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
For the given reaction: CaCO_3+2 HCl longrightarrow 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…
JEE Main 2026 — Chemistry Physical Chemistry
For the given reaction:
CaCO3+2HCl⟶CaCl2+H2O+CO2
If 90 gCaCO3 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.
Held on 28 Jan 2026 · Verified 1 Jul 2026.
Options
- A
64.97 g of HCl remains unreacted
- B
60.32 g of HCl remains unreacted
- C
97.30 g of HCl reacted
- D
32.85 g of CaCO3 remains unreacted
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