Q. For the given reaction;

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

If 90 g CaCO₃ 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.

Correct Answer: 64.97 g of HCl remains unreacted

Solution

Mass of HCl solution = 300 × 1.13 = 339 g

Mass of pure HCl present = 38.55% of 339 = (38.55 / 100) × 339 = 130.7 g

Moles of HCl = 130.7 / 36.5 = 3.58 mol

Molar mass of CaCO₃ = 40 + 12 + 48 = 100 g mol⁻¹

Moles of CaCO₃ = 90 / 100 = 0.9 mol

From the balanced equation, 1 mol CaCO₃ reacts with 2 mol HCl

HCl required for 0.9 mol CaCO₃ = 1.8 mol

Since available HCl = 3.58 mol, HCl is in excess and CaCO₃ is the limiting reagent.

Unreacted HCl = 3.58 − 1.8 = 1.78 mol

Mass of unreacted HCl = 1.78 × 36.5 = 64.97 g

Hence, 64.97 g of HCl remains unreacted.

Related JEE Main Questions