1 / 19

CP Chemistry

CP Chemistry. Unit 8 Stoichiometry Chapter 9. Unit 6b Stoichiometry Standards. 3.2.10 B Apply process knowledge and organize scientific and technological phenomena in varied ways. 3.2.10 C Apply the elements of scientific inquiry to solve problems.

khoi
Download Presentation

CP Chemistry

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. CP Chemistry Unit 8 Stoichiometry Chapter 9

  2. Unit 6b Stoichiometry Standards • 3.2.10 B Apply process knowledge and organize scientific and technological phenomena in varied ways. • 3.2.10 C Apply the elements of scientific inquiry to solve problems. • 3.4.10 A Explain concepts about the structure and properties of matter.

  3. Chapter Themes • The Power of a Balanced Equation • Stoichiometry • Limiting Reactants • Yield (actual / theoretical / percent) • Day One Assignment • Day Two … Class Information • Day Two Assignment

  4. The Power of a Balanced Equation • Discover that a balanced equation provides much information … 2H2O  2H2 + O2 • 2 molecules of water = 2 molecules of hydrogen plus 1 molecule of oxygen • 2 moles of water = 2 moles of hydrogen plus 1 mole of oxygen • 2 dozen molecules of water = 2 dozen molecules of hydrogen plus 1 dozen molecules of oxygen • The decomposition of water yields twice as much hydrogen as oxygen.

  5. Stoichiometry • Definition: The process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction. • In ordinary English: If you know one mass, you can calculate any other mass!

  6. Stoichiometry • We do this everyday … • Recipe … Betty Crocker Cake Mix … Double Chocolate Delight 1 egg 1/3 cup oil + 8 X 12 greased cake pan 1 Double Chocolate Delight Cake 1 egg + 1/3 cup oil  1 D.C.D.C.

  7. Stoichiometry • We do this everyday … • Recipe … • Milage … 11 gallons gas + 1 ’97 Neon 382 miles driven 11 gallons gas + 1 Neon  382 miles

  8. Limiting Reactants • Consider sandwiches … • 1 sandwich = 2 pieces of bread 1 piece cheese 3 slices of meat • How many sandwiches can you make with 7 pieces of bread; 5 pieces of cheese; and 25 slices of meat?

  9. Yield (actual / theoretical / percent) • Back to the ’97 Neon … - Do I always get 382 miles / 11 gallons? • What do you think these words mean? • Actual yield … • Theoretical yield … • Percent yield …

  10. Day One Assignment p. 282: 3a&d; 4b&d; 10b&d • 3a 2NO + O2 2NO2 2 molecules NO + 1 molecule O2 yields 2 molecules NO2 2 moles NO + 1 mole O2 yields 2 moles NO2 • 3d C2H6 + Cl2 C2H5Cl + HCl 1 molecule C2H6 + 1 moleculeCl2 yields 1 molecule C2H5Cl + 1 molecule HCl 1 mole C2H6 + 1 moleCl2 yields 1 mole C2H5Cl + 1 mole HCl

  11. Day One Assignment p. 282: 3a&d; 4b&d; 10b&d • 4b B2O3 + 3CaF2 2BF3 + 3CaO 1 molecule B2O3 + 3 molecule CaF2 yields 2 molecules BF3 + 3 molecules CaO 1 mole B2O3 + 3 moles CaF2 yields 2 moles BF3 + 3 moles CaO • 4d C6H6 + 3H2 C6H12 1 molecule C6H6 + 3 molecules hydrogen gas yields 1 molecule C6H12 1 mole C6H6 + 3 moles hydrogen gas yields 1 mole C6H12

  12. Day One Assignment p. 282: 3a&d; 4b&d; 10b&d • 10b SeO2 + 2H2Se  3Se + 2H2O 0.625 mole H2O X 3 mole Se/2 mole H2O = 0.938 mole Se • 10d Fe2O3 + 2Al  2Fe + Al2O3 0.625 mole Al2O3 X 2 moles Fe/1 mole Al2O3 = 1.25 mole Fe

  13. Day Two … Class Information • The Meaning of a Balanced Chemical Equation 2H2O  2H2 + O2 • 2 molecules of water = 2 molecules of hydrogen plus 1 molecule of oxygen • 2 moles of water = 2 moles of hydrogen plus 1 mole of oxygen • 2 dozen molecules of water = 2 dozen molecules of hydrogen plus 1 dozen molecules of oxygen • The decomposition of water yields twice as much hydrogen as oxygen … MOLE RATIOs

  14. Day Two … Class Information 2H2O  2H2 + O2 2 moles of H2O = 2 moles H2 + 1 mole O2 • 6 moles of H2O = __ moles H2 + __ moles O2 • 6 moles of H2O = 6 moles H2 + 3 moles O2 • ½ mole H2O = __ mole H2 + __ mole O2 • ½ mole H2O = ½ mole H2 + ¼ mole O2

  15. Day Two … Class Information 2NH3 N2 + 3H2 2 moles NH3  1 moles N2 + 3 moles H2 4 moles NH3  __ moles N2 + __ moles H2 4 moles NH3 (1 mole N2 / 2 moles NH3) = 2 moles N2 4moles NH3 (3 mole H / 2 moles NH3) = 6 mole H2

  16. Day Two … Class Information 3MnO2 + 4Al  3Mn + 2Al2O3 3 moles MnO2 + 4 moles Al  3 moles Mn + 2 moles Al2O3 0.685 moles Al = ______ moles Mn 0.685 moles Al (3 moles Mn / 4 moles Al) = 0.514 moles Mn

  17. Day Two Assignment p. 282: 8b&d; 13d&g; 14b&d • 8b Cl2 + 2KI 2KCl + I2 0.125 mole KI X 2 mole KCl/2 mole KI = 0.125 mole KCl 0.125 mole KI X 1 mole I2 /2 mole KI = 0.0625 mole I2 • 8d CaC2 + 2H2O  Ca(OH)2 + C2H2 0.125 mole CaC2X 1 mole Ca(OH)2/1 mole CaC2 = 0.125 mole Ca(OH)2 0.125 mole CaC2 X 1 mole C2H2 / 1 mole CaC2 = 0.125 mole C2H2

  18. Day Two Assignment p. 282: 8b&d; 13d&g; 14b&d • 13d molar mass of SO2 = 64.07 g/mole (5.23 kg SO2)(1000g/1 kg)(1 mole/64.07g) = 81.6 mole SO2 • 13g molar mass of LiOH = 23.95 g/mole (6.91 X 103 g LiOH)(1 mole/23.95 g) = 288.5 = 289 mole LiOH

  19. Day Two Assignment p. 282: 8b&d; 13d&g; 14b&d • 14b molar mass of He = 4.003 g/mole (2.75 mole He)(4.003 g/1 mole He) = 11.0 g He • 14d molar mass of CO2 = 44.01 g/mole (7.21 X 10-3 mole CO2)(44.01 g/1 mole CO2) = 0.317 g CO2

More Related