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Round 2 Safety Quiz Fails
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1. Round 2 Safety Quiz Fails

2. Figure 3.11: Steps in a stoichiometric calculation.

3. Limiting Reactant The limiting reactant is the reactant in a chemical reaction which limits the amount of products that can be formed. The limiting reactant in a chemical reaction is present in insufficient quantity to consume the other reactant(s). This situation arises when reactants are mixed in non-stoichiometric ratios.

4. Conceptual Problem 3.13

5. Conceptual Problem 3.18

6. Practice Problem 3.86

7. O2 0.7291 mol SO2 O2 is limiting  46.7 g SO2 CS2  0.7879 mol SO2 2.23 g CS2 remaining

8. Limiting Reactant Example 1 76.15 g/mol 32.00 g/mol

9. Using Stoichiometry Stoichiometry is the basis for answering two fundamental questions in chemical analysis: What is the theoretical yield? What is the limiting reactant? REMEMBER: stoichiometry shows molar ratios not mass ratios

10. Percent yield actual yield % yield = actual yield: observed yield of product theoretical yield: calculated assuming 100% conversion of the LIMITING REAGENT x 100 theoretical yield

11. Theoretical Yield: Which Reactant is Limiting? 1) calculate moles (or mass) of product formed by complete reaction of each reactant. 2) the reactant that yields the least product is the limiting reactant (or limiting reagent). 3) the theoretical yield for a reaction is the maximum amount of product that could be generated by complete consumption of the limiting reagent.

12. Limiting Reactant Example 2 • When 66.6 g of O2 gas is mixed with 27.8 g of NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is produced by the following reaction: • 2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O • 16.04 17.03 32.00 27.03 g/mol • What is the % yield of HCN in this reaction? • How many grams of NH3 remain?

13. Mass to moles 66.6 g of O2→ 2.08 mol O2 27.8 g of NH3 → 1.63 mol NH3 25.1 g of CH4→ 1.56 mol CH4 Which reactant is limiting? 2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN 1.63 mol NH3 can yield 1.63 mol (or 44.1 g) HCN 1.56 mol CH4 can yield 1.56 mol (or 42.2 g) HCN Conclusion? O2 is the limiting reagent.

14. % yield = actual yield O2 is the limiting reagent. So, the theoretical yield is based on 100% consumption of O2. 2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN x 100 theoretical yield % yield = 36.4 g HCN x 100 = 97.1% 37.5 g HCN

15. 2. How many grams of NH3 remain? 36.4 g (or 1.35 mol) of HCN gas is produced 2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O Since the reaction stoichiometry is 1:1, 1.35 mol of NH3 is consumed: 1.63 mol NH3initially present – 1.35 mol NH3 consumed 0.28 mol NH3 remaining 0.28 mol NH3 x (17.03 g NH3/mol) = 4.8 g NH3 remain

16. Limiting Reactant Example 3 4NH3 + 5O2→ 4NO + 6H2O Add: 14 mol 20 mol Could make 16 mol NO Could make 14 mol NO NH3 is the limiting reagent. (Use this as basis for all further calculations)

17. Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature: 2NH3 + CO2(g)  CH4N2O + H2O In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions?

18. Molar masses NH3 1(14.01) + 3(1.008) = 17.02 g CO2 1(12.01) + 2(16.00) = 44.01 g CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = 60.06 g CO2 is the limiting reactant. 13.6 g CH4N2O will be produced.

19. To find the excess NH3, we find how much NH3 reacted: Now subtract the amount reacted from the starting amount: 10.0 at start -7.73 reacted 2.27 g remains 2.3 g NH3 is left unreacted. (1 decimal place)

20. 2NH3 + CO2(g)  CH4N2O + H2O When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield? Theoretical yield = 13.6 g Actual yield = 9.3 g = 68% yield (2 significant figures)

21. Other Resources Visit the student website at college.hmco.com/pic/ebbing9e

22. Chapter 4 Chemical Reactions: An Introduction

23. Figure 4.1: Reaction of potassium iodide solution and lead (II) nitrate solution. Photo courtesy of James Scherer.

24. Electrolytes

25. Figure 4.2: Motion of ions in solution.

26. Figure 4.3: Testing the electrical conductivity of a solution: water.Photo courtesy of American Color.

27. Figure 4.3: Testing the electrical conductivity of a solution: sodium chloride.Photo courtesy of American Color.

28. Figure 4.4: Comparing strong and weak electrolytes: HCl. Photo courtesy of American Color.

29. Figure 4.4: Comparing strong and weak electrolytes: NH3. Photo courtesy of American Color.

30. Methanol Li

31. The Role of Water as a Solvent: The Solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called strong Eeectrolytes. NaCl(s) + H2O(l) Na+(aq) + Cl -(aq) When sodium chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

32. Fig. 4.2

33. The Solubility of Ionic Compounds in Water The solubility of ionic compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20oC = 365 g/L Solubility of MgCl2 in water at 20oC = 542.5 g/L Solubility of AlCl3 in water at 20oC = 699 g/L Solubility of PbCl2 in water at 20oC = 9.9 g/L Solubility of AgCl in water at 20oC = 0.009 g/L Solubility of CuCl in water at 20oC = 0.0062 g/L

34. Precipitation Reactions: Will a Precipitate Form? If we add a solution containing potassium chloride to a solution containing ammonium nitrate, will we get a precipitate? KCl(aq) + NH4NO3 (aq) = K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) By exchanging cations and anions we see that we could have potassium chloride and ammonium nitrate, or potassium nitrate and ammonium chloride. In looking at the solubility table it shows all possible products as soluble, so there is no net reaction! KCl(aq) + NH4NO3 (aq) = No Reaction! If we mix a solution of sodium sulfate with a solution of barium nitrate, will we get a precipitate? From the solubility table it shows that barium sulfate is insoluble, therefore we will get a precipitate! Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

35. Solubility • Soluble = ability to dissolve in a liquid • Insoluble = inability to dissolve in a liquid • Not all Ionic Compounds are water soluble • Not all molecular compounds are insoluble!

36. 1. Group IA and ammonium compounds are soluble. • 2. Acetates and nitrates are soluble. • 3. Most chlorides are soluble. • Exceptions: AgCl, Hg2Cl2, PbCl2; AgBr, Hg2Br2, HgBr2, PbBr2; AgI, Hg2I2, HgI2, PbI2 • 4. Most sulfates are soluble. • Exceptions: CaSO4, SrSO4, BaSO4, Ag2SO4, Hg2SO4, PbSO4 Solubility Rules

37. 5. Most carbonates are insoluble. Exceptions: Group IA carbonates and (NH4)2SO4 6. Most phosphates are insoluble. Exceptions: Group IA phosphates and (NH4)3PO4 7. Most sulfides are insoluble. Exceptions: Group IA sulfides and (NH4)2S 8. Most hydroxides are insoluble. Exceptions: Group IA hydroxides, Ca(OH)2, Sr(OH)2, Ba(OH)2

38. Song For Solubility!! (Taken from Cornell University – Adapted by Daley – Sing to Rhythm of 99 Bottles)  Potassium, sodium, and ammonium salts,  Whatever they may be, Can always be depended on For solubility. Asked about the nitrates or acetates The answer is always clear, They each and all are soluble, Is all we want to hear.  Most every chloride's soluble At least we've always read Save silver, mercurous mercury And (slightly) chloride of lead. Take the Bromide and iodide salts There soluble as can be Save silver, mercury, and lead That precipitate as you see Every single sulfate Is soluble ,  'Tis said 'Cept barium and strontium And calcium and lead.  Hydroxides of metals won't dissolve That is, all but three Potassium, sodium and ammonium Dissolve quite readily. And then you must remember That you must not "forgit" Calcium, barium, strontium Dissolve a little bit.  The carbonates are insoluble,  It's lucky that it's so, Or else, our marble buildings Would melt away like snow.  (Repeat with feeling)  Only note is that all Lithium salts are Soluble too!!!

39. Molecular Equation • A chemical equation in which the reactants and products are written as if they were molecular substances, even though they may actually exist in solution as ions. • State symbols are include: (s), (l), (g), (aq). • For example: • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • Although AgNO3, NaCl, and NaNO3 exist as ions in aqueous solutions, they are written as compounds in the molecular equation.

40. Complete Ionic Equation • A chemical equation in which strong electrolytes are written as separate ions in the solution. Other reactants and products are written in molecular form. State symbols are included: (s), (l), (g), (aq). • For example: • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) • In ionic form: • Ag+(aq) + NO3-(aq) + Na+(aq)Cl-(aq)  • AgCl(s) + Na+(aq) + NO3-(aq)

41. Spectator Ion • An ion in an ionic equation that does not take part in the reaction. It appears as both a reactant and a product.

42. Net Ionic Equation • A chemical equation in which spectator ions are omitted. It shows the reaction that actually occurs at the ionic level. • For example: • Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  • AgCl(s) + Na+(aq) + NO3-(aq) • In net ionic form: • Ag+(aq) + Cl-(aq)  AgCl(s)

43. Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. • KBr + MgSO4 • Determine the product formulas: • K+ and SO42- make K2SO4 • Mg2+and Br- make MgBr2 • Determine whether the products are soluble: • K2SO4 is soluble • MgBr2 is soluble • KBr + MgSO4 no reaction

44. Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. • NaOH + MgCl2 • Determine the product formulas: • Na+ and Cl- make NaCl • Mg2+and OH- make Mg(OH)2 • Determine whether the products are soluble: • NaCl is soluble • Mg(OH)2 is insoluble

45. Molecular Equation • (Balance the reaction and include state symbols) • 2NaOH(aq) + MgCl2(aq)  • 2NaCl(aq) + Mg(OH)2(s) • Ionic Equation • 2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)  • 2Na+(aq) + 2Cl-(aq) + Mg(OH)2(s) • Net Ionic Equation • 2OH-(aq) + Mg2+(aq)  Mg(OH)2(s)

46. Decide whether the following reaction occurs. If it does, write the molecular, ionic, and net ionic equations. • K3PO4 + CaCl2 • Determine the product formulas: • K+ and Cl- make KCl • Ca2+and PO43- make Ca3(PO4)2 • Determine whether the products are soluble: • KCl is soluble • Ca3(PO4)2 is insoluble