Question

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

Answer 1

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