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CP Chemistry

CP Chemistry

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CP Chemistry

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  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