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UT Arlington CHEM 1465 - Limiting Reagent

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CHEM 1465 1st Edition Lecture 6Outline of Last Lecture 1. Fundamentals of stoichiometryA. Combustion reactionsB. Reaction stoichiometryC. Limiting reagentD. Theoretical and percent yieldE. Titration Outline of Current Lecture 1. Review for possible test questionsA. Physical and chemical changesB. Prefix multiplierC. Significant figuresD. Ionic forms of elements chart E. Atomic weight F. Naming ionic compoundsG. Combustion reactionH. Limiting reagentI. Molarity Current Lecture1. Review for possible test questionsThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.A. Physical and chemical changes: state whether the following are physical or chemical changes.- Gas burning on a stove: chemical- Liquid propane evaporating: physical- Liquid propane burns a flame: chemical- A bike rusting after exposure to rain: chemicalB. Prefix multiplier: use a prefix multiplier to express each measurement without any exponents.- 1.2x10-9 m: 1.2 nm- 22x10-15 s: 22fs- 1.5x109 g: 1.5 Gg- 3.5x106 L: 3.5 MLC. Significant figures: calculate each value using the correct number of significant figures. - (-24.6681 x 2.38) + 332.58 = 391.3- (85.3 – 21.489) / 0.0059 = 1.1x104- (512 / 986.7) + 5.44 = 5.96D. Ionic forms of elements chartSymbol Ion formed Number of electrons Number of protonCa Ca2+18 20Be Be2+2 4Se Se2-36 34P P2-17 15E. Atomic weight: Find the atomic weight of the element and then identify the elementIsotope Mass Abundance (%)1 135.90714 0.192 137.90599 0.253 139.90543 88.434 141.90924 11.13- (135.90714)(.0019)+(137.90599)(.0025)+(139.90543)(.8843)+(141.90924)(.1113)Weight = 140.12Element = CeF. Naming ionic compounds- CuNO3 : copper(I) nitrate- Mg(C2H3O2)2 : magnesium acetate- KClO2 : potassium chlorite- ZnCl2 : zinc chlorideG. Combustion reactions: consider the unbalanced equation for the combustion of hexane- C6H14 + O2 → CO2 + H2O. Determine how many moles of O2 are needed to react with 7.2 moles of hexane. 2C6H14 + 19O2 → 12CO2 + 14H2O7.2 mol C6H14 (19 mol O 22mol C 6 H 14)=¿68 mol O2H. Limiting reagent: Find the limiting reagent and determine in grams the mass of the product. 2Na + Br2 → 2NaBr- 45.9g Na and 319.6g Br2If Na was LR: 45.9g Na(1 mol Na22.9 g Na)(2 mol NaBr2 mol Na)(102.8 g NaBr1 mol NaBr)=¿ 206g NaBrIf Br was LR: 319.6g Br2(1 mol Br 2159.8 g Br2)(2 mol NaBr1 mol Br 2)(102.8 g NaBr1mol NaBr)=¿4.11g NaBrLimiting reagent is Na, which formed 206 grams of NaBr- 41.39g Na and 111.89g Br2If Na was LR: 41.39g Na(1 mol N a22.9 g Na)(2 mol NaBr2 mol Na)(102.8 g NaBr1 mol NaBr)=¿185.g NaBrIf Br was LR: 111.89g Br2(1 mol Br 2159.8 g Br2)(2 mol NaBr1 mol Br 2)(102.8 g NaBr1mol NaBr)=¿143.92g NaBrLimiting reagent is Br, which formed 143.2 grams of NaBrI. Molarity: A lab calls for making 400mL of 1.1M NaNO3 solution. What mass of NaNO3 in grams is needed? 400mL(10−3L1 mL)(1.1 mol NaNO 31 L)(84.19 g NaNO 31 mol NaNO 3)=¿ 37.39g


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