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### Chapter08_Fall11

Course: 160 161, Fall 2009
School: Rutgers
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Word Count: 2634

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Ratios 10/27/2011 8 Stoichiometry: of Combination 8.1 Chemical Equations 8.2 Combustion Analysis 8.3 Calculations with Balanced Chemical Equations 8.4 Limiting Reactants 8.5 Periodic Trends in Reactivity of the Main Group Elements Chemical Equations A chemical equation uses chemical symbols to denote what occurs in a chemical reaction. Ammonia and hydrogen chloride react to produce ammonium chloride....

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Ratios 10/27/2011 8 Stoichiometry: of Combination 8.1 Chemical Equations 8.2 Combustion Analysis 8.3 Calculations with Balanced Chemical Equations 8.4 Limiting Reactants 8.5 Periodic Trends in Reactivity of the Main Group Elements Chemical Equations A chemical equation uses chemical symbols to denote what occurs in a chemical reaction. Ammonia and hydrogen chloride react to produce ammonium chloride. Balancing Chemical Equations Chemical equations must be balanced so that the law of conservation of mass is obeyed. NH3 + HCl NH4Cl reactants products Chemical formulas indicate elements or compounds that are reacting or being formed Labels in parentheses indicate the physical state (s) (l) (g) (aq) Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas. solid liquid gas aqueous (dissolved in water) NH3 (g) + HCl (g) NH4Cl (s) SO3 (g) + H2O (l) H2SO4 (aq) Balancing Chemical Equations 1. Change the coefficients of compounds before changing the coefficients of elements. 2. Treat polyatomic ions that appear on both sides of the equation as units. 3. Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient. Example Write the balanced chemical equation that represents the combustion of propane. Step 1: Write the unbalanced equation: C3H8(g) + O2(g) CO2(g) + H2O(l) Step 2: Leaving O2 until the end, balance each of the atoms: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Step 3: Double check to make sure there are equal numbers of each type on atom on both sides of the equation. 1 10/27/2011 Examples O2(g) + 2 H(g) H2H2O(g) O (g) + H2 2(g) 2 O(g) SiO2(s) + C(s) iCiC(s) + O(g) (g) 3 C(s) S S (s) + C 2 CO C8H18(l) + O2(g) CO2(g) + H2O(g) C8H18+ O2 CO2 + H2O C8H18+ O2 8 CO2 + H2O C8H18+ O2 8 CO2 + 9 H2O C8H18+ 25/2 O2 8 CO2 + 9 H2O 2 C8H18+ 25 O2 16 CO2 + 18 H2O Examples O2(g) + 2 H2(g) 2 H2O(g) SiO2(s) + 3 C(s) SiC(s) + 2 CO(g) 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 2 (s) + 3 H2 4(aq) A Al2(SO (aq) + 3 (g) AlAl(s) +H2SOSO4(aq) l2(SO4)34)3(aq) + H2H2(g) 2 Al + 6 HCl A 2 AlCl3( + + (g) Al(s)(s) +HCl(aq()aq) lCl3(aq) aq) H23 H2(g) H |C |O 18 8 2 21 3 18 8 2 2 8 17 18 8 2 18 8 25 18 8 25 18 8 25 36 16 50 36 16 50 Patterns of Chemical Reactivity Three of the most commonly encountered reaction types are combination, decomposition, and combustion. Combination two or more reactants combine to form a single product NH3(g) + HCl(g) NH4Cl(s) Decomposition two or more products form from a single reactant CaCO3 CaO + CO2 Combustion a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and water. CH2O(l) + O2(g) CO2(g) + H2O(l) Combustion Analysis Combustion Analysis The experimental determination of an empirical formula is carried out by combustion analysis. In the combustion of 18.8 g of glucose, 27.6 g of CO2 and 11.3 g of H2O are produced. Calculate the empirical formula of glucose. It is possible to determine the mass of carbon and hydrogen in the original sample as follows: mass of C = 27.6 CO2 mass of H = 11.3 H2 O 1 CO2 1 C 12.01 C = 7.53 C 44.01 CO2 1 CO2 1 C 1 H2 O 2 H 1.008 H = 1.26 H 18.01 H2 O 1 H2 O 1 H The remaining mass is oxygen: 18.8 glucose 7.53 C + 1.26 H = 10.0 O 2 10/27/2011 Combustion Analysis Combustion Analysis It is now possible to calculate the empirical formula. The molar mass of glucose is 180 g/mol. What is its molecular formula? Step 1: Determine the number of moles of each element. C = 7.53 C 1 C = 0.627 C 12.01 C 1 H = 1.25 H 1.008 H Empirical formula of glucose: CH2O 1 O = 0.626 O 16.00 H Empirical formula mass = 12.01 H = 1.26 H O = 7.53 O Step 2: Write the empirical formula and divide by the smallest subscript to find the whole number ratio. C0.627H1.25O0.626 simplifies to CH2O 0.626 To determine the molecular formula, divide the molar mass by the empirical formula mass. + 2 1.008 + 16.00 30 Molar mass: 180 g/mol Molar mass/Empirical mass: 180/30 = 6 Molecular formula = [CH2O] x 6 = C6H12O6 0.626 0.626 Calculations with Balanced Chemical Equations Calculations with Balanced Chemical Equations Making Pancakes: Relationships Making Pancakes: Relationships Similar to a recipe 5 Pancakes Numerical relationship 1 cup of flour 2 eggs tsp baking powder 2 Eggs 5 Pancakes Two eggs are required to produce 5 pancakes. 8 eggs are required to produce 20 pancakes 1 cup Flour 5 Pancakes tsp Baking Powder 5 Pancakes 1 cup Flour tsp Baking Powder If any of your ingredients runs out in proportion to the rest you cant make any more pancakes Calculations with Balanced Chemical Equations A balanced chemical equation is like a recipe: atoms must combine in precise mole ratios. The recipe (mole ratios) are given by the coefficients of the balanced chemical equation. The ratio of coefficients is called stoichiometry Calculations with Balanced Chemical Equations If we have 6 moles of O2 and more than enough CO, how much CO2 can we make? 2 CO (g) + O2(g) 2 CO2 (g) Given: 6 mol O2 Find: mol CO2 Conversion Factors: 1 mol O2 = 2 mol CO2 6 O2 2 moles of CO combine with 1 mole of O2 to produce 2 moles of CO2. 2 CO2 = 12 CO2 O2 2 moles of CO is stoichiometrically equivalent to 2 moles of CO2. 3 10/27/2011 Calculations with Balanced Chemical Equations Calculations with Balanced Chemical Equations Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of CO2 produced. Mass of Reactant to Mass of Product Mass A Moles A Moles B Mass B Moles A Mass A Equivalence from balanced chemical equation CO2 produced = 3.82 CO 2 CO2 = 3.82 CO2 2 CO Making Molecules: Mass to Mass Calculate the mass of CO2 emitted upon the combustion of 5.0 x 102 grams of octane (C8H18). mass C8H18 moles C8H18 moles CO2 mass CO2 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 5.0 102 C8 H18 Molar mass B Moles B More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield Our recipe: 1 cup of Flour + 2 Eggs + tsp Baking Powder 5 pancakes 3 cups of flour, 10 eggs, and 4 tsp baking powder 3 cups of flour 15 pancakes 10 eggs 25 pancakes 4 tsp baking powder 40 pancakes You can only make 15 pancakes 1 C8 H18 = 4.4 C8 H18 114.2 C8 H18 4.4 C8 H18 35 CO2 Mass B 16 CO2 = 35 CO2 2 C8 H18 44.01 CO2 = 1.5 103 CO2 1 CO2 Note: Do not actually round until the final step More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield 1 cup of Flour + 2 Eggs + tsp Baking Powder 5 pancakes More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield Summary Theoretical Yield Limiting Reactant Actual Yield Excess reactants 15 pancakes Most that can be produced Reactant Completely consumed in Reaction What was actually produced 11 pancakes Burned a pancake Dropped a pancake Spilled some batter Dog ate one Reactants present in quantities greater than necessary to react with the quantity of the limiting reactant. Theoretical Yield Amount of product that can be produced based on Limiting Reactant Actual Yield Percent Yield Percent of the theoretical yield actually attained 11 = 100 = 15 100 = 73% Amount of product actually produced in the reaction Percent Yield Actual Yield x 100 = % Yield Theoretical Yield 4 10/27/2011 Limiting Reactants Example Consider the reaction between 5 moles of CO and 8 moles of H2 to produce methanol. CO(g) + 2H2(g) CH3OH(l) How many moles of H2 are necessary in order for all the CO to react? moles of H2 = 5 mol CO 2 mol H2 1 mol CO = 10 mol H 2 If we begin with 1.8 moles of titanium and 3.2 moles of chlorine, what is the limiting reactant in moles of TiCl4? Ti(s) + 2 Cl2(g) TiCl4(s) Given: 1.8 moles Ti and 3.2 moles Cl2 Find: Limiting reactant Conversions: 1 mole Ti 1 mole TiCl4 2 mole Cl2 1 mole TiCl4 1 TiCl4 = 1.8 TiCl4 1 Ti 1 TiCl4 3.2 Cl2 = 1.6 TiCl4 2 Cl2 How many moles of CO are necessary in order for all of the H2 to react? moles CO of = 8 mol H 2 1 mol CO 2 mol H2 = 4 mol CO 10 moles of H2 required; 8 moles of H2 available; limiting reactant. 1.8 Ti Limiting Reactant Cl2 Least Amount of Product 4 moles of CO required; 5 moles of CO available; excess reactant. The 3.2 mol of Cl2 will run out before the 1.8 mol of Ti is used up, and hence Cl2 is the limiting reactant, and will produce 1.6 mol of TiCl4. Therefore, the theoretical yield is 1.6 mol of TiCl4. Example Example In the following synthesis reaction: 2 Na(s) + Cl2(g) 2 NaCl(s) If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant? In the following synthesis reaction: 2 Na(s) + Cl2(g) 2 NaCl(s) If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant? Given: 53.2 g Na (atomic mass 22.99 g/mol) 65.8 g Cl2 (molar mass 70.91 g/mol) Find: Limiting Reactant mass Na moles NaCl mass NaCl 1 mol Na 2 mol NaCl 58 .44 g NaCl 22 .99 g Na 2 moles Na 2 moles NaCl 1 mole Cl2 2 moles NaCl Conversion: moles Na 2 mol Na 1 mol NaCl mass Cl2 moles Cl2 moles NaCl mass NaCl 1 mol Cl 2 58 .44 g NaCl 1 mol Cl 2 1 mol NaCl Example In the following synthesis reaction: 2 Na(s) + Cl2(g) 2 NaCl(s) If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant? 53.2 Na = 2.31 Na 22.99 / 2 NaCl = 2.31 NaCl 2 Na 65.8 Cl2 = 0.928 Cl2 70.91 / 2.31 Na Limiting Reactant Cl2 2 mol NaCl 70 .90 g Cl 2 Example Smallest amount determine the limiting reactant 0.928 Cl2 Least Amount 2 NaCl = 1.86 NaCl of Product 1 Cl2 Add Another Step Percent Yield In the following synthesis reaction: 2 Na(s) + Cl2(g) 2 NaCl(s) If we have 53.2 g of Na and 65.8 g of Cl2, what is the limiting reactant? What is the Percent Yield if the actual yield of NaCl was 86.4 grams? Given: 53.2 grams Na 65.8 grams Cl2 Find: Limiting Reactant, Theoretical Yield, and Percent Yield (you need theoretical yield plus the actual yield to get the percent yield.) 5 10/27/2011 Example Example 10.4 g iron metal reacts with 11.8 g liquid sulfur to produce 14.2 g iron(III) sulfide. Find the limiting reactant, theoretical yield and percent yield. Iron metal: Fe(s), 10.4 g Liquid sulfur: S(l), 11.8 g Iron(III) sulfide: Fe2S3(s), 14.2 g Theoretical Yield is 108 g of NaCl (least amount of product) Percent Yield = Actual Yield 86.4 100 = 100 = 80.0% Theoretical Yield 108 Step 1: W rite the balanced chemical equation: 2 Fe(s) + 3 S(l) Fe2S3(s) Step 2: Convert to moles: Atomic mass Fe = 55.85 g/mol Atomic mass S = 32.06 g/mol Molar mass Fe2S3 = (2 x 55.85 g/mol + 3 x 32.06 g/mol) = 207.9 g/mol Example 10.4 g iron metal reacts with 11.8 g liquid sulfur to produce 14.2 g iron(III) sulfide. Find the limiting reactant, theoretical yield and percent yield. The limiting reactant is Fe(s) The theoretical yield is 19.4 g Fe2S3(s) The actual yield is 14.2 g Fe2S3(s) 14.2 = 100 = 19.4 100 = 73.2 % Example Example Lakes that have been acidified by acid rain can be neutralized by limestone (CaCO3). How much limestone in kg is required to neutralize a lake with volume 5.2x109 L that contains 5.0x10-3 g of H2SO4 per L? Aspirin can be synthesized by reacting acetic anhydride (C4H6O3) with salicylic acid (C7H6O3) to form aspirin (C9H8O4) and acetic acid (C2H4O2) according to the reaction: C4H6O3 + C7H6O3 C9H8O4 + C2H4O2 Molar mass CaCO3 = 100.09 g/mol Molar mass H2SO4 = 98.09 g/mol If 5.00 mL of acetic anhydride (density=1.08 g/mL) is reacted with 2.08 g of salicylic acid, 2.01 g of aspirin is produced. What is the limiting reactant, theoretical yield and percent yield for the reaction? Balanced chemical equation: H2SO4(aq) + CaCO3(s) CaSO4(aq) + H2O(l) + CO2(g) 3 mol H 2 SO4 5.2 109 L 5.0 10 g H 2 SO4 L 1 mol H 2 SO4 98.09 g H 2 SO4 Molar mass C7H6O3 = 138.13 g/mol 2.65 105 mol H 2 SO4 1 mol CaCO3 100.09 g CaCO3 1 kg kg CaCO3 2.65 10 mol H 2 SO4 1 mol H 2 SO4 1 mol CaCO3 1000 g 5 2.7 10 4 kg CaCO3 Molar mass C4H6O3 = 102.10 g/mol ANS: 2.7104 kg Molar mass C9H8O4 = 180.17 g/mol 5.00 mL C4 H 6O3 1.08 g C4 H 6O3 mL C4 H 6O3 2.08 g C7 H 6O3 1 mol C4 H 6O3 1 mol C9 H 8O4 102.10 g C4 H 6O3 1 mol C4 H 6O3 1 mol C7 H 6O3 1 mol C9 H 8O4 138.13 g C7 H 6O3 1 mol C7 H 6O3 6 10/27/2011 Example Practice Aspirin can be synthesized by reacting acetic anhydride (C4H6O3) with salicylic acid (C7H6O3) to form aspirin (C9H8O4) and acetic acid (C2H4O2) according to the reaction: How much salicylic acid, in an excess of acetic anhydride, would be required to produce 10.0 g aspirin if the percent yield was 74.1%? C4H6O3 + C7H6O3 C9H8O4 + C2H4O2 C4H6O3 + C7H6O3 C9H8O4 + C2H4O2 If 5.00 mL of acetic anhydride (density=1.08 g/mL) is reacted with 2.08 g of salicylic acid, 2.01 g of aspirin is produced. What is the limiting reactant, theoretical yield and percent yield for the reaction? Molar mass C4H6O3 = 102.10 g/mol Molar mass C7H6O3 = 138.13 g/mol Thus, if we want to produce 10.0 g of aspirin, it will require more salicylic acid (1/0.741 = 1.350 times more). Molar mass C9H8O4 = 180.17 g/mol mol C9 H 8O4 180.17 g C9 H 8O4 1 mol C9 H 8O4 Percent Yield According to the equation, 1 mol of acetic anhydride is required to produce 1 mol of aspirin (the theoretical yield). But if the percent yield is only 74.1%, then the amount of aspirin that is actually produced is only 0.741 times the theoretical yield. 2.08 g C9 H 8O4 2.71 g C9 H 8O4 2.71 g C9 H 8O4 ANS: 10.3 g of salicylic acid 100 74.1% ANS: salicylic acid, 2.71 g, 74.1 % Periodic Trends in Chemical Properties of the Main Group Elements Ionization Energy and Electron Affinity enable us to understand types of reactions that elements undergo and the types of compounds formed. General Trends in Reactivity Hydrogen (1s1) Grouped by itself Forms a cation with a +1 charge (H+) Forms an anion with a -1 charge (H-) Hydrides react with water to produce hydrogen gas and a base. CaH2(s) + H2O(l) Ca(OH)2(aq) + H2(g) Reactions of the Active Metals Group 1A Elements (ns1, Reactions of the Active Metals n 2) Group 2A Elements (ns2, n 2) Low IE Less reactive than 1A Never found in nature in pure elemental state Some react with H2O to produce H2 React with oxygen to form metal oxides Some react with acid to produce H2 sodium Alkali metals reacting with water potassium cesium Barium reacting with water 7 10/27/2011 Reactions of Other Main Group Elements Group 3A elements (ns2np1, n 2) Reactions of Other Main Group Elements Group 4A elements (ns2np2, n 2) Metalloid (B) and metals (all others) Nonmetal (C); metalloids (Si, Ge) and others metals Al forms Al2O3 with oxygen Form +2 and +4 oxidation states Al forms +3 ions in acid Sn, Pb react with acid to produce H2 Others form +1 and +3 Finely-divided aluminum sprinkled into a flame to form Al2O3 Reactions of Other Main Group Elements Group 5A elements (ns2np3, n 2) Nonmetal (N2, P), metalloid (As, Sb), and metal (Bi) Reactions of Other Main Group Elements Group 6A elements (ns2np4, n 2) Nonmetals (O, S, Se) Nitrogen, N2, forms variety of oxides Oxygen, O2 Phosphorus, P4 Sulfur, S8 As, Sb, Bi (crystalline) HNO3 and H3PO4 important industrially Selenium, Se8 Metalloids (Te, Po) Te, Po (crystalline) SO2, SO3, H2S, H2SO4 Forest damaged by acid rain Nonmetal oxides added to water produce an acid Reactions of Other Main Group Elements Reactions of Other Main Group Elements Group 7A elements (ns2np5, n 2) Group 8A elements (ns2np6, n 2) All diatomic All monatomic Do not exist in elemental form in nature Filled valence shells Form ionic salts Considered inert until 1963 when Xe and Kr were used to form compounds Form molecular compounds with each other No major commercial use React with hydrogen to form hydrogen halides 8 10/27/2011 Comparison of Group 1A and Group 1B Elements Have single valence electron Properties differ Group 1B much less reactive than 1A High IE of 1B - incomplete shielding of nucleus by inner d outer s electron of 1B strongly attracted to nucleus 1B metals often found elemental in nature (coinage metals) 9
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PHYS 1122Exp. 4 Magnetic ForcesHomeworkNAME:_1. A current I is passing through a wire. What is the magnitude and direction of themagnetic field at point A?IrA2. The rails of a commuter train are used to carry the electric current to power the eng
U. Houston - PHYS - 1101
PHYS 1122Exp. 5 RC CircuitsHomeworkNAME_1. A student built a simple RC circuit consisting of a 1000 ohms resistor, a 500 F capacitor, and a 5 voltsbattery connected in series. If the capacitor was initially uncharged, how long does it take the voltag
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
PHYS 1122Exp. 7 Electromagnetic InductionHomeworkNAME:_1. State Faradays law and Lenzs law.2. Using Eq. (2) as a guide, state three ways to change the magnetic flux.3. A copper disk swings like simple pendulum inand out of a magnetic field, as show
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
PHYS 1122Exp. 9 Diffraction and InterferenceHomeworkNAME:_1. State the Huygens-Fresnel principle2. Light with wavelength is incident on a single slit whose width is adjustable. The slitinitially has a width and increases to 1000 . Describe qualitati
U. Houston - PHYS - 1101
PHYS 1122Exp. 10 PolarizationHomeworkNAME:_1. What is the difference between polarized and unpolarized light?2. If the angle between the incident electric field and a polarizers transmission axis is400, what fraction of the incident intensity is tra
U. Houston - PHYS - 1101
PHYS 1122Exp.11 Thermal ConductivityHomeworkNAME:_1) Why portable coolers are often made of styrene foam?2) A bar with cross-sectional area A = 0.5cm2 is used to conduct heat between tworegions at different temperature. The bar is made of silver, k
U. Houston - PHYS - 1101
PHYS 1122Exp. 12 Ideal Gas LawHomeworkNAME:_1. How is an ideal gas different from a real gas?2. A boy places his inflated birthday balloon inside the refrigerator (so nobody steals it).What will happen to the balloon? Explain.3. A 5 liters cylinder
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101
U. Houston - PHYS - 1101