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Bell Ringer

Bell Ringer. Aluminum + Sulfuric Acid  Aluminum Sulfate + Hydrogen Gas. Which of the following is the balanced chemical equation for the reaction shown above?. A Al + H 2 SO 4  Al 2 (SO 4 ) 3 + H 2.

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Bell Ringer

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  1. Bell Ringer Aluminum + Sulfuric Acid  Aluminum Sulfate + Hydrogen Gas Which of the following is the balanced chemical equation for the reaction shown above? A Al + H2SO4Al2(SO4)3 +H2 B 2Al +3H2SO4Al2(SO4)3 +3H2 C 2Al +3H2SO4Al2(SO4)3 +H2 D 2Al +H2SO4Al2(SO4)3 +H2 2004 SOL

  2. 2.25 cups flour 8 Tbsp butter 0.5 cups shortening 0.75 cups sugar 0.75 cups brown sugar 1 tsp salt 1 tsp baking soda 1 tsp vanilla 0.5 cups Egg Beaters Chocolate Chip Cookies

  3. 2.25cupsflour 8Tbspbutter 0.5cupsshortening 0.75cupssugar 0.75cupsbrown sugar 1tspsalt 1tspbaking soda 1tspvanilla 0.5cupsEgg Beaters How much? Of what? What units? Chocolate Chip Cookies

  4. How much? Of what? Chocolate Chip Cookies • 2.25flour • 8butter • 0.5shortening • 0.75sugar • 0.75brown sugar • 1salt • 1baking soda • 1vanilla • 0.5Egg Beaters

  5. How much? What units? Chocolate Chip Cookies • 2.25cups • 8Tbsp • 0.5cups • 0.75cups • 0.75cups • 1tsp • 1tsp • 1tsp • 0.5cups

  6. How much? Of what? What units? Chocolate Chip Cookies • 2.25cupsflour • 8Tbspbutter • 0.5cupsshortening • 0.75cupssugar • 0.75cupsbrown sugar • 1tspsalt • 1tspbaking soda • 1tspvanilla • 0.5cupsEgg Beaters

  7. Get on with it! What does this have to do with CHEMISTRY?

  8. (177ºC) 1 batch of chocolate chip cookies! 2.25 cups flour + 8 Tbsp butter + 0.5 cups shortening + 0.75 cups sugar + 0.75 cups brown sugar + 1 tsp salt + 1 tsp baking soda + 1 tsp vanilla + 0.5 cups Egg Beaters unit substance coefficient (a synthesis reaction)

  9. Welcome to STOICHIOMETRY Ms. Besal 2/23/2006

  10. What is Stoichiometry? • The study of quantitative relationships within chemical reactions • A balanced equation is the key to stoichiometry! • Tools you’ll need for this chapter: • Writing proper formulas and balanced reactions • Converting from mass to moles and vice versa

  11. Let’s Revisit the Cookies… For 1 batch: The Egg Beaters I have are close to expiring! I’d like to use the rest of them in this recipe. I have 1.5 cups of Egg Beaters. • 2.25 cups flour • 8 Tbsp butter • 0.5 cups shortening • 0.75 cups sugar • 0.75 cups brown sugar • 1 tsp salt • 1 tsp baking soda • 1 tsp vanilla • 0.5 cups Egg Beaters How many batches of cookies can I make with that many Egg Beaters?

  12. Let’s Revisit the Cookies… For 1 batch: I have 1.5 cups of Egg Beaters. • 2.25 cups flour • 8 Tbsp butter • 0.5 cups shortening • 0.75 cups sugar • 0.75 cups brown sugar • 1 tsp salt • 1 tsp baking soda • 1 tsp vanilla • 0.5 cups Egg Beaters How many batches of cookies can I make with that many Egg Beaters? 1 batch cookies 1.5 cups E.B. x = 0.5 cups E.B. 3.0 batches of cookies

  13. Let’s Revisit the Cookies… For 1 batch: I have 1.5 cups of Egg Beaters. • 2.25 cups flour • 8 Tbsp butter • 0.5 cups shortening • 0.75 cups sugar • 0.75 cups brown sugar • 1 tsp salt • 1 tsp baking soda • 1 tsp vanilla • 0.5 cups Egg Beaters How much butter do I need to deplete (use up) the Egg Beaters? 8 Tbsp butter 1.5 cups E.B. x = 0.5 cups E.B. 24 Tablespoons of butter

  14. … Back to Chemistry • There are three types of stoichiometry problems we will deal with today: • Mole-Mole problems (1 conversion) • Mass-Mole problems (2 conversions) • Mass-Mass problems (3 conversions) given required

  15. Baby Steps… Mole-Mole Problems • Step 1: Write a BALANCED EQUATION • Step 2: Determine the mole ratio from the coefficients in the equation. • Mole ratio = moles of required substance moles of given substance • Step 3: Set up the problem like a unit conversion and solve!

  16. Mole-Mole Problems Example: 2 H2O 2 H2 + O2 How many moles of water can be formed from 0.5 mol H2? 2 mol H2O 0.5 mol H2 0.5 mol H2O x = 2 mol H2

  17. 3 2 3 Mole-Mole Practice CuSO4 + Al Al2(SO4)3 + Cu Mole ratio 3 mol CuSO4 1. a. 0.5 mol Al 0.8 mol CuSO4 x = 2 mol Al 1 mol Al2(SO4)3 b. 0.5 mol Al 0.3 mol Al2(SO4)3 x = 2 mol Al 3 mol Cu c. 0.5 mol Al 0.8 mol Cu x = 2 mol Al

  18. 3 2 2 Mole-Mole Practice Ca + AlCl3 3 CaCl2 + Al 2 mol AlCl3 2. a. 2.5 mol Ca 1.7 mol AlCl3 x = 3 mol Ca 3 mol CaCl2 b. 2.5 mol Ca 2.5 mol CaCl2 x = 3 mol Ca 2 mol Al c. 2.5 mol Ca 1.7 mol Al x = 3 mol Ca

  19. Mass-Mole Problems • Step 1: Write a BALANCED EQUATION • Step 2: Calculate the molar mass of your given substance and convert from mass to moles • Step 3: Determine the mole ratio from the coefficients in the equation • Step 4: Set up the conversion and solve!

  20. Mass-Mole Problems Example: 2 H2O 2 H2 + O2 How many moles of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 48.0 g O2 3.00 mol H2O x x = 32.00 g O2 1 mol O2

  21. 3 2 3 Mass-Mole Practice CuSO4 + Al Al2(SO4)3 + Cu Mole ratio 1 mol Al 3 mol CuSO4 1. a. 13.5 g Al 0.751 mol CuSO4 x x = 26.98 g Al 2 mol Al 1 mol Al 1 mol Al2(SO4)3 b. 13.5 g Al x x 0.250 mol Al2(SO4)3 = 26.98 g Al 2 mol Al 1 mol Al 3 mol Cu c. 13.5 g Al x 0.751 mol Cu x = 26.98 g Al 2 mol Al

  22. 3 2 2 Mass-Mole Practice Ca + AlCl3 3 CaCl2 + Al 1 mol Ca 2 mol AlCl3 2. a. 5.7 g Ca x x 0.095 mol AlCl3 = 40.08 g Ca 3 mol Ca 1 mol Ca 3 mol CaCl2 b. 5.7 g Ca x x 0.14 mol CaCl2 = 40.08 g Ca 3 mol Ca 1 mol Ca 2 mol Al c. 5.7 g Ca x x 0.095 mol Al = 40.08 g Ca 3 mol Ca

  23. Mass-Mass Problems Example: 2 H2O 2 H2 + O2 How many grams of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 18.02 g H2O 48.0 g O2 54.1 g H2O x x x = 32.00 g O2 1 mol O2 1 mol H2O

  24. 3 2 3 Mass-Mass Practice CuSO4 + Al Al2(SO4)3 + Cu Mole ratio 1 mol Al 3 mol CuSO4 159.61 g CuSO4 1. a. 8.5 g Al = x x x 26.98 g Al 2 mol Al 1 mol CuSO4 75 g CuSO4 1 mol Al 1 mol Al2(SO4)3 342.14 g Al2(SO4)3 b. 8.5 g Al x x x = 26.98 g Al 2 mol Al 1 mol Al2(SO4)3 54 g Al2(SO4)3

  25. Mass-Mass Practice 1 mol Al 3 mol Cu 63.55 g Cu c. 8.5 g Al x x x = 26.98 g Al 2 mol Al 1 mol Cu 30. g Cu

  26. 3 2 2 Mass-Mass Practice Ca + AlCl3 3 CaCl2 + Al 1 mol Ca 2 mol AlCl3 133.33 g AlCl3 2. a. 1.9 g Ca x x x = 40.08 g Ca 3 mol Ca 1 mol AlCl3 4.2 g AlCl3 1 mol Ca 3 mol CaCl2 110.98 g CaCl2 b. 1.9 g Ca x x x = 40.08 g Ca 3 mol Ca 1 mol CaCl2 5.3 g CaCl2

  27. 3 2 2 3 points 2 points 2 points 3 points Mass-Mass Practice Ca + AlCl3 3 CaCl2 + Al 1 mol Ca 2 mol Al 26.98 g Al c. 1.9 g Ca x x x = 40.08 g Ca 3 mol Ca 1 mol Al 0.85 g Al • For Monday’s Quiz: • 2 Mass-Mass problems, 10 points each • Watch for Significant Figures! • Label EVERYTHING!

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