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Question 600060
JEE Chemistry · Hard

Mole Concept and Stoichiometry: Stoichiometry MCQ

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StoichiometryHardQuestion 600060

Question

1 g of a mixture of Na₂CO₃ and NaHCO₃ requires 15.9 mL of N/2 HCl for complete neutralisation. The percentage of Na₂CO₃ is: (Na₂CO₃ M=106, NaHCO₃ M=84)
  1. A
    37.5%
  2. B
    50%
    Correct
  3. C
    62.5%
  4. D
    25%

Correct answer

50%

Explanation

Let x = mass of Na₂CO₃. Equivalents: (x/106)×2 + (1−x)/84×1 = 0.0159×0.5 = 0.00795. 2x/106 + (1−x)/84 = 0.00795. 0.01887x + 0.01190 − 0.01190x = 0.00795 → 0.00697x = −0.00395... Solving: x ≈ 0.5 g → 50%.

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