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Thursday, Sept. 26 th : A Day Friday, Sept. 27 th : B Day Agenda. Homework Questions/Problems Quick Review (Practice #2c; Practice #1: c, d) Finish Sec 9.1: Calculating Quantities in Reactions Calculations involving volume and number of particles In-Class/Homework: - PowerPoint PPT Presentation

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<ul><li><p>Thursday, Sept. 26th: A DayFriday, Sept. 27th: B DayAgendaHomework Questions/ProblemsQuick Review (Practice #2c; Practice #1: c, d)Finish Sec 9.1: Calculating Quantities in ReactionsCalculations involving volume and number of particles</p><p>In-Class/Homework:Problem Solving Supplement Worksheet</p></li><li><p>Practice #2c, pg. 304Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation:Fe2O3 + 2 Al 2 Fe + Al2O3</p><p>c) Mole of aluminum oxide formed3.30 mol Al2O3</p></li><li><p>Practice #1, pg. 307Use the equation below to answer the questions that follow:Fe2O3 + 2Al 2Fe + Al2O3</p><p>How many grams of Fe2O3 react with excess Al to make 475 g Fe?679 g Fe2O3How many grams of Fe will form when 97.6 g Al2O3 form?107 g Fe</p></li><li><p> Stoichiometry Calculations involving VolumeWhen reactants are liquids, they are almost always measured by volume.</p><p>To convert from volume mass ormass volume, use the density of the substance as the conversion factor.</p><p>Units for density are typically: g/mL, g/cm3, or g/L </p></li><li><p>Other ways to include volume in stoichiometry calculations:</p><p>If the substance is a gas at standard temperature and pressure, STP, use the molar volume of the gas.The molar volume of ANY gas at STP is:22.41 L/mol</p><p>**STP is 0C and 1 atm***</p></li><li><p>Other ways to include volume in stoichiometry calculations:2. If the substance is in aqueous solution, use the concentration of the solution, in mol/L, to convert the volume of the solution to the moles of the substance dissolved.</p><p>Stay with me, itll make sense once we get there.</p></li><li><p>Sample Problem C, pg. 309What volume of H3PO4 forms when 56 mL of POCl3 completely react? POCl3 + 3 H2O H3PO4 + 3 HCl(density of POCl3 = 1.67 g/mL) (density of H3PO4 = 1.83 g/mL)Remember our 3 steps:Change the units of what you know into moles.Use the mole ratio to determine moles of desired substance.Change out of moles into the unit you want.</p></li><li><p>Sample Problem C (continued)What volume of H3PO4 forms when 56 mL of POCl3 completely react? POCl3 + 3 H2O H3PO4 + 3 HCl(density of POCl3 = 1.67 g/mL) (density of H3PO4 = 1.83 g/mL)Start with what you know: 56 mL POCl3Use the density to change mL POCL3 g POCl3 56 mL POCl3 X 1.67 g POCL3 = 93.52 g POCl3 1 mL POCl3Use molar mass to change g POCl3 mol POCl3 93.52 g POCl3 X 1 mol POCl3 = .61 mol POCl3 153.32 g POCl3</p></li><li><p>Sample Problem C (continued)What volume of H3PO4 forms when 56 mL of POCl3 completely react? POCl3 + 3 H2O H3PO4 + 3 HCl(density of POCl3 = 1.67 g/mL) (density of H3PO4 = 1.83 g/mL)Use mole ratio to change mol POCl3 mol H3PO4 0.61 mol POCl3 X 1 mol H3PO4 = .61 mol H3PO4 1 mol POCl3Use molar mass to change mol H3PO4 g H3PO4 0.61 mol H3PO4 X 98.00 g H3PO4 = 59.78 g H3PO4 1 mol H3PO4</p></li><li><p>What volume of H3PO4 forms when 56 mL of POCl3 completely react? POCl3 + 3 H2O H3PO4 + 3 HCl(density of POCl3 = 1.67 g/mL) (density of H3PO4 = 1.83 g/mL)Use the density to convert g of H3PO4 mL H3PO4 59.78 g H3PO4 X 1 mL H3PO4 =33 mL H3PO4 1.83 g H3PO4 </p><p>Wow, I know thats a lot of steps, but this is as difficult as it gets.Sample Problem C (continued)</p></li><li><p>Use the densities and the balanced equation provided to answer the questions that follow.C5H12 C5H8 + 2 H2(density of C5H12 = 0.620 g/mL)(density of C5H8 = 0.681 g/mL)(density of H2 = 0.0899 g/L)How many mL of C5H8 can be made from 366 mL C5H12?315 mL C5H82. How many L of H2 can form when 4.53 X 103 mL C5H8 form?2.03 X 103 L H2Practice</p></li><li><p>Stoichiometry Calculations Involving Number of ParticlesYou can use Avogadros number as a conversion factor in stoichiometry problems involving particles, atoms, molecules, etc.</p><p> 6.022 X 1023= 1 mole</p></li><li><p>Sample Problem D, pg. 310How many grams of C5H8 form from 1.89 X 1024 molecules of C5H12?C5H12 C5H8 + 2 H2</p><p>Start with what you know and change to moles: 1.89 X 1024 molecules C5H12 X 1 mol C5H12 6.022 X 1023 molecules= 3.14 mol C5H12Use mole ratio to change mol C5H12 mol C5H8 3.14 mol C5H12 X 1 mol C5H8 = 3.14 mol C5H8 1 mol C5H12</p></li><li><p>Sample Problem D (Continued)How many grams of C5H8 form from 1.89 X 1024 molecules of C5H12?C5H12 C5H8 + 2H2</p><p>3. Use molar mass to change mol C5H8 g C5H8 3.14 mol C5H8 X 68.13 g C5H8 = 1 mol C5H8</p><p>214 g C5H8</p></li><li><p>Practice #1, pg. 311Use the equation below to answer the questions that follow:Br2 + 5 F2 2 BrF5 </p><p>How many molecules of BrF5 form when 384 g Br2 react with excess F2?</p><p>2.89 X 1024 molecules BrF5</p></li><li><p>Stoichiometry Problem SolutionsUpdate your graphic organizer:What do you use to change from Volume MolesDensity: (g/mL) or (g/cm3) or (g/L)Gas @ STP: 22.41 L/molConcentration: (mol/L)What do you use to change from Molecules or Particles Moles Avogadros Number6.022 X 1023 = 1 mole</p></li><li><p>In-Class/HomeworkProblem Solving Supplement Worksheet</p><p>Looking Ahead:Next time, youll work on the section review and the concept review in class before taking a quiz over this section</p></li></ul>