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Intro Chemistry - Ch.3 notes - Ch.4 notes

by: Alexis Tate

Intro Chemistry - Ch.3 notes - Ch.4 notes 13699

Marketplace > Appalachian State University > Chemistry > 13699 > Intro Chemistry Ch 3 notes Ch 4 notes
Alexis Tate

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These notes are a continuation of chapter 3 and begin chapter 4
Intro Chemistry I
Alexander Schwab
Class Notes
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This 5 page Class Notes was uploaded by Alexis Tate on Friday September 30, 2016. The Class Notes belongs to 13699 at Appalachian State University taught by Alexander Schwab in Fall 2016. Since its upload, it has received 9 views. For similar materials see Intro Chemistry I in Chemistry at Appalachian State University.


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Date Created: 09/30/16
Chapter 3 Cont. Friday, September 30, 2016 Limiting Reagent  The limiting reagent determines how much of a substance you can make  Example: o You have 14.0 moles of H and 2.7 moles of O . Howm2ny moles of H O can 2 you make basedoff of this reaction? 2H 2 O 22H O 2 * How many moles ofO do y2u need to react with 14.0 moles of H ? 2 - One way to find the limiting reagent is to seewhich reactant cannot completely react with the other 14.0 mol H 2 1 mol O 2 Since you do not have = 7.0mol O 2 enough moles ofO to 2 2 mol H 2 react with 14.0 moles of H 2O i2 the limiting 3.7 mol O 2 2 mol H 2 = 7.4 molH O2 reagent 1 mol O 2 o How many moles of H O cou2d you produce with 14.0 moles of H and 8.0 m2les of O 2 Which is the limiting reactant and which is the excess reactant? 2H 2 O 22H O 2 - Another way to find the limiting reagent is to convert both reactants to the desired product and seewhich produces a fewer amount 14.0 mol H 2 1 mol H 2 = 14.0 molH O2 2 mol H 2 8.0 mol O 2 2 mol H 2 = 16mol H O2 1 mol O 2 * Thelimiting reactant is H and the excess reactantis O 2 2 o Ammonia is produced by reacting hydrogen and nitrogen gas: N 2g) + 3H (2)  2NH (g)3 How many moles of NH can3e made ifgiven 3.0moles of N and 7.0m2les of H 2 3.0 mol N 3 mol H 2 2 = 9.0mol H 2 1 mol N 2 7.0 mol H 2 2 mol NH 3 3 mol H 2 = 4.7 molNH 3 o 2.00 g of Zn (s) is placedin an aqueous solution containing 2.50 g of silver nitrate. How many grams of Ag will be produced ifthe following chemical equation is obeyed? Zn (s) + 2AgNO (3q)  2Ag (s) +Zn(NO ) (a3 2 2.00 g Zn 1 mol Zn 2 mol 169.874 AgNO 3 = 10.39 g AgNO 3 AgNO 3 65.38 g Zn 1 mol Zn 1 mol AgNO 3 2.50 g 1 mol 2 mol Ag 107.87 g Ag AgNO 3 AgNO 3 = 1.59 g Ag 169.874 g 2 mol 1 mol Ag AgNO 3 AgNO 3  To find amount excess reagentleftover: o First, convert the limiting reagent’s amount (usuallymoles or grams) to the excess reagent’s amount (moles or grams) o Then, subtract that answer from the amount of excess giveninthe problem o Example (Using the ammonia example):  How many moles of excess N is 2eftover from the reaction? 7.0 mol H 2 1 mol N 2 = 2.5mol N 2 3 mol H 2 3.0 mol N 22.5 mol N = 2.50 molofN left 2ver Reaction Yield  Actual Yield o Amount of desired product actuallyproduced by a chemical reaction (mass)  Theoretical Yield o Amount of desired product that is predicted to be produced by a chemical reaction  Percent Yield o Percent yield = Actual Yield x100 Theoretical Yield  Example: o Basedoffof lastexample’s answer, the predicted amount of silverthat would be produced was 1.59 g.The reaction actually produced 1.41 g of Ag (s). What was the theoretical yieldand percent yieldfor this reaction? 1.41 g x100 = 88.7% 1.59 g Percent Yield Theoretical Yield o Iron, the main component of steel,is produced from iron ore through reduction with carbon monoxide as inthe following equation: Fe 2 3s) + 3CO (g) 2Fe (s) +2CO (g)2 If you start with 300. g of iron ore, how much iron should be produced assuming there is aninfinite supply of CO (g). Molar Mass: 2(55.845) 3(15.999) 159.687 g/mol 300. g Fe2O 3 1 mol Fe 2 3 2 mol Fe 55.845 g Fe = 210. g Fe 159.687 g 1 mol Fe 2 3 1 mol Fe Fe2O 3 o If 176g of Fe (s)was produced from 300. g of iron ore, what was the percent yield of the reaction? 176 g x100 =83.8% 210 g Chapter 4 AqueousSolutions  Solution – homogenous mixture o Canbe vapor, liquid, or solid  Aqueous solution o Homogenous mixture inwhich water is the dissolving medium o Salt inwater, sugarin water, vodka o Water is the solvent (thing that does the dissolving),thestaffdissolvedinwater is the solute (thing that is dissolved)  What happens… o I (2) o Water separates the I mol2cules from each other o Individual I 2olecules are dispersedin solvent o Water does not necessarilyseparateIatoms from one another Solution Concentration  The concentration of a solution describes how much solute there is for a givenamount of solvent (or solution) o Concentrated – means highconcentration o Dilute –means low concentration o Saturated – means concentration is as highas possible o Unsaturated – not saturated, means concentration is lower than a saturated solution  Supersaturated Solution o Are unstable  Too much stuff is dissolvedandcan leavethe solution atany moment  The precipitation canbe started by dust, vibration, or other things Electrolytes  H2o(l)causes ionic solids to dissociateintotheir respective ions CaCl 2s)  Ca (aq)+2Cl (aq)-  Substances that form hydrated ions in aqueous solutions are called electrolytes o Ionic substances andsome molecular substances (acids) o Write the chemical equations for the dissociationof the following:  Ammonium sulfateand sodium perchlorate + - (NH 4 2O 4aq)  2NH (aq4 + SO (aq)4 NaClO 4aq)  Na (aq) +ClO (aq)4  When ions are present in a solution, the solution can conduct electricity  The conductivity of the solution increases withincreasing ion concentration Nonelectrolytes  Some compounds do not dissociateinto ions when dissolvedin water o Most molecular compounds  Methanol (CH OH3molecules aredispersed by H O (l) but 2re not dissociateinto ions CH 3H(l) CH OH(aq3 Precipitation Reactions  Pb(NO )3 2q)reacts with KI (aq)and produces PbI (s): 2 Pb(NO ) 3 2)+ 2KI (aq)  PbI (s)+22NO (aq) 3 o Keep in mind that KNO exists3as twoseparate ions K and NO inthe solutio3- Ionic Equations:  This is calledthemolecular equation: Pb(NO ) 3 2)+ 2KI (aq)  PbI (s)+22NO (aq) 3 Calleda precipitate  This is the ionic equation: Pb (aq) +2NO (aq) 32K (aq) + 2I (aq)  PbI (s)+ 2K (aq2 + 2NO (aq) 3- o Emphasizes that compounds existas ions ina solution o K and NO are s3ectator ions  Spectator ions canbe removed by yielding a net ionic equation 2+ - Pb (aq) +2I (aq)  PbI (s) 2 *To write a net ionic equation all you need are the ions that make up the precipitate


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