This is basically asking if CaCO3 is the excess reactant, correct? If it is the excess, then how much CaCO3 remains?
Roger the Mole
Favorite Answer
Yes, it is a limiting reactant problem.
2 HCl + CaCO3 → CaCl2 + CO2 + H2O
(2.56 g CaCO3) / (100.0875 g CaCO3/mol) = 0.0255776 mol CaCO3
(0.2500 L) x (0.125 mol/L HCl) = 0.03125 mol HCl
0.03125 mole of HCl would react completely with 0.03125 x (1/2) = 0.015625 mole of CaCO3, but there is more CaCO3 present than that, so CaCO3 is in excess and HCl is the limiting reactant. So, yes, some calcium carbonate remains.
If you are interested in how much:
((0.0255776 mol CaCO3 initially) - (0.015625 mol CaCO3 reacted)) x (100.0875 g CaCO3/mol) = 0.996 g CaCO3 left over