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Lecture_4_Exp5_Spr12

Course: CHEM 1122, Spring 2012
School: Texas San Antonio
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Chemistry General I Laboratories CHE 1120 CHE Limiting Reactants Limiting 02/13 02/14 Experiment 5 Experiment Objective: Determine the theoretical yield Determine the percent yield Make CuI2 Develop an understanding of Limiting Reactant Six Types of Reactions Six Combustion: C10H8 + 12 O2 10 CO2 + 4 H2O Synthesis: A + B AB ; 8 Fe + S8 8 FeS Decomposition: AB A + B ; 2 H2O 2 H2 + O2 Single...

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Chemistry General I Laboratories CHE 1120 CHE Limiting Reactants Limiting 02/13 02/14 Experiment 5 Experiment Objective: Determine the theoretical yield Determine the percent yield Make CuI2 Develop an understanding of Limiting Reactant Six Types of Reactions Six Combustion: C10H8 + 12 O2 10 CO2 + 4 H2O Synthesis: A + B AB ; 8 Fe + S8 8 FeS Decomposition: AB A + B ; 2 H2O 2 H2 + O2 Single Displacement: A + BC AC + B ; Mg + 2 H2O Mg(OH)2 + H2 Double displacement: AB + CD AD + CB Pb(NO3)2 + 2 KI PbI2 + 2 KNO3 Acid/Base: HA + BOH H2O + BA; HBr + NaOH NaBr + H2O Limiting Reagent Limiting -reagent that is all used up in the reaction 5 Easy Steps 1. Balance the equation 2. Convert all reactants to moles 3. Use balanced equation to convert reactants to product (if more than one product do the same one for all the reactants) 4. Which reactant produced the smallest amount of product? This is your limiting reactant! 5. Take the moles of the product formed by the limiting reactant and figure out how much of the other reactant is left over. Limiting Reagent Cont Limiting Recipe: 2 T.S + 2 Chz + 1Crust 1 pizza Have: 3 T.S, 2 Chz, and 3 crust How many pizzas can each ingredient make? 3 T.S 1 pizza = 1.5 pizzas 2 T.S 2 Chz 1 pizza = 1 pizzas Which ingredient makes the fewest number of pizzas? How many cans of T.S. used? 1 pizza 2 Chz 2 T.S. = 2 T.S. used 1 pizza 3 Crust 1 pizza 1 Crust = 3 pizzas How much T.S. was in excess? Excess = initial - used Excess T.S. = 3 -2 = 1 T.S. excess Example Example Start with: 75.0 mL of a 3.0 M solution of Na2CrO4(aq) and 125 mL of a 2.5 M solution of AgNO3(aq). What is the limiting reagent and how much excess reagent is left? Na2CrO4 + AgNO3 Ag2CrO4 + NaNO3 I. Balance Equation: Na2CrO4 + 2AgNO3 II. Convert reactants to moles Na2CrO4: 3.0 mols Ag2CrO4 + 2NaNO3 0.075 L 0.225 mols L 2.5 mols AgNO3 : L 0.125 L 0.312 mols Example Cont. Example III. Convert reactants to products - Ag2CrO4 0.225 mols Na2CrO4 Na2CrO4: 0.225 mols Ag2CrO4 1 mol Na2CrO4 0.312 mols AgNO3 AgNO3 : 1 mol Ag2CrO4 1 mol Ag2CrO4 0.156 molsAg2CrO4 2 mols AgNO3 IV. Which reactant is the limiting reagent? AgNO3 is the limiting reagent Example Cont. Example V. Determine the amount of excess reagent - 0.156 mols of Ag2CrO4 will be formed - How many moles of Na2CrO4 will be used? 0.156 mols Ag2CrO4 1 mol Na2CrO4 0.156 mols Na2CrO4 (used) 1 mol Ag2CrO4 - Excess Na2CrO4 = 0.225 mols 0.156 mols = 0.069 mols Na2CrO4 (excess) Type of Reaction in the Experiment Type -Precipitation Reaction (i.e. double displacement): a reaction that results in the formation of a solid substance (precipitate) AgNO3(aq) + NaCl(aq) AgNO3(aq) + NaOH(aq) AgCl(s) + NaNO3(aq) AgOH(s) + NaNO3(aq) 2NaI(aq) + HgCl2(aq) HgI2(s) + 2NaCl(aq) *2CuSO4(aq) + 4KI(aq) 2CuI(s) + I2(aq) + 2K2SO4 (aq) Lab Info Lab - Use the following equation: 2Cu(NO3)2(aq) + 4KI(aq) 2CuI(s) + I2(aq) + 4KNO3(aq) - Safety note: reaction will produce iodine - leaves a yellow stain on skin/clothes - if allergic please notify TA, IMMEDITELY - Determine which reactant is the limiting reagent Lab Procedure TA will assign certain volumes of reagents to each group 1 1a 1b 2 2a 2b test 2, labeled 2a ,2b are duplicates of 1, 1a, 1b respectively -wipe off any fingerprints -weigh to the nearest 0.0001g -do not touch them with fingers 1 KI Cu(NO3)2 -Add reagents -mix Let sit for 5mins 1b 1a 1 1 Remove liquid Using a pipet Place to the side Lab Procedure Cont 1 Wash with ~3mL of MeOH and shake 1 Cloudy 5 mins Weigh test tube, to nearest 0.0001g 1 dry in oven ~20 mins place in beaker Clear 1 Lab Procedure Cont Lab ** Testing the Supernatants 1a 1b add 1mL Cu(NO3)2 add 1mL KI Record observations Record observations Use a (+) for more precipitate formed Use a (-) for no more precipitate formed Prelab Tables Prelab CuI Vol. Cu(NO3)2 0.50M Vol. KI 0.50M A 1.0mL 6.0mL B 2.0mL 5.0mL C 3.0mL 4.0mL D 4.0mL 3.0mL E 5.0mL 2.0mL F 6.0mL Moles of Cu(NO3)2 1.0mL Remember: # mols = M * V Moles of KI Moles of Moles of ppt based ppt based on moles on moles of of KI Cu(NO3)2 Class Data Class Vol. Cu(NO3)2 0.50M Vol. KI 0.50M A 1.0mL 1.0mL 6.0mL 6.0mL B 2.0mL 2.0mL 5.0mL 5.0mL C 3.0mL 3.0mL 4.0mL 4.0mL D 4.0mL 4.0mL 3.0mL 3.0mL E 5.0mL 5.0mL 2.0mL 2.0mL F 6.0mL 6.0mL Mass of ppt in test tubes Cu(NO3)2 test KI test 1.0mL 1.0mL Make sure to complete(in lab manual) and attach to post lab report! Post Lab Table Get from prelab Balanced Equation: Group A moles of Cu(NO3)2 Theoretical moles of CuI based on Cu(NO3)2 moles of KI Theoretical moles of CuI based on KI Avg. mass of CuI* Avg. moles of CuI* Limiting Reactant (Cu(NO3)2 or KI) % error of moles of CuI Group B Group C Group D Group E Group F Table for Results for Limiting Reagent Lab Data Anaylsis A student was assigned the combination in Group A. Group A consists of 1.0mL of 0.5M Cu(NO3)2 and 6.0mL of 0.5M KI. The student obtained 0.0910 and 0.0940 g for trial 1 and 2, respectively. A (+) result occurred when performing the Cu(NO3)2 test. Balanced Equation: MM(KI) = 166.1 2Cu(NO3)2 (aq) + 4KI(aq) 2CuI(s) + I2(aq) + 4KNO3 (aq) Group A MM(CuI) = 190.5 MM(Cu(NO3)2) = 187.6 moles of Cu(NO3)2 0.0005 Theoretical moles of CuI based on Cu(NO3)2 moles of KI 0.0005 Theoretical moles of CuI based on KI Avg. mass of CuI* 0.0015 Avg. moles of CuI* 0.000485 Limiting Reactant (Cu(NO3)2 or KI) % error of moles of CuI Cu(NO3)2 0.003 0.0925 3.09% Group B Group C Group D Group E Group F Additional Problems -Read over sections 3.8-3.10 (pgs 99-108) in Chang book. -Problems 3.84, 3.86, 3.90, 3.92 (pg 114) in Chang book -Aluminum reacts with sulfur gas to form aluminum sulfide. Initially 1.18 mol of Al and 2.25 mols of S are combined. a. write a balance equation for reaction 2Al(s) + 3S(g) Al2O3 b. What is the limiting reagent? Al c What is the theoretical yield of aluminum sulfide in mols? 0.590 mols d. How many moles of excess reactant remain unreacted? 0.48 mols Lab 4: Exp 5 Checklist Completed online quiz # 5 By 10pm Sunday (02/19/12) Prelab for Lab 4 Exp. 5 completed Hand written in lab notebook. Complete Prelab questions for Exp. 5 Postlab for Lab 3 Exp. 3 completed, Typed. Exp. 4 parts B,C,D is due as well. Complete prelab questions 1-6 on for exp. 4 as well. Attach to post lab for exp. 3. Bring Lab Coat, goggles, lab notebook to lab
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