Question:

At STP, 1.54 g of MnO\(_2\) reacted with excess HCl. What is the volume of Cl\(_2\) produced at STP? (At. wt: Mn = 55 u, O = 16 u)

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At STP, always remember: \(1\) mole of gas occupies \(22.4\) L and check stoichiometric coefficients carefully.
Updated On: Jun 20, 2026
  • 0.396 L
  • 0.224 L
  • 0.336 L
  • 0.112 L
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The Correct Option is A

Solution and Explanation

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: \[ \boxed{0.396 \, L} \]
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