the volume of HCl of specific gravity 1.2 g `mL^(-1)` and `4%` nature by weight, needed to produce 1.78 L of `Cl_(2)` at STP. The reaction involved is:
`MnO_(2)+4"HCl"toMnCl_(2)+2H_(2)O+Cl_(2)`

Options

A. `0.48 L`
B. `0.24 L`
C. `0.12 L`
D. `0.06 L`

Explanation

To solve the problem, we need to determine the volume of HCl solution required to produce 1.78 L of Cl₂ at STP, given the specific gravity and concentration of the HCl solution. Here’s a step-by-step breakdown:

### Step 1: Calculate the moles of Cl₂ produced
To find the moles of Cl₂ produced, we can use the formula:
\[ \text{Moles of Cl₂} = \frac{\text{Volume of Cl₂ at STP}}{\text{Molar Volume at STP}} \]
Given:
- Volume of Cl₂ = 1.78 L
- Molar Volume at STP = 22.4 L/mol

Calculating the moles:
\[ \text{Moles of Cl₂} = \frac{1.78 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.0794643 \, \text{mol} \]

### Step 2: Use stoichiometry to find moles of HCl required
From the balanced chemical equation:
\[ \text{MnO₂} + 4 \text{HCl} \rightarrow \text{MnCl₂} + 2 \text{H₂O} + \text{Cl₂} \]
We see that 4 moles of HCl produce 1 mole of Cl₂. Therefore, the moles of HCl required can be calculated as follows:
\[ \text{Moles of HCl} = 4 \times \text{Moles of Cl₂} = 4 \times 0.0794643 \approx 0.3178572 \, \text{mol} \]

### Step 3: Calculate the weight of HCl required
Now, we need to find the weight of HCl using its molar mass (HCl = 36.5 g/mol):
\[ \text{Weight of HCl} = \text{Moles of HCl} \times \text{Molar Mass of HCl} \]
Calculating the weight:
\[ \text{Weight of HCl} = 0.3178572 \, \text{mol} \times 36.5 \, \text{g/mol} \approx 11.6 \, \text{g} \]

### Step 4: Calculate the weight of the HCl solution needed
Given that the HCl solution is 4% by weight, we can set up the equation:
\[ \text{Weight of HCl} = 0.04 \times \text{Weight of solution} \]
Let the weight of the solution be \(X\):
\[ 11.6 \, \text{g} = 0.04 \times X \]
Solving for \(X\):
\[ X = \frac{11.6 \, \text{g}}{0.04} = 290 \, \text{g} \]

### Step 5: Calculate the volume of the HCl solution
Using the specific gravity of the HCl solution (1.2 g/mL), we can find the volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{290 \, \text{g}}{1.2 \, \text{g/mL}} \approx 241.67 \, \text{mL} \]

### Final Answer
Thus, the volume of HCl solution needed is approximately **240 mL**.

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