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Question 600093
JEE Chemistry · Hard

Mole Concept and Stoichiometry: Stoichiometry MCQ

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StoichiometryHardQuestion 600093

Question

In the reaction: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O, if 4.35 g MnO₂ (M=87) is taken with excess HCl, volume of Cl₂ at STP is:
  1. A
    0.56 L
  2. B
    1.12 L
    Correct
  3. C
    2.24 L
  4. D
    0.28 L

Correct answer

1.12 L

Explanation

n(MnO₂) = 4.35/87 = 0.05 mol. 1 mol MnO₂ → 1 mol Cl₂. n(Cl₂) = 0.05 mol. V = 0.05 × 22.4 = 1.12 L.

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