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Question 600088
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Mole Concept and Stoichiometry: Equivalent Concept MCQ

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Equivalent ConceptHardQuestion 600088

Question

The equivalents of Ca(OH)₂ (M=74) in 7.4 g when reacting with H₃PO₄ to form Ca₃(PO₄)₂ are:
  1. A
    0.1 equiv
  2. B
    0.2 equiv
    Correct
  3. C
    0.3 equiv
  4. D
    0.4 equiv

Correct answer

0.2 equiv

Explanation

n(Ca(OH)₂) = 7.4/74 = 0.1 mol. Ca(OH)₂ donates 2 OH⁻ → n-factor = 2. Equivalents = 0.1 × 2 = 0.2 equiv.

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