When 80.0cm^3 of 0.500 M hydrochloric acid was added to 1.75g of impure CaCO3, not all HCl reacts. The unreacted HCl required 22.4 cm^3 of a 0.500 M solution of NaOH for complete reaction. Calculate percentage by mass of CaCO3 in the impure sample.

Balanced equations: CaCO₃ + 2 HCl --> CaCl₂ + H₂O + CO₂ , HCl + NaOH --> NaCl + H₂O
HCl moles that react overall: (80.0 cm³ x 0.500 mol dm³) / 1000 dm³ cm^-3 = 4.00 x 10^-2 mol

HCl moles that react with just NaOH:(22.4 cm³ x 0.500 mol dm^-3) / 1000 dm³ cm^-3 = 1.12 x 10^-2 mol

Ratios of the two balanced equations, work out the moles of HCl that reacted with CaCO₃:
1 : 1, HCl : NaOH --> 1.12 x 10^-2 mol HCl reacted with NaOH
2 : 1, HCl : CaCO₃ --> (4.00 - 1.12) x 10^-2 mol / 2 = 1.44 x 10^-2 mol HCl reacted with CaCO_3.
Determine the Mr of CaCO₃ :(40.078 + 12.011 + (3 x 15.999) g mol^-1 = 100.086 g mol^-1
Determine the mass of CaCO₃ that reacted with HCl:1.44 x 10^-2 mol x 100.086 g mol^-1 = 1.44 g
Determine percentage by mass of CaCO₃ in the impure sample:(1.44 g / 1.75 g) x 100% = 82.4 %