Chapter 8 Reactions in Aqueous Solutions

Chapter 8 Reactions in Aqueous Solutions1. Water is the most universal of all liquids. Water has a relatively large heat capacity, and a relatively large liquid range, which means it can absorb the heat liberated by many reactions while still remaining in the liquid state. Water is very polar and dissolves well both ionic solutes and solutes with which it can hydrogen-bond (this is especially important to the biochemical reactions of the living cell).2. Driving forces are types of changes in a system that pull a reaction in the direction of product formation; driving forces include formation of a solid, formation of water, formation of a gas, and transfer of electrons.3. The net charge of a precipitate must be zero. The total number of positive charges equals the total number of negative charges.4. When an electrolyte such as NaCl (sodium chloride) is dissolved in water, the resulting solution consists of separate, individual, discrete sodium ions (Na+) and separate, individual, discrete chloride ions (Cl–). There are no identifiable NaCl units in such a solution. 5. A substance is said to be a strong electrolyte if each unit of the substance produces separated, distinct ions when the substance is dissolved in water. NaCl and KNO3 are both strong electrolytes.6. NaNO3 must be soluble in water.7. For most practical purposes, “insoluble” and “slightly” soluble mean the same thing. The difference between “insoluble” and “slightly soluble” could be crucial if, for example, a substance were highly toxic and were found in a water supply.8. a. soluble (Rule 1: most nitrate salts are soluble) b. soluble (Rule 2: most potassium salts are soluble) c. soluble (Rule 2: most sodium salts are soluble) d. insoluble (Rule 5: most hydroxide compounds are insoluble) e. insoluble (Rule 3: exception for chloride salts) f. soluble (Rule 2: most ammonium salts are soluble) g. insoluble (Rule 6: most sulfide salts are insoluble) h. insoluble (Rule 4: exception for sulfate salts)9. a. Rule 5: most hydroxides are only slightly soluble b. Rule 6: most carbonates are only slightly soluble c. Rule 6: most phosphates are only slightly soluble d. Rule 3: exception to the rule for chlorides10. a. CaSO4. Rule 4: exception to the rule for sulfates b. AgI. Rule 3: although the text does not mention it explicitly, as you might expect from your knowledge of the periodic table, bromide and iodide + 2+ 2+ compounds of Ag , Pb , and Hg2 are insoluble. c. Pb3(PO4)2. Rule 6: most phosphate salts are only slightly soluble. d. Fe(OH)3. Rule 5: most hydroxides are only slightly soluble. e. No precipitate is likely: rules 1, 2, and 4. f. BaCO3. Rule 6: most carbonate salts are only slightly soluble.11. The precipitates are marked in boldface type. a. No precipitate: Ba(NO3)2 and HCl are each soluble. b. Rule 6: most sulfide salts are insoluble.(NH4)2S(aq) + CoCl2(aq)  CoS(s) + 2NH4Cl(aq) c. Rule 4: lead sulfate is a listed exception.H2SO4(aq) + Pb(NO3)2(aq)  PbSO4(s) + 2HNO3(aq) d. Rule 6: most carbonate salts are insoluble.CaCl2(aq) + K2CO3(aq)  CaCO3(s) + 2KCl(aq) e. No precipitate: NaNO3 and NH4C2H3O2 are each soluble. f. Rule 6: most phosphate salts are insoluble.Na3PO4(aq) + CrCl3(aq)  3NaCl(aq) + CrPO4(s)12. Hint: when balancing equations involving polyatomic ions, especially in precipitation reactions, balance the polyatomic ions as a unit, not in terms of the – atoms the polyatomic ions contain (e.g., treat nitrate ion, NO3 , as a single entity, not as one nitrogen and three oxygen atoms). When finished balancing, however, do be sure to count the individual number of atoms of each type on each side of the equation. a. AgNO3(aq) + H2SO4(aq)  Ag2SO4(s) + HNO3(aq)Balance silver: 2AgNO3(aq) + H2SO4(aq)  Ag2SO4(s) + HNO3(aq)Balance nitrate: 2AgNO3(aq) + H2SO4(aq)  Ag2SO4(s) + 2HNO3(aq)Balanced equation: 2AgNO3(aq) + H2SO4(aq)  Ag2SO4(s) + 2HNO3(aq) b. Ca(NO3)2(aq) + H2SO4(aq)  CaSO4(s) + HNO3(aq)Balance nitrate: Ca(NO3)2(aq) + H2SO4(aq)  CaSO4(s) + 2HNO3(aq)Balanced equation: Ca(NO3)2(aq) + H2SO4(aq)  CaSO4(s) + 2HNO3(aq) c. Pb(NO3)2(aq) + H2SO4(aq)  PbSO4(s) + HNO3(aq)Balance nitrate: Pb(NO3)2(aq) + H2SO4(aq)  PbSO4(s) + 2HNO3(aq)Balanced equation: Pb(NO3)2(aq) + H2SO4(aq)  PbSO4(s) + 2HNO3(aq)13. The products are determined by having the ions “switch partners.” For example, for a general reaction AB + CD , the possible products are AD and CB if the ions switch partners. If either AD or CB is insoluble, then a precipitation reaction has occurred. In the following reaction, the formula of the precipitate is given in boldface type. a. (NH4)2S(aq) + CoCl2(aq)  CoS(s) + 2NH4Cl(aq) Rule 6: most sulfide salts are only slightly soluble. b. FeCl3(aq) + 3NaOH(aq)  Fe(OH)3(s) + 3NaCl(aq) Rule 5: most hydroxide compounds are only slightly soluble. c. CuSO4(aq) + Na2CO3(aq)  CuCO3(s) + Na2SO4(aq) Rule 6: most carbonate salts are only slightly soluble.14. The net ionic equation for a reaction indicates only those ions that go to form the precipitate, and does not show the spectator ions present in the solutes mixed. The identity of the precipitate is determined from the Solubility Rules.2+ 2– a. Ca (aq) + SO4 (aq)  CaSO4(s) Rule 4: exception to rule about sulfate salts.3+ 2– b. 2Fe (aq) + 3CO3 (aq)  Fe2(CO3)3(s) Rule 6: most carbonate salts are only slightly soluble. c. Ag+(aq) + I–(aq)  AgI(s) Rule 3: AgI, like AgCl, is insoluble.2+ 3– d. 3Co (aq) + 3PO4 (aq)  Co3(PO4)2(s) Rule 6: most phosphate salts are only slightly soluble.2+ – e. Hg2 (aq) + 2Cl (aq)  Hg2Cl2(s) Rule 3: listed exception to the general rule about chlorides.2+ – f. Pb (aq) + 2Br (aq)  PbBr2(s)Rule 3: like PbCl2, PbBr2 and PbI2 are also insoluble. 15. Ag+(aq) + Cl–(aq)  AgCl(s)2+ – Pb (aq) + 2Cl (aq)  PbCl2(s)2+ – Hg2 (aq) + 2Cl (aq)  Hg2Cl2(s)2+ 2– 16. Ca (aq) + C2O4 (aq)  CaC2O4(s)17. Strong acids are acids that ionize completely in water. The strong acids are also strong electrolytes.18. Strong bases are bases that fully produce hydroxide ions when dissolved in water. The strong bases are also strong electrolytes.19. 1000; 1000 + – + – 20. HBr(aq)  H (aq) + Br (aq) HClO4(aq)  H (aq) + ClO4 (aq)21. RbOH(s)  Rb+(aq) + OH–(aq) CsOH(s)  Cs+(aq) + OH–(aq)22. The formulas of the salts are marked in boldface type. Remember that in an acid– base reaction in aqueous solution, water is always one of the products: keeping this in mind makes predicting the formula of the salt produced easy to do. a. HCl(aq) + RbOH(aq)  H2O(l) + RbCl(aq) b. HClO4(aq) + NaOH(aq)  H2O(l) + NaClO4(aq) c. HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) d. H2SO4(aq) + 2CsOH(aq)  2H2O(l) + Cs2SO4(aq)23. In general, the salt formed in an aqueous acid–base reaction consists of the positive ion of the base involved in the reaction, combined with the negative ion of the acid. The hydrogen ion of the strong acid combines with the hydroxide ion of the strong base to produce water, which is the other product of the acid–base reactions. a. 2NaOH(aq) + H2SO4(aq)  2H2O(l) + Na2SO4(aq) b. RbOH(aq) + HNO3(aq)  H2O(l) + RbNO3(aq) c. KOH(aq) + HClO4(aq)  H2O(l) + KClO4(aq) d. KOH(aq) + HCl(aq)  H2O(l) + KCl(aq)24. A driving force, in general, is an event that tends to help to convert the reactants of a process into the products. Some elements (metals) tend to lose electrons, while other elements (nonmetals) tend to gain electrons. A transfer of electrons from atoms of a metal to atoms of a nonmetal would be favorable, and would result in a chemical reaction. A simple example of such a process is the reaction of sodium with chlorine: sodium atoms tend to each lose one electron (to form Na+), whereas chlorine atoms tend to each gain one electron (to form Cl–). The reaction of sodium metal with chlorine gas represents a transfer of electrons from sodium atoms to chlorine atoms to form sodium chloride.25. The metallic element loses electrons and the nonmetallic element gains electrons.26. Each potassium atom would lose one electron to become a K+ ion. Each sulfur atom would gain two electrons to become a S2– ion. Two potassium atoms would have to be oxidized to provide the two electrons needed to reduce one sulfur atom.3+ – 27. AlBr3 is made up of Al ions and Br ions. Aluminum atoms each lose three electrons to become Al3+ ions. Bromine atoms each gain one electron to become – – Br ions (so each Br2 molecule gains two electrons to become two Br ions).28. a. K + F2  KFBalance fluorine: K + F2  2KFBalance potassium: 2K + F2  2KFBalanced equation: 2K(s) + F2(g)  2KF(s) b. K + O2  K2OBalance oxygen: K + O2  2K2OBalance potassium: 4K + O2  2K2OBalanced equation: 4K(s) + O2(g)  2K2O(s) c. K + N2  K3NBalance nitrogen: K + N2  2K3NBalance potassium: 6K + N2  2K3NBalanced equation: 6K(s) + N2(g)  2K3N(s) d. K + C  K4CBalance potassium: 4K + C  K4CBalanced equation: 4K(s) + C(s)  K4C(s)29. a. Fe(s) + S(s)  Fe2S3(s)Balance iron: 2Fe + S  Fe2S3Balance sulfur: 2Fe + 3S  Fe2S3 Balanced equation: 2Fe(s) + 3S(s)  Fe2S3(s) b. Zn(s) + HNO3(aq)  Zn(NO3)2(aq) + H2(g)Balance nitrate ions: Zn + 2HNO3  Zn(NO3)2 + H2Balanced equation: Zn(s) + 2HNO3(aq)  Zn(NO3)2(aq) + H2(g) c. Sn(s) + O2(g)  SnO(s)Balance oxygen: Sn + O2  2SnOBalance tin: 2Sn + O2  2SnOBalanced equation: 2Sn(s) + O2(g)  2SnO(s) d. K(s) + H2(g)  KH(s)Balance hydrogen: K + H2  2KHBalance potassium: 2K + H2  2KHBalanced equation: 2K(s) + H2(g)  2KH(s) e. Cs(s) + H2O(l)  CsOH(aq) + H2(g)Balance hydrogen: Cs + 2H2O  2CsOH + H2Balance cesium: 2Cs + 2H2O  2CsOH + H2Balanced equation: 2Cs(s) + 2H2O(l)  2CsOH(aq) + H2(g)30. A double-displacement reaction has the form AB + CD  AD + CB. In a double- displacement reaction, when two solutions of ionic solutes are mixed, the positive ions of the two solutes exchange anions (this presupposes that some driving force is present that causes a detectable reaction to occur). Two examples are: Pb(NO3)2(aq) + 2HCl(aq)  PbCl2(s) + 2HNO3(aq)BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq) A single-displacement reaction has the form A + BC  AC + B. In a single- displacement reaction, a new element replaces a less active element in its compound. Two examples are:Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)31. Examples of formation of water:HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)H2SO4(aq) + 2KOH(aq)  2H2O(l) + K2SO4(aq) Examples of formation of a gaseous product:Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)2KClO3(s)  2KCl(s) + 3O2(g) 32. For each reaction, the type of reaction is first identified, followed by some of the reasoning that leads to this choice (there may be more than one way in which you can recognize a particular type of reaction). a. precipitation (BaSO4 is insoluble) b. oxidation–reduction (Zn changes from the elemental to the combined state; hydrogen changes from the combined to the elemental state) c. precipitation (AgCl is insoluble) d. acid–base (HCl is an acid; KOH is a base; water and a salt are produced) e. oxidation–reduction (Cu changes from the combined to the elemental state; Zn changes from the elemental to the combined state)– f. acid–base (the H2PO4 ion behaves as an acid; NaOH behaves as a base; a salt and water are produced) g. precipitation (CaSO4 is insoluble); acid–base [Ca(OH)2 is a base; H2SO4 is an acid; a salt and water are produced] h. oxidation–reduction (Mg changes from the elemental to the combined state; Zn changes from the combined to the elemental state) i. precipitation (BaSO4 is insoluble)33. For each reaction, the type of reaction is first identified, followed by some of the reasoning that leads to this choice (there may be more than one way in which you can recognize a particular type of reaction). a. oxidation–reduction (oxygen changes from the combined state to the elemental state) b. oxidation–reduction (copper changes from the elemental to the combined state; hydrogen changes from the combined to the elemental state) c. acid–base (H2SO4 is a strong acid and NaOH is a strong base; water and a salt are formed) d. acid–base, precipitation (H2SO4 is a strong acid, and Ba(OH)2 is a base; water and a salt are formed; an insoluble product forms) e. precipitation (AgCl is only slightly soluble) f. precipitation (Cu(OH)2 is only slightly soluble) g. oxidation–reduction (chlorine and fluorine change from the elemental to the combined state) h. oxidation–reduction (oxygen changes from the elemental to the combined state) i. acid–base (HNO3 is a strong acid and Ca(OH)2 is a strong base; a salt and water are formed)34. A synthesis reaction represents the production of a given compound from simpler substances (either elements or simpler compounds). For example,O2(g) + 2F2(g)  2OF2(g) represents a simple synthesis reaction. Synthesis reactions may often (but not necessarily always) also be classified in other ways. For example, the reactionC(s) + O2(g)  CO2(g) could also be classified as an oxidation–reduction reaction, or as a combustion reaction (a special subclassification of oxidation–reduction reaction that produces a flame). As another example, the reaction2Fe(s) + 3Cl2(g)  2FeCl3(s) is a synthesis reaction that also is an oxidation–reduction reaction.35. A decomposition reaction is one in which a given compound is broken down into simpler compounds or constituent elements. The reactionsCaCO3(s)  CaO(s) + CO2(g)2HgO(s)  2Hg(l) + O2(g) both represent decomposition reactions. Such reactions often (but not necessarily always) may be classified in other ways. For example, the reaction of HgO(s) is also an oxidation–reduction reaction. 36. Compounds like those in parts b and c of this problem, containing only carbon and hydrogen, are called hydrocarbons. When a hydrocarbon is reacted with oxygen (O2), the hydrocarbon is almost always converted to carbon dioxide and water vapor. Since water molecules contain an odd number of oxygen atoms, whereas O2 contains an even number of oxygen atoms, it is often difficult to balance such equations. For this reason, it is simpler to balance the equation using fractional coefficients if necessary, and then to multiply by a factor that will give whole number coefficients for the final balanced equation. a. C2H5OH(l) + O2(g)  CO2(g) + H2O(g)Balance carbon: C2H5OH(l) + O2(g)  2CO2(g) + H2O(g)Balance hydrogen: C2H5OH(l) + O2(g)  2CO2(g) + 3H2O(g)Balance oxygen: C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)Balanced equation: C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) b. C6H14(l) + O2(g)  CO2(g) + H2O(g)Balance carbon: C6H14(l) + O2(g)  6CO2(g) + H2O(g)Balance hydrogen: C6H14(l) + O2(g)  6CO2(g) + 7H2O(g)Balance oxygen: C6H14(l) + (19/2)O2(g)  6CO2(g) + 7H2O(g)Balanced equation: 2C6H14(l) + 19O2(g)  12CO2(g) + 14H2O(g) c. C6H12(l) + O2(g)  CO2(g) + H2O(g)Balance carbon: C6H12(l) + O2(g)  6CO2(g) + H2O(g)Balance hydrogen: C6H12(l) + O2(g)  6CO2(g) + 6H2O(g)Balanced equation: C6H12(l) + 9O2(g)  6CO2(g) + 6H2O(g)37. a. C2H6(g) + O2(g)  CO2(g) + H2O(g)Balance carbon: C2H6(g) + O2(g)  2CO2(g) + H2O(g)Balance hydrogen: C2H6(g) + O2(g)  2CO2(g) + 3H2O(g)Balance oxygen: C2H6(g) + (7/2)O2(g)  2CO2(g) + 3H2O(g)Balanced equation: 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g) b. C2H6O(l) + O2(g)  CO2(g) + H2O(g)Balance carbon: C2H6O(l) + O2(g)  2CO2(g) + H2O(g)Balance hydrogen: C2H6O(l) + O2(g)  2CO2(g) + 3H2O(g)Balance oxygen: C2H6O(l) + 3O2(g)  2CO2(g) + 3H2O(g)Balanced equation: C2H6O(l) + 3O2(g)  2CO2(g) + 3H2O(g) c. C2H6O2(l) + O2(g)  CO2(g) + H2O(g) Balance carbon: C2H6O2(l) + O2(g)  2CO2(g) + H2O(g)Balance hydrogen: C2H6O2(l) + O2(g)  2CO2(g) + 3H2O(g)Balance oxygen: C2H6O2(l) + (5/2)O2(g)  2CO2(g) + 3H2O(g)Balanced equation: 2C2H6O2(l) + 5O2(g)  4CO2(g) + 6H2O(g)38. a. 2Co(s) + 3S(s)  Co2S3(s) b. 2NO(g) + O2(g)  2NO2(g) c. FeO(s) + CO2(g)  FeCO3(s) d. 2Al(s) + 3F2(g)  2AlF3(s) e. 2NH3(g) + H2CO3(aq)  (NH4)2CO3(s)39. a. 2NI3(s)  N2(g) + 3I2(s) b. BaCO3(s)  BaO(s) + CO2(g) c. C6H12O6(s)  6C(s) + 6H2O(g) d. Cu(NH3)4SO4(s)  CuSO4(s) + 4NH3(g) e. 3NaN3(s)  Na3N(s) + 4N2(g) 40. A molecular equation uses the normal, uncharged formulas for the compounds involved. The complete ionic equation shows the compounds involved broken up into their respective ions (all ions present are shown). The net ionic equation shows only those ions that combine to form a precipitate, a gas, or a nonionic product such as water. The net ionic equation shows most clearly the species that are combining with each other. 41. In several cases, the given ion may be precipitated by many reactants. The following are only three of the possible examples. a. Chloride ion would precipitate when treated with solutions containing silver ion, lead(II) ion, or mercury(I) ion. Ag+(aq) + Cl–(aq)  AgCl(s) 2+ – Pb (aq) + 2Cl (aq)  PbCl2(s) 2+ – Hg2 (aq) + 2Cl (aq)  Hg2Cl2(s) b. Calcium ion would precipitate when treated with solutions containing sulfate ion, carbonate ion, and phosphate ion. 2+ 2– Ca (aq) + SO4 (aq)  CaSO4(s) 2+ 2– Ca (aq) + CO3 (aq)  CaCO3(s) 2+ 3– 3Ca (aq) + 2PO4 (aq)  Ca3(PO4)2(s) c. Iron(III) ion would precipitate when treated with solutions containing hydroxide, sulfide,or carbonate ions. 3+ – Fe (aq) + 3OH (aq)  Fe(OH)3(s) 3+ – 2Fe (aq) + 3S (aq)  Fe2S3(s) 3+ 2– 2Fe (aq) + 3CO3 (aq)  Fe2(CO3)3(s) d. Sulfate ion would precipitate when treated with solutions containing barium ion, calcium ion, or lead(II) ion. 2+ 2– Ba (aq) + SO4 (aq)  BaSO4(s) 2+ 2– Ca (aq) + SO4 (aq)  CaSO4(s) 2+ 2– Pb (aq) + SO4 (aq)  PbSO4(s) e. mercury(I) ion would precipitate when treated with solutions containing chloride ion, sulfide ion, or carbonate ion.2+ – Hg2 (aq) + 2Cl (aq)  Hg2Cl2(s)2+ 2– Hg2 (aq) + S (aq)  Hg2S(s)2+ 2– Hg2 (aq) + CO3 (aq)  Hg2CO3(s) f. silver ion would precipitate when treated with solutions containing chloride ion, sulfide ion, or carbonate ion. Ag+(aq) + Cl–(aq)  AgCl(s)+ 2– 2Ag (aq) + S (aq)  Ag2S(s)+ 2– 2Ag (aq) + CO3 (aq)  Ag2CO3(s)42. a. no precipitate b. Cu2+(aq) + S2–(aq)  CuS(s)2+ – c. Pb (aq) + 2Cl (aq)  PbCl2(s)2+ 2– d. Ca (aq) + CO3 (aq)  CaCO3(s)3+ – e. Au (aq) + 3OH (aq)  Au(OH)3(s)43. The formulas of the salts are indicated in boldface type. a. HNO3(aq) + KOH(aq)  H2O(l) + KNO3(aq) b. H2SO4(aq) + Ba(OH)2(aq)  2H2O(l) + BaSO4(s) c. HClO4(aq) + NaOH(aq)  H2O(l) + NaClO4(aq) d. 2HCl(aq) + Ca(OH)2(aq)  2H2O(l) + CaCl2(aq)44. For each cation, the precipitates that form with the anions listed are given below. If no formula is listed, it should be assumed that that anion does not form a precipitate with the particular cation.+ Ag ion: AgCl, Ag2CO3, AgOH, Ag3PO4, Ag2S, Ag2SO4 2+ Ba ion: BaCO3, Ba(OH)2, Ba3(PO4)2, BaS, BaSO4 2+ Ca ion: CaCO3, Ca(OH)2, Ca3(PO4)2, CaS, CaSO4 3+ Fe ion: Fe2(CO3)3, Fe(OH)3, FePO4, Fe2S3 2+ Hg2 ion: Hg2Cl2, Hg2CO3, Hg2(OH)2, (Hg2)3(PO4)2, Hg2S Na+ ion: all common salts are soluble2+ Ni ion: NiCO3, Ni(OH)2, Ni3(PO4)2, NiS 2+ Pb ion: PbCl2, PbCO3, Pb(OH)2, Pb3(PO4)2, PbS, PbSO445. The precipitates are marked in boldface type. a. Rule 3: AgCl is listed as an exception.AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq) b. Rule 6: most cabonate salts are only slightly soluble.CuSO4(aq) + (NH4)2CO3(aq)  CuCO3(s) + (NH4)2SO4(aq) c. Rule 6: most carbonate salts are only slightly soluble.FeSO4(aq) + K2CO3(aq)  FeCO3(s) + K2SO4(aq) d. no reaction e. Rule 6: most carbonate salts are only slightly soluble.Pb(NO3)2(aq) + Li2CO3(aq)  PbCO3(s) + 2LiNO3(aq) f. Rule 5: most hydroxide compounds are only slightly soluble.SnCl4(aq) + 4NaOH(aq)  Sn(OH)4(s) + 4NaCl(aq)46. a. Rule 3: Ag+(aq) + Cl–(aq)  AgCl(s)2+ 3– b. Rule 6: 3Ca (aq) + 2PO4 (aq)  Ca3(PO4)2(s)2+ – c. Rule 3: Pb (aq) + 2Cl (aq)  PbCl2(s)3+ – d. Rule 6: Fe (aq) + 3OH (aq)  Fe(OH)3(s)47. For simplicity, the physical states of the substances are omitted. 2Ba + O2  2BaO Ba + S  BaSBa + Cl2  BaCl2 3Ba + N2  Ba3N2Ba + Br2  BaBr2 4K + O2  2K2O2K + S  K2S 2K + Cl2  2KCl6K + N2  2K3N 2K + Br2  2KBr 2Mg + O2  2MgO Mg + S  MgSMg + Cl2  MgCl2 3Mg + N2  Mg3N2Mg + Br2  MgBr2 4Rb + O2  2Rb2O2Rb + S  Rb2S 2Rb + Cl2  2RbCl6Rb + N2  2Rb3N 2Rb + Br2  2RbBr2Ca + O2  2CaO Ca + S  CaSCa + Cl2  CaCl2 3Ca + N2  Ca3N2Ca + Br2  CaBr2 4Li + O2  2Li2O2Li + S  Li2S 2Li + Cl2  2LiCl6Li + N2  2Li3N 2Li + Br2  2LiBr48. a. one b. one c. two d. two e. three49. a. two; O + 2e–  O2– b. one; F + e–  F– c. three; N + 3e–  N3– d. one; Cl + e–  Cl– e. two; S + 2e–  S2– 50. a. 2C3H8O(l) + 9O2(g)  6CO2(g) + 8H2O(g) oxidation–reduction, combustion b. HCl(aq) + AgC2H3O2(aq)  AgCl(s) + HC2H3O2(aq) precipitation, double-displacement c. 3HCl(aq) + Al(OH)3(s)  AlCl3(aq) + 3H2O(l) acid–base, double-displacement d. 2H2O2(aq)  2H2O(l) + O2(g) oxidation–reduction, decomposition e. N2H4(l) + O2(g)  N2(g) + 2H2O(g) oxidation–reduction, combustion51. a. acid–base b. precipitation c. oxidation–reduction, combustion, synthesis d. oxidation–reduction, decomposition

Chapter 8 Reactions in Aqueous Solutions

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