Chemistry Chapter 9. Unit 6 Stoichiometry

Transcription

1 Chemistry Chapter 9 Unit 6 Stoichiometry

2 The arithmetic of equations Equations are recipes. They tell chemists what amounts of reactants to mix and what amounts of products to expect. What is Stoichiometry? The calculation of quantities in chemical reactions

3 A quick example If we consider chemical equations to be recipes: When you bake cookies, you probably follow a recipe The recipe tells you what ingredients to mix and in what ratio The end result of the combination of ingredients is cookies If you want more cookies you can double or triple the recipe If you want fewer cookies, the recipe can be cut in half or quartered The ingredients are the reactants and the cookies are the products In a way, the cookie recipe gives the same kind of information that a balanced chemical equation does

4 Stoichiometry: A Closer Look Lets examine stoichiometry a little closer by looking at the production of ammonia from it s elements: Write the equation below: N 2(g) + 3H 2(g) 2NH 3(g)

5 N 2(g) + 3H 2(g) 2(g) 2NH 2NH3(g) Valuable information can be derived from this equation: Write sentences to describe the equation - 1. In terms of PARTICLES One molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas.

6 N 2(g) + 3H 2(g) 2(g) 2NH 2NH3(g) 2. In terms of MOLES One mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas. The coefficients of a balanced chemical equation indicate the relative number of moles of reactants to the moles of products in a reaction. This is the most important information obtained from a balanced chemical equation.

7 N 2(g) + 3H 2(g) 2(g) 2NH 3. In terms of MASS Reactants: N = 14 x 2 = 28g H = 1 x 6 = + 6g 34g total Products: N = 14 x 2 = 28g H = 1 x 6 = + 6g 34g total 2NH3(g) Balanced chemical equations must obey the law of conservation of mass. Remember that mass is related to number of atoms. Even though the number of moles is different, the mass of reactants and products is equal.

8 N 2(g) + 3H 2(g) 2(g) 2NH 2NH3(g) 4. In terms of VOLUME (for gases only!) Remember that one mole of any gas at STP is 22.4 L. Therefore, the volume (as well as number of moles, molecules, and formula units) of gases may not be the same.

9 N 2(g) + 3H 2(g) 2(g) 2NH 2NH3(g) What was conserved (stayed the same from left to right) in this equation? (2 answers) 1. Mass 2. Number of Atoms

10 Another quick example Imagine that you are in charge of manufacturing for Tiny Tyke Tricycle Company. The business plan for Tiny Tike requires the production of 128 custom made tricycles each day. One of your responsibilities is to be sure that there are enough parts available at the start of each day to make these tricycles. To simplify this discussion, assume that the major components of the tricycle are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P). The finished tricycle has a formula of FSW 3 HP 2. The balanced equation for the production of a tricycle is: F + S + 3W + H + 2P FSW 3 HP 2

11 F + S + 3W + H + 2P FSW 3 HP 2 In a five day workweek, Tiny Tike is scheduled to make 640 tricycles. How many wheels should be in the plant on Monday morning to make these tricycles? # of tricycles to be made: 640 tricycles 1 FSW 3 HP 2 = 3W (from balanced equation) Use the conversion factor: 3W 1 FSW 3 HP FSW 3 HP 2 x 3W 1 FSW 3 HP 2 = 1920W

12 F + S + 3W + H + 2P FSW 3 HP 2 Here is another one: How many tricycle seats, wheels, and pedals are needed to make 288 tricycles? Seats: 288 Wheels: 864 Pedals: 576

13 Practice Problems: 1. Try this: a. Write the equation for the complete combustion of liquid ethanol, C 2 H 5 OH: b. Interpret the equation in terms of numbers of molecules c. Interpret the equation in terms of moles d. Interpret the equation in terms of mass e. Why can t this problem be interpreted in terms of volume?

14 Practice Problem 1 Write the equation for the complete combustion of liquid ethanol, C 2 H 5 OH: C 2 H 5 OH (l) + 3O 2(g) 3H 2 O (l) + 2CO 2(g)

15 C 2 H 5 OH (l) + 3O 2(g) 3H 2 O (l) + 2CO 2(g) Interpret the equation in terms of numbers of molecules. 1 molecule of liquid ethanol reacts with 3 molecules of oxygen gas to yield 3 molecules of liquid water and 2 molecules of carbon dioxide gas.

16 C 2 H 5 OH (l) + 3O 2(g) 3H 2 O (l) + 2CO 2(g) Interpret the equation in terms of moles 1 mole of liquid ethanol reacts with 3 moles of oxygen gas to yield 3 moles of liquid water and 2 moles of carbon dioxide gas. 4 moles total reactants 5 moles total products

17 C 2 H 5 OH (l) + 3O 2(g) 3H 2 O (l) + 2CO 2(g) Interpret the equation in terms of mass Reactants: 1 molecule of ethanol = 46g 3 molecules of oxygen = 96g 142g total reactants Products: 3 molecules of water = 54g 2 molecules of carbon dioxide = 88g 142g total products

18 C 2 H 5 OH (l) + 3O 2(g) 3H 2 O (l) + 2CO 2(g) Why can t this problem be interpreted in terms of volume? Ethanol is in liquid form. The volume terms of writing is based on gases.

19 Practice Problem 2 Write the balanced equation for the single replacement reaction of solid potassium metal reacting with liquid water: 2K (s) + H 2 O (l) H 2(g) + K 2 O (aq)

20 2K + (s) H 2 O (l) H + 2(g) K 2 O (aq) Interpret this equation in terms of: Interaction particles- moles mass volume (if allowed)

21 2K + (s) H 2 O (l) H + 2(g) K 2 O (aq) Interaction particles: Two atoms of solid potassium react with one molecule of liquid water to produce one molecule of hydrogen gas and one formula unit of aqueous potassium oxide.

22 2K + (s) H 2 O (l) H + 2(g) K 2 O (aq) Moles: Two moles of solid potassium metal reacts with one mole of liquid water to produce one mole of hydrogen gas and one mole of aqueous potassium oxide. 3 moles total reactants 2 moles total products

23 2K + (s) H 2 O (l) H + 2(g) K 2 O (aq) Mass: Reactants: K = 39g (2) = 78g H 2 O = 18g (1) = + 18g 96g total Products: H 2 = 1g (2) = 2g K 2 O = 94g (1) = + 94g 96g total

24 2K + (s) H 2 O (l) H + 2(g) K 2 O (aq) volume (if allowed) This is not allowed because not all are gases.

25 Practice Test A Objective 1 1. Write the balanced equation for the formation of gaseous fluorine trioxide from its elements. Include the adjectives. F 2(g) 2(g) + 3O 2(g) 2FO 2FO 3(g)

26 Practice Test A Objective 1 2. Interpret this equation in terms of the number of representative particles. F 2(g) 2(g) + 3O 2(g) 2FO 2FO 3(g) 1 molecule of fluorine gas reacts with 3 molecules of oxygen gas to produce 2 molecules of fluorine trioxide gas.

27 Practice Test A Objective 1 3. Interpret this equation in terms of number of moles. F 2(g) 2(g) + 3O 2(g) 2FO 2FO 3(g) 1 mole of fluorine gas reacts with 3 moles of oxygen gas to produce 2 moles of fluorine trioxide gas.

28 Practice Test A Objective 1 4. Interpret this equation in terms of mass. F 2(g) Reactants: 2(g) + 3O 2(g) 2FO F = 19g (2) = 38g O = 16g (6) = + 96g 134g total Products: F = 19g (2) = 38g O = 16g (6) = + 96g 134g total 2FO 3(g)

29 Practice Test A Objective 1 5. Interpret this equation in terms of volume. F 2(g) + 3O 2(g) 2FO 3(g) Reactants: F 2 = 1 mol x 22.4 L/mol = 22.4 L O 2 = 3 mol x 22.4 L/mol = L 89.6 L total Products: FO 3 = 2 mol x 22.4 L/mol = 44.8 L total

30 Chemical Calculations Section 9.2

31 Mole-Mole Calculations The heart of the stoichiometry problem.

32 Think WANT OVER GIVEN Example Problem: Rewrite the balanced equation for the production of ammonia from its elements: N 2(g) 2(g) + 3H 2(g) 2NH 2NH 3(g)

33 N 2(g) + 3H 2(g) 2(g) 2NH 2NH3(g) How many moles of ammonia are produced when 0.6 mol of nitrogen react with hydrogen? 0.6 mol N 0.6 mol N 2 x 2 mol NH 2 mol NH 3 = 1.2 mol NH = 1.2 mol NH 3 1mol N 1mol N 2

34 Example Problem Write the equation of aluminum oxide from its elements. 4Al (s) + 3O 2(g) 2Al 2 O 3(s)

35 Example Problem 4Al (s) + 3O 2(g) 2Al 2 O 3(s) 1. How many moles of aluminum are needed to form 3.7 moles of aluminum oxide? 3.7 mol Al 2 O x 3 4mol Al 2mol Al 2 O 3 = 7.4 mol Al 2. How many moles of oxygen are required to react completely with 14.8 moles of aluminum? 14.8 mol Al x 3 mol O 2 = 11.1 mol O 2 4 mol Al 3. How many moles of aluminum oxide are formed when 0.78 moles of oxygen reacts with aluminum? 0.78 mol O 2 x 2 mol Al 2 O 3 3 mol O 2 = 0.52 mol Al 2 O 3

36 Mass-Mass Calculations Rewrite the equation for the production of ammonia: N 2(g) + 3H 2(g) 2NH 3(g) Calculate the number of grams of ammonia produced by the reaction of 5.4 grams of hydrogen with an excess of nitrogen. 5.4g H 2 x 1 mol H 2 = 2.7 mol H 2 2g H mol H 2 x 2 mol NH 3 = 1.8 mol NH 3 3 mol H mol NH 1.8 mol NH 3 x 17.0g NH 17.0g NH 3 = 31g NH = 31g NH 3 1 mol NH 1 mol NH 3

37 Example Problem Write the balanced chemical equation for the decomposition for potassium chlorate. Calculate the number of grams of reactant necessary to produce 50 grams of potassium chloride. How many grams of oxygen gas are produced when you decompose 15 grams of potassium chlorate? How many MOLES of each product are produced when 10 grams of potassium chlorate are heated? How many grams of each product are produced when 2.5 moles of reactant are used?

38 Example Problem Write the balanced chemical equation for the decomposition for potassium chlorate. 2KClO 3 3O 2 + 2KCl

39 2KClO 3 3O 2 + 2KCl Calculate the number of grams of reactant necessary to produce 50 grams of potassium chloride. 50g KCl x 1 mol KCl x 2 mol KClO 3 x 122.5g KClO 3 = 82.21g KClO g KCl 2 mol KCl 1 mol KClO 3

40 2KClO 3 3O 2 + 2KCl How many grams of oxygen gas are produced when you decompose 15 grams of potassium chlorate? 15g KClO 3 x 1mol KClO x 3 3mol O g KClO 3 2mol KClO 3 x 32g O 2 1mol O 2 = 5.88g O 2

41 2KClO 3 3O 2 + 2KCl How many MOLES of each product are produced when 10 grams of potassium chlorate are heated? 10g KClO x 3 1mol KClO 3 x 3mol O g KClO 3 2mol KClO 3 = 0.12mol O 2 10g KClO 3 x 1mol KClO x 3 2mol KCl 122.5g KClO 3 2mol KClO 3 = 0.08mol KCl

42 2KClO 3 3O 2 + 2KCl How many grams of each product are produced when 2.5 moles of reactant are used? 2.5 mol KClO x 3 3 mol O 2 = 3.75 mol O 2 2 mol KClO mol O 2 x 32g O 2 = 120g O 2 1 mol O mol KClO x 3 2 mol KCl 2 mol KClO 3 = 2.5 mol KCl 2.5 mol KCl x 74.5g KCl = g KCl 1 mol KCl

43 Other Stoichiometric Calculations The Stoichiometry Road Map

44 With your knowledge of conversion factors and the problem-solving diagram, you can solve a variety of stoichiometric problems. And Have lots of fun doing it!

45 Example Problem How many molecules of oxygen are produced when a sample of 29.2 grams of water is decomposed by electrolysis? Balanced Equation: 2H 2 O (l) 2H 2(g) + O 2(g) 29.2g H 2 O x 1mol H 2 O X 1mol O 2 X 6.02 e^23 molecules O 2 = 18g H 2 0 2mol H 2 O 1 mol X 10^23 molecules O 2

46 Example Problem How many liters of oxygen are produced by the decomposition of 6.54 grams of potassium chlorate? Balanced Equation: 2KClO 3(s) 3O 2(g) + 2KCl (s) 6.54g KClO 3 x 1mol KClO 3 x 3mol O g KClO 3 2mol KClO 3 x 22.4L O 2 = 1mol O L O 2

47 Example Problem Assuming STP, how many liters of oxygen gas are needed to produce 19.8 L sulfur trioxide according to this balanced equation? 2SO 2(g) + O 2(g) 2SO 2SO 3(g) 19.8L SO 3 x 1mol SO 3 x 1mol O L SO 3 2mol SO 3 x 22.4L O 2 = 1mol O 2 9.9L O 2

48 Example Problem Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide. How many milliliters of nitrogen dioxide are produced when 3.4 milliliters of oxygen react with an excess of nitrogen monoxide? Assume STP conditions. Balanced Equation: 2NO + O 2 2NO L O 2 x 1mol O 2 x 2mol NO L O 2 1mol O 2 x 22.4L NO 2 x 1000ml NO = 2 1mol NO 2 1L NO 2

49 Practice Problems See Note Pack and Transparencies And most of all.. Have Fun

50 Section 9.3 Limiting Reagent

51 Limiting Reagent The limiting reagent limits or determines the amount of product that can be formed in a reaction In contrast, the excess reagent is the reagent that is not used up in a reaction (there is some left over)

52 Ammonia Production (again) N 2(g) 2(g) + 3H 2(g) 2NH 2NH 3(g) This represents the most efficient recipe that a chemist can follow What is the limiting reagent if the reaction is run with 2 moles of nitrogen and 3 moles of hydrogen? HYDROGEN

53 Example Problem Sodium chloride can be prepared by the reaction of sodium with chlorine gas according to the following equation: 2Na (s) + Cl 2(g) 2NaCl Suppose that 6.7 mol of sodium reacts with 3.2 mol of chlorine. How many moles of sodium chloride are produced? What is the limiting reagent? Follow along on page 254 of the book

54 Example Problem 2Na (s) + Cl 2(g) 2NaCl Suppose that 6.7 mol of sodium reacts with 3.2 mol of chlorine. How many moles of sodium chloride are produced? Choose one known and compare it to the required amount of the other reactant. 6.7 mol Na x 1mol Cl 2 / 2 mol Na = 3.35 mol Cl 2 (Limiting) 3.20 mol Cl 2 x 2 mol NaCl / 1 mol Cl 2 = 6.40 mol NaCl (Answer) What is the limiting reagent? The limiting reagent becomes Cl 2 because we do not have enough to completely react all of the Na

55 Example Problem 2 Write the equation for the complete combustion of ethene,, C 2 H 4 If 2.7 mol of ethene is reacted with 6.3 mol of oxygen Calculate the moles of water produced What is the limiting reagent

56 Practice Problem 3 Write the equation for the incomplete combustion of ethene,, C 2 H 4 If 2.7 mol of ethene is reacted with 6.3 mol of oxygen Calculate the moles of water produced What is the limiting reagent

57 Practice Problem 4 When copper reacts with sulfur, solid copper (I) sulfide is produced. Write this reaction. What is the limiting reagent when 80.0 grams of copper reacts with 25.0 grams of sulfur? What is the maximum number of grams of product that can be formed?

58 Practice Problem 5 Hydrogen gas can be produced in the lab by the reaction of magnesium metal with hydrochloric acid. Write the COMPLETE balanced equation. Identify the limiting reagent when 6 grams of HCl react with 5 grams of Mg. How many grams of hydrogen can be produced? The End