Chapter3 - Chapter 3: Stoichiometry Chem 6A Michael J....

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Unformatted text preview: Chapter 3: Stoichiometry Chem 6A Michael J. Sailor, UC San Diego 1 Announcements: • Thursday (Sep 29) quiz: – – – – Bring student ID or we cannot accept your quiz! No notes, no calculators Covers chapters 1 and 2 Need to know your name, PID, and section # Chem 6A Michael J. Sailor, UC San Diego 2 day ’s q uiz Chem 6A 2010 (Sailor) QUIZ Name: Student ID Number: Section Number: 1 18 1 H 2 2 Th urs 1.0079 3 Li 13 4 5 Be B 6.941 9.01218 11 12 Na Mg 22.9898 19 of K 39.0983 37 pag e Rb 85.4578 55 Cs 4 21 Sc 40.078 44.9559 38 39 Sr Y 87.62 88.9059 56 57 Ba La 132.905 137.327 138.906 87 88 89 Co ver Fr Ra 5 22 6 23 Ti 7 24 V 8 25 Cr Hf W Co 55.847 58.9332 44 45 Ru 95.94 98.9063 74 75 Ta 178.49 180.948 104 105 10 27 Fe Nb Mo Tc 91.224 92.9064 72 73 9 26 Mn 47.88 50.9415 51.9961 54.9381 40 41 42 43 Zr 6 C 16 7 N He 17 8 O 9 F 4.0026 10 Rh 101.07 102.906 76 77 Re Os 183.85 186.207 106 107 Ir 190.2 108 192.22 109 11 28 Ni 58.69 46 Pd 29 Cu 63.546 47 Ag 12 Ne 30 Zn 65.39 48 Cd 12.011 14.0067 15.9994 18.9984 20.1797 14 15 16 17 18 Al Si P 26.9815 28.0855 30.9738 31 32 33 Ga 69.723 49 Ge As 72.61 74.9216 50 51 In Sn Sb 106.42 107.868 112.411 78 79 80 114.82 81 118.71 82 121.75 83 Pt Tl Pb Au Hg 195.08 196.967 200.59 204.383 207.2 Bi S Cl Ar 32.066 35.4527 34 35 39.948 36 Se Kr 78.96 52 Te Br 79.904 53 I 127.6 126.905 84 85 Po 83.8 54 Xe At 131.29 86 Rn 208.98 208.982 209.987 222.018 Ac 223.02 226.025 227.028 - 58 59 60 - - 61 62 63 64 65 Ce Pr Nd Pm Sm Eu Gd Tb 140.12 90 Lanthanides Actinides 15 10.811 13 3 24.305 20 Ca 14 140.91 91 144.24 92 157.25 96 158.92 97 Th 232.038 Pa 231.04 U 238.03 146.92 93 150.35 94 151.96 95 66 67 Dy Ho 162.5 98 71 Lu 174.97 103 Fm Md No Pu Am Cm Bk Cf Es 239.05 251.08 254.09 249.08 70 168.93 101 237.05 247.07 69 Tm Yb 167.26 100 Np 241.06 68 Er 164.93 99 257.1 258.1 173.04 102 255 Lr 262.1 Some useful constants and relationships: Ideal gas constant: 0.08206 L.atm.mol-1 .K-1 = 8.31451 J.mol-1 .K-1 Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s speed of light: 3.00 x 108 m/s 1J = 1kg.m2/s2 1 atm = 760 Torr 101.325 J = 1 L.atm E = -RHh/n2 RH = 3.29 x 1015 Hz 1 eV = 1.6022 x 10-19 J C2 = second radiation constant = 1.44 x 10-2 K.m hc 1 Emitted power (W) E= T"max = C2 = (cons tan t )T 4 2 " 5 Surface area (m ) ! ! ! Chem 6A Michael J. Sailor, UC San Diego 3 The Periodic Table of the Elements QUIZ THURS Oct 20 (front page) Chem 6A Michael J. Sailor, UC San Diego 4 The Periodic Table of the Elements QUIZ THURS Oct 20 (back page) Provide the names of the following elements: Element Name H Li Fr Pt Ag Sb hydrogen lithium francium platinum silver antimony Etc… 20 elements total Chem 6A Michael J. Sailor, UC San Diego 5 Problem: Mass-to-mass calculations Since the bronze age (4000 b.c.), copper metal has been produced by “smelting,” in which Cu2O ore is reduced with excess carbon (charcoal). How much copper can be produced from smelting of 1.00 kg of pure Cu2O? a) 222 g b) 888 g c) 444 g d) 2252 g e) none of the above Chem 6A Michael J. Sailor, UC San Diego 6 Solution: Mass-to-mass calculations (1) The balanced equation for the reaction is: 2Cu 2O + C → 4Cu + CO2 (2) Find out how many moles of Cu are there, then convert to grams Cu: 1000 g Cu2O Mol Cu2O 143.1 g Cu2O 4 mol Cu 2 mol Cu2O 63.546 g Cu 1 mol Cu = 888 g Cu Chem 6A Michael J. Sailor, UC San Diego 7 Problem: Limiting reactant In the Haber process, hydrogen (H2) reacts with nitrogen (N2) in a reactor to make ammonia. The reactor is initially charged with 2 mol of H2 and 3 mol of N2 and the reaction is allowed to go to completion. Fill out the table below indicating the amounts of reactants and products present after the reaction is complete. 3 H2 + N2 → 2 NH3 Moles H2 Moles N2 Moles NH3 Chem 6A Michael J. Sailor, UC San Diego 8 Solution: Limiting reactant The reactor is initially charged with 2 mol of H2 and 3 mol of N2 Balanced equation: 3 H2 + N2 → 2 NH3 If all 2 mol of H2 reacts, how many moles of N2 are needed? Moles H2 Before reaction 2 Moles N2 3 Moles NH3 After reaction 0 Chem 6A Michael J. Sailor, UC San Diego 9 Solution: Limiting reactant The reactor is initially charged with 2 mol of H2 and 3 mol of N2 Balanced equation: 3 H2 + N2 → 2 NH3 If all 2 mol of H2 reacts, how many moles of N2 are needed? = 2/3 mol N2 needed We have 3 mole of N2, so N2 is in excess and H2 is limiting Chem 6A Michael J. Sailor, UC San Diego 10 Solution: Limiting reactant The reactor is initially charged with 2 mol of H2 and 3 mol of N2 and the reaction is allowed to go to completion. Balanced equation: 3 H2 + N2 → 2 NH3 Before reaction What’s left? Moles H2 2 2 0 Moles N2 3 2/3 2.33 Moles NH3 After reaction, What’s used? 0 - 1.33 Moles H2 = 2-2 = 0 Moles N2 = 3 - 2/3 = 7/3, or 2.33 2 mol H2 2 mol NH3 Moles NH3 = 0 + 3 mol H2 = 4/3, or 1.33 Chem 6A Michael J. Sailor, UC San Diego 11 Solution: Limiting reactant What if we chose to calculate it assuming the N2 is limiting? Balanced equation: 3 H2 + N2 → 2 NH3 If all 3 mol of N2 reacts, how many moles of H2 are needed? Moles H2 Before reaction 2 Moles N2 3 Moles NH3 0 Chem 6A Michael J. Sailor, UC San Diego After reaction 12 Solution: Limiting reactant If all 3 mol of N2 reacts, how many moles of H2 are needed? Balanced equation: 3 H2 + N2 → 2 NH3 = 9 mol H2 needed We only have 2 mole of H2, so H2 is limiting Chem 6A Michael J. Sailor, UC San Diego 13 Problem: Limiting reactant Since the bronze age (4000 b.c.), copper metal has been produced by “smelting,” in which Cu2O ore is reduced with excess carbon (charcoal). How much copper can be produced from smelting of 1.00 kg of pure Cu2O with 25 g of charcoal? a) 222 g b) 888 g c) 444 g d) 2252 g e) none of the above Chem 6A Michael J. Sailor, UC San Diego 14 Solution: Limiting reactant The balanced equation for the reaction is: 2Cu 2O + C → 4Cu + CO2 Find out how many moles of Cu2O we have: 1000 g Cu2O Mol Cu2O 143.1 g Cu2O = 6.99 mol Cu2O Find out how many moles of C we have: 25 g C Mol C 12.011 g C = 2.08 mol C how many moles of C are needed: 1000 g Cu2O Mol Cu2O 143.1 g Cu2O 1 mol C 2 mol Cu2O = 3.49 mol C needed Chem 6A Michael J. Sailor, UC San Diego 15 Solution: Limiting reactant We have 2.08 mol C. We need 3.49, so all of the C will be used up before all of the Cu2O has reacted. So C (carbon) is the limiting reactant. how many moles of Cu will be produced using this much C? 25 g C Mol C 12.011 g C 4 mol Cu 1 mol C 63.546 g Cu Mol Cu = 529 g Cu Chem 6A Michael J. Sailor, UC San Diego 16 Problem: Combustion analysis Combustion analysis is carried out on 1.621 g of a compound that contains only carbon, hydrogen, and oxygen. The masses of water and carbon dioxide produced are 1.902 g and 3.095 g, respectively. What is the empirical formula of the compound? See problem 3.34 from the book Chem 6A Michael J. Sailor, UC San Diego 17 Solution: Combustion analysis Like excess-limiting reagent problem except the oxygen is never limiting: CxHyOz + xs O2 → x CO2 + y/2 H2O Figure out number of moles and 3.095 g CO2 mass of carbon, hydrogen: 1mol CO2 1 mole C 44.01 g CO2 1 mol CO2 = 0.07032 moles C 0.07032 moles C 12.011 g C 1 mol C = 0.8446 g C 1.902 g H2O 1mol H2O 2 mole H 18.02 g H2O 1 mol H2O = 0.2111 moles H 0.2111 moles H 1.0079 g H 1 mol H = 0.2128 g H Chem 6A Michael J. Sailor, UC San Diego 18 Solution: Combustion analysis CxHyOz + xs O2 → x CO2 + y/2 H2O Mass left over = 1.621 - 0.2128 - 0.8446 = 0.5636 g mass CxHyOz mass C mass H So 0.5636 g is the mass of O (NOT O2) in the sample. The number of moles of O is: 0.5636 g O 1mol O 15.9994 g O = 0.0352 mol O Chem 6A Michael J. Sailor, UC San Diego 19 Solution: Combustion analysis CxHyOz + xs O2 → x CO2 + y/2 H2O Dividing by the least common denominator (O in this case): Element moles C 0.07032 H 0.2128 fraction 2.00 6.05 O 1 0.0352 So the empirical formula is C2H6O Chem 6A Michael J. Sailor, UC San Diego 20 Water: H !+ 2!O 105° H !+ Chem 6A Michael J. Sailor, UC San Diego 21 Sodium Chloride (NaCl) Crystal Cl- Na+ Unit Cell Chem 6A Michael J. Sailor, UC San Diego 22 Dissolution of NaCl in water ion-dipole interactions replace ion-ion interactions Solvent Solute = Na+ = H2O = Cl- Chem 6A Michael J. Sailor, UC San Diego 23 Ethanol in water dipole-dipole interactions: “Like Dissolves Like” Hydrogen Bonds Chem 6A Michael J. Sailor, UC San Diego 24 Problem: Determine molar concentration What is the molarity of a solution made by adding enough water to 20.0 g of NaCl to make 50.0 mL of solution? MW of NaCl is 58.44 g/mol. Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 25 Solution: Determine molar concentration Molarity is defined as moles of solute/liter of solution: 20.0 g NaCl mol NaCl 58.44 g NaCl 0.050 L solution (= 6.84 M) Chem 6A Michael J. Sailor, UC San Diego 26 Problem: Dilution calculations What volume of 0.146 M sucrose solution is needed to make 2.00 L of a 0.05 M sucrose solution? Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 27 Solution: Dilution calculations 1. How many moles of sucrose are needed? 0.05 mol sucrose 2 Liters Liter = 0.1 mol sucrose 2. What volume of solution do I need? 0.1 mol sucrose Liter 0.146 mol sucrose = 0.685 L of solution Chem 6A Michael J. Sailor, UC San Diego 28 Problem: Dilution calculations Calculate the volume (in liters) of a 0.1 M KMnO4(aq) stock solution that is needed to prepare 250 mL of 1.50 x 10-3 M KMnO4(aq) (set up but do not solve). Chem 6A Michael J. Sailor, UC San Diego 29 Solution: Dilution calculations Volume = 1.5 x 10-3 mol KMnO4 L 250 mL L 1000 mL L 0.1 mol KMnO4(aq) (= 3.75 x 10-3 L) Chem 6A Michael J. Sailor, UC San Diego 30 Problem: Titration calculations A 0.202 g sample of iron ore (mixture of Fe2O3 and SiO2) is dissolved in hydrochloric acid and 22.3 mL of 0.0118 M KMnO4 are required to reach the stoichiometric point. What is the percent by mass of iron present in the sample? The balanced net ionic equation for the titration reaction is given below: 5Fe2+ + MnO4- + 8H+→ 5Fe3+ + Mn2+ + 4H2O a) 63.6 % b) 7.28 % c) 36.4 % d) 100 % e) none of the above Chem 6A Michael J. Sailor, UC San Diego 31 Solution: Titration calculations Find out how many moles of Fe are there: 1.18 x 10-2 mol KMnO4 L 22.3 mL L 1000 mL 5 mol Fe2+ 1 mol MnO4(aq)- = 1.316 x 10-3 mol Fe2+ = 1.316 x 10-3 mol Fe. How many grams of Fe were there? 1.316 x 10-3 mol Fe 55.847 g Fe Mol Fe = 0.07347 grams Fe. So the % by mass Fe is: 0.07347g/0.202g x 100% = 36.4% Chem 6A Michael J. Sailor, UC San Diego 32 ...
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