Question

# A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.24 M NaI...

A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.24 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 28.7 g Find the mass of AgI in the precipitate.

Ag+(aq) + Hg2+(aq) + 3I-(aq) AgI(s) + HgI2(s)

Note: Mercury(II) iodide is HgI2 and not HgI.

moles I- added = (Molarity I-)(L of I-) = (1.24)*(0.100) = 0.124 moles I- added
Hg2+ consumes twice as much I- in the reaction as does Ag+, since the formula HgI2 contains 2 I- atoms while the formula AgI contains only 1 I- atom.

If Ag+ consumes x moles of I-, then Hg2+ consumes 2x moles of I-. The total moles of I- consumed = 3x = 0.124 moles I-.

3x = 0.124; x = 0.124 / 3 = 0.0413 moles I-
0.0413moles I- x (126.9 g I- / 1 mole I-) = 5.240 g I-

The formula AgI tells us that the mole ratio of Ag+ to I- is 1:1.

0.0413 moles I- x (1 mole Ag+ / 1 mole I-) = 0.0413 moles Ag+
0.0413 moles Ag+ x (107.86 g Ag / 1 mole Ag) = 4.454 g Ag+

So the mass of AgI = mass of Ag+ + mass of I- = 5.240 + 4.454 = 9.694 g AgI

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