Molarity and Stoichiometry V P mol M = mol L mol = M L M L M L The coefficients in the balanced chemical equation indicate the number of moles of each reactant that is needed and the number of moles of each product that can be produced. It doesn’t matter if you are dealing with volumes of solutions of reactants or masses of reactants. __ 1 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq) 2 1 2 What volume of 4.0 M KI solution is required to yield 89 g PbI2?

Stoichiometry for Reactions in Solution 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

1 Pb(NO3)2(aq) + 2 KI (aq)  1 PbI2(s) + 2 KNO3(aq) ? L 4.0 M 89 g What volume of 4.0 M KI solution is required to yield 89 g PbI2? Strategy: (1) Find mol KI needed to yield 89 g PbI2. (2) Based on (1), find volume of 4.0 M KI solution. 1 mol PbI2 2 mol KI X mol KI = 89 g PbI2 = 0.39 mol KI 461 g PbI2 1 mol PbI2 M = mol L L = mol M = 0.39 mol KI 4.0 M KI = 0.098 L of 4.0 M KI

How many mL of a 0.500 M CuSO4 solution will react w/excess Al to produce 11.0 g Cu? Al3+ SO42– __CuSO4(aq) + __Al (s)  __Cu(s) + __Al2(SO4)3(aq) 3 CuSO4(aq) + Al (s)  Cu(s) + Al2(SO4)3(aq) 2 3 1 x mol 11 g 1 mol Cu 3 mol CuSO4 X mol CuSO4 = 11 g Cu = 0.173 mol CuSO4 63.5 g Cu 3 mol Cu M = mol L L = mol M 0.173 mol CuSO4 0.500 M CuSO4 = 0.346 L 1000 mL 0.346 L = 346 mL 1 L

Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 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? The concept of limiting reactants applies to reactions that are carried out in solution and reactions that involve pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant can be calculated. When the limiting reactant is not known, one can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation. Use volumes and concentrations of solutions of reactants to calculate the number of moles of reactants. Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 22.4 L H2 1 mol = 27.1 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 0.90 2.5 mol HCl 1 L 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 left over zinc Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem