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# Answer to Question #190631 in Chemistry for Karla

Question #190631

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.

1
Expert's answer
2021-05-08T23:36:21-0400

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%

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