1. Balancing Equations Balancing, Writing, and Naming Equations

2. 2 Mass:proton = 1.00728 amu neutron = 1.0086 amu electron = 0.0005486 12 C atom = 12.00000 amu 13 C atom = 13.00335 amu amu = Atomic Mass Units amu = Atomic Mass Units Atomic and Molecular Mass

3. 3 The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes. The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes. Mass of C = average of 12 C and 13 C Mass of C = average of 12 C and 13 C = 0.9889 x 12 amu + 0.0111 x 13.0034 amu = 0.9889 x 12 amu + 0.0111 x 13.0034 amu = 12.011 amu = 12.011 amu

4. 4 Atomic and Molecular Mass The mass of a molecule is just the sum of the masses of the atoms making up the molecule. The mass of a molecule is just the sum of the masses of the atoms making up the molecule. m(C 2 H 4 O 2 ) = 2·m C + 4·m H + 2·m O m(C 2 H 4 O 2 ) = 2·m C + 4·m H + 2·m O = 2·(12.01) + 4·(1.01) + 2·(16.00) = 2·(12.01) + 4·(1.01) + 2·(16.00) = 60.06 amu = 60.06 amu

7. 7 Avogadro and the Mole One mole of a substance is the gram mass value equal to the amu mass of the substance. One mole of a substance is the gram mass value equal to the amu mass of the substance. One mole of any substance contains 6.02 x 10 23 units of that substance. One mole of any substance contains 6.02 x 10 23 units of that substance. Avogadro’s Number (N A, 6.022 x 10 23 ) is the numerical value assigned to the unit, 1 mole. Avogadro’s Number (N A, 6.022 x 10 23 ) is the numerical value assigned to the unit, 1 mole.

8. 8 Avogadro and the Mole

9. 9 Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H) Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

10. 10 Avogadro and the Mole The Mole: Allows us to make comparisons between substances that have different masses. The Mole: Allows us to make comparisons between substances that have different masses.

11. LecturePLUS Timberlake 9911 Writing Mole Factors 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mole Fe and 3 mole O 2 3 mole O 2 4 mole Fe 3 mole O 2 4 mole Fe Fe and Fe 2 O 3 4 mole Feand2 mole Fe 2 O 3 2 mole Fe 2 O 3 4 mole Fe 2 mole Fe 2 O 3 4 mole Fe O 2 and Fe 2 O 3 3 mole O 2 and 2 mole Fe 2 O 3 2 mole Fe 2 O 3 3 mole O 2 2 mole Fe 2 O 3 3 mole O 2

12. LecturePLUS Timberlake 9912 Learning Check S1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole factor for H 2 and N 2 is 1) 3 mole N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 mole H 2 2 mole H 2 1 mole H 2 3 mole H 2 2 mole H 2 B. A mole factor for NH 3 and H 2 is 1) 1 mole H 2 2) 2 mole NH 3 3) 3 mole N 2 2 mole NH 3 3 mole H 2 2 mole NH 3 2 mole NH 3 3 mole H 2 2 mole NH 3

13. LecturePLUS Timberlake 9913 Solution S1 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole factor for H 2 and N 2 is 2) 1 mole N 2 3 mole H 2 3 mole H 2 B. A mole factor for NH 3 and H 2 is 2) 2 mole NH 3 3 mole H 2 3 mole H 2

14. LecturePLUS Timberlake 9914 Chemical Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? 6.0 mole O 2 x 2 mole Fe 2 O 3 = 4.0 mole Fe 2 O 3 3 mole O 2 3 mole O 2

15. LecturePLUS Timberlake 9915 Learning Check S2 4 Fe + 3 O 2 2 Fe 2 O 3 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 mole of O 2 ? 1) 3.00 mole Fe 2) 9.00 mole Fe 3) 16.0 mole Fe

16. LecturePLUS Timberlake 9916 Solution S2 4 Fe + 3 O 2 2 Fe 2 O 3 12.0 mole O 2 x 4 mole Fe = 16.0 mole Fe 3 mole O 2 3 mole O 2

17. LecturePLUS Timberlake 9917 Learning Check S 3 4 Fe + 3 O 2 2 Fe 2 O 3 4 Fe + 3 O 2 2 Fe 2 O 3 How many grams of O 2 are needed to produce 0.400 mole of Fe 2 O 3 ? 1) 38.4 g O 2 2) 19.2 g O 2 3) 1.90 g O 2

18. LecturePLUS Timberlake 9918 Solution S 3 2) 19.2 g O 2 0.400 mole Fe 2 O 3 x 3 mole O 2 x 32.0 g O 2 2 mole Fe 2 O 3 1 mole O 2 = 19.2 g O 2

19. LecturePLUS Timberlake 9919 Mass of A Reaction The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g)H 2 O (g) Balance the equation 2 H 2 (g) + O 2 (g)2 H 2 O (g) 2 H 2 (g) + O 2 (g)2 H 2 O (g)

20. LecturePLUS Timberlake 9920 Organize data mole bridge mole bridge 2 H 2 (g) + O 2 (g)2 H 2 O (g) 2 H 2 (g) + O 2 (g)2 H 2 O (g) ? g 13.1 g ? g 13.1 g Plan g H 2 O mole H 2 O mole O 2 O 2 Setup 13.1 g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O 2 18.0 g H 2 O2 mole H 2 O 1 mole O 2 = 11.6 g O 2 = 11.6 g O 2

21. LecturePLUS Timberlake 9921 Learning Check S 4 How many grams of O 2 are need to react 50.0 grams of Na in the reaction 4 Na + O 2 2 Na 2 O Complete the set up: 50.0 g Na x 1 mole Na x ________ x _______ 50.0 g Na x 1 mole Na x ________ x _______ 23.0 g Na 23.0 g Na

22. LecturePLUS Timberlake 9922 Solution S 4 4 Na + O 2 2 Na 2 O 50.0 g Na x 1 mole Na x 1 mole O 2 x 32.0 g 23.0 g Na 4 mole Na 1 mole O 2 23.0 g Na 4 mole Na 1 mole O 2 = g O 2

23. LecturePLUS Timberlake 9923 Learning Check S5 Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2 ? 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O

24. LecturePLUS Timberlake 9924 Solution S5 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 75.0 g CO 2 x 1 mole CO 2 x 2 mole C 2 H 2 x 26.0 g C 2 H 2 44.0 g CO 2 4 mole CO 2 1 mole C 2 H 2 = 22.2 g C 2 H 2

25. 25 Balancing Chemical Equations A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved. A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved. reactants  products limestone  quicklime + gas Calcium carbonate  calcium oxide + carbon dioxide CaCO 3 (s)  CaO(s) + CO 2 (g)

26. 26 Balancing Chemical Equations CaCO 3 (s)  CaO(s) + CO 2 (g) The letters in parentheses following each substance are called State Symbols (g) → gas(l) → liquid(s) → solid(aq) → aqueous

27. 27 Balancing Chemical Equations A balanced equation MUST have the same number of atoms of each element on both sides of the equation. H 2 + O 2 → H 2 ONot Balanced H 2 + ½O 2 → H 2 OBalanced 2H 2 + O 2 → 2H 2 OBalanced

28. 28 Balancing Chemical Equations The numbers multiplying chemical formulas in a chemical equation are called: Stoichiometric Coefficients (S.C.) 2H 2 + O 2 → 2H 2 OBalanced Here 2, 1, and 2 are stoichiometric coefficients.

29. 29 Balancing Chemical Equations Hints for Balancing Chemical Equations: 1) Save single element molecules for last. 2) Try not to change the S.C. of a molecule containing an element that is already balanced. 3) If possible, begin with the most complex molecule that has no elements balanced.

30. 30 Balancing Chemical Equations Example 1:CH 4 + O 2 → CO 2 + H 2 O Balance O 2 last C is already balanced Start by changing S.C. of H 2 O to balance H CH 4 + O 2 → CO 2 + 2H 2 O

31. 31 Balancing Chemical Equations Example 1: CH 4 + O 2 → CO 2 + 2H 2 O Now C and H are balanced Balance O by changing the S.C. of O 2 CH 4 + 2O 2 → CO 2 + 2H 2 O BALANCED!

32. 32 Balancing Chemical Equations Example 2: B 2 H 6 + O 2 → B 2 O 3 + H 2 O Balance O last B is already balanced Start by changing S.C. of H 2 O: B 2 H 6 + O 2 → B 2 O 3 + 3H 2 O

33. 33 Balancing Chemical Equations Example 2: B 2 H 6 + O 2 → B 2 O 3 + 3H 2 O B and H are balanced Balance O by changing S.C. of O 2 B 2 H 6 + 3O 2 → B 2 O 3 + 3H 2 O BALANCED!

34. 34 Balancing Chemical Equations Example 3:MnO 2 + KOH + O 2 → K 2 MnO 4 + H 2 O Balance O last Mn is already balanced Change S.C. of KOH to balance K MnO 2 + 2KOH + O 2 → K 2 MnO 4 + H 2 O

35. 35 Balancing Chemical Equations Example 3:MnO 2 + 2KOH + O 2 → K 2 MnO 4 + H 2 O Mn, K, and H are balanced (H was balanced by chance) Balance O MnO 2 + 2KOH + ½O 2 → K 2 MnO 4 + H 2 O or 2MnO 2 + 4KOH + O 2 → 2K 2 MnO 4 + 2H 2 O 2MnO 2 + 4KOH + O 2 → 2K 2 MnO 4 + 2H 2 O

36. 36 Balancing Chemical Equations Example 4:NaNO 2 + H 2 SO 4 → NO + HNO 3 + H 2 O + Na 2 SO 4 Hard one (no single element molecules) S is balanced Start with NaNO 2 to balance Na 2NaNO 2 + H 2 SO 4 → NO + HNO 3 + H 2 O + Na 2 SO 4

37. 37 Balancing Chemical Equations Example 4:2NaNO 2 + H 2 SO 4 → NO + HNO 3 + H 2 O + Na 2 SO 4 S, Na, and N are balanced Cannot balance H without changing S.C. for H 2 SO 4 ! Boo!Option 1: trial and error Option 2: Go on to next problem!

38. 38 Balancing Chemical Equations Balance the following equations: Balance the following equations: C 6 H 12 O 6 → C 2 H 6 O + CO 2 C 6 H 12 O 6 → C 2 H 6 O + CO 2 Fe + O 2 → Fe 2 O 3 Fe + O 2 → Fe 2 O 3 NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O

39. 39 Balancing Chemical Equations Balance the following equations: Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 Fe + O 2 → Fe 2 O 3 Fe + O 2 → Fe 2 O 3 NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O

40. 40 Balancing Chemical Equations Balance the following equations: Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 4Fe + 3O 2 → 2Fe 2 O 3 (balance O first) 4Fe + 3O 2 → 2Fe 2 O 3 (balance O first) NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O

41. 41 Balancing Chemical Equations Balance the following equations: Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 4Fe + 3O 2 → 2Fe 2 O 3 (balance O first) 4Fe + 3O 2 → 2Fe 2 O 3 (balance O first) NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl N:H is 1:3 on left, must get 1:3 on right! N:H is 1:3 on left, must get 1:3 on right!

42. 42 Balancing Chemical Equations NH 3 + Cl 2 → N 2 H 4 + NH 4 Cl N:H is 1:3 on left, must get 1:3 on right! N:H is 1:3 on left, must get 1:3 on right! 4NH 3 + Cl 2 → N 2 H 4 + 2NH 4 Cl

43. 43 Balancing Chemical Equations Balance the following equations: Balance the following equations: C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 C 6 H 12 O 6 → 2C 2 H 6 O + 2CO 2 4Fe + 3O 2 → 2Fe 2 O 3 4Fe + 3O 2 → 2Fe 2 O 3 4NH 3 + Cl 2 → N 2 H 4 + 2NH 4 Cl 4NH 3 + Cl 2 → N 2 H 4 + 2NH 4 Cl KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O (tough!) KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O (tough!)

44. 44 Balancing Chemical Equations Balance the following equations: Balance the following equations: KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O KClO 3 + C 12 H 22 O 11 → KCl + CO 2 + H 2 O balance C balance C KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + H 2 O KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + H 2 O balance H balance H KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + 11H 2 O KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + 11H 2 O balance O balance O 8KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + 11H 2 O 8KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + 11H 2 O

45. 45 Balancing Chemical Equations Balance the following equations: Balance the following equations: 8KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + 11H 2 O 8KClO 3 + C 12 H 22 O 11 → KCl + 12CO 2 + 11H 2 O balance K (and hope Cl is balanced) balance K (and hope Cl is balanced) 8KClO 3 + C 12 H 22 O 11 → 8KCl + 12CO 2 + 11H 2 O 8KClO 3 + C 12 H 22 O 11 → 8KCl + 12CO 2 + 11H 2 O Balanced! Balanced!

46. 46 Balancing Chemical Equations Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below: Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below:

47. 47 Stoichiometry Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities. Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

48. 48Stoichiometry Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl 2 : 2 NaOH(aq) + Cl 2 (g)  NaOCl(aq) + NaCl(aq) + H 2 O(l) 2 NaOH(aq) + Cl 2 (g)  NaOCl(aq) + NaCl(aq) + H 2 O(l) How many grams of NaOH are needed to react with 25.0 g of Cl 2 ?

49. 49 Stoichiometry 2 NaOH + Cl 2 → NaOCl + NaCl + H 2 O 25.0 g Cl 2 reacts with ? g NaOH

50. 50 Avogadro and the Mole Calculate the molar mass of the following: Calculate the molar mass of the following: Fe 2 O 3 (Rust) Fe 2 O 3 (Rust) C 6 H 8 O 7 (Citric acid) C 6 H 8 O 7 (Citric acid) C 16 H 18 N 2 O 4 (Penicillin G) C 16 H 18 N 2 O 4 (Penicillin G) Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe 2 O 3. Fe 2 O 3 (s) + CO(g) Fe(s) + CO 2 (g) Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe 2 O 3. Fe 2 O 3 (s) + CO(g) Fe(s) + CO 2 (g)

51. 51 Avogadro and the Mole Fe 2 O 3 + CO → Fe + CO 2 Balance (not a simple one) Save Fe for last C is balanced, but can’t balance O In the products the ratio C:O is 1:2 and can’t change Make the ratio C:O in reactants 1:2 Fe 2 O 3 + 3CO → 2Fe + 3CO 2

52. 52 Avogadro and the Mole Fe 2 O 3 + 3CO → 2Fe + 3CO 2

53. 53 Stoichiometry

54. 54 Stoichiometry Aspirin is prepared by reaction of salicylic acid (C 7 H 6 O 3 ) with acetic anhydride (C 4 H 6 O 3 ) to form aspirin (C 9 H 8 O 4 ) and acetic acid (CH 3 CO 2 H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by- product? Aspirin is prepared by reaction of salicylic acid (C 7 H 6 O 3 ) with acetic anhydride (C 4 H 6 O 3 ) to form aspirin (C 9 H 8 O 4 ) and acetic acid (CH 3 CO 2 H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by- product?

55. 55 Stoichiometry Salicylic acid + Acetic anhydride → Aspirin + acetic acid C 7 H 6 O 3 + C 4 H 6 O 3 → C 9 H 8 O 4 + CH 3 CO 2 H C 7 H 6 O 3 + C 4 H 6 O 3 → C 9 H 8 O 4 + C 2 H 4 O 2 Balanced! Equal # moles for all

56. 56 Stoichiometry 4.50 g Salicylic acid (C 7 H 6 O 3 ) = ? moles MW C 7 H 6 O 3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00 = 138.12 g/mole

57. 57 Stoichiometry Since all compounds have the same S.C., there must be 0.0326 moles of all 4 of them involved in the reaction. g Aspirin (C 9 H 8 O 4 ) = 0.0326 moles x MW Aspirin =.0326 x [9x12.01 + 8x1.008 + 4x16.00] =.0326 mole x 180.15 g/mole 5.87 g Aspirin

58. 58 Stoichiometry Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield. Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield.

59. 59Stoichiometry Dichloromethane (CH 2 Cl 2 ) is prepared by reaction of methane (CH 4 ) with chlorine (Cl 2 ) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%? Dichloromethane (CH 2 Cl 2 ) is prepared by reaction of methane (CH 4 ) with chlorine (Cl 2 ) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%?

60. 60 Stoichiometry CH 4 + Cl 2 → CH 2 Cl 2 + HCl Balance CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 1.85 kg CH 4 = ? moles CH 4

61. 61 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 1.85 kg CH 4 = ? moles CH 4 MW CH 4 = 1x12.01 + 4x1.008 = 16.04 g/mole

62. 62 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 115 moles CH 4 in theory we should produce: 115 moles of CH 2 Cl 2 and 230 moles of HCl And use up 230 moles of Cl 2

63. 63 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl 115 moles of CH 2 Cl 2 = ? g MW CH 2 Cl 2 = 12.01 + 2x1.008 + 2x35.45 = 84.93 115 moles x (84.03 g/mole) = 9770 g

64. 64 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl Expect 9770 g CH 2 Cl 2 but the yield is 43.1% So we produced just 0.431 x 9770 g 4.21 kg CH 2 Cl 2

65. 65 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl Suppose the reaction went to completion (100% yield) Is mass conserved?

66. 66 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl Start with 115 moles CH 4 and 230 moles Cl 2 total mass = 115x16.04 + 230x70.90 = 1850 + 16300 = 18150 only 3 sig. figs. → 18.2 kg

67. 67 Stoichiometry CH 4 + 2Cl 2 → CH 2 Cl 2 + 2HCl End with 115 moles CH 2 Cl 2 and 230 moles HCl total mass = 115x84.93 + 230x36.46 = 9770 + 8390 = 18160 only 3 sig. figs → 18.2 kg

68. Balancing Equations Hydrogen and oxygen are diatomic elements. Their subscripts cannot be changed. The subscripts on water cannot be changed. Hydrogen + oxygen water Hydrogen + oxygen water H 2 + O 2 H 2 O H 2 + O 2 H 2 O

69. Balancing Equation Count the atoms on each side. Reactant side: 2 atoms H and 2 atoms O Product side: 2 atoms H and 1 atom O H 2 + O 2 H 2 O H 2 + O 2 H 2 O

70. Balancing Equations H 2 + O 2 H 2 O H 2 + O 2 H 2 O If the subscripts cannot be altered, how can the atoms be made equal? If the subscripts cannot be altered, how can the atoms be made equal? Adjust the number of molecules by changing the coefficients. Adjust the number of molecules by changing the coefficients.

71. Balancing Equations Reactants: 2 atoms of H and 2 atoms of O Products: 4 atoms of H and 2 atoms of O H is no longer balanced! H 2 + O 2 2H 2 O H 2 + O 2 2H 2 O

72. Balancing Equations Reactant side: 4 atoms of H and 2 atoms of O Product side: 4 atoms of H and 2 atoms of O It’s Balanced! 2H 2 + O 2 2H 2 O 2H 2 + O 2 2H 2 O

73. Balancing Equations Count atoms. Reactants: 2 atoms N and 2 atoms H Products: 1 atom N and 3 atoms of NH 3 N 2 + H 2 NH 3 N 2 + H 2 NH 3 Nitrogen + hydrogen ammonia Nitrogen + hydrogen ammonia

74. Balancing Equations Nothing is balanced. Nothing is balanced. Balance the nitrogen first by placing a coefficient of 2 in front of the NH 3. Balance the nitrogen first by placing a coefficient of 2 in front of the NH 3. N 2 + H 2 2NH 3 N 2 + H 2 2NH 3

75. Balancing Equations Hydrogen is not balanced. Hydrogen is not balanced. Place a 3 in front of H 2. Place a 3 in front of H 2. Reactant side: 2 atoms N, 6 atoms H Reactant side: 2 atoms N, 6 atoms H Product side: 2 atoms N, 6 atoms H Product side: 2 atoms N, 6 atoms H N 2 + 3H 2 2NH 3 N 2 + 3H 2 2NH 3

76. Balancing Equations Count atoms. Reactants: Ca – 3 atoms, P – 2 atoms, O – 8 atoms; H – atoms, S – 1 atom, O – 4 atoms Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4

77. Balancing Equations Side note on Ca 3 (PO 4 ) 2 Side note on Ca 3 (PO 4 ) 2 The subscript after the phosphate indicates two phosphate groups. The subscript after the phosphate indicates two phosphate groups. This means two PO 4 3- groups with two P and eight O atoms. This means two PO 4 3- groups with two P and eight O atoms.

78. Balancing Equations Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 Count atoms in the product. Ca atoms – 1, S atom – 1, O atoms – 4; H atoms – 3, P atom – 1, O atoms - 4

79. Balancing Equations In this equation, the ion groups do not break up. In this equation, the ion groups do not break up. Instead of counting individual atoms, ion groups may be counted. Instead of counting individual atoms, ion groups may be counted. Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4

80. Balancing Equations Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 Reactants: Ca 2+ – 3, PO 4 3- - 2, H + – 2, SO 4 2+ - 1 Products: Ca 2+ - 1, SO 4 2- - 1, H + - 3, PO 4 3- - 1

81. Balancing Equations Balance the metal first by placing a coefficient of 3 in front of CaSO 4. Balance the metal first by placing a coefficient of 3 in front of CaSO 4. Products: Ca – 3 atoms, SO 4 2- - 3 groups Products: Ca – 3 atoms, SO 4 2- - 3 groups Ca 3 (PO 4 ) 2 + H 2 SO 4 3CaSO 4 + H 3 PO 4

82. Balancing Equations Three sulfate groups are needed on the reactant side so place a coefficient of 3 in front of H 2 SO 4. Three sulfate groups are needed on the reactant side so place a coefficient of 3 in front of H 2 SO 4. 3H 2 SO 4 gives 6 H + and 3 SO 4 2-. 3H 2 SO 4 gives 6 H + and 3 SO 4 2-. Neither phosphate nor calcium is balanced. Neither phosphate nor calcium is balanced. Ca 3 (PO 4 ) 2 + 3H 2 SO 4 3CaSO 4 + H 3 PO 4

83. Balancing Equations A coefficient of 2 placed in front of H 3 PO 4 which balances both hydrogen and phosphate. A coefficient of 2 placed in front of H 3 PO 4 which balances both hydrogen and phosphate. Ca 3 (PO 4 ) 2 + 3H 2 SO 4 3CaSO 4 + 2H 3 PO 4

84. Balancing Equations The sulfate group breaks up. Each atom must be counted individually. Ugh! Reactants: Cu – 1, H – 2, S – 1, O – 4 Products: Cu – 1, S – 1, O - 4, H – 2, O – 1, S – 1, O - 2 Cu + H 2 SO 4 Cu + H 2 SO 4 CuSO 4 + H 2 O + SO 2 CuSO 4 + H 2 O + SO 2

85. Balancing Equations Sulfur is not balanced. Sulfur is not balanced. Place a two in front of sulfuric acid. Place a two in front of sulfuric acid. Count atoms: 2 H 2 SO 4 H – 4, S – 2, O - 8 Count atoms: 2 H 2 SO 4 H – 4, S – 2, O - 8 Cu + 2H 2 SO 4 CuSO 4 + H 2 O + SO 2

86. Balancing Equations Hydrogen needs to be balanced so place a 2 in front of the H 2 O. Hydrogen needs to be balanced so place a 2 in front of the H 2 O. Count the number of atoms. Count the number of atoms. Cu + 2H 2 SO 4 CuSO 4 + 2H 2 O + SO 2

87. Balancing Equations Reactants: Cu – 1, H – 4, S – 2, O – 8 Reactants: Cu – 1, H – 4, S – 2, O – 8 Products: Cu – 1, S – 1, O – 4, H – 4, O – 2, S – 1, O – 2 = Cu – 1, S – 2, H – 4, O – 8 Products: Cu – 1, S – 1, O – 4, H – 4, O – 2, S – 1, O – 2 = Cu – 1, S – 2, H – 4, O – 8 It’s balanced! It’s balanced! Cu + 2H 2 SO 4 CuSO 4 + 2H 2 O + SO 2

88. Balancing Equations Balancing hints: Balancing hints: Balance the metals first. Balance the metals first. Balance the ion groups next. Balance the ion groups next. Balance the other atoms. Balance the other atoms. Save the non ion group oxygen and hydrogen until the end. Save the non ion group oxygen and hydrogen until the end.

89. Balancing Equations This method of balancing equations is the inspection method. This method of balancing equations is the inspection method. The method is trial and error. The method is trial and error. Practice. Practice.

90. Writing and Naming Write the corresponding formula equation and then balance the equation. Nickel + hydrochloric acid Nickel + hydrochloric acid Nickel(II) chloride + hydrogen Nickel(II) chloride + hydrogen

91. Writing and Naming Write each formula independently. Write each formula independently. Ignore the rest of the equation. Ignore the rest of the equation. Balance the equation after writing the formulas. Balance the equation after writing the formulas. Ni + HCl NiCl 2 + H 2 Ni + HCl NiCl 2 + H 2 Ni + 2HCl NiCl 2 + H 2 Ni + 2HCl NiCl 2 + H 2

92. Writing and Naming Remember the diatomic elements: H 2, N 2, O 2, F 2, Cl 2, Br 2, and I 2. Remember the diatomic elements: H 2, N 2, O 2, F 2, Cl 2, Br 2, and I 2.

93. Writing and Naming Balance the formula equation. Write the word equation. Cu + H 2 SO 4 Cu + H 2 SO 4 CuSO 4 + H 2 O + SO 2 CuSO 4 + H 2 O + SO 2

94. Writing and Naming Cu + 2H 2 SO 4 Cu + 2H 2 SO 4 CuSO 4 + 2H 2 O + SO 2 CuSO 4 + 2H 2 O + SO 2 Write the names: Cu by itself is just copper. Copper(I) or copper(II) would be incorrect. H 2 SO 4 should be named as an acid. Sulfuric acid

95. Writing and Naming CuSO 4 has a SO 4 2- group so Cu must be 2+. Some metals must have Roman Numerals. Copper(II) sulfate CuSO 4 has a SO 4 2- group so Cu must be 2+. Some metals must have Roman Numerals. Copper(II) sulfate H 2 O is known as water. H 2 O is known as water. SO 2 is a nonmetal compound. Its name is either sulfur dioxide or sulfur(IV) oxide. SO 2 is a nonmetal compound. Its name is either sulfur dioxide or sulfur(IV) oxide.

96. Writing and Naming Copper + sulfuric acid  Copper(II) sulfate + water + sulfur dioxide Cu + 2H 2 SO 4 Cu + 2H 2 SO 4 CuSO 4 + 2H 2 O + SO 2 CuSO 4 + 2H 2 O + SO 2