Answer: A, B, C
CaCO3 + 2HCl gives CaCl2 + H2O + CO2, so 2 mol HCl per mol; dissolved CO2 (carbonic acid) would affect a phenolphthalein end point, hence boiling. Back titration with excess reagent suits insoluble or slow reactants. NaOH equals the excess HCl, not the HCl consumed by CaCO3, so the third statement is wrong.