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Types of Chemical Reactions and Solution Stoichiometry

Types of Chemical Reactions and Solution Stoichiometry

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Types of Chemical Reactions and Solution Stoichiometry

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  1. Types of Chemical Reactions and Solution Stoichiometry Chapter 4 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

  2. Types of Chemical Reactions and Solution Stoichiometry – ch 4 1. Distinguish between solute, solvent and solution. 2. What is an electrolyte?

  3. Types of Chemical Reactions and Solution Stoichiometry – ch 4

  4. Types of Chemical Reactions and Solution Stoichiometry – ch 4 3. Determine if the following substances are strong, weak or non electrolytes in water. a. KC2H3O2 b. CH2O c. HNO3 d. H2S

  5. Types of Chemical Reactions and Solution Stoichiometry – ch 4 4. A solution is prepared by dissolving 50 g of barium fluoride in enough water to make 750 mL of solution. a. What is the molarity of the barium fluoride? b. What is the concentration for each of the ions in solution? c. What is the concentration of the fluoride ion if 350 mL of water is added? d. What volume of the initial solution would you use in order to make 80 mL of 0.10 M Ba2+?

  6. Types of Chemical Reactions and Solution Stoichiometry – ch 4 Molarity (M) ⇒ a unit of concentration that measures the moles of solute per liter of solution M = nsolute/Lsolution

  7. Types of Chemical Reactions and Solution Stoichiometry – ch 4 5. What is the resulting concentration of bromide ion if 28 mL of 0.15 M sodium bromide solution were mixed with 45 mL of 0.12 M calcium bromide solution?

  8. Types of Chemical Reactions and Solution Stoichiometry – ch 4 6. What volume of 0.150 M Li3PO4 in milliliters is required to precipitate all of the mercury(II) ions from 250 mL of 0.32 M Hg(NO3)2?

  9. Types of Chemical Reactions and Solution Stoichiometry – ch 4 7. A precipitate will form when 10 mL of 0.25 M calcium chloride is mixed with 8 mL of 0.4 M silver nitrate. a. Write the molecular and net ionic equation for this reaction b. How many grams of precipitate form? c. What is the concentration of the ions remaining in solution?

  10. Types of Chemical Reactions and Solution Stoichiometry – ch 4 8. What type of solution results (acidic, basic or neutral) when 110 mL of 0.40 M HNO3 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the molecular and net ionic equations.

  11. Types of Chemical Reactions and Solution Stoichiometry – ch 4 9. What volume of 0.15 M H2SO4 is required to neutralize 60 mL of 0.80 M KOH?

  12. Types of Chemical Reactions and Solution Stoichiometry – ch 4 10. Assign oxidation states for the following: a. S in SO3 b. P in PF5 c. Cl in Cl2 d. Cr in Cr2O72- e. Mn in KMnO4 f. H in CaH2 g. O in H2O2

  13. Types of Chemical Reactions and Solution Stoichiometry – ch 4 11. Identify the oxidizing and reducing agents for the following reactions: a. Ba2+ + 2 Li  Ba + 2 Li+ b. 8 H+ + MnO4- + 5 Fe2+ Mn2+ + 5 Fe3+ + 4 H2O

  14. Types of Chemical Reactions and Solution Stoichiometry – ch 4 12. Balance the following red-ox reactions: a. MnO2 + NO3- N2O4 + MnO4- (acidic) b. CNO- + As2O3 CN- + HAsO42- (basic) c. Cr3+ + SO42– Cr2O72– + H2SO3 (acidic)

  15. Types of Chemical Reactions and Solution Stoichiometry – ch 4 13. It required 85 mL of 0.47 M K2Cr2O7 solution to oxidize all of the Fe2+ ions in a 200 mL solution. What was the initial concentration of Fe2+ ions? Cr2O72- (aq) + Fe2+ (aq)  Cr3+ (aq) + Fe3+ (aq) (acidic)

  16. Answer Key – Ch. 4 1. Distinguish between solute, solvent and solution. solute – a substance dissolved in a liquid to form a solution solvent – the dissolving medium in a solution solution – homogenous mixture 2. What is an electrolyte? electrolyte – a substance that produces ions upon dissolving in water thus causing a solution that can conduct electricity

  17. Answer Key – Ch. 4 3. Determine if the following substances are strong, weak or non electrolytes in water. a. KC2H3O2 ⇒ soluble ionic ⇒ strong electrolyte b. CH2O ⇒ molecular ⇒ non electrolyte c. HNO3 ⇒ strong acid ⇒ strong electrolyte d. H2S ⇒ weak acid ⇒ weak electrolyte

  18. Answer Key – Ch. 4 4. A solution is prepared by dissolving 50 g of barium fluoride in enough water to make 750 mL of solution. a. What is the molarity of the barium fluoride? (50 g BaF2)/(175.33 g/mol) = 0.285 mol BaF2 (750 mL)(1L/1000mL) = 0.75 L M = (0.285 mol)/(0.75 L) = 0.38 M BaF2 b. What is the concentration for each of the ions in solution? BaF2(s) Ba2+(aq) + 2F– (aq) 0.38 mol/L BaF2 x 1 mol Ba2+ = 0.38 M Ba2+ 0.38 mol/L BaF2 x 2 mol F– = 0.76 M F– 1 mol BaF2 1 mol BaF2

  19. Answer Key – Ch. 4 c. What is the concentration of the fluoride ion if 350 mL of water is added? since the volume will go up the solution will get diluted – if you’re only adding solvent the moles of solute will remain constant ⇒ M1V1 = M2V2 (0.76 M F–)(750 mL) = (M2)(750 mL + 350 mL) M2 = 0.52 M F– d. What volume of the initial solution would you use in order to make 80 mL of 0.10 M Ba2+? M1V1 = M2V2 (0.38 M Ba2+)(V1) = (0.10 M Ba2+)(80 mL) V1 = 21 mL

  20. Answer Key – Ch. 4 5. What is the resulting concentration of bromide ion if 28 mL of 0.15 M sodium bromide solution were mixed with 45 mL of 0.12 M calcium bromide solution? NaBr (s)  Na+ (aq) + Br– (aq) 0.15 M NaBr ⇒0.15 M Br– (0.15 mol/L)(28 mL) = 4.2 mmol Br– (before mixing) CaBr2 (s)  Ca2+ (aq) + 2 Br– (aq) 0.12 M CaBr2⇒ 0.24 M Br– (0.24 mol/L)(45 mL) = 10.8 mmol Br– (before mixing) After mixing ⇒ 4.2 mmol Br– + 10.8 mmol = 15 mmol Br– Final concentration ⇒ (15 mmol Br– )/(28 mL + 45 mL) = 0.21 M Br–

  21. Answer Key – Ch. 4 6. What volume of 0.150 M Li3PO4 in milliliters is required to precipitate all of the mercury(II) ions from 250 mL of 0.32 M Hg(NO3)2? First write a balanced chemical reaction 2 Li3PO4(aq) + 3 Hg(NO3)2 (aq) 6 LiNO3(aq) + Hg3(PO4)2(s) M1V1 = M2V2 coeff coeff (0.15 M)(V1)/2 = (0.32 M)(250 mL)/3 V1 = 356 mL of Li3PO4

  22. Answer Key – Ch. 4 7. A precipitate will form when 10 mL of 0.25 M calcium chloride is mixed with 8 mL of 0.4 M silver nitrate. a. Write the molecular and net ionic equation for this reaction Molecular ⇒ CaCl2(aq)+ 2 AgNO3(aq) Ca(NO3)2(aq)+ 2 AgCl(s) Net ionic ⇒ Ca2+and NO3– are the spectator ions Ag+(aq)+Cl–(aq) AgCl(s) b. How many grams of precipitate form? First determine the LR …continue to next slide

  23. Answer Key – Ch. 4 7. …continued CaCl2⇒ (10 mL)(0.25 M) = 2.5 mmols/1 = 2.5 AgNO3⇒ (8 mL)(0.4 M) = 3.2 mmols/2 = 1.6 AgNO3 is the LR 3.2 mmols AgNO3 x 1mol AgCl x 143.32 g AgCl = 459 mg AgCl c. What is the concentration of the ions remaining in solution? Spectator ions don’t get consumed ⇒ [Ca2+] = (2.5 mmol)/(10 mL + 8 mL) = 0.14 M Ca2+ [NO3–] = (3.2 mmol)/(18 mL) = 0.18 M NO3– Reacting ions get consumed ⇒ [Ag+] = (3.2 mmol – 3.2 mmol) = 0 M (consumed entirely b/c it’s the LR) [Cl–] = (5 mmol – 3.2 mmol)/(18 mL) = 0.1 M Cl– 1 mol AgNO3 1 mol AgCl or 0.459 g AgCl

  24. Answer Key – Ch. 4 8. What type of solution results (acidic, basic or neutral) when 110 mL of 0.40 M HNO3 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the molecular and net ionic equations. Molecular ⇒ 2 HNO3(aq)+ Ba(OH)2(aq) 2 H2O(l)+ Ba(NO3)2(aq) Net Ionic ⇒ H+(aq) + OH–(aq) H2O (l) H+⇒ (110 mL)(0.4 M) = 44 mmols OH–⇒ (42 mL)(1.8 M) = 75.6 mmols The solution will be basic because the moles of OH – exceed the moles of H+

  25. Answer Key – Ch. 4 9. What volume of 0.15 M H2SO4 is required to neutralize 60 mL of 0.80 M KOH? H2SO4 + 2 KOH  2 H2O + K2SO4 M1V1 = M2V2 Coeff Coeff (0.15 M)(V1)/1 = (0.8 M)(60 mL)/2 V1 = 160 mL

  26. Answer Key – Ch. 4 10. Assign oxidation states for the following: a. S in SO3⇒ S + 3(-2) = 0 ⇒ S = +6 b. P in PF5 ⇒ P + 5(-1) = 0 ⇒ P = +5 c. Cl in Cl2 ⇒ Cl = 0 d. Cr in Cr2O72 - ⇒2(Cr) + 7(-2) = -2 ⇒ Cr = +6 e. Mn in KMnO4 ⇒ (+1) + Mn + 4(-2) = 0 ⇒ Mn = +7 f. H in CaH2 ⇒ (+2) + 2(H) = 0 ⇒ H = -1 g. O in H2O2 ⇒ 2(+1) + 2(O) = 0 ⇒ O = -1

  27. Answer Key – Ch. 4 11. Identify the oxidizing and reducing agents for the following reactions: a. Ba2+ + 2 Li  Ba + 2 Li+ Ba2+ is getting reduced ⇒ oxidizing agent Li is getting oxidized ⇒ reducing agent b. 8H+ + MnO4- + Fe2+ Mn2+ + Fe3+ 4H2O Mn in MnO4– is getting reduced ⇒ oxidizing agent Fe2+ is getting oxidized ⇒ reducing agent

  28. Answer Key – Ch. 4 12. Balance the following red-ox reactions: a. MnO2 + NO3- N2O4 + MnO4- (acidic) Reduction ½ : 2x(3 e– + 2 H2O + MnO2 MnO4– + 4 H+) Oxidation ½ : 3x(4 H+ + 2 NO3–  N2O4 + 2H2O + 2 e–) 2 MnO2 + 6 NO3– 4 H+  2 MnO4– + 3 N2O4 + 2 H2O b. CNO- + As2O3 CN- + HAsO42- (basic) Reduction ½ : 2x(2 e– + H2O + CNO– CN– + 2 OH–) Oxidation ½ : 8 OH– + As2O3 2 HAsO42– + 3 H2O + 4 e– 2 CNO– + As2O3 + 4 OH–  2 CN– + 2 HAsO42– + H2O …continue to next slide

  29. Answer Key – Ch. 4 12. …continued c. Cr3+ + SO42– Cr2O72– + H2SO3 (acidic) Reduction ½ : 3x(2 e– + 4 H+SO42- H2SO3 + H2O) Oxidation ½ : 7 H2O + 2 Cr3+ 6 e– + 14 H+ + Cr2O72– 3 SO42– + 2 Cr3+ + 4 H2O  2 H+ + 3 H2SO3 + CrO72–

  30. Answer Key – Ch. 4 13. It required 85 mL of 0.47 M K2Cr2O7 solution to oxidize all of the Fe2+ ions in a 200 mL solution. What was the initial concentration of Fe2+ ions? Cr2O72- (aq) + Fe2+ (aq)  Cr3+ (aq) + Fe3+ (aq) (acidic) First balance the reaction⇒ Reduction ½ : 6 e– + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O Oxidation ½ : 6x(Fe2+ Fe3+ + e–) 14 H+ + 6 Fe2+ + Cr2O72- 2 Cr3+ + 6 Fe3+ 7 H2O M1V1/coeff = M2V2/coeff (0.47 M)(85 mL)/1 = (M2)(200 mL)/6 M2 = 1.2 M Fe2+