CaCO3 reacts with HC1 to produce CaCl2 , CO2 and H2O . The approximate mass (in g) of CaCO3 required to react completely with 25 mL of 0.75 M HCl is (atomic mass of Ca = 40 , C = 12 , O = 16 , Cl = 35.5 and H= 1)
The reaction is shown below. 2HCl+CaCO3→CaCl2+CO2+H2O 25mL of 0.075NHCl is equal to 0.01875‌mole of HCl. According to the above reaction, two moles of mathrm‌HCl is needed to neutralize CaCO3. Therefore, Mass of CaCO3 neutralised =
0.01875
2
=0.009375‌mol Now, mass of CaCO3, required can be calculated as: 0.009375‌mol×100g∕mol=0.9383g ≈0.94gCaCO3