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SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P1 Quantity relationships: How much? How many? How many dragonflies’ wings are required for 4 dragonflies to be able to fly? 1 Dragonfly → 4 wings 0 0 ? SCC-CH110/UCD-CH41C Same Units Chapter: 9 Instructor: J.T. P2 1 DF 4 wings = 5 DFs ? Same Units SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P3 Quantity Relationships in Chemical Reactions: How to solve stoichiometry problems: What we need to know are two subjects as: 1) A balanced chemical reaction. 2) Relationship between mass and mole. EXAMPLE: How many moles of oxygen are required to burn 2.40 moles of ethane gas? 2 C2H6 (g) + 7 O2 (g) Given → 4 CO2 (g) + 6 H2O (l) Wanted 2 mol C2H6 2.40 mol C 2H 6 0 7 mol O2 ? 0 SCC-CH110/UCD-CH41C Chapter: 9 2 mol 2.4 mol Instructor: J.T. P4 7 mol ? = 2 mol (C2H6) × ? = 7 mol (O2) × 2.4 mol (C2H6) 7 mol ?= × 2.4 mol 2 mol *********************************************** EXAMPLE: Calculate the number of grams of oxygen that are required to burn 155 g of ethane. 2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (l) Given 2 mol C 2H 6 155 g g 30.06 ( ) mol 155 g C 2H 6 Same Units Wanted 0 7 mol O2 ? 0 Same Units SCC-CH110/UCD-CH41C Chapter: 9 2 mol 5.15 mol Same Units Instructor: J.T. = 7 mol ? Same Units 7 ? = 5.15 × 2 ? = 18.03 mol (oxygen) Wanted in gram! mass mole = molar mass molar mass of oxygen = 2 × 16.00 (g/mol) ? = 18.03 mol × 32.00 (g/mol) ? = 577 g O2 P5 SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P6 EXAMPLE: How many milligrams of Nickel Chloride are in a solution if 503 mg of Silver Chloride is precipitated in the reaction of Silver Nitrate and Nickel Chloride solution? 2 AgNO3 (aq) + NiCl2 (aq) → 2 AgCl (s) + Ni(NO3)2 (aq) Wanted 1 mol NiCl2 Given 2 mol 0 ? 0 AgCl 503 mg mg (AgCl) AgCl mol (AgCl) ? Same Units ¾ (mg) to (g): 503 mg × (1 g/ 1000 mg) = 0.503 g SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P7 ¾ (mass) to (mole) mole = mass / molar mass mole of AgCl = 0.503 g ÷ 143.35 (g/mol) mole of AgCl = 3.51 × 10-3 mol 1 mol 2 mol = ? 72.1 mol ? = 3.51 × 10 −3 mol AgCl × 1 mol NiCl2 2 mol AgCl ? = 1.75 × 10-3 mol NiCl2 Wanted in milligram! mass mole = molar mass molar mass of NiCl2= 129.59 (g/mol) Mass of NiCl2 = 1.75 × 10-3 mol × 129.59 (g/mol) =0.227 g Mass of NiCl2 = 227 mg SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P8 Gas Stoichiometry at Standard Temperature and Pressure: One mole of any gas at STP occupies 22.4 L EXAMPLE: What volume of hydrogen, at STP, can be released by 42.7 g of zinc as it reacts with hydrochloric acid? Zn(s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq) Given 1 mol Zn 42.7 g g 65.39 ( ) mol Wanted 22.4 L 0 ? 42.7 g 0 Same Units 1 mol 0.653 mol = 22.4 L ? SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P9 EXAMPLE: 1.30L of gaseous ethylene is burned. What volume of oxygen is required if both gas volume are measured at STP? → 2CO2 (g) + 2 H2O(l) C2H4 (g) + 3O2 (g) Given Wanted 1 mol C 2H 4 22.4 L 3 mol O2 0 3 × 22.4 L 0 ? 1.30 L 22.4 L 1.30 L 0 = 67.2 ? SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P10 Gas Stoichiometry at nonstandard condition: In STP condition: T= 273 K, and P =760 torr P1 V1 T1 = STP = P2 V2 T2 nonstandard In the previous example for the reaction: Zn (s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq) We calculated the volume of hydrogen as 14.6 L, now let’s change the problem. Instead of measuring the hydrogen volume at STP, let’s use t = 21 °C and P = 748 torr. TK=21+273 =294 760 × 14.6 748 V2 = 2731 294 STP V2= 16.0 L = nonstandard SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P11 14 transistors 22 resistors 18 capacitors How many of the above electronic circuits (ec) can you assemble from 22 resistors, 18 capacitors, and 14 transistors? 2 +4 → +3 Calculate the theoretical yield of ec: 1 ec 14 T × 2T = 7 ec 1 ec 22 R × = 5.5 ec ≈ 5 ec 4T 1 ec 18 C × = 6 ec 3C Resistor is the limiting item, because 22 of resistors give 5 ec. SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P12 Limiting Reactants: Example: The fertilizer ammonium nitrate can be made by direct combination of ammonia with nitric acid. If 74.4 g of ammonia is reacted with 159 g of nitric acid, how many grams of ammonium nitrate can be produced? Also calculate the mass of unreacted reactant that is in excess. NH3 + HNO3 → NH4NO3 Convert the number of grams of each reactant to moles. NH3: 74.4 g = 4.37 mol g 17.03 mol 159 g HNO3: 63.02 g = 2.52 mol mol 1 mol NH 4 NO3 4.37 mol NH 3 × = 4.37 mol NH 4 NO3 1 mol NH 3 1 mol NH 4 NO3 2.52 mol HNO3 × = 2.52 mol NH 4 NO3 1 mol HNO3 ∴ The reactant that yields the smaller amount of product is limiting reactant. 1 mol of NH3 ≡ 1 mol of HNO3 2.52 mol ≡ ? ? =2.52 mol of HNO3 4.37 mol – 2.52 mol = 1.85 mol NH3 left SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P13 1.85 × 17.03 = 31.51 g NH3 left A mixture of 5.0 g of H2 (g) and 10.0 g of O2(g) is ignited. Water forms according to the following combination reaction: 2H2(g) +O2(g) → 2H2O(g) Which reactant is limiting? How much water will the reaction produce? SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P14 Percent Yield: The calculated amount of product if is based on the assumption that all of the reactant is converted into product is called the theoretical yield. In laboratory or in industrial production, the actual amount of product isolated from a reaction is usually less than the theoretical yield, and it is called actual yield. General solution: αA + βB → γC Given: n gram of A Given: actual yield Wanted: %yield Step 1: Convert mass of A into the mole of A. Step 2: Calculate the theoretical yield: γ mol C Number of moles A × α mol A Step 3: Convert mole of C into the mass of C. SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P15 Step 4: actual yield %Yield = × 100 theoretical yield Example: A general chemistry student, preparing copper metal by the reaction of 1.274 g of copper sulfate with zinc metal, obtained a yield of 0.392 g copper. What was the percent yield? To calculate %yield, we need both the theoretical yield of copper and the actual yield. CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq) Mass to Mole: 1.274 g CuSO 4 = 7.982 × 10 −3 mol CuSO 4 159.6 g mol The theoretical yield: 7.982 × 10 −3 mol CuSO 4 × 1 mol Cu 1 mol CuSO 4 = 7.982 × 10 −3 mol Cu Mole to Mass: 7.982 × 10 −3 mol Cu × 63.55 g % Yield = mol = 0.5072 g Cu actual yield 0.392 g × 100 = × 100 = 77.3 % theoretical yield 0.5072 g SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. Liquid Oxygen Liquid Hydrogen Main Engines P16 SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P17 Thermochemical Equations: H2 (l) + O2 (l) → H2O (g) + energy System Gives off heat 2 NH3 (g) + energy System Exothermic -ΔH → N2(g) + 3 H2 (g) Absorbs heat Endothermic +ΔH The SI unit for energy: joule (J) How many kilojoule of energy are released when 73.0 g C2H6 burns according to: 2C2H6 (g) +7 O2 (g) → 4 CO2(g) + 6H2O (l)+ 3119KJ 2 mol C 2H 6 0 3119 KJ ? 73.0 g g 30.05 ( ) mol 73.0 g 0 SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P18 ΔH =2.82 × 103 KJ for the photosynthesis reaction by which plants use energy from the sun to form one mole sugar from carbon dioxide and water. How much energy is required to form 454g (1lb) of simple sugar C6H12 O6? 6 CO2 (g) +6 H2O (g) + → C6H12O6 (s) + 6O2 (g) 1 mol Suga 0 2.82 × 103 KJ ? 454 g g 180 ( ) mol 454 g 0 ∴ 7.11 × 103 KJ SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P19 SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P20