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Unit 6: Balancing Chemical Equations, Chemical Reactions & Stoichiometry

Unit 6: Balancing Chemical Equations, Chemical Reactions & Stoichiometry. The GowerHour. a A + b B c C + d D. Atoms. Atoms. A. B. Molecules. Molecules. Formula Units. Formula Units. Mol to Mol Ratio Using the Coefficients from the balanced chemical equation.

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Unit 6: Balancing Chemical Equations, Chemical Reactions & Stoichiometry

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  1. Unit 6: Balancing Chemical Equations, Chemical Reactions & Stoichiometry The GowerHour

  2. aA + bB cC + dD Atoms Atoms A B Molecules Molecules Formula Units Formula Units Mol to Mol Ratio Using the Coefficients from the balanced chemical equation VII. Stoichiometric “road map” (Use the balanced chemical equation) Mass B Mass A Molar mass Molar mass Mol A Mol B 6.022 x 1023 6.022 x 1023

  3. 10.1 atoms; particles; molecules; formula units D. Converting Number of Particles to Moles 1. One mole (mol) of a substance is 6.022  1023 representative particles of that substance and is the SI unit for measuring the amount of a substance. 2. The number of representative particles in a mole, 6.022  1023, is called Avogadro’s number.

  4. X 3. Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium is 1.25 x 1023 atoms of magnesium? 1 mol Mg atoms = 6.022 x 1023 atoms Mg 1 mol Mg atoms 1.25 x 1023 atoms Mg 6.022 x 1023 atoms Mg = 0.208 mol Mg = 2.08 x 10-1 mol Mg

  5. 10.1 x atoms; particles; molecules; formula units = 9.40 x 1023 molec. CH3OH E. Converting Moles to Number of Particles 1. 1.56 moles of methanol, CH3OH, contain how many molecules? 1 mol molec CH3OH = 6.022 x 1023 molec CH3OH 6.022 x 1023 molec CH3OH 1.56 mol CH3OH 1 mol CH3OH

  6. Calculating Molecular Mass Calculate the molecular mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mol

  7. x 2. Aluminum satellite dishes are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass of 9.45 mol of aluminum oxide? 1 mol Al2O3 = 101. 96 g Al2O3 101. 96 g Al2O3 9.45 mol Al2O3 = 964 g Al2O3 1 mol Al2O3

  8. x 4. When iron is exposed to air, it corrodes to form red-brown rust. Rust is iron(III) oxide (Fe2O3). How many moles of iron(III) oxide are contained in 92.2 g of pure Fe2O3? 1 mol Fe2O3 = 159.70 g Fe2O3 1 mol Fe2O3 92.2 g Fe2O3 = 0.577 mol Fe2O3 159.70 g Fe2O3 = 5.77 x 10-1 mol Fe2O3

  9. Calculations molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!!

  10. mass IV. Stoichiometry = The _____ relationships among reactants and products in chemical reactions. A chemical equation is a statement of ___________ fact. On the left side of the reaction are the _________, and on the right side the _________ of the reaction. Since no atoms are created nor destroyed in a ___________ chemical reaction, the equation must be _________. This means that the combined weight of the reactants is exactly ______ to the combined weight of the products. In terms of chemical laws, this is the Law of _____________ of matter (__________). experimental reactants products non-nuclear balanced equal conservation Lavoisier

  11. V. Limiting and Excess Reagents A. In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. B. The _____________ is the reagent that determines the amount of product that can be formed by a reaction. C. In the reaction of nitrogen and hydrogen, hydrogen is the limiting reagent. Nitrogen is the reagent that is not completely used up in the reaction. The reagent that is not used up is called the ____________. limiting reagent excess reagent

  12. D. Analogy: How to make a cheese sandwich. 2 slices of bread + 1 slice of cheese → 1 cheese sandwich   If you have 8 slices of bread and 6 slices of cheese, how many sandwiches can you make? __ (theoretical yield)   What is the limiting reagent? ______   What is the excess reagent? _______   How much of the excess reagent is left at the end of the rxn? ______________ 4 bread cheese 2 slices cheese

  13. efficiency VI. Percent Yield A. The percent yield is a measure of the __________ of a reaction carried out in the laboratory. B. A batting average is actually a percent yield.

  14. theoretical C. The ____________ yield is the maximum amount of product that could be formed from given amounts of reactants. D. In contrast, the amount of product that actually forms when the reaction is carried out in the __________ is called the actual yield. E. The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. laboratory

  15. aA + bB cC + dD Atoms Atoms A B Molecules Molecules Formula Units Formula Units Mol to Mol Ratio Using the Coefficients from the balanced chemical equation VII. Stoichiometric “road map” (Use the balanced chemical equation) Mass B Mass A Molar mass Molar mass Mol A Mol B 6.022 x 1023 6.022 x 1023

  16. Ca + O2 CaO x x x x ER used Theoretical Yield = 112 g CaO x = 0.998 mol O2 Examples: A. Calcium reacts with oxygen to form calcium oxide. If you are given 80.0 g of calcium, (1) Calculate the theoretical yield of calcium oxide. (2) How many moles of your excess reagent (ER) are used? 2 2 LR ER 1 mol Ca 2 mol CaO 56.08 g CaO (1) 80.0 g Ca 40.08 g Ca 2 mol Ca 1 mol CaO LR ER 1 mol Ca 1 mol O2 (2) 80.0 g Ca 40.08 g Ca 2 mol Ca LR

  17. N2 + O2 N2O5 x x x x ER used Theoretical Yield = 432 g N2O5 x = 9.99 mol O2 (112 g) 2 5 2 B. Nitrogen and oxygen react to form dinitrogen pentoxide. If 112 g of nitrogen react with unlimited oxygen, (1) how many moles of the excess reagent (ER) react and (2) how many grams of dinitrogen pentoxide (theoretical yield) are formed? LR ER 1 mol N2 5 mol O2 (1) 112 g N2 28.02 g N2 2 mol N2 LR 1 mol N2 2 mol N2O5 108.02 g N2O5 (2) 112 g N2 28.02 g 2 mol N2 1 mol N2O5 LR

  18. NaCl (aq) + AgNO3 (aq) AgCl (s) + NaNO3 (aq) x x x Theoretical Yield = 84.37 g AgCl C. Aqueous sodium chloride and aqueous silver nitrate react forming solid silver chloride (a precipitate) and aqueous sodium nitrate. If one solution contains 100.0 g ofsilver nitrate and it reacts with unlimited sodium chloride, how many grams of silver chloride (Theoretical Yield) can be made? ER LR LR 1 mol AgNO3 1 mol AgCl 143.32 g 100.0 g AgNO3 169.88 g 1 mol AgNO3 1 mol AgCl

  19. x x x x x x = 274.2 g CO2 Theoretical Yield = 3.752 x 1024 molec CO2 2 2 LR CH4 + O2 CO2 + H2O ER D. How many (1) grams (theoretical yield) and (2) molecules of carbon dioxide gas are produced when 100.0 g of methane gas (CH4) are burned. Methane reacts with oxygen to form carbon dioxide and water. 1 mol CH4 1 mol CO2 44.01 g CO2 (1) 100.0 g CH4 16.05 g 1 mol CH4 1 mol CO2 LR 1 mol CH4 1 mol CO2 6.022 x 1023 molec CO2 (2) 100.0 g CH4 16.05 g 1 mol CH4 1 mol CO2 LR

  20. = 80.7 g H2 = 13.3 mol N2 x x x x x x x 3 2 N2 + H2 NH3 E. Nitrogen reacts with hydrogen gas forming ammonia (nitrogen trihydride). How many (1) grams of hydrogen and (2) moles of nitrogen are required to form 1.00 pound of ammonia? 453.6 g NH3 1 mol NH3 3 mol H2 2.02 g H2 (1) 1.00 lb NH3 1 lb 17.04 g 2 mol NH3 1 mol H2 453.6 g NH3 1 mol NH3 1 mol N2 (2) 1.00 lb NH3 1 lb 17.04 g 2 mol NH3

  21. x x x = 135 g N2 1 mol N2H4 1 mol N2 28.02 g N2 155 g N2H4 32.06 g 1 mol N2H4 1 mol N2 Review Question: Nitrogen and hydrogen gas combine to form hydrazine (dinitrogen tetrahydride) which is used for rocket fuel. How many grams of nitrogen gas are needed to form 155 g of hydrazine? N2 + 2 H2 N2H4

  22. Theoretical yield = 168 g H2O2 = 133 g H2O2 L.R. = O2 x x x x x x H2 + O2 H2O2 F. How much hydrogen peroxide can be formed (theoretical yield) from 10.0 g of hydrogen and 125 g of oxygen? What is the limiting reagent? 1 mol H2 1 mol H2O2 34.02 g H2O2 10.0 g H2 2.02 g 1 mol H2 1 mol H2O2 1 mol O2 1 mol H2O2 34.02 g H2O2 125 g O2 32.00 g 1 mol O2 1 mol H2O2

  23. = 89.2 g H2O 100.0 g O279.2 g O2 = 20.8 g O2 remaining = 79.2 g O2 = 112.6 g H2O Theoretical yield L.R. = H2 ER used x x x x x x x x x ER 2 2 H2 + O2 H2O G. Determine the theoretical yield of water that can be formed from 10.0 g of hydrogen and 100.0 g of oxygen. What is the limiting reagent? How much excess reagent is remaining? 1 mol H2 1 mol H2O 18.02 g H2O 10.0 g H2 2.02 g 1 mol H2 1 mol H2O 1 mol O2 2 mol H2O 18.02 g H2O 100.0 g O2 32.00 g 1 mol O2 1 mol H2O 1 mol H2 1 mol O2 32.00 g O2 10.0 g H2 2.02 g 2 mol H2 1 mol O2

  24. 100.0 g O2 - 88.95 g O2 = 11.05 g O2 Remaining L.R. = Al = 212.4 g Al2O3 = 88.95 g O2 Theoretical yield = 189.0 g Al2O3 x x x x x x x x x 4 3 2 Al + O2 Al2O3 H. How many grams of aluminum oxide can be formed (theoretical yield) from 100.0 g of aluminum and 100.0 g of oxygen gas? How much excess reagent is remaining? 1 mol Al 2 mol Al2O3 101.96 g 100.0 g Al 26.98 g 4 mol Al 1 mol Al2O3 1 mol O2 2 mol Al2O3 101.96 g 100.0 g O2 32.00 g 3 mol O2 1 mol Al2O3 1 mol Al 3 mol O2 32.00 g O2 100.0 g Al 26.98 g 4 mol Al 1 mol O2

  25. (4) Ea = 2.55 g – 2.17 g = 0.38 g I. A student does a lab in which she makes 2.17 g of Lead (II) nitrate. From her limiting reagent she calculated that she should have made 2.55 g. Determine: (1) theoretical yield, (2) actual yield, (3) % yield(4) absolute and relative error. (1) 2.55 g Pb(NO3)2 (2) 2.17 g Pb(NO3)2 Note: % yield + Er = 100 %

  26. = 112.4 g H2O = 105.1 g H2O % error = 100% - 92.5% = 7.5% Theo. yield LR x x x x x x 3 2 2 N2H4 + O2 H2O + NO2 J. A sample of 100.0 g of hydrazine (N2H4) burns in 280.0 g of oxygen gas and 97.2 g of water are formed. Calculate the % yield of water. (Nitrogen dioxide is also produced.) Calculate the % error (relative error). 1 mol N2H4 2 mol H2O 18.02 g 100.0 g N2H4 32.06 g 1 mol N2H4 1 mol H2O 1 mol O2 2 mol H2O 18.02 g 280.0 g O2 32.00 g 3 mol O2 1 mol H2O

  27. 1 mol AgNO3 2 mol AgCl 143.32 g 1.70 g AgNO3 169.88 g 2 mol AgNO3 1 mol AgCl = 1.43 g AgCl = 2.15 g AgCl 1 mol CaCl2 2 mol AgCl 143.32 g 0.832 g CaCl2 110.98 g 1 mol CaCl2 1 mol AgCl Theo. yield x x x x x x 2 AgNO3 + CaCl2 Ca(NO3)2 + 2 AgCl (s) K. In lab a student mixes a solution that contains 1.70 g silver nitrate and a solution that contains 0.832 g calcium chloride. Aqueous calcium nitrate and solid silver chloride are produced. After washing, filtering, and drying the precipitate, he finds that he has obtained 1.25 g of silver chloride. What is the % yield?

  28. IQ1 4 Fe + 3 O2→ 2 Fe2O3 • Determine the mass of Iron III oxide formed when 25.1 g of iron reacts with oxygen. Balanced equations: ______________________________ 2. What mass of oxygen is needed to react with 6.8 x 1023 atoms of iron? (Use the above equations). 25.1 g Fe 2 mol Fe2O3 159.70 g Fe2O3 1 mol Fe 55.85 g Fe 4 mol Fe 1 mol Fe2O3 = 35.9 g Fe2O3 6.8 x 1023 atoms Fe 1 mol Fe 3 mol O2 32.00 g O2 6.022 x 1023 atoms Fe 4 mol Fe 1 mol O2 = 27 g O2

  29. IQ 2 3 2 3 Fe2O3 + CO → Fe + CO2 Determine the (a) limiting reagent and (b) theoretical yield of iron when 82.3 g of iron (III) oxide reacts with 79.3 g of carbon monoxide. (c) Grams of excess reagent remaining. Balance equ.: __________________________________ (a) (b) • Grams excess reagent remaining? 3 mol CO 28.01 g CO 82.3 g Fe2O3 1 mol Fe2O3 159.70 g Fe2O3 1 mol Fe2O3 1 mol CO Fe2O3 – Limiting Reagent 43.3 g CO needed 82.3 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.85 g Fe 159.70 g Fe2O3 1 mol Fe2O3 1 mol Fe 57.6 g Fe Excess = grams @ start – grams reacted Excess = 79.3 g CO – 43.3 g CO = 36.0 g CO

  30. Stoichiometry WS I • a. 6.92 mol O2 b. 6.20 g KCl • a. 4.07g H2O b. 104 g CO2 • a. 0.026 mol FeCl3 b. 15.3 mol Cu

  31. Stoichiometry WS II • a. 6.92 mol O2 b. 6.20 g KCl • a. 4.07g H2O b. 104 g CO2 • a. 0.026 mol FeCl3 b. 15.3 mol Cu

  32. Sample Test Part A • (1) Syn (2) Decomp (3) DD (4) SD (5) Combustion • a. 1, 4, 3, 4 b. 1, 1, 1, 1 c. 6, 1, 2, 3 d. 2, 3, 1, 3 • a. 4, 3, 2 (Syn) b. 2, 2, 2, 1 (SD) c. 2, 7, 4, 6 (Comb) d. 1, 2, 1, 2 (DD) e. 2, 1, 1 (Decomp) • a. 1, 1, 2 AgBr, 1 CaSO4 b. 2, 1 H2, 1 Br2 c. 1, 5, 3 CO2, 4 H2O d. 2, 3, 1 Al2S3 e. 2, 1 N2, 3 H2

  33. Sample Test Part A • a. 2, 3, 1 Al2(SO4)3 Cu b. 1, 2, 1 Ca(OH)2, 1H2 c. 1, 2, 1 Fe(OH)2, 1 H2 d. 1, 2, 1 ZnCl2, 1 H2 e. NR f. 1, 2, 1 Cu(NO3)2, 2 Ag • HOFBrINCl • Speeds up rxn w/o being used up in the rxn. • C, CO, CO2, H2O • See Metal activities series.

  34. Sample Test Part B • a. Fe b. O2 • a. 1.87 x 1023 molec b. 138 g NaCl c. 6.33 x 1023 molec Cl2 • a. AgNO3 b. 69.9 g Ag c. 89 g Cu • 74.3 % • When LR runs out, the rxn stops. • 145 g NH3 • F, A , E, C , D, B

  35. Unit 6 HW Packet • Schedule • Unit 6 Study Guide • Stoichiometry WS I • Stoichiometry WS II • Limiting Reagent & Percent Yield WS • Theoretical Yield/Percent Yield WS • Reactions WS I • Reactions WS II • Reactions WS III • Stoichiometry Sample Test

  36. Chemistry WS • a. 6.92 mol O2 b. 6.20 g KCl • a. 4.07g H2O b. 104 g CO2 • a. 0.026 mol FeCl3 b. 15.3 mol Cu

  37. Stoichiometry WS • 1.81 x 1023 molecules H2 • 7.3 g H2 • 3.09 mol Al2(SO4)3 • 22 g Al • 6.7 x 1022 atoms Al • 8.1 g H2

  38. Limiting Reagent & Percent Yield WS • a. L.R. = HCl b. L.R. = O2 c. L.R. = Cl2 • a. L.R. = HCl; 0.26 mol H2 produced b. 0.26 mol Zn used; 0.04 mol Zn excess • a. L.R. = HCl; 0.12 mol AlCl3 produced b. 0.12 mol Al used; 0.03 mol Al excess

  39. Limiting Reagent & Percent Yield • 87.9 % • 75.2 % Theoretical Yield/Percent Yield • 54.2 % • Actual yield = 1660 g NH3 • 37.0 % • Actual yield = 58.7 g MgO

  40. Reaction Types • DD: Al2(SO4)3 + Ca3(PO4)2 2 AlPO4 + 3 CaSO4 • DD: MgCl2 + 2 AgNO3 Mg(NO3)2 + 2 AgCl • Syn: 2 H2 + O2  2 H2O • SD: Zn + Cu(NO3)2  Zn(NO3)2 + Cu • DD: CuO + H2SO4  CuSO4 + H2O • SD: 2 NaI + Cl2  2 NaCl + I2 • Decomp: Cu(OH)2  CuO + H2O 12. SD: K + H2O  KOH + H2

  41. Sample Test Part A • (1) Syn (2) Decomp (3) DD (4) SD (5) Combustion • a. 1, 4, 3, 4 b. 1, 1, 1, 1 c. 6, 1, 2, 3 d. 2, 3, 1, 3 • a. 4, 3, 2 (Syn) b. 2, 2, 2, 1 (SD) c. 2, 7, 4, 6 (Comb) d. 1, 2, 1, 2 (DD) e. 2, 1, 1 (Decomp) • a. 1, 1, 2 AgBr, 1 CaSO4 b. 2, 1 H2, 1 Br2 c. 1, 5, 3 CO2, 4 H2O d. 2, 3, 1 Al2S3 e. 2, 1 N2, 3 H2

  42. Sample Test Part A • a. 2, 3, 1 Al2(SO4)3 Cu b. 1, 2, 1 Ca(OH)2, 1H2 c. 1, 2, 1 Fe(OH)2, 1 H2 d. 1, 2, 1 ZnCl2, 1 H2 e. NR f. 1, 2, 1 Cu(NO3)2, 2 Ag • HOFBrINCl • Speeds up rxn w/o being used up in the rxn. • C, CO, CO2, H2O • See Metal activities series.

  43. Sample Test Part B • a. Fe b. O2 • a. 1.87 x 1023 molec b. 138 g NaCl c. 6.33 x 1023 molec Cl2 • a. AgNO3 b. 69.9 g Ag c. 89 g Cu • 74.3 % • When LR runs out, the rxn stops. • 145 g NH3 • F, A , E, C , D, B

  44. Stoichiometry Review WS • a. N2O5 + H2O  2 HNO3 b.111.8 g HNO3 c. 15.7 g H2O 2. a. 4 Al + 3 PbO2 2 Al2O3 + 3 Pb b. 2.113 g Al 3. a. LR = H2S b. 8.257 g (NH4)2S c. 2.709 g NH3 4. a. Yes; SD; Cl2 + 2KBr  2 KCl + Br2 b. LR = Cl2 c. 1.93 g Br2 d. 0.335 g KBr 5. a. LR = As2O3 b. 79.3 % 6. a. 6, 6, 1, 6 b. 6.32 x 103 g O2 7. a. 71 % b. 26 metric tons

  45. Unit 6 HW Packet • Schedule • Unit 6 Study Guide • Stoichiometry WS I • Stoichiometry WS II • Limiting Reagent & Percent Yield WS • Theoretical Yield/Percent Yield WS • Reactions WS I • Reactions WS II • Reactions WS III • Stoichiometry Sample Test

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