what volume of 0.1mol/dm3 hydrochloric acid would be required to react completely with 1.50g of calciumtrioxocarbonate (4) salt.
If your reaction is
CaCO3 + 2HCl → CaCl2 + H2O + CO2
then since
1.5g CaCO3 = 0.015 moles
you need 0.03 moles HCl = 0.3L of 0.1M HCl