) + 4HCl rightarrow MnCl_(2) + Cl_(2) + 2H_(2)O`

Options

A. 336ml
B. 112ml
C. 134.4ml
D. 44.8ml

Answer

To calculate the volume of Cl₂ gas liberated when 1.74 g of MnO₂ reacts with 2.19 g of HCl, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O} \]

### Step 2: Calculate the moles of MnO₂ and HCl
1. **Molar mass of MnO₂**:
- Mn: 54.94 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
- Total: 54.94 + 32.00 = 86.94 g/mol

\[ \text{Moles of MnO}_2 = \frac{1.74 \text{ g}}{86.94 \text{ g/mol}} \approx 0.0200 \text{ mol} \]

2. **Molar mass of HCl**:
- H: 1.01 g/mol
- Cl: 35.45 g/mol
- Total: 1.01 + 35.45 = 36.46 g/mol

\[ \text{Moles of HCl} = \frac{2.19 \text{ g}}{36.46 \text{ g/mol}} \approx 0.0600 \text{ mol} \]

### Step 3: Determine the limiting reactant
From the balanced equation, 1 mole of MnO₂ reacts with 4 moles of HCl. Therefore, the required moles of HCl for 0.0200 moles of MnO₂ is:
\[ 0.0200 \text{ mol MnO}_2 \times 4 \text{ mol HCl/mol MnO}_2 = 0.0800 \text{ mol HCl} \]

We have only 0.0600 moles of HCl, which is less than the required amount. Thus, HCl is the limiting reactant.

### Step 4: Calculate the moles of Cl₂ produced
From the balanced equation, 4 moles of HCl produce 1 mole of Cl₂. Therefore, the moles of Cl₂ produced from 0.0600 moles of HCl is:
\[ \text{Moles of Cl}_2 = \frac{0.0600 \text{ mol HCl}}{4} = 0.0150 \text{ mol Cl}_2 \]

### Step 5: Adjust for % yield
Given that the % yield is 40%, the actual moles of Cl₂ produced will be:
\[ \text{Actual moles of Cl}_2 = 0.0150 \text{ mol} \times \frac{40}{100} = 0.0060 \text{ mol} \]

### Step 6: Calculate the volume of Cl₂ at STP
At standard temperature and pressure (STP: 273 K and 1 atm), 1 mole of gas occupies 22.4 L. Therefore, the volume of Cl₂ produced is:
\[ \text{Volume of Cl}_2 = 0.0060 \text{ mol} \times 22.4 \text{ L/mol} = 0.1344 \text{ L} \]

### Step 7: Convert volume to mL
\[ \text{Volume in mL} = 0.1344 \text{ L} \times 1000 \text{ mL/L} = 134.4 \text{ mL} \]

### Final Answer
The volume of Cl₂ gas liberated at 1 atm and 273 K is approximately **134.4 mL**.

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