(L10)stoichimetry part 2

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(L10)stoichimetry part 2

  1. 1. © 2008 Brooks/Cole 2 2 C2H6 (g) + 7 O2 (g) 4 CO2(g) + 6 H2O (l) The Mole and Chemical Reactions Mole ratios: 2 mol C2H6 7 mol O2 =1 7 mol O2 2 mol C2H6 =1 2 moles of C2H6 react with 7 moles of O2 2 moles of C2H6 produce 4 moles of CO2 2 mol C2H6 ≡ 7 mol O2 2 mol C2H6 ≡ 4 mol CO2 etc.
  2. 2. © 2008 Brooks/Cole 3 The Mole and Chemical Reactions What mass of O2 and Br2 is produced by the reaction of 25.0 g of TiO2 with excess BrF3? Notes: • Check the equation is balanced! • Stoichiometric ratios: 3TiO2 ≡ 3O2 ; 3TiO2 ≡ 2Br2 ; and many others • Excess BrF3 = enough BrF3 to react all the TiO2. 3 TiO2(s) + 4 BrF3(l) 3 TiF4(s) + 2 Br2(l) + 3 O2(g)
  3. 3. © 2008 Brooks/Cole 4 = 25.0 g x = 0.3130 mol TiO2 1 mol 79.88 g nTiO2 = mass TiO2 / FM TiO2 What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 with excess BrF3? 3 TiO2(s) + 4 BrF3(l) 3 TiF4(s) + 2 Br2(l) + 3 O2 (g) 3 mol TiO2 ≡ 3 mol O2 0.3130 mol TiO2 = 0.3130 mol O2 3 mol O2 3 mol TiO2 The Mole and Chemical Reactions
  4. 4. © 2008 Brooks/Cole 5 Mass of O2 produced = nO2 (mol. wt. O2) = 0.3130 mol x 32.00 g/mol = 10.0 g 3 TiO2 ≡ 2 Br2 nBr2 = 0.3130 mol TiO2 = 0.2087 mol Br2 2Br2 3 TiO2 Mass of Br2 = 0.2087 mol = 33.4 g Br2 159.81 g mol Br2 What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 with excess BrF3? 3 TiO2(s) + 4 BrF3(l) 3 TiF4(s) + 2 Br2(l) + 3 O2(g) The Mole and Chemical Reactions
  5. 5. © 2008 Brooks/Cole 6 Practice Problem 4.8 The purity of Mg can be found by reaction with excess HCl (aq), evaporating the water from the resulting solution and weighing the solid MgCl2 formed. Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) Calculate the % Mg in a 1.72-g sample that produced 6.46 g of MgCl2 when reacted with excess HCl. More difficult – What should you calculate? – How much pure Mg will make 6.46 g of MgCl2? – Express as a % of the original mass.
  6. 6. © 2008 Brooks/Cole 7 Practice Problem 4.8 Mg +2 HCl MgCl2 + H2 FW of MgCl2 = 24.31 + 2(35.45) = 95.21 g/mol nMgCl2 = 6.46 g MgCl2 1 mol 95.21 g Use mole ratio 1 mol Mg ≡ 1 mol MgCl2 = 0.06785 mol MgCl2 Mg required: 0.06785 mol MgCl2 1 Mg 1 MgCl2 = 0.06785 mol of pure Mg
  7. 7. © 2008 Brooks/Cole 8 0.06785 mol Mg x = 1.649 g Mg24.31 g 1 mol Practice Problem 4.8 Mg + 2 HCl MgCl2 + H2 Calculate mass of pure Mg needed 1.649 g 1.72 g Given 1.72 g of impure Mg. Purity (as mass %) = x 100% = 95.9 %
  8. 8. © 2008 Brooks/Cole 9 Reactions with Reactant in Limited Supply Given 10 slices of cheese and 14 slices of bread. How many sandwiches can you make? Balanced equation 1 cheese + 2 bread 1 sandwich 1 cheese ≡ 2 bread 1 cheese ≡ 1 sandwich 2 bread ≡ 1 sandwich
  9. 9. © 2008 Brooks/Cole 10 Two methods can be used: Product Method Calculate the product from each starting material.  The reactant giving the smallest number is limiting. 10 cheese x = 10 sandwiches1 sandwich 1 cheese 14 bread x = 7 sandwiches1 sandwich 2 bread Correct answer Bread is limiting. It will be used up first Reactions with Limited Reactants
  10. 10. © 2008 Brooks/Cole 11 Reactant Method Pick a reactant; calculate the amount of the other(s) needed. Enough? e.g. choose bread cheese needed: 14 bread (1 cheese /2 bread ) = 7 Available …. bread is limiting. e.g. choose cheese bread needed: 10 cheese (2 bread/1 cheese) = 20 Not available …. bread is limiting. • Yes = Your choice is the limiting reactant. • No = Another reactant is limiting. Reactions with Limited Reactants
  11. 11. © 2008 Brooks/Cole 12 …base all other calculations on the limiting reactant. Bread is limiting… Sandwiches made 14 bread (1 sandwich / 2 bread ) = 7 sandwiches Cheese remaining 14 bread (1 cheese / 2 bread ) = 7 cheese used. Started with 10 cheese. Cheese remaining 10 – 7 = 3 slices Reactions with Limited Reactants
  12. 12. © 2008 Brooks/Cole 13 How much water will be produced by the combustion of 25.0 g of H2 in the presence of 100. g of O2? Write a balanced equation: nH2 = 25.0 g = 12.40 mol H2 1 mol H2 2.016 g 2 H2(g) + O2(g) 2 H2O(l) nO2 = 100. g = 3.125 mol O2 1 mol O2 32.00 g Reactions with Limited Reactants
  13. 13. © 2008 Brooks/Cole 14 2 H2 + O2 2 H2O Moles available: 12.40 3.125 Using H2 12.40 mol H2 (2H2O /2H2 ) = 12.40 mol H2O Using O2 3.125 mol O2 (2 H2O /1 O2 ) = 6.250 mol H2O O2 gave less water. O2 is limiting. Base all calculations on O2 6.250 mol H2O x (18.02 g/ 1 mol ) = 113. g water How much water will be produced? Product Method Reactions with Limited Reactants
  14. 14. © 2008 Brooks/Cole 15 2 H2 + O2 2 H2O Moles available: 12.40 3.125 e.g. choose H2 O2 needed: 12.40 mol H2 (1 O2 /2 H2)= 6.20 mol Not available …. O2 is limiting. You only need one calculation. Had you chosen O2 H2 needed: 3.125 mol O2 (2 H2 /1 O2)= 6.250 mol Available …. O2 is limiting. H2O formed: 3.125 mol O2 (2H2O/1O2) = 6.250 mol. = 113. g Reactant Method Reactions with Limited Reactants
  15. 15. © 2008 Brooks/Cole 16 Consider : 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) If 374 g of NH3 and 768 g of O2 are mixed, what mass of NO will form? 1 mol 17.03 g nNH3 = 374 g = 21.96 mol 1 mol 32.00 g nO2 = 768g = 24.00 mol Balanced equation? yes Reactions with Limited Reactants
  16. 16. © 2008 Brooks/Cole 17 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Mol available: 21.96 24.00 From NH3 NO formed: 21.96 mol NH3 = 21.96 mol NO From O2 NO formed: 24.00 mol O2 = 19.20 mol NO Smallest amount…. O2 is limiting. 4 NO 4 NH3 4 NO 5 O2 Reactions with Limited Reactants
  17. 17. © 2008 Brooks/Cole 18 Mass of NO = 19.20 mol NO x = 576 g NO O2 is limiting. Base all calculations on O2. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Mass of NO formed? 30.01g 1 mol NO formed = 19.20 mol NO 21.96 mol 24.00 mol 19.20 mol Reactions with Limited Reactants
  18. 18. © 2008 Brooks/Cole 19 What mass of MgI2 is made by the reaction of 75.0 g of Mg with 75.0 g of I2? Mg + I2 → MgI2 • Balanced? YES • Calculate moles  75.0 g of Mg = 75.0g/(24.31 g mol-1) = 3.085 mol Mg  75.0 g of I2 = 75.0g/(253.9 g mol-1) = 0.2955 mol I2 Limiting reactant? 1Mg ≡ 1I2 so I2 = limiting • Since 1MgI2 ≡ 1I2 produce 0.2955 mol MgI2 • Mass of MgI2 = 0.2955 mol x 278.2 g/mol = 82.2 g Reactions with Limited Reactants
  19. 19. © 2008 Brooks/Cole 20 Theoretical yield The amount of product predicted by stoichiometry. Actual yield The quantity of desired product actually formed. Percent yield % yield = x 100% Actual yield Theoretical yield Percent Yield
  20. 20. © 2008 Brooks/Cole 21 Percent Yield Few reactions have 100% yield. Possible reasons Side reactions may occur that produce undesired product(s). Product loss during isolation and purification. Incomplete reaction due to poor mixing or reaching equilibrium…
  21. 21. © 2008 Brooks/Cole 22 Percent Yield You heat 2.50 g of copper with an excess of sulfur and synthesize 2.53 g of copper(I) sulfide 16 Cu(s) + S8(s) 8 Cu2S(s) What was the percent yield for your reaction? nCu used: = 0.03934 mol Cu 1 mol 63.55g 2.50 g 16 mol Cu used 8 mol Cu2S made Theoretical yield: 0.03934 mol Cu = 0.01967 mol Cu2S8 Cu2S 16 Cu
  22. 22. © 2008 Brooks/Cole 23 Percent Yield Heat 2.50 g of Cu with excess S8 and make 2.53 g of copper(I) sulfide: 16 Cu(s) + S8(s) → 8 Cu2S(s). What was the %-yield for your reaction? Theoretical yield = 0.01967 mol Cu2S 159.2 g 1 mol = 0.01967 mol Cu2S = 3.131 g Cu2S Actual yield = 2.53 g Cu2S (in problem) Percent yield = x 100% = 80.8% 2.53 g 3.131 g

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