Asolution contains ag and hg2 ions. the addition of 0.100 l of 1.10m nai solution is just enough to precipitate all the ions as agi and hgi2. the total mass of the precipitate is 25.5 g. find the mass of agi in the precipitate.

First we write out the balanced equation for this reaction. Note: Mercury(II) iodide is HgI2 and not HgI.

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

moles I^- added = (Molarity I^-)(L of I^-) = (1.10)(0.100) = 0.11moles I^- addedhe balanced equation indicates that Hg^2+ uses up twice as much I^- in the reaction as does Ag+, since the formula for HgI2 contains 2 I^- atoms while the formula for AgI contains only 1 I^- atom. Hence if Ag^+ consumes x moles of I-, then Hg2+ consumes 2x moles of I-. The total moles of I^- consumed = 3x = 0.11 moles I-.

3x = 0.11; x = 0.11/ 3 = 0.0367 moles I^-

0.0367moles I^- x (126.9 g I- / 1 mole I-) = 4.657 g I^-

The formula for AgI shows the that the mole ratio of Ag^+ to I^- is 1:1.

0.0367 moles I- x (1 mole Ag^+ / 1 mole I-) = 0.0367 moles Ag+

0.0367 moles Ag^+ x (107.9 g Ag / 1 mole Ag) = 3.960g Ag^+

So therefore the mass of AgI = mass of Ag^+ + mass of I- = 3.96 + 4.657 = 8.607 g AgI

which statement about this sum is true? −245+2.7

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