Solving limiting reactant problems in solution

Question

Suppose 2.77g of iron(II) iodide is dissolved in 200.mL of a 68.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iron(II) cation in the solution. You can assume the volume of the solution doesn't change when the iron(II) iodide is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Iron(II) iodide = FeI₂ (molar mass = 310 g / mol

Silver nitrate = AgNO₃ (molar mass = 170 g / mol

FeI₂(aq) + 2AgNO₃(aq) ==> Fe(NO₃)₂(aq) + 2AgI(s) ... balanced equation

Because Fe²⁺ is NOT involved in the formation of the precipitate, its concentration will remain constant, and so the final molarity of Fe²⁺ will be the same as the initial concentration. That is calculated as follows:

2.77 g x 1 mol / 310 g = 8.935x10^-3 mols FeI₂ = 8.935x10^-3 mols Fe²⁺ / 0.2 L = 0.0447 M Fe²⁺

Now, if you wanted to go through the trouble of finding the limiting reactant, and then determining the final concentration of iron(II), you could do so as follows:

To find the limiting reactant, one simple way is to divide moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For FeI2: 2.77 g x 1 mol / 310 g = 8.935x10^-3 mols (÷1->8.9x10^-3)

For AgNO3: 200 ml x 1 L / 1000 ml x 0.068 mol/L = 0.0136 mols (÷2->6.8x10^-3)

Thus AgNO3 is limiting.

FeI₂(aq) + 2AgNO₃(aq) ==> Fe(NO₃)₂(aq) + 2AgI(s)

0.00894........0.0136......................0.....................0................Initial

-0.0068.......-0.0136.................+0.0068............+0.0136........Change

0.00214............0........................0.0068...........0.0136..........Equilibrium

Final volume = 200 ml = 0.200 L

Final concentrations:

[FeI₂] = 0.00214 mols / 0.200 L = 0.0107 M thus [Fe²⁺] = 0.0107 M

[Fe(NO₃)₂] = 0.0068 mols / 0.200 L = 0.034 M thus [Fe²⁺] = 0.0340 M

Total [Fe²⁺] = 0.0107 + 0.0340 = 0.0447 M Fe²⁺