1. Ch 4. Chemical Quantities and Aqueous Reactions

2. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) 1 mol2 mol1 mol2 mol Stoichiometry of the reaction FIXED ratio for each reaction Can calculate how much other chemicals are required or produced if the amount of one chemical is known.

3. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) 1 mol2 mol1 mol2 mol 4 mol2 mol4 mol 3 mol6 mol3 mol6 mol 5.22 mol10.44 mol5.22 mol10.44 mol

4. 4 NH 3 (g) + 5 O 2 (g)  4 NO (g) + 6 H 2 O (g) 5.02 mol of NH 3 is used in the above reaction. How many moles of O 2 is required to react with all the NH 3 ? How many moles of H 2 O will be produced?

5. 2NH 3 (g) + 3CuO(s)  N 2 (g) + 3Cu(s) + 3H 2 O(g) 2 mol 3 mol1 mol3 mol 5.02 mol xyzu

6. 2NH 3 (g) + 3CuO(s)  N 2 (g) + 3Cu(s) + 3H 2 O(g) 2 mol 3 mol1 mol3 mol 6.04 g xy 6.04 g ÷ (14.01 g/mol x 1 + 1.008 g/mol x 3) = 0.355 mol 0.355 mol mass? Mass of CuO = 0.533 mol x 79.55 g/mol = 42.4 g Mass of N 2 = 0.178 mol x 28.02 g/mol = 4.99 g

7. Other Examples: page 131 ― 132

8. CH 4 + 2O 2  CO 2 + 2H 2 O 0 mol 1 mol2 mol initial: 0 mol 1 mol 2 mol final: 1 mol 0 mol initial: ? mol final: The actual amount of reactants consumed and actual amount of products generated agree with the stoichiometry.

9. CH 4 + 2O 2  CO 2 + 2H 2 O 1 mol 0 mol initial: 1 mol CH 4 requires 2 mol O 2, available O 2 is 1 mol: limiting reagent. (1 − 0.5) mol 0 mol = 0.5 mol= 1 mol Result: 1 mol O 2 will be consumed completely and CH 4 will have leftover: excess reagent. = 0.5 mol consumed: 1 mol x x = 0.5 mol final: yz

10. The reactant of which there are fewer moles than the stoichiometry requires is the limiting reagent. The reactant of which there are more moles than the stoichiometry requires is the excess reagent. Chemical reactions always occur according to the stoichiometry, therefore the limiting reagent is consumed and the excess reagent has leftover. The amount of products is determined by the amounts of reagents that are actually consumed.

11. CH 4 + 2O 2  CO 2 + 2H 2 O limiting reagentexcess reagent 1 mol 0 mol initial: consumed: 0.5 mol1 mol (1 − 0.5) mol final: 0 mol 0.5 mol1 mol 0.5:1 : 0.5 :1 1:2 : 1 :2=

12. + + + + + + + +

13. 2 slices of bread + 1 slice if ham  1 sandwich 4 slices of bread + 1 slice if ham  excess reagent leftover limiting reagentamount of product 1 sandwich + 2 slices of bread

14. For the following reaction, if a sample containing 18.1 g of NH 3 is reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N 2 will be formed? How many grams of excess reagent will be leftover? If 6.63 g of N 2 is actually produced, what is the percent yield? 2NH 3 (g) + 3CuO(s)  N 2 (g) + 3Cu(s) + 3H 2 O(g)

15. 1)Make sure the equation is balanced. 2)Find the moles of each reactant: moles = mass in gram / molar mass 3)Pick up any reactant, say A, and use the stoichiometry to calculate the required amount of the other reactant B. 4)Compare the required amount of B with the available amount of B. a) If required > available, then B is the limiting reagent and A is the excess reagent. b) If required < available, then B is the excess reagent and A is the limiting reagent. 5)Use the amount of the limiting reagent and the stoichiometry to calculate the amount of any product and the amount of the excess reagent that has been consumed. 6)Leftover excess reagent = available − consumed 7)If actual yield is given percent yield = (actually yield / theoretical yield) x 100% Procedure for limiting/excess reagent calculations aA + bB  cC + dD

16. 68.5 g CO reacts with 8.60 g H 2 in the following reaction. What is the limiting reagent? How many grams of excess reagent is leftover? What is the theoretical yield of CH 3 OH? If 35.7 g CH 3 OH is actually produced, what is the percent yield of CH 3 OH? H 2 (g) + CO(g)  CH 3 OH(g)

17. 1)Make sure the equation is balanced. 2)Find the moles of each reactant: moles = mass in gram / molar mass 3)Pick up any reactant, say A, and use the stoichiometry to calculate the required amount of the other reactant B. 4)Compare the required amount of B with the available amount of B. a) If required > available, then B is the limiting reagent and A is the excess reagent. b) If required < available, then B is the excess reagent and A is the limiting reagent. 5)Use the amount of the limiting reagent and the stoichiometry to calculate the amount of any product and the amount of the excess reagent that has been consumed. 6)Leftover excess reagent = available − consumed 7)If actual yield is given percent yield = (actually yield / theoretical yield) x 100% Procedure for limiting/excess reagent calculations aA + bB  cC + dD

18. 1.50 g of ammonia reacts with 2.75 g of oxygen gas to produce nitrogen monoxide and water. NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) a) Balance the equation. b) What is the mass of O 2 in grams required by NH 3 ? c) Which reactant is the limiting reagent? d) How many grams of NO will be produced in theory? e) How many grams of H 2 O will be produced in theory? f) How many grams of the excess reagent remain unreacted? g) If only 1.80 g of NO are produced, what is the percent yield?

19. Classification of Matter Matter Elements Compounds Mixtures (multiple components) Pure Substances (one component) Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) (Solutions)

20. Solute + Solvent  Solution Solvent = water, aqueous solution Water can dissolve many substances

21. O H H H2OH2O

24. C 12 H 22 O 11

27. electrolytes nonelectrolytes Based on the electrical conductivity in aqueous solution strong electrolytes weak electrolytes salts strong acids strong bases weak acids weak bases solutes

28. strong electrolytes: dissociate 100 % into ions weak electrolytes: only a small fraction dissociate into ions nonelectrolytes: no dissociation

30. salts: NaCl, Na 2 SO 4, Fe(ClO 4 ) 3 …… strong acids: HCl, HNO 3, H 2 SO 4, HClO 4 strong bases: NaOH, KOH Base: compounds that give OH − when dissolved in water. weak acids: acetic acid: HC 2 H 3 O 2 weak bases: ammonia: NH 3 remember

31. HCl is Completely Ionized

32. Aqueous Solution of NaOH

33. Reaction of NH 3 in Water

34. concentrations no unit

35. 10. g of sugar is dissolved in 40. g of water. What is the mass percent of sugar in this solution?

36. Unit: mol/L or M

37. 0.50 mol of KBr is dissolved in water and forms a solution of 12 L. What is the molarity of the solution? Example 4.5, page 141 25.5 g of KBr is dissolved in water and forms a solution of 1.75 L. What is the molarity of the solution? Example 4.6, page 142 How many liters of a 0.125 mol/L NaOH solution contains 0.255 mol of NaOH?

38. How to prepare 1.00 L of NaCl aqueous solution with a molarity of 1.00 mol/L? 1.00 mol NaCl + 1.00 L of H 2 O = 1.00 mol/L NaCl (aq)

40. Solution Dilution Concentrated solutions for storage, called stock solutions stock solution + water  desired solution

41. moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2 M 1 : molarity of concentrated solution V 1 : volume of concentrated solution M 2 : molarity of diluted solution V 2 : volume of diluted solution Example on page 143 A lab procedure calls for 3.00 L of a 0.500 mol/L CaCl 2 solution. How should we prepare it from a 10.0 mol/L stock solution?

43. Example 4.7, page 144 To what volume should you dilute 0.200 L of a 15.0 mol/L NaOH solution to obtain a 3.00 mol/L NaOH solution?

44. Types of reactions Precipitation reactions

45. NaCl(aq) + AgNO 3 (aq)  AgCl(s) + NaNO 3 (aq) formula equation Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq)  AgCl(s) + Na + (aq) + NO 3 − (aq) complete ionic equation Cl − (aq) + Ag + (aq)  AgCl(s) net ionic equation Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq)  AgCl(s) + Na + (aq) + NO 3 − (aq) spectator ions

47. EXAMPLE 4.9 Predicting whether an Ionic Compound Is Soluble Predict whether each compound is soluble or insoluble. (a) PbCl 2 (b) CuCl 2 (c)Ca(NO 3 ) 2 (d) BaSO 4

49. BaCl 2 (aq) + K 2 SO 4 (aq)  BaCl 2 (aq)  Ba 2+ (aq) + 2Cl − (aq) K 2 SO 4 (aq)  2K + (aq) + SO 4 2− (aq) BaSO 4 (s) + 2KCl(aq) Ba 2+ (aq) + 2Cl − (aq) + 2K + (aq) + SO 4 2− (aq)  BaSO 4 (s) + 2Cl − (aq) + 2K + (aq) Ba 2+ (aq) + SO 4 2− (aq)  BaSO 4 (s)

50. Fe(NO 3 ) 3 (aq) + KOH(aq)  Fe(NO 3 ) 3 (aq)  Fe 3+ (aq) + 3NO 3 − (aq) KOH(aq)  K + (aq) + OH − (aq) Fe 3+ (aq) + 3NO 3 − (aq) +3K + (aq) +3OH − (aq)  Fe(OH) 3 (s) + 3NO 3 − (aq) + 3K + (aq) Fe 3+ (aq) + 3OH − (aq)  Fe(OH) 3 (s) 3KOH(aq)  3K + (aq) + 3OH − (aq) Fe(NO 3 ) 3 (aq) + 3KOH(aq)  Fe(OH) 3 (s) + 3KNO 3 (aq)

51. BaCl 2 (aq) + KNO 3 (aq)  BaCl 2 (aq)  Ba 2+ (aq) + 2Cl − (aq) KNO 3 (aq)  K + (aq) + NO 3 − (aq) BaCl 2 (aq) + 2KNO 3 (aq)  Ba(NO 3 ) 2 (aq) + 2KCl(aq) Ba 2+ (aq) + 2Cl − (aq) + 2K + (aq) + 2NO 3 − (aq)  Ba 2+ (aq) + 2NO 3 − (aq) + 2Cl − (aq) + 2K + (aq) 2KNO 3 (aq)  2K + (aq) + 2NO 3 − (aq)

52. Types of reactions Precipitation reactions Acid-base reactions

53. Acid: Substance that produces H + ions in aqueous solution Base: Substance that produces OH − ions in aqueous solution

57. NaOH(aq)  Na + (aq) + OH − (aq) H + (aq) + Cl − (aq) +Na + (aq) +OH − (aq)  H 2 O(l) + Na + (aq) + Cl − (aq) H + (aq) + OH − (aq)  H 2 O(l) HCl(aq)  H + (aq) + Cl − (aq) HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) H + (aq) + OH − (aq)  H 2 O(l) acidic basicneutral neutralization

58. HCl(aq)  H + (aq) + Cl − (aq) What is the molarity of HCl(aq) or H + (aq)?

59. When reaction completes n NaOH = n HCl M NaOH V NaOH = M HCl V HCl HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq) prepared, known measured by buret, known unknownmeasured by pippet, known

60. M NaOH V NaOH = M HCl V HCl

61. Read acid-base titration starting on page 158 and the online instruction for next week’s titration

62. Example 4.14 The titration of a 10.00-mL sample of an HCl solution of unknown concentration requires 12.54 mL of a 0.100 M NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution in M? M NaOH V NaOH = M HCl V HCl HCl(aq) + NaOH(aq)  H 2 O(l) + NaCl(aq)

63. Types of reactions Precipitation reactions Acid-base reactions Oxidation-Reduction reactions

64. Reactions that involve electron transfer are called oxidation-reduction reactions, or redox reactions. 2Mg(s) + O 2 (g)  2MgO(s)

65. Oxidation numbers (states) 1) For atoms in its elemental form, oxidation number = 0 A way to keep track of the electrons gained or lost Na, Ag, Ar, O 2, N 2, P 4 2) For monatomic ion, oxidation number = charge of the ion Na +, Ca 2+, Co 2+, Co 3+, Cl −, O 2− NaCl, Na 2 O, CaCl 2, CaO, CoCl 2, CoCl 3, Co 2 O 3, CoO

66. O H H H2OH2O

67. 3) In covalent compounds O: −2H: +1F: −1 In a neutral compound, the sum of the oxidation number = 0 In a polyatomic ion, the sum of the oxidation number = ion charge CO, CO 2, SF 6, SF 4, H 2 S, NH 3, P 2 O 5, N 2 O 3 NO 3 −, SO 4 2−, NH 4 +, Cr 2 O 7 2−, MnO 4 −