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TABLE OF CONTENTS Page How to Use This Guide v Chapter Atoms, Molecules, and Ions Chapter Stoichiometry 19 Chapter Types of Chemical Reactions and Solution Stoichiometry 59 Chapter Gases 108 Chapter Chemical Equilibrium 161 Chapter Acids and Bases 207 Chapter Applications of Aqueous Equilibria 273 Chapter Energy, Enthalpy, and Thermochemisty 375 Chapter 10 Spontaneity, Entropy, and Free Energy .394 Chapter 11 Electrochemistry 442 Chapter 12 Quantum Mechanics and Atomic Theory 489 Chapter 13 Bonding: General Concepts 527 Chapter 14 Covalent Bonding: Orbitals 583 Chapter 15 Chemical Kinetics 624 Chapter 16 Liquids and Solids 677 Chapter 17 Properties of Solutions 723 Chapter 18 The Representative Elements .763 Chapter 19 Transition Metals and Coordination Chemistry 791 Chapter 20 The Nucleus: A Chemist’s View 823 Chapter 21 Organic Chemistry .843 iii CHAPTER ATOMS, MOLECULES, AND IONS Development of the Atomic Theory 18 Law of conservation of mass: mass is neither created nor destroyed The total mass before a chemical reaction always equals the total mass after a chemical reaction Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass For example, water is always g hydrogen for every g oxygen Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with g of the first element always can be reduced to small whole numbers For CO2 and CO discussed in section 2.2, the mass ratios of oxygen that react with g carbon in each compound are in a : ratio 19 From Avogadro’s hypothesis (law), volume ratios are equal to molecule ratios at constant temperature and pressure Therefore, we can write a balanced equation using the volume data, Cl2 + F2 → X Two molecules of X contain 10 atoms of F and two atoms of Cl The formula of X is ClF5 for a balanced equation 20 a The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed b Avogadro’s hypothesis (law) implies that volume ratios are proportional to molecule ratios at constant temperature and pressure H2 + Cl2 → HCl From the balanced equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted 21 Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at constant temperature and pressure Here, volume of N2 reacts with volumes of H2 to produce volumes of the gaseous product or in terms of molecule ratios: N2 + H2 → product In order for the equation to be balanced, the product must be NH3 22 For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with g of carbon From the formulas (two oxygen atoms per carbon atom in CO2 versus one oxygen atom per carbon atom in CO), CO2 will have twice the mass of oxygen that combines per gram of carbon as compared to CO For CO2 and C3O2, it is easiest to concentrate on the CHAPTER ATOMS, MOLECULES, AND IONS mass of carbon that combines with g of oxygen From the formulas (three carbon atoms per two oxygen atoms in C3O2 versus one carbon atom per two oxygen atoms in CO2), C3O2 will have three times the mass of carbon that combines per gram of oxygen as compared to CO2 As expected, the mass ratios are whole numbers as predicted by the law of multiple proportions 23 Hydrazine: 1.44 × 10 −1 g H/g N; ammonia: 2.16 × 10 −1 g H/g N; hydrogen azide: 2.40 × 10 −2 g H/g N Let's try all of the ratios: 0.144 0.216 0.216 0.0240 = 6.00; = 9.00; = 1.00; = 1.50 = 0.0240 0.0240 0.0240 0.144 All the masses of hydrogen in these three compounds can be expressed as simple wholenumber ratios The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios : : 24 Compound 1: 21.8 g C and 58.2 g O (80.0 – 21.8 = mass O) Compound 2: 34.3 g C and 45.7 g O (80.0 – 34.3 = mass O) The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: 21.8 g C = 0.375 g C/g O 58.2 g O Compound 2: 34.3 g C = 0.751 g C/g O 45.7 g O 0.751 = ; this 0.375 supports the law of multiple proportions because this carbon ratio is a small whole number The ratio of the masses of carbon that combine with g of oxygen is 25 To get the atomic mass of H to be 1.00, we divide the mass that reacts with 1.00 g of oxygen by 0.126, that is, 0.126/0.126 = 1.00 To get Na, Mg, and O on the same scale, we the same division Na: 1.500 1.00 2.875 = 22.8; Mg: = 11.9; O: = 7.94 0.126 0.126 0.126 H O Na Mg Relative value 1.00 7.94 22.8 11.9 Accepted value 1.0079 15.999 22.99 24.31 The atomic masses of O and Mg are incorrect The atomic masses of H and Na are close Something must be wrong about the assumed formulas of the compounds It turns out that the correct formulas are H2O, Na2O, and MgO The smaller discrepancies result from the error in the assumed atomic mass of H CHAPTER ATOMS, MOLECULES, AND IONS The Nature of the Atom 26 Deflection of cathode rays by magnetic and electrical fields led to the conclusion that they were negatively charged The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electrical field β particles are electrons A cathode ray is a stream of electrons (β particles) 27 From section 2.6, the nucleus has “a diameter of about 10−13 cm” and the electrons “move about the nucleus at an average distance of about 10−8 cm from it.” We will use these statements to help determine the densities Density of hydrogen nucleus (contains one proton only): 4 Vnucleus = π r = (3.14) (5 × 10 −14 cm) = × 10 − 40 cm 3 d = density = 1.67 × 10 −24 g = × 1015 g/cm 3 − 40 × 10 cm Density of H atom (contains one proton and one electron): Vatom = d= 28 (3.14) (1 × 10 −8 cm) = × 10 − 24 cm 3 1.67 × 10 −24 g + × 10 −28 g = 0.4 g/cm 3 − 24 × 10 cm From Section 2.6 of the text, the average diameter of the nucleus is approximately 10−13 cm, and the electrons move about the nucleus at an average distance of approximately 10 −8 cm From this, the diameter of an atom is about × 10 −8 cm × 10 −8 cm × 10 −13 cm = × 105; mi 5280 ft 63,360 in = = grape grape grape Because the grape needs to be × 105 times smaller than a mile, the diameter of the grape would need to be 63,360/(2 × 105) ≈ 0.3 in This is a reasonable size for a small grape 29 First, divide all charges by the smallest quantity, 6.40 × 10−13 2.56 × 10 −12 = 4.00; 6.40 × 10 −13 7.68 = 12.00; 0.640 3.84 = 6.00 0.640 Because all charges are whole-number multiples of 6.40 × 10−13 zirkombs, the charge on one electron could be 6.40 × 10−13 zirkombs However, 6.40 × 10−13 zirkombs could be the charge of two electrons (or three electrons, etc.) All one can conclude is that the charge of an electron is 6.40 × 10−13 zirkombs or an integer fraction of 6.40 × 10−13 30 The proton and neutron have similar mass, with the mass of the neutron slightly larger than that of the proton Each of these particles has a mass approximately 1800 times greater than that of an electron The combination of the protons and the neutrons in the nucleus makes up the bulk of the mass of an atom, but the electrons make the greatest contribution to the chemical properties of the atom 31 CHAPTER ATOMS, MOLECULES, AND IONS If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout), then α particles should have traveled through the thin foil with very minor deflections in their path This was not the case because a few of the α particles were deflected at very large angles Rutherford reasoned that the large deflections of these α particles could be caused only by a center of concentrated positive charge that contains most of the atom’s mass (the nuclear model of the atom) Elements, Ions, and the Periodic Table 32 a A molecule has no overall charge (an equal number of electrons and protons are present) Ions, on the other hand, have electrons added to form anions (negatively charged ions) or electrons removed to form cations (positively charged ions) b The sharing of electrons between atoms is a covalent bond An ionic bond is the force of attraction between two oppositely charged ions c A molecule is a collection of atoms held together by covalent bonds A compound is composed of two or more different elements having constant composition Covalent and/or ionic bonds can hold the atoms together in a compound Another difference is that molecules not necessarily have to be compounds H2 is two hydrogen atoms held together by a covalent bond H2 is a molecule, but it is not a compound; H2 is a diatomic element d An anion is a negatively charged ion, for example, Cl−, O2−, and SO42− are all anions A cation is a positively charged ion, for example, Na+, Fe3+, and NH4+ are all cations 33 The atomic number of an element is equal to the number of protons in the nucleus of an atom of that element The mass number is the sum of the number of protons plus neutrons in the nucleus The atomic mass is the actual mass of a particular isotope (including electrons) As is discussed in Chapter 3, the average mass of an atom is taken from a measurement made on a large number of atoms The average atomic mass value is listed in the periodic table 34 a Metals: Mg, Ti, Au, Bi, Ge, Eu, and Am Nonmetals: Si, B, At, Rn, and Br b Si, Ge, B, and At The elements at the boundary between the metals and the nonmetals are B, Si, Ge, As, Sb, Te, Po, and At Aluminum has mostly properties of metals, so it is generally not classified as a metalloid 35 a The noble gases are He, Ne, Ar, Kr, Xe, and Rn (helium, neon, argon, krypton, xenon, and radon) Radon has only radioactive isotopes In the periodic table, the whole number enclosed in parentheses is the mass number of the longest-lived isotope of the element b promethium (Pm) and technetium (Tc) 36 Carbon is a nonmetal Silicon and germanium are called metalloids as they exhibit both metallic and nonmetallic properties Tin and lead are metals Thus metallic character increases as one goes down a family in the periodic table The metallic character decreases from left to right across the periodic table CHAPTER 37 ATOMS, MOLECULES, AND IONS Use the periodic table to identify the elements a Cl; halogen b Be; alkaline earth metal c Eu; lanthanide metal d Hf; transition metal e He; noble gas f U; actinide metal g Cs; alkali metal 38 The number and arrangement of electrons in an atom determine how the atom will react with other atoms, i.e., the electrons determine the chemical properties of an atom The number of neutrons present determines the isotope identity and the mass number 39 For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number of neutrons When the number of protons and neutrons is equal to each other, the mass number (protons + neutrons) will be twice the atomic number (protons) Therefore, for lighter isotopes, the ratio of the mass number to the atomic number is close to For example, consider 28Si, which has 14 protons and (28 – 14 =) 14 neutrons Here, the mass number to atomic number ratio is 28/14 = 2.0 For heavier isotopes, there are more neutrons than protons in the nucleus Therefore, the ratio of the mass number to the atomic number increases steadily upward from as the isotopes get heavier and heavier For example, 238U has 92 protons and (238 – 92 =) 146 neutrons The ratio of the mass number to the atomic number for 238U is 238/92 = 2.6 40 a transition metals b alkaline earth metals d noble gases e halogens 41 42 a 235 92 U: d 208 82 Pb: 82 p, 126 n, 82 e b 27 13 Al: 13 p, 14 n, 13 e c 57 26 Fe: 26 p, 31 n, 26 e e 86 37 Rb: 37 p, 49 n, 37 e f 41 20 Ca: 20 p, 21 n, 20 e a Cobalt is element 27 A = mass number = 27 + 31 = 58; 58 Co 27 b 43 92 p, 143 n, 92 e c alkali metals 10 B c 23 Mg 12 d 132 I 53 19 F e f a 24 Mg: 12 b 24 Mg2+: 12 12 p, 12 n, 10 e c 59 Co2+: 27 d 59 Co3+: 27 27 p, 32 n, 24 e e 59 Co: 27 f 79 Se: 34 34 p, 45 n, 34 e g 79 2− Se : 34 34 p, 45 n, 36 e h 63 Ni: 28 28 p, 35 n, 28 e i 59 2+ Ni : 28 28 p, 31 n, 26 e 12 protons, 12 neutrons, 12 electrons 27 p, 32 n, 25 e 27 p, 32 n, 27 e 65 Cu 29 CHAPTER ATOMS, MOLECULES, AND IONS 44 Number of Protons in Nucleus Number of Neutrons in Nucleus Number of Electrons Net Charge 238 92 U 92 146 92 40 2+ Ca 20 20 20 18 2+ 51 3+ V 23 23 28 20 3+ 89 Y 39 39 50 39 79 − Br 35 35 44 36 1− 31 3− P 15 15 16 18 3− Symbol 45 Atomic number = 63 (Eu); net charge = +63 − 60 = 3+; mass number = 63 + 88 = 151; 3+ symbol: 151 63 Eu Atomic number = 50 (Sn); mass number = 50 + 68 = 118; net charge = +50 − 48 = 2+; 2+ symbol: 118 50 Sn 46 Atomic number = 16 (S); net charge = +16 − 18 = 2−; mass number = 16 + 18 = 34; symbol: 34 S2− 16 Atomic number = 16 (S); net charge = +16 −18 = 2−; Mass number = 16 + 16 = 32; symbol: 32 S2− 16 47 48 In ionic compounds, metals lose electrons to form cations, and nonmetals gain electrons to form anions Group 1A, 2A, and 3A metals form stable 1+, 2+, and 3+ charged cations, respectively Group 5A, 6A, and 7A nonmetals form 3−, 2−, and 1− charged anions, respectively a Lose e − to form Ra2+ b Lose e − to form In3+ c Gain e − to form P 3− d Gain e − to form Te 2− e Gain e − to form Br− f Lose e − to form Rb+ See Exercise 47 for a discussion of charges various elements form when in ionic compounds a Element 13 is Al Al forms 3+ charged ions in ionic compounds Al3+ b Se2− c Ba2+ d N3− e Fr+ f Br− CHAPTER ATOMS, MOLECULES, AND IONS Nomenclature 49 AlCl3, aluminum chloride; CrCl3, chromium(III) chloride; ICl3, iodine trichloride; AlCl3 and CrCl3 are ionic compounds following the rules for naming ionic compounds The major difference is that CrCl3 contains a transition metal (Cr) that generally exhibits two or more stable charges when in ionic compounds We need to indicate which charged ion we have in the compound This is generally true whenever the metal in the ionic compound is a transition metal ICl3 is made from only nonmetals and is a covalent compound Predicting formulas for covalent compounds is extremely difficult Because of this, we need to indicate the number of each nonmetal in the binary covalent compound The exception is when there is only one of the first species present in the formula; when this is the case, mono- is not used (it is assumed) 50 a Dinitrogen monoxide is correct N and O are both nonmetals resulting in a covalent compound We need to use the covalent rules of nomenclature The other two names are for ionic compounds b Copper(I) oxide is correct With a metal in a compound, we have an ionic compound Because copper, like most transition metals, forms at least a couple of different stable charged ions, we must indicate the charge on copper in the name Copper oxide could be CuO or Cu2O, hence why we must give the charge of most transition metal compounds Dicopper monoxide is the name if this were a covalent compound, which it is not c Lithium oxide is correct Lithium forms 1+ charged ions in stable ionic compounds Because lithium is assumed to form 1+ ions in compounds, we not need to indicate the charge of the metal ion in the compound Dilithium monoxide would be the name if Li2O were a covalent compound (a compound composed of only nonmetals) 51 52 53 54 a mercury(I) oxide b iron(III) bromide c cobalt(II) sulfide d titanium(IV) chloride e Sn3N2 f g HgO h CrS3 a barium sulfite b sodium nitrite c potassium permanganate d potassium dichromate e Cr(OH)3 f g Pb(CO3)2 h NH4C2H3O2 CoI3 Mg(CN)2 a sulfur difluoride b dinitrogen tetroxide c iodine trichloride d tetraphosphorus hexoxide a sodium perchlorate b magnesium phosphate c aluminum sulfate d sulfur difluoride 55 56 57 58 CHAPTER e sulfur hexafluoride f g sodium dihydrogen phosphate h lithium nitride i sodium hydroxide j magnesium hydroxide k aluminum hydroxide l silver chromate sodium hydrogen phosphate a copper(I) iodide b copper(II) iodide d sodium carbonate e sodium hydrogen carbonate or sodium bicarbonate f tetrasulfur tetranitride g selenium tetrabromide i barium chromate j c cobalt(II) iodide h sodium hypochlorite ammonium nitrate a acetic acid b ammonium nitrite c colbalt(III) sulfide d iodine monochloride e lead(II) phosphate f potassium chlorate g sulfuric acid h strontium nitride i aluminum sulfite j k sodium chromate l hypochlorous acid tin(IV) oxide a SO2 b SO3 c Na2SO3 d KHSO3 e Li3N f g Cr(C2H3O2)2 h SnF4 Cr2(CO3)3 i NH4HSO4: composed of NH4+ and HSO4− ions j (NH4)2HPO4 k KClO4 l NaH m HBrO n HBr a Na2O b Na2O2 c KCN d Cu(NO3)2 e SiCl4 f PbO g PbO2 h CuCl i GaAs: We would predict the stable ions to be Ga3+ and As3− j CdSe m HNO2 59 ATOMS, MOLECULES, AND IONS k ZnS l Hg2Cl2: Mercury(I) exists as Hg22+ n P2O5 a Pb(C2H3O2)2; lead(II) acetate b CuSO4; copper(II) sulfate c CaO; calcium oxide d MgSO4; magnesium sulfate e Mg(OH)2; magnesium hydroxide f CaSO4; calcium sulfate g N2O; dinitrogen monoxide or nitrous oxide (common name) 60 a Iron forms 2+ and 3+ charged ions; we need to include a Roman numeral for iron Iron(III) chloride is correct b This is a covalent compound so use the covalent rules Nitrogen dioxide is correct c This is an ionic compound, so use the ionic rules Calcium oxide is correct Calcium only forms stable 2+ ions when in ionic compounds, so no Roman numeral is needed 84 CHAPTER SOLUTION STOICHIOMETRY 16 H2O + S2− + MnO4− → S + MnS + H2O + 16 OH− Reducing gives: H2O(l) + S2−(aq) + MnO4−(aq) → S(s) + MnS(s) + 16 OH−(aq) e CN− → CNO− (H2O + CN− → CNO− + H+ + e−) × MnO4− → MnO2 (3 e + H + MnO4− → MnO2 + H2O) × − + Common factor is a transfer of electrons H2O + CN− → CNO- + H+ + e− e + H+ + MnO4− → MnO2 + H2O − OH− + H+ + CN− + MnO4− → CNO− + MnO2 + H2O + OH− Reducing gives: H2O(l) + CN−(aq) + MnO4−(aq) → CNO−(aq) + MnO2(s) + OH−(aq) 83 a HCl(aq) dissociates to H+(aq) + Cl−(aq) For simplicity, let's use H+ and Cl− separately H+ → H2 (2 H+ + e− → H2) × Fe → HFeCl4 (H + Cl + Fe → HFeCl4 + e−) × + − H+ + e− → H2 H + Cl− + Fe → HFeCl4 + e− + H+ + Cl− + Fe → HFeCl4 + H2 or b HCl(aq) + Fe(s) → HFeCl4(aq) + H2(g) IO3− → I3− IO3− → I3− IO3− → I3− + H2O − + 16 e + 18 H + IO3− → I3− + H2O I− → I3− (3 I− → I3− + e−) × 16 e− + 18 H+ + IO3− → I3− + H2O 24 I− → I3− + 16 e− 18 H+ + 24 I− + IO3− → I3− + H2O Reducing: H+(aq) + I−(aq) + IO3−(aq) → I3−(aq) + H2O(l) CHAPTER SOLUTION STOICHIOMETRY 85 c (Ce4+ + e− → Ce3+) × 97 Cr(NCS)64− → Cr3+ + NO3− + CO2 + SO42− 54 H2O + Cr(NCS)64− → Cr3+ + NO3− + CO2 + SO42− + 108 H+ Charge on left = −4 Charge on right = +3 + 6(−1) + 6(−2) + 108(+1) = +93 Add 97 e− to the product side, and then add the two balanced half-reactions with a common factor of 97 e− transferred 54 H2O + Cr(NCS)64− → Cr3+ + NO3− + CO2 + SO42− + 108 H+ + 97 e− 97 e− + 97 Ce4+ → 97 Ce3+ 97 Ce4+(aq) + 54 H2O(l) + Cr(NCS)64−(aq) → 97 Ce3+(aq) + Cr3+(aq) + NO3−(aq) + CO2(g) + SO42−(aq) + 108 H+(aq) This is very complicated A check of the net charge is a good check to see if the equation is balanced Left: charge = 97(+4) −4 = +384 Right: charge = 97(+3) + + 6(−1) + 6(−2) + 108(+1) = +384 d CrI3 → CrO42− + IO4− (16 H2O + CrI3 → CrO42− + IO4− + 32 H+ + 27 e−) × Cl2 → Cl− (2 e− + Cl2 → Cl−) × 27 Common factor is a transfer of 54 e− 54 e− + 27 Cl2 → 54 Cl− 32 H2O + CrI3 → CrO42− + IO4− + 64 H+ + 54 e− 32 H2O + CrI3 + 27 Cl2 → 54 Cl− + CrO42− + IO4− + 64 H+ Add 64 OH− to both sides and convert 64 H+ into 64 H2O 64 OH− + 32 H2O + CrI3 + 27 Cl2 → 54 Cl− + CrO42− + IO4− + 64 H2O Reducing gives: 64 OH−(aq) + CrI3(s) + 27 Cl2(g) → 54 Cl−(aq) + CrO42− (aq) + IO4−(aq) + 32 H2O(l) e Ce4+ → Ce(OH)3 (e + H2O + Ce4+ → Ce(OH)3 + H+) × 61 − Fe(CN)64− → Fe(OH)3 + CO32− + NO3− Fe(CN)64− → Fe(OH)3 + CO32− + NO3− There are 39 extra O atoms on right Add 39 H2O to left, then add 75 H+ to right to balance H+ 39 H2O + Fe(CN)64− → Fe(OH)3 + CO32− + NO3− + 75 H+ net charge = 4− net charge = 57+ Add 61 e− to the product side, and then add the two balanced half-reactions with a common factor of 61 e− transferred 86 CHAPTER SOLUTION STOICHIOMETRY 39 H2O + Fe(CN)64− → Fe(OH)3 + CO3− + NO3− + 75 H+ + 61 e− 61 e− + 183 H2O + 61 Ce4+ → 61 Ce(OH)3 + 183 H+ 222 H2O + Fe(CN)64− + 61 Ce4+ → 61 Ce(OH)3 + Fe(OH)3 + CO32− + NO3− + 258 H+ Adding 258 OH− to each side, and then reducing gives: 258 OH−(aq) + Fe(CN)64−(aq) + 61 Ce4+(aq) → 61 Ce(OH)3(s) + Fe(OH)3(s) + CO32−(aq) + NO3−(aq) + 36 H2O(l) 84 Mn → Mn2+ + e− HNO3 → NO2 HNO3 → NO2 + H2O (e− + H+ + HNO3 → NO2 + H2O) × Mn → Mn2+ + e− e− + H+ + HNO3 → NO2 + H2O H+(aq) + Mn(s) + HNO3(aq) → Mn2+(aq) + NO2(g) + H2O(l) or H+(aq) + Mn(s) + NO3−(aq) → Mn2+(aq) + NO2(g) + H2O(l) (HNO3 is a strong acid.) (4 H2O + Mn2+ → MnO4− + H+ + e−) × (2 e− + H+ + IO4− → IO3− + H2O) × H2O + Mn2+ → MnO4− + 16 H+ + 10 e− 10 e + 10 H+ + IO4− → IO3− + H2O − H2O(l) + Mn2+(aq) + IO4−(aq) → MnO4−(aq) + IO3−(aq) + H+(aq) 85 (H2C2O4 → CO2 + H+ + e−) × (5 e− + H+ + MnO4− → Mn2+ + H2O) × H2C2O4 → 10 CO2 + 10 H+ + 10 e− 10 e− + 16 H+ + MnO4− → Mn2+ + H2O H+(aq) + H2C2O4(aq) + MnO4−(aq) → 10 CO2(g) + Mn2+(aq) + H2O(l) − 0.1058 g H2C2O4 × mol H C O mol MnO × = 4.700 × 10−4 mol MnO4− 90.034 g mol H C O − Molarity = 86 4.700 × 10 −4 mol MnO 1000 mL = 1.622 × 10−2 M MnO4− × 28.97 mL L a (Fe2+ → Fe3+ + e−) × 5 e− + H+ + MnO4− → Mn2+ + H2O The balanced equation is: H+(aq) + MnO4−(aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq) + H2O(l) CHAPTER SOLUTION STOICHIOMETRY 87 − 0.0216 mol MnO mol Fe 2+ × = 2.23 × 10−3 mol Fe2+ 20.62 × 10 L soln × − L so ln mol MnO −3 Molarity = 2.23 × 10 −3 mol Fe 2+ = 4.46 × 10−2 M Fe2+ 50.00 × 10 −3 L b (Fe2+ → Fe3+ + e−) × 6 e− + 14 H+ + Cr2O72− → Cr3+ + H2O The balanced equation is: 14 H+(aq) + Cr2O72−(aq) + Fe2+(aq) → Fe3+(aq) + Cr3+(aq) + H2O(l) 2− 50.00 × 10−3 L × 4.46 × 10 −2 mol Fe 2+ mol Cr2 O × L mol Fe 2+ × 1L 0.0150 mol Cr2 O 2− = 2.48 × 10−2 L or 24.8 mL (Fe2+ → Fe3+ + e−) × 5 e + H + MnO4− → Mn2+ + H2O 87 − + H+(aq) + MnO4−(aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq) + H2O(l) From the titration data we can get the number of moles of Fe2+ We then convert this to a mass of iron and calculate the mass percent of iron in the sample − 38.37 × 10−3 L MnO4− × 3.80 × 10−3 mol Fe × Mass % Fe = 88 0.0198 mol MnO mol Fe 2+ × = 3.80 × 10−3 mol Fe2+ − L mol MnO = 3.80 × 10−3 mol Fe present 55.85 g Fe = 0.212 g Fe mol Fe 0.212 g × 100 = 34.6% Fe 0.6128 g The unbalanced reaction is: VO2+ + MnO4− → V(OH)4+ + Mn2+ This is a redox reaction in acidic solution and must be balanced accordingly The two halfreactions to balance are: VO2+ → V(OH)4+ and MnO4− → Mn2+ Balancing by the half-reaction method gives: MnO4−(aq) + VO2+(aq) + 11 H2O(l) → V(OH)4+(aq) + Mn2+(aq) + H+(aq) − 0.02645 L × 0.581 = 0.02250 mol MnO mol VO 2+ mol V 50.94 g V × × × = 0.1516 g V − 2+ L mol V mol VO mol MnO 0.1516 g V , 0.1516/0.581 = 0.261 g ore sample mass of ore sample 88 89 CHAPTER Mg(s) + HCl(aq) → MgCl2(aq) + H2(g) 3.00 g Mg × 90 SOLUTION STOICHIOMETRY a mol HCl L HCl mol Mg × × = 0.0494 L = 49.4 mL HCl 24.31 g Mg mol Mg 5.0 mol HCl 16 e− + 18 H+ + IO3− → I3− + H2O (3 I− → I3− + e−) × 24 I− → I3− + 16 e− 16 e + 18 H + IO3− → I3− + H2O − + 18 H+ + 24 I− + IO3− → I3− + H2O Reducing: H+(aq) + I−(aq) + IO3−(aq) → I3−(aq) + H2O(l) b 0.6013 g KIO3 × mol KIO3 = 2.810 × 10−3 mol KIO3 214.0 g KIO3 2.810 × 10−3 mol KIO3 × mol KI 166.0 g KI × = 3.732 g KI mol KIO3 mol KI 2.810 × 10−3 mol KIO3 × mol HCl 1L × = 5.62 × 10−3 L = 5.62 mL HCl mol KIO3 3.00 mol HCl c I3− + e− → I− S2O32− → S4O62− + e− Adding the balanced half-reactions gives: S2O32−(aq) + I3−(aq) → I−(aq) + S4O62−(aq) − d 25.00 × 10−3 L KIO3 × 0.0100 mol KIO3 mol I mol Na 2S2 O × × = − L mol KIO3 mol I 1.50 × 10−3 mol Na2S2O3 M Na 2S2O3 = e 0.5000 L × 1.50 × 10 −3 mol = 0.0468 M Na2S2O3 32.04 × 10 −3 L 0.0100 mol KIO3 214.0 g KIO3 × = 1.07 g KIO3 L mol KIO3 Place 1.07 g KIO3 in a 500-mL volumetric flask; add water to dissolve the KIO3; continue adding water to the 500.0-mL mark, with mixing along the way Additional Exercises 91 Mol CaCl2 present = 0.230 L CaCl2 × 0.275 mol CaCl = 6.33 × 10−2 mol CaCl2 L CaCl The volume of CaCl2 solution after evaporation is: CHAPTER SOLUTION STOICHIOMETRY 6.33 × 10−2 mol CaCl2 × 89 L CaCl = 5.75 × 10−2 L = 57.5 mL CaCl2 1.10 mol CaCl Volume H2O evaporated = 230 mL − 57.5 mL = 173 mL H2O evaporated 92 There are other possible correct choices for the following answers We have listed only three possible reactants in each case a AgNO3, Pb(NO3)2, and Hg2(NO3)2 would form precipitates with the Cl− ion Ag+(aq) + Cl−(aq) → AgCl(s); Pb2+(aq) + Cl-(aq) → PbCl2(s) Hg22+(aq) + Cl−(aq) → Hg2Cl2(s) b Na2SO4, Na2CO3, and Na3PO4 would form precipitates with the Ca2+ ion Ca2+(aq) + SO42−(aq) → CaSO4(s); Ca2+(aq) + CO32−(aq) → CaCO3(s) Ca2+(aq) + PO43−(aq) → Ca3(PO4)2(s) c NaOH, Na2S, and Na2CO3 would form precipitates with the Fe3+ ion Fe3+(aq) + OH−(aq) → Fe(OH)3(s); Fe3+(aq) + S2−(aq) → Fe2S3(s) Fe3+(aq) + CO32−(aq) → Fe2(CO3)3(s) d BaCl2, Pb(NO3)2, and Ca(NO3)2 would form precipitates with the SO42− ion Ba2+(aq) + SO42−(aq) → BaSO4(s); Pb2+(aq) + SO42−(aq) → PbSO4(s) Ca2+(aq) + SO42−(aq) → CaSO4(s) e Na2SO4, NaCl, and NaI would form precipitates with the Hg22+ ion Hg22+ (aq) + SO42−(aq) → Hg2SO4(s); Hg22+(aq) + Cl−(aq) → Hg2Cl2(s) Hg22+ (aq) + I−(aq) → Hg2I2(s) f NaBr, Na2CrO4, and Na3PO4 would form precipitates with the Ag+ ion Ag+(aq) + Br−(aq) → AgBr(s); Ag+(aq) + CrO42−(aq) → Ag2CrO4(s) Ag+(aq) + PO43−(aq) → Ag3PO4(s) 93 a MgCl2(aq) + AgNO3(aq) → AgCl(s) + Mg(NO3)2(aq) 0.641 g AgCl × mol MgCl mol AgCl 95.21 g × × = 0.213 g MgCl2 143.4 g AgCl mol AgCl mol MgCl 0.213 g MgCl × 100 = 14.2% MgCl2 1.50 g mixture mol MgCl b 0.213 g MgCl2 × 95.21 g × mol AgNO mol MgCl × 1L 0.500 mol AgNO × 1000 mL 1L = 8.95 mL AgNO3 90 94 CHAPTER SOLUTION STOICHIOMETRY Mol CoCl2 = 0.0500 L × 0.250 mol CoCl = 0.0125 mol CoCl2 L Mol NiCl2 = 0.0250 L × 0.350 mol NiCl = 0.00875 mol NiCl2 L Both CoCl2 and NiCl2 are soluble chloride salts by the solubility rules A 0.0125-mol aqueous sample of CoCl2 is actually 0.0125 mol Co2+ and 2(0.0125 mol) = 0.0250 mol Cl− A 0.00875-mol aqueous sample of NiCl2 is actually 0.00875 mol Ni2+ and 2(0.00875) = 0.0175 mol Cl− The total volume of solution that these ions are in is 0.0500 L + 0.0250 L = 0.0750 L M Co + = M Cl − = 95 0.00875 mol Ni 2+ 0.0125 mol Co 2+ = 0.167 M ; M Ni + = = 0.117 M 0.0750 L 0.0750 L 0.0250 mol Cl − + 0.0175 mol Cl − = 0.567 M 0.0750 L 35.45 g Cl 0.0761 g = 0.0761 g Cl; % Cl = × 100 = 29.7% Cl 143.4 g AgCl 0.256 g a 0.308 g AgCl × Cobalt(III) oxide, Co2O3: 2(58.93) + 3(16.00) = 165.86 g/mol 0.103 g 117.86 g Co = 0.103 g Co; % Co = × 100 = 24.8% Co 165.86 g Co O 0.416 g 0.145 g Co2O3 × The remainder, 100.0 − (29.7 + 24.8) = 45.5%, is water Assuming 100.0 g of compound: 45.5 g H2O × 2.016 g H 5.09 g H = 5.09 g H; % H = × 100 = 5.09% H 18.02 g H O 100.0 g compound 45.5 g H2O × 16.00 g O 40.4 g O = 40.4 g O; % O = × 100 = 40.4% O 18.02 g H O 100.0 g compound The mass percent composition is 24.8% Co, 29.7% Cl, 5.09% H, and 40.4% O b Out of 100.0 g of compound, there are: 24.8 g Co × 5.09 g H × mol mol = 0.421 mol Co; 29.7 g Cl × = 0.838 mol Cl 58.93 g Co 35.45 g Cl mol mol = 5.05 mol H; 40.4 g O × = 2.53 mol O 1.008 g H 16.00 g O Dividing all results by 0.421, we get CoCl2•6H2O for the empirical formula, which is also the molecular formula c CoCl2•6H2O(aq) + AgNO3(aq) → AgCl(s) + Co(NO3)2(aq) + H2O(l) CHAPTER SOLUTION STOICHIOMETRY 91 CoCl2•6H2O(aq) + NaOH(aq) → Co(OH)2(s) + NaCl(aq) + H2O(l) Co(OH)2 → Co2O3 This is an oxidation-reduction reaction Thus we also need to include an oxidizing agent The obvious choice is O2 Co(OH)2(s) + O2(g) → Co2O3(s) + H2O(l) 96 a Fe3+(aq) + OH−(aq) → Fe(OH)3(s) Fe(OH)3: 55.85 + 3(16.00) + 3(1.008) = 106.87 g/mol 0.107 g Fe(OH)3 × 55.85 g Fe = 0.0559 g Fe 106.9 g Fe(OH) b Fe(NO3)3: 55.85 + 3(14.01) + 9(16.00) = 241.86 g/mol 0.0559 g Fe × 241.9 g Fe( NO ) = 0.242 g Fe(NO3)3 55.85 g Fe c Mass % Fe(NO3)3 = 97 0.242 g × 100 = 53.1% 0.456 g Ag+(aq) + Cl−(aq) → AgCl(s); let x = mol NaCl and y = mol KCl (22.90 × 10−3 L) × 0.1000 mol/L = 2.290 × 10−3 mol Ag+ = 2.290 × 10−3 mol Cl− total x + y = 2.290 × 10−3 mol Cl−, x = 2.290 × 10−3 − y Because the molar mass of NaCl is 58.44 g/mol and the molar mass of KCl is 74.55 g/mol: (58.44)x + (74.55)y = 0.1586 g 58.44(2.290 × 10−3 − y) + (74.55)y = 0.1586, (16.11)y = 0.0248, y = 1.54 × 10−3 mol KCl Mass % KCl = 1.54 × 10 −3 mol × 74.55 g / mol × 100 = 72.4% KCl 0.1586 g % NaCl = 100.0 − 72.4 = 27.6% NaCl 98 a Assume 100.00 g of material 42.23 g C × 2.11 g B × mol C mol F = 3.516 mol C; 55.66 g F × = 2.929 mol F 12.011 g C 19.00 g F mol B = 0.195 mol B 10.81 g B Dividing by the smallest number: 3.516 2.929 = 18.0; = 15.0 0.195 0.195 92 CHAPTER SOLUTION STOICHIOMETRY The empirical formula is C18F15B b 0.3470 L × 0.01267 mol = 4.396 × 10 −3 mol BARF L Molar mass of BARF = 2.251 g = 512.1 g/mol 4.396 × 10 −3 mol The empirical formula mass of BARF is 511.99 g Therefore, the molecular formula is the same as the empirical formula, C18F15B 99 Cr(NO3)3(aq) + NaOH(aq) → Cr(OH)3(s) + NaNO3(aq) Mol NaOH used = 2.06 g Cr(OH)3 × to form precipitate mol Cr (OH) 3 mol NaOH × = 6.00 × 10−2 mol 103.02 g mol Cr (OH) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Mol NaOH used = 0.1000 L × to react with HCl MNaOH = 100 0.400 mol HCl mol NaOH × = 4.00 × 10−2 mol L mol HCl total mol NaOH 6.00 × 10 −2 mol + 4.00 × 10 −2 mol = = 2.00 M NaOH volume 0.0500 L (NH4)2CrO4(aq) + Cr(NO2)3(aq) → NH4NO2(aq) + Cr2(CrO4)3(s) 0.203 L × 0.307 mol = 6.23 × 10 −2 mol (NH4)2CrO4 L 0.137 L × 0.269 mol = 3.69 × 10 −2 mol Cr(NO2)3 L 0.0623 mol/0.0369 mol = 1.69 (actual); the balanced reaction requires a 3/2 = 1.5 to mole ratio between (NH4)2CrO4 and Cr(NO2)3 Actual > required, so Cr(NO2)3 (the denominator) is limiting 3.69 × 10 −2 mol Cr(NO2)3 × 0.880 = 101 mol Cr2 (CrO ) 452.00 g Cr2 (CrO ) × = 8.34 g Cr2(CrO4)3 mol Cr ( NO ) mol Cr2 (CrO ) actual yield , actual yield = (8.34 g)(0.880) = 7.34 g Cr2(CrO4)3 isolated 8.34 g Mol of KHP used = 0.4016 g × mol = 1.967 × 10−3 mol KHP 204.22 g Because mole of NaOH reacts completely with mole of KHP, the NaOH solution contains 1.967 × 10−3 mol NaOH Molarity of NaOH = 1.967 × 10 −3 mol 7.849 × 10 −2 mol NaOH = L 25.06 × 10 −3 L CHAPTER SOLUTION STOICHIOMETRY Maximum molarity = 1.967 × 10 −3 mol 7.865 × 10 −2 mol NaOH = −3 L 25.01 × 10 L Minimum molarity = 1.967 × 10 −3 mol 7.834 × 10 −2 mol NaOH = L 25.11 × 10 −3 L 93 We can express this as 0.07849 ±0.00016 M An alternate way is to express the molarity as 0.0785 ±0.0002 M This second way shows the actual number of significant figures in the molarity The advantage of the first method is that it shows that we made all our individual measurements to four significant figures 102 Desired uncertainty is 1% of 0.02, or ±0.0002 So we want the solution to be 0.0200 ± 0.0002 M, or the concentration should be between 0.0198 and 0.0202 M We should use a 1L volumetric flask to make the solution They are good to ±0.1% We want to weigh out between 0.0198 mol and 0.0202 mol of KIO3 Molar mass of KIO3 = 39.10 + 126.9 + 3(16.00) = 214.0 g/mol 0.0198 mol × 214.0 g 214.0 g = 4.237 g; 0.0202 mol × = 4.323 g (carrying extra sig figs.) mol mol We should weigh out between 4.24 and 4.32 g of KIO3 We should weigh it to the nearest milligram or 0.1 mg Dissolve the KIO3 in water, and dilute to the mark in a 1-liter volumetric flask This will produce a solution whose concentration is within the limits and is known to at least the fourth decimal place 103 Mol C6H8O7 = 0.250 g C6H8O7 × mol C H O = 1.30 × 10−3 mol C6H8O7 192.1 g C H O Let HxA represent citric acid, where x is the number of acidic hydrogens The balanced neutralization reaction is: HxA(aq) + x OH−(aq) → x H2O(l) + Ax−(aq) Mol OH− reacted = 0.0372 L × x= 0.105 mol OH − = 3.91 × 10−3 mol OH− L 3.91 × 10 −3 mol mol OH − = = 3.01 mol citric acid 1.30 × 10 −3 mol Therefore, the general acid formula for citric acid is H3A, meaning that citric acid has three acidic hydrogens per citric acid molecule (citric acid is a triprotic acid) 104 Using HA as an abbreviation for the monoprotic acid acetylsalicylic acid: HA(aq) + NaOH(aq) → H2O(l) + NaA(aq) Mol HA = 0.03517 L NaOH × 0.5065 mol NaOH mol HA × = 1.781 × 10−2 mol HA L NaOH mol NaOH 94 CHAPTER SOLUTION STOICHIOMETRY From the problem, 3.210 g HA was reacted, so: molar mass = 105 3.210 g HA 1.781 × 10 − mol HA = 180.2 g/mol Let H2A = formula for the unknown diprotic acid H2A(aq) + NaOH(aq) → H2O(l) + Na2A(aq) Mol H2A = 0.1375 L × = 0.0516 mol 6.50 g = 126 g/mol 0.0516 mol Molar mass of H2A = 106 mol H A 0.750 mol NaOH × L mol NaOH a MgO(s) + HCl(aq) → MgCl2(aq) + H2O(l) Mg(OH)2(s) + HCl(aq) → MgCl2(aq) + H2O(l) Al(OH)3(s) + HCl(aq) → AlCl3(aq) + H2O(l) b Let's calculate the number of moles of HCl neutralized per gram of substance We can get these directly from the balanced equations and the molar masses of the substances mol HCl mol MgO × mol MgO 40.31 g MgO mol HCl mol Mg (OH) mol HCl mol Al(OH) × × = 4.962 × 10 −2 mol HCl g MgO mol Mg (OH) 58.33 g Mg (OH) mol Al(OH) 78.00 g Al(OH) = = 3.429 × 10 −2 mol HCl g Mg (OH) 3.846 × 10 −2 mol HCl g Al(OH) Therefore, gram of magnesium oxide would neutralize the most 0.10 M HCl 107 0.104 g AgCl × mol AgCl 35.45 g Cl − mol Cl − = 2.57 × 10−2 g Cl− × × mol AgCl 143.4 g AgCl mol Cl − All of the Cl− in the AgCl precipitate came from the chlorisondamine chloride compound in the medication So we need to calculate the quantity of C14H20Cl6N2 which contains 2.57 × 10−2 g Cl− Molar mass of C14H20Cl6N2 = 14(12.01) + 20(1.008) + 6(35.45) + 2(14.01) = 429.02 g/mol There are 6(35.45) = 212.70 g chlorine for every mole (429.02 g) of C14H20Cl6N2 2.57 × 10−2 g Cl− × 429.02 g C14 H 20 Cl N 212.70 g Cl − Mass % chlorisondamine chloride = = 5.18 × 10−2 g C14H20Cl6N2 5.18 × 10 −2 g × 100 = 4.05% 1.28 g CHAPTER SOLUTION STOICHIOMETRY 95 ChemWork Problems 108 0.160 g MgCl2 × Molarity = mol = 1.68 × 10 −3 mol MgCl2 95.21 g 1.68 × 10 −3 mol 1000 mL = 0.0168 M MgCl2 × 100.0 mL L MgCl2(s) → Mg2+(aq) + Cl−(aq); M Mg + = 0.0168 M; M Cl − = 2(0.0168) = 0.0336 M 109 Stock solution = 6.706 × 10 −3 g oxalic acid 0.6706 g = 100.0 mL mL 10.00 mL stock × 6.706 × 10 −3 g oxalic acid = 6.706 × 10−2 g oxalic acid mL This is diluted to a final volume of 250.0 mL 6.706 × 10 −2 g H C O 1000 mL mol H C O = 2.979 × 10 −3 M H2C2O4 × × 250.0 mL L 90.04 g 110 In the first reaction, Sr3(PO4)2(s) is the precipitate, while Ag2CO3(s) is the precipitate for the second reaction The third reaction produces no precipitate AgCl(s) is the precipitate for the fourth reaction and PbCl2(s) is the precipitate for the last reaction 111 NaOH(aq) + Ni(NO3)2(aq) → Ni(OH)2(s) + NaNO3(aq) 0.1500 L × 112 0.249 mol Ni(NO ) L NaOH mol NaOH × × = 0.747 L L mol Ni(NO ) 0.100 mol NaOH = 747 mL NaOH The balanced equation is BaCl2(aq) + Na3PO4(aq) → Ba3(PO4)2(s) + NaCl(aq) Assuming BaCl2 is limiting: 0.4000 L BaCl2 × mol Ba (PO ) 601.8 g Ba (PO ) 0.289 mol BaCl × × L mol BaCl mol Ba (PO ) = 23.2 g Ba3(PO4)2 Assuming Na3PO4 is limiting: 0.5000 L Na3PO4 × mol Ba (PO ) 0.200 mol Na PO × mol Na PO L 601.8 g Ba (PO ) × = 30.1 g Ba3(PO4)2 mol Ba (PO ) The BaCl2 reagent produces the smaller quantity of precipitate, so BaCl2 is limiting and 23.2 g Ba3(PO4)2(s) can form 96 113 CHAPTER SOLUTION STOICHIOMETRY AgNO3(aq) + CaCl2(aq) → AgCl(s) + Ca(NO3)2(aq) 0.4500 L × 0.257 mol AgNO mol AgCl 143.4 g AgCl = 1.66 g AgCl × × L mol AgNO mol AgCl 0.4000 L × 0.200 mol CaCl 2 mol AgCl 143.4 g AgCl × × = 2.29 g AgCl mol CaCl mol AgCl L AgNO3 is limiting (it produces the smaller mass of AgCl) and 1.66 g AgCl(s) can form Note that we did this calculation for your information It is typically asked in this type of problem The net ionic equation is Ag+(aq) + Cl−(aq) → AgCl(s) The ions remaining in solution after precipitation is complete will be the unreacted Cl− ions and the spectator ions NO3− and Ca2+ (all Ag+ is used up in forming AgCl) The moles of each ion present initially (before reaction) can be determined from the moles of each reactant We have 0.4500 L(0.257 mol AgNO3/L) = 0.116 mol AgNO3, which dissolves to form 0.116 mol Ag+ and 0.116 mol NO3− We also have 0.4000 L(0.200 mol CaCl2/L) = 0.0800 mol CaCl2, which dissolves to form 0.0800 mol Ca2+ and 2(0.0800) = 0.160 mol Cl− To form the 1.66 g of AgCl precipitate, 0.116 mol Ag+ will react with 0.116 mol of Cl− to form 0.116 mol AgCl (which has a mass of 1.66 g) Mol unreacted Cl− = 0.160 mol Cl− initially − 0.116 mol Cl− reacted to form the precipitate Mol unreacted Cl− = 0.044 mol Cl− M Cl− = 114 0.044 mol Cl − 0.044 mol Cl − = 0.052 M Cl− in excess after reaction = total volume 0.4500 L + 0.4000 L Zn2P2O7: 2(65.38) + 2(30.97) + 7(16.00) = 304.70 g/mol All of the zinc in Zn2P2O7 came from the zinc in the foot powder 0.4089 g Zn2P2O7 × Mass % Zn = 115 130.76 g Zn = 0.1755 g Zn in the foot powder 304.70 g Zn P2 O 0.1755 g Zn × 100 = 13.07% Zn 1.343 g foot powder HNO3(aq) + Ca(OH)2(aq) → H2O(l) + Ca(NO3)2(aq) 34.66 × 10 −3 L HNO3 × Molarity of Ca(OH)2 = 116 0.944 mol HNO L HNO 1.64 × 10 −2 mol 50.00 × 10 −3 L × mol Ca(OH) = 1.64 × 10 −2 mol Ca(OH)2 mol HNO = 0.328 M Ca(OH)2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) CHAPTER SOLUTION STOICHIOMETRY 0.02844 L NaOH × mol H SO 0.1000 mol NaOH × = 1.422 × 10 −3 mol H2SO4 L NaOH mol NaOH Mass sulfur = 1.422 × 10 −3 mol H2SO4 × Mass % S = 117 97 32.07 g S mol S × = 4.560 × 10−2 g S mol S mol H SO 4.560 × 10 −2 g S × 100 = 3.502% S 1.302 g coal MgSO4: +2 + x + 4(−2) = 0, x = +6 = oxidation state of S PbSO4: The sulfate ion has a 2− charge (SO42−), so +2 is the oxidation state (charge) of lead O2: O has an oxidation state of zero in O2; Ag: Ag has an oxidation state of zero in Ag CuCl2: Copper has a +2 oxidation state since each Cl has a -1 oxidation state (charge) Challenge Problems 118 2(6 e− + 14 H+ + Cr2O72− → Cr3+ + H2O) H2O + C2H5OH → CO2 + 12 H+ + 12 e− 16 H+ + Cr2O72− + C2H5OH → Cr3+ + CO2 + 11 H2O 0.0600 mol Cr2 O 2− mol C H OH 46.07 g = 0.0429 g C2H5OH 0.03105 L mol Cr O 2− mol C H OH L 0.0429 g C H OH × 100 = 0.143% C2H5OH 30.0 g blood 119 a Let x = mass of Mg, so 10.00 − x = mass of Zn Ag+(aq) + Cl−(aq) → AgCl(s) From the given balanced equations, there is a : mole ratio between mol Mg and mol Cl− The same is true for Zn Because mol Ag+ = mol Cl− present, one can setup an equation relating mol Cl− present to mol Ag+ added x g Mg × mol Mg mol Cl − mol Zn mol Cl − × + (10.00 − x) g Zn × × mol Mg 24.31 g Mg 65.38 g Zn mol Zn + − 3.00 mol Ag mol Cl = 0.156 L × = 0.468 mol Cl− × L mol Ag + 20.00 − x 2x 2(10.00 − x) 2x = 0.468 + + = 0.468, 24.31 × 65.38 24 31 65 38 24.31 65.38 (130.8)x + 486.2 − (48.62)x = 743.8 (carrying extra significant figure) (82.2)x = 257.6, x = 3.13 g Mg; % Mg = 3.13 g Mg × 100 = 31.3% Mg 10.00 g mixture 98 CHAPTER b 0.156 L × MHCl = 120 SOLUTION STOICHIOMETRY 3.00 mol Ag + mol Cl − = 0.468 mol Cl− = 0.468 mol HCl added × + L mol Ag 0.468 mol = 6.00 M HCl 0.0780 L Let x = mass of NaCl, and let y = mass K2SO4 So x + y = 10.00 Two reactions occur: Pb2+(aq) + Cl−(aq) → PbCl2(s) and Pb2+(aq) + SO42−(aq) → PbSO4(s) Molar mass of NaCl = 58.44 g/mol; molar mass of K2SO4 = 174.27 g/mol; molar mass of PbCl2 = 278.1 g/mol; molar mass of PbSO4 = 303.3 g/mol x y = moles NaCl; = moles K2SO4 58.44 174.27 mass of PbCl2 + mass PbSO4 = total mass of solid y x (1/2)(278.1) + (303.3) = 21.75 58.44 174.27 We have two equations: (2.379)x + (1.740)y = 21.75 and x + y = 10.00 Solving: x = 6.81 g NaCl; 121 6.81 g NaCl × 100 = 68.1% NaCl 10.00 g mixture Zn(s) + AgNO2(aq) → Ag(s) + Zn(NO2)2(aq) Let x = mass of Ag and y = mass of Zn after the reaction has stopped Then x + y = 29.0 g Because the moles of Ag produced will equal two times the moles of Zn reacted: (19.0 − y) g Zn × mol Zn mol Ag mol Ag × = x g Ag × 65.38 g Zn mol Zn 107.9 g Ag Simplifying: 3.059 × 10−2(19.0 − y) = (9.268 × 10−3)x Substituting x = 29.0 − y into the equation gives: 3.059 × 10−2(19.0 − y) = 9.268 × 10−3(29.0 − y) Solving: 0.581 − (3.059 × 10−2)y = 0.269 − (9.268 × 10−3)y, (2.132 × 10−2)y = 0.312, y = 14.6 g Zn 14.6 g Zn are present, and 29.0 − 14.6 = 14.4 g Ag are also present after the reaction is stopped ... respectively a Lose e − to form Ra2+ b Lose e − to form In3+ c Gain e − to form P 3− d Gain e − to form Te 2− e Gain e − to form Br− f Lose e − to form Rb+ See Exercise 47 for a discussion of charges... atoms of H Octane formula = C8H18 and the ratio of C:H = 8:18 or 4:9 94 Let Xa be the formula for the atom/molecule X, Yb be the formula for the atom/molecule Y, XcYd be the formula of compound... empirical formula 50 a The molecular formula is N2O4 The smallest whole number ratio of the atoms (the empirical formula) is NO2 b Molecular formula: C3H6; empirical formula: CH2 c Molecular formula: