a.) The mass of CaCO₃ is 2.28 g.
b.) The mass % of CaCO₃ is 91.3 %.
Part a.) is a 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 × ##(1″mol CaCO₃”)/(2″mol HCl”)× (100.1″g CaCO₃”)/(1″mol CaCO₃”)## =
“2.2823” g CaCO₃ = 2.28 g CaCO₃ (3 significant figures)
Part b.) is a mass percent problem.
% by mass = ##”mass of component”/”total mass”## × 100 %
% of CaCO₃ = ##”mass of CaCO₃”/”mass of eggshell”## × 100 % = ##(2.2823″g”)/(2.50″g”)## × 100 % = 91.3 % by mass
