1. Unit 5 Reactions and Stoichiometry Chemistry

2. Ch 11 Describing Chemical Reactions Introduction to Chemical Equations ALL chemical reactions follow 2 laws! Law of Conservation of mass: – In a chemical reaction, the total mass of the reactants and the total mass of the products must be the same. Law of Conservation of Atoms: – In a chemical reaction, atoms can not be created or destroyed.

3. Chemical Reactions are about the arrangement of Atoms, Not the changing of Atoms The only thing about atoms that changes to drive chemistry is the location of electrons

4. Reaction Vocabulary: Reactants: Starting Atoms, Molecules, or Compounds Products: Ending Atoms, Molecules, or Compounds Yields: Progress of reaction – Gives “→” Catalyst: Substance that speeds up chemical reactions – Additional Information States of Matter – Shown as (s), (l), (g), (aq) after formulas Catalyst – Shown above Yield arrow – If Heat is added show with ∆ or ∆H above Yield arrow

5. Example: – Aqueous hydrogen peroxide will decompose into gaseous forms of oxygen and water in the presence of manganese dioxide as a catalyst

6. Balancing Chemical Equations Following the Law of conservation of ATOMS You must NEVER change the # of Atoms in a formula by changing the subscript numbers To change # of atoms (or molecules or compounds), add a coefficient Coefficients are numbers placed in front of a formula. Coefficients change the # of atoms by a multiplying effect

7. Example: – Hydrogen gas combusts with Oxygen gas to form water. 1.Get Formulas correct 2.Organize REACTANTS vs. PRODUCTS 3.BALANCE

8. Balancing Reaction HINTS – Look for Polyatomic Ions! – Save elements O2 and H2 for the end. – If you assume the coefficient of a large particle is 1 and come to a dead end, change it to a 2. Mg 3 (PO 4 ) 2 + NaCl  MgCl 2 + Na 3 PO 4 C 2 H 6 + O 2  CO 2 + H 2 O

9. Reaction Types 1.Combination Reaction: – 2 or more substances as Reactants combine to form 1 Product – Also Called “Synthesis” 2.Decomposition Reaction: – 1 Reactant breaks down into 2 or more Reactants – Most require energy to take place (take in Energy) 3.Single-Replacement Reaction: – 1 element replaces another element in a compound – Reactivity of Metals and Non-Metals is important: – Use activity series (Page 361).

10. 4.Double-Replacement Reaction: – An Exchange of positive ions between 2 Ionic Compounds – Both Reactants are Ionic Compounds – Often takes place in solution with water. – Sometimes a Product is a solid that will not dissolve in water and forms a PRECIPITATE 5.Combustion Reactions – A Reactant combines with Oxygen (other Reactant) – Usually produce energy (give off energy) – Usually involves a Molecular Compounds with (C,H, & or O) and Oxygen – Often have Water and Carbon dioxide as Products “Burning Hydrocarbons” Wood, Gas, Fuel, Sugars (Remember Cellular Respiration?) – May also be a Combination (synthesis) Reaction if only one Product forms “Burning Elements”

11. Combination: A + B  AB Decomposition: AB  A + B Single Replacement: AB + C  CB + A Double Replacement: AB + CD  AD + CB Combustion: ? + O 2  H 2 O + CO 2

12. Predicting Reaction Types Examples 1.A single Reactant: H 2 O 2 2.A metal (Na) and an Ionic Compound (MgCl 2 ) React: 3.C 2 H 6 reacts with O 2 : 4.Two Ionic compounds react [KCl & Mg(NO 3 ) 2 ] 5.A single product: H 2 O

13. 12.1 The Arithmetic of Equations Everyday equations

14. What’s a good equation for a tricycle made of the following components? – Frame (F) – Seat (S) – Wheels (W) – Handbars (H) – Pedals (P) F + S + 3 W + H + 2 P  FSW 3 HP 2 But what if Wheels and Pedals only came in pairs where the manufacturer of the tricycles get their parts? F + S + W 2 + H + P 2 

15. What’s a good equation for a tricycle made of the following components? – Frame (F) – Seat (S) – Wheels (W) – Handbars (H) – Pedals (P) F + S + 3 W + H + 2 P  FSW 3 HP 2 But what if Wheels and Pedals only came in pairs where the manufacturer of the tricycles get their parts? 2F + 2S + 3W 2 + 2H + 2P 2  2FSW 3 HP 2

16. Chemical Equations Stoichiometry: – The calculation of quantities in chemical reactions – Based on a balanced chemical equation What does a balanced chemical reaction REALLY MEAN? – Example: Photosynthesis 6 H 2 O + 6 CO 2 → C 6 H 12 O 6 + 6 O 2 Mole Ratio: – Ratio of moles of substances in a balanced chemical reaction. – Based on Coefficients in front of formula

17. 6 H 2 O (l) + 6 CO 2 (g) → C 6 H 12 O 6 (s) + 6 O 2 (g) For every 1 Glucose _Molecule__there needs to be? ___ H 2 O _______ ___ CO 2 _______ ___ O 2 _______

18. 6 H 2 O (l) + 6 CO 2 (g) → C 6 H 12 O 6 (s) + 6 O 2 (g) For every 1 Glucose _Molecule__there needs to be? _6_ H 2 O Molecules _6_ CO 2 Molecules _6_ O 2 Molecules

19. 6 H 2 O (l) + 6 CO 2 (g) → C 6 H 12 O 6 (s) + 6 O 2 (g) For every 1 Glucose __Mole__there needs to be? ___ H 2 O _______ ___ CO 2 _______ ___ O 2 _______

20. 6 H 2 O (l) + 6 CO 2 (g) → C 6 H 12 O 6 (s) + 6 O 2 (g) For every 1 Glucose __Mole__there needs to be? _6_ H 2 O __ Mole _ _6_ CO 2 __ Mole _ _6_ O 2 __ Mole _

21. 1 mole of Glucose is ____________ g 6 (12.01) = 12 (1.01) = 6 (16.00) = So for every _________ g of Glucose _________ H 2 O _________ CO 2 _________ O 2

22. 1 mole of Glucose is ____________ g 6 (12.01) = 72.06 g 12 (1.01) = 12.12 g 6 (16.00) = 96.00 g So for every _________ g of Glucose _________ H 2 O _________ CO 2 _________ O 2

23. 1 mole of Glucose is __ 180.18 g 6 (12.01) = 72.06 g 12 (1.01) = 12.12 g 6 (16.00) = 96.00 g So for every _________ g of Glucose _________ H 2 O _________ CO 2 _________ O 2

24. 1 mole of Glucose is __ 180.18 g 6 (12.01) = 72.06 g 12 (1.01) = 12.12 g 6 (16.00) = 96.00 g So for every ___ 180.18 g of Glucose _________ H 2 O _________ CO 2 _________ O 2

25. 1 mole of Glucose is __ 180.18 g 6 (12.01) = 72.06 g 12 (1.01) = 12.12 g 6 (16.00) = 96.00 g So for every ___ 180.18 g of Glucose _108.12 g_ H 2 O _264.06 g_ CO 2 _192.00 g_ O 2

26. So 6 H 2 O (l) + 6 CO 2 (g) → C 6 H 12 O 6 (s) + 6 O 2 (g)

27. Sample Problem 12.2 (page 388) Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is: 2 H 2 S (g) + 3 O 2 (g)  2 SO 2 (g) + 2 H 2 O (g) Interpret this equation in terms of: a. numbers of representative particles and moles. b. masses of reactants and products.

28. 12.2 Chemical Calculations Writing and Using Mole Ratios: Page 390 Mole Ratio: – Ratio of moles of substances in a balanced chemical reaction. – Based on Coefficients in front of formula

29. 2H 2 O 2  2H 2 O + O 2 What are all the mole ratios of this equation? If 4.3 moles of Oxygen gas is produced, how many moles of water will also be produced?

30. Sample Problem 12.3 (page 391) How many moles of NH 3 are produced when 0.60 mol of nitrogen reacts with hydrogen in the equation? N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

31. Stoichiometry: Mass - Mass Calculations Using balanced reactions and mole ratios to predict measurable quantities in a chemical reaction. Example: MASS – How many grams of glucose (C 6 H 12 O 6 ) are produced if 120 g of CO 2 react completely with enough water in photosynthesis?

32. Steps 1.Make sure equation is balanced, balance if need be 2.Write down given value 3.Convert given value to moles 4.Convert moles substance to moles other substance USE Mole Ratio 5.Convert moles to mass 6 H 2 O + 6 CO 2 → C 6 H 12 O 6 + 6 O 2

33. Sample Problem 12.4 (page 393) Calculate the number of grams of NH 3 produced by the reactions of 5.40 g of H 2 with an excess of N 2. The balanced equation is: N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

34. # of Particles Mass # Moles Substance A # Moles Substance B # of Particles Mass Other Stoichiometric Calculations

35. How many moles of water are produced from 2.5 moles copper? How many moles of NO are produced from 33.5g of HNO 3 ? How many grams of copper are needed to produce 1.25g of water? How many grams of NO are produced if 3.20x10 23 molecules of water are produced? 3 Cu (s) + 8 HNO 3(aq)  3 Cu(NO 3 ) 2(aq) + 4 H 2 O (l) + 2 NO (g)

36. Sample Problem 12.5 (page 395) How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2 H 2 O (l)  2 H 2 (g) + O 2 (g)

37. 12.3 Limiting Reagent and Percent Yield Limiting Reagent: – Determines the amount of product that can/will form – Reactant that is completely used up in a reaction Excess Reagent: – Reactant that is NOT completely used up in a reaction

38. Limiting Reagent Analogy: HAM SANDWICH 1 slice ham + 1 slice cheese + 1 slice tomato + 2 slices bread  1 ham sandwich – Actual Ingredients: 4 slices of bread 4 slices of tomato 10 slices of ham 1 slice of cheese What is the Limiting Reagent?

39. How to find limiting reagents – Need to analyze MOLES – Find the # of Moles for each reactant based on the # of moles of the other reactants. – Compare the # of Moles needed to the actual # of Moles present. – If amounts are given in other units (g’s?), covert to Moles first.

40. Example: C 2 H 4 + 3O 2  2CO 2 + 2 H 2 O – If 2.70 mol C 2 H 4 is reacted with 6.30 mol O 2 : – Find the limiting reagent – Calculate the moles of water produced Must use LIMITING REAGENT!

41. Another Example 2Al (s) + 3CuSO 4(aq) → 3Cu (s) + Al 2 (SO 4 ) 3(aq) 1.Which reactant is the limiting reagent? 2.What mass of copper is produced 3.How much (mass) excess reagent is left over?

42. 1.Which reactant is the limiting reagent? Using the actual amount of Al (25.0 g), find how much CuSO 4 would be needed. Compare 222 g CuSO 4 needed to actual 75.0g CuSO 4 is the LIMITING REAGENT

43. 2.What mass of copper is produced Use the 75.0 g CuSO 4 Limiting Reagent 29.8 g Cu should be formed

44. 3.How much (mass) excess reagent is left over? Need to solve for actual amount of excess reagent that reacted. Use Limiting reagent. Started with 25.0 g Al and used 8.46 g Al 16.5 g Al will remain after the reaction is complete

45. Sample Problem 12.8 (page 402) Copper reacts with sulfur to form copper (I) sulfide according to the following balanced equation: 2 Cu (s) + S (s)  Cu 2 S (s) What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

46. Percent yield The actual yield is the amount of a product that is experimentally collected. The theoretical yield is the amount of a product that should have formed. – Theoretical Yield: Found from stoichiometry of the LIMITING REAGENT Percent yield is a comparison of the actual amount of product collected to what should have formed.

47. Sample Problem 12.10 (page 406) Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is: CaCO 3 (s)  CaO (s) + CO 2 (g) What is the theoretical yield of CaO if 24.8 g CaCO 3 is heated?

48. Sample Problem 12.11 (page 408) What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO 3 is heated? CaCO 3 (s)  CaO (s) + CO 2 (g)

49. More Difficult Problems: Mixing Limiting Reagent with Percent Yield What is the percent yield if 17.5 g of SO 3 is collected when 20.0 g of FeS 2 is mixed with 16.0 g O 2 ? 4FeS 2 + 15O 2  2Fe 2 O 3 + 8SO 3 Identify your given information:  17.5 g SO 3 is the ____________________  20.0 g FeS 2 is _______________________  16.0 g O 2 is _______________________ Find the LIMITING REAGENT: Find the THEORETICAL YIELD: Find the PERCENT YIELD:

50. Find the LIMITING REAGENT:  Choose one reactant to start with:  Since 20.0 g O 2 is greater than the starting 16.0 g, O 2 is the LIMITING REAGENT

51. Find the THEORETICAL YIELD  SO 3 is the Product we are trying to find  Use 16.0 g O 2 Limiting Reagent  21.4 g SO 3 is the Theoretical Yield

52. Find the PERCENT YIELD  Use the Theoretical Yield and the Actual Yield