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Solving Mass  Moles Reaction Problems

Solving Mass  Moles Reaction Problems. grams. molar mass. moles. mass. M m =. mole. Generic Reaction: aA + bB → cC + dD. mole ratio. d a. mass A (g). moles A. moles D. mass D (g). Limiting Reactant.

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Solving Mass  Moles Reaction Problems

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  1. Solving Mass  Moles Reaction Problems grams molar mass moles mass Mm = mole Generic Reaction:aA + bB→cC + dD mole ratio d a mass A (g) moles A moles D mass D (g)

  2. 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.

  3. Figure 3.14: Limiting reactant analogy using cheese sandwiches.

  4. Ethanol combusts to form CO2 & water. __C2H5OH + __O2 __CO2 + __H2O • Balance the chemical equation • Identify as many mole ratios as you can • How many moles of oxygen (O2) react with 15.0 moles of ethanol • How many grams of O2 react with 15.0 moles of ethanol • How many g of CO2 are formed when 1.00 kg of ethanol is burned with 1.00 kg of O2

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

  6. Using Stoichiometry Stoichiometry is used to answer two fundamental questions in chemical analysis: What is the theoretical yield? What is the limiting reactant? REMEMBER: stoichiometry shows molar ratios not mass ratios

  7. Percent yield actual yield % yield = actual yield: observed yield of product theoretical yield: calculated assuming 100% conversion of the LIMITING REAGENT Both yields can be in moles or grams x 100 theoretical yield

  8. 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.

  9. 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: • 16.04 17.03 32.00 27.03 g/mol • 2CH4 + 2NH3 + 3O2→ 2HCN + 6H2O • What is the % yield of HCN in this reaction? • How many grams of NH3 remain?

  10. 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 HCN (or 37.5 g) 1.63 mol NH3 can yield 1.63 mol HCN (or 44.1 g) 1.56 mol CH4 can yield 1.56 mol HCN (or 42.2 g) Conclusion? O2 is the limiting reagent.

  11. % yield = actual yield O2 is the limiting reagent. Thus, theoretical yield is based on 100% consumption of O2. 2.08 mol O2can 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 moles could also be used

  12. 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

  13. An Ice Cream Sundae Analogy for Limiting Reactions Fig. 3.10

  14. Conceptual Problem 3.13

  15. Conceptual Problem 3.18

  16. Practice Problem 3.86

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

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

  19. 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?

  20. 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.

  21. % 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

  22. 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

  23. 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)

  24. 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?

  25. 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.

  26. 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)

  27. 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)

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

  29. Chapter 4 Chemical Reactions: An Introduction

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

  31. Electrolytes

  32. Figure 4.2: Motion of ions in solution.

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

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

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

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

  37. Methanol Li

  38. 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

  39. Fig. 4.2

  40. 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

  41. 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)

  42. 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!

  43. 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

  44. 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

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