Question

An aqueous solution of silver acetate, AgC2H3O2, 50.00 mL and .300 M, is mixed with 25.00 mL of .300 M iron (III) chloride, FeCl3. Calculate the mass of any precipitate formed and the concentration of each ion rremaining in solution.

Answer 1

3AgC2H3O2 + FeCl3 ------------> 3AgCl_(s) + Fe(C2H3O2)3

moles of AgC2H3O2 = 0.300 x 0.05 = 0.015

moles of FeCl3 = 0.300 x 0.025 = 0.0075

3 moles AgC2H3O2 reacts with 1 mole FeCl3

0.015 moles AgC2H3O2 reacts with 0.015 x 1 / 3 = 0.005 moles FeCl3

we have 0.0075 moles FeCl3 so exess FeCl3 present

AgC2H3O2 is limiting reagent

3 moles AgC2H3O2 gives 3 moles AgCl precipitate

0.015 moles AgC2H3O2 gives 0.015 x 3 . 3 = 0.015 moles AgCl precipitate.

mass of precipitate = 0.015 x 143.32 = 2.15 g

FeCl3 remaining = 0.0075 - 0.005 = 0.0025 moles

[FeCl3] = 0.025/ 0.075 = 0.33 M

[Fe⁺³] remaining = 0.33 M

[Cl^-] remaining = 0.33 x 3 = 0.99 M