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CHEMICAL REACTIONS CHAPTER 4

Reactants: Zn + I 2. Product: ZnI 2. CHEMICAL REACTIONS CHAPTER 4. Chapter 4 Outline. Chemical Equations Stoichiometry Limiting Reactants Chemical Analysis. Chemical Equations. Depict the kind of reactants and products and their relative amounts in a reaction.

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CHEMICAL REACTIONS CHAPTER 4

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  1. Reactants: Zn + I2 Product: ZnI2 CHEMICAL REACTIONSCHAPTER 4

  2. Chapter 4 Outline • Chemical Equations • Stoichiometry • Limiting Reactants • Chemical Analysis

  3. Chemical Equations Depict the kind of reactants and products and their relative amounts in a reaction 4 Al(s) + 3 O2(g) 2Al2O3(s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

  4. Chemical Equations 4 Al(s) + 3 O2(g) 2 Al2O3(s) This equation means 4 Al atoms + 3 O2 molecules 2 molecules of Al2O3 4 moles of Al + 3 moles of O2 2 moles of Al2O3

  5. Chemical Equations • Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. • The Law of the Conservation of Matter

  6. Chemical Equations Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Lavoisier, 1788

  7. Balancing Equations

  8. Balancing Equations

  9. Balancing Equations ___C3H8 (g) + ___ O2 (g) ----> ___CO2 (g) + ___ H2O (g) C3H8 (g) + 5 O2 (g) ----> 3 CO2 (g) + 4 H2O (g)

  10. Balancing Equations ___B4H10 (g) + ___ O2 (g) ---> ___ B2O3 (g) + ___ H2O (g) 2 B4H10 (g) + 11 O2 (g) ---> 4 B2O3 (g) + 10 H2O (g)

  11. 11 STOICHIOMETRY - the study of the quantitative aspects of chemical reactions.

  12. PROBLEM: If 454 g of NH4NO3 decomposes, how much H2O and N2O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH4NO3 N2O + 2 H2O

  13. 1 mol 454 g • = 5.68 mol NH NO 4 3 80.04 g 454 g of NH4NO3 --> N2O + 2 H2O STEP 2 Convert reactant mass to moles (454 g) --> moles

  14. 2 mol H O produced 2 1 mol NH NO used 4 3 454 g of NH4NO3 --> N2O + 2 H2O STEP 3 Convert moles reactant --> moles product. Relate moles NH4NO3 to moles product. 1 mol NH4NO3 --> 2 mol H2O Express this relation as the STOICHIOMETRIC FACTOR

  15. 2 mol H Oproduced 2 5.68 mol NH NO • 4 3 1 mol NH NO used 4 3 454 g of NH4NO3 --> N2O + 2 H2O STEP 3 Convert moles reactant (5.68 mol) moles product = 11.4 mol H2O produced

  16. 454 g of NH4NO3 --> N2O + 2 H2O STEP 4 Convert moles product (11.4 mol) to mass product. This is called the THEORETICAL YIELD This is the “Expected” # of moles.

  17. 454 g of NH4NO3 --> N2O + 2 H2O STEP 4 Convert moles prod. (11.4 mol) to mass prod. This is the “Expected” Mass! ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS! Repeat to find the grams of N2O formed. 18 . 02 g 11.4 mol H • = 204 g H O O 2 2 1 mol

  18. 454 g of NH4NO3 --> N2O + 2 H2O STEP 5 How much N2O is formed? Total mass of reactants=total mass of products 454 g NH4NO3 = ___ g N2O + 204 g H2O mass of N2O = 250. g This is an alternate method.

  19. 454 g of NH4NO3 --> N2O + 2 H2O STEP 6 Calculate the percent yield. If you isolated only 131 g of N2O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

  20. actual yield % yield = • 100% theoretical yield 131 g % yield = • 100% = 52.4% 250. g 454 g of NH4NO3 --> N2O + 2 H2O STEP 6 Calculate the percent yield.

  21. Mass product Mass reactant Stoichiometric factor Moles reactant Moles product General Plan For Stoichiometry Calculations

  22. PROBLEM: Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained? 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Reaction is catalyzed by MnO2 Step 1: moles of H2O2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O2 Step 3: mass of O2 Repeat for H2O.

  23. Reactions Involving aLIMITING REACTANT • In a given reaction, there is not enough of one reagent to use up the other reagent completely. • The reagent in short supply LIMITS the quantity of product that can be formed.

  24. 2NO(g) +O2(g)2NO2(g) LIMITING REACTANTS Reactants Products NO Limiting reactant = ___________ Excess reactant = ____________ O2

  25. LIMITING REACTANTS

  26. LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl---> ZnCl2 + H2 Rxn 1 Rxn 2 Rxn 3 mass Zn (g) 7.00 3.27 1.31 mol Zn 0.107 0.050 0.020 mol HCl 0.100 0.100 0.100 mol HCl/mol Zn 0.93 2.00 5.00

  27. PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. How many grams of Al2Cl6 can form?

  28. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

  29. mol Cl 3 2 = mol Al 2 • Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. 2 Al + 3 Cl2 ---> Al2Cl6 Reactants must be in the mole ratio

  30. mol Cl 3 2 > mol Al 2 Deciding on the Limiting Reactant 2 Al + 3 Cl2 Al2Cl6 If then there is not enough Al to use up all the Cl2, and the limiting reagent is Al.

  31. mol Cl 3 2 < mol Al 2 Deciding on the Limiting Reactant 2 Al + 3 Cl2---> Al2Cl6 If then there is not enough Cl2 to use up all the Al, and the limiting reagent is Cl2

  32. 1 mol 5 . 40 g Al • = 0.200 mol Al 27.0 g Step 2 of LR problem: Calculate moles of each reactant We have 5.40 g of Al and 8.10 g of Cl2 1 mol 8 .10 g Cl • = 0.114 mol Cl 2 2 70.9 g

  33. mol Cl 0.114 mol 2 = = 0.57 mol Al 0.200 mol Find mole ratio of reactants This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl2

  34. grams Cl2 grams Al2Cl6 moles Cl2 moles Al2Cl6 Limiting reactant = Cl2 Base all calculations on Cl2 Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?

  35. 1 mol Al Cl 2 6 0 . 114 mol Cl • = 0.0380 mol Al Cl 2 2 6 3 mol Cl 2 2 66.4 g Al Cl 2 6 0 . 0380 mol Al Cl • = 10.1 g Al Cl 2 6 2 6 mol CALCULATIONS: calculate mass of Al2Cl6 expected. Step 1: Calculate moles of Al2Cl6 expected based on LR. Step 2: Calculate mass of Al2Cl6 expected based on LR.

  36. How much of which reactant will remain when reaction is complete? • Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? • First find how much Al was required. • Then find how much Al is in excess.

  37. Calculating Excess Al 2 Al + 3 Cl2 products 0.114 mol = LR 0.200 mol

  38. Calculating Excess Al 2 Al + 3 Cl2 products 0.114 mol = LR 0.200 mol 2 mol Al 0 .114 mol Cl • = 0.0760 mol Al req' d 2 3 mol Cl 2 Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

  39. Nitrogen and iodine react to form nitrogentriiodide. If 50.0 g of nitrogen is mixed with 350.0 g iodine, calculate the number of grams of product formed and the grams of reactant remaining. N2 + 3 I2 2 NI3 28.0 g/mol 253.8 g/mol 394.7 g/mol 1.79 mole 1.38 mole L.R. (0.460 S.U.) 50.0g -12.9g 0g left 363 g 37.1g left 350.0g + 50.0g = 400.0g = 363g + 37.1g

  40. Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. What is the empirical formula of CxHy? CxHy + oxygen 0.379 g CO2 + 0.1035 g H2O 40

  41. Chemical Molecular Analysis in the lab

  42. Using Stoichiometry to Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate moles of C in CO2 8.61 x 10-3 mol C 2. Calculate moles of H in H2O 1.149 x 10 -2 mol H

  43. Using Stoichiometry to Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C3H4

  44. Chemical AnalysisCombustion-Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O Recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate moles of C in CO2 0.379 g CO2 mole CO2 mole C 44.0 g CO2 mole CO2 =0.00861 mole C

  45. Chemical AnalysisCombustion-Determine a Formula CxHy+some oxygen 0.379gCO2+0.1035 g H2O 2. Calculate moles of H in H2O 0.1035 g H2O mole H2O 2 mole H 18.0 g H2O mole H2O =0.0115 mole H

  46. Chemical AnalysisCombustion-Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 0.0115mol H/ 0.00861 mol C = 1.34 mol H / 1.00 mol C = 4.02 mol H / 3.00 mol C Empirical formula = C3H4

  47. Sample Problems • 1) A 0.537 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced 1.030 g of CO2 and 0.632 g of H2O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula. 1.030 g CO2 mole CO2 mole C 12.0 g C 44.0g CO2 mole CO2 mole C = 0.281 g C 0.632 g H2O mole H2O 2 mole H 1.0 g H 18.0 g H2O mole H2O mole H = 0.070 g H

  48. Sample Problems • 1) A 0.537 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced 1.030 g of CO2 and 0.632 g of H2O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula. 0.537 g C, H, O - 0.281 g C - 0.070 g H 0.186 g O

  49. .281 g .070 g .186 g 1 mole 12.0 g 1 mole 1.0 g 1 mole 16.0 g Sample Problems C H O 0.0234 0.070 0.0116 0.0116 0.0116 0.0116 2.02 6.0 1.00 Empirical formula C2H6O

  50. 90 = 2 46 Sample Problems Empirical formula C2H6O Molecular formula C4H12O2 Alternate method. Use grams of cmpd. and its molar mass to determine moles of cmpd. Divide moles of C and H by these moles to find the subscripts for C and H. The subscript for O can be determinded by difference.

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