11/18/2013111A188A11H1.0079722APeriodic 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 109Wed, Nov 20• Lecture 21 (Change)– Last example from L20– Mass Calculations (9.3)– The Concept of Limiting Reactants (9.4)• Questions we’ll answer:– What is the maximum mass of product we can obtain when one of the reactants is completely consumed?– What happens when we have limited amounts of each reactant?11/18/20132Example. One method of producing nitric acid (HNO3) involves a three step process:1. 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)2. 2 NO(g) + O2(g)  2 NO2(g)3. 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g)How much HNO3can be produced from 16.37 mol of NH3? Assume all other reactants are present in excess.4Stoichiometry and Mass• In the lab, we measure quantities of substances in grams, milliliters, etc rather than in moles directly. We can use molar masses to convert between the macroscopic and microscopic representations.• Example. Powdered aluminum and finely ground iodine react violently to produce aluminum iodide How many grams of ground iodine is required to react completely with 35.0 g of aluminum powder? (http://youtu.be/ELcZduuAx9k, http://youtu.be/SKSU72‐1ERc) • We need to construct the mole ratios from our balanced chemical equation.2Al(s) + 3I2(s)  2AlI3(s)2 mol Al 3 mol I2~2 mol Al3 mol I211/18/20133Stoichiometry and Mass (cont)Example, cont. How many grams of I2are required to react with 35.0 g Al?2Al(s) + 3I2(s)  2AlI3(s)1. Convert grams of Al to moles of Al.2. Convert moles of Al to moles of I2.3. Convert moles of I2to grams of I2.1 mol Al26.98 g Al35.0 g Al 1.297 mol Al=1.297 mol Al2 mol Al3 mol I21.945 mol I2=1 mol I2253.8 g I2493.87 g I2=494 g I2=1.945 mol I2Stoichiometry and Mass (cont)Example, cont. How many grams of I2are required to react with 35.0 g Al?2Al(s) + 3I2(s)  2AlI3(s) Alternatively, we can write the solution in one line:? g I22 mol Al3 mol I2=1 mol Al26.98 g Al35.0 g Al1 mol I2253.8 g I2493.87 g I2=494 g I2=11/18/20134Exercise. How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? 1. Convert grams of Cr to moles of Cr.2. Convert moles of Cr to moles of Cr2O3.3. Convert moles of Cr2O3to grams of Cr2O3.223Cr( ) + O ( ) Cr O ( )sgs423? g produced15.0 g151.990 g/mol51.996 g/molexcessHow many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? Exercise223Cr( ) + O ( ) Cr O ( )sgs423? g produced15.0 g151.990 g/mol51.996 g/molexcess11/18/201359Limiting Reactants• Given a balanced chemical equation, the amount of reactants available to react determines the amount of products we can produce. one whole chocolate cakeone half of a chocolate cake1 ¾ cup cake flour3 tsp baking powder2 oz. baking chocolate1 ½ cup sugar ½ cup butter4 eggs½ cup milk1 ¾ cup cake flour3 tsp baking powder2 oz. baking chocolate1 ½ cup sugar ½ cup butter2 eggs½ cup milkThe same kind of thing happens in chemical reactions. Often one reactant is completely used up before the other reactant(s) are.Balloon Rxn: H2(g) + 2 O2(g)  2 H2O(g)A stoichiometric mixture contains exactly enough moles of each reactant, relative to the balanced chemical equation, for each reactant to be completely consumed. Stoichiometric vs. LimitedA limited reactant mixture contains fewer moles of one of the reactants than is necessary, relative to the balanced chemical reaction, to completely consume the other reactant.Consider the reaction: N2(g) + 3H2(g)  2NH3(g)11/18/20136KeyA2B2KeyO2H2Stoichiometric or Limited? Microscopic Scale KeyB2AB2KeyN2H2N2+ 3 H2 2 NH32 H2+ O2 2 H2OA2+ 3 B2  2 AB32AB2+ B2  2 AB3Stoichiometric 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 molLimiting Reactant?Exp. 1Exp. 2Exp. 3Initial ChangeFinalInitial ChangeFinalInitial ChangeFinal11/18/20137Stoichiometric or Limited? Macroscopic Scale N2(g) + 3H2(g) 