Home/
Competitive Exams/
JEE Main & Advanced/
Chemistry/
Mole Concept and Stoichiometry/
Question 600086
JEE Chemistry · Hard

Mole Concept and Stoichiometry: Stoichiometry MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
StoichiometryHardQuestion 600086

Question

In a double displacement reaction, 100 mL of 0.3 M BaCl₂ is mixed with 150 mL of 0.2 M Na₂SO₄. Moles of BaSO₄ precipitate formed are:
  1. A
    0.03 mol
  2. B
    0.03 mol (BaCl₂ limits)
    Correct
  3. C
    0.02 mol (Na₂SO₄ limits)
  4. D
    0.05 mol

Correct answer

0.03 mol (BaCl₂ limits)

Explanation

n(BaCl₂) = 0.1×0.3 = 0.03 mol. n(Na₂SO₄) = 0.15×0.2 = 0.03 mol. Equal moles → both exactly consumed. BaSO₄ = 0.03 mol.

Review Mole Concept and Stoichiometry notesStart timed chapter practice