1. Essential Questions: What are the parts of a chemical equation? What is a catalyst ? Word Equations: Methane (g) + Oxygen (g) [Yield to produce]  Carbon dioxide (g) + Water (l) Reactants  Product Phases: Gas (g) Liquid (1) Solid (s) Other symbols are used in equations.

5. Catalyst: Substance that speeds up a rxn. Written above the  Not a reactant or a product Ex. H 2 0 2(g)  H 2 0 (1) + 0 2(g)

6. Essential Question: How do you balance a chemical equation? Balancing Equations: 1. D etermine correct formulas for reactants/products. 2. B alance each compound using oxidation numbers. 3. B alance each compound using coefficients. 4. Make sure coefficients are in the lowest ratio.

7. Exs: A. Solid zinc reacts with a water solution of hydrogen sulfate to produce hydrogen gas, a water solution of zinc sulfate, and heat. Zinc (s y + hydrogen sulfate (g)  hydrogen (g) + zinc sulfate (g) Zn (s) + H 2 SO 4 (g)  H 2 (g) + Zn+1SO 4 therefore is balanced.

10. B. Calcium hydroxide + Sulfuric acid  Calcium sulfate + Water. Balance each compound using coefficients. Balance each compound using coefficients. Calcium hydroxide + Sulfuric acid  Calcium sulfate + water Ca (OH)2 + H 2 S0 4  Ca (SO 4 ) + H 2 O

11. Using Rule #4: Inspect the three equations below. Are they balanced? 2 CH 4 + 4O 2  2CO 2 + 4H 2 O 6 CH 4 + 12O 2  6CO 2 + 12H 2 O 1/2 CH 4 + O 2  1/2 CO 2 + H 2 O The first two are balanced but are they in the smallest whole-numbers ratio? (Rule 4) *NO, it should be CH 4 + 2O 2  CO 2 + 2H 2 O

12. Essential Questions: What are the four types of reactions? What is general formula for each type? What is the relationship of Table J to writing equations? Types of Reactions:  Combination (synthesis)  Decomposition  Single replacement  Double replacement

13. Synthesis: Two or more reactants combine to form a single product General Formula: A +B  AB H +Cl  HCl Na + Cl  NaCl

14. Decomposition: Reverse of synthesis A single compound is broken down (decomposed into two or more simpler substances) General Formula: AB  A + B Exs: 2 H 2 O 2(1 )  2 H 2 O( 1 ) + O 2 2 NH 3(g )  N 2(g ) + 3 H 2(g )

15. Single Replacement

16. Single replacement: Rxn, in which one element replaces another element in a compound. General formula: A + BCAC + B Exs: Zn (s) + CuS04    ZnS04(g) + Cu(s) Mg + 2 HClH2 + MgCl2 ** Use table J to determine if rxn will occur.

18. Table J ** Use table J to determine if rxn will occur. A reactive metal will replace any other metal below it. Exs: a. Zn + Cu(NO3) 2  Cu + ZnNO3) 2 b. Cu + Zn(NO3) 2  no rxn c. F 2 + NaCl  Cl 2 + NaF d. C 12 + NaF  no rxn

20. Double Replacement: Chemical change involving an exchange of positive ions between two compounds. (Between two ionic compounds) Generally occur in an aqueous solution. Often produces a gas, precipitate or a molecular compound. (H 2 O) General formula: AB + CD  AD + CB

21. Exs. Ag(NO 3 ) (aq) + NaCl (aq)  NaNO 3(aq) + AgCl (s) Ba(N0 3 ) 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2 Na(NO 3 ) (aq)

22. Essential Question: What are the three types of situation in which a double replacement reaction will occur? Three situations in which a double replacement will occur: 1. One product is a solid (precipitate) Table F Table F AgNO 3(aq) + NaCl (aq)  AgCl (s) + NaNO 3(aq) NaCl is a solid (insoluble) NaCl is a solid (insoluble) NaNO 3 is a soluble therefore, rxn occurs. NaNO 3 is a soluble therefore, rxn occurs.

23. 2. A rxn in which a gas is produced NaS (g) + 2 HCl (aq)  H 2 S (g) + 2 NaCl (aq) 3. A rxn in which a molecular substance is formed(H 2 O) NaOH (aq) + HCl (aq)  H2O (l) + NaOH (aq)

24. Essential Questions: What is a mole? What is Avogadro’s number? Mole: Is a specific number of particles Is a specific number of particles 1 mole of any substance = 1 mole of any substance = 6.02 x10 23 particles (Avogadro’s number) 6.02 x10 23 particles (Avogadro’s number)

25. **A mole of any substance contains Avogadro's number of particles (6.02 x l0 23 ) ** 1 mole of Ag = 6.02 x10 23 atoms of Ag 1 mole of Ag = 6.02 x10 23 atoms of Ag 1 mole of CO 2 = 6.02 x10 23 molecules of CO 2 1 mole of CO 2 = 6.02 x10 23 molecules of CO 2 1 mole of Ca ions = 6.02 x10 23 ions 1 mole of Ca ions = 6.02 x10 23 ions 1 mole of N 2 = 6.02 xl0 23 molecules of N 2 1 mole of N 2 = 6.02 xl0 23 molecules of N 2

26. Types of Mole Problems: Atoms -Moles Atoms -Moles Moles-Atoms Moles-Atoms Grams-Moles Grams-Moles Moles-Grams Moles-Grams Mole-Mole (equations) Mole-Mole (equations) Mass—Mass (equations) Mass—Mass (equations) Mass-Volume (equations) Mass-Volume (equations) Volume-Volume (equations) Volume-Volume (equations) # of Particles (equations) # of Particles (equations)

27. 1. Atoms-Mole: Ex. Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium are in 1.25x10 23 atoms of magnesium? a. List knows/unknowns: Known: Unknown: 1 mole Mg = 6.02x10 23 atoms Mg # of moles of Mg # of atoms Mg = 1.25 x l0 23 atoms atoms Mg  moles Mg b. Solve problem: 1.25 xl 0 23 atoms Mg x 1 mole Mg 6.02xlO 23 atomsMg 6.02xlO 23 atomsMg =1.25xl0 23 x lmoleMg 6.02x10 23 6.02x10 23 =2.08 x lO -1 mole Mg =.208 mole Mg

28. B. How many moles are 2.08 x 10 23 atoms of silicon? a. List known/unknown: Known: Unknown: 1 mole Si =6.02xl0 23 atoms moles of Si 2.80x10 23 = # of atoms Atoms Si  mole Si b. Solve 2.80xl0 23 atoms Si x 1 mole Si 6.02x10 23 atoms Si 6.02x10 23 atoms Si =4.65 mole Si

29. 2. Moles  Atoms : How many atoms are in a mole of CO 2 ? a. List known/unknown Known : Unknown: 1 mole CO 2 =6.02xl0 23 molecules #of atoms of CO 2 CO 2 = 1 atom C 2 atoms O 2 atoms O Therefore 3 atoms in CO 2 b. Solve: 1 mole CO 2 x 3(6.02x10 23 atoms CO 2 ) 1 Mole CO 2 1 Mole CO 2 =18.06xlO 23 atoms

30. B. Propane is a gas used for cooking and heat. How many atoms are in 2.12 moles of propane (C3H8)? a. List Known/Unknown Known: Unknown: 2.12 moles of C3H8 a atoms of C3H8 1 mole of C3H8 = 6.02x1023 molecules C3H8 (3C and 8H) 2.12moles C3H8 x 11(6.02 x 1023) 1mole C3H8 =1.4039 x l025 atoms C3H8 =1.40 x l025atoms C3H8

32. 3. Grams  Moles: Formula Mass- the sum of masses of all the atoms in a given formula Find the formula mass of O2 A. O2 = 2 atoms x 16amu=32 amu(formula mass) *1 mole of any substance = formula mass therefore, 1 mole of O2= 32g/mole B. Find the formula mass of CuSO4 Cu=l atom x 63.55amu=63.55amu S= 1 atom x 32.07amu= 32.07 amu O= 4 atom x 16amu = 64.00 amu 159.62 amu

33. Gram Formula Mass- The formula mass expressed in grams. ** 1 mole of any substance= Formula mass

34. Ex. How many moles are equivalent to 4.75g NaOH? a. Known Unknown 4.75g NaOH moles of NaOH b. Calculate GFM (gram formula mass) of NaOH Na=23amu O=16 amu H=l amu 40amu= 40g/mole NaOH C. Solve: 4.75g NaOH x 1 mole NaOH 40gNaOH 40gNaOH =0.119 moles of NaOH

35. 4. Moles  Grams: Ex. How many grams are present in 40.5 moles of H 2 SO 4 ? A. Known Unknown 40.5 moles H 2 SO 4 # grams ofH 2 SO 4 B Calculate GFM: H= 2amu x 2 =4 S= 32amu O= 16amu x 4 = 64 =98 amu= 98 g/mole Solve: 40.5 mole H 2 SO 4 x 98gH 2 SO 4 1 mole H 2 SO 4 1 mole H 2 SO 4 =3970g H 2 SO 4

36. 5. Mole  Mole a. In the equation N2 + 3H22NH3, how many moles of N2 are needed to produce 5.0 moles of NH3? Moles NH3 moles of N2 5.0 mole NH3 x 1 mole N2 2 Moles NH3 = 2.5 mole N2 b. In the equation 2NO + O2  2 NO2, How many moles of O2 are needed to produce 3.5 moles of NO2? Moles of NO2 Moles of O2 3.5 moles of NO2 x 1 mole O2 2 moles NO2 = 1.75 mole O2

37. Mole  Mass: a. In the equation: 4Al +3O 2  2 A1 2 O 3, how many grams of aluminum will combine with 1.50 moles of oxygen? Mole  Grams 1.50 mole O 2 x 4 mole Al x 27g Al 3 mole O 2 1 mole Al =54 g Al

38. b. Based on the previous equation, how many moles of O 2 are needed to produce 51.0 g of Al 2 O 3 ? Grams  Moles 51.O g A1 2 O 3 x 1 mole Al 2 O 3 x 3 mole O 2 102g Al 2 O 3 2 moleAl 2 O 3 =.75 mole O 2

39. 8. Mass  Mass: a. In the equation, CH4 + 2 O2 2 H2O, how many grams of CO2 are formed when 8.0g of CH4 reacts with an excess of CO2? 8.0g CH4 x 1 mole CH4 x 1 mole CO2 x 44 g CO 16gCH4 1 mole CH4 1 1 mole CO2 = 22g C02

40. b. In the equation 2 H 2 O 2  2H 2 O+ O 2, how many grams of 0 2 will be formed from the decomposition of 17.0g of H 2 O 2 ? 17.0g H 2 O 2 x 1 mole H 2 O 2 x 1 mole O 2 x 32O 2 34g H 2 O 2 2 mole H 2 O 2 1 mole O 2 =8.0g O 2

41. 9. Mass  Volume: **1 mole of any substance = 22.4L(STP) a. In the equation: 2 CO (g) + O 2 (g)  2 CO 2 (g), how many liters of CO2, at STP are produced by the reaction of 64.0 g of O 2 (g) 64g O 2 x 1 mole O 2 x 2 mole CO 2 _ x 22.4L CO 2 32g O 2 1 mole O 2 1 mole CO 2 32g O 2 1 mole O 2 1 mole CO 2 = 89.6 L CO 2

42. b. In the equation: 2 CO (g ) + O 2 (g)  2 CO 2 (g) How many liters of O2 (g) at STP are able to react with 28.0g of CO(g)? 28.0gCO x 1 mole CO 2 x 1 mole O 2 x 22.4L O 2 28gCO 2 mole CO 1 moleO 2 28gCO 2 mole CO 1 moleO 2 = 11.2 L O 2

43. 10. Volume  Volume: In the equation 4 NH 3 (g) + 5 O 2 (g)  4NO(g) + 6 H 2 0(l). How many liters of NH 3 (g) at STP are needed to react with 200 L of O 2 (g) at STP? 2.0 L O 2 x 1 mole O 2 x 4 mole NH 3 x 22.4 L NH 3 22.4 L O 2 5 mole O 2 1 mole NH 3 =160L NH 3

44. b. In the equation: 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(l) How many liters of NH 3 (g) at STP are needed to produce 1000 L of NO(g) at STP? 1OOL NO x l mole NO x 4 mole NH 3 x 22.4LNH 3 22.4L NO 4 mole NO 1 mole NH 3 22.4L NO 4 mole NO 1 mole NH 3 = 1,OOOL NH 3

45. Number of particles : a. In the equation: 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O, how many molecules of C 2 H 6 are needed to produce 27.Og of H 2 O? 27.0g H2O x 1 mole H 2 O x 2 mole C 2 H 6 x 18gH 2 O 6moleH 2 O 6.02xl0 23 mole 1 mole C 2 H 6 = 3.01xl0 23 molecules of C 2 H 6

46. b. in the equation 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O, how many molecules of CO 2 are produced when 3.5 moles of O 2 react? 3.5 mole O 2 x 4 mole CO 2 x 6.02x1023 molecules of CO 2 7 mole O 2 1 mole CO 2 =12.04x 10 23 molecules of C0 2 =1.204x 10 24 molecules of C0 2 Limiting Reactant (regent)- a reactant that limits the amount of production of product formed in a reaction.