Answer
$1.73\ g$
Work Step by Step
Number of moles of $CaCO_3$:
$2.56\ g\div 100.09\ g/mol=0.0256\ mol$
Number of moles of $HCl$:
$0.125\ M\times 0.250\ L=0.0313\ mol$
Ratio: $1.22$, HCl is the limiting reactant, so calcium carbonate remains after the reaction is complete.
Mass of calcium chloride:
$0.0313\ mol \times 1/2\times 110.98\ g/mol=1.73\ g$
