Answer:
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
