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2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g)

Acid - Metal Limiting Reactant - I. 2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed?

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2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g)

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  1. Acid - Metal Limiting Reactant - I • 2Al(s) + 6HCl(g) 2AlCl3(s) + 3H2(g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? • 30.0 g Al / (26.98 g Al / mol Al) = 1.11 mol Al 1.11 mol Al / 2 = 0.555 equivalents Al • 20.0g HCl / (36.5gHCl / mol HCl) = 0.548 mol HCl O.548 mol HCl / 6 = 0.0913 equivalents HCl • HCl is smaller therefore the Limiting reactant!

  2. Acid - Metal Limiting Reactant - II • Since 6 moles of HCl yield 2 moles of AlCl3 ________ moles of HCl will yield:

  3. Acid - Metal Limiting Reactant - II • Since 6 moles of HCl yield 2 moles of AlCl3 0.548 moles of HCl will yield: 0.548 mol HCl x (2 moles of AlCl3 / 6 mol HCl) = 0.183 mol of AlCl3

  4. Ostwald Process Limiting Reactant Problem • What mass of NO could be formed by the reaction 30.0g of ammonia gas and 40.0g of oxygen gas w/ the rxn below? 4NH3 (g) + 5 O2 (g) 4NO(g) + 6 H2O(g) • 30.0g NH3 • 40.0g O2 • _______ fewer, therefore ______ is the Limiting Reagent! • Moles NO formed = • Mass NO =

  5. Ostwald Process Limiting Reactant Problem • What mass of NO could be formed by the reaction 30.0g of ammonia gas and 40.0g of oxygen gas w/ the rxn below? 4NH3 (g) + 5 O2 (g) 4NO(g) + 6 H2O(g) • 30.0g NH3 / (17.0g NH3/mol NH3) = 1.76 mol NH3 1.76 mol NH3 / 4 = 0.44 eq. NH3 • 40.0g O2 / (32.0g O2 /mol O2) = 1.25 mol O2 1.25 mol O2 / 5 = 0.25 eq O2 • Oxygen fewer, therefore oxygen is the Limiting Reagent! • 1.25 mol O2 x = 1.00 mol NO • mass NO = 1.00 mol NO x = 30.0 g NO 4 mol NO 5 mol O2 30.0 g NO 1 mol NO

  6. Fig. 3.10

  7. Actual Yield (mass or moles) Theoretical Yield (mass or moles) % Yield = x 100% Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form smaller amounts of different products that take away from the theoretical yield of the main product. Actual yield: The actual amount of product that is obtained. Percent yield (%Yield): or Actual yield = Theoretical yield x (% Yield / 100%)

  8. Actual Yield Theoretical Yield Percent Yield = x 100% = Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is reacted with sufficent water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe(s) + 4 H2O(l) Fe3O4 (s) + 4 H2 (g) Moles Fe = Theoretical moles Fe3O4 = Theoretical mass Fe3O4 =

  9. 1 mol Fe3O4 3 mol Fe 0.0815 mol Fe x = 0.0272 mol Fe3O4 Actual Yield Theoretical Yield 6.02 g Fe3O4 6.30 g Fe3O4 Percent Yield = x 100% = x 100% = 95.6 % Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is reacted with sufficent water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe(s) + 4 H2O(l) Fe3O4 (s) + 4 H2 (g) 4.55 g Fe 55.85 g Fe mol Fe = 0.081468 mol Fe = 0.0815 mol Fe 231.55 g Fe3O4 1 mol Fe3O4 0.0272 mol Fe3O4 x = 6.30 g Fe3O4

  10. Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen Gas. If 85.90g of nitrogen are reacted with 21.66 g hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: moles N2 = moles H2 = eqs N2 = eqs H2 =

  11. Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen Gas. If 85.90g of nitrogen are reacted with 21.66 g hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: Divide by coefficient to get eqs. of each: 3.066 g N2 1 10.74 g H2 3 85.90 g N2 28.02 g N2 1 mole N2 moles N2 = = 3.066 mol N2 = 3.066eq 21.66 g H2 2.016 g H2 1 mole H2 moles H2 = = 10.74 mol H2 = 3.582eq

  12. Percent Yield/Limiting Reactant Problem - II N2 (g) + 3 H2 (g) 2 NH3 (g) Solution Cont. We have 3.066 moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: mol NH3 = 3.066 mol N2 x mass NH3 = Actual Yield Theoretical Yield Percent Yield = x 100% 98.67 g NH3 g NH3 Percent Yield = x 100% =

  13. Percent Yield/Limiting Reactant Problem - II N2 (g) + 3 H2 (g) 2 NH3 (g) Solution Cont. We have 3.066 moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: 2 mol NH3 1 mol N2 3.066 mol N2 x = 6.132 mol NH3 (Theoretical Yield) 6.132 mol NH3 x = 104.427 g NH3 (Theoretical Yield) 17.03 g NH3 1 mol NH3 Actual Yield Theoretical Yield Percent Yield = x 100% 98.67 g NH3 104.427 g NH3 Percent Yield = x 100% = 94.49 %

  14. 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%)

  15. Fig. 3.11

  16. 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)

  17. Preparing a Solution - II • Molar mass of Na3PO4 = g / mol • mol Na3PO4 = • dissolve and dilute to 300.0 mL = volume of solution • M in Na3PO4(aq) = • M in PO4-3 ions = • M in Na+ ions =

  18. Preparing a Solution - II • Molar mass of Na3PO4 = 163.94 g / mol • 3.95 g / (163.94 g/mol) = 0.0241 mol Na3PO4 • dissolve and dilute to 300.0 mL = volume of solution • M = 0.0241 mol Na3PO4 / 0.300 L = 0.0803 mol / L = 0.0803 M in Na3PO4 • for PO4-3 ions = 0.0803 M inPO4-3 ions • for Na+ ions = 3 x 0.0803 M = 0.241 M inNa+ ions

  19. Fig. 3.12

  20. Make a Solution of Potassium Permanganate Potassium permanganate is KMnO4 and has a molecular mass of 158.04 g / mole. Problem: Prepare a 0.0400 M solution of KMnO4 by dissolving solid KMnO4 into water until the final volume of solution is 250. mL. What mass of KMnO4 is needed? moles KMnO4 = mass KMnO4 = Molarity of K+ ion = [K+] ion = [MnO4-] ion = 0.0400 M since 1:1

  21. Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO4 and has a molecular mass of 158.04 g / mole. Problem: Prepare a 0.0400 M solution of KMnO4 by dissolving solid KMnO4 into water until the final volume of solution is 250. mL. What mass of KMnO4 is needed? 0.0400 mole KMnO4 1.000 L soln. moles KMnO4 = x 0.250 L soln. = 0.0100 mol KMnO4 mass KMnO4 = 0.0100 moles KMnO4 x (158.04 g / mole) = 1.58 g KMnO4 Molarity of K+ ion = [K+] ion = [MnO4-] ion = 0.0400 M

  22. 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 (M) of the diluted solution? • # moles KMnO4 = Vol1 x M1 = • M2 = final M KMnO4 = # moles / Vol2 = Note: V1 x M1 = moles solute = V2 x M2

  23. 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 KMnO4 = Vol1 x M1 = • 0.0250 L x 0.0400 M = 0.00100 Moles • M2 = final M KMnO4 = # moles / Vol2 = 0.00100 Mol / 1.000 L = 0.00100 M Note: V1 x M1 = moles solute = V2 x M2

  24. Fig. 3.13 V1 x M1 = moles solute = V2 x M2

  25. Chemical Equation Calc - III Mass Atoms (Molecules) Molecular Mass (amu) Avogadro’s Number Molar mass M = g/mol 6.02 x 1023 Molecules Reactants Products Moles Molarity M = moles/liter Solutions (Volume)

  26. Calculating Mass of Solute from a Given Volume of Solution Volume (L) of Solution x Molarity (M = mol solute / Liters of solution = M/L) Moles of Solute x Molar Mass (M = mass / mole = g/mol) Mass (g) of Solute

  27. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl(aq) 3 H2O(l) + AlCl3 (aq) Problem: Given 10.0 g Al(OH)3(s), what volume of 1.50 M HCl(aq) is required to neutralize the base? Mass (g) of Al(OH)3 ÷M (g/mol) 10.0 g Al(OH)3 Moles of Al(OH)3 x molar ratio Moles of HCl ÷ M (mol/L) Volume (L) of HCl

  28. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH)3 (s) + 3 HCl(aq) 3 H2O(l) + AlCl3 (aq) Problem: Given 10.0 g Al(OH)3(s), what volume of 1.50 M HCl(aq) is required to neutralize the base? Mass (g) of Al(OH)3 ÷M (g/mol) 10.0 g Al(OH)3 78.00 g/mol = 0.128 mol Al(OH)3 Moles of Al(OH)3 3 moles HCl moles Al(OH)3 0.128 mol Al(OH)3 x = x molar ratio 0.385 Moles HCl Moles of HCl 1.00 L HCl 1.50 Moles HCl Vol HCl = x 0.385 Moles HCl Vol HCl(aq) = 0.256 L = 256 mL ÷ M (mol/L) Volume (L) of HCl

  29. Solving Limiting Reactant Problems in Solution - Precipitation Problem - I Like Sample Prob. #3.16 Problem: Lead has been used as a glaze for pottery for years, and can be a problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by lead precipitated as a sulfide. If 257.8 ml of a 0.0468 M solution of lead nitrate is added to 156.00 ml of a 0.095 M solution of sodium sulfide, what mass of solid lead sulfide will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Write the formulas for each ionic compound using the names of ions and their charges in Tables 2.3-2.5. Write the balanced equation: Pb(NO3)2 (aq) + Na2S(aq) PbS(s) 2 + NaNO3 (aq)

  30. Volume (L) of Pb(NO3)2 solution Volume (L) of Na2S solution Amount (mol) of Pb(NO3)2 Amount (mol) of Na2S Amount (mol) of PbS Amount (mol) of PbS Mass (g) of PbS

  31. Volume (L) of Pb(NO3)2 solution Volume (L) of Na2S solution Multiply by M (mol/L) Multiply by M (mol/L) Amount (mol) of Pb(NO3)2 Amount (mol) of Na2S x Molar Ratio x Molar Ratio Amount (mol) of PbS Amount (mol) of PbS Choose the lower number of PbS and multiply by M(g/mol) Mass (g) of PbS

  32. Solving Limiting Reactant Problems in Solution - Precipitation Problem #I, cont. # Moles Pb(NO3)2 = V x M = # Moles Na2S = V x M = # equivalents = # moles for both since stoich. coeff. = 1 for both. Therefore is the Limiting Reactant! Calculation of product yield: Moles PbS = Mass of PbS =

  33. Solving Limiting Reactant Problems in Solution - Precipitation Problem #I, cont. # Moles Pb(NO3)2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb+2 # Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2 # equivalents = # moles for both since stoich. coeff. = 1 for both. Therefore Lead Nitrate is the Limiting Reactant! Calculation of product yield: 1 mol PbS 1 mol Pb(NO3)2 Moles PbS = 0.012065 Mol Pb+2x = 0.012065 Mol Pb+2 0.012065 Mol Pb+2 = 0.012065 Mol PbS 0.012065 Mol PbS x = 2.89 g PbS 239.3 g PbS 1 Mol PbS

  34. Solving Limiting Reactant Problems in Solution - Precipitation Problem #II Problem: When aqueous silver nitrate and sodium chromate solutions are mixed, a reaction occurs that forms solid silver chromate and a solution of sodium nitrate. If 257.8 ml of a 0.0468 M solution of silver nitrate is added to 156.00 ml of a 0.095 M solution of sodium chromate, what mass of solid silver chromate (M = 331.8 g/mol) will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Write the formulas for each ionic compound using the names of ions and their charges in Tables 2.3-2.5. Write the balanced equation: Tables=> Ag ion = , nitrate = , Na ion = , chromate = Therefore balanced reaction is: AgNO3(aq) + Na2CrO4(aq) Ag2CrO4(s) + NaNO3(aq)

  35. Solving Limiting Reactant Problems in Solution - Precipitation Problem #II Problem: When aqueous silver nitrate and sodium chromate solutions are mixed, a reaction occurs that forms solid silver chromate and a solution of sodium nitrate. If 257.8 ml of a 0.0468 M solution of silver nitrate is added to 156.00 ml of a 0.095 M solution of sodium chromate, what mass of solid silver chromate (M = 331.8 g/mol) will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Write the formulas for each ionic compound using the names of ions and their charges in Tables 2.3-2.5. Write the balanced equation: Tables=> Ag ion = Ag+, nitrate = NO3-, Na ion = Na+, chromate = CrO42- Therefore balanced reaction is: 2AgNO3(aq) + Na2CrO4(aq) Ag2CrO4(s) + 2 NaNO3(aq)

  36. Volume (L) of AgNO3 solution Volume (L) of Na2CrO4 solution Multiply by M (mol/L) Multiply by M (mol/L) Amount (mol) of AgNO3 Amount (mol) of Na2CrO4 x Molar Ratio x Molar Ratio Equivalent Amount (mol) of Ag2CrO4 Equivalent Amount (mol) of Ag2CrO4 Choose the lower number of Ag2CrO4 and multiply by M(g/mol) Mass (g) of Ag2CrO4

  37. Solving Limiting Reactant Problems in Solution - Precipitation Problem #II - cont. • Moles AgNO3= V x M = Moles Ag2CrO4 = • Moles Na2CrO4= V x M = Moles Ag2CrO4 = __________________ is the Limiting Reactant (i.e., less max yield)! Calculation of product yield: Mass Ag2CrO4 = (moles Ag2CrO4 from limiting reactant) x MAg2CrO4 =

  38. Solving Limiting Reactant Problems in Solution - Precipitation Problem #II - cont. • Moles AgNO3= V x M = 0.2578 L x (0.0468 mol/L) = 0.012065 mol AgNO3 Moles Ag2CrO4 = 0.012065 mol AgNO3 x (molar ratio in rxn) = 0.012065 mol AgNO3 x (1 mol Ag2CrO4 / 2 mol AgNO3) = 0.006033 mol (or eqvs.) Ag2CrO4 = max possible yield • Moles Na2CrO4= V x M = 0.156 L x (0.095 mol/L) = 0.0148 mol Na2CrO4 Moles Ag2CrO4 = 0.012065 mol Na2CrO4 x (molar ratio in rxn) = 0.0148 mol (or eqvs.) Ag2CrO4 = max possible yield = more eqvs. than from AgNO3. Therefore: Silver Nitrate is the Limiting Reactant (i.e., less max yield)! Calculation of product yield: Mass Ag2CrO4 = (moles Ag2CrO4 from limiting reactant) x MAg2CrO4 = 0.006033 mol x (331.8 g/mol) = 2.00 grams Ag2CrO4

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