2. A Study of Chemical Reactions Equations, Mole Conversions, & Stoichiometry

3. Types of Reactions Many chemical reactions have defining characteristics which allow them to be classified as to type. Many chemical reactions have defining characteristics which allow them to be classified as to type.

4. Types of Chemical Reactions The five types of chemical reactions in this unit are: The five types of chemical reactions in this unit are: Combination/Synthesis Combination/Synthesis Decomposition/Analysis Decomposition/Analysis Single Replacement/Displacement Single Replacement/Displacement Double Replacement/Metathesis Double Replacement/Metathesis Combustion Combustion

5. Combination Reactions Two or more substances combine to form one substance. Two or more substances combine to form one substance. The general form is A + X  AX The general form is A + X  AX Example: Example: Magnesium + oxygen  magnesium oxide Magnesium + oxygen  magnesium oxide 2Mg + O 2  2MgO 2Mg + O 2  2MgO

6. Magnesium + Oxygen

7. Combination Reactions Combination reactions may also be called composition or synthesis reactions. Combination reactions may also be called composition or synthesis reactions. Some types of combination reactions: Some types of combination reactions: Combination of elements Combination of elements K + Cl 2  K + Cl 2  One product will be formed One product will be formed

8. Combination Reactions K + Cl 2  K + Cl 2  Write the ions: K + Cl - Write the ions: K + Cl - Balance the charges: KCl Balance the charges: KCl Balance the equation: 2K + Cl 2  2KCl Balance the equation: 2K + Cl 2  2KCl

9. Combination Reactions Some types of combination reactions: Some types of combination reactions: Oxide + water  Oxide + water  Nonmetal oxide + water  acid Nonmetal oxide + water  acid SO 2 + H 2 O  H 2 SO 3 SO 2 + H 2 O  H 2 SO 3 Metal oxide + water  base Metal oxide + water  base BaO + H 2 O  Ba(OH) 2 BaO + H 2 O  Ba(OH) 2

10. Combination Reactions Some types of combination reactions: Some types of combination reactions: Metal oxides + nonmetal oxides Metal oxides + nonmetal oxides Na 2 O + CO 2  Na 2 CO 3 Na 2 O + CO 2  Na 2 CO 3 CaO + SO 2  CaSO 3 CaO + SO 2  CaSO 3

11. Decomposition Reactions One substance reacts to form two or more substances. One substance reacts to form two or more substances. The general form is AX  A + X The general form is AX  A + X Example: Example: Water can be decomposed by electrolysis. Water can be decomposed by electrolysis. 2H 2 O  2H 2 + O 2 2H 2 O  2H 2 + O 2

12. Electrolysis of Water

13. Decomposition Reactions Types of Decomposition Reactions: Types of Decomposition Reactions: Decomposition of carbonates Decomposition of carbonates When heated, some carbonates break down to form an oxide and carbon dioxide. When heated, some carbonates break down to form an oxide and carbon dioxide. CaCO 3  CaO + CO 2 CaCO 3  CaO + CO 2 H 2 CO 3  H 2 O + CO 2 H 2 CO 3  H 2 O + CO 2

14. Decomposition Reactions Types of decomposition reactions: Types of decomposition reactions: Some metal hydroxides decompose into oxides and water when heated. Some metal hydroxides decompose into oxides and water when heated. Ca(OH) 2  CaO + H 2 O Ca(OH) 2  CaO + H 2 O Note that this is the reverse of a similar combination reaction.

15. Decomposition Reactions Types of decomposition reactions: Types of decomposition reactions: Metal chlorates decompose into chlorides and oxygen when heated. Metal chlorates decompose into chlorides and oxygen when heated. 2KClO 3  2KCl + 3O 2 2KClO 3  2KCl + 3O 2 Zn(ClO 3 ) 2  ZnCl 2 + 3O 2 Zn(ClO 3 ) 2  ZnCl 2 + 3O 2 Some of these reactions are used in explosives. Some of these reactions are used in explosives.

16. Decomposition Reactions Some substances can easily decompose: Some substances can easily decompose: Ammonium hydroxide is actually ammonia gas dissolved in water. Ammonium hydroxide is actually ammonia gas dissolved in water. NH 4 OH  NH 3 + H 2 O NH 4 OH  NH 3 + H 2 O Some acids decompose into water and an oxide. Some acids decompose into water and an oxide. H 2 SO 3  H 2 O + SO 2 H 2 SO 3  H 2 O + SO 2

17. Decomposition Reactions Some decomposition reactions are difficult to predict. Some decomposition reactions are difficult to predict. The decomposition of nitrogen triiodide, NI 3, is an example of an interesting decomposition reaction. The decomposition of nitrogen triiodide, NI 3, is an example of an interesting decomposition reaction.

18. Nitrogen triiodide

19. Single Replacement Reactions Cationic: A metal will replace a metal ion in a compound. Cationic: A metal will replace a metal ion in a compound. The general form is A + BX  AX + B The general form is A + BX  AX + B Anionic: A nonmetal will replace a nonmetal ion in a compound. Anionic: A nonmetal will replace a nonmetal ion in a compound. The general form is Y + BX  BY + X The general form is Y + BX  BY + X

20. Single Replacement Reactions Examples: Examples: Ni + AgNO 3  Ni + AgNO 3  Nickel replaces the metallic ion Ag +. Nickel replaces the metallic ion Ag +. The silver becomes free silver and the nickel becomes the nickel(II) ion. The silver becomes free silver and the nickel becomes the nickel(II) ion. Ni + AgNO 3  Ag + Ni(NO 3 ) 2 Ni + AgNO 3  Ag + Ni(NO 3 ) 2 Balance the equation: Balance the equation: Ni + 2AgNO 3  2Ag + Ni(NO 3 ) Ni + 2AgNO 3  2Ag + Ni(NO 3 )

21. Activity Series

22. Single Replacement Reactions Not all single replacement reactions that can be written actually happen. Not all single replacement reactions that can be written actually happen. The metal must be more active than the metal ion. The metal must be more active than the metal ion. Aluminum is more active than iron in Al + Fe 2 O 3 in the following reaction: Aluminum is more active than iron in Al + Fe 2 O 3 in the following reaction:

23. Thermite Reaction

24. Al + Fe 2 O 3  Al + Fe 2 O 3  Aluminum will replace iron(III) Aluminum will replace iron(III) Iron(III) becomes Fe and aluminum metal becomes Al 3+. Iron(III) becomes Fe and aluminum metal becomes Al 3+. 2Al + Fe 2 O 3  2Fe + Al 2 O 3 2Al + Fe 2 O 3  2Fe + Al 2 O 3

25. Single Replacement Reactions An active nonmetal can replace a less active nonmetal. An active nonmetal can replace a less active nonmetal. The halogen (F 2, Cl 2, Br 2, I 2 ) reactions are good examples. The halogen (F 2, Cl 2, Br 2, I 2 ) reactions are good examples. F 2 is the most active and I 2 is the least. F 2 is the most active and I 2 is the least. Cl 2 +2 NaI  2 NaCl + I 2 Cl 2 +2 NaI  2 NaCl + I 2

26. Double Replacement Reactions Ions of two compounds exchange places with each other. Ions of two compounds exchange places with each other. The general form is AX + BY  AY + BX The general form is AX + BY  AY + BX Metathesis is an alternate name for double replacement reactions. Metathesis is an alternate name for double replacement reactions.

27. NaOH + CuSO 4

28. Metathesis (sink or float?) NaOH + CuSO 4  NaOH + CuSO 4  The Na + and Cu 2+ switch places. The Na + and Cu 2+ switch places. Na + combines with SO 4 2- to form Na 2 SO 4. Na + combines with SO 4 2- to form Na 2 SO 4. Cu 2+ combines with OH - to form Cu(OH) 2 Cu 2+ combines with OH - to form Cu(OH) 2 NaOH + CuSO 4  Na 2 SO 4 + Cu(OH) 2 NaOH + CuSO 4  Na 2 SO 4 + Cu(OH) 2 2NaOH + CuSO 4  Na 2 SO 4 + Cu(OH) 2 2NaOH + CuSO 4  Na 2 SO 4 + Cu(OH) 2

29. CuSO 4 + Na 2 CO 3

30. Double Replacement CuSO 4 + Na 2 CO 3  CuSO 4 + Na 2 CO 3  Cu 2+ combines with CO 3 2- to form CuCO 3. Cu 2+ combines with CO 3 2- to form CuCO 3. Na + combines with SO 4 2- to form Na 2 SO 4. Na + combines with SO 4 2- to form Na 2 SO 4. CuSO 4 + Na 2 CO 3  CuCO 3 + Na 2 SO 4 CuSO 4 + Na 2 CO 3  CuCO 3 + Na 2 SO 4

31. Na 2 CO 3 + HCl

32. Double Replacement Na 2 CO 3 + HCl  Na 2 CO 3 + HCl  Notice that gas bubbles were produced rather than a precipitate. Notice that gas bubbles were produced rather than a precipitate. What was the gas? What was the gas? Write the double replacement reaction first. Write the double replacement reaction first.

33. Double Replacement Na 2 CO 3 + HCl  Na 2 CO 3 + HCl  Na + combines with Cl - to form NaCl. Na + combines with Cl - to form NaCl. H + combines with CO 3 2- to form H 2 CO 3. H + combines with CO 3 2- to form H 2 CO 3. Na 2 CO 3 + 2HCl  2NaCl + H 2 CO 3 Na 2 CO 3 + 2HCl  2NaCl + H 2 CO 3 H 2 CO 3 breaks up into H 2 O and CO 2. H 2 CO 3 breaks up into H 2 O and CO 2.

34. Double Replacement The gas formed was carbon dioxide. The gas formed was carbon dioxide. The final balanced reaction is: Na 2 CO 3 + HCl  NaCl + H 2 O + CO 2. The final balanced reaction is: Na 2 CO 3 + HCl  NaCl + H 2 O + CO 2. Balance the equation. Balance the equation. Na 2 CO 3 + 2HCl  2NaCl + H 2 O + CO 2 Na 2 CO 3 + 2HCl  2NaCl + H 2 O + CO 2

35. Combustion Reaction When a substance combines with oxygen, a combustion reaction results. When a substance combines with oxygen, a combustion reaction results. The combustion reaction may also be an example of an earlier type such as 2Mg + O 2  2MgO. The combustion reaction may also be an example of an earlier type such as 2Mg + O 2  2MgO. The combustion reaction may be burning of a fuel. The combustion reaction may be burning of a fuel.

36. Combustion Reaction Methane, CH 4, is natural gas. Methane, CH 4, is natural gas. When hydrocarbon compounds are burned in oxygen, the products are water and carbon dioxide. When hydrocarbon compounds are burned in oxygen, the products are water and carbon dioxide. CH 4 + O 2  CO 2 + H 2 O CH 4 + O 2  CO 2 + H 2 O CH 4 + 2O 2  CO 2 + 2H 2 O CH 4 + 2O 2  CO 2 + 2H 2 O

37. Combustion Reactions Combustion reactions involve light and heat energy released. Combustion reactions involve light and heat energy released. Natural gas, propane, gasoline, etc. are burned to produce heat energy. Natural gas, propane, gasoline, etc. are burned to produce heat energy. Most of these organic reactions produce water and carbon dioxide. Most of these organic reactions produce water and carbon dioxide.

38. Practice Classify each of the following as to type: Classify each of the following as to type: H 2 + Cl 2  2HCl H 2 + Cl 2  2HCl Combination Combination Ca + 2H 2 O  Ca(OH) 2 + H 2 Ca + 2H 2 O  Ca(OH) 2 + H 2 Single replacement Single replacement

39. Practice 2CO + O 2  2CO 2 2CO + O 2  2CO 2 Combination and combustion Combination and combustion 2KClO 3  2KCl + 3O 2 2KClO 3  2KCl + 3O 2 Decomposition Decomposition

40. Practice FeS + 2HCl  FeCl 2 + H 2 S FeS + 2HCl  FeCl 2 + H 2 S Double replacement Double replacement Zn + HCl  ? Zn + HCl  ? Single replacement Single replacement Zn + 2HCl  ZnCl 2 + H 2 Zn + 2HCl  ZnCl 2 + H 2

42. How molecules are symbolized Cl 2 2Cl 2Cl 2 Molecules may also have brackets to indicate numbers of atoms. E.g. Ca(OH) 2 O H O H Ca Notice that the OH is a group The 2 refers to both H and O How many of each atom are in the following? a) NaOH b) Ca(OH) 2 c) 3Ca(OH) 2 Na = 1, O = 1, H = 1 Ca = 1, O = 2, H = 2 Ca = 3, O = 6, H = 6

43. Balancing equations: MgO The law of conservation of mass states that matter can neither be created or destroyed The law of conservation of mass states that matter can neither be created or destroyed Thus, atoms are neither created or destroyed, only rearranged in a chemical reaction Thus, atoms are neither created or destroyed, only rearranged in a chemical reaction Thus, the number of a particular atom is the same on both sides of a chemical equation Thus, the number of a particular atom is the same on both sides of a chemical equation Example: Magnesium + Oxygen (from lab) Example: Magnesium + Oxygen (from lab) Mg + O 2  MgO Mg + O 2  MgO O MgO +  O However, this is not balanced Left: Mg = 1, O = 2 Right: Mg = 1, O = 1

44. Balance equations by “inspection” Hints: start with elements that occur in one compound on each side. Treat polyatomic ions that repeat as if they were a single entity. 5 233.5 2746 222 263 C 2 H 6 + O 2  CO 2 + H 2 O a)P 4 + O 2  P 4 O 10 b)Li + H 2 O  H 2 + LiOH c)Bi(NO 3 ) 3 + K 2 S  Bi 2 S 3 + KNO 3 d)C 2 H 6 + O 2  CO 2 + H 2 O From Mg + O 2  MgO 2Mg + O 2  2MgO is correct Mg + ½O 2  MgO is incorrect Mg 2 + O 2  2MgO is incorrect 4Mg + 2O 2  4MgO is incorrect

45. a)Mg + 2HCl  MgCl 2 + H 2 b)3Ca + N 2  Ca 3 N 2 c)NH 4 NO 3  N 2 O + 2H 2 O d)2BiCl 3 + 3H 2 S  Bi 2 S 3 + 6HCl e)2C 4 H 10 + 13O 2  8CO 2 + 10H 2 O f)6O 2 + C 6 H 12 O 6  6CO 2 + 6H 2 O g)3NO 2 + H 2 O  2HNO 3 + NO h)Cr 2 (SO 4 ) 3 + 6NaOH  2Cr(OH) 3 + 3Na 2 SO 4 i)Al 4 C 3 + 12H 2 O  3CH 4 + 4Al(OH) 3 Balance these skeleton equations:

46. The Mole Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second?

47. Background: atomic masses Look at the “atomic masses” on the periodic table. What do these represent? Look at the “atomic masses” on the periodic table. What do these represent? E.g. the atomic mass of C is 12 (atomic # is 6) E.g. the atomic mass of C is 12 (atomic # is 6) We know there are 6 protons and 6 neutrons We know there are 6 protons and 6 neutrons Protons and neutrons have roughly the same mass. So, C weighs 12 u (atomic mass units). Protons and neutrons have roughly the same mass. So, C weighs 12 u (atomic mass units). What is the actual mass of a C atom? What is the actual mass of a C atom? Answer: approx. 2 x 10 -23 grams (protons and neutrons each weigh about 1.7 x10 -24 grams) Answer: approx. 2 x 10 -23 grams (protons and neutrons each weigh about 1.7 x10 -24 grams) Two problems 1. Atomic masses do not convert easily to grams 2. They can’t be weighed (they are too small)

48. The Mole With these problems, why use atomic mass at all? 1. Masses give information about # of p +, n 0, e – 2. It is useful to know relative mass E.g. Q - What ratio is needed to make H 2 O? A - 2:1 by atoms, but 2:16 by mass It is useful to associate atomic mass with a mass in grams. It has been found that It is useful to associate atomic mass with a mass in grams. It has been found that 1 g H, 12 g C, or 23 g Na have 6.02 x 10 23 atoms 6.02 x 10 23 is a “mole” or “Avogadro’s number” 6.02 x 10 23 is a “mole” or “Avogadro’s number” “mol” is used in equations, “mole” is used in writing; one gram = 1 g, one mole = 1 mol. “mol” is used in equations, “mole” is used in writing; one gram = 1 g, one mole = 1 mol.

49. Mollionaire Q: how long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A: $ 6.02 x 10 23 / $1 000 000 000 = 6.02 x 10 14 payments = 6.02 x 10 14 seconds 6.02 x 10 14 seconds / 60 = 1.003 x 10 13 minutes 1.003 x 10 13 minutes / 60 = 1.672 x 10 11 hours 1.672 x 10 11 hours / 24 = 6.968 x 10 9 days 6.968 x 10 9 days / 365.25 = 1.908 x 10 7 years A: It would take 19 million years

50. Comparing sugar (C 12 H 22 O 11 ) & H 2 O No, sugar has more (45:3 ratio) Yes (6.02 x 10 23 in each) Yes. No, molecules have dif. masses No, molecules have dif. sizes. 1 mol each Yes, that’s what grams are. mass? No, they have dif. molar masses # of moles? No, they have dif. molar masses # of molecules? No # of atoms? No, they have dif. densities. volume? 1 gram eachSame

51. Molar mass The mass of one mole is called “molar mass” The mass of one mole is called “molar mass” E.g. 1 mol Li = 6.94 g Li E.g. 1 mol Li = 6.94 g Li This is expressed as 6.94 g/mol This is expressed as 6.94 g/mol What are the following molar masses? What are the following molar masses? SSO 2 Cu 3 (BO 3 ) 2 32.06 g/mol 64.06 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl 2 (NH 4 ) 2 CO 3 O 2 Pb 3 (PO 4 ) 2 C 6 H 12 O 6 Cux 3 = 63.55 x 3= 190.65 Bx 2 = 10.81 x 2= 21.62 Ox 6 = 16.00 x 6= 96.00 308.27

52. Molar mass The mass of one mole is called “molar mass” The mass of one mole is called “molar mass” E.g. 1 mol Li = 6.94 g Li E.g. 1 mol Li = 6.94 g Li This is expressed as 6.94 g/mol This is expressed as 6.94 g/mol What are the following molar masses? What are the following molar masses? SSO 2 Cu 3 (BO 3 ) 2 32.06 g/mol64.06 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl 2 (NH 4 ) 2 CO 3 O 2 Pb 3 (PO 4 ) 2 C 6 H 12 O 6 110.98 g/mol (Ca x 1, Cl x 2) 96.11 g/mol (N x 2, H x 8, C x 1, O x 3) 32.00 g/mol (O x 2) 811.54 g/mol (Pb x 3, P x 2, O x 8) 180.18 g/mol (C x 6, H x 12, O x 6)

53. Converting between grams and moles If we are given the # of grams of a compound we can determine the # of moles, & vise-versa If we are given the # of grams of a compound we can determine the # of moles, & vise-versa In order to convert from one to the other you must first calculate molar mass In order to convert from one to the other you must first calculate molar mass 0.25HCl 53.15H 2 SO 4 3.55NaCl 1.27Cu mol (n)gg/molFormula 9.136.46 0.541998.08 20758.44 0.020063.55

54. The Mole Convert 36.0 grams of carbon into atoms. Convert 30 molecules of methane into Liters of gas.

55. Simplest and molecular formulae Consider NaCl (ionic) vs. H 2 O 2 (covalent) Cl Na Cl Na Chemical formulas are either “simplest” (a.k.a. “empirical”) or “molecular”. Ionic compounds are always expressed as simplest formulas. Covalent compounds can either be molecular formulas (I.e. H 2 O 2 ) or simplest (e.g. HO) Q - Write simplest formulas for propene (C 3 H 6 ), C 2 H 2, glucose (C 6 H 12 O 6 ), octane (C 8 H 14 ) Q - Identify these as simplest formula, molecular formula, or both H 2 O, C 4 H 10, CH, NaCl H O O H H O O H H O O H

56. Answers Q - Write simplest formulas for propene (C 3 H 6 ), C 2 H 2, glucose (C 6 H 12 O 6 ), octane (C 8 H 14 ) Q - Identify these as simplest formula, molecular formula, or both H 2 O, C 4 H 10, CH, NaCl A - CH 2 A - H 2 O is both simplest and molecular C 4 H 10 is molecular (C 2 H 5 would be simplest) CH is simplest (not molecular since CH can’t form a molecule - recall Lewis diagrams) NaCl is simplest (it’s ionic, thus it doesn’t form molecules; it has no molecular formula) CHCH 2 OC4H7C4H7

57. Calculating percentage mass If you can work out M r then this bit is easy… Calculate the percentage mass of magnesium in magnesium oxide, MgO: A r for magnesium = 24Ar for oxygen = 16 M r for magnesium oxide = 24 + 16 = 40 Therefore percentage mass = 24/40 x 100% = 60% Percentage mass (%) = Mass of element A r Relative formula mass M r x100% Calculate the percentage mass of the following: 1)Hydrogen in hydrochloric acid, HCl 2)Potassium in potassium chloride, KCl 3)Calcium in calcium chloride, CaCl 2 4)Oxygen in water, H 2 O

58. 25/08/2015 Empirical formulae Empirical formulae is simply a way of showing how many atoms are in a molecule (like a chemical formula). For example, CaO, CaCO 3, H 2 0 and KMnO 4 are all empirical formulae. Here’s how to work them out: A classic exam question: Find the simplest formula of 2.24g of iron reacting with 0.96g of oxygen. Step 1: Divide both masses by the relative atomic mass: For iron 2.24/56 = 0.04For oxygen 0.96/16 = 0.06 Step 2: Write this as a ratio and simplify: 0.04:0.06 is equivalent to 2:3 Step 3: Write the formula: 2 iron atoms for 3 oxygen atoms means the formula is Fe 2 O 3

59. 25/08/2015 Example questions 1)Find the empirical formula of magnesium oxide which contains 48g of magnesium and 32g of oxygen. 2)Find the empirical formula of a compound that contains 42g of nitrogen and 9g of hydrogen. 3)Find the empirical formula of a compound containing 20g of calcium, 6g of carbon and 24g of oxygen.

60. Stoichiometry

61. Molar Mass of Compounds The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl 2 Ex. Molar mass of CaCl 2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl 2 Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl 2 20 Ca 40.08a 17 Cl 35.45 Cl

62. Mole Roadmap Atoms or Molecules Moles Mass (grams) 6.02 X 10 23 molar mass from periodic table Liters of Gas 22.4

63. Practice Calculate the Molar Mass of calcium phosphate Calculate the Molar Mass of calcium phosphate Formula = Formula = Masses elements: Masses elements: Molar Mass = Molar Mass = Ca 3 (PO 4 ) 2

64. molar mass Avogadro’s number Grams Moles particles molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

65. Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

66. Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

67. Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Be sure you have a balanced reaction before you start! Example: 2 Na + Cl 2  2 NaCl Example: 2 Na + Cl 2  2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles? What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

68. Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O 2 H 2 + O 2  2 H 2 O How many moles of reactants are needed? How many moles of reactants are needed? What if we wanted 4 moles of water? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

69. Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2  2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

70. Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

71. Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl 5.00 moles Na 1 mol Cl 2 70.90g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

72. Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

73. Mass-Mole We can also start with mass and convert to moles of product or another reactant We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 10.0 g H 2 O 1 mol H 2 O 2 mol C 2 H 6 18.0 g H 2 O 6 mol H 2 0 = 0.185 mol C 2 H 6

74. Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

75. Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

76. Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N 2 + 3 H 2  2 NH 3 N 2 + 3 H 2  2 NH 3 2.00g N 2 1 mol N 2 2 mol NH 3 17.06g NH 3 28.02g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

77. Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

78. Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

79. Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant. The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

80. Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

81. Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Start with Al: Now Cl 2 : Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

82. LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete. We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

83. Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

84. Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem? Can we find the amount of excess potassium in the previous problem?

85. Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

86. Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Convert ALL of the reactants to the SAME product (pick any product you choose.) 3. The lowest answer is the correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!