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For the given reaction: CaCO ₃+2 HCl CaCl ₂+ H ₂ O + CO ₂ If 90 ~g CaCO ₃ is added to 300 mL of HCl which contains 38.55 % HCl by mass and has density 1.13 ~g ~mL ⁻¹ , then which of the following option is correct ? Given molar mass of H , Cl , Ca and O are 1, 35.5, 40 and 16 ~g ~mol ⁻¹ respectively.

Options

  1. A64.97 g of HCl remains unreacted
  2. B60.32 g of HCl remains unreacted
  3. C97.30 ~g of HCl reacted
  4. D32.85 g of CaCO ₃ remains unreacted

Correct answer

A. 64.97 g of HCl remains unreacted

Step-by-step solution

Molar masses: HCl = 36.5, CaCO₃ = 100 g/mol HCl solution: mass = 300 × 1.13 = 339 g HCl present: 339 × 0.3855 = 130.74 g = 130.74/36.5 = 3.58 mol CaCO₃: 90/100 = 0.9 mol Reaction: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ HCl required: 0.9 × 2 = 1.8 mol HCl available: 3.58 mol > 1.8 mol CaCO₃ is limiting. HCl remaining: (3.58 - 1.8) × 36.5 = 1.78 × 36.5 = 64.97 g

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