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Stoichiometry: Calculation of Quantities in Chemical Reactions

This chapter explains the calculation of quantities in chemical reactions using stoichiometry, including mole-to-mole calculations, mass-to-mass calculations, and limiting reagent calculations. It also covers the concept of percent yield.

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Stoichiometry: Calculation of Quantities in Chemical Reactions

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  1. Chapter 9

  2. Stoichiometry • The calculation of quantities in chemical reactions

  3. Particles N2(g) + 3H2(g) 2NH3(g) • 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3 • Ratio : 1:3:2 particles

  4. Moles 1 mole = 6.02 x 1023 representative particles Based on the reaction – mole ratio is 1:3:2 * Coefficients are the relative # of moles

  5. Mass • Law of conservation of mass 1:3 2 28g + 6 g 34 g N = 2x14 H = 2x3 NH3 = 17x2

  6. Volume • Assume STP • 22.4 L/mol

  7. Example 2H2S + 3O2 2SO2 + 2H2O

  8. Mass of Reactants 2 mol x 34.1 g = 68.2g mol 3 mol x 32 g = 96g mol 164.2 g

  9. Mass of Products 2 mol x 64.1g = 128.2 g mol 2 mol x 18g = 36 g mol 164.2 g

  10. Volume of gases at STP Reactants 2 mol H2S x 22.4 L = 44.8 L H2S 1 mol 3 mol O2 x 22.4 L = 67.2 L O2 1 mol

  11. Products 2 mol SO2 x 22.4 L = 44.8 L SO2 1 mol 2 mol H2O x 22.4 L = 44.8 L H2O 1 mol

  12. Mole to Mole Calculations • How many moles of ammonia are produced when 0.06 mol of nitrogen reacts with hydrogen?

  13. N2 + 3H2 2NH3 0.06 mol N2 x 2mol NH3 1 mol N2 = .12mol NH3

  14. Example Aluminum Oxide is formed from aluminum and oxygen a) Write a balanced equation 4Al + 3O2 2Al2O3

  15. b) How many moles of Al are needed to form 2.3 mol of Al2O3? 2.3 mol Al2O3 x 4 mol Al 2 mol Al2O3 = 4.6mol Al

  16. c) How many moles of Oxygen are required to react completely with 0.84 mol of Al ? .84mol Al x 3mol O2 4mol Al = .63 mol O2

  17. d) Calculate the # of moles of Al2O3 formed when 17.2 mol of O2 reacts with Al ? 17.2 mol O2 x 2mol Al2O3 3mol O2 = 11.47 mol Al2O3

  18. Mass – Mass Calculations • GFM • Ex) H2 1 mol or 2g 2g 1 mol

  19. Example • Calculate the number of grams of ammonia produced by the reaction of 5.40 g of hydrogen with nitrogen. N2(g) + 3H2(g) 2NH3(g)

  20. 5.40g H2 x 1mol H2 x 2mol NH3 x 17g NH3 2g H2 3mol H2 1mol NH3 = 30.6g NH3 These problems are solved the same way as mole – mole problems

  21. Steps • Change mass to moles G (given) • Change moles of G to moles of W • Change moles of W to grams of W

  22. Example • 2C2H2 + 5O2 4CO2 + 2H2O a) How many grams of O2 are required to burn 13g of C2H2 ? = 40.0g O2

  23. b) How many grams of CO2 and H2O are produced when 13g of C2H2 react with the oxygen. 13g C2H2 x 1mol C2H2 x 4mol CO2 x 44g CO2 26g C2H2 2mol C2H2 1mol CO2 = 44g CO2

  24. Continued 13g C2H2 x 1mol C2H2 x 2mol H2O x 18g H2O 26g C2H2 2mol C2H2 1mol H2O = 9g H2O

  25. Other Stoichiometric Calculations • Mass – Volume • Volume – Volume • Particle – Mass

  26. Example • Tin plus hydroflouric acid yields Tin(II)flouride and hydrogen gas Sn(s) + 2HF(g) SnF2(s) + H2(g)

  27. a) How many grams of SnF2 can be made by reacting 7.42 x 1024 molecules of HF with tin? 966g SnF2

  28. b) How many L of H2 (STP) are produced by reacting 23.4g of Sn with HF? 4.42 L H2

  29. c) How many L of HF are needed to produce 14.2 L of H2(STP) ? 28.4 L HF

  30. d) How many molecules of H2 are produced by the reaction of Sn with 80 L of HF (STP) ? 1.08 x 1024 molecules H2

  31. Limiting Reagent • Example • 1lb of swiss & 1lb of pastrami and 14 slices of bread • You can only make 7 sandwiches • Limiting reagent - bread • Excess reagent – luncheon meats

  32. Example • Sodium Chloride is prepared by the reaction of sodium metal with chlorine gas 2Na(s) + Cl2(g) 2NaCl(s) • What will occur when 6.70mol of Na reacts with 3.20 mol of Cl2 ?

  33. 6.70mol Na x 1mol Cl2 2mol Na a) What is the limiting reagent ? = 3.35 mol Cl2 required * 3.35 mol Cl2 are required to completely react with 6.70mol Na but you only have 3.20 mol Cl2. Cl2 is the limiting reagent.

  34. b) How many moles of NaCl are produced? 3.20mol Cl2 x 2mol NaCl 1mol Cl2 = 6.40 mol NaCl

  35. c) How much of the excess reagent remains un-reacted ? 3.20mol Cl2 x 2mol Na 1mol Cl2 = 6.40mol Na 6.70mol Na - 6.40mol Na .30mol Na is excess

  36. Example C2H4 + 3O2 2CO2 + 2H2O If 2.70 mol C2H4 reacts with 6.30 mol 02 a) What is the limiting reagent?

  37. 2.7mol C2H4 x 3mol O2 = 8.1mol O2 1mol C2H4 * You need 8.1mol O2 to react with 2.7 mol C2H4 the limiting reagent is O2

  38. 6.3 mol O2 x 2mol H2O 3mol O2 b) Calculate moles of H2O produced = 4.2mol H2O

  39. c) Calculate moles of excess reagent 6.30mol O2 x 1mol C2H4 3mol O2 = 2.1mol C2H4

  40. Final Step 2.7mol C2H4 - 2.1mol C2H4 0.60mol C2H4

  41. Example a) How many grams of H2 can be produced when 4g HCl is added to 3g of Mg?

  42. 4g HCl x 1mol HCl = 0.11mol HCl 36g HCl 3g Mg x 1mol Mg = .12mol Mg 24.3g Mg .11mol HCl x 1mol H2 x 2g H2 2mol HCl 1mol H2 =0.11g H2

  43. b) Assuming STP, what is the volume of H2? 0.11g H2 x 1mol x 22.4L 2g 1mol = 1.23 L H2

  44. Percent Yield • Theoretical Yield – the maximum amount of product that could be formed from a given amount of reactant

  45. Actual Yield – Amount of product that forms when the reaction is carried out in the lab.

  46. Percent Yield – Ratio of actual yield to theoretical yield % Yield = actual yield x 100 theoretical yield

  47. Example CaCO3 CaO + CO2 *Calculate the % yield of CaO if 24.89g CaCO3 is heated to give 13.1g of CaO

  48. 24.8g CaCO3 x 1mol CaCO3 x 100.1g CaCO3 1mol CaO x 56.1g CaO 1mol CaCO3 1mol CaO = 13.9g CaO

  49. Percent Yield 13.1 13.9 94.2% x 100 =

  50. Example • What is the % yield of copper if 3.74g of copper is produced when 1.87g of Aluminum is reacted with an excess of copper(II)sulfate?

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