1. Chapter 6 Chemical Reactions: Classification and Mass Relationships

2. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations (chemical reactions) reactants  products limestone  quicklime + gas Calcium carbonate  calcium oxide + carbon dioxide CaCO 3 (s)  CaO(s) + CO 2 (g)

3. Balancing Chemical Equations Chemical reactions include –Reactants –Products –Balanced – Law of Conservation of Mass # of atoms of an element on the reactant side must equal the # of atoms of that element on the product side. –Indicate the state of matter of each chemical in the reaction (Chapter 4)

4. Balancing Chemical Equations Write the equation without coefficients List the elements in each equation –Secret: if the same polyatomic ion exists on both sides, keep it together Determine the # of each kind of atom on both sides Balance atoms one element at a time by adjusting coefficients –DO NOT ALTER THE FORMULA OF THE COMPOUND!!!!! Only coefficients can be altered –Secret: Balance atoms appearing only once on each side first. Save compounds comprised of only one type of element till last. Reduce to lowest terms if necessary

5. Examples Balance the following equations: –Al(s) + Fe 2 O 3 (s) → Al 2 O 3 (s) + Fe (l) –Solid copper reacts with aqueous silver nitrate to form aqueous copper (II) nitrate and silver solid –H 3 PO 4 (l) → H 2 O (l) + P 4 O 10 (s) –C 4 H 10 (g) + O 2 (g) → CO 2 (g) + H 2 O (g)

6. Avogadro’s Number and the Mole Meaning of a chemical reaction 2 C 4 H 10 (g) + 13 O 2 (g) → 8 CO 2 (g) + 10 H 2 O (g) –2 molecules of C 4 H 10 (g) reacts with 13 molecules of O 2 (g) to form 8 molecule of CO 2 (g) and 10 molecules of H 2 O(g)

7. Avogadro’s Number and the Mole Molecule’s mass = the sum of the atomic masses of the atoms making up the molecule. 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) » = 60.06 amu

8. Avogadro’s Number of the Mole One mole (mol) of any substance contains 6.02 x 10 23 (Avogadro’s Number) units of that substance. One mole (mol) of a substance is the gram mass value equal to the amu mass of the substance. –Calculated the same as amu’s for a molecule

9. Avogadro’s Number and the Mole Calculate the molar mass of the following: –Fe 2 O 3 (Rust) –C 6 H 8 O 7 (Citric acid) –C 16 H 18 N 2 O 4 (Penicillin G)

10. Avogadro’s Number and the Mole 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)

11. Stoichiometry 4 Conversion units –Chemical formula –Balanced chemical equation Coefficients can read as; –# of molecules –# of moles of that molecules Allows conversion between compounds in an equation –Avogadro’s # - 6.02 x 10 23 of X = 1 mole of X –Molar mass – how many grams of a substance = 1 mole of that substance

12. Stoichiometric Calculations

13. Avogadro’s Number and the Mole How many grams of oxygen are present in 5.961 x 10 20 molecules of KClO 3 ? How many atoms of oxygen are present?

14. Avogadro’s Number and the Mole Calculate the number of oxygen atoms in 29.34 g of sodium sulfate, Na 2 SO 4. –A. 1.244 × 10 23 O atoms –B. 4.976 × 10 23 O atoms –C. 2.409 × 10 24 O atoms –D. 2.915 × 10 24 O atoms –E. 1.166 × 10 25 O atoms

15. Problem Potassium dichromate, K 2 Cr 2 O 7, is used in tanning leather, decorating porcelain and water proofing fabrics. Calculate the number of chromium atoms in 78.82 g of K 2 Cr 2 O 7. –A. 9.490 × 10 25 Cr atoms –B. 2.248 × 10 24 Cr atoms –C. 1.124 × 10 24 Cr atoms –D. 3.227 × 10 23 Cr atoms –E. 1.613 × 10 23 Cr atoms

16. Stoichiometry: Chemical Arithmetic

17. Stoichiometry: Equation Arithmetic 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)

18. Stoichiometry: Chemical Arithmetic Aqueous sodium hydroxide and chlorine gas are combined to form aqueous sodium hypochlorite (household bleach), aqueous sodium chloride and liquid water. –How many grams of NaOH are needed to react with 25.0 g of Cl 2 ?

19. Problem Sulfur dioxide reacts with chlorine to produce thionyl chloride (used as a drying agent for inorganic halides) and dichlorine monoxide (used as a bleach for wood, pulp and textiles). SO 2 (g) + 2Cl 2 (g) → SOCl 2 (g) + Cl 2 O(g) If 0.400 mol of Cl 2 reacts with excess SO 2, how many moles of Cl 2 O are formed? –A. 0.800 mol –B. 0.400 mol –C. 0.200 mol –D. 0.100 mol –E. 0.0500 mol

20. Problem Nitrogen gas and hydrogen gas are combined to form ammonia (NH 3 ), an important source of fixed nitrogen that can be metabolized by plants, using the Haber process. How many grams of nitrogen are needed to produce 325 grams of ammonia? –A. 1070 g –B. 535 g –C. 267 g –D. 178 g –E. 108 g

21. Percent Yields 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.

22. Problem What is the percent yield for the reaction PCl 3 (g) + Cl 2 (g) → PCl 5 (g) if 119.3 g of PCl 5 ( MM = 208.2 g/mol) are formed when 61.3 g of Cl 2 ( MM = 70.91 g/mol) react with excess PCl 3 ? –A. 195% –B. 85.0% –C. 66.3% –D. 51.4% –E. 43.7%

23. Types of Chemical Reactions Chemical Reactions discussed in College Chemistry can be broken down into 3 main categories –Precipitation reactions –Acid-Base reactions –Oxidation-Reduction (redox) reactions

24. Types of Chemical Reactions Precipitation Reactions: A process in which an insoluble solid (precipitate) drops out of the solution.Precipitation Reactions: –Clear solutions of two ionic compounds when mixed form a cloudy solution (cloudiness indicates solid)

25. Types of Reactions Acid–Base Neutralization: A process in which an acid reacts with a base to yield water plus an ionic compound called a salt. –The driving force of this reaction is the formation of the stable water molecule.

26. Types of Reaction Metathesis Reactions (Double Displacement Reaction): These are reactions where two reactants just exchange parts. AX + BY  AY + BX

27. Types of Reactions Oxidation–Reduction (Redox) Reaction: A process in which one or more electrons are transferred between reaction partners. –The driving force of this reaction is the decrease in electrical potential.

28. Precipitation Reactions Develop the reaction equation Balance the reaction equation Predict the state of matter of each species present

29. Precipitation Reactions and Solubility Rules To predict whether a precipitation reaction will occur must be able to predict whether a compound is soluble or not –Solubility rules

30. Solubility Rules Salts - soluble: All alkali metal and ammonium ion salts All salts of the NO 3 –, ClO 3 –, ClO 4 –, C 2 H 3 O 2 –, and HCO 3 – ions

31. Solubility Rules Salts which are soluble with exceptions: Cl –, Br –, I – ion salts except with Ag +, Pb 2+, & Hg 2 2+ SO 4 2– ion salts except with Ag +, Pb 2+, Hg 2 2+, Ca 2+, Sr 2+, & Ba 2+

32. Solubility Rules Salts which are insoluble with exceptions: O 2– & OH – ion salts except with the alkali metal ions, and Ca 2+, Sr 2+, & Ba 2+ ions CO 3 2–, PO 4 3–, S 2–, CrO 4 2–, & SO 3 2– ion salts except with the alkali metal ions and the ammonium ion

33. Precipitation Reactions and Solubility Rules Predict the solubility of: –(a) CdCO 3 (b) MgO(c) Na 2 S (d) PbSO 4 (e) (NH 4 ) 3 PO 4 (f) HgCl 2

34. Precipitation Reaction Precipitation reactions only occur if a solid is produced as a product. If all products are aqueous compounds then no reaction has taken place.

35. Precipitation Reactions and Solubility Guidelines Predict whether a precipitate will form for: –(a) NiCl 2 (aq) + (NH 4 ) 2 S(aq)  –(b) Na 2 CrO 4 (aq) + Pb(NO 3 ) 2 (aq)  –(c) AgClO 4 (aq) + CaBr 2 (aq) 

36. Problem Select the precipitate that forms when aqueous ammonium sulfide reacts with aqueous copper(II) nitrate. –A. CuS –B. Cu 2 S –C. NH 4 NO 3 –D. NH 4 (NO 3 ) 2 –E. CuSO 4

37. Problem Select the precipitate that forms when the following reactants are mixed. Mg(CH 3 COO) 2 (aq) + LiOH(aq) → –A. LiCH 3 COO –B. Li(CH 3 COO) 2 –C. MgOH –D. Mg(OH) 2 –E. CH 3 OH

38. Acids, Bases and Neutralization Reactions Acid / Base Definitions –Arrhenius Acid – donates a H + (H 3 O + ) Base – donates an OH - –Bronsted-Lowry Acid – donates a H + Base – H + acceptor

39. Acids, Bases and Neutralization Reactions Strong Acids (all) Hydrochloric acid – HCl (aq) Hydrobromic acid – HBr (aq) Hydroiodic acid – HI (aq) Sulfuric acid – H 2 SO 4 (aq) Nitric acid – HNO 3 (aq) Perchloric acid – HClO 4 (aq) Weak Acids (examples) Acetic acid – HC 2 H 3 O 2 (aq) Cyanic acid – HCN (aq) Phosphoric acid – H 3 PO 4 (aq) Organic acids – contain ending group –COOH Strong Bases (all) Lithium hydroxide – LiOH (aq) Sodium hydroxide – NaOH (aq) Potassium hydroxide – KOH (aq) Calcium hydroxide – Ca(OH) 2 (aq) Strontium hydroxide – Sr(OH) 2 (aq) Barium hydroxide – Ba(OH) 2 (aq) Weak Bases (examples) Ammonia (ammonium hydroxide) – NH 3 (aq) actually NH 4 OH (aq) Organic amines – contain ending group -NH x

40. Acids, Bases and Neutralization Reactions Neutralization Reaction: produces salt & water. –HA(aq) + MOH(aq)  H 2 O(l) + MA(aq) Write a balanced chemical equation for the following: –(a) HBr(aq) + Ba(OH) 2 (aq)  –(b) HCl(aq) + NH 3 (aq) 

41. Oxidation-Reduction Reactions Redox reactions are those involving the oxidation and reduction of species (element or ion of an element). Oxidation and reduction must occur together. They cannot exist alone. Two important types –Single displacement reactions (activity series) –Combustions – reaction of a substance with O 2

42. Oxidation Reduction Reactions Oxidation Is Loss (of electrons) Anode Oxidation Reducing Agent

43. Oxidation Reduction Reactions Reduction Is Gain (of electrons) Cathode Reduction Oxidizing Agent

44. Redox Reactions Assigning Oxidation Numbers: All atoms have an “oxidation number” regardless of whether it carries an ionic charge. 1.An atom in its elemental state has an oxidation number of zero. 2. An atom in a monatomic ion has an oxidation number identical to its charge.

45. Redox Reactions 3. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. –A. Hydrogen can be either +1 or –1. –B. Oxygen usually has an oxidation number of –2. In peroxides, oxygen is –1. –C. Halogens usually have an oxidation number of –1. When bonded to oxygen, chlorine, bromine, and iodine have positive oxidation numbers.

46. Redox Reactions 4. The sum of the oxidation numbers must be zero for a neutral compound and must be equal to the net charge for a polyatomic ion. –A. H 2 SO 4 2(+1) + (?) + 4(–2) = 0 net charge ? = 0 – 2(+1) – 4(–2) = +6 –B. ClO 4 – (?) + 4(–2) = –1 net charge ? = –1 – 4(–2) = +7

47. Problem Sodium tripolyphosphate is used in detergents to make them effective in hard water. Calculate the oxidation number of phosphorus in Na 5 P 3 O 10. –A. +3 –B. +5 –C. +10 –D. +15 –E. none of these is the correct oxidation number

48. Problem The oxidation numbers of P, S and Cl in H 2 PO 2 -, H 2 S and KClO 4 are, respectively –A. -1, -1, +3 –B. +1, -2, +7 –C. +1, +2, +7 –D. -1, -2, +7 –E. -1, -2, +3

49. Redox Reactions Whenever one atom loses electrons (is oxidized), another atom must gain those electrons (be reduced). –A substance which loses electrons (oxidized) is called a reducing agent. Its oxidation number increases. –A substance which gains electrons (reduced) is called the oxidizing agent. Its oxidation number decreases.

50. Redox Reactions For each of the following, identify which species is the reducing agent and which is the oxidizing agent. Ca(s) + 2 H + (aq)  Ca 2+ (aq) + H 2 (g) 2 Fe 2+ (aq) + Cl 2 (aq)  2 Fe 3+ (aq) + 2 Cl – (aq) SnO 2 (s) + 2 C(s)  Sn(s) + 2 CO(g) Sn 2+ (aq) + 2 Fe 3+ (aq)  Sn 4+ (aq) + 2 Fe 2+ (aq)

51. Problem Identify the oxidizing agent in the following redox reaction. Hg 2+ (aq) + Cu(s) → Cu 2+ (aq) + Hg(l) –A. Hg 2+ (aq) –B. Cu(s) –C. Cu 2+ (aq) –D. Hg(l) –E. Hg 2+ (aq) and Cu 2+ (aq)

52. Problem Sodium thiosulfate, Na 2 S 2 O 3, is used as a "fixer" in black and white photography. Identify the reducing agent in the reaction of thiosulfate with iodine. 2S 2 O 3 2- (aq) + I 2 (aq) → S 4 O 6 2- (aq) + 2I - (aq) –a. I 2 (aq) –b. I - (aq) –c. S 2 O 3 2- (aq) –d. S 4 O 6 2- (aq) –e. S 2 O 3 2- (aq) and I - (aq)

53. Optional Homework Text – 6.28, 6.29, 6.30, 6.33, 6.34, 6.38, 6.40, 6.42, 6.50, 6.54, 6.56, 6.60, 6.62, 6.66, 6.68, 6.72, 6.74, 6.76, 6.80, 6.82, 6.88, 6.90,6.92, 6.98, 6.100, 6.102, 6.106 Chapter 6 Homework - website

54. Required Homework Assignment 6