Solution Limiting Reactants What is produced? What is left?

A Problem: When 70. 0 mL 0. 200 M Pb(NO₃)₂(aq) are mixed with 80 A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? Determine the net ionic equation: Pb(NO₃)₂ + 2 KI → PbI₂(s) + 2 KNO₃ Pb²⁺ + 2 NO₃⁻ + 2 K⁺ + 2 I⁻ → PbI₂(s) + 2 K⁺ + 2 NO₃⁻ back Pb²⁺ + 2 I⁻ → PbI₂(s) This is all that happens. The spectator ions stay in the solution and have nothing to do with the amount of precipitate. So we really just care about starting amounts of Pb²⁺ and I⁻ in the solution.

It is common to use [A] to mean . A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? To determine amounts of Pb²⁺ and I⁻ in the solution it will be useful to know the molarities of these ions. It is common to use [A] to mean . So to calculate [Pb⁺] in the original Pb(NO₃)₂ solution let [Pb²⁺] = = 0.200 M Pb²⁺ Since Pb(NO₃)₂ → Pb²⁺ + 2 NO₃⁻

Likewise, if we calculate [NO₃⁻] in the initial Pb(NO₃)₂ solution, Repeat: [Pb²⁺] = = 0.200 M Pb²⁺ Likewise, if we calculate [NO₃⁻] in the initial Pb(NO₃)₂ solution, [NO₃⁻] = = 0.400 M NO₃⁻ Notice that the molarity of the ions are equal to The molarity of the compound times the subscript of the ion Since Pb(NO₃)₂ → Pb²⁺ + 2 NO₃⁻ [Pb²⁺] = 0.200 M x 1 and [NO₃⁻] = 0.200 M x 2 = 0.400 M

A Problem: When 70. 0 mL 0. 200 M Pb(NO₃)₂(aq) are mixed with 80 A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? Pb²⁺ + 2 I⁻ → PbI₂(s) Moles of Pb²⁺ and I⁻ are needed to solve the problem. It is very convenient to calculate millimoles of each, where 1 mmol = 0.00100 mol. Using n = VM = 70.0 mL = 14.0 mmol Pb²⁺ Notice the quantity in brackets [ ] always equals 1, so volume in mL x molarity = mmol So moles I⁻ = 80.0 mL x 0.400 M = 32.0 mmol I⁻

So write the starting quantities under the net ionic equation: A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? So write the starting quantities under the net ionic equation: Pb²⁺ + 2 I⁻ → PbI₂(s) 14.0 32.0 Before solving this, consider a corresponding example: 1 frame + 2 wheels → 1 bicycle How many bicycles can be made from 14 frames and 32 wheels

How many bicycles can be made from 14 frames and 32 wheels? 1 frame + 2 wheels → 1 bicycle Start line: 14 32 0 Change Line: −14 −28 +14 Final line: 0 4 14 By this simple chart one can see immediately the complete answer to the problem. Frames are limiting so all 14 are used. Each frame uses 2 wheels, so 28 wheels get used up and 14 bicycles are produced. The change line is always in the ratio of the coefficients 1, 2, 1 → 14, 28, 14 with negatives on what is used and positives on what is produced. Final values are obtained by adding start and change lines together.

This problem can be solved the same way. Pb²⁺ + 2 I⁻ → PbI₂(s) A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? This problem can be solved the same way. Pb²⁺ + 2 I⁻ → PbI₂(s) Start line: 14.0 32.0 0 Change line: −14.0 −28.0 +14.0 Final line: 0.0 4.0 14.0 mmol PbI₂ This is easily converted to moles: 14.0 mmol PbI₂ = 0.014 mol PbI₂

Start line: 14 32 0 Change Line: −14 −28 +14 Final line: 0 4 14 Notice a simple procedure to determine the limiting reactant and calculate the change line. Remember that the number of processes for a reactant is determined this way: 32 wheels = ? processes 32 wheels = 32 wheels = 16 processes That is, the number of processes is determined by dividing the starting quantity (32) by the coefficient in the equation (2 in this case.) Divide starting quantities by coefficients and squeeze them into a line above the start line. 1 frame + 2 wheels → 1 bicycle 14 16 The smallest is limiting Start line: 14 32 0 Change Line: −14 −28 +14 Final line: 0 4 14

Start line: 14 32 0 Change Line: −14 −28 +14 Final line: 0 4 14 Circle the limiting reactant and determine the change line by multiplying the limiting by the coefficient for that reactant or product. For frames: 14 x 1 = 14 and change it to −14 For wheels: 14 x 2 = 28 becomes −28 For bicycles: 14 x 1 = 14 becomes +14 since it is produced not used 1 frame + 2 wheels → 1 bicycle 14 16 The smallest is limiting Start line: 14 32 0 Change Line: −14 −28 +14 Final line: 0 4 14

Final line: 0.0 4.0 14.0 all in millimoles A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? Expand the problem: Determine the molarity of each ion in solution when the reaction is complete. Pb²⁺ + 2 I⁻ → PbI₂(s) Final line: 0.0 4.0 14.0 all in millimoles Since iodide is in excess, there must be an I⁻ molarity, [I⁻], in the solution after reaction. The total volume of solution must be the sum: 70.0 mL + 80.0 mL = 150.0 mL The molarity of the iodide will be: 4 mmol I⁻ in 150 mL solution So needs to be converted to molarity.

The mmol and mL are easily converted to moles and liters by the conversion: [I⁻] = = 0.027 M I⁻ Again the quantity in brackets [ ] is always equal to 1, so mmol per mL is the same as moles per liter. So it is very convenient to do the limiting calculations in millimoles. Since Pb²⁺ is limiting, all lead(II) is precipitated so that [Pb²⁺] ≈ 0 It is only approximately zero because such reactions as these always have an equilibrium, Pb²⁺ + 2 I⁻  PbI₂(s) so there will be a tiny bit of Pb²⁺ in solution. It will be less than 2x10⁻⁵ M = 0.00002, which we will count as approximately 0.

Final line: 0.0 4.0 14.0 all in millimoles A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? Expand the problem: Determine the molarity of each ion in solution when the reaction is complete. Pb²⁺ + 2 I⁻ → PbI₂(s) Final line: 0.0 4.0 14.0 all in millimoles So [Pb²⁺] ≈0 and [I⁻] = 0.027 M I⁻ PbI₂ is an insoluble precipitate so not in solution. Are there other ions in the solution?

Final line: 0.0 4.0 14.0 all in millimoles A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? Expand the problem: Determine the molarity of each ion in solution when the reaction is complete. Pb²⁺ + 2 I⁻ → PbI₂(s) Final line: 0.0 4.0 14.0 all in millimoles So [Pb²⁺] ≈0 and [I⁻] = 0.027 M I⁻ PbI₂ is an insoluble precipitate so not in solution. Are there other ions in the solution? Surely there are. The spectator ions, K⁺ and NO₃⁻, from the reaction must still be in solution. They are not used up at all. So however many were placed in the solution must still be there. [K⁺] = = 0.213 M K⁺

A Problem: When 70. 0 mL 0. 200 M Pb(NO₃)₂(aq) are mixed with 80 A Problem: When 70.0 mL 0.200 M Pb(NO₃)₂(aq) are mixed with 80.0 mL 0.400 M KI(aq), how many moles of PbI₂(s) are precipitated? Expand the problem: Determine the molarity of each ion in solution when the reaction is complete. [K⁺] = = 0.213 M K⁺ Likewise for NO₃⁻ = [NO₃⁻] = = 0.187 M NO₃⁻ Remember [NO₃⁻] in the 70.0 mL solution was 0.200 M x 2 =0.400 M Notice that for spectator ions, mixing solutions is equivalent to a dilution problem: = 0.213 M the first K⁺ molarity diluted to a new volume.