2. Chemical Bonding

3. Indicators of chemical reactions Formation of a gas Emission of light or heat Formation of a precipitate Color change Emission of odor

4. All chemical reactions: have two parts Reactants - the substances you start with Products- the substances you end up with The reactants turn into the products. Reactants  Products

5. Symbols used in equations (s) after the formula –solid Cu (s) (g) after the formula –gas H 2 (g) (l) after the formula -liquid H 2 O (l) (aq) after the formula - dissolved in water, an aqueous solution. CaCl 2 (aq)

6. Summary of Symbols

7. What is a catalyst? A substance that speeds up a reaction without being changed by the reaction. Enzymes are biological or protein catalysts.

8. All chemical reactions are accompanied by a change in energy. Exothermic - reactions that release energy to their surroundings (usually in the form of heat) oΔH (enthalpy) is negative – energy leaving system Endothermic - reactions that need to absorb heat from their surroundings to proceed. oΔH (enthalpy) is positive – energy coming into the system Reaction Energy

9. Diatomic elements There are 8 elements that never want to be alone. They form diatomic molecules. H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2, and At 2 The –ogens and the –ines 1 + 7 pattern on the periodic table

11. Convert this to an equation Fe 2 S 3 (s) + HCl (g)  FeCl 2 (s) + H 2 S (g)

12. HNO 3 (aq) + Na 2 CO 3 (s)  NaNO 3 (aq) + H 2 O (l) Convert this to an equation

13. The other way Fe (s) + O 2 (g)  Fe 2 O 3 (s) Solid iron reacts with oxygen gas to form solid iron oxide (rust).

14. A silver spoon tarnishes. The solid silver reacts with sulfur in the air to make solid silver sulfide, the black material we call tarnish. Ag (s) + H 2 S (g) + O 2 (g)  Ag 2 S (s) + H 2 O

15. Aluminum metal reacts with liquid bromine to form solid aluminum bromide Translate Equation ___ Al(s) + ___ Br 2 (l) →___ AlBr 3 (s) 2 32

16. Synthesis Reactions Also called combination reactions 2 elements, or compounds combine to make one compound. A + B  AB Na (s) + Cl 2 (g)  NaCl (s) Ca (s) +O 2 (g)  CaO (s) We can predict the products if they are two elements. Mg (s) + N 2 (g)  Mg 3 N 2 (s)

17. A simulation of the reaction: 2H 2 + O 2  2H 2 O

18. Decomposition Reactions decompose = fall apart one compound (reactant) falls apart into two or more elements or compounds. Usually requires energy AB  A + B NaCl Na + Cl 2 CaCO 3 CaO + CO 2

20. Can predict the products if it is a binary compound Made up of only two elements Falls apart into its elements H 2 O HgO Decomposition Reactions H 2 (g) + O 2

21. Single Replacement Also referred to as single displacement One element replaces another Reactants must be an element and a compound. Products will be a different element and a different compound. A + BC  AC + B 2Na + SrCl 2  Sr + 2NaCl

22. Double Replacement Two things replace each other. Reactants must be two ionic compounds or acids. Usually in aqueous solution AB + CD  AD + CB AgNO 3 + NaCl  AgCl + NaNO 3 ZnS + 2HCl  ZnCl + H 2 S

23. Combustion A reaction in which a compound (often carbon) reacts with oxygen CH 4 + O 2  CO 2 + H 2 O C 3 H 8 + O 2  CO 2 + H 2 O C 6 H 12 O 6 + O 2  CO 2 + H 2 O

24. The charcoal used in a grill is basically carbon. The carbon reacts with oxygen to yield carbon dioxide. The chemical equation for this reaction is C + O 2  CO 2

25. 24 STOICHIOMETRY The measurement system in chemistry

26. 25 (1-2-3) General Approach For Problem Solving: 1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?) WHAT IS ASKED 2. Determine what is given and the UNITS. WHAT IS GIVEN!!! 3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

27. 26 Sample problem for general problem solving. Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race. 5280 ft = 1 mile; 12 inches = 1 ft 1. What is the goal and what units are needed? Goal = ______ inches 2. What is given and its units? 10 miles 3. Convert using factors (ratios). 10 miles = inches633600 Units match Goal Menu

28. 27 Converting grams to moles. Determine how many moles there are in 5.17 grams of Fe(C 5 H 5 ) 2. Goal = moles Fe(C 5 H 5 ) 2 Given 5.17 g Fe(C 5 H 5 ) 2 Use the molar mass to convert grams to moles. Fe(C 5 H 5 ) 2 2 x 5 x 1.001 = 10.01 2 x 5 x 12.011 = 120.11 1 x 55.85 = 55.85 0.0278 units match

29. 28 Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 1 1 2 moles of HCl react with 1 mole of Ba(OH) 2 to form 2 moles of H 2 O and 1 mole of BaCl 2 coefficients give MOLAR RATIOS

30. 29 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

31. 30 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

32. 31 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Gram to Gram Conversions

33. 32 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g)2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

34. 33 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g)2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

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36. 35 Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) When working problems, it is a good idea to change M into its units.

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38. 37 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution. Solutions

39. 38 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1 st : = mol L 3.73 g 200.0 x 10 - 3 L 0.140 2 nd : M 1 V 1 = M 2 V 2 (0.140 M)(10.0 mL) = (? M)(100.0 mL) 0.0140 M = M 2 molar mass of AlCl 3 dilution formula final concentration Solutions

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41. 40 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry

42. 41 50.0 mL 6.0 M ? g Look! A conversion factor! 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal

43. 42 50.0 mL 6.0 M ? g 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal = g NaHCO 3 H 2 SO 4 50.0 mL 1 mol H 2 SO 4 NaHCO 3 2 mol NaHCO 3 84.0 g mol NaHCO 3 50.4

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45. 44 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. First write a balanced Equation. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1

46. 45 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Our Goal

47. 46 Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now let’s get to work converting. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL = mL NaOH H 2 SO 4 35.0 mL H 2 SO 4 0.125 mol 1000 mL H 2 SO 4 NaOH 2 mol 1 mol H 2 SO 4 1000 mL NaOH 0.102 mol NaOH 85.8 Units Match Solution Stoichiometry: shortcut

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49. 48 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 1st write out a balanced chemical equation Solution Stoichiometry

50. 49 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 2HCl(aq) + Ba(OH) 2 (aq)  2H 2 O(l) + BaCl 2 0.40 M 47.1 mL 0.75 M ? mL = mL HCl Ba(OH) 2 47.1 mL 1 mol Ba(OH) 2 HCl 2 mol 0.40 mol HCl HCl 1000 mL 176 Units match Solution Stoichiometry

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53. 52 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? First write a balanced chemical reaction. ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L

54. 53 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L = mol Ba(OH) 2 L Ba(OH) 2 25.00 x 10 -3 L Ba(OH) 2 Units Already Match on Bottom! 0.0629 Units match on top!

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56. 55 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3. Determine the molarity of the Ca(OH) 2 solution. We must first write a balanced equation. Solution Stochiometry Problem:

57. 56 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3. Determine the molarity of the Ca(OH) 2 solution. Ca(OH) 2 (aq) + HNO 3 (aq)  H 2 O(l) + Ca(NO 3 ) 2 (aq) 2 2 48.0 mL19.2 mL 0.385 M = mol (Ca(OH) 2) L (Ca(OH) 2 ) 19.2 mL HNO 3 48.0 x 10 -3 L ? M units match! 0.0770 Solution Stochiometry Problem:

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59. 58 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) First copy down the the BALANCED equation! Now place numerical the information below the compounds.

60. 59 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Hide one

61. 60 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Hide Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125

62. 61 Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Hide Based on: H 2 O = mol O 2 0.10 mol H2OH2O 0.150 Limiting/Excess/ Reactant and Theoretical Yield Problems :

63. 62 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Based on: H 2 O = mol O 2 0.10 mol H 2 O 0.150 What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant? It was limited by the amount of KO 2. H 2 O = excess (XS) reactant!

64. 63 Theoretical yield vs. Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

65. 64 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Hide one Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Limiting/Excess Reactant Problem with % Yield

66. 65 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Based on: H 2 O = g O 2 Question if only 35.2 g of O 2 were recovered, what was the percent yield? Hide 47.0 g H 2 O 125.3 Limiting/Excess Reactant Problem with % Yield

67. 66 If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Based on: H 2 O = g O 2 47.0 g H 2 O 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H 2 O. 125.3 - 40.51 = 84.79 g of O 2 that could have been formed from the XS water. = g XS H 2 O 84.79 g O 2 31.83

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69. 68 Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO 3 ) 3 in 455 mL of solution. After you have worked the problem, click here to see setup answer Try this problem (then check your answer):

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