Answer
$84.28\%$
Work Step by Step
Number of moles of $CO_2$:
$0.558\ g/44.001\ g/mol=0.0127\ mol$
From stoichiometry, that's the number of moles of calcium carbonate.
$0.0127\ mol\times 100.087\ g/mol=1.269\ g$
Mass fraction:
$1.269\ g/1.506\ g=84.28\%$
