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