1. When copper (II) reacts with silver nitrate, the number of grams of copper required to produce 432 grams of silver is: A 31.5 g B 127 g C 216 g D 252 g With your lab partner, work on the following question in pairs: Bell Ringer 2002 VA Chemistry SOL CuAgNO 3 +Ag22+Cu(NO 3 ) 2 ? 432 g Ag x 1 mol Ag 107.87 g Ag x 2 mol Ag 1 mol Cu x 63.55 g Cu 1 mol Cu = 127.25 g Cu

2. Mass-Mass Quiz Good job on: - Calculating molar mass - Balancing reactions with coefficients We need to fix: - remembering decomposition reactions - making units cancel in stoichiometry work - using the mole ratio

3. Decomposition of Carbonates Common mistake: BaCO 3  BaO + C How to balance? What are we missing? To remember: think of a soda why do we call them “carbonated” ? H 2 CO 3 CO 2 +H2OH2O

4. Homework Answers 13. 150 L O 2 14. 0.73 g Zn 15. 8.36 g NaClO 3 16. 16.70 g KCl 17. 93 L H 2 18. 14.2 L CO 2 19. 30.0 L CO 2 45.0 L H 2 O

5. The Wisdom of Gallagher Why are there Interstate Highways in Hawaii? Why are there floatation devices under plane seats instead of parachutes? Why do we drive on parkways and park on driveways? Why do hot dogs come ten to a package and hot dog buns only eight?

6. Limiting Factors and Percent Yield Ms. Besal 3/1/06

7. Hot Dogs in the News Takeru Kobayashi of Japan downed 44½ hot dogs in 12 minutes. Source: CNN.com WHAT IF… One hot dog = one hot dog + one bun. Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10 per package) and 5 packs of hot dog buns (8 per package). How many hot dogs (according to the official formula) could he have eaten?

8. Hot Dogs in the News Source: CNN.com WHAT IF… One hot dog = one hot dog + one bun. Mr. Kobayashi didn’t do his math correctly. He bought 5 packs of hot dogs (10 per package) and 5 packs of hot dog buns (8 per package). How many hot dogs (according to the official formula) could he have eaten? 5 hot dog packs x 10 hot dogs 1 hot dog pack 50 hot dogs = 5 bun packs8 buns 1 bun pack 40 buns = x 40 possible hot dogs

9. 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 12 oz. Chocolate chips For 1 batch: In my pantry, I have: 5 cups of flour 16 Tbsp of butter lots of everything else How many batches of cookies can I make? Let’s Revisit the Cookies (again)…

10. 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 12 oz. Chocolate chips For 1 batch: How many batches of cookies can I make? Let’s Revisit the Cookies (again)… 5.5 c flour x 2.25 c flour 1 batch cookies = 2.4 batches 16 Tbsp butter x 1 batch cookies 8 Tbsp butter = 2.0 batches EXCESS LIMITING

11. Now I Want to Bake a Cake! But do I have all the ingredients I need? How much flour do I have left after baking all those cookies? 5.5 c flour x 2.25 c flour 1 batch cookies = 16 Tbsp butter x 1 batch 8 Tbsp butter = 2.4 batches of cookies 2.0 batches of cookies GONE! SOME FLOUR LEFT OVER… 2.0 batchesx 2.25 cups flour 1 batch cookies = 4.5 cups flour used 5.5 cups – 4.5 cups = 1.0 cups left

12. Limiting Reactants in Chemistry 5.0 moles of chlorine gas react with 5.0 moles of sodium to produce sodium chloride. Which reagent is the limiting factor? How much of the excess reactant is left over? Cl 2 (g)+NaNaCl22 2 givens = 2 equations! 5.0 mol Cl 2 x 1 mol Cl 2 2 mol NaCl =10. mol NaCl 5.0 mol Na x 2 mol NaCl 2 mol Na =5.0 mol NaCl EXCESS LIMITING 5.0 mol Na x 2 mol Na 1 mol Cl 2 = 2.5 mol Cl 2 5.0 mol Cl 2 given 2.5 mol Cl 2 used 2.5 mol Cl 2 left

13. Practice Problems 1.3 CuSO 4 +2 AlAl 2 (SO 4 ) 3 3 Cu+ 20.0 g CuSO 4 20.0 g Al x x 26.98 g Al 1 mol Al 1 mol CuSO 4 159.61 g CuSO 4 3 mol Cu 3 mol CuSO 4 2 mol Al x x= =0125 mol Cu 1.11 mol Cu 20.0 g CuSO 4 x 159.61 g CuSO 4 1 mol CuSO 4 x 26.98 g Al 1 mol Al = 2.25 g Al USED 20.0 g Al – 2.25 g Al =17.8 g AlEXCESS 3 mol CuSO 4 2 mol Al x

14. Practice Problems 2.2 H 2 (g)+O 2 (g)2 H 2 O 5.0 g H 2 x 1 mol H 2 2.02 g H 2 2 mol H 2 O 2 mol H 2 x=2.5 mol H 2 O USED 5.0 g H 2 – 0.63 g H 2 =4.37 g H 2 EXCESS 5.0 g O 2 x 32.00 g O 2 1 mol O 2 2 mol H 2 O x= 0.31 mol H 2 O 1 mol O 2 5.0 g O 2 x 32.00 g O 2 1 mol O 2 x 2.02 g H 2 1 mol H 2 =0.63 g H 2 1 mol O 2 2 mol H 2 x

15. On Perfection “Perfection never exists in reality, but only in our dreams.” - Dr. Rudolf Dreikurs “Perfection is our goal, excellence will be tolerated.” - J. Yahl

16. Johnny took a quiz yesterday. He missed 4 questions and earned 63 points out of 70. -Was he perfect? -What was his possible score? -What was his actual percent score? Get Real!

17. Ms. Besal ran a reaction in her lab yesterday. She predicted that 183 grams of product would be formed. The reaction only yielded 162 grams of product. But she looked really cool in her lab coat. -Was her reaction perfect? -What was the percent yield? 162 grams 183 grams x 100 = 88.5 %

18. Practice #20 on Homework 100 L N 2 x 1 mol N 2 22.4 L N 2 2 mol NH 3 22.4 L NH 3 xx 1 mol N 2 1 mol NH 3 = 200 L NH 3 THEORETICAL YIELD 200 L NH 3 x 90 % = 180 L NH 3 ACTUAL YIELD What volume of ammonia can be obtained by reacting 100 L of nitrogen gas with an excess of hydrogen, if the yield is 90%? H2H2 NH 3 N2N2 +23

19. Next Class: Quiz on Volume Conversions –Mass-Volume –Volume-Mass –Volume-Volume 2 questions, 10 points each Watch significant figures & labels!