biochemical calculations

Question

We have the solution of calcium chloride. The sodium carbonate was added to this solution, which resulted in precipitation of calcium carbonate. this precipitate was then calcinated, and at the end 0.56 g of a white substance was obtained. Calculate the mass (in grams) of calcium chloride that was in an analyses volume of this solution.

ChatGPT

We have the solution of calcium chloride. The sodium carbonate was added to this solution, which resulted in precipitation of calcium carbonate. This precipitate was then calcinated, and at the end 0.56 g of a white substance was obtained. Calculate the mass (in grams) of calcium chloride that was in an analyses volume of this solution.

Answer 1

CaCl₂(aq) + Na₂CO₃(aq) ==> CaCO₃(s) + 2NaCl(aq) .. balanced equation

CaCO₃(s) ==> CaO(s) + CO₂(g) .. calcination reaction

molar mass CaO = 56 g / mole

moles CaO formed = 0.56 g x 1 mol CaO / 56 g = 0.10 moles CaO

moles CaCO₃ formed = 0.10 mols CaO x 1 mol CaCO₃ / mol CaO = 0.10 mols CaCO₃ formed

moles CaCl₂ = 0.10 mols CaCO₃ x 1 mol CaCl₂ / mol CaCO₃ = 0.10 mols CaCl₂

grams CaCl₂ present in original sample = 0.10 g CaCl₂ x 111 g / mol = 11 g CaCl₂ present (2 sig.figs.)