a.) The mass of CaCO₃ is 2.28 g.
b.) The mass % of CaCO₃ is 91.3 %.
Part a.) is a stoichiometry problem.
You must make the following conversions:
moles of HCl → moles of CaCO₃ → mass of CaCO₃
The balanced equation is
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Mass of CaCO₃ = 0.0456 mol HCl ×
"2.2823" g CaCO₃ = 2.28 g CaCO₃ (3 significant figures)
Part b.) is a mass percent problem.
% by mass =
% of CaCO₃ =