3. 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: Problem Solving Supplement Worksheet

4. Practice #2c, pg. 304  Calculate the amounts requested if 3.30 mol Fe 2 O 3 completely react according to the following equation: Fe 2 O 3 + 2 Al 2 Fe + Al 2 O 3 c) Mole of aluminum oxide formed 3.30 mol Al 2 O 3

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

6. 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/cm 3, or g/L

7. Other ways to include volume in stoichiometry calculations: 1.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***

8. 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….

9. Sample Problem C, pg. 309 What volume of H 3 PO 4 forms when 56 mL of POCl 3 completely react? POCl 3 + 3 H 2 O H 3 PO 4 + 3 HCl (density of POCl 3 = 1.67 g/mL) (density of H 3 PO 4 = 1.83 g/mL) Remember our 3 steps: 1.Change the units of what you know into moles. 2.Use the mole ratio to determine moles of desired substance. 3.Change out of moles into the unit you want.

10. Sample Problem C (continued)  What volume of H 3 PO 4 forms when 56 mL of POCl 3 completely react? POCl 3 + 3 H 2 O H 3 PO 4 + 3 HCl (density of POCl 3 = 1.67 g/mL) (density of H 3 PO 4 = 1.83 g/mL) 1.Start with what you know: 56 mL POCl 3 2.Use the density to change mL POCL 3 g POCl 3 56 mL POCl 3 X 1.67 g POCL 3 = 93.52 g POCl 3 1 mL POCl 3 3.Use molar mass to change g POCl 3 mol POCl 3 93.52 g POCl 3 X 1 mol POCl 3 =.61 mol POCl 3 153.32 g POCl 3

11. Sample Problem C (continued)  What volume of H 3 PO 4 forms when 56 mL of POCl 3 completely react? POCl 3 + 3 H 2 O H 3 PO 4 + 3 HCl (density of POCl 3 = 1.67 g/mL) (density of H 3 PO 4 = 1.83 g/mL) 4.Use mole ratio to change mol POCl 3 mol H 3 PO 4 0.61 mol POCl 3 X 1 mol H 3 PO 4 =.61 mol H 3 PO 4 1 mol POCl 3 5.Use molar mass to change mol H 3 PO 4 g H 3 PO 4 0.61 mol H 3 PO 4 X 98.00 g H 3 PO 4 = 59.78 g H 3 PO 4 1 mol H 3 PO 4

12. What volume of H 3 PO 4 forms when 56 mL of POCl 3 completely react? POCl 3 + 3 H 2 O H 3 PO 4 + 3 HCl (density of POCl 3 = 1.67 g/mL) (density of H 3 PO 4 = 1.83 g/mL) 6.Use the density to convert g of H 3 PO 4 mL H 3 PO 4 59.78 g H 3 PO 4 X 1 mL H 3 PO 4 = 33 mL H 3 PO 4 1.83 g H 3 PO 4 Wow, I know that’s a lot of steps, but this is as difficult as it gets…. Sample Problem C (continued)

13. Use the densities and the balanced equation provided to answer the questions that follow. C 5 H 12 C 5 H 8 + 2 H 2 (density of C 5 H 12 = 0.620 g/mL) (density of C 5 H 8 = 0.681 g/mL) (density of H 2 = 0.0899 g/L) 1.How many mL of C 5 H 8 can be made from 366 mL C 5 H 12 ? 315 mL C 5 H 8 2. How many L of H 2 can form when 4.53 X 10 3 mL C 5 H 8 form? 2.03 X 10 3 L H 2 Practice

14. 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 10 23 = 1 mole

15. Sample Problem D, pg. 310  How many grams of C 5 H 8 form from 1.89 X 10 24 molecules of C 5 H 12 ? C 5 H 12 C 5 H 8 + 2 H 2 1.Start with what you know and change to moles: 1.89 X 10 24 molecules C 5 H 12 X 1 mol C 5 H 12 6.022 X 10 23 molecules = 3.14 mol C 5 H 12 2.Use mole ratio to change mol C 5 H 12 mol C 5 H 8 3.14 mol C 5 H 12 X 1 mol C 5 H 8 = 3.14 mol C 5 H 8 1 mol C 5 H 12

16. Sample Problem D (Continued) How many grams of C 5 H 8 form from 1.89 X 10 24 molecules of C 5 H 12 ? C 5 H 12 C 5 H 8 + 2H 2 3. Use molar mass to change mol C 5 H 8 g C 5 H 8 3.14 mol C 5 H 8 X 68.13 g C 5 H 8 = 1 mol C 5 H 8 214 g C 5 H 8

17. Practice #1, pg. 311 Use the equation below to answer the questions that follow: Br 2 + 5 F 2 2 BrF 5 1.How many molecules of BrF 5 form when 384 g Br 2 react with excess F 2 ? 2.89 X 10 24 molecules BrF 5

18. Stoichiometry Problem Solutions  Update your graphic organizer: 1.What do you use to change from Volume Moles Density: (g/mL) or (g/cm 3 ) or (g/L) Gas @ STP: 22.41 L/mol Concentration: (mol/L) 2.What do you use to change from Molecules or Particles Moles Avogadro’s Number 6.022 X 10 23 = 1 mole

19. 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…