initial moles of HCl = 200 x 0.1220 / 1000 = 0.0244
moles of NaOH = 0.1050 x 32.20 / 1000 = 3.38 x 10^-3
moles of HCl = moles of NaOH
so moles of HCl excess reacted with NaOH = 3.38 x 10^-3
moles of HCl reacted with antacid = 0.0244 - 3.38 x 10^-3 =
0.02102
CaCO3 + 2HCl -------------------> CaCl2 + H2O + CO2
1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
2
x             Â
0.02102
moles of CaCO3 = 0.02102 / 2 = 0.01051
moles of CaCO3 = 0.01051
mass of CaCO3 = moles x molar mass
                        Â
= 0.01051 x 100.09
                         Â
= 1.052 g
mass of CaCO3 =1.052 g
