1. 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? 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! Acid - Metal Limiting Reactant - I

2. Since 6 moles of HCl yield 2 moles of AlCl 3 ________ moles of HCl will yield: Acid - Metal Limiting Reactant - II

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

4. 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? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6 H 2 O (g) 30.0g NH 3 40.0g O 2 _______ fewer, therefore ______ is the Limiting Reagent! Moles NO formed = Mass NO = Ostwald Process Limiting Reactant Problem

5. 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? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6 H 2 O (g) 30.0g NH 3 / (17.0g NH 3 /mol NH 3 ) = 1.76 mol NH 3 1.76 mol NH 3 / 4 = 0.44 eq. NH 3 40.0g O 2 / (32.0g O 2 /mol O 2 ) = 1.25 mol O 2 1.25 mol O 2 / 5 = 0.25 eq O 2 Oxygen fewer, therefore oxygen is the Limiting Reagent! 1.25 mol O 2 x = 1.00 mol NO mass NO = 1.00 mol NO x = 30.0 g NO 4 mol NO 5 mol O 2 30.0 g NO 1 mol NO Ostwald Process Limiting Reactant Problem

6. Fig. 3.10

7. 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): % Yield = x 100% Actual Yield (mass or moles) Theoretical Yield (mass or moles) or Actual yield = Theoretical yield x (% Yield / 100%)

8. Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe 3 O 4 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 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) Moles Fe = Theoretical moles Fe 3 O 4 = Theoretical mass Fe 3 O 4 = Percent Yield = x 100% = Actual Yield Theoretical Yield

9. Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe 3 O 4 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 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) 4.55 g Fe 55.85 g Fe mol Fe = 0.081468 mol Fe = 0.0815 mol Fe 0.0815 mol Fe x = 0.0272 mol Fe 3 O 4 1 mol Fe 3 O 4 3 mol Fe 0.0272 mol Fe 3 O 4 x = 6.30 g Fe 3 O 4 231.55 g Fe 3 O 4 1 mol Fe 3 O 4 Percent Yield = x 100% = x 100% = 95.6 % Actual Yield Theoretical Yield 6.02 g Fe 3 O 4 6.30 g Fe 3 O 4

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. N 2 (g) + 3 H 2 (g) 2 NH 3 (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 N 2 = moles H 2 = eqs N 2 = eqs H 2 =

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. N 2 (g) + 3 H 2 (g) 2 NH 3 (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: moles N 2 = = 3.066 mol N 2 85.90 g N 2 28.02 g N 2 1 mole N 2 moles H 2 = = 10.74 mol H 2 21.66 g H 2 2.016 g H 2 1 mole H 2 Divide by coefficient to get eqs. of each: 3.066 g N 2 1 10.74 g H 2 3 = 3.066eq = 3.582eq

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

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

14. Molarity (Concentration of Solutions)= M M = = Moles of Solute Moles Liters of Solution L 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? Na 3 PO 4 (s) + H 2 O (solvent) = 3 Na + (aq) + PO 4 -3 (aq)

17. Preparing a Solution - II Molar mass of Na 3 PO 4 = g / mol mol Na 3 PO 4 = dissolve and dilute to 300.0 mL = volume of solution M in Na 3 PO 4 (aq) = M in PO 4 -3 ions = M in Na + ions =

18. Preparing a Solution - II Molar mass of Na 3 PO 4 = 163.94 g / mol 3.95 g / (163.94 g/mol) = 0.0241 mol Na 3 PO 4 dissolve and dilute to 300.0 mL = volume of solution M = 0.0241 mol Na 3 PO 4 / 0.300 L = 0.0803 mol / L = 0.0803 M in Na 3 PO 4 for PO 4 -3 ions = 0.0803 M in PO 4 -3 ions for Na + ions = 3 x 0.0803 M = 0.241 M in Na + ions

19. Fig. 3.12

20. Make a Solution of Potassium Permanganate Potassium permanganate is KMnO 4 and has a molecular mass of 158.04 g / mole. Problem: Prepare a 0.0400 M solution of KMnO 4 by dissolving solid KMnO 4 into water until the final volume of solution is 250. mL. What mass of KMnO 4 is needed? moles KMnO 4 = mass KMnO 4 = Molarity of K + ion = [K + ] ion = [MnO 4 - ] ion = 0.0400 M since 1:1

21. Make a Solution of Potassium Permanganate Potassium Permanganate is KMnO 4 and has a molecular mass of 158.04 g / mole. Problem: Prepare a 0.0400 M solution of KMnO 4 by dissolving solid KMnO 4 into water until the final volume of solution is 250. mL. What mass of KMnO 4 is needed? moles KMnO 4 = x 0.250 L soln. = 0.0100 mol KMnO 4 mass KMnO 4 = 0.0100 moles KMnO 4 x (158.04 g / mole) = 1.58 g KMnO 4 0.0400 mole KMnO 4 1.000 L soln. Molarity of K + ion = [K + ] ion = [MnO 4 - ] ion = 0.0400 M

22. Dilution of Solutions Take 25.00 mL of the 0.0400 M KMnO 4 Dilute the 25.00 mL to 1.000 L. - What is the resulting molarity (M) of the diluted solution? # moles KMnO 4 = Vol 1 x M 1 = M 2 = final M KMnO 4 = # moles / Vol 2 = Note: V 1 x M 1 = moles solute = V 2 x M 2

23. Dilution of Solutions Take 25.00 mL of the 0.0400 M KMnO 4 Dilute the 25.00 mL to 1.000 L. - What is the resulting Molarity of the diluted solution? # moles KMnO 4 = Vol 1 x M 1 = 0.0250 L x 0.0400 M = 0.00100 Moles M 2 = final M KMnO 4 = # moles / Vol 2 = 0.00100 Mol / 1.000 L = 0.00100 M Note: V 1 x M 1 = moles solute = V 2 x M 2

24. Fig. 3.13 V 1 x M 1 = moles solute = V 2 x M 2

25. Chemical Equation Calc - III ReactantsProducts Molecules Moles Mass Molecular Mass (amu) Molar mass M = g/mol Atoms (Molecules) Avogadro’s Number 6.02 x 10 23 Solutions (Volume) Molarity M = moles/liter

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

27. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH) 3 (s) + 3 HCl (aq) 3 H 2 O(l) + AlCl 3 (aq) Mass (g) of Al(OH) 3 Moles of Al(OH) 3 Moles of HCl Volume (L) of HCl ÷M (g/mol) x molar ratio ÷ M (mol/L) Problem: Given 10.0 g Al(OH) 3 (s), what volume of 1.50 M HCl(aq) is required to neutralize the base? 10.0 g Al(OH) 3

28. Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH) 3 (s) + 3 HCl (aq) 3 H 2 O(l) + AlCl 3 (aq) Mass (g) of Al(OH) 3 Moles of Al(OH) 3 Moles of HCl Volume (L) of HCl ÷M (g/mol) x molar ratio ÷ M (mol/L) Problem: Given 10.0 g Al(OH) 3 (s), what volume of 1.50 M HCl(aq) is required to neutralize the base? 10.0 g Al(OH) 3 78.00 g/mol = 0.128 mol Al(OH) 3 0.128 mol Al(OH) 3 x 3 moles HCl moles Al(OH) 3 = 0.385 Moles HCl Vol HCl = x 0.385 Moles HCl Vol HCl(aq) = 0.256 L = 256 mL 1.00 L HCl 1.50 Moles HCl

29. Solving Limiting Reactant Problems in Solution - Precipitation Problem - I 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(NO 3 ) 2 (aq) + Na 2 S (aq) PbS (s) 2 + NaNO 3 (aq) Like Sample Prob. #3.16

30. Volume (L) of Pb(NO 3 ) 2 solution Mass (g) of PbS Amount (mol) of Pb(NO 3 ) 2 Volume (L) of Na 2 S solution Amount (mol) of Na 2 S Amount (mol) of PbS Amount (mol) of PbS

31. Volume (L) of Pb(NO 3 ) 2 solution Mass (g) of PbS Amount (mol) of Pb(NO 3 ) 2 Volume (L) of Na 2 S solution Amount (mol) of Na 2 S Amount (mol) of PbS Amount (mol) of PbS Multiply by M (mol/L) Multiply by M (mol/L) x Molar Ratio Choose the lower number of PbS and multiply by M (g/mol)

32. Solving Limiting Reactant Problems in Solution - Precipitation Problem #I, cont. # Moles Pb(NO 3 ) 2 = V x M = # Moles Na 2 S = V x M = Therefore is the Limiting Reactant! Calculation of product yield: Moles PbS = Mass of PbS = # equivalents = # moles for both since stoich. coeff. = 1 for both.

33. Solving Limiting Reactant Problems in Solution - Precipitation Problem #I, cont. # Moles Pb(NO 3 ) 2 = V x M = 0.2578 L x (0.0468 Mol/L) = = 0.012065 Mol Pb +2 # Moles Na 2 S = 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: Moles PbS = 0.012065 Mol Pb +2 x = 0.012065 Mol Pb +2 1 mol PbS 1 mol Pb(NO 3 ) 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: AgNO 3 (aq) + Na 2 CrO 4 (aq) Ag 2 CrO 4 (s) + NaNO 3 (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 = NO 3 -, Na ion = Na +, chromate = CrO 4 2- Therefore balanced reaction is: 2AgNO 3 (aq) + Na 2 CrO 4 (aq) Ag 2 CrO 4 (s) + 2 NaNO 3 (aq)

36. Volume (L) of AgNO 3 solution Mass (g) of Ag 2 CrO 4 Amount (mol) of AgNO 3 Volume (L) of Na 2 CrO 4 solution Amount (mol) of Na 2 CrO 4 Equivalent Amount (mol) of Ag 2 CrO 4 Equivalent Amount (mol) of Ag 2 CrO 4 Multiply by M (mol/L) Multiply by M (mol/L) x Molar Ratio Choose the lower number of Ag 2 CrO 4 and multiply by M (g/mol)

37. Solving Limiting Reactant Problems in Solution - Precipitation Problem #II - cont. Moles AgNO 3 = V x M = Moles Ag 2 CrO 4 = Moles Na 2 CrO 4 = V x M = Moles Ag 2 CrO 4 = __________________ is the Limiting Reactant (i.e., less max yield)! Calculation of product yield: Mass Ag 2 CrO 4 = (moles Ag 2 CrO 4 from limiting reactant) x M Ag2CrO4 =

38. Solving Limiting Reactant Problems in Solution - Precipitation Problem #II - cont. Moles AgNO 3 = V x M = 0.2578 L x (0.0468 mol/L) = 0.012065 mol AgNO 3 Moles Ag 2 CrO 4 = 0.012065 mol AgNO 3 x (molar ratio in rxn) = 0.012065 mol AgNO 3 x (1 mol Ag 2 CrO 4 / 2 mol AgNO 3 ) = 0.006033 mol (or eqvs.) Ag 2 CrO 4 = max possible yield Moles Na 2 CrO 4 = V x M = 0.156 L x (0.095 mol/L) = 0.0148 mol Na 2 CrO 4 Moles Ag 2 CrO 4 = 0.012065 mol Na 2 CrO 4 x (molar ratio in rxn) = 0.0148 mol (or eqvs.) Ag 2 CrO 4 = max possible yield = more eqvs. than from AgNO 3. Therefore: Silver Nitrate is the Limiting Reactant (i.e., less max yield)! Calculation of product yield: Mass Ag 2 CrO 4 = (moles Ag 2 CrO 4 from limiting reactant) x M Ag2CrO4 = 0.006033 mol x (331.8 g/mol) = 2.00 grams Ag 2 CrO 4