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Chapter 9 Stoichiometry

Chapter 9 Stoichiometry. Stoichiometry. Greek Translation Stoicheion – means “element” Metron – means “measure” Element Measuring. Stoichiometry. EQ: How do we compare elements so that we can measure them? For example, does… 23 grams of Na = 23 grams of Cl

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Chapter 9 Stoichiometry

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  1. Chapter 9 Stoichiometry

  2. Stoichiometry • Greek Translation • Stoicheion – means “element” • Metron – means “measure” • Element Measuring

  3. Stoichiometry EQ: How do we compare elements so that we can measure them? For example, does… 23 grams of Na = 23 grams of Cl 23g Na x 1 mol = 1 mol Na or 6.022E23 atoms 23g 23g Cl x 1mol = 0.65 mol Cl or 3.85E23 atoms 35.5g

  4. Stoichiometry So how much Chlorine do I need since NaCl is a one to one ratio? 23g Na x 1 mol = 1 mol Na or 6.022E23 atoms 23g 35.5g Cl x 1mol = 1 mol Cl or 6.022E23 atoms 35.5g 23g Na =35.5g Cl

  5. Stoichiometry What if all I had was 23 grams of Chlorine? How much NaCl could l make? 23g Na x 1 mol = 1 mol Na or 6.022E23 atoms 23g 23g Cl x 1mol = 0.65 mol Cl or 3.85E23 atoms 35.5g Limiting Reagent

  6. Chemical Stoichiometry Stoichiometry- The study ofquantitiesof materialsconsumedandproducedin chemical reactions. Consumed  Produced Mole  Mole Mass Mole Mole Mass Mass Mass

  7. Review: The Mole • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. • 1 moleof anything = 6.022 ´ 1023units of that thing

  8. Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = 44.01 grams per mole H2O = 18.02 grams per mole Ca(OH)2 = 74.10 grams per mole

  9. Review: Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 2CO2 + 3H2O reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

  10. Teacher Examples

  11. Mole to Mole Problems Ammonia, NH3, is widely used as fertilizer and in many household cleaners. How many moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas? Step 1: Step 2: Step 3: Step 4: 6 mol H2(g) X 2 mol NH3 = 4 mol NH3 3 mol H2 mol unknown mol known 2=N=1 2=H=3 N2 + H2NH3 2 3

  12. Mole to Mole Problems The decomposition of potassium chlorate, KClO3, is used as a source of oxygen in the laboratory. How many moles of potassium chlorate are needed to produce 15 mol of oxygen?

  13. Mole to Mass Problems In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide? 6CO2 + 6H2OC6H12O6 +6O2 3.00 mol H2O X 1 mol C6H12O6X 6 mol H2O 180.18g= 1 mol C6 X 12.01 = 72.06 H12 X 1.01 = 12.12 O6 X 16.00 = 96.00 180.18g/mol 90.09g C6H12O6

  14. Mole-Mass Problem Solution of iron (III) chloride, FeCl3, are used in photoengraving and to make ink. This compound can be made by the following reaction.                  2 Fe + 3 Cl2 ---------> 2 FeCl3 How many grams of FeCl3 form from 2.4 moles of Fe?

  15. Mass to Mole Problems • The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. • NH3 + O2 NO + H2O (unbalanced) • The reaction is run using 824 g of NH3and excess oxygen. • How many moles of NO are formed? • How many moles of H2O are formed? 4 5 4 6 A. 824g NH3X 1 mol X 4 mol NO= 48.39 mol NO 17.03g NH34 mol NH3 B. 824g NH3X 1 mol X 6 mol H2O= 72.58 mol H2O 17.03g NH34 mol NH3 N X 14.00 = 14.00 H3 X 1.01 = 3.03 17.03g/mol

  16. Mass to Mole Problems • Oxygen was discovered by Joseph Priestley in 1774 when he heated mercury (II) oxide to decompose it to form its constituent elements. • How many moles of mercury (II) oxide are needed to produce 125 g of oxygen? • How many moles of mercury are produced?

  17. Mass to Mass Problems When copper metal is added to silver nitrate in solution, silver metal and copper (II) nitrate are produced. What mass of silver is produced from 100. g of Cu? +2 -1 Cu + Ag(NO3) 2  Ag + Cu(NO3) 2 2 100g Cu X 1 mol Cu X 2 mol AgX 107.87g = 339.48g Ag 63.55g 1 mol Cu 1 mol Ag Molar Mass Molar Mass

  18. Mass to Mass Problems Tin (II) fluoride is used in some toothpastes. It is made by the reaction of tin (II) with hydrogen fluoride according to the following equation. Sn + 2 HF  SnF2 + H2 How many grams of SnF2 are produced from the reaction of 30.00 g of HF with Sn?

  19. Student In-Class Practice

  20. Mole-Mole How many moles of carbon dioxide, CO2, can be made from Fe2O3 by the use of 1.90 moles of carbon monoxide, CO, in the following reaction?              Fe2O3 + 3 CO -----------> 2 Fe + 3 CO2

  21. Mole- Mass Problem The Solvay process is used to make sodium carbonate, Na2CO3, a chemical that ranked 11th among all chemicals in annual production in 1986. The process begins with the passing of ammonia and carbon dioxide through a solution of sodium chloride. This makes sodium bicarbonate and ammonium chloride: H2O + NaCl + NH3 + CO2 ---------> NH4Cl + NaHCO3 How many grams of sodium bicarbonate could, in theory, be made from 5 moles of ammonia, NH3?

  22. Mass-Mole Using the equation: Pb(SO4)2 + 4 LiNO3 Pb(NO3)4 + 2 Li2SO4 , How many moles of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction?

  23. Mass to Mass Problems What mass of aluminum is produced by the decomposition of 5.0 kg of Al2O3?

  24. Practice Assignment

  25. × Mole to Mole Problems 20 mol X mol CO2 + 2 LiOH  Li2CO3 + H2O Use the mole ratio, Duh! Mole Train How many moles of lithium hydroxide are required to react with 20 moles of carbon dioxide? Given: 20 mol CO2 How can we go from moles of one compound to mole of another compound? Unknown: X mol LiOH 2 mol LiOH = 20 mol CO2 40 mol LiOH 1 mol CO2 The MOLE TRAIN!!!!!!

  26. How many moles of water are produced in the complete combustion of 1.00 moles of glucose according to the following reaction: C6H12O6 + 6O2 ----> 6CO2 + 6H2O 6 mole H2O 1.00 moles C6H12O6 X -------------------- 1 moles C6H12O6 6.00 moles of water

  27. What mass of H2SO4 can be formed from 5.0 moles of SO2 and an unlimited quantity of H2O and O2? 2SO2 + O2 + 2H2O ---- 2H2SO4 2 moles H2SO4 98 grams H2SO4 5.0 moles SO2 X --------------- X ---------------------- 2 moles SO2 1 mole H2SO4 490 grams

  28. The newly discovered element Takalahium (symbol Tak; molecular mass = 411 g/mol) combines with oxygen to form Takalahium Oxide. The unbalanced equation is: Tak + O2 ---> Tak2O3 How many grams of Tak Oxide are formed when burning 8.00 moles of Tak? 2 mole Tak2O3 870grams Tak2O3 8.00 moles Tak X --------------- X ---------------------- 4 mole Tak 1 mole Tak2O3 3480 grams

  29. A piece of zinc metal weighing 2.00 grams is placed in an aqueous solution of silver nitrate. How many moles of silver can be recovered from this single replacement reaction? • 1 Zn + 2 AgNO3 1 Zn(NO3)2 + 2 Ag 1 mole Zn 2 moles Ag 2.00 grams Zn X --------------- X ---------------------- 65.39 grams Zn 1 mole Zn .061 moles

  30. Mass- Mole Problem The octane present in gasoline burns according to the following equation:                     2 C8H18 + 25 O2 ---------> 16 CO2 + 18 H2OHow many moles of CO2 are produced when 30 grams of octane react with an excess of oxygen? 1 mole C8H18 16 moles CO2 30 grams C8H18 X --------------- X ---------------------- 114 grams C8H18 2 mole C8H18 2.10 moles

  31. The formation of water from hydrogen and oxygen gas is: 2H2 (g) + O2 (g) ---> 2H2O (l). What mass of water will form from 12.0 grams of hydrogen and excess oxygen? 1 mole H2 2 mole H2O 18 grams H2O 12.0 grams H2 X --------------- X ------------------ X ---------------- 2 grams H2 2 moles H2 1 mole H2O 108 grams

  32. In the reaction: KClO3 + 5KCl + 6HNO3 --> 6KNO3 + 3Cl2 + 3H2O,how many grams of KClO3 are required to prepare 10.0 g of Cl2? 1 mole Cl2 1 mole KClO3 122.55 grams KClO3 10 grams Cl2 X --------------- X ------------------ X ----------------------- 70.9 grams Cl2 3 moles Cl2 1 mole KClO3 5.76 grams

  33. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2 2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 = ? g Al2O3 4 mol Al 1 mol Al2O3 26.98 g Al 6.50 x 2 x 101.96÷ 26.98 ÷ 4 = 12.3 g Al2O3

  34. Limiting Reactant and Percent Yield

  35. Limiting Reactant Video The limiting reactantis the reactant that is consumed first,limiting the amounts of products formed.

  36. Analogy • You need to bake some cakes for the school bake sale. The cake recipe calls for 2 cups of flour and 4 eggs. If you have 4 cups of flour and a dozen of eggs, how many cakes can you make? 2 cups flour 4 eggs 4 cups of flour X = 2 cakes 2 cups flour = 1 cake 4 eggs Which ingredient ran out first (limiting)?

  37. Silicon dioxide is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. • SiO2 + 4 HF  SiF4 + 2 H2O • If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant? 4.5 mol SiO2 2.0 mol HF 1 mol SiO2 4 mol HF Look at the balanced equation and mole-to-mole ratio

  38. How do you know which one to use? If 12.0 g CH4 and 24.0 g O2 react to form CO2 and H2O, how many grams of CO2 will be formed? • 1 CH4 + 2 O2 1 CO2 + 2 H2O 1 mole CH4 2 mole O2 32 grams O2 12 grams CH4 X --------------- X ------------------ X ----------------------- 16 grams CH4 1 mole CH4 1 mole O2 Need: 48 grams- O2 Limited: O2 will run out first 1 mole O2 1 mole CO2 44 grams CO2 24 grams O2 X --------------- X --------------- X ------------------- 32 grams O2 2 moles O2 1 mole CO2 16.5 grams CO2

  39. Practice Methane (CH4) reacts with O2 to form CO2 and water. If 30.0 grams of CH4 and 64.0 grams of O2 are mixed and the reaction goes to completion, how many grams of the reactant "in excess" remain? • 1 CH4 + 2 O2 1 CO2 + 2 H2O 1 mole O2 1 mole CH4 16 grams CH4 64 grams O2 X --------------- X ------------------ X -------------------- 32 grams O2 2 mole O2 1 mole CH4 Excess Remaining = 30.0 grams – 16 grams = Will Use 16 grams CH4 14 grams

  40. Now It's Your Turn

  41. Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, H2O2, as the propellant. The reaction is given by the following equation. N2H4 + 2 H2O2  N2 + 4 H2O a. Which is the limiting reactant in this reaction when 0.750 mol of N2H4 is mixed with 0.500 mol of H2O2? b. How much of the excess reactant, in moles, remains unchanged? c. How much of each product, in moles, is formed?

  42. The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. This substance can also be made in the laboratory by the reaction between red-hot iron and steam according to the following equation. • 3 Fe + 4 H2O  Fe3O4 + 4 H2 • When 36.0 g of H2O is mixed with 167 g of Fe, which is the limiting reactant? • b. What mass in grams of black iron oxide is produced? • c. What mass in grams of excess reactant remains when the reaction is completed?

  43. In the reaction PCl5 --> PCl3 + Cl2, 104.3 g of PCl5 are used, and 30.0 g of Cl2 are produced. What is the % yield of Cl2 in this reaction? % Yield = Actual Yield (given in problem) X 100 Theoretical Yield (find with stoichiometry) 104.3 grams PCl5 X 1 mole PCl5 X 1 mole Cl2 X 70.9 grams Cl2 208.25 grams PCl5 1 mole PCl5 1 mole Cl2 Ans: 35.5 Theoretical Yield= 35.5 grams Actual Yield = 30.0 grams Percent Yield = 30.0 grams Cl2 X 100 35.5 grams Cl2 Ans: 84.5 %

  44. How many grams of CaF2 is formed when 35.8mL of 0.678 M NaF is treated with excess of Ca(NO3)2(aq)? NaF + Ca(NO3)2  CaF2 + Na2(NO3)2 M = 0.678 35.8 mL X __1 L_ X 0.678 mol X 1 mol CaF2X 78g CaF2 1000mL 1L 2 mol NaF 1 mol 0.947 g CaF2

  45. Determine which compound is limiting The reaction needs 14.4g of AgNO3 But we only have 6.00g. The reaction will stop when AgNO3 is gone Now that you know which is the limiting reagent we can use it’s mass to solve for how many grams of Ag2CO3

  46. The End! The End! The End! The End! The End! The End! The End!

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