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