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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.

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Presentation on theme: "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."— Presentation transcript:

1 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?

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3 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?

4 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

5 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: 5.5 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)…

6 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

7 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

8 In the laboratory, a reaction is rarely carried out with exactly the required amounts of each reactant. In most cases, one or more of the reactants is present in excess, that is, there is more than the exact amount required to react. Once one of the reactants is used up, no more product can be formed. The substance that is completely used up first in a reaction is the limiting reactant. The limiting reactant (reagent) is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. The excess reactant is the substance that is not used up completely in a reaction.

9 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 reactant? 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 mol Cl 2 given 2.5 mol Cl 2 used 2.5 mol Cl 2 left

10 Practice Problems 1.3 CuSO 4 +2 AlAl 2 (SO 4 ) 3 3 Cu g CuSO g Al x x g Al 1 mol Al 1 mol CuSO g CuSO 4 3 mol Cu 3 mol CuSO 4 2 mol Al x x= =0.125 mol Cu 1.11 mol Cu 20.0 g CuSO 4 x g CuSO 4 1 mol CuSO 4 x 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 In this reaction, 20.0 g of CuSO 4 reacts with 20.0 g of Al. Which reactant is limiting? How much of the excess reactant is left over?

11 Practice Problems 2.2 H 2 (g)+O 2 (g)2 H 2 O 5.0 g H 2 x 1 mol H g H 2 2 mol H 2 O 2 mol H 2 x=2.5 mol H 2 O 5.0 g O 2 x g O 2 1 mol O 2 2 mol H 2 O x= 0.31 mol H 2 O 1 mol O 2 In this reaction, 5.0 g of H 2 reacts with 5.0 g O 2. Which reactant is limiting? How many grams of H 2 O are produced? 0.31 mol H 2 O smallest number of moles x g H 2 O 1 mol H 2 O =g H 2 O5.6

12 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

13 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!

14 Mrs. Wilson 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. -Was her reaction perfect? -What was the percent yield? 162 grams 183 grams x 100 = 88.5 % 162 grams = actual yield 183 grams = theoretical yield

15 Percentage Yield The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. In most chemical reactions, the amount of product obtained is less than the theoretical yield. The measured amount of product obtained from a reaction is called the actual yield of the product. The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

16 Percentage Yield, continued Sample Problem H Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C 6 H 6, with chlorine, as represented by the following equation. C 6 H 6 (l) + Cl 2 (g) → C 6 H 5 Cl(l) + HCl(g) When 36.8 g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percentage yield of C 6 H 5 Cl?

17 Percentage Yield, continued Sample Problem H Solution C 6 H 6 (l) + Cl 2 (g) → C 6 H 5 Cl(l) + HCl(g) Given:mass of C 6 H 6 = 36.8 g mass of Cl 2 = excess actual yield of C 6 H 5 Cl = 38.8 g Unknown: percentage yield of C 6 H 5 Cl Solution: Theoretical yield molar mass mol ratio molar mass

18 Percentage Yield, continued Sample Problem H Solution, continued C 6 H 6 (l) + Cl 2 (g) → C 6 H 5 Cl(l) + HCl(g) Theoretical yield Percentage yield


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