A mixture contains Mg(NO₃)₂ and KBr. A 1.85 g sample of this mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of Mg(OH)₂.

Question

The precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.265 g. What is the mass percent of Mg(NO₃)₂ in the original mixture?

Answer 1

Mg(NO₃)₂(aq) + 2NaOH(aq) ==> Mg(OH)₂(s) + 2NaNO₃(aq) ... balanced equation

mass Mg(OH)₂ = 0.265 g

mols Mg(OH)₂ = 0.265 g x 1 mol / 58.32 g = 0.00454 mols

mols Mg(NO₃)₂ = 0.00454 mols Mg(OH)₂ x 1 mol Mg(NO₃)₂ / 1 mol Mg(OH)₂ = 0.00454 mols Mg(NO₃)₂

mass Mg(NO₃)₂ = 0.00454 mols Mg(NO₃)₂ x 148.3 g / mol = 0.673 g Mg(NO₃)₂

mass % Mg(NO₃)₂ = 0.673 g / 1.85 g (x100%) = 36.4%