How many milliliters of 0.564 m hcl are required to react with 6.03 grams of caco3?

First write up a balanced equation

CaCO3 + 2 HCl = CO2 + CaCl2 + H2O

Find how many moles of CaCO3 you are dealing with

6.03 g x 1 mol / 100 g =.0603 moles

However, you need twice as much HCl

.0603 x 2 moles /.564 M =.21 L