Learning Stoichiometry in Chemistry

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Explore the fundamentals of stoichiometry in chemistry, where ratios in balanced reactions are used to manipulate chemical variables like mass, volume, and numbers of particles. Understand the concepts of mole ratios and learn how to calculate the required amounts of reactants and products in chemical reactions. Dive into examples to practice applying stoichiometry principles effectively.

  • Chemistry
  • Stoichiometry
  • Balanced Reactions
  • Mole Ratios
  • Chemical Variables
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  1. OB: intro to STOICHIOMETRY Using ratios in balanced reactions to manipulate the chemical variables in these reactions. (mass, volume, and numbers of particles) Calculators, reference tables, notes. Get your thinking caps on!

  2. 1. What is stoichiometry? Chemistry math that follows the Law of Conservation of Matter and starts with a balanced equation.

  3. 4Al (S) + 3O2(G) 2Al2O3(S) 2. How many atoms and molecules and formula units are in ratio here?

  4. 4Al (S) + 3O2(G) 2Al2O3(S) 2. How many atoms and molecules and formula units are in ratio here? 4 atoms Al : 3 molecules O2 : 2FU s Al2O3 4:3:2 ratio

  5. 4Al (S) + 3O2(G) 2Al2O3(S) 3. What is the mole ratio for this balanced equation?

  6. 4Al (S) + 3O2(G) 2Al2O3(S) 3. What is the mole ratio for this balanced equation? 4 moles Al : 3 moles O2 : 2 moles Al2O3 The Mole Ratio here is 4:3:2

  7. 4Al (S) + 3O2(G) 2Al2O3(S) 4. If you used up 8 moles of Aluminum in a big reaction, how many moles of oxygen would you need to complete the reaction? Do this all here

  8. 4Al (S) + 3O2(G) 2Al2O3(S) 4. If you used up 8 moles of Aluminum in a big reaction, how many moles of oxygen would you need to complete the reaction? 4 Moles Al 3 Moles O2 8 Moles Al X Moles O2 = 4X = 24 X = 6 moles O2

  9. 4Al (S) + 3O2(G) 2Al2O3(S) 5. If you used up only two moles of Al, how many moles of O2(G) would form? Do this all here

  10. 4Al (S) + 3O2(G) 2Al2O3(S) 5. If you used up only two moles of Al, how many moles of O2(G) would form? 2 Moles Al 3 Moles O2 1 Moles Al X Moles O2 = 2X = 3 X = 1.5 Al2O3

  11. Take out table H and draw in the entire Stoich Mole Map now. It will be your guide for all stoichiometry math, now & in college. All problems are on this map.

  12. Volume Island Volume Island 2 1 mole = 22.4 Liters of Gas 1 mole = 22.4 Liters of Gas Mole Ratio Tunnel Mole Island Mole Island 2 Mass Island Mass Island 2 Particle Island Particle Island 2

  13. 6. Stoichiometry is the math based upon a balanced chemical equation as a starting point to figure out the ratio of grams (or liters of gas, or numbers of particles) with any substance compared to another in the same reaction.

  14. 7. If you react 56.9 grams of Al, how many liters of O2 would be necessary to complete the reaction? 4Al (S) + 3O2(G) 2Al2O3(S) Do this all here

  15. 7. If you react 56.9 grams of Al, how many liters of O2 would be necessary to complete the reaction? 4Al (S) + 3O2(G) 2Al2O3(S) 1 mole Al 27 g Al 56.9 grams of Al 1 = 2.11 moles Al X MRAl 4x = 3.87 4 3 =1.72 O2 x X = 8.790 moles O2 1.58 moles O2 1 22.4 Liters O2 1 mole O2 = 35.4 Liters O2 X

  16. 8. Aluminum and chlorine form aluminum chloride. If you use 75.0 g of metal, how many grams of product form? Write balanced equation first. ____________________________________________ Do this all here

  17. 8. Aluminum and chlorine form aluminum chloride. If you use 75.0 g of metal, how many grams of product form? Write balanced equation first. 2Al(S) + 3Cl2(G) 2AlCl3(S) 75.0 g Al 1 1 mole Al 27 g Al = 2.78 moles Al 1st X Al 2 2 = 2.78 MR X = 2.78 moles AlCl3 2nd AlCl3 X 2.78 moles AlCl3 1 132 grams AlCl3 1 mole AlCl3 X = 367 g AlCl3 3rd

  18. Stoich Class #2 more stoichiometry math problems 2-step sized (Smaller & quicker) You need a calculator, reference tables, and Your notes.

  19. 9. When 8.50 moles of propanol combusts, how many grams of water forms? 2C3H7OH(G) + 9O2(G) 6CO2(G) + 8H2O(G) Do this all here

  20. 9. When 8.50 moles of propanol combusts, how many grams of water forms? 2C3H7OH(G) + 9O2(G) 6CO2(G) + 8H2O(G) 2 8 = 8.50 propanol water 2X = 68.0 MR 1st X = 34.0 moles water X 34.0 moles water 1 18 grams water 1 mole water X = 612 g AlCl3 2nd

  21. 10. If you want to use up 23.1 moles of HCl, how many molecules of hydrogen form? Just 2 steps. 2Al(S) + 6HCl(AQ) 2AlCl3(AQ) + 3H2(G) Do this all here

  22. 10. If you want to use up 23.1 moles of HCl, how many molecules of hydrogen form? Just 2 steps. 2Al(S) + 6HCl(AQ) 2AlCl3(AQ) + 3H2(G) 6 3 = 23.1 HCl H2 6X = 69.3 MR X = 11.6 moles H2 X X = 69.8 x 1023 6.02 x 1023 molecules H2 1 mole H2 11.6 moles H2 1 6.98 x 1024 molecules H2

  23. 11. 371.5 grams of candle wax (C21H44) combusts. How many liters of CO2 gas form? Assume STP for the gas. Start with a balanced equation. 3 steps. Do this all here

  24. 11. 371.5 grams of candle wax (C21H44) combusts. How many liters of CO2 gas form? Assume STP for the gas. Start with a balanced equation. 3 steps. C21H44 + 32O2 21CO2 + 22H2O 1 mole wax 371.5 g wax 1 X = 1.255 moles wax 296 g wax wax 1 1.255 MR = CO2 21 x x = 26.36 moles CO2 x22.4 L CO2 1 mole CO2 26.36 moles CO2 1 = 590.5 Liters CO2

  25. 12. Using the same wax combustion reaction, if you consume 23.9 moles of oxygen, how many moles of water form? One step. (sometimes it s easy) C21H44 + 32O2 21CO2 + 22H2O Do this all here

  26. 12. Using the same wax combustion reaction, if you consume 23.9 moles of oxygen, how many moles of water form? One step. (sometimes it s easy) C21H44 + 32O2 21CO2 + 22H2O oxygen 32 23.9 water 22 x MR = 32x = 525.8 x = 16.4 moles wata

  27. 14. You have 4.56 x 1025 atoms of Zn that you put into H3PO4(AQ) to make them fizz away. How many grams of hydrogen gas form? _Zn + _H3PO4(AQ) _Zn3(PO4)2(AQ) +_H2(G) This needs to be balanced first.

  28. 14. You have 4.56 x 1025 atoms of Zn that you put into H3PO4(AQ) to make them fizz away. How many grams of hydrogen gas form? 3Zn + 2H3PO4(AQ) Zn3(PO4)2(AQ) + 3H2(G) Do this all here

  29. 14. You have 4.56 x 1025 atoms of Zn that you put into H3PO4(AQ) to make them fizz away. How many grams of hydrogen gas form? 3Zn + 2H3PO4(AQ) Zn3(PO4)2(AQ) + 3H2(G) 1 mole Zn 4.56 6.02x1025 4.56 x 1025 atoms 1 = x 1023 =75.7 moles Zn 6.02 x 1023 atoms Zn Zn H2 3 3 75.7 X MR = X = 75.7 moles hydrogen 75.7 moles H2 1 2 grams H2 1 mole H2 x = 151 grams H2

  30. 15. How many liters of nitrogen gas are required to combine with 809 liters of hydrogen when ammonia forms. 3H2 + N2 2NH3 Do this all here

  31. 15. How many liters of nitrogen gas are required to combine with 809 liters of hydrogen when ammonia forms. 3H2 + N2 2NH3 809 L H2 1 x 1 mole hydrogen 22.4 L H2 = 36.1 moles H2 1 3 N2 H2 X 3X = 36.1 moles X = 12.0 moles N2 MR = 36.1 22.4 L N2 1 mole N2 12.0 moles N2 1 x = 269 Liters N2(G)

  32. 16. If exactly 15.6 moles of ethane gas combusts, how many moles of oxygen are used? 2C2H6(G) + 7O2(G) 4CO2(G) + 6H2O(L) Do this all here

  33. 16. If exactly 15.6 moles of ethane gas combusts, how many moles of oxygen are used? 2C2H6(G) + 7O2(G) 4CO2(G) + 6H2O(L) 2 15.6 7 x Ethane oxygen MR MR = 2X = 109.2 X = 54.6 moles O X = 54.6 moles O2 2

  34. 17 If exactly 649.6 L of NO(G) form, how many liters of O2 are used? 4NH3(G) + 5O2(G) 4NO(G) + 6H2O(L) Do this all here

  35. 17 If exactly 649.6 L of NO(G) form, how many liters of O2 are used? 4NH3(G) + 5O2(G) 4NO(G) + 6H2O(L) 649.6 L NO 1 1 mole NO 22.4 L NO X = 29.00 mole NO(G)(4 SF) 4X = 145.0 NO O2 4 29.00 5 X = MR X = 36.25 moles O2 36.25 mole O2 1 X22.4 L O2 1 mole O2 = 812.0 Liters O2 (4 SF)

  36. 18. In this combustion, 125 g of oxygen are used up. How many g of H2O are produced? 2C8H18(L) + 25O2(G) 16CO2(G) +18H2O(G) Do this all here

  37. 18. In this combustion, 125 g of oxygen are used up. How many g of H2O are produced? 2C8H18(L) + 25O2(G) 16CO2(G) +18H2O(G) 1 mole O2 32 g O2 125 g O2 1 = 3.91 mole O2(G) (3 SF) X 25X = 70.38 18 X 25 3.91 H2O O2 = MR X = 2.82 moles water 2.82 moles water 1 X18 g water 1 mole H2O = 50.8 g water 3 SF

  38. 19. 105 g of N2 react with oxygen to form dinitrogen pentoxide. How many molecules of O2 are required in this reaction? Write a balanced equation first. Do this all here

  39. 19. 105 g of N2 react with oxygen to form dinitrogen pentoxide. How many molecules of O2 are required in this reaction? 2N2(G) + 5O2(G) 2N2O5(G) X1 mole N2 28 g N2 105 g N2 1 = 3.75 mole N2(G) (3 SF) 2X = 18.75 moles O2 X = 9.38 moles O2 2 3.75 5 X N2 O2 MR = X6.02 x 1023 molecules 1 mole 9.38 moles O2 1 = 56.4676 x 1023 56.5 x 1023 5.65 x 1024 molecules O2

  40. 20. In an odd chemical reaction, 0.135 moles of H2 reacts. How many grams of NH3 form in this reaction? 2NO2(G) + 7H2(G) 2NH3(G) + 4H2O(L) Do this all here

  41. 20. In an odd chemical reaction, 0.135 moles of H2 reacts. How many grams of NH3 form in this reaction? 2NO2(G) + 7H2(G) 2NH3(G) + 4H2O(L) 7 0.135 2 X 7X = 0.270 H2 NH3 MR = X = 0.0386 moles NH3 17 g NH3 1 mole NH3 0.0386 moles NH3 1 = 0.656 grams NH3 X

  42. 21. When 9.42 x 1025 atoms of phosphorous react with sufficient chlorine to make phosphorous pentachloride, how many molecules of chlorine gas are necessary? This is hardest stoich problem in the history of stoich problems, and it s the last one you have to do. Start with a balanced equation! _P(S) + _Cl2(G) _PCl5(G)

  43. 21. When 9.42 x 1025 atoms of phosphorous react with sufficient chlorine to make phosphorous pentachloride, how many molecules of chlorine gas are necessary? 2P(S) + 5Cl2(G) 2PCl5(G) Do this all here

  44. 21. When 9.42 x 1025 atoms of phosphorous react with sufficient chlorine to make phosphorous pentachloride, how many molecules of chlorine gas are necessary? 2P(S) + 5Cl2(G) 2PCl5(G) 9.42 6.02X 102 moles P = 156 moles P 1 mole P 9.42 x 1025 atoms P 1 = X 6.02 x 1023 atoms P P 2 5 MR 156 X 2X = 780 X = 390. moles Cl2 = Cl2 = 2347.8 x 1023 molecules Cl2 390. moles Cl2 1 6.02 x 1023 molecules Cl2 1 mole Cl2 x 2350 x 1023 molecules Cl2 2.35 x 1026 molecules Cl2

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