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Chemistry 100(02) Fall 2011

Chemistry 100(02) Fall 2011. Instructor: Dr. Upali Siriwardane e-mail : upali@chem.latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F 9:00 - 10:00 a.m.   Test Dates : March 25, April 26, and May 18; Comprehensive Fina

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Chemistry 100(02) Fall 2011

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  1. Chemistry 100(02) Fall 2011 Instructor: Dr. UpaliSiriwardane e-mail: upali@chem.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F9:00 - 10:00 a.m.   Test Dates: March 25, April 26, and May 18; Comprehensive Fina Exam: 9:30-10:45 am, CTH 328. October 3, 2011 (Test 1): Chapter 1 & 2 October 26, 2011 (Test 3): Chapter 3 & 4 November 16, 2011 (Chapter 5 & 6) November 17, 2011 (Make-up test) comprehensive: Chapters 1-6 9:30-10:45:15 AM, CTH 328

  2. Chapter 4. Quantities of Reactants and Products

  3. Types of Reactions • Synthesis reactions or combination reactions • Decomposition reactions • Displacement reactions • Single • Double(Exchange reactions) • Combustion Reactions • Formation Reactions

  4. Types of Chemical Reactions

  5. Reaction of H2 and I2

  6. Synthesis or Combination Reactions Formation of a compound from simpler compounds or elements. Decomposition reactions Compound breaks up to from simpler compounds or elements. Displacement reactions Single displacement: In a compound an element is replaced by another element. Exchange reactions Double displacement: In a compound group or ion is replaced by another group or ion in another compound. .

  7. Formation Reactions Formation of a compound from elements at standard state. Combustion Reactions Compound reacts with oxygen to produce oxides: water and carbon dioxide for organic compounds Acid/Base (Neutralization)Reactions An acid and a base react to form water and salt ( most ionic compounds are salts) Precipitation Reactions when two aqueous salt solutions are mixed an insoluble salt is formed when two aqueous salt solutions are mixed.

  8. Combination Reaction

  9. Decomposition Reactions

  10. Dynamite

  11. Electrolysis Decomposition caused by an electric current Anode • electrode where oxidation occurs Cathode • electrode where reduction occurs

  12. Electrolysis

  13. Displacement Reactions

  14. Exchange Reactions They are double displacement or exchange reactions of ionic compounds where an insoluble salt is formed (Precipitation Reactions) when two aqueous salt solutions are mixed. Ba(NO3)2(aq) +Na2SO4(aq)= BaSO4(s) + 2 NaNO3(aq)

  15. Chemical Equations P4O10 (s) + 6H2O (l) = 4 H3PO4(l) reactants enter into a reaction. products are formed by the reaction. Paranthesesrepresent physical state Stoichiometric Coefficients are numbers in front of chemical formula formula gives the amounts (moles) of each substance used and each substance produced. • Equation Must be balanced!

  16. Chemical Reactions Could be described in words Chemical equation: • Reactants? • Products? • Reaction conditions? =, ---> , <==> or ? • Stoichiometric coefficients? Number in front of substances representing moles, atoms, molecules

  17. Balanced Chemical Equation • Representation of a chemical reaction which uses coefficients (prefix numbers known as stoichiometric coefficients) to represent the relative amounts of reactants and products

  18. Writing and BalancingChemical Equations • Write a word equation. • Convert word equation into formula equation. • Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type of atom on the reactant and product sides of the equation.

  19. Balancing Chemical Equations 1. Check for Diatomic Molecules - H2 - N2- O2 - F2 - Cl2 - Br2 - I2 If these elements appear By Themselves in an equation, they Must be written with a subscript of 2 2. Balance Metals 3. Balance Nonmetals 4. Balance Oxygen 5. Balance Hydrogen 6.Recount All Atoms 7. If EVERY coefficient will reduce, rewrite in the simplest whole-number ratio.

  20. Example Hydrogen gas reacts with oxygen gas to produce water. Step 1. hydrogen + oxygen  water Step 2. H2 + O2 H2O Step 3. 2 H2 + O2 2 H2O

  21. Example Iron(III) oxide reacts with carbon monoxide to produce the iron oxide (Fe3O4) and carbon dioxide. iron(III) oxide + carbon monoxide  Fe3O4 + carbon dioxide Fe2O3 + CO  Fe3O4 + CO2 3 Fe2O3 + CO  2 Fe3O4 + CO2

  22. Stoichiometry stoi·chi·om·e·trynoun Calculations of the quantitative relationships between reactants and products in a chemical reaction.

  23. Stoichiometric Relationships

  24. Stoichiometric Converstion Factors The large numbers in front of chemical formulas. Coefficients represent the number of molecules of the substance in the reaction. They provide the convestion factor to conver moles of reactants to products or vice versa. 4 NH3(g) + 5 O2(g) ------> 4 NO(g) + 6 H2O(g)   4 mol NH3 = 5 mol O2 5 mol O2 = 6 mol H2O 4 mol NH3 = 4 mol NO; 1mole NH3 = 1 mole NO 1 mol NH3 = 1 mol NO

  25. Examples • Calculate the following using the chemical equation given below: • 4 NH3(g) + 5 O2(g) ----> 4 NO(g) + 6 H2O(g) • a) moles of NO(g) from 2 moles of NH3(g) and excess O2(g). • b) moles of H2O(g) from 3 moles of O2(g) and excess NH3(g).

  26. The Mole and Chemical Reactions:The Macro-Nano Connection 2 H2 + O2 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules 2 moles H2 molecules 1 mole O2 molecules 2 moles H2O molecules 2 kmoles H2 molecules 1 kmole O2 molecules 2 kmoles H2O molecules 2 mmoles H2 molecules 1 mmole O2 molecules 2 mmoles H2O molecules 4 g H2 32 g O2 36 g H2O

  27. Examples • How many moles of H2O will be produced by 0.80 mole of O2 with excess H2 according to the equation? • 2H2(g) + O2(g) = 2 H2O(l)

  28. Stoichiometric Reactions • Reactions where mole ratio of the products and reactants are the same as mole ratio from the stoichiometric coefficients in the balanced chemical equation. All reactants will be completely converted into products.

  29. Limiting Reactant

  30. Limiting Reactant

  31. Analogy in Recipe: Making Cheese Sandwiches You were given 20 slices bread, 5 slices of cheese, 4 slices of ham If you want to make sandwiches containing two slices bread and one slice of cheese and one slice of ham, How many sandwiches you could make? What is the limiting ingredient?

  32. What is the limiting reagent? • Limiting reagent is the reactant, which is used up first. • To find the limiting reactant you have to compare the amounts of reactants in moles.

  33. Ways to find Limiting Reactant There two ways to find the limiting ingredient: • 1) Comparing mole ratio of the chemical equation to mole ratio calculated from the grams of reactants 2) Calculate moles of product and whichever reactant produces lowest moles of product is the limiting reactant

  34. Examples A 300.0 g sample of phosphorus ( M.W. 123.88 g/mol) burns in 500.0 g of oxygen (M.W. 32.00 g/mol) according to following equation: P4(s) + 5O2(g) = P4O10(s) What is the limiting reagent?

  35. 2.42 mol P4 1 mol P4O10 = 2.42 mole P4O10 1 mol P4 15.63 mole O2 1 mol P4O10 = 3.13 mole P4O10 5 mole O2 First Method moles of            P4               and               O2 300                 500              ---------- = 2.42 mol P4;          ----------= 15.63 mol O2            123.88                                      32.00 P4               and      O2 theoretical mole ratio   1                    :           5 actually mole ratio       1        :     6.46 P4 is present in lower amount than required, it is used up first. Therefore, P4 is the limiting reagent. Second Method

  36. EXAMPLE How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2? 2 H2 + O2  2 H2O (3.3 mol O2) (2 mol H2O) #mol H2O = = 6.6 mol H2O (1 mol O2) Mole ratio from balanced chemical equation

  37. Theoretical yield • Theoretical yield is the amount (grams) of products formed according to chemical equation. • Use the limiting reagent to calculate the moles of the product and then convert moles to grams.

  38. Actual Yield • Actual yieldis the grams of the product obtained by an experiment. • Actual yield should be less than the theoretical yield if the experiment was carried out meticulously. If the products are contaminated with impurities or the formula of product was wrong the actual yield could be higher.

  39. % Yield • actual yield • % yield = ------------------------ x 100 • theoretical yield • If the products are contaminated with impurities or the formula of product was wrong the % yield could be higher than 100%.

  40. Question • Sulfur trioxide, SO3 , is made from the oxidation of SO2 and the reaction is represented by the equation • 2SO2 + O2 -----> 2SO3 • A 16.0-g sample of SO2 gives 18.0 g of SO3. The percent yield of SO3 is

  41. Examples • A 300.0 g sample of phosphorus ( M.W. 123.88 g/mol) burns in 500.0 g of oxygen (M.W. 32.00 g/mol) according to following equation: • P4(s) + 5O2(g) = P4O10(s) • a) What is the limiting reagent? • b) How many moles of P4O10 are produced theoretically? • c) If 612 g of P4O10 is actually produced in this reaction, calculate the percent yield.

  42. Steps in Stoichiometric Calculations • Check whether chemical equation is balanced • get the moles from grams of materials • find the limiting reactant • calculate moles of products from the limiting reactant • convert moles of the products to grams • find the actual yield of the reaction • calculate % yield of the reaction

  43. Question • How much hydrogen gas is produced when 1 kg of sodium reacts with water?

  44. Question • 2Al(s) + 6HCl(aq)--> 2AlCl3(aq) +3H2(g) According to the equation above, how many grams of aluminum are needed to react with 0.582 mol of hydrochloric acid?

  45. 5.23 g Al

  46. EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. weld Photo by Mike Condren

  47. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron. The mass of iron in 1 inch of this rail is: #g/in = (132 #/yard) (1 yard/36 in) (454 g/#) = 1.67  103 g/in The mass of iron in a weld adding 10% mass: #g = (1.67  103 g) (0.10) = 167 g

  48. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g = (1.67  103 g) (0.10) = 167 g Balanced chemical equation: Fe2O3 + 2 Al  2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? (1 mol Fe) (1 mol Fe2O3) (159.7 g Fe2O3) #g Fe2O3 = (167 g Fe) (55.85 g Fe) (2 mol Fe) (1 mol Fe2O3) = 238 g Fe2O3

  49. EXAMPLEHow much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron. The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe Balanced chemical equation: Fe2O3 + 2 Al  2 Fe + Al2O3 What mass of Fe2O3 is required for the thermite process? #g Fe2O3 = 238 g Fe2O3 What mass of Al is required for the thermite process? (1 mol Fe) (2 mol Al) (26.9815 g Al) #g Al = (167 g Fe)  (55.85 g Fe) (2 mol Fe) (1 mol Al) = 80.6 g Al

  50. EXAMPLEWhat is the number of moles of CaSO4 (S) that can be produced by allowing 1.0 mol SO2, 2.0 mol CaCO3, and 3.0 mol O2 to react? 2SO2(g) + 2CaCO3(s) + O2(g) 2CaSO4(S) + 2CO2(g) balanced equation relates: 2SO2(g) 2CaCO3(s)O2(g) have only: 1SO2(g)2CaCO3(s) 3O2(g) not enough SO2 to use all of the CaCO3 or the O2 not enough CaCO3 to use of the O2 SO2 is the limiting reactant

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