Answer to Question #190630 in Chemistry for Karla

Question

The mercury in a 0.7152 g sample was precipitated with an excess of paraperiodic acid, H5IO6: 5Hg2+ + 2H5IO6 → Hg5(IO6)2 + 10H+ The precipitate was filtered, washed free of precipitating agents, dried and weighed, 0.3408 g being recovered. Calculate the percentage of Hg2Cl2 in the sample.

Answer 1

1 mol (Hg₅(IO₆)₂) = 1451.0 g

No.mol (Hg₅(IO₆)₂) = 0.7152 g Hg₅(IO₆)₂ x (1 mol (Hg₅(IO₆)₂) / 1451 g (Hg₅(IO₆)₂)) = 0.0005 mol

2 mol (H₅IO₆) = 5 mol (HgCl₂)

No.mol Hg₂Cl₂= 0.0005 mol Hg₅(IO₆)₂ x (5 mol Hg₂Cl₂ / 2 mol Hg₅(IO₆)₂) = 0.001 mol

1 mol Hg₂Cl₂ ≡ 473.0 g

Mass of Hg₂Cl₂ = 0.001 mol x 473.0 g = 0.5829 g

% Hg₂Cl₂ = 0.5829 / 0.7152 x 100 = 81.50%