CHECAL1Dr. Susan A RocesAY 2011-2012Stoichiometry and Composition RelationshipsObjectives:1. Write and balance chemical reaction equations.2. Know the products of common reaction given the reactants.3. Calculate the stoichiometric quantities of reactants and products given the chemical reactions.4. Define excess reactant, limiting reactant, conversion, degreeofcompletion, selectivity, yieldinareactionand extent of reaction.5. Identify the limiting and excess reactants and calculate the percent excess reactants, the percent conversion, the percent completion, and yield for a chemical reaction with the reactants being in non-stoichiometric proportions.Definitions:Stoichiometry deals with combining of elements and compoundsChemical Equation relates the moles of reactants to the moles of products and it does not indicate the true mechanism of the reaction or how fast or to what extent the reaction will take place.Potassium chloratePotassium chloride+Oxygen gasKClO3KCl+ O21CHECAL1Dr. Susan A RocesAY 2011-2012When written in this manner, however, the chemical equation does not satisfy the Law of Conservation of Matter or Mass.Hence, wemustbalanceitbycountingtheatomsoneach side of the arrow and making sure that the number of one kind of atom is the same on both sides.Stoichiometric ratios ratios obtained from the numerical coefficients in the chemical equation that permits us to calculate the moles of one substance as related to the moles of another substance in the chemical equation.Stoichiometric coefficients coefficient or the number before any reactant or product in chemical reaction. Subscript is the number written after the symbol of the element on its lower right side.We start by assigning a trial coefficient to any one of the substancesintheequation, multiplythiscoefficientbythe subscript of one of its elements.2KClO32KCl +3O22 moles ofpotassium chlorate yields ( or produce )2 moles of potassium chloride and 3 moles of oxygen2CHECAL1Dr. Susan A RocesAY 2011-2012Units:mole,lb mole,g mole,kg moleAnybalancedchemical equationisthereforeanalgebraic equation which indicates all of the reactants on the left side and all of the products on the right side of the arrow.Note:arrow pointing upward () escapes as gas arrow pointing downward ( ) formation of precipitatesdouble arrow() reverse reaction takes placeThe variables temperature, pressure, catalysts, etc. may be written above or below the equation arrow.Quantitative Information from Chemical Equations:1.Molecular and Molar relations Mole is a certain number of molecules, atoms, electrons, or other specified types of particles. is defined as the amount of substance equal to its formula weight (molar mass)Example:The equation for the preparation of phosphorous in an electric furnace:Ca3(PO4)2+3SiO2+5C 3CaSiO3 + 5CO+2P1 mole3 moles 5 moles3 moles5 moles2moles3CHECAL1Dr. Susan A RocesAY 2011-2012The relative numbers of reacting and resulting molecules are indicated by the coefficients of the formulas representing these molecules.Some possible conversion factors:etc ,) (PO Ca mole 1P moles 2 ,CO moles 5SiO moles 32 4 32Note:1. SI units:1 mol=1 g-mole=6.02 x 1023 molecules2. Other units :1 lb mol=6.02 x 1023x453.6 molecules1kg mol=1kilo mol=1k mol =1000 x 6.02 x 1023 molecules2.Weight Relations To convert the number of moles to mass, we make use of the molecular weightmass in gram ( g )=( MW ) (g mol )mass in pound ( lb ) =( MW ) ( lb mol )Examples: 1. How many grams of NH4Cl are there in 5 mol?Solution:Molar mass of NH4Cl = 14 + 4 + 35.5=53.5 gFive (5) mol of NH4Cl=5 x 53.5=267.5 g NH4Cl4CHECAL1Dr. Susan A RocesAY 2011-20122. The preparation of commercial hydrochloric acid by heating NaCl with concentrated sulfuric acid.2NaCl + H2SO4 Na2SO4+ 2HClSolution:2 lb moles 1 lb mole1 lb mole2 lb moles2 ( 58.5 )1 ( 98.1 )1 ( 142.1 ) 2 ( 36.5 )117 lb98.1 lb142.1 lb 73 lbThe Law of Conservation of matter is satisfied since: Mass of reactants= Mass of products 215.1 lb= 215.1 lbPossible conversion factors:4 24 2 4 2SO H lb 98.1SO Na lb 142.1 ' NaCl lb 117SO H lb 98.13. Convert 499 g CuSO4.5H2O into mol. Find equivalent mole of CuSO4in the crystals.Solution:Molar mass of CuSO4 =159.5 gMolar mass of CuSO4.5H2O = 159.5 + 5(1 x 2 +16) = 249.5 g5CHECAL1Dr. Susan A RocesAY 2011-2012mol gmol g ggO HO H gO H25 . 2494995 . CuSOmol gg5 . CuSO5 . CuSO of moles2 42 42 4 In the formula CuSO4.5H2O, the moles of CuSO4 are equal (one in each), hence the equivalent moles of CuSO4 in the crystals are also 2 mol. 3. Gas Volume Relations 4 NH3 (g)+ 3 O22 N2 (g)+6 H2O(vapor)4 moles3 moles 2 moles6 moles4 volumes 3 volumes 2 volumes6 volumes4 liters3 liters 2 liters 6 liters4 ft3 3 ft32 ft3 6 ft3Definition of terms in Industrial Chemical Calculations: In industrial reactors you will rarely find exact stoichiometric mixture of reactants.To make a desired reaction takeplaceortouseupacostlyreactant, excessreactantsare nearlyalwaysused. Thisexcessmaterial comesout together with, or perhaps separately from the product and sometimes can be used again.Even if stoichiometric quantities of reactants are used, the reaction may not be complete or side reaction may occur so that the products will be accompaniedby unused reactants as well as side products.In these circumstances some newdefinitions must be understood:6CHECAL1Dr. Susan A RocesAY 2011-20121.Stoichiometric mixture of reactants a mixture in which the molal proportions of the reactantsareexactlythesameasthoseindicatedbythe stiochiometric coefficients in the balanced chemical equation such that no excess of any constituents is present.2.Limiting reactantthe reactant which is present in the smallest/least stoichiometric proportion. inother words, if two or more reactants are mixed and if the reaction were to proceed according to the chemical equation, whether it does or not , the reactant that would first disappear is termed the limiting reactant.How to determine the limiting reactant:1. Calculatethemoleratio(s)ofthereactantsinthefeedand compare each ratio with the corresponding ratio of the coefficients of the reactants in the chemical equation.C7H16+11 O27 CO2+ 8 H2OGiven : 1 g mol of C7H16and 12 g mol ofO2 are mixedRatio in feed Ratio in chemical equation7CHECAL1Dr. Susan A RocesAY 2011-201211 1 H CO1 211 2 H COt h e o .1 6 72G i v e n1 6 721]1

1]1

Therefore: C7H16 is the limiting reactant2. If more than two reactants are present, you have to use one reactant as the reference substance, calculate the mole ratios of theother reactantsinthefeedrelativetothereference, make pairwisecomparisons versus the analogous ratios in the chemical equation, and rank each compound.A+ 3B +2C productsGiven:1.1 moles of A, 3.2 moles of B, 2.4 moles of C are fed as reactants in the reactorReference substance A3 13 AB2 . 9 11 . 13 . 2 ABT h e o . G i v e n 1]1

1]1

Therefore: Bis the limiting reactant relative to A, B< A2 12 AC2 . 1 81 . 12 . 4 ACT h e o . G i v e n 1]1

1]1

8CHECAL1Dr. Susan A RocesAY 2011-2012Therefore: Ais the limiting reactant relative to C,A < CTherefore:B < A < C, B is the limiting reactant among the set of three reactants.3. Excess reactant the one present in excess of the fixed amount needed to combine with the limiting reactant.Note: Moles in excess can be calculated as the total available moles of a reactant less the moles required to react with the limiting reactant.Excess air=a common term used incombustion reactions=it means the amount of air available to react that is in excess of the air theoretically required to completely burn the combustible material.4. Percent Excess of a Reactant the true excess expressed as a percentage of the theoretically required amount of that reactant. is based on the amount of any excess reactant above theamount requiredtoreact withthelimitingreactant according to the chemical equation. % excess reactant=100x value l theoreticareactant excess of value% excess reactant =100x value l theoretica value al theoretic - amountoriginalExample 3.( Refer to Ex.1 )% excess O2=% 9.1 100x1111 129CHECAL1Dr. Susan A RocesAY 2011-20125.Theoretically required amount of reactant is present in a stoichiometric proportion with the limiting reactant.established by the limiting reactant and can be calculated from the chemicalequation for all other reactants.Note: Evenif onlypart of thelimitingreactant actually reacts, the required and excess quantities are based on theentireamountofthelimitingreactantasifithad reacted completely.6.Conversion the fraction of the feed or some key material in the feedthat is convertedinto products.Percent conversion of any reactant percentage of the amount originally present which reacts.% Conversion = 100 xintroduced feed in the compound /feed of mass /molereact thatfeed in the compound /feed of mass /moleExample. 4 (Refer to example 1)C7H16 +11O27CO2 +8H2O 10 kg C7H6 react completely with O2, and 14.4 kg of CO2 are formed in the reaction (C7H6) reacted 10CHECAL1Dr. Susan A RocesAY 2011-2012= 16 7216 7222H C mol kg 0.0468CO mol kg 7H C mol kg 1 CO kg 44CO mol kg 1CO kg 4 . 14 (C7H6)original 16 716 716 716 7H C mol kg 0.0999H C kg 100.1H C mol kg 1H C kg 10 Therefore:% Conversion= % 46.8 100x 0999 . 00468 . 07.Degree of completion of a reaction is usually the percentage or fraction of the limiting reactant converted into product percentage of limiting reactant which reacts8.Selectivity is the ratio of the moles of a particular ( usually the desired) product produced to the moles of another (usually undesired or by-product) product produced in a set of reactions.9.Yield ( for a single reactant and product) is the weight (mass) or moles of final product divided bytheweight(mass)ormolesofinitialorkeyreactant either fed or consumed.Example:1. A mixture containing 100 g H2 and 100 g O2 is sparked so that water is formed according to the reaction2H2 + O22H2O MWs: 2.023218Required: a)limiting reactant b)% excess H211CHECAL1Dr. Susan A RocesAY 2011-2012Solution:a)moles ofH2=moles 49.5 g/mole 2.02g 100

MWM moles ofO2=moles 3.13mole /g 32g 100 if all the H2will react :moles of requiredO2 2222O moles 24.8H moles 2O mole 15 . 49 x H molesSince this moles of required O2 is greater than the availableO2 (3.13moles), we will not be able touse all the hydrogen. Therefore,O2is the limiting reactant andH2is the reactant in excess. In view of this, we must base the calculations on the amount of O2.Moles of hydrogen that actually reacted : (theoretically required H2)2Hn=3.13 moles O2x222H moles 6.26 O mole 1H moles 2unreacted H2=49.50 6.26=43.24( excess H2 )b)% excess H2=% 690.73 100 x26 . 624 . 43Note: Conversion factors obtained from the balanced equation2222O mole 1H moles 2andH moles 2O mole 112CHECAL1Dr. Susan A RocesAY 2011-20122. Suppose the 10 lb-moles of nitrogen and 32 lb-moles of H2 are reacted at 515 0C and 300 atm but only tow lb-mole of N2 reacts. Determinethedegreeofcompletionofthereaction and the percentage conversion of hydrogen to ammonia.Solution: N2+ 3 H22NH3Before reaction: 10 lb-moles 32 lb-moles 0After reaction:2 x 1 = 22 x 3 = 62 x 2 = 42 lb-molex O moles 1H mole - lb 3222 lb-molex N moles 1NH mole - lb 223At equil.:10 2 = 832 6 = 260 + 4 = 4 a) 3 3 . 031 HN0 . 3 13 21 0 HNt h e o .22G i v e n22 1]1

1]1

Therefore:N2is the limiting reactant % 20 100N moles - lb 10N moles - lb 2 completion%22 x

% 75 . 18 100H moles - lb 32H moles - lb 6 H Conversion % b)222 x3. Diborane, B2H6, a possible propellant for solid fuel rockets, canbemadeinavarietyof different ways. Oneof the 13CHECAL1Dr. Susan A RocesAY 2011-2012simplest but not thecheapest istouselithiumhydrideas follows:6 LiH + 2 BCl3B2H6+ 6 LiClSuppose 200 lb of LiH are mixed with 1000 lb of BCl3 and 45 lb of B2H6 are recovered.a) what is the limiting reactant:b) what is the excess reactant?What is the % excess?c) What is the % conversion of LiH to B2H6?d) How many pounds of LiCl are formed?Given: 6 LiH + 2 BCl3B2H6+ 6 LiCl200 lbs. 1000 lbs. 45 lbs.MWs: 7.948 117.179 27.688 42.393Solution:a)MWW nmoles lb nLiH 164 . 257.948200

moles lb nBCl 534 . 8117.1791000

3moles lb nH B 6253 . 127.68845

6 2Ntotal=27.688Theo GivenBClLiHBClLiH

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3 3??14CHECAL1Dr. Susan A RocesAY 2011-20120 . 3269 5 . 25 3 4 . 81 6 4 . 2 5

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T h e o G i v e nTherefore: LiH is the limiting reactant and BaCl3 is the excess reactantAnother solution:If nBCl 3=8.534:( )moles lbnrequired LiH

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602 . 25BCl moles - lb 2LiH moles - lb 6 BCl moles - lb 8.53433 ) (Since: nLiH (required)>nLiH (available)If nLiH=25.164:( )moles lbnrequired BCl

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388 . 8LiH moles - lb 6BCl moles - lb 2 LiH moles - lb 25.1643) (3Since: nBCl3 (required)