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Bell Ringer

With your lab partner, work on the following question in pairs :. 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. 2. 2. Bell Ringer. Cu. +. AgNO 3. Ag. +.

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Bell Ringer

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  1. With your lab partner, work on the following question in pairs: 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 2 2 Bell Ringer Cu + AgNO3 Ag + Cu(NO3)2 ? 432 g Ag 1 mol Ag 1 mol Cu 63.55 g Cu x x x 107.87 g Ag 2 mol Ag 1 mol Cu = 127.25 g Cu 2002 VA Chemistry SOL

  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: BaCO3 BaO + C How to balance? What are we missing? CO2 To remember: think of a soda why do we call them “carbonated” ? H2CO3 CO2 + H2O

  4. Homework Answers • 150 L O2 • 0.73 g Zn • 8.36 g NaClO3 • 16.70 g KCl • 93 L H2 • 14.2 L CO2 • 30.0 L CO2 • 45.0 L H2O

  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. One hot dog = one hot dog + one bun. WHAT IF… 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? Source: CNN.com

  8. Hot Dogs in the News One hot dog = one hot dog + one bun. WHAT IF… 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? Excess 5 hot dog packs 10 hot dogs 50 hot dogs x = 1 hot dog pack Limiting Factor 40 possible hot dogs 5 bun packs 8 buns 40 buns x = 1 bun pack Source: CNN.com

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

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

  11. 1 batch cookies 5.5 c flour x = 2.4 batches of cookies 2.25 c flour 1 batch 16 Tbsp butter x = 2.0 batches of cookies 8 Tbsp butter 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? SOME FLOUR LEFT OVER… GONE! 2.25 cups flour 4.5 cups flour used 2.0 batches x = 1 batch cookies 5.5 cups – 4.5 cups = 1.0 cups left

  12. 2 2 LIMITING 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? Cl2 (g) + Na NaCl EXCESS 2 givens = 2 equations! 2 mol NaCl 5.0 mol Cl2 x = 10. mol NaCl 1 mol Cl2 2 mol NaCl 5.0 mol Na x = 5.0 mol NaCl 5.0 mol Cl2 given 2 mol Na 2.5 mol Cl2 used 1 mol Cl2 2.5 mol Cl2 5.0 mol Na x = 2.5 mol Cl2 left 2 mol Na

  13. Practice Problems 1. 3 CuSO4 + 2 Al Al2(SO4)3 + 3 Cu 1 mol CuSO4 3 mol Cu 20.0 g CuSO4 x = 0125 mol Cu x 159.61 g CuSO4 3 mol CuSO4 20.0 g Al 1 mol Al 3 mol Cu 1.11 mol Cu x x = 26.98 g Al 2 mol Al 1 mol CuSO4 2 mol Al 26.98 g Al 20.0 g CuSO4 x x = x 159.61 g CuSO4 3 mol CuSO4 1 mol Al 2.25 g Al 20.0 g Al – 2.25 g Al = 17.8 g Al EXCESS USED

  14. 1 mol H2 2 mol H2O 5.0 g H2 x = 2.5 mol H2O x 2.02 g H2 2 mol H2 5.0 g O2 1 mol O2 2 mol H2O 0.31 mol H2O x x = 1 mol O2 32.00 g O2 1 mol O2 2 mol H2 2.02 g H2 = 0.63 g H2 5.0 g O2 x x x 32.00 g O2 1 mol O2 1 mol H2 Practice Problems 2. 2 H2 (g) + O2 (g) 2 H2O USED 5.0 g H2 – 0.63 g H2 = 4.37 g H2 EXCESS

  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. Get Real! 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? 70 points = 100 % 90 %

  17. Get Real! 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 100 88.5 % x = 183 grams

  18. 3 2 Practice #20 on Homework What volume of ammonia can be obtained by reacting 100 L of nitrogen gas with an excess of hydrogen, if the yield is 90%? N2 + H2 NH3 100 L N2 1 mol N2 2 mol NH3 22.4 L NH3 x x x = 22.4 L N2 1 mol N2 1 mol NH3 200 L NH3 THEORETICAL YIELD 200 L NH3 x 90 % = 180 L NH3 ACTUAL YIELD

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

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