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Bell Ringer Aluminum + Sulfuric Acid  Aluminum Sulfate + Hydrogen Gas Which of the following is the balanced chemical equation for the reaction shown.

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Presentation on theme: "Bell Ringer Aluminum + Sulfuric Acid  Aluminum Sulfate + Hydrogen Gas Which of the following is the balanced chemical equation for the reaction shown."— Presentation transcript:

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 + H 2 SO 4   Al 2 (SO 4 ) 3 + H 2 B 2Al + 3H 2 SO 4   Al 2 (SO 4 ) 3 + 3H 2 C 2Al + 3H 2 SO 4   Al 2 (SO 4 ) 3 + H 2 D2Al + H 2 SO 4   Al 2 (SO 4 ) 3 + H 2 2004 SOL

2 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

3 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 How much? What units? Of what?

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

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

6 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 How much? What units? Of what?

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

8 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 (a synthesis reaction) (177ºC) 1 batch of chocolate chip cookies! coefficient unit substance

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… 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 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. How many batches of cookies can I make with that many Egg Beaters?

12 Let’s Revisit the 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 For 1 batch: I have 1.5 cups of Egg Beaters. How many batches of cookies can I make with that many Egg Beaters? 1.5 cups E.B. x 1 batch cookies 0.5 cups E.B. = 3.0 batches of cookies

13 Let’s Revisit the 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 For 1 batch: I have 1.5 cups of Egg Beaters. How much butter do I need to deplete (use up) the Egg Beaters? 1.5 cups E.B. x 8 Tbsp butter 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 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! Baby Steps… Mole-Mole Problems

16 Mole-Mole Problems Example: 2 H 2 + O 2 2 H 2 O How many moles of water can be formed from 0.5 mol H 2 ? 0.5 mol H 2 x 2 mol H 2 2 mol H 2 O = 0.5 mol H 2 O

17 Mole-Mole Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu323 ++ 1. a.0.5 mol Al 3 mol CuSO 4 2 mol Al =x 0.8 mol CuSO 4 b. c. 0.5 mol Al 1 mol Al 2 (SO 4 ) 3 2 mol Al =x 0.3 mol Al 2 (SO 4 ) 3 0.5 mol Al 3 mol Cu 2 mol Al =x 0.8 mol Cu Mole ratio

18 Mole-Mole Practice 2. a.2.5 mol Ca 2 mol AlCl 3 3 mol Ca =x 1.7 mol AlCl 3 b. c. 2.5 mol Ca 3 mol CaCl 2 3 mol Ca =x 2.5 mol CaCl 2 2.5 mol Ca 2 mol Al 3 mol Ca =x 1.7 mol Al CaAlCl 3 CaCl 2 Al322 ++ 3

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 H 2 + O 2 2 H 2 O How many moles of water can be formed from 48.0 g O 2 ? 48.0 g O 2 x 1 mol O 2 2 mol H 2 O = 3.00 mol H 2 O 32.00 g O 2 1 mol O 2 x

21 Mass-Mole Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu323 ++ 1. a.13.5 g Al 1 mol Al 26.98 g Al =x 0.751 mol CuSO 4 b. c. 13.5 g Al 1 mol Al 2 (SO 4 ) 3 2 mol Al = x0.250 mol Al 2 (SO 4 ) 3 13.5 g Al 3 mol Cu 2 mol Al = x0.751 mol Cu Mole ratio x 3 mol CuSO 4 2 mol Al 1 mol Al 26.98 g Al x 1 mol Al 26.98 g Al x

22 Mass-Mole Practice 2. a.5.7 g Ca 2 mol AlCl 3 3 mol Ca = x0.095 mol AlCl 3 b. c. 5.7 g Ca 3 mol CaCl 2 3 mol Ca = x0.14 mol CaCl 2 5.7 g Ca 2 mol Al 3 mol Ca = x0.095 mol Al CaAlCl 3 CaCl 2 Al322 ++ 1 mol Ca x x x 40.08 g Ca 1 mol Ca 40.08 g Ca 1 mol Ca 40.08 g Ca 3

23 Mass-Mass Problems Example: 2 H 2 + O 2 2 H 2 O How many grams of water can be formed from 48.0 g O 2 ? 48.0 g O 2 x 1 mol O 2 2 mol H 2 O = 32.00 g O 2 1 mol O 2 xx 18.02 g H 2 O 1 mol H 2 O 54.1 g H 2 O

24 Mass-Mass Practice CuSO 4 AlAl 2 (SO 4 ) 3 Cu323 ++ 1. a.8.5 g Al 1 mol Al 26.98 g Al = x 75 g CuSO 4 b.8.5 g Al 1 mol Al 2 (SO 4 ) 3 2 mol Al = x 54 g Al 2 (SO 4 ) 3 Mole ratio x 3 mol CuSO 4 2 mol Al 1 mol Al 26.98 g Al x 1 mol CuSO 4 342.14 g Al 2 (SO 4 ) 3 1 mol Al 2 (SO 4 ) 3 x x 159.61 g CuSO 4

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

26 Mass-Mass Practice 2. a.1.9 g Ca 2 mol AlCl 3 3 mol Ca = x 4.2 g AlCl 3 b.1.9 g Ca 3 mol CaCl 2 3 mol Ca = x 5.3 g CaCl 2 CaAlCl 3 CaCl 2 Al322 ++ 1 mol Ca x x 40.08 g Ca 1 mol Ca 40.08 g Ca x 133.33 g AlCl 3 1 mol AlCl 3 x 1 mol CaCl 2 110.98 g CaCl 2 3

27 Mass-Mass Practice c.1.9 g Ca 2 mol Al 3 mol Ca = x 0.85 g Al x 1 mol Ca 40.08 g Ca1 mol Al 26.98 g Al x For Monday’s Quiz: 2 Mass-Mass problems, 10 points each Watch for Significant Figures! Label EVERYTHING! CaAlCl 3 CaCl 2 Al322 ++ 3 points 2 points3 points 2 points 3


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