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UVM CHEM 023 - Quantity Relationships in Chemical Reactions

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8Quantity Relationships inChemical ReactionsChapter 10Stoichiometry:• The quantitative relationships between thesubstances involved in a chemical reaction areestablished by the balanced equation for the reaction• A stoichiometry problem asks, “How much or howmany?”• Example: Designing and optimizing chemicalreactions (greater efficiency, better yields…….)9Examples of stoichiometry problems• How many moles of oxygen are required to burn 2.40moles of ethane?• How many mg of nickel chloride are in solution if 503mg of silver chloride is precipitated in the reaction ofsilver nitrate and nickel chloride?• How many mL of 0.321 M HCl solution is required toneutralize 5.0 mL 1.284 M KOH solution?• A solution containing 3.18g barium chloride is addedto a solution containing excess sodium sulfate, 3.37gof barium sulfate is produced, what is the percentageyield?• How may moles of KNO3 can be made from 7.94 moleKCl and 9.96 mole HNO3?Writing chemical formulasCh 6Calculating molar massesfrom chemical formulasCh 7.4Using molar masses to changemass to moles and moles to massCh 7.5Writing and balancingchemical equationsCh 8Use equations to change from moles ofone species to moles of anotherCh 10.1Stoichiometry problems bring togethermany of the topics we have studied10Conversion factors from achemical equation• Example:How many moles of oxygen are needed tocompletely react with 2.34 moles of methane (CH4) ina combustion reaction?• Carbon dioxide and steam are the productsCH4 + O2CO2 + H2O22• The balanced equation tells us that 2 moles of O2 arerequired to react with one mole of CH4• PER path: 1 mole of CH4 PER 2 moles of O2GIVEN: 2.34 mol CH4WANTED: mol O2PER/PATH: mol CH4mol O21 mol CH4/2 mol O22.34 mol CH4X 2 mol O21 mol CH4= 4.68 mol O2We need 4.68 moles of oxygen to completely react with2.34 moles of methane (CH4) in a combustion reaction11Example:• NO is made from ammonia in the first stage of theOstwald process for manufacturing nitric acid:• How many moles of ammonia will react with 20 molesof oxygen?• The equation shows that we use 4 moles of ammoniafor every 5 moles of oxygen4 moles NH35 moles O220 moles O2 x 4 moles NH35 moles O2= 16 moles NH3 3 2 24NH 5O 4NO + 6H O+ !PER =Example:• The first stage of the Ostwald process formanufacturing nitric acid• How many moles of nitrogen monoxide will resultfrom the reaction of 8 moles ammonia?• The equation shows that we obtain 4 moles ofnitrogen monoxide for every 4 moles of ammonia4 moles NO4 moles NH38 mole NH3 x 1 mole NO1 mole NH3= 8 moles NO 3 2 24NH 5O 4NO + 6H O+ !PER =1 mole NO1 mole NH312GivenMassMoles ofGivenQuantityMoles ofWantedQuantityWantedMassMolar Massg PER molEquationmol PER molMolar Massg PER molMass–Mass StoichiometryStep 1: Change the mass of thegiven species to moles.Step 2: Change the moles of the given speciesto moles of the wanted species.Step 3: Change the moles of the wantedspecies to mass.The Stoichiometry Path13GIVEN: 10.0 g CH4WANTED: g CO2PER/PATH: g CH416.04 g CH4/mol CH4mol CH41 mol CO2/1 mol CH4mol CO244.01 g CO2/mol CO2g CO24 2 2 2CH 2O CO 2H O+ ! +Example: How much product• How many grams of carbon dioxide are producedwhen 10.0 g of methane, CH4, is burned?10.0 g CH4x1 mol CH416.04 g CH4X1 mol CO21 mol CH4x44.01 g CO2mol CO2= 27.4 g CO2PER/PATH: g CH416.04 g CH4/mol CH4mol CH41 mol CO2/1 mol CH444.01 g CO2/mol CO2g CO214M =moles soluteliter solutionStoichiometry in solutions - Molarity• Molarity (M):• Provides a direct conversion between a macroscopicmeasurable quantity, volume in liters, and aparticulate-level quantity, number of moles of soluteparticles (e.g. molecules)How many moles of methanol arein 45.3 mL of 0.550 M CH3OH?Key concept:0.550 M CH3OH =0.550 mol methanolL solution=0.550 mol methanol1000 mL solution45.3 mL solution x0.550 mol methanol1000 mL solution=0.0249 mol methanol1513.0 g C12H22O11x1 mol C12H22O11342.30 g C12H22O11 = 0.0380 mol C12H22O11Example• Calculate the molarity of a solution made bydissolving 13.0 grams of sugar, C12H22O11, in avolume of 4.00 X 102 mL of solution• First (always!) convert the mass into moles:=x1000 mLL0.0380 mol C12H22O114.00 X 102 mL0.0950 mol C12H22O11 / L = 0.0950 M C12H22O11M =moles soluteliter solutionCalculate the molarityConvert into L16Working with molarity• How many mL of 0.321 M HCl solution is required toneutralize 5.0 mL 1.284 M KOH solution?• First write and balance the equation!!• The molar ratio of HCl to KOH is 1:12HCl + KOH KCl + H O!GIVEN: 5.0 mL 1.284 M KOH solutionWANTED: mL of 0.321 M HCl solution How many mL of 0.321 M HCl solution is requiredto neutralize 5.0 mL 1.284 M KOH solution?2HCl + KOH KCl + H O!GIVEN: 5.0 mL 1.284 M KOH solutionWANTED: mL of 0.321 M HCl solution PER/PATH:ml KOH solution 1.284 M KOH / 1000 mLmol KOH1 mol HCL /1 mol KOHmol HCl1000mL / 0.321 mol HCl ml HCL solution17PER/PATH:ml KOH solution 1.284 M KOH / 1000 mLmol KOH1 mol HCL /1 mol KOHmol HCl1000mL / 0.321 mol HCl ml HCL solution5.0 mL KOH solutionx1.284 M KOH1000 mLx1 mol HCl1 mol KOHx1000 mL0.321 mol HCl= 20 mL0.321 M HCl Percentage yield expresses the ratio ofactual yield to theoretical yield:% yield = actual yield x 100theoretical yield18Percentage yield• The actual yield of a chemical reaction is usuallyless than the theoretical yield predicted by astoichiometry calculation because:• reactants may be impure• the reaction may not go to completion• other reactions may occur (side reactions)2H2 + O2 2H2OA + B C + DX + YDetermine the percentage yield if 6.97 grams of ammonia isproduced from the reaction of 6.22 grams of nitrogen with excesshydrogen.Example: Calculate yields in theHaber process• The Haber Process is a method for producing ammoniadeveloped by Germany during World War I. The Germans usedthe ammonia as a source of nitrogen for making explosives.• The Haber process now produces 100 million tons of nitrogenfertilizer per year, mostly in the form of anhydrous ammonia,ammonium nitrate, and urea. 3-5% of world natural gasproduction is consumed in the Haber process (~1-2% of theworld's annual energy supply)2 2 3N 3H 2NH+ !19GIVEN: 6.22 g N2WANTED: g NH3 (theo)PER/PATH:g N228.02 g N2/mol N2mol N22 mol NH3/1 mol N2mol NH317.03 g NH3/mol NH3g NH3STEP 1: Calculate the theoretical yield2 2 3N 3H 2NH+ !PER/PATH:g N228.02 g N2/mol N2mol N22 mol NH3/1 mol N2mol NH317.03 g NH3/mol NH3g

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