Step 1: Write the balanced chemical equation.
The reaction of MnO\(_2\) with HCl is:
\[
MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O
\]
From the equation, 1 mole of MnO\(_2\) produces 1 mole of Cl\(_2\).
Step 2: Calculate molar mass of MnO\(_2\).
\[
\text{Molar mass} = 55 + 2 \times 16 = 87 \, g/mol
\]
Step 3: Find moles of MnO\(_2\).
Given mass = 1.54 g:
\[
n = \frac{1.54}{87} \approx 0.0177 \, mol
\]
Step 4: Use stoichiometry for Cl\(_2\).
From equation:
\[
1 \, mol \, MnO_2 \rightarrow 1 \, mol \, Cl_2
\]
So:
\[
n(Cl_2) = 0.0177 \, mol
\]
Step 5: Convert moles to volume at STP.
At STP:
\[
1 \, mol = 22.4 \, L
\]
So:
\[
V = 0.0177 \times 22.4
\]
Step 6: Final calculation.
\[
V \approx 0.396 \, L
\]
Final Answer:
\[
\boxed{0.396 \, L}
\]