Unit 3C: Stoichiometry Review

Mole Island Diagram

Unit 3C: Stoichiometry Review1The Mole Atoms are so small, it is impossible to count them by the dozens, thousands, or even millions.

To count atoms, we use the concept of the mole1 mole of atoms = That is, 1 mole of atoms = __________ atoms

The mole is the SI unitfor amount of substance.602,213,673,600,000,000,000,000 atoms 6.02 x 1023

If I had a mole of dollars, I could give every person on EarthHow Big is a Mole? about the size of a chipmunk,weighing about 5 oz. (140 g), andhaving a length of about 7 inches (18 cm).

I Meant, How Big is 6.022 x 1023?BIG.6.022 x 1023 marbles would cover theentire Earth (including the oceans) to a height of 2 miles.

There are ~ 6,880,900,000 people on Earth.$87.5 trillion or $87.5 x 1012

Welcome to Mole Island1 mole = 22.4 L @ STP1 mol = molar mass 1 mol = 6.02 x 1023 particles

4Mole Island Diagram(gases)(gases)MassParticlesVolumeMoleMassVolumeParticles Substance A Substance B1 mole = molar mass (g)Use coefficientsfrom balancedchemical equation1 mole = 22.4 L @ STP1 mole = 6.02 x 1023 particles(atoms or molecules)Mole1 mole = molar mass (g)1 mole = 22.4 L @ STP1 mole = 6.02 x 1023 particles(atoms or molecules)5Mole Island DiagramMassParticlesVolumeMoleMassVolumeParticles Substance A Substance B1 mole = molar mass (g)Use coefficientsfrom balancedchemical equation1 mole = 22.4 L @ STP1 mole = 6.022 x 1023 particles(atoms or molecules)(gases)(gases)Mole1 mole = molar mass (g)1 mole = 22.4 L @ STP1 mole = 6.022 x 1023 particles(atoms or molecules)6Mole Island Diagram Substance A Substance B7White Board OverlayMassParticlesVolumeMoleMoleMassVolumeParticles Known Unknown

Substance A Substance B82__TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO 1. How many mol chlorine will react with 4.55 mol carbon? 3 mol C4 mol Cl24.55 mol C= 6.07 mol Cl22. What mass titanium (IV) oxide will react with 4.55 mol carbon?= 242 g TiO243221CCl2CTiO23 mol C2 mol TiO24.55 mol C1 mol TiO279.9 g TiO2Stoichiometry Practice Problems3. How many molecules titanium (IV) chloride can be madefrom 115 g titanium (IV) oxide?TiCl4TiO2()= 8.66 x 1023 mc TiCl4115 g TiO21 mol TiO279.9 g TiO2()2 mol TiO22 mol TiCl4()1 mol TiCl46.02 x 1023 mc TiCl4Island Diagram helpful reminders: 1 mol1 mol1 mol1 molcoeff.1 mol1 mol1. Use coefficients from the equation only when crossing the middle bridge. The other six bridges always have 1 mol = as a part of the conversion.2. The middle bridge conversion factor is the only one that has two different substances in it. The conversion factors for the other six bridges have the same substance in both the numerator and denominator.3. The units on the islands at each end of the bridge being crossed must appear in the conversion factor for that bridgeHow many molecules titanium (IV) chloride can be madefrom 115 g titanium (IV) oxide?TiCl4TiO2()= 8.7 x 1023 mc TiCl4115 g TiO21 mol TiO279.9 g TiO2()2 mol TiO22 mol TiCl4()1 mol TiCl46.02 x 1023 mc TiCl4Island Diagram helpful reminders: 1. Use coefficients from the equation only when crossing the middle bridge. The other six bridges always have 1 mol = as a part of the conversion.1 mol1 mol1 mol1 molcoeff.1 mol1 molHow many molecules titanium (IV) chloride can be madefrom 115 g titanium (IV) oxide?TiCl4TiO2()= 8.7 x 1023 mc TiCl4115 g TiO21 mol TiO279.9 g TiO2()2 mol TiO22 mol TiCl4()1 mol TiCl46.02 x 1023 mc TiCl4Island Diagram helpful reminders: 2. The middle bridge conversion factor is the only one that has two different substances in it. The conversion factors for the other six bridges have the same substance in both the numerator and denominator.1 mol1 mol1 mol1 molcoeff.1 mol1 mol3. The units on the islands at each end of the bridge being crossed must appear in the conversion factor for that bridgeHow many molecules titanium (IV) chloride can be madefrom 115 g titanium (IV) oxide?TiCl4TiO2()= 8.7 x 1023 mc TiCl4115 g TiO21 mol TiO279.9 g TiO2()2 mol TiO22 mol TiCl4()1 mol TiCl46.02 x 1023 mc TiCl4Island Diagram helpful reminders: 1 mol1 mol1 mol1 molcoeff.1 mol1 mol2 Ir + Ni3P2 3 Ni + 2 IrP 1. If 5.33 x 1028 mcules nickel (II) phosphidereact w/excess iridium, what mass iridium (III)phosphide is produced?Ni3P2IrP= 3.95 x 107 g IrP1 mol IrP223.2 g IrP1 mol Ni3P22 mol IrP1 mol Ni3P26.02 x 1023 mc Ni3P25.33 x 1028 mc Ni3P22. How many grams iridium will react with465 grams nickel (II) phosphide?= 751 g Ir1 mol Ir192.2 g Ir1 mol Ni3P22 mol Ir1 mol Ni3P2238.1 g Ni3P2465 g Ni3P2Ni3P2Ir2 mol Ir3. How many moles of nickel are producedif 8.7 x 1025 atoms of iridium are consumed?IrNi= 220 mol Ni3 mol Ni1 mol Ir6.02 x 1023 at. Ir8.7 x 1025 at. Ir2 Ir + Ni3P2 3 Ni + 2 IrP iridium (Ir)

nickel (Ni)Zn4. What volume hydrogen gas is liberated(at STP) if 50 g zinc react w/excesshydrochloric acid (HCl)? __ Zn + __ HCl__ H2+ __ ZnCl2 1211H2= 20 L H21 mol H222.4 L H21 mol Zn1 mol H21 mol Zn65.4 g Zn50 g Zn50 gexcess? L

5. At STP, how many mcules oxygen reactwith 632 dm3 butane (C4H10)? C4H10O2= 1.10 x 1026 mc O21 mol O22 mol C4H1013 mol O21 mol C4H10632 dm3 C4H1022.4 dm3 C4H10__ C4H10 + __ O2__ CO2 + __ H2O 1452810136.02 x 1023 mc O2

Suppose the question had been,How many ATOMS of oxygen1.10 x 1026 mc O21 mc O22 atoms O= 2.20 x 1026 at. O1 mol CH4A balanced eq. gives the ratios of moles-to-molesEnergy and Stoichiometry CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ 1. How many kJ of energy are releasedwhen 54 g methane are burned? AND moles-to-energy.= 3.0 x 103 kJ1 mol CH4891 kJ54 g CH416.04 g CH4EECH4

10,540 kJ1 mol H2O3. What mass of water is made if10,540 kJ are released?CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ 2. At STP, what volume oxygen is consumedin producing 5430 kJ of energy? 2 mol O2= 273 L O21 mol O222.4 L O25430 kJ891 kJE2 mol H2O= 426.2 g H2O18.02 g H2O891 kJEO2EEH2OLimiting ReactantsA balanced equation for makinga Big Mac might be:3 B + 2 M + EE BM30 B and excess EE30 Mexcess M and excess EE30 Bexcess B and excess EE30 Mone can makeandWith15 BM10 BM10 BM

50 PA balanced equation for makinga big wheel might be:one can makeandWith50 S + excess of all other reactantsexcess of all other reactants50 Sexcess of all other reactants50 P25 Bw3 W + 2 P + S + H + F Bw 50 Bw25 BwSolid aluminum reacts w/chlorine gas to yield solidaluminum chloride. 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) If 125 g aluminum react w/excess chlorine,how many g aluminum chloride are made?= 618 g AlCl32 mol Al1 mol AlCl31 mol Al26.98 g Al125 g AlAlAlCl3133.34 g AlCl32 mol AlCl3If 125 g chlorine react w/excess aluminum,how many g aluminum chloride are made?= 157 g AlCl33 mol Cl21 mol AlCl31 mol Cl270.91 g Cl2125 g Cl2Cl2AlCl3133.34 g AlCl32 mol AlCl32 Al(s) + 3 Cl2(g) 2 AlCl3(s) If 125 g aluminum react w/125 g chlorine,how many g aluminum chloride are made?157 g AlCl3 (Were out of Cl2) limiting reactant (LR): the reactant that runs out first Any reactant you dont run outof is an excess reactant (ER). amount of product is limited by the LR

when pouring liquid into a funnel, it doesnt matter how much you pour into the top, the bottom of the funnel limits how much you get out.Think of this analogyHow to Find the Limiting ReactantFor the generic reaction RA + RB P,assume that the amountsof RA and RB are given.Should you use RA or RBin your calculations? 1. Using Stoichiometry, calculate the amount of product possible from both RA and RB (2 separate calculations)2. Whichever reactant produces the smaller amount of product is the LR

3. The smaller amount of product is the maximum amount producedFor the Al / Cl2 / AlCl3 example: 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) If 125 g aluminum react w/excess chlorine,how many g aluminum chloride are made?= 618 g AlCl32 mol Al1 mol AlCl31 mol Al26.98 g Al125 g AlAlAlCl3133.34 g AlCl32 mol AlCl3If 125 g chlorine react w/excess aluminum,how many g aluminum chloride are made?= 157 g AlCl33 mol Cl21 mol AlCl31 mol Cl270.91 g Cl2125 g Cl2Cl2AlCl3133.34 g AlCl32 mol AlCl3LRER2 Fe(s) + 3 Cl2(g) 2 FeCl3(s) 223 g Fe 179 L Cl2 Which is the limiting reactant: Fe or Cl2? 1 mol Fe55.85 g Fe223 g Fe1 mol Cl222.4 L Cl2179 L Cl2How many g FeCl3 are produced?= 648 g FeCl32 mol FeCl32 mol Fe162.21 g FeCl31 mol FeCl32 mol FeCl33 mol Cl2162.21 g FeCl31 mol FeCl3= 864 g FeCl3648 g FeCl3*Remember that the LR limits how much product can be made!Limiting Reactant Practice2 H2(g) + O2(g) 2 H2O(g) 13 g H2 80 g O2Which is the limiting reactant: H2 or O2? 1 mol H22.02 g H213 g H21 mol O232.00 g O280 g O2How many g H2O are produced?= 120 g H2O

2 mol H2O2 mol H218.02 g H2O1 mol H2O2 mol H2O

1 mol O218.02 g H2O1 mol H2O= 90 g H2O90 g H2O*Notice that the LR doesnt always have the smaller amount (13 v. 80)How many g O2 are left over?How many g H2 are left over?zero; O2 is the LR and therefore is all used up We know how much H2 we HAD (i.e. 13 g) To find how much is left over, we first need to figure out how much was USED in the reaction.H2O2 HAD 13 g, USED 10 g

Start with the LR and relate to the other1 mol O232.00 g O280 g O22 mol H2

1 mol O22.02 g H21 mol H2 10 g H2 USED=3 g H2 left over181 g Fe 96.5 L Br2Which is the limiting reactant: Fe or Br2? 1 mol Fe55.85 g Fe181 g Fe1 mol Br222.4 L Br296.5 L Br2How many g FeBr3 are produced?= 958 g FeBr3

2 mol FeBr32 mol Fe295.55 g FeBr31 mol FeBr32 mol FeBr3

3 mol Br2295.55 g FeBr31 mol FeBr3= 849 g FeBr3849 g FeBr32 Fe(s) + 3 Br2(g) 2 FeBr3(s) How many g of the ER are left over?FeBr2181 g Fe 96.5 L Br22 Fe(s) + 3 Br2(g) 2 FeBr3(s) HAD 181 g, USED 160.4 g1 mol Br222.4 L Br296.5 L Br22 mol Fe

3 mol Br255.85 g Fe1 mol Fe 160. g Fe USED=21 g Fe left over101.96 g Al2O3Percent Yield moltensodiumsolidaluminumoxidemoltenaluminumsolidsodiumoxideFind mass of aluminum produced if you start w/575 g sodiumand 357 g aluminum oxide. + Al2O3(s)Al(l)6Na(l)+ Na2O(s)123Na+O2Al3+O2Al1 mol Na22.99 g Na575 g Na1 mol Al2O3357 g Al2O3= 225 g Al

2 mol Al6 mol Na26.98 g Al1 mol Al2 mol Al

1 mol Al2O326.98 g Al1 mol Al= 189 g Al189 g AlNow suppose that we performthis reaction and get only 172grams of aluminum. Why? This amt. of product (______)is the theoretical yield. amt. we get if reaction is perfect found by calculation using Stoich 189 g couldnt collect all Al not all Na and Al2O3 reacted some reactant or product spilled and was lost

Actual yield= 91.0%Find % yield for previous problem. Note: % yield should never be > 100%

Batting averageFG %GPA= 41,384 g Li2CO3On NASA spacecraft, lithium hydroxide scrubbers removetoxic CO2 from cabin. 1. For a seven-day mission, each of four individuals exhales880 g CO2 daily. If reaction is 75% efficient, how many g Li2CO3 will actually be produced? CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l)CO2Li2CO3()880 g CO2person-dayx (4 p) x (7 d)= 24,640 g CO2percent yield1 mol CO224, 640 g CO21 mol Li2CO3

1 mol CO273.9 g Li2CO31 mol Li2CO344.0 g CO2theo yieldA = 31,000 g Li2CO3

2. Automobile air bags inflate with nitrogen via the decompositionof sodium azide: 2 NaN3(s) 3 N2(g) + 2 Na(s)At STP and a % yield of 85%, what mass sodium azideis needed to yield 74 L nitrogen?N2NaN3percent yieldact yieldx = 87.1 L N2 Theo yield1 mol N287.1 L N22 mol NaN3

3 mol N265 g NaN31 mol NaN3= 170 g NaN322.4 L N2

___ZnS + ___O2 ___ZnO + ___SO2100 g100 g X g ? (assuming 81% yield)Strategy: 1. 2. 3. Balance and find LR Use LR to calc. X g ZnO (theo. yield) Actual yield is 81% of theo. yield 2322ZnO2 mol ZnS1 mol ZnS97.5 g ZnS100 g ZnS1 mol O2100 g O2= 83.5 g ZnO

2 mol ZnO81.4 g ZnO1 mol ZnO2 mol ZnO

3 mol O281.4 g ZnO1 mol ZnO= 169.6 g ZnO

83.5 g ZnO

32 g O2x = 67.6 g ZnO

___Al + ___Fe2O3 ___Fe + ___Al2O3 X g? X g? 800. g needed **Rxn. has an80.% yield. 2112FeAlFeFe2O3act yieldtheo = 1000 g Fe2 mol Fe1 mol Fe55.85 g Fe1000 g Fe= 480 g Al

2 mol Al26.98 g Al1 mol Al2 mol Fe1 mol Fe55.85 g Fe1000 g Fe= 1400 g Fe2O3

1 mol Fe2O3159.7 g Fe2O31 mol Fe2O3

Reaction that powers space shuttle is: 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ From 100 g hydrogen and 640 g oxygen, what amount ofenergy is possible?EE

1 mol H2= 14300 kJ2 mol H2572 kJ100 g H22 g H21 mol O2= 11440 kJ1 mol O2572 kJ640 g O232 g O211440 kJReview QuestionsWhat mass of excess reactant is left over? 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ H2O2Started with 100 g, used up 80 g20 g H2 left over

1 mol O2640 g O22 mol H2

1 mol O22 g H21 mol H2= 80 g H232 g O2100 g640 g