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Theoretical Yield: Which Reactant is Limiting?

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).

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Theoretical Yield: Which Reactant is Limiting?

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  1. 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.

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

  3. Limiting Reactant Example 2 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)

  4. Limiting Reactant Example 3 • 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?

  5. 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.

  6. % 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

  7. 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

  8. 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?

  9. 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.

  10. 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)

  11. 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)

  12. Acid-Base

  13. Two Definitions of Acids & Bases • 1) Arrhenius: When dissolved in water, • acids produce H+ HCl(aq)→ H+(aq) + Cl-(aq) • bases produce OH- NaOH(aq)→ Na+(aq) + OH-(aq) • 2) Brønsted-Lowry: Proton transfer • acids are proton donors • bases are proton acceptors NaOH(aq) + HCl(aq)→? Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)→ ? NaOH(aq) + HCl(aq)→ H2O(l) + NaCl(aq)

  14. Acids - A Group of Covalent Molecules Which Lose Hydrogen Ions to Water Molecules in Solution When gaseous hydrogen iodide dissolves in water, the attraction of the oxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the iodide ion for the hydrogen atom, and it is lost to the water molecule to form an hydronium ion and an iodide ion in solution. We can write the hydrogen atom in solution as either H+(aq) or as H3O+(aq) they mean the same thing in solution. The presence of a hydrogen atom that is easily lost in solution is an “Acid” and is called an “acidic” solution. The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved. HI(g) + H2O(L) H+(aq) + I -(aq) HI(g) + H2O(L) H3O+(aq) + I -(aq) H2O HI(g) H+(aq) + I -(aq)

  15. Figure 4.8B: Red cabbage juice added to solutions in the beakers.Photo courtesy of James Scherer.

  16. Molecular representation of ammonium hydroxide. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

  17. Reaction of nitric acid with water. HNO3(aq)+ H2O(l)  NO3-(aq) + H3O+(aq)

  18. “salt” water Two Types of Acid-Base Reactions 1) A-B Neutralization: LiOH(aq) + HCN(aq)→ LiCN(aq) + H2O(l) H2SO4(aq) + Ca(OH)2(aq)→? gas 2) A-B Reactions with Gas Formation: Na2CO3(aq) + 2HBr(aq)→ 2NaBr(aq) + H2O(l) + CO2(g) Li2SO3(aq) + NaOH(aq)→? Which salts?carbonates  CO2 sulfites  SO2 sulfides  H2S

  19. Beaker with Na+(aq), C2H302-(aq), and SrS04 (solid). Na2SO4 (aq)+ Sr(C2H3O2)2(aq)  SrSO4 (s)+ NaC2H3O2 (aq) SO4-2 (aq)+ Sr+2(aq) SrSO4 (s) 2 Na+(aq) + SO4-2 (aq)+ Sr+2(aq) + 2C2H3O2-(aq)  SrSO4 (s)+ 2 Na+(aq) + 2C2H3O2-(aq)

  20. Figure 4.9: Reaction of a carbonate with an acid.Photo courtesy of American Color.

  21. Oxidation-Reduction 2Na (s) + Cl2(g) 2NaCl(s)

  22. Figure 4.10: Iron nail and copper ( II) sulfate.Photo courtesy of American Color.

  23. Figure 4.10: Fe reacts with Cu2+(aq) and makes Cu(s).

  24. Figure 4.10: The copper metal plates out on the nail. Write a net ionic equation for this reaction! Cu+2(aq) + Fe(s) Cu(s) + Fe+2(aq)

  25. Redox: changing oxidation numbers Oxidation number: a real charge (ionic compounds) or hypothetical charge (molecular compounds, polyatomic ions) associated with an individual atom in a compound. The oxidation number allows for electron accounting N2(g) + 3H2(g)→2NH3(g) CH4(g) + 2O2(g)→CO2(g)+ 2H2O(g)

  26. Rules for Assigning Oxidation Numbers

  27. Molar concentration “Molarity” or “M” mol of solute liter of solution NOT: mol solute liter of solvent = M 0.2500 L mark

  28. Tips for Molarity-Based Calculations • Use molarity to convert volume of solution to moles of solute: Mi × Vi = mol solute • Use Mi × Vi = Mf × Vfto calculate concentrations of solutions after dilution • Never use this for reactions(e.g. neutralization) • Use Mi × Vi =mol solute& stoichiometry to calculate concentrations of sample solutions in reactions (e.g. titrations).

  29. Molarity (Concentration of Solutions)= M Moles of Solute Moles Liters of Solution L M = = solute = material dissolved into the solvent In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water , Water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

  30. Fig. 3.11

  31. Preparing a Solution - I Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l ! What is the Molarity of the salt and each of the ions? Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

  32. Preparing a Solution - II Mol wt of Na3PO4 = 163.94 g / mol 3.95 g / 163.94 g/mol = 0.0241 mol Na3PO4 dissolve and dilute to 300.0 ml M = 0.0241 mol Na3PO4 / 0.300 l = 0.0803 M Na3PO4 for PO4-3 ions = 0.0803 M for Na+ ions = 3 x 0.0803 M = 0.241 M

  33. Concept Check Stoichiometry & Ion Dissociation For a 0.27 M aq. solution of sodium carbonate: • Write the dissociation reaction & identify solute(s). • Find the molarity of Na+(aq) & CO32-(aq). • If you had 185 mL of this solution, how many moles of Na+(aq) & CO32-(aq) would be present? • If you added excess MgBr2(aq), would you expect a rxn.? If so, how many moles of solid would form?

  34. Converting a Concentrated Solution to a Dilute Solution

  35. mol solute liter of solution x = liter of sample mol solute MixVi = mol solute=MfxVf i = initial f = final Add more solvent (“Dilution”) Mass of solute does not change! Constant!

  36. The Dilution Dogma: NEVER FORGET IT! M1V1=M2V2

  37. Dilution of Solutions Take 25.00 ml of the 0.0400 M KMnO4 Dilute the 25.00 ml to 1.000 l - What is the resulting Molarity of the diluted solution? # moles = Vol x M 0.0250 l x 0.0400 M = 0.00100 Moles 0.00100 Mol / 1.00 l = 0.00100 M

  38. Concept Check Dilution Calculation How much 0.20 M HCl is needed to make 50 mL of 10 mM HCl solution? Mi×Vi = Mf×Vf (0.20 M HCl)×(L) = (0.010 M HCl)×(0.050 L)  (L) = (0.010 M HCl)×(0.050 L) = 2.5×10-3L 0.20 M HCl

  39. Concept Check Dilution & Solution Examples A) We have a 3.0 M aqueous solution of H2SO4. How do you make 100. mL of 1.4 M H2SO4(aq)? B) Determine how you would make 250. mL of 0.56 M CaCl2(aq)? What is [Ca2+]? [Cl-]? (Reagent is solid is CaCl2·2H20; 147.01 g/mol) C) Predict what would happen in you mixed solutions A and B together. H2SO4(aq) + CaCl2(aq) →?

  40. How could you make 5.0 L of 0.025 M sucrose from a solution which is 0.100 M sucrose? Mix 1.250 L of 0.100 M sucrose with 3.75 L water.

  41. “salt” water Two Types of Acid-Base Reactions 1) A-B Neutralization: LiOH(aq) + HCN(aq)→ LiCN(aq) + H2O(l) H2SO4(aq) + Ca(OH)2(aq)→? gas 2) A-B Reactions with Gas Formation: Na2CO3(aq) + 2HBr(aq)→ 2NaBr(aq) + H2O(l) + CO2(g) Li2SO3(aq) + NaOH(aq)→? Which salts?carbonates  CO2 sulfites  SO2 sulfides  H2S

  42. Concept Check Neutralization Calculation • How much 2.0 M HCl is needed to “neutralize” 2.3 liters of 0.15 M NaOH? HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l) acid + base → “salt” + water or….. How much 2.0 M HCl should be added such that the mol of H+(aq)=mol OH–(aq)? N.I.E.: H+(aq) + OH-(aq)→ H2O(l)

  43. Neutralization Calculations Determine the volume of 0.10 M KOH(aq)solution required to neutralize 25.00 mL samples of three different acids, all at 0.20 M. Acids: 1) Nitric acid 2) Carbonic acid 3) Phosphoric acid • 1) __ HNO3 (aq) + __ KOH (aq)→ • 2) + __ KOH (aq)→ • 3) + __ KOH (aq)→

  44. Titration Calculation 4.49 mL of 0.2500 M NaOH is required to titrate a 25.00 mL sample of H2SO4 to the endpoint. What is the molar concentration of H2SO4 in the sample? NaOH(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l) 2NaOH(aq) + 1H2SO4(aq) Na2SO4(aq) + 2H2O(l) Stoichiometry not 1:1 !!!

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