Answer
a) $CaCO_3(s)+2\ HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+ CO_2(g)$
b) $20.0\ mL$
Work Step by Step
a) $CaCO_3(s)+2\ HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+ CO_2(g)$
b) Number of moles of calcium carbonate:
$500.\ mg/100.09\ mg/mmol=5.00\ mmol$
Number of moles of HCl:
$5.00\ mmol\times \dfrac{2\ mmol\ HCl}{1\ mmol\ CaCO_3}=10.0\ mmol\ HCl$
Volume:
$10.0\ mmol\div0.500\ M=20.0\ mL$