11/20/2013111A188A11H1.0079722APeriodic Table133A144A155A166A177A2He4.002623Li6.9394Be9.01225B10.8116C12.01127N14.00678O15.99949F18.998410Ne20.179311Na22.989812Mg24.30533B44B55B66B77B898B10 111B122B13Al26.981514Si28.08615P30.973816S32.06417Cl35.45318Ar39.948419K39.10220Ca40.0821Sc44.95622Ti47.9023V50.94224Cr51.99625Mn54.938026Fe55.84727Co58.933228Ni58.7129Cu63.5430Zn65.3731Ga65.3732Ge72.5933As74.921634Se78.9635Br79.90936Kr83.80537Rb85.4738Sr87.6239Y88.90540Zr91.2241Nb92.90642Mo95.9443Tc[99]44Ru101.0745Rh102.90546Pd106.447Ag107.87048Cd112.4049In114.8250Sn118.6951Sb121.7552Te127.6053I126.90454Xe131.30655Cs132.90556Ba137.3457La138.9172Hf178.4973Ta180.94874W183.8575Re186.276Os190.277Ir192.278Pt195.0979Au196.96780Hg200.5981Tl204.3782Pb207.1983Bi208.98084Po[210]85At[210]86Rn[222]787Fr[223]88Ra[226]89Ac[227 ]104Ku[260]105 106 107 108 109Fri, Nov 22• Lecture 22 (Change)– Concept of LR example from end of L21– Calculations Involving a Limiting Reactant (9.5)– Percent Yield (9.6)• Questions we’ll answer:– What is the maximum mass of product we can obtain when we have limited amounts of each reactant?– What is the maximum mass of product we can obtain under real experimental conditions?11/20/20132Stoichiometric or Limited? Macroscopic Scale CO(g) + 2 H2(g)  CH3OH(l)1.0 mol0.5 mol1.0 mol2.0 mol2.0 mol1.0 mol0.0 mol0.0 mol0.0 molExp. 1Exp. 2Exp. 3Initial ChangeFinalInitial ChangeFinalInitial ChangeFinalLimiting Reactant?Stoichiometric or Limited? Macroscopic Scale N2(g) + 3H2(g)  2 NH3(g)Limiting Reactant?Exp. 1Exp. 2Exp. 3Initial ChangeFinalInitial ChangeFinalInitial ChangeFinal2.0 mol2.5 mol1.5 mol8.0 mol6.0 mol4.5 mol0.0 mol0.0 mol0.0 mol11/20/20133Stoichiometric or Limited? Macroscopic Scale N2(g) + 3H2(g)  2 NH3(g)Limiting Reactant?Exp. 1Exp. 2Exp. 3Initial ChangeFinalInitial ChangeFinalInitial ChangeFinal2.0 mol‐ 2.0 mol0.0 mol2.5 mol‐ 2.0 mol0.5 mol1.5 mol‐ 1.5 mol0.0 mol8.0 mol‐ 6.0 mol2.0 mol6.0 mol‐ 6.0 mol0.0 mol4.5 mol‐ 4.5 mol0.0 mol0.0 mol+ 4.0 mol4.0 mol0.0 mol+ 4.0 mol4.0 mol0.0 mol+ 3.0 mol3.0 molBoth N2and H2completely consumed…stoichiometric mixture.CO consumed, excess H2remains… CO is limiting.N2consumed, excess H2 remains… N2is limiting.LR Method 1: Suppose 25.0 g of nitrogen gas and 5.00 g of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion.22 3N ( ) + H ( ) NH ( )gg g11/20/20134LR Method 2: Suppose 25.0 g of nitrogen gas and 5.00 g of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion.22 3N ( ) + H ( ) NH ( )gggExercise: When exposed to fluorine gas, white phosphorus (P4) undergoes rapid and violent oxidation to form phosphorus trifluoride gas according to the chemical equation below. In a certain experiment, you react 184 g of white phosphorus with 546 g of fluorine gas. What mass of phosphorus trifluoride is produced in this experiment? P4(s) + F2(g)  PF3(g)11/20/20135Exercise: When exposed to fluorine gas, white phosphorus (P4) undergoes rapid and violent oxidation to form phosphorus trifluoride gas according to the chemical equation below. In a certain experiment, you react 184 g of white phosphorus with 546 g of fluorine gas. What mass of phosphorus trifluoride is produced in this experiment? P4(s) + F2(g)  PF3(g)10Making the LR work for YOU.• Chemistry happens at the particulate level……so the limiting reactant in a chemical equation must be determined by comparing numbers of moles.• To determine which reactant is the LR, we must compare the following quantities… – what we have (based on the problem set up) – what we need (based on the balanced chemical equation)• The mole ratios in a chemical equation are crucial for determining the LR…• The reactant for which you have fewer moles is not necessarily the LR!!11/20/2013611Theoretical, Actual, and Percent Yield• Theoretical yield: the amount of product that would be formed under ideal reaction conditions in which starting materials are completely consumed (up to the LR). This is a calculated number.• Actual yield: the amount of product that is actually produced in the experiment. This is a measured number.• The actual yieldis always less than the theoretical yield because…– starting materials may not be completely consumed– side reactions may occur– the reverse reaction may occur– there may be loss of product during purification steps • Percent Yield: ratio of how much we actually produced to how much we could theoretically produce. actual yield% Yield = 100theoretical yield12N2(g) + 3H2(g)  2NH3(g)25.0 g0.8925 mol5.00 g2.480 mol28.2 gLIMITING REAGENTTheoretical YieldIn a certain experiment, the actual yield of this reaction was only 26.7 g NH3. What was the percent yield?% yield =26.7 g NH328.2 g NH3100= 94.68%Percent yield should never be greater than 100%. If it is, there is something wrong with your theoretical yield calculation, or with the actual yield you obtained experimentally.= 94.7%11/20/20137Exercise: