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Question 600064
JEE Chemistry · Hard

Mole Concept and Stoichiometry: Limiting Reagent MCQ

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Limiting ReagentHardQuestion 600064

Question

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. If 228 g C₈H₁₈ (M=114) and 400 g O₂ are combusted, how many grams of CO₂ form?
  1. A
    352 g
    Correct
  2. B
    528 g
  3. C
    704 g
  4. D
    440 g

Correct answer

352 g

Explanation

n(C₈H₁₈)=2, n(O₂)=12.5. Need 2×(25/2)=25 mol O₂ for 2 mol octane. Only 12.5 mol O₂ → O₂ limits. From 12.5 mol O₂: CO₂ = 12.5×(16/25) = 8 mol. Mass = 8×44 = 352 g.

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