Question

A 0.450-gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3+2HCl>>>CaCl2+H2O+CO2 The excess HCl(aq) is titrated by 7.95 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample.

Answer 1

first calculate the exess HCl reacted with NaOH

NaOH + HCl -----------> NaCl + H2O

moles of NaOH = 0.125 x 7.95 / 1000 = 0.000994

0.000994 moles HCl reacted with NaOH

total HCl added = 0.15 x 50 / 1000 = 0.0075 moles

moles of HCl reacts with CaCO3 = 0.0075 - 0.000994 = 0.006506

2 moles HCl reacts with 1 mole CaCO3

0.006506 moles HCl reacts with 0.006506 x 1 / 2 = 0.003253

mass of CaCO3 = 0.003253 x 100.09 = 0.325 g

% mass = (0.325 / 0.450) x 100

% mass = 72.2