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Thursday, Sept. 26 th : “A” Day Friday, Sept. 27 th : “B” Day Agenda

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:

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Thursday, Sept. 26 th : “A” Day Friday, Sept. 27 th : “B” Day Agenda

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  1. Thursday, Sept. 26th: “A” DayFriday, Sept. 27th: “B” DayAgenda • 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: • Problem Solving Supplement Worksheet

  2. Practice #2c, pg. 304 • Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation: Fe2O3 + 2 Al 2 Fe + Al2O3 c) Mole of aluminum oxide formed 3.30 mol Al2O3

  3. Practice #1, pg. 307 • Use the equation below to answer the questions that follow: Fe2O3 + 2Al 2Fe + Al2O3 • How many grams of Fe2O3 react with excess Al to make 475 g Fe? 679 g Fe2O3 • How many grams of Fe will form when 97.6 g Al2O3 form? 107 g Fe

  4. Stoichiometry Calculations involving Volume • When reactants are liquids, they are almost always measured by volume. • To convert from volume mass or mass volume, use the density of the substance as the conversion factor. • Units for density are typically: g/mL, g/cm3, or g/L

  5. Other ways to include volume in stoichiometry calculations: • 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 **STP is 0˚C and 1 atm***

  6. 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. • Stay with me, it’ll make sense once we get there….

  7. Sample Problem C, pg. 309 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) • 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.

  8. 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 POCl3 • Use the density to change mL POCL3 g POCl3 56 mL POCl3 X 1.67 g POCL3 = 93.52 g POCl3 1 mL POCl3 • Use molar mass to change g POCl3 mol POCl3 93.52 g POCl3 X 1 mol POCl3 = .61 mol POCl3 153.32 g POCl3

  9. 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 POCl3 • Use molar mass to change mol H3PO4 g H3PO4 0.61 mol H3PO4 X 98.00 g H3PO4 = 59.78 g H3PO4 1 mol H3PO4

  10. 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 the density to convert g of H3PO4 mLH3PO4 59.78 g H3PO4 X 1 mL H3PO4=33 mL H3PO4 1.83 g H3PO4 Wow, I know that’s a lot of steps, but this is as difficult as it gets….

  11. Practice 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 C5H8 2. How many L of H2 can form when 4.53 X 103 mL C5H8 form? 2.03 X 103 L H2

  12. Stoichiometry Calculations Involving Number of Particles • You can use Avogadro’s number as a conversion factor in stoichiometry problems involving particles, atoms, molecules, etc. 6.022 X 1023= 1 mole

  13. Sample Problem D, pg. 310 • How many grams of C5H8 form from 1.89 X 1024 molecules of C5H12? C5H12C5H8+ 2 H2 • 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 C5H12 • Use mole ratio to change mol C5H12 mol C5H8 3.14 mol C5H12 X 1 mol C5H8 = 3.14 mol C5H8 1 mol C5H12

  14. Sample Problem D (Continued) How many grams of C5H8 form from 1.89 X 1024 molecules of C5H12? C5H12C5H8+ 2H2 3. Use molar mass to change mol C5H8 g C5H8 3.14 mol C5H8 X 68.13 g C5H8 = 1 mol C5H8 214 g C5H8

  15. Practice #1, pg. 311 Use the equation below to answer the questions that follow: Br2+ 5 F2 2 BrF5 • How many molecules of BrF5 form when 384 g Br2 react with excess F2? 2.89 X 1024 molecules BrF5

  16. Stoichiometry Problem Solutions • Update your graphic organizer: • What do you use to change from Volume Moles • Density: (g/mL)or(g/cm3) or (g/L) • Gas @ STP: 22.41 L/mol • Concentration: (mol/L) • What do you use to change from Molecules or Particles Moles Avogadro’s Number 6.022 X 1023 = 1 mole

  17. In-Class/Homework • Problem Solving Supplement Worksheet Looking Ahead: Next time, you’ll work on the section review and the concept review in class before taking a quiz over this section…

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