1 Pb(NO 3 ) 2 (aq) + KI (aq)  PbI 2 (s) + KNO 3 (aq) __ 212 Molarity and Stoichiometry M M V V P P mol M L M = mol L mol = M L What volume of 4.0 M KI.

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1 Pb(NO 3 ) 2 (aq) + KI (aq)  PbI 2 (s) + KNO 3 (aq) __ 212 Molarity and Stoichiometry M M V V P P mol M L M = mol L mol = M L What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ?

Step 1) Identify the species present in the combined solution, and determine what reaction occurs. Step 2) Write the balanced net ionic equation for the reaction. Step 3) Calculate the moles of reactants. Step 4) Determine which reactant is limiting. Step 5) Calculate the moles of product or products, as required. Step 6) Convert to grams or other units, as required. Stoichiometry steps for reactions in solution Stoichiometry for Reactions in Solution

461 g PbI 2 Strategy: 1 Pb(NO 3 ) 2 (aq) + 2 KI (aq)  1 PbI 2 (s) + 2 KNO 3 (aq) X mol KI = 89 g PbI 2 1 mol PbI 2 2 mol KI = 0.39 mol KI (1) Find mol KI needed to yield 89 g PbI 2. (2) Based on (1), find volume of 4.0 M KI solution. What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ? 89 g ? L 4.0 M M = mol L L = mol M = 0.39 mol KI 4.0 M KI = L of 4.0 M KI

= mol CuSO 4 How many mL of a M CuSO 4 solution will react w /excess Al to produce 11.0 g Cu? __CuSO 4 (aq) + __Al (s)  Al 3+ SO 4 2– CuSO 4 (aq) + Al (s)  Cu(s) + Al 2 (SO 4 ) 3 (aq)3231 x mol11 g __Cu(s) + __Al 2 (SO 4 ) 3 (aq) X mol CuSO 4 = 11 g Cu 1 mol Cu 63.5 g Cu 3 mol CuSO 4 3 mol Cu M = mol L L = mol M mol CuSO M CuSO 4 = L L 1000 mL 1 L = 346 mL

63.55 g Cu 1 mol Cu Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M Courtesy Christy Johannesson

Limiting Reactants 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

Limiting Reactants 79.1 g Zn 1 mol Zn g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

Limiting Reactants 22.4 L H 2 1 mol H L 2.5 mol HCl 1 L = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H g ? L 0.90 L 2.5M Courtesy Christy Johannesson

Limiting Reactants Zn: 27.1 L H 2 HCl: 25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc Courtesy Christy Johannesson

Molarity and Stoichiometry Keys Molarity and Stoichiometry