CHAPTER 9StoichiometryOnline ChemistrySection 1Introduction toStoichiometrySection 2Ideal StoichiometricCalculationsSection 3Limiting Reactants andPercentage YieldOnline Labs include:Stoichiometry and GravimetricAnalysisPremium ContentWhy It Matters VideoHMDScience.comStoichiometry(c) Richard Megna/Fundamental PhotographsHMDScience.com
Section 1Introduction toStoichiometryKey TermsMain IdeaRatios of substances inchemical reactions can be usedas conversion factors. composition stoichiometryreaction stoichiometrymole ratioMuch of our knowledge of chemistry is based on the careful quantitative analysisof substances involved in chemical reactions. Composition stoichiometry dealswith the mass relationships of elements in compounds. Reaction stoichiometryinvolves the mass relationships between reactants and products in a chemicalreaction. Reaction stoichiometry, the subject of this chapter, is based on chemicalVirginia standardsCH.4 The student will investigateand understand that quantities in achemical reaction are based on molarrelationships. Key concepts include:CH.4.a Avogadro’s principle andmolar volume.CH.4.b stoichiometric relationships.CH.1.EKS-19; CH.1.EKS-26; CH.4.EKS-2equations and the law of conservation of mass. All reaction stoichiometrycalculations start with a balanced chemical equation. This equation gives therelative numbers of moles of reactants and products.Main IdeaRatios of substances in chemical reactions can beused as conversion factors.The reaction stoichiometry problems in this chapter can be classifiedaccording to the information given in the problem and the informationyou are expected to find, the unknown. The given and the unknown mayboth be reactants, they may both be products, or one may be a reactantand the other a product. The masses are generally expressed in grams,but you will encounter both large-scale and micro scale problems withother mass units, such as kg or mg. Stoichiometric problems are solved byusing ratios from the balanced equation to convert the given quantity.Problem Type 1: Given and unknown quantities are amounts in moles.When you are given the amount of a substance in moles and asked tocalculate the amount in moles of another substance in the chemicalreaction, the general plan isamount ofgiven substance (mol) amount ofunknown substance (mol)Problem Type 2: Given is an amount in moles and unknown is a massthat is often expressed in grams.When you are given the amount in moles of one substance and asked tocalculate the mass of another substance in the chemical reaction, thegeneral plan isamount ofgiven substance(mol)amount ofmass of unknown substance unknown substance(mol)(g)Stoichiometry283
Problem Type 3: Given is a mass in grams and unknown is an amountin moles.When you are given the mass of one substance and asked to calculate theamount in moles of another substance in the chemical reaction, thegeneral plan ismass ofgivensubstance(g) amount ofgivensubstance(mol) amount ofunknownsubstance(mol)Problem Type 4: Given is a mass in grams and unknown is a massin grams.When you are given the mass of one substance and asked to calculate themass of another substance in the chemical reaction, the general plan ismass ofgivensubstance(g) amount ofgivensubstance(mol) amount ofunknownsubstance(mol) mass ofunknownsubstance(g)Mole RatioChemical TechnicianChemical technicians are highly skilledscientific professionals who bringvaluable skills to the developmentof new products, the processingof materials, the managementof hazardous waste, regulatorycompliance, and many other aspectsof getting products and services tothe consumer. Chemical techniciansmust have a solid background inapplied chemistry and mathematicsand be highly skilled in laboratorymethods. Earning an associate’sdegree in applied science or chemicaltechnology is one good way toprepare for this career. Many chemicaltechnicians have a bachelor’s degreein chemical technology, chemistry, orother sciences.Solving any reaction stoichiometry problem requires the use of a moleratio to convert from moles or grams of one substance in a reaction tomoles or grams of another substance. A mole ratio is a conversion factorthat relates the amounts in moles of any two substances involved in achemical reaction. This information is obtained directly from the balancedchemical equation. Consider, for example, the chemical equation for theelectrolysis of melted aluminum oxide to produce aluminum and oxygen.2Al2O3(l) 4Al(s) 3O2(g)Recall from your study of equations and reactions that the coefficients ina chemical equation satisfy the law of conservation of matter and represent the relative amounts in moles of reactants and products. Therefore,2 mol of aluminum oxide decompose to produce 4 mol of aluminum and3 mol of oxygen gas. These relationships can be expressed in the following mole ratios.2 mol Al2O34 mol Al or 4 mol Al2 mol Al2O32 mol Al2O33 mol O2 or 3 mol O22 mol Al2O33 mol O 4 mol Al or 2 3 mol O24 mol AlFor the decomposition of aluminum oxide, the appropriate mole ratio isused as a conversion factor to convert a given amount in moles of onesubstance to the corresponding amount in moles of another substance.284Chapter 9
To determine the amount in moles of aluminum that can be producedfrom 13.0 mol of aluminum oxide, the mole ratio needed is that of Al toAl2O3.13.0 mol Al2O3 4 mol Al 26.0 mol Al2 mol Al2O3Mole ratios are exact, so they do not limit the number of significantfigures in a calculation. The number of significant figures in the answeris therefore determined only by the number of significant figures of anymeasured quantities in a particular problem.Molar MassRecall that the molar mass is the mass, in grams, of one mole of asubstance. The molar mass is the conversion factor that relates themass of a substance to the amount in moles of that substance. To solvereaction stoichiometry problems, you will need to determine molarmasses using the periodic table.Returning to the previous example, the decomposition of aluminumoxide, the rounded masses from the periodic table are the following.1 mol Al2O3 101.96 g 1 mol Al 26.98 g 1 mol O2 32.00 gThese molar masses can be expressed by the following conversion factors.1 mol Al2O3101.96 g Al2O3 or 1 mol Al2O3101.96 g Al2O326.98 g Al1 mol Al or 1 mol Al26.98 g Al32.00 g O21 mol O2 or 32.00 g O21 mol O2To find the number of grams of aluminum equivalent to 26.0 mol ofaluminum, the calculation would be as follows.26.98 g Al 701 g Al26.0 mol Al 1 mol AlSection 1 Formative ASSESSMENTReviewing Main Ideas1. What is stoichiometry?2. For each equation, write all possible mole ratios.a. 2HgO(s) 2Hg(l) O2(g)b. 4NH3(g) 6NO(g) 5N2(g) 6H2O(l)Critical Thinking4. relating ideas What step must beperformed before any stoichiometryproblem is solved? Explain.3. How is a mole ratio used in stoichiometry?Stoichiometry285
Chemistry EXPLORERSThe Case ofCombustionPeople throughout history have transformedsubstances by burning them in air. Yet at the dawnof the scientific revolution, very little was knownabout the process of combustion. In attempting to explain thiscommon phenomenon, chemists of the 18th century developedone of the first universally accepted theories in their field.But as one man would show, scientific theories do not alwaysstand the test of time.Changing AttitudesShunning the ancient Greek approach of logical argument basedon untested premises, investigators of the 17th century began tounderstand the laws of nature by observing, measuring, andperforming experiments on the world around them. However, thisscientific method was incorporated into chemistry slowly.Although early chemists experimented extensively, mostconsidered measurement to be unimportant. This viewpointhindered the progress of chemistry for nearly a century.A Flawed TheoryThe phlogiston theory was used to explain many chemicalobser vations of the day. For example, a lit candle under a glassjar burned until the surrounding air became satu rated withphlogiston, at which time the flame died because the air insidecould not absorb more phlogiston.A New Phase of StudyBy the 1770s, the phlogiston theory had gained universalacceptance. At that time, chemists also began to experimentwith air, which was generally believed to be an element.286Antoine Laurent Lavoisier helped establishchemistry as a science.In 1772, when Daniel Rutherford found that a mouse kept ina closed container soon died, he explained the results based onthe phlogiston theory. Like a burning candle, the mouseemitted phlogiston; when the air could hold no morephlogiston, the mouse died. Thus, Ruther ford figured that theair in the container had become “phlogisticated air.”A couple of years later, Joseph Priestley obtained a reddishpowder when he heated mercury in the air. He assumed thatthe powder was mercury devoid of phlogiston. But when heheated the powder, an unexpected result occurred: Metallicmercury, along with a gas that allowed a candle to burn,formed. Following the phlogiston theory, he believed this gasthat supports combustion to be “dephlogisti cated air.” Stefano Bianchetti/CorbisBy 1700, combustion was assumed to be the decompositionof a material into simpler substances. People saw burningsubstances emitting smoke and energy as heat and light.To account for this emission, scientists proposed a theorythat combustion depended on the emission of a substancecalled phlogiston (Greek, meaning “burned”), which appearedas a combination of energy as heat and light while thematerial was burning but which could not be detectedbeforehand.
Nice Try, But . . .Antoine Laurent Lavoisier was a meticulous scientist. He realizedthat Rutherford and Priestley had carefully observed anddescribed their experiments but had not measured the mass ofanything. Unlike his colleagues, Lavoisier knew the importanceof using a balance. He measured the masses of reactants andproducts and compared them. He observed that the total mass ofthe reactants equaled the total mass of the products. Based onthese observations, which supported what would become knownas the law of conservation of mass, Lavoisier endeavored toexplain the results of Rutherford and Priestley.Lavoisier put some tin in a closed vessel and weighed theentire system. He then burned the tin. When he opened thevessel, air rushed into it as if something had been removedfrom the air in the vessel during combustion. He thenmeasured the mass of the burned metal and observed thatthis mass was greater than the mass of the original tin.Curiously, this increase in mass equaled the mass of the airthat had rushed into the vessel. To Lavoisier, this change inmass did not support the idea of phlogis ton escaping theburning material. Instead, it indicated that duringcombustion, part of the air reacted with the tin.After obtaining similar results by using various substances,Lavoisier concluded that air was not an element but amixture composed principally of two gases, Priestley’s“dephlogisticated air” (which Lavoisier renamed oxygen) andRuther ford’s “phlogisticated air” (which was mostly nitrogenbut had traces of other nonflammable atmospheric gases).When a substance burned, it chemically combined withoxygen, resulting in a product Lavoisier named an oxide.Lavoisier’s t heory of combustion persists today. He used thename oxygen because he thought that all acids containedoxygen. Oxygen means “acid former.”The Father of ChemistryBy emphasizing the importance of quantitative analysis,Lavoisier helped establish chemistry as a science. His work oncombustion laid to rest the phlogiston theory and the theorythat air is an element. He also explained why hydrogen burnedin oxygen to form water, or hydrogen oxide. He later publishedone of the first chemistry textbooks, which established acommon naming system of compounds and elements andhelped unify chemistry worldwide. These accomplishmentsearned Lavoisier the reputation of being the father of chemistry.Questions1. Why does the mass of tin increase when tin is heated in air?TABLE OF SIMPLE SUBSTANCESSimple substances belonging to all the kingdomsof nature, which may be considered as the elements of bodies.New NamesLightCaloricOxygenAzoteHydrogenCorrespondent old Names2. What was the composition of Priestley’s “dephlogisticatedair” and Rutherford’s “phlogisticated air”?Light.Heat.Principle or element of heat.Fire. Igneous fluid.Matter of fire and of heat.Dephlogisticated air.Empyreal air.Vital air, orBase of vital air.Phlogisticated air or gas.Mephitis, or its base.Inflammable air or gas,or the base of inflammable air.Lavoisier’s concept of simple substances was publishedin his book Elements of Chemistry in 1789.287Linda Wilbourn
Section 2Main IdeasBalanced equations giveamounts of reactants andproducts under ideal conditions.Mole-to-gram calculationsrequire two conversion factors.Gram-to-mole conversionsrequire the molar mass ofthe given substance andthe mole ratio.Mass-to-mass calculations usethe mole ratio and the molarmasses of the given andunknown substances. Virginia standardsCH.4.b The student will investigateand understand that chemical quantities arebased on molar relationships. Key conceptsinclude: stoichiometric relationships.CH.1.EKS-19; CH.4.EKS-2Ideal StoichiometricCalculationsA balanced chemical equation is the key step in all stoichiometric calculations,because the mole ratio is obtained directly from it. Solving any reactionstoichiometry problem must begin with a balanced equation.Chemical equations help us plan the amounts of reactants to use in a chemicalreaction without having to run the reaction in the laboratory. The reactionstoichiometry calculations described in this chapter are theoretical. They tell usthe amounts of reactants and products for a given chemical reaction under idealconditions, in which all reactants are completely converted into products. However,many reactions do not proceed such that all reactants are completely converted intoproducts. Theoretical stoichiometric calculations allow us to determine the maximumamount of product that could be obtained in a reaction when the reactants are notpure or when by-products are formed in addition to the expected products.Solving stoichiometric problems requires practice. Practice by working thesample problems in the rest of this chapter. Using a logical, systematic approachwill help you successfully solve these problems.Main IdeaBalanced equations give amounts of reactants andproducts under ideal conditions.In these stoichiometric problems, you are asked to calculate the amountin moles of one substance that will react with or be produced from thegiven amount in moles of another substance. The plan for a simple moleconversion problem isamount ofgiven substance (mol) amount ofunknown substance (mol)This plan requires one conversion factor—the stoichiometric mole ratioof the unknown substance to the given substance from the balancedequation. To solve this type of problem, simply multiply the knownquantity by the appropriate conversion factor (see Figure 2.1).given quantity conversion factor unknown quantityFigure 2.1Mole-to-Mole Conversions Thisis a solution plan for problems in whichthe given and unknown quantities areexpressed in moles.Amount ofgivensubstance(mol)GIVEN INTHE PROBLEM288Chapter 9Mole ratio(Balanced equation)mol unknownmol givenCONVERSION FACTOR Amount ofunknownsubstance(mol)CALCULATED
Stoichiometric Calculations Using Mole RatiosSample Problem A In a spacecraft, the carbon dioxide exhaled by astronauts can beremoved by its reaction with lithium hydroxide, LiOH, according to the followingchemical equation.CO2(g) 2LiOH(s) Li2CO3(s) H2O(l)How many moles of lithium hydroxide are required to react with 20 mol CO2, the averageamount exhaled by a person each day?AnalyzeGiven:amount of CO2 20 molUnknown:amount of LiOH (mol)amount of CO2 (mol) amount of LiOH (mol)PLANThis problem requires one conversion factor—the mole ratio of LiOH to CO2.The mole ratio is obtained from the balanced chemical equation. Because youare given moles of CO2, select a mole ratio that will cancel mol CO2 and giveyou mol LiOH in your final answer. The correct ratio has the following units.mol LiOH mol CO2This ratio cancels mol CO2 and gives the units mol LiOH in the answer.mol ratio mol LiOH mol LiOHmol CO2 mol CO2Substitute the values in the equation in step 2, and compute the answer.Solve 2 mol LiOH 40 mol LiOH20 mol CO2 1 mol CO2CHECK YOURWORKThe answer is written correctly with one significant figure to match thenumber of significant figures in the given 20 mol CO2, and the units cancel toleave mol LiOH, which is the unknown. The balanced equation shows thattwice the amount in moles of LiOH reacts with CO2. Therefore, the answershould be 2 20 40.Answers in Appendix E1. Ammonia, NH3, is widely used as a fertilizer and as an ingredient in many householdcleaners. How many moles of ammonia are produced when 6 mol of hydrogen gas reactwith an excess of nitrogen gas?2. The decomposition of potassium chlorate, KClO3, into KCl and O2 is used as a source ofoxygen in the laboratory. How many moles of potassium chlorate are needed to produce15 mol of oxygen gas?Stoichiometry289
Main IdeaMole-to-gram calculations require two conversionfactors.In these stoichiometric calculations, you are asked to calculate the mass(usually in grams) of a substance that will react with or be produced froma given amount in moles of a second substance. The plan for thesemole-to-gram conversions isamount ofgiven substance(mol)check for understandingExplain Why might it be importantfor a chemist to be able to calculate,in advance, how much product achemical reaction would yield?amount ofmass of unknown substance unknown substance(mol)(g)This plan requires two conversion factors—the mole ratio of the unknownsubstance to the given substance and the molar mass of the unknownsubstance for the mass conversion. To solve this kind of problem, yousimply multiply the known quantity, which is the amount in moles, by theappropriate conversion f actors (see Figure 2.2 on the next page).Stoichiometric Calculations Using Mole RatiosSample Problem B In photosynthesis, plants use energy from the sun to produceglucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, ingrams, of glucose is produced when 3.00 mol of water react with carbon dioxide?AnalyzePLANGiven:amount of H2O 3.00 molUnknown:mass of C6H12O6 produced (g)You must start with a balanced equation.6CO2(g) 6H2O(l) C6H12O6(s) 6O2(g)Given the amount in mol of H2O, you need to get the mass of C6H12O6 ingrams. Two conversion factors are needed—the mole ratio of C6H12O6 to H2Oand the molar mass of C6H12O6.mol ratiomolar mass factorO6 g C6H12 mol C6H12O6 g C6H12O6mol H2O mol C6H12O6mol H2OSolveUse the periodic table to compute the molar mass of C6H12O6.C6H12O6 180.18 g/mol180.18 g C6H12O61 mol C6H12O63.00 mol H2O 90.1 g C6H12O6 1 mol C6H12O66 mol H2OCHECK YOURWORK290Chapter 9The answer is correctly rounded to three significant figures, to match those in3.00 mol H2O. The units cancel in the problem, leaving g C6H12O6 as the unitsfor the answer, which matches the unknown. The answer is reasonablebecause it is one-half of 180.
Figure 2.2Mole-to-Gram Conversions This is a solution plan for problems in which thegiven quantity is expressed in moles, and the unknown quantity is expressed in grams.Amount ofgivensubstance(in mol)Mole ratio(Equation)Molar mass(Periodic table)mol givenmol unknownMolar mass of unknown(in g/mol) CONVERSION FACTORSMass ofunknownsubstance(in g)CALCULATEDGIVEN INTHE PROBLEMStoichiometric Calculations Using Mole RatiosSample Problem C What mass of carbon dioxide, in grams, is needed to react with 3.00mol H2O in the photosynthetic reaction described in Sample Problem B?AnalyzeGiven:Linda Wilbournamountof H2O 3.00 molUnknown:mass of5thCOPASS2 (g)8/20/97MC99PEC09000003AThe chemical equation from Sample Problem B isPLAN6CO2(g) 6H2O(l) C6H12O6(s) 6O2(g).Two conversion factors are needed—the mole ratio of CO2 to H2O and themolar mass factor of CO2.mol ratioSolvemolar mass factorg CO mol CO2 2 g CO2mol H2O mol CO2mol H2OUse the periodic table to compute the molar mass of CO2.CO2 44.01 g/mol44.01 g CO26 mol CO2 132 g CO23.00 mol H2O 6 mol H2O1 mol CO2CHECK YOURWORKThe answer is rounded correctly to three significant figures to match those in3.00 mol H2O. The units cancel to leave g CO2, which is the unknown. Theanswer is close to an estimate of 120, which is 3 40.Answers in Appendix E1. When magnesium burns in air, it combines with oxygen to form magnesium oxideaccording to the following equation.2Mg(s) O2(g) 2MgO(s)What mass in grams of magnesium oxide is produced from 2.00 mol of magnesium?2. What mass of glucose can be produced from a photosynthesis reaction that occurs using10 mol CO2?6CO2(g) 6H2O(l) C6H12O6(aq) 6O2(g)Stoichiometry291
Main IdeaGram-to-mole conversions require the molar mass ofthe given substance and the mole ratio.In these stoichiometric calculations, you are asked to calculate theamount in moles of one substance that will react with or be producedfrom a given mass of another substance. In this type of problem, you arestarting with a mass (probably in grams) of some substance. The plan forthis conversion isamount ofamount ofmass ofgiven substance given substance unknown substance(g)(mol)(mol)This route requires two additional pieces of data: the molar mass ofthe given substance and the mole ratio. The molar mass is determined byusing masses from the periodic table. We will follow a procedure muchlike the one used previously by using the units of the molar mass conversion factor to guide our mathematical operations. Because the knownquantity is a mass, the conversion factor will need to be 1 mol divided bymolar mass. This conversion factor cancels units of grams and leavesunits of moles (see Figure 2.3 below).You should take time at this point to look at each type of stoichiometric conversion and make note that in every type, you must begin with acorrectly balanced chemical equation. It is important to remember thatwithout a balanced equation, you will not have an accurate molar ratioand will not be able to calculate the correct molar mass to use in yourconversions.If you are still struggling with balancing equations, it is highlyrecommended that you continue to practice the process. The skill ofbalancing chemical equations will be used continuously throughout theremainder of the course.Figure 2.3Gram-to-Mole Conversions This is a solution plan for problems in which thegiven quan tity is expressed in grams, and the unknown quantity is expressed in moles.Mass ofgivensubstance(g)Molar mass factor(Periodic table)Mole ratio(Balanced equation)1 mol givenMolar mass ofgiven (g)mol unknownmol givenCONVERSION FACTORSGIVEN INTHE PROBLEM292Chapter 9 Amount ofunknownsubstance(mol)CALCULATED
Premium ContentStoichiometric Calculations Using Mole RatiosLearn It! VideoSample Problem D The first step in the industrial manufacture ofnitric acid is the catalytic oxidation of ammonia.Solve It! CardsHMDScience.comHMDScience.comNH3(g) O2(g) NO(g) H2O(g) (unbalanced)The reaction is run using 824 g NH3 and excess oxygen.a. How many moles of NO are formed?b. How many moles of H2O are formed?AnalyzeGiven:mass of NH3 824 gUnknown:a. amount of NO produced (mol)b. amount of H2O produced (mol)First, write the balanced chemical equation.PLAN4NH3(g) 5O2(g) 4NO(g) 6H2O(g)Two conversion factors are needed to solve part (a)—the molar mass factor forNH3 and the mole ratio of NO to NH3. Part (b) starts with the same conversionfactor as part (a), but then the mole ratio of H2O to NH3 is used to convert tothe amount in moles of H2O. The first conversion factor in each part is themolar mass factor of NH3.molar mass factormol ratio 1 mol NHmol NO 3 mol NH mol NOa. g NH3 g NH33molar mass factormol ratio 1 mol NHmol H2O 3 mol H2O b. g NH3 g NH3mol NH3Use the periodic table to compute the molar mass of NH3.Solve1 mol NH3 17.04 g /mol1 mol NH3a. 824 g NH3 4 mol NO 48.4 mol NO17.04 g NH34 mol NH31 mol NH36 mol H2Ob. 824 g NH3 72.5 mol H2O 17.04 g NH34 mol NH3CHECK YOURWORKThe answers are correctly given to three significant figures. The units cancelin the two problems to leave mol NO and mol H2O, respectively, which arethe unknowns.Answers in Appendix EOxygen was discovered by Joseph Priestley in 1774 when he heated mercury(II) oxide to decompose it toform its constituent elements.1. How many moles of mercury(II) oxide, HgO, are needed to produce 125 g of oxygen, O2?2. How many moles of mercury are produced?Stoichiometry293
Figure 2.4Mass-to-Mass Conversions This is a solution plan for problems in which thegiven quantity is expressed in grams, and the unknown quantity is also expressed in grams.Mass ofgivensubstance(g)Molar mass factor(Periodic table)Mole ratio(Balanced equation)Molar mass factor(Periodic table)1 mol givenMolar mass ofgiven (g)mol unknownmol givenMolar massof unknown (g)1 mol unknownMass ofunknownsubstance(g)CONVERSION FACTORSGIVEN INTHE PROBLEMCALCULATEDMain IdeaMass-to-mass calculations use the mole ratio andthe molar masses of the given and unknownsubstances.Mass-mass calculations are more practical than other mole calculationsyou have studied. You can never measure moles directly. You are generally required to calculate the amount in moles of a substance from itsmass, which you can measure in the lab. Mass-mass problems can beviewed as the combination of the other types of problems. The plan forsolving mass-mass problems ismass ofgivensubstance(g) amount ofgivensubstance(mol) amount ofunknownsubstance(mol) mass ofunknownsubstance(g)Three additional pieces of data are needed to solve mass-mass problems:the molar mass of the given substance, the mole ratio, and the molarmass of the unknown substance (see Figure 2.4 above).Stoichiometric Calculations Using Mole RatiosSample Problem E Tin (II) fluoride, SnF2, is used in some toothpastes. It is made by thereaction of tin with hydrogen fluoride according to the following equation.Sn(s) 2HF(g) SnF2(s) H2(g)How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?AnalyzePLANGiven:amount of HF 3.00 gUnknown:mass of SnF2 produced (g)The conversion factors needed are the molar masses of HF and SnF2 and themole ratio of SnF2 to HF.molar mass factormol ratiomolar mass factor g SnF mol SnF2 mol HF2 g SnF2 g HF g HF mol SnFmol HF2Continued294Chapter 9
Stoichiometric Calculations Using Mole Ratios(continued)Use the periodic table to compute the molar masses of HF and SnF2.Solve1 mol HF 20.01 g1 mol SnF2 156.71 g156.71 g SnF21 mol SnF21 mol HF 117.5 g SnF230.00 g HF 20.01g HF2 mol HF1 mol SnF2CHECK YOURWORKThe answer is correctly rounded to four significant figures. The units cancel toleave g SnF2, which matches the unknown. The answer is close to an estimatedvalue of 120.Answers in Appendix E1. Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry. It isproduced when ammonium nitrate is decomposed according to the following reaction.NH4NO3(s) N2O(g) 2H2O(l)a. How many grams of NH4NO3 are required to produce 33.0 g N2O?b. How many grams of water are produced in this reaction?2. When copper metal is added to silver nitrate in solution, silver metal and copper(II) nitrateare produced. What mass of silver is produced from 100. g Cu?3. What mass of aluminum is produced by the decomposition of 5.0 kg Al2O3?Section 2 Formative ASSESSMENTReviewing Main Ideas1. Balance the following equation. Then, given themoles of reactant or product below, determinethe corresponding amount in moles of each ofthe other reactants and products.NH3 O2 N2 H2Oa. 4 mol NH3 b. 4 mol N2 c. 4.5 mol O22. One reaction that produces hydrogen gas can berepresented by the following unbalanced chemical equation:Mg(s) HCl(aq) MgCl2(aq) H2(g)a. W hat mass of HCl is consumed by thereaction of 2.50 moles of magnesium?b. W hat mass of each product is produced inpart (a)?3. Acetylene gas, C2H2, is produced as a result ofthe following reaction:CaC2(s) 2H2O(l) C2H2(g) Ca(OH)2(aq)a. I f 32.0 g CaC2 are consumed in this reaction,how many moles of H2O are needed?b. H ow many moles of each product wouldform?4. When sodium chloride reacts with silver nitrate,silver chloride precipitates. What mass of AgCl isproduced from 75.0 g AgNO3?Critical Thinking5. RELATING IDEAS Carbon and oxygen react toform carbon monoxide: 2C O2 2CO. Whatmasses of carbon and oxygen are needed tomake 56.0 g CO? Which law does this illustrate?Stoichiometry295
Section 3Main IdeasOne reactant limits the productof a reaction.Comparing the actual andtheoretical yields helpschemists determine thereaction’s efficiency. Limiting Reactantsand Percentage YieldKey Termslimiting reactantexcess reactantVirginia standardsCH.1 The student will investigate andunderstand that experiments in which variablesare measured, analyzed, and evaluatedproduce observations and verifiable data. Keyconcepts include:CH.1.a designated laboratory techniques.CH.1.b safe use of chemicals and equipment.CH.4.b The student will investigateand understand that chemical quantities arebased on molar relationships. Key conceptsinclude: stoichiometric relationships.CH.4.EKS-3; CH.4.EKS-4PremiumContenttheoretical yieldactual yieldpercentage yieldIn the laboratory, a reaction is rarely carried out with exactly the required amount ofeach of the reactants. In many cases, one or more reactants is present in excess;that is, there is more than the exact amount requir
HMDScience.com Online labs include: Stoichiometry and Gravimetric Analysis toichiometr Section 1 Introduction to Stoichiometry Section 2 Ideal Stoichiometric Calculations Section 3 Limiting Reactants and Percentage Yield Why it Matters Video HMDScience.com Premium Content Stoichiometry CHAPTER 9