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Molarity and Stoichiometry

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

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Molarity and Stoichiometry

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  1. M M mol M = L mol mol V V P P __ __ __ __ Molarity and Stoichiometry mol = M L M L M L 1 2 1 2 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq) What volume of 4.0 M KI solution is required to yield 89 g PbI2?

  2. 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

  3. mol L M = mol M 0.39 mol KI 4.0 M KI L = = 1 Pb(NO3)2(aq) + 2 KI (aq) 1 PbI2(s) + 2 KNO3(aq) 89 g ? L 4.0 M 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 = 0.098 L of 4.0 M KI

  4. mol M mol L M = L = 0.173 mol CuSO4 0.500 M CuSO4 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) CuSO4(aq) + Al (s)  Cu(s) + Al2(SO4)3(aq) 3 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 = 0.346 L 1000 mL 0.346 L = 346 mL 1 L

  5. 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 1.5L 0.10M ? g .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu 1.5 L = 4.8 g Cu Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  6. 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  ZnCl2 + H2 ? L 79.1 g 0.90 L 2.5M Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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

  8. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 0.90 L 2.5M 0.90 L 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

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

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