1 / 60

REACTION STOICHIOMETRY

REACTION STOICHIOMETRY. 1792 JEREMIAS RICHTER The amount of substances produced or consumed in chemical reactions can be quantified. 4F-1 (of 14). INFORMATION FROM CHEMICAL EQUATIONS. 2H 2 + O 2 → 2H 2 O. 2 molecules 2 moles 0.84 moles

edna
Télécharger la présentation

REACTION STOICHIOMETRY

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. REACTION STOICHIOMETRY 1792 JEREMIAS RICHTER The amount of substances produced or consumed in chemical reactions can be quantified 4F-1 (of 14)

  2. INFORMATION FROM CHEMICAL EQUATIONS 2H2 + O2 → 2H2O 2 molecules 2 moles 0.84 moles 0.028 moles 1 molecule 1 moles 0.42 moles 0.014 moles 2 molecules 2 moles 0.84 moles 0.028 moles The moles that react and form do so in the ratio of the balanced equation 4F-2 (of 14)

  3. INFORMATION FROM CHEMICAL EQUATIONS 2H2 + O2 → 2H2O starting reacting ending 0.60 moles -0.60 moles 0.00 moles 0.40 moles -0.30 moles 0.10 moles 0.00 moles +0.60 moles 0.60 moles 2H2 + O2 → 2H2O starting reacting ending 0.50 moles -0.40 moles 0.10 moles 0.20 moles -0.20 moles 0.00 moles 0.00 moles +0.40 moles 0.40 moles The moles that react and form do so in the ratio of the balanced equation 4F-3 (of 14)

  4. MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O. 4½ C3H8O + O2→ CO2 + H2O 3 4 4F-4 (of 14)

  5. MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O. 9 C3H8O + O2→ CO2 + H2O 2 6 8 x g 5.00 g 2 mol 9 mol 2 mol C3H8O = 9 mol O2 5.00 g C3H8O x 1 mol C3H8O ____________________ 60.094g C3H8O x 9 mol O2 _________________ 2 mol C3H8O x 32.00 g O2 ______________ 1 mol O2 = 12.0 g O2 4F-5 (of 14)

  6. Calculate the mass of carbon dioxide produced from the 5.00 g of propanol. 9 C3H8O + O2→ CO2 + H2O 2 6 8 5.00 g x g 2 mol 6 mol 2 mol C3H8O = 6 mol CO2 5.00 g C3H8O x 1 mol C3H8O ____________________ 60.094g C3H8O x 6 mol CO2 _________________ 2 mol C3H8O x 44.01 g CO2 ________________ 1 mol CO2 = 11.0 g CO2 4F-6 (of 14)

  7. 9 C3H8O + O2→ 2 6 CO2 + H2O 8 5.0 g 6.0 g 12.0 g 11.0 g 4F-7 (of 14)

  8. LIMITING REACTANT CALCULATIONS LIMITING REACTANT – The reactant that is completely used up in a reaction 4F-8 (of 14)

  9. x 12 pieces 22 slices 12 pieces x 1 sandwich _______________ 1 piece = 12 sandwiches 22 slices x 1 sandwich _______________ 2 slices = 11 sandwiches actual amount produced limiting reactant 4F-9 (of 14)

  10. Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. Ba + WF3→ 3 2 3 BaF2 + W 2 26.0 g 25.0 g x g 2 mol 3 mol 2 mol 4F-10 (of 14)

  11. Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. Ba + WF3→ 3 2 3 BaF2 + W 2 26.0 g 25.0 g x g 2 mol 3 mol 2 mol 25.0 g Ba x 1 mol Ba _______________ 137.3 g Ba x 2 mol W ____________ 3 mol Ba x 183.8 g W ______________ 1 mol W = 22.3 g W 4F-11 (of 14)

  12. Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. Ba + WF3→ 3 2 3 BaF2 + W 2 26.0 g 25.0 g x g 2 mol 3 mol 2 mol 25.0 g Ba x 1 mol Ba _______________ 137.3 g Ba x 2 mol W ____________ 3 mol Ba x 183.8 g W _______________ 1 mol W = 22.3 g W 26.0 g WF3 x 1 mol WF3 _________________ 240.8 g WF3 x 183.8 g W _______________ 1 mol W = 19.8 g W x 2 mol W _____________ 2 mol WF3 WF3 is the limiting reactant 19.8 g W are produced 4F-12 (of 14)

  13. Determine the percentage of magnesium and silver in an alloy of the two metals. A 6.50 gram sample of the alloy reacts with 14.5 grams of hydrogen chloride. 2 Mg + HCl → MgCl2 + H2 x g 14.5 g 1 mol 2 mol 1 mol Mg = 2 mol HCl 14.5 g HCl x 1 molHCl ________________ 36.458g HCl x 1 mol Mg ______________ 2 molHCl x 24.31 g Mg ______________ 1 mol Mg = 4.834 g Mg 4.834 g Mg x 100 _______________ 6.50 g alloy = 74.4% Mg 100% - 74.4% = 25.6% Ag 4F-13 (of 14)

  14. Determine the percentage by mass of iodide in a solid unknown. A 1.17 gram sample of the unknown is dissolved in water, treated with lead (II) ions, and 1.42 grams of precipitate are collected. Pb2+ + I-→ 2 PbI2 x g 1.42 g 2 mol 1 mol 1.42 g PbI2 x 1 mol PbI2 _________________ 461.0 g PbI2 x 126.9 g I- _____________ 1 mol I- = 0.7818 g I- x 2 mol I- _____________ 1 mol PbI2 x 100 = 66.8% I- in the sample 0.7818 g I- ___________________ 1.17 g sample 4F-14 (of 14)

  15. MOLES FROM SOLUTION DATA Find the moles of potassium carbonate contained in 275 mL of a 0.300 M potassium carbonate solution. M = n ___ V MV = n 0.300 mol K2CO3 ______________________ L solution x 0.275 L solution = 0.0825 mol K2CO3 4G-1 (of 12)

  16. Find the moles of each ion in the 0.300 M potassium carbonate solution. 0.0825 mol K2CO3 x 2 = 0.165 mol K+ 0.0825 mol K2CO3 x 1 = 0.0825 mol CO32- 4G-2 (of 12)

  17. 10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4. (a) Find the moles of each ion in the solution. 0.450 mol BaCl2 ______________________ L solution x 0.0100 L solution = 0.004500 mol BaCl2  0.00450 mol Ba2+ and 0.00900 mol Cl- 0.300 mol K2SO4 ______________________ L solution x 0.0200 L solution = 0.006000 mol K2SO4  0.0120 mol K+ and 0.00600 mol SO42- 4G-3 (of 12)

  18. 10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4. (b) Find the moles of each ion after any reaction. Ba2+ + SO42-→ BaSO4 0.00450 0.00600 0 -0.00450 -0.00450 +0.00450 0 0.00150 0.00450 Initial moles Reacting moles Final moles  0 mol Ba2+ 0.00150 mol SO42- 0.00900 mol Cl- 0.0120 mol K+ 4G-4 (of 12)

  19. 10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4. (c) Find the final molarities of each ion in the solution. 0 mol Ba2+ ______________________ 0.0300 L solution = 0 M Ba2+ 0.00150 mol SO42- _______________________ 0.0300 L solution = 0.0500 M SO42- 0.00900 mol Cl- _______________________ 0.0300 L solution = 0.300 M Cl- 0.0120 mol K+ _______________________ 0.0300 L solution = 0.400 M K+ 4G-5 (of 12)

  20. 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (a) Find the moles of each ion in the solution. 0.350 mol HCl ___________________ L solution x 0.0200 L solution = 0.007000 mol HCl  0.00700 mol H+ and 0.00700 mol Cl- 0.250 mol NaOH ______________________ L solution x 0.0300 L solution = 0.007500 mol NaOH  0.00750 mol Na+ and 0.00750 mol OH- 4G-6 (of 12)

  21. 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (b) Find the moles of each ion after any reaction. H+ + OH-→ H2O 0.00700 0.00750 0 -0.00700 -0.00700 +0.00700 0 0.00050 0.00700 Initial moles Reacting moles Final moles  0 mol H+ 0.00050 mol OH- 0.00700 mol Cl- 0.00750 mol Na+ 4G-7 (of 12)

  22. 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (c) Find the final molarities of each ion in the solution. 0 mol H+ ______________________ 0.0500 L solution = 0 M H+ 0.00050 mol OH- _______________________ 0.0500 L solution = 0.010 M OH- 0.00700 mol Cl- _______________________ 0.0500 L solution = 0.140 M Cl- 0.00750 mol Na+ _______________________ 0.0500 L solution = 0.150 M Na+ 4G-8 (of 12)

  23. 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (d) Find the mass of water produced by the reaction. 0.007000 mol H2O x 18.016g H2O __________________ 1 mol H2O = 0.126 g H2O 4G-9 (of 12)

  24. DILUTION CALCULATIONS When a solution is diluted, only solvent is added , so the moles of solute are unchanged mol solute (concentrated) = mol solute (diluted) MCVC = MDVD 4G-10 (of 12)

  25. Calculate the volume of 6.00 M ammonia needed to prepare 250. mL of a 0.100 M ammonia solution. MCVC= MDVD MC= VC = 6.00 M ? MD= VD = 0.100 M 250. mL VC= MDVD _______ MC = (0.100 M)(250. mL) ________________________ (6.00 M) = 4.17 mL 4G-11 (of 12)

  26. Calculate the volume of water that must be added to 5.00 mL of concentrated hydrochloric acid (12.1 M) to make the acid 3.00 M. MCVC= MDVD MC= VC = 12.1 M 5.00 mL MD= VD = 3.00 M ? MCVC= VD _______ MD = (12.1 M)(5.00 mL) _______________________ (3.00 M) = 20.2 mL 20.2 mL - 5.00 mL ____________ Volume of dilute solution Volume of concentrated solution 15.2 mL Water that must be added 4G-12 (of 12)

  27. REACTIONS IN SOLUTION TITRATION – A technique in which one solution is used to analysis another Buret: a solution of 1 reactant of known concentration Flask: another reactant of unknown concentration, mass, etc. STANDARD SOLUTION – A solution of known concentration 4H-1 (of 13)

  28. The mass of sodium bicarbonate in an antacid tablet is to be determined. ACID-BASE INDICATOR – A weak organic acid or base that changes color in acidic or basic solutions The tablet is dissolved in water, an acid-base indicator added, and 21.5 mL of a 0.300 M hydrochloric acid solution produces a color change. NaHCO3 + HCl → NaCl + H2O + CO2 21.5 mL 0.300 M x g 1 mol 1 mol 0.300 mol HCl x 0.0215 L solution __________________ L solution x 1 mol NaHCO3 _________________ 1 mol HCl x 84.008g NaHCO3 _______________________ molNaHCO3 = 0.542 g NaHCO3 4H-2 (of 13)

  29. A sodium hydroxide solution is to be standardized. 34.2 mL of the sodium hydroxide solution are required to neutralize a solution made with 0.619 grams of solid H2C2O4.2H2O (m = 126.08 g/mol). _ 4 2 NaOH + H2C2O4→ 2 .2H2O Na2C2O4 + H2O 34.2 mL x M 0.619 g 1 mol 2 mol 0.619 g O.A.D. x 1 mol O.A.D. ____________________ 126.08 g O.A.D. x 1 _______________________ 0.0342 L solution x 2 mol NaOH _________________ 1 mol O.A.D. = 0.287 M NaOH 4H-3 (of 13)

  30. A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL of a 0.115 M MnO4- solution produces a light purple color. If the V2+ was oxidized to V5+, determine the molarity of the V2+ in the original solution. MnO4- (aq) + V2+ (aq) → Mn2+ + V5+ +7 -2 +2 +2 +5 ( ) x 3 5e- + 8H+ + MnO4-→ Mn2+ + 4H2O V2+→ V5+ ( ) x 5 + 3e- 4H-4 (of 13)

  31. A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL of a 0.115 M MnO4- solution produces a light purple color. If the V2+ was oxidized to V5+, determine the molarity of the V2+ in the original solution. MnO4- (aq) + V2+ (aq) → Mn2+ + V5+ +7 -2 +2 +2 +5 15e- + 24H+ + 3MnO4-→ 3Mn2+ + 12H2O 5V2+→ 5V5+ + 15e- 15e- + 24H+ + 3MnO4- + 5V2+→ 3Mn2+ + 12H2O + 5V5+ + 15e- 4H-5 (of 13)

  32. A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL of a 0.115 M MnO4- solution produces a light purple color. If the V2+ was oxidized to V5+, determine the molarity of the V2+ in the original solution. 24H+(aq) + 3MnO4- (aq) + 5V2+(aq)→ 3Mn2+ (aq) + 12H2O(l) + 5V5+(aq) 32.7 mL 0.115 M 10.0 mL x M 3 mol 5 mol 0.115 mol MnO4- x 0.0327 L solution _____________________ L solution x 5 mol V2+ ________________ 3 mol MnO4- x 1 _______________________ 0.0100 L solution = 0.627 M V2+ 4H-6 (of 13)

  33. Calculate the molar mass of a diprotic acid if 0.409 grams of it are neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution. molar mass H2X = grams H2X _______________ moles H2X molar mass H2X = 0.109 grams H2X ______________________ ? moles H2X 4H-7 (of 13)

  34. Calculate the molar mass of a diprotic acid if 0.409 grams of it are neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution. H2X + NaOH → 2 H(OH) + Na2X 2 0.409 g x mol 19.50 mL 0.287 M 2 mol 1 mol 0.287 mol NaOH x 0.01950 L sol’n _____________________ L sol’n x 1 mol H2X _______________ 2 mol NaOH = 0.002798 mol H2X 0.409 g H2X ________________________ 0.002798 mol H2X = 146 g/mol 4H-8 (of 13)

  35. The molarity of an aluminum hydroxide solution is to be determined. An acid-base indicator is added to 10.0 mL of the aluminum hydroxide solution, and 12.5 mL of 0.300 M hydrochloric acid produces a color change. Al(OH)3 + HCl → 3 AlCl3 + H(OH) 3 12.5 mL 0.300 M 10.0 mL x M 1 mol 3 mol 0.300 mol HCl x 0.0125 L solution __________________ L solution x 1 mol Al(OH)3 _________________ 3 mol HCl x 1 ____________ 0.0100 L = 0.125 M Al(OH)3 4H-9 (of 13)

  36. Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O72- solution. Determine the percentage of S2- in cinnabar ore. K2Cr2O7 (aq) + S2- (aq)→ K+(aq) + Cr2O72- (aq) + S2- (aq) → Cr3+ + S8 +1 +6 -2 -2 +3 0 ( ) x 8 6e- + 14H+ + Cr2O72-→ Cr3+ 2 + 7H2O 8 S2-→ S8 ( ) x 3 + 16e- 4H-10 (of 13)

  37. Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O72- solution. Determine the percentage of S2- in cinnabar ore. K2Cr2O7 (aq) + S2- (aq)→ K+(aq) + Cr2O72- (aq) + S2- (aq) → Cr3+ + S8 +1 +6 -2 -2 +3 0 48e- + 112H+ + 8Cr2O72-→ 16Cr3+ + 56H2O 24S2-→ 3S8 + 48e- 48e- + 112H+ + 8Cr2O72- + 24S2-→ 16Cr3+ + 56H2O + 3S8 + 48e- 4H-11 (of 13)

  38. Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O72- solution. Determine the percentage of S2- in cinnabar ore. 112H+(aq) + 8Cr2O72- (aq) + 24S2-(aq)→ 16Cr3+ (aq) + 56H2O(l) + 3S8(s) 20.4 mL 0.110 M x g 8 mol 24 mol 0.110 mol Cr2O72- x 0.0204 L solution _______________________ L solution x 24 mol S2- _________________ 8 mol Cr2O72- x 32.07 g S2- _____________ mol S2- = 0.2159 g S2- 0.2159 g S2- x 100 _______________ 1.534 g ore = 14.1% S2- in the ore 4H-12 (of 13)

  39. Calculate the molarity of a sulfuric acid solution if 25.0 mL of it are neutralized by 33.5 mL of a 0.240 M potassium hydroxide solution. H2SO4 + KOH → 2 H(OH) + K2SO4 2 33.5 mL 0.240 M 25.0 mL x M 1 mol 2 mol 0.240 mol KOH x 0.0335 L solution ___________________ L solution x 1 mol H2SO4 _________________ 2 mol KOH x 1 ____________ 0.0250 L = 0.161 M H2SO4 4H-13 (of 13)

  40. MOLAR MASSES AND STOICHIOMETRIC CONVERSIONS Calculate the molar mass of ethanol, C2H5OH • 2 mol C (12.01 g/mol) = 24.02 g • 6 mol H (1.008 g/mol) = 6.048 g • 1 mol O (16.00 g/mol) = 16.00 g • 46.068 g 46.068 g C2H5OH = 1 molC2H5OH 4I-1 (of 8)

  41. Calculate the number of ethanol molecules in 25.0 mL of pure ethanol. The density of the ethanol is 0.789 g/mL. 0.789 g C2H5OH = 1 mL C2H5OH 25.0 mL C2H5OH x 0.789 g C2H5OH ____________________ 1 mL C2H5OH x 1 mol C2H5OH ______________________ 46.068g C2H5OH x 6.022 x 1023 molecules C2H5OH ________________________________________ 1 mol C2H5OH = 2.58 x 1023 molecules C2H5OH 4I-2 (of 8)

  42. Calculate the number of carbon atoms in a 10.0 mL sample of pure ethanol. 10.0 mL C2H5OH x 1 mol C2H5OH ______________________ 46.068g C2H5OH x 2 mol C __________________ 1 mol C2H5OH x 0.789 g C2H5OH ____________________ 1 mL C2H5OH x 6.022 x 1023 atoms C ___________________________ 1 mol C = 2.06 x 1023 atoms C 4I-3 (of 8)

  43. Calculate the number of ethanol molecules in 45.0 mL of 80. proof vodka. The density of the vodka is 0.92 g/mL. 80. Proof vodka = 40.% C2H5OH by volume 100 mL vodka = 40. mL C2H5OH 45.0 mL vodka x 40. mL C2H5OH ____________________ 100 mL vodka x 0.789 g C2H5OH ____________________ 1 mL C2H5OH x 1 mol C2H5OH ______________________ 46.068g C2H5OH x 6.022 x 1023 molecules C2H5OH ________________________________________ 1 mol C2H5OH = 1.9 x 1023 molecules C2H5OH 4I-4 (of 8)

  44. Calculate the mass of one ethanol molecule, in grams. 1 molecule C2H5OH x 46.068g C2H5OH _______________________ molC2H5OH x mol C2H5OH ________________________________________ 6.022 x 1023 molecules C2H5OH = 7.650 x 10-23g 4I-5 (of 8)

  45. g M mol M = 71.0 g M ? mol M A metal oxide with the formula M2O3 is 29.0% oxygen by mass. Calculate the molar mass of metal M. Molar Mass of M = 29.0 g O = 1.208 mol M x 1 mol O ____________ 16.00 g O x 2 mol M __________ 3 mol O 71.0 g M _________________ 1.208mol M = 58.8 g/mol 4I-6 (of 8)

  46. THEORETICAL PERCENT COMPOSITION OF COMPOUNDS BY MASS Calculate the percent composition by mass of sodium nitrate NaNO3 • 1 mol Na (22.99 g/mol) = 22.99 g • 1 mol N (14.01 g/mol) = 14.01 g • 3 mol O (16.00 g/mol) = 48.00 g • 85.00 g % Na = 22.99 g Na  100 ___________________ 85.00 g NaNO3 = 27.05 % Na % N = 14.01 g N  100 ___________________ 85.00 g NaNO3 = 16.48 % N % O = 48.00 g O  100 ___________________ 85.00 g NaNO3 = 56.47 % O 4I-7 (of 8)

  47. Calculate the percentage by mass of water in barium chloride dihydrate • BaCl2.2H2O • 1 mol Ba (137.3 g/mol) = 137.3 g • 2 molCl (35.45 g/mol) = 70.90 g • 2 mol H2O (18.016g/mol) = 36.032 g • 244.232 g % H2O = 36.032g H2O  100 ____________________________ 244.232g BaCl2.2H2O = 14.75 % H2O 4I-8 (of 8)

More Related