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  • In photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H 12 O 6 ) according to the following reaction: CO 2 (g) + H 2 O (l)  O 2 (g) + C 6 H 12 O 6 (aq) Suppose you determine that a particular plant consumes 37.8 g CO 2 in one week. Assuming that there is more than enough water present to react with all of the CO 2 , what mass of glucose (in grams) can the plant synthesize from the CO 2 ?
  • Usually A is a reactant and B is a product. However, A and B can both be reactants or products. Sometimes A is a product and B is a reactant.
  • The first thing you do with stoichiometry is balance the equation.
  • It’s an expensive experiment, but drop some pearls into vinegar sometime to see this reaction.
  • What is the reaction taking place here? The big spheres are nitrogen and the small ones are hydrogen. Which one is the limiting reactant?
  • Ammonia can be synthesized by reaction nitrogen monoxide with hydrogen, a byproduct will be water.
  • Almost always the reactions do not go to completion, use up all of the reactants. An important tool for chemists if percent yield. The amount of product determined for a certain reaction based on the amounts of reactant given is called the theoretical yield. The actual yield is determined experimentally. To determine the percent yield we take the actual yield and divide it by the theoretical yield and multiply by 100 to get the percentage. Titanium metal can be obtained from titanium (iV) oxide by reaction with elemental C, carbon monoxide is a by product. When 28.6 kg of C is allowed to react with 88.2 kg of titanium (IV) oxide, 42.8 kg of Ti is produced. Find the limiting reactant, theoretical yield (in kg), and the percent yield.
  • Ammonia can be synthesized by reaction nitrogen monoxide with hydrogen, a byproduct will be water. Starting with 86.3 g NO and 25.6 g H 2 , determine the amount of ammonia that can be produced. Which is the limiting reactant? How much of the other reactant is left over?
  • Solvent: the thing doing the dissolving, retains its state Solute: the thing being dissolved, becomes the state of the solvent Hydrophilic: likes water Hydrophobic: does not like water
  • Moles solute/ liters solution This is how to make solution in lab. A  weigh out the appropriate amount of solute B  Add enough water to dissolve all of the solid C  Fill to the mark with water.
  • Solubility: the amount of a substance that dissolves in a given volume of solvent at a given temperature Electrolyte: a material that dissolves in water to give an electrical current Dilution: the process of adding solvent to lower the concentration of solute in the solution Polar Molecule: a molecule that has a permanent dipole moment (a property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge
  • Phosphoric acid is a weak electrolyte, because it doesn’t dissociate 100%. Solubility: the amount of a substance that dissolves in a given volume of solvent at a given temperature. Electrolyte: a material that dissolves in water to give an electrical current Dilution: the process of adding solvent to lower the concentration of solute in the solution Polar Molecule: a molecule that has a permanent dipole moment (a property of a molecule whose charge distribution can be represented by a center of positive charge and a center of negative charge
  • Reaction of iron (III) oxide with carbon monoxide yields pure iron and carbon dioxide. The reaction of 167 g iron (III) oxide with 85.8 g carbon monoxide produces 72.3 g of iron. Find the limiting reactant, theoretical yield, and percent yield.
  • Just another picture. An example of a strong electrolyte: NaCl An example of a weak electrolyte: acetic acid, HC 2 H 3 O 2 An example of a nonelectrolyte: sucrose, C 12 H 22 O 11
  • Balance the equation What is the limiting reagent? Usually, stoichiometry goes m A → mol A → mol B → m B and we solve for mass of B. In this problem, we’re given m B and have to solve for molar mass of A.
  • Stock: routinely used solutions prepared in concentrated form Concentrated: relatively large ratio of solute to solvent (10.0 M NaCl) Dilute: relatively small ratio of solute to solvent (0.01 M NaCl) Saturated: any more solute will not dissolve
  • Theories and laws are very different kinds of knowledge. A law holds in all circumstances. If there were an exception or a loophole, the law would be discarded. We understand the law of gravity, but there is no accepted theory of gravity. A theory explains why something happens. These are very difficult to prove. A law is much more simple to prove. The fact that we have a pretty good understanding of how stars explode doesn't necessarily mean we could predict the next supernova; we have a theory but not a law. Myth 1: Hypotheses Become Theories Which Become Laws
  • Theories and laws are very different kinds of knowledge. A law holds in all circumstances. If there were an exception or a loophole, the law would be discarded. We understand the law of gravity, but there is no accepted theory of gravity. A theory explains why something happens. These are very difficult to prove. A law is much more simple to prove. The fact that we have a pretty good understanding of how stars explode doesn't necessarily mean we could predict the next supernova; we have a theory but not a law. Myth 1: Hypotheses Become Theories Which Become Laws
  • AgNO 3 ( aq ) + NaCl( aq )   AgCl( s ) + NaNO 3 ( aq )
  • Practice assigning oxidation states to elements:
  • 4 to post 1

    1. 1. Chapter 4STOICHIOMETRY
    2. 2. §4.1 StoichiometryStoichiometry - calculating the quantities of reactants and products in a chemical reaction.Stoichiometry is entirely based on the Law of Conservation of Mass.
    3. 3. §4.2 Stoichiometric CalculationsThe amounts of any other substance in a chemical reaction can be determined from the amount of just one substance: (where A and B are in the same reaction)
    4. 4. §4.2 StoichiometrySulfuric acid dissolves aluminum metal according to thefollowing reaction. If you had 15.2 g aluminum, what amountof sulfuric acid would you need to complete the reaction? 3 H2SO4 (aq) + 2 Al(s) → Al2(SO4)3 (aq) + 3 H2(g)15.2 g Al 1 mol Al 3 mol H2SO4 98.09 g H2SO4 = 82.9 g 26.98 g Al 2 mol Al 1 mol H2SO4 H2SO4
    5. 5. §4.2 StoichiometryPearls and baby teeth dissolve in vinegar according tothis reaction. CaCO3 (s) + 2 HC2H3O2 (aq) → Ca(C2H3O2)2 (aq) + H2CO3 (aq) 10 g CaCO3 1 mol CaCO3 2 mol HC2H3O2 60 g HC2H3O2 100 g CaCO3 1 mol CaCO3 1 mol HC2H3O2 = 12 g HC2H3O2
    6. 6. §4.3 Limiting ReactantSometimes all reactants are completely consumed; sometimes one or more is in excess.A limiting reactant is present in lesser molar amounts and is consumed first, limiting the amount of product that can form.Always determine which reactant is limiting when performing stoichiometry calculations!
    7. 7. N2(g) + 3H2(g) → 2NH3(g) §4.3 Limiting Reactantin excess limiting reagent
    8. 8. §4.3 Limiting Reactant  Reaction of ammonia and oxygen gives nitrogen monoxide and water. Determine the mass of nitrogen monoxide produced from 10.0 g ammonia and 10.0 g oxygen. 4NH3(g) ++ 5O2(g)  NH3(g) O 4NO(g) ++ H2O(l) (l) NO(g) 6H2O Limiting Reagent10.0 g NH3 mol NH3 1. Balance the NO =30.01 gmol NO 17.7 g NO 4 mol equation 0.587 NO = 17.034 g NH3 4 mol NH3 1 mol NO 2. Limiting reagent? 10.0 g O2 mol O2 4 mol NO =30.01 gmol NO7.50 g NO 0.250 NO = 32.00 g O2 5 mol O2 1 mol NO
    9. 9. §4.3 Percent YieldTheoretical Yield (calculated) – the maximum amount of product that can form during a reactionActual Yield (measured) – the amount of pure product collected from a reaction (actual yield) ≤ (theoretical yield) Percent Yield = Actual Yield (100%) Theoretical Yield
    10. 10. §4.3 Percent Yield  Reaction of titanium (IV) oxide with graphite produces titanium metal and carbon monoxide. When 28.6 kg of C is allowed to react with 88.2 kg of titanium (IV) oxide, 42.8 kg of Ti is 42.8 kg Ti (100%) produced. Find the percent yield of Ti from this reaction. TiO2(s) ++ 2C(s)  TiO2(s) C (s)  Ti(s) + Ti(s) + CO(g)(g) 2CO Limiting Reagent 1. Balance the equation88.2 x 103 g TiO2 mol TiO2 1 mol Ti 47.88 gmol Ti (100%) = 1104 Ti 42.8 kg Ti 79.88 g TiO2 1 mol TiO2 mol Ti = 52.9 kg Ti 2. Limiting reagent?28.6 x 103 g C = = 5.29 x 104 g Ti 80.9% mol C 1 mol Ti 12.01 g C 2 mol C = 1191 mol Ti
    11. 11. §4.4 Aqueous SolutionsWater is one of the most important substances on earth.All chemistry that makes life possible occurs in water.Stoichiometry also applies to chemistry in aqueous solutions.
    12. 12. §4.4 Some DefinitionsSolution – a homogenous mixtureSolvent – the majority solution component, retains its stateSolute – the minority solution component, becomes the state of the solventSalt – 1) any ionic compound; 2) NaCl
    13. 13. §4.3 Solution ConcentrationMolarity (M) = moles of solute liters of solution 0.5 M 0.3 M 0.1 M CuSO4 CuSO4 CuSO4
    14. 14. §4.4 Molarity CalculationsDetermine the molarity of a solution where 45.5 mg of NaCl is dissolved in 154.4 mL of solution.45.5 mg NaCl 1000 mL soln 1 g NaCl 1 mol NaCl154.4 mL soln 1 L soln 1000 mg NaCl 58.44 g NaCl = 5.04 x 10-3 M NaClHow many moles of Cl- ion are contained in 114 mL of a 1.85 M CaCl2 solution? CaCl2(s) -> Ca2+(aq) + 2Cl-(aq)114 mL soln 1 L soln 1.85 mol CaCl2 2 mol Cl- 1000 mL soln 1 L soln 1 mol CaCl2 = 0.422 mol Cl-
    15. 15. §4.5 ElectrolytesAn electrolyte is a substance that when dissolved in water produces a solution that conducts electricity.Nonelectrolytes dissolve in water but dont ionize.
    16. 16. strong weak non- electrolyte electrolyte electrolyte§4.5ElectrolytesElectricalconductivity isa qualitativemeasure of thedegree towhich a soluteionizes.
    17. 17. §4.5 Strong ElectrolytesStrong electrolytes completely ionize in water. Thereare three types:1) Soluble salts – NaCl, KBr, LiNO3, etc.2) Strong acids – HCl, HNO3, H2SO4, HClO43) Strong bases – NaOH, KOH H2OExample: NaOH(s) → Na+(aq) + OH-(aq) 100% ionized
    18. 18. §4.5 Weak ElectrolytesWeak electrolytes slightly ionize in water.Examples include weak acids such as HC2H3O2 and weak bases such as NH3.For instance, acetic acid is ~1% ionized in water: HC2H3O2(l) → ← H+(aq) + C2H3O2-(aq) 99% 1%In contrast, strong electrolytes are 100% ionized.
    19. 19. §4.5 SolubilityDissolution occurs if interactions between solvent and solute particles overcome interactions of the solute to itself.
    20. 20. §4.5 SolubilityAn ionic compound dissolving in water:Just because a compound is ionic doesn’t automatically mean it dissolves in water. (Chapter 16)
    21. 21. §4.5 SolubilityA nonelectrolyte (sucrose) dissolved in water:
    22. 22. §4.3 Percent Yield If 18.20 g of NH3 (mm 17.03 g/mol) are produced by a reaction mixture that initially contains 6.00 g of H 2 and an excess of N2, what is the percent yield of the reaction? N2(g) + 3H2(g) → 2NH3(g) Calculate theoretical yield of NH3 :6.00 g H2 1 mol H2 2 mol NH3 17.03 g NH3 = 33.8 g NH3 2.016 g H2 3 mol H2 1 mol NH3 Percent yield: 18.20 g NH3 = 53.8% (100%) 33.8 g NH3
    23. 23. §4.6 Three Types of Reactions1. Precipitation reactions2. Acid-base reactions3. Oxidation-reduction reactionsWe will not be covering gas-evolution reactions (§4.6).
    24. 24. Learn§4.5 Solubility Rules these.
    25. 25. §4.6 Precipitation ReactionsHow to predict a precipitation reaction:1) Swap the anions and cations.2) Consult the solubility rules. If any of the “products” are insoluble in water, a precipitation reaction has occurred. Otherwise, there is no reaction. KKSO4(aq) ++ BaI2(aq) → BaSO4(?) ++ 2KI(?)(aq) 2 2SO4(aq) 2 4(aq) BaI2(aq) → ??? 4(s) 2(aq) BaSO 2KI
    26. 26. §4.6 Demo – Precipitation Reactions Pb(NO3)2 + 2KI --> PbI2 + 2KNO3soluble soluble insoluble soluble• NO3- salts are soluble.• Li+, Na+, and K+ salts are soluble.• Cl-, Br- and I- salts are soluble unless the cation is Ag +, Pb2+ or Hg22+.
    27. 27. §4.6 Demo – Precipitation Reactions AgNO3 + NaCl --> NaNO3 + AgClsoluble soluble soluble insoluble• NO3- salts are soluble.• Li+, Na+, and K+ salts are soluble.3.Cl-, Br- and I- salts are soluble unless the cation is Ag+, Pb2+ and Hg22+.
    28. 28. §4.6 Water-soluble or -insoluble?Sr3(PO4)2 insolubleNa3PO4 solubleCaSO4 insolubleRbCl solubleBaS soluble(NH4)2CrO4 solubleAg2SO4 insolubleCdBr2 solubleNH4OH solubleNiCO3 insoluble
    29. 29. §4.6 Precipitation Reactions A 1.000 g sample of a metal dichloride was dissolved in water and treated with excess AgNO3(aq) to give 1.286 g of a white solid. What is the metal? MCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + M(NO3)2(aq)1.000 g MCl2 mol MCl2 2 mol AgCl 143.32 g AgCl = 1.286 g AgCl x g MCl2 1 mol MCl2 1 mol AgClx = 222.89 g/mol MCl2 M + 2(35.45 g/mol) = 222.89 g/mol M = 152.0 g/mol M = europium
    30. 30. §4.4 DilutionIt is very common to make dilute solutions from more concentrated ones (stock solutions).For dilutions: more solvent is added but the amount of solute remains constant. Hence: mol solute before dilution = mol solute after dilution M1V11 = M2V2 of initial solution M = molarity V1 = volume of initial solution Molarity = mol solute M2 = molarity of final solution L solution V2 = volume of final solution (Molarity) (L solution) = mol solute
    31. 31. §4.6 Solubility100.0 mL of a Pb(NO3)2 solution was concentrated to 80.0 mL. To2.00 mL of the concentrated solution was added an excess ofNaCl(aq), and 3.407 g of a solid was obtained. What was the molarityof the original Pb(NO3)2 solution? M1V1 = M2V2
    32. 32. §4.6 Solubility 100.0 mL of a Pb(NO3)2 solution was concentrated to 80.0 mL. To 2.00 mL of the concentrated solution was added an excess of NaCl(aq), and 3.407 g of a solid was obtained. What was the molarity of the original Pb(NO3)2 solution? Pb(NO33)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) Pb(NO )2(aq) + NaCl (aq) 2(s) NaNO3(aq)2.00 x 10-3 L x mol LN 1 mol PbCl2 278.10 g PbCl2 = 3.407 g PbCl2 LN soln 1 L soln 1 mol LN 1 mol PbCl2 LN = Pb(NO3)= 6.125 M Pb(NO3)2 x 2 Use the same equation for concentration as for dilution:V1M1 = V2M2 (100 mL) · M1 = (80.0 mL)(6.125 M) M1 = 4.90 M
    33. 33. §4.6 StoichiometryA 2.25 g sample of scandium metal is reacted with excess hydrochloricacid to produce 0.1502 g hydrogen gas. What is the formula of the 2 Sc Sc(s) ++xxHCl(aq) → ScClx x(aq) + +H2 (g)2(g) 2 Sc(s)(s) 2xHCl (aq) → 2 ScCl(aq) + x/2 H2 (g) 2Sc + + HCl (aq) → ScCl x x (aq)scandium chloride produced in the reaction? (s) 2 H x0.1502 g H2 1 mol H2 2x mol HCl 1 mol Cl- = 0.1487 mol Cl- 2.018 g H2 x mol H2 1 mol HCl2.25 g Sc 1 mol Sc = 0.0500 mol Sc 44.96 g Sc
    34. 34. §4.6 StoichiometryA 2.25 g sample of scandium metal is reacted with excess hydrochloricacid to produce 0.1502 g hydrogen gas. What is the formula of the 2 Sc (s) + 2x HCl (aq) → 2 ScClx (aq) + x H2 (g)scandium chloride produced in the reaction? = 0.1487 mol Cl- ≈ 3.0 = 0.0500 mol Sc ScCl3
    35. 35. §4.7 Writing Chemical Equations for SolutionsMolecular Equation – reactants and products are written as compounds. Cs2S(aq) + HgBr2(aq) → 2CsBr(aq) + HgS(s)Complete Ionic Equation – all strong electrolytes are written as ions. 2Cs+(aq) + S2-(aq) + Hg2+(aq) + 2Br -(aq) → 2Cs+(aq) + 2Br -(aq) + HgS(s) In this reaction, Cs+ and Br- are called spectator ions.Net Ionic Equation – species that are the same on both sides of the equation are removed. Hg2+(aq) + S2-(aq) → HgS(s)
    36. 36. Write the net ionic equation (if any) for these reactions. FeSO4(aq) + 2 (aq) → KCl FeCl2(?) + K2SO4(?) 2(aq) + K2SO4(aq) No precipitate = No reactionAl(NO3)3(aq) + 3 LiOH(aq) → Al(OH)3(s) + 3LiNO3(aq) Al(OH)3(?) + 3LiNO3(?) Al3+(aq) + 3 OH-(aq) → Al(OH)3(s)
    37. 37. §4.8 Acid-Base ReactionsAcid – produces H+ in waterBase – produces -OH in waterAcid-base reactions often have the format: acid + base → water + saltExample: HBr(aq) + LiOH(aq) → H2O(l) + LiBr(aq)The net ionic equation is: H+(aq) + OH-(aq) → H2O(l)
    38. 38. §4.8 Acid-Base Titrations Titration – analysis of the amount of a substance by reaction with a solution of known concentration. Equivalence point –the reaction is complete with neither acid nor base in excess. For titration stoichiometry: mole ratios are mols acid/mols base or vice versa.
    39. 39. §4.8 Acid-Base Reactions Titration of a 20.0 mL solution of H2SO4 (a diprotic acid) required 22.87 mL of 0.158 M KOH solution. What is the concentration of the H2SO4 solution? H2H2SO4(aq) + 2KOH(aq) → 2H(l) (l) + 2SO4(aq) SO4(aq) + 2KOH(aq) → 2H2O2O+ K K2SO4(aq)0.02287 L KOH soln 0.158 mol KOH 1 mol H2SO4 = 1.81 x 10-3 mol H22SO44 1.81x 10-3 mol H SO 1 L KOH soln 2 mol KOH = 0.0903 M H2SO4 0.0200 L H2SO4 soln
    40. 40. §4.9 Oxidation-Reduction ReactionsOxidation – loss of electronsReduction – gain of electronsOxidizing agent – gains electrons, is itself reducedReducing agent – loses electrons, is itself oxidizedOxidation and reduction always go together. One species transfers electrons to another.
    41. 41. §4.9 Oxidation-Reduction Reactions “LEO says GER” “OIL RIG” Loss of Electrons: Loss Oxidation Is Oxidation Gain Reduction Is Reduction of Electron: Gain
    42. 42. §4.9 Oxidation-Reduction ReactionsOxidation used to mean reaction with oxygen.Anything that reacts with O2 (except F2) is oxidized (loses electrons), but many redox reactions don’t involve oxygen at all: 2 K(s) + F2(g) → 2 KF(s)
    43. 43. §4.9 Oxidation-Reduction ReactionsReactions of metals and nonmetals to give ionic compounds are typical redox reactions. 4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s)The metal is oxidized, the nonmetal is reduced.
    44. 44. §4.9 Oxidation StateAssigning elements in a reaction an oxidation state allows us to trace electron flow during the reaction.Except for monoatomic ions, oxidation states are not ionic charges.Oxidation states are assigned by “giving” electrons in a bond (for counting purposes, not in reality) to the more electronegative element.
    45. 45. §4.9 Oxidation State Rules1. Free elements have an oxidation state of 0.  Na = 0 and Cl = 0 in 2 Na(s) + Cl (g) 2 21. Monoatomic ions have an oxidation state equal to their charge.  For NaCl, Na = +1 and Cl = -11. a. The sum of the oxidation states for all the atoms in a compound is 0.  For NaCl, (+1) + (-1) = 0b. The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion.  For NO –, N = +5 and O = -2, (+5) + 3(-2) = -1 3
    46. 46. §4.9 Oxidation State Rules4. a. Group I metals have an oxidation state of +1 in all their compounds.  Na = +1 in NaCl b. Group II metals have an oxidation state of +2 in all their compounds.  Mg = +2 in MgCl 2
    47. 47. §4.9 Oxidation State Rules5. In their compounds, nonmetals have oxidation states in the table below.  Elements higher on the table take priority. Nonmetal Oxidation State Example F -1 CF4 H +1 CH4 O -2 CO2
    48. 48. §4.9 Oxidation StatesWhat are the oxidation states of the elements in thefollowing species?F2 0Mg2P2O7 +2, +5, -2Hg2Cl2 +1, -1S2O32- +2, -2
    49. 49. §4.9 Oxidation StateConsider the following reaction: 2 F2 + 2 NaOH → OF2 + 2 NaF + H2O 0 +1, -2, +1 +2, -1 +1, -1 +1, -2What is reduced? FWhat is oxidized? OWhat is the reducing agent? NaOHWhat is the oxidizing agent? F2
    50. 50. §4.6 Precipitation StoichiometryWhen 159 mL of a 0.423 M Al(NO3)3 solution of is added to 378 mL of a0.296 M Ba(OH)2 solution, a solid forms. What mass of solid forms andwhat is the concentration of each ion in solution after the reaction?2Al(NO3)3(aq) + 3Ba(OH)2(aq) → 2Al(OH)3(s) + 3Ba(NO3)2(aq) Al(NO 3 3(aq) Ba(OH)2(aq) Al(OH) Ba(NO 3 2(aq) Limiting reagent0.159 L Al(NO3)3 soln 0.423 mol Al(NO3)3 2 mol Al(OH)3 1 L Al(NO3)3 soln 2 mol Al(NO3)3 78.004 g Al(OH)3 = 6.73 x 10-2 mol Al(OH)3 = 5.25 g Al(OH)3 1 mol Al(OH)30.378 L Ba(OH)2 soln 0.296 mol Ba(OH)2 2 mol Al(OH)3 1 L Ba(OH)2 soln 3 mol Ba(OH)2 = 7.46 x 10-2 mol Al(OH)3
    51. 51. §4.6 Precipitation Stoichiometry When 159 mL of a 0.423 M Al(NO3)3 solution of is added to 378 mL of a 0.296 M Ba(OH)2 solution, a solid forms. What mass of solid forms and what is the concentration of each ion in solution after the reaction? 2Al(NO3)3(aq) + 3Ba(OH)2(aq) → 2Al(OH)3(s) + 3Ba(NO3)2(aq) Limiting reagent [Al3+] = ? M 0 [NO3-]0.159 L Al(NO3)3 soln 0.423 mol Al(NO3)3 3 mol NO3- = 0.376 M NO3- (0.159 L + 0.378 L) 1 L Al(NO3)3 soln 1 mol Al(NO3)3 [Ba+]0.378 L Ba(OH)2 soln 0.296 mol Ba(OH)2 1 mol Ba2+ = 0.208 M Ba2+ (0.159 L + 0.378 L) 1 L Ba(OH)2 soln 1 mol Ba(OH)2
    52. 52. §4.6 Precipitation Stoichiometry When 159 mL of a 0.423 M Al(NO3)3 solution of is added to 378 mL of a 0.296 M Ba(OH)2 solution, a solid forms. What mass of solid forms and what is the concentration of each ion in solution after the reaction? 2Al(NO3)3(aq) +- 3Ba(OH)2(aq)) /→ 2Al(OH)3(s) L) =3Ba(NO3)2(aq) ( (0.378 L + 0.159 + Limiting reagent = 3.91 x 10-2 M OH- Initial mol OH-0.378 L Ba(OH)2 soln 0.296 mol Ba(OH)2 2 mol OH- = 0.223 mol OH- 1 L Ba(OH)2 soln 1 mol Ba(OH)2 mol OH- consumed0.159 L Al(NO3)3 soln 0.423 mol Al(NO3)3 2 mol Al(OH)3 3 mol OH- 1 L Al(NO3)3 soln 2 mol Al(NO3)3 1 mol Al(OH)3 = 0.202 mol OH-

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