Question

Question #ee3a8

Answer 1

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 × `(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