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Unit 3C: Stoichiometry Review The Mole • The mole is the SI unit for “amount of substance.” • 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 mole 1 mole of atoms = 602,213,673,600,000,000,000,000 atoms 6.02 x 1023 atoms That is, 1 mole of atoms = __________ How Big is a Mole? …about the size of a chipmunk, weighing about 5 oz. (140 g), and having 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 the entire Earth (including the oceans) …to a height of 2 miles. There are ~ 6,880,900,000 people on Earth. If I had a mole of dollars, I could give every person on Earth… $87.5 trillion or $87.5 x 1012 Welcome to Mole Island 1 mol = molar mass 1 mole = 22.4 L @ STP 1 mol = 6.02 x 1023 particles Mole Island Diagram Substance A Substance B Mass Use coefficients from balanced chemical equation Volume (gases) 1 mole = 22.4 L @ STP Particles Mole Mole Mass 1 mole = 22.4 L @ STP Volume (gases) Particles Stoichiometry Practice Problems 2 4 2 + __C 3 __TiO 2 + __Cl 2 1 2 __TiCl 4 + __CO 2 + __CO 1. How many mol chlorine will react with 4.55 mol carbon? 4.55 mol C C Cl2 4 mol Cl2 = 6.07 mol Cl2 3 mol C 2. What mass titanium (IV) oxide will react with 4.55 mol carbon? 4.55 mol C 2 mol TiO2 C TiO2 3 mol C 79.9 g TiO2 1 mol TiO2 = 242 g TiO2 3. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? coeff. 1 mol TiO2 ( 115 g TiO2 )( 1 mol TiO2 79.9 g TiO2 1 mol TiCl4 )( ) 2 mol TiCl4 6.02 x 1023 m’c TiCl4 2 mol TiO2 1 mol TiCl4 = 8.66 x 1023 m’c TiCl4 Island Diagram helpful reminders: 2. The middle bridge conversion factorend is the onlybridge one 3. The units on the islands at each of the 1. Use coefficients from the equation only when crossing that has two different substances inconversion it. The conversion being crossed mustThe appear the factor the middle bridge. otherinsix bridges always havefor factors for the other six bridges have the same that bridge “1 mol = “ as a part of the conversion. substance in both the numerator and denominator. 2 Ir + Ni3P2 3 Ni + 2 IrP 1. If 5.33 x 1028 m’cules nickel (II) phosphide react w/excess iridium, what mass iridium (III) phosphide is produced? 5.33 x 1028 m’c Ni3P2 1 mol Ni3P2 6.02 x 1023 m’c Ni3P2 2 mol IrP 1 mol Ni3P2 Ni3P2 IrP 223.2 g IrP 1 mol IrP = 3.95 x 107 g IrP 2. How many grams iridium will react with 465 grams nickel (II) phosphide? 465 g Ni3P2 1 mol Ni3P2 238.1 g Ni3P2 2 mol Ir 1 mol Ni3P2 = 751 g Ir Ni3P2 Ir 192.2 g Ir 1 mol Ir 2 Ir + Ni3P2 3 Ni + 2 IrP 3. How many moles of nickel are produced if 8.7 x 1025 atoms of iridium are consumed? Ir 8.7 x 1025 at. Ir 1 mol Ir 6.02 x 1023 at. Ir 3 mol Ni 2 mol Ir = 220 mol Ni iridium (Ir) nickel (Ni) Ni 4. What volume hydrogen gas is liberated (at STP) if 50 g zinc react w/excess hydrochloric acid (HCl)? Zn H2 1 Zn + __ 2 HCl __ 50 g excess 50 g Zn 1 mol Zn 65.4 g Zn 1 H2 + __ 1 ZnCl2 __ ?L 1 mol H2 1 mol Zn 22.4 L H2 1 mol H2 = 20 L H2 5. At STP, how many m’cules oxygen react with 632 dm3 butane (C4H10)? C4H10 1 C4H10 + __ 2 13 O2 __ 632 dm3 C4H10 O2 4 CO2 + __ 5 H2O 8 10 __ 1 mol C4H10 22.4 dm3 C4H10 13 mol O2 2 mol C4H10 6.02 x 1023 m’c O2 1 mol O2 = 1.10 x 1026 m’c O2 Suppose the question had been,“How many ATOMS of oxygen…” 1.10 x 1026 m’c O2 2 atoms O 1 m’c O2 = 2.20 x 1026 at. O Energy and Stoichiometry CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ A balanced eq. gives the ratios of moles-to-moles AND moles-to-energy. E CH4 1. How many kJ of energy are released when 54 g methane are burned? 54 g CH4 1 mol CH4 891 kJ 16.04 g CH4 1 mol CH4 = 3.0 x 103 kJ CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ 2. At STP, what volume oxygen is consumed in producing 5430 kJ of energy? 5430 kJ 2 mol O2 891 kJ 22.4 L O2 1 mol O2 3. What mass of water is made if 10,540 kJ are released? 10,540 kJ 2 mol H2O 891 kJ 18.02 g H2O 1 mol H2O E O2 = 273 L O2 E H2O = 426.2 g H2O Limiting Reactants A balanced equation for making a Big Mac® might be: 3 B + 2 M + EE With… 30 M 30 B 30 M BM …and… excess B and excess EE excess M and excess EE 30 B and excess EE …one can make… 15 BM 10 BM 10 BM A balanced equation for making a big wheel might be: 3W+2P+S+H+F Bw With… …and… …one can make… 50 P excess of all other reactants 25 Bw 50 S excess of all other reactants 50 Bw 50 P 50 S + excess of all other reactants 25 Bw Solid aluminum reacts w/chlorine gas to yield solid aluminum 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? Al 125 g Al 1 mol Al 26.98 g Al 2 mol AlCl3 2 mol Al AlCl3 133.34 g AlCl3 1 mol AlCl3 = 618 g AlCl3 If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? Cl2 AlCl3 125 g Cl2 1 mol Cl2 2 mol AlCl3 133.34 g AlCl3 70.91 g Cl2 3 mol Cl2 1 mol AlCl3 = 157 g AlCl3 2 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 (We’re out of Cl2…) limiting reactant (LR): the reactant that runs out first • amount of product is “limited” by the LR • Any reactant you don’t run out of is an excess reactant (ER). Think of this analogy…when pouring liquid into a funnel, it doesn’t matter how much you pour into the top, the bottom of the funnel limits how much you get out. How to Find the Limiting Reactant For the generic reaction RA + RB P, assume that the amounts of RA and RB are given. Should you use RA or RB in 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 produced For the Al / Cl2 / AlCl3 example: ER 2 Al(s) + 3 Cl2(g) LR 2 AlCl3(s) If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? 125 g Al 1 mol Al 26.98 g Al 2 mol AlCl3 2 mol Al Al AlCl3 133.34 g AlCl3 1 mol AlCl3 = 618 g AlCl3 If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? Cl2 AlCl3 125 g Cl2 1 mol Cl2 2 mol AlCl3 133.34 g AlCl3 70.91 g Cl2 3 mol Cl2 1 mol AlCl3 = 157 g AlCl3 Limiting Reactant Practice 2 Fe(s) + 3 Cl2(g) 223 g Fe 179 L Cl2 2 FeCl3(s) Which is the limiting reactant: Fe or Cl2? 223 g Fe 1 mol Fe 55.85 g Fe 2 mol FeCl3 2 mol Fe 162.21 g FeCl3 1 mol FeCl3 = 648 g FeCl3 179 L Cl2 1 mol Cl2 22.4 L Cl2 2 mol FeCl3 162.21 g FeCl3 3 mol Cl2 1 mol FeCl3 = 864 g FeCl3 How many g FeCl3 are produced? *Remember that the LR “limits” how much product can be made! 2 H2(g) + O2(g) 13 g H2 2 H2O(g) 80 g O2 Which is the limiting reactant: H2 or O2? 13 g H2 1 mol H2 2.02 g H2 2 mol H2O 2 mol H2 18.02 g H2O 1 mol H2O = 120 g H2O 80 g O2 1 mol O2 2 mol H2O 18.02 g H2O 32.00 g O2 1 mol O2 1 mol H2O = 90 g H22O How many g H2O are produced? *Notice that the LR doesn’t always have the smaller amount (13 v. 80) How many g O2 are left over? zero; O2 is the LR and therefore is all used up How many g H2 are left over? 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. Start with the LR and relate to the other… 80 g O2 O2 1 mol O2 2 mol H2 2.02 g H2 32.00 g O2 1 mol O2 1 mol H2 HAD 13 g, USED 10 g… H2 = 10 g H2 USED 3 g H2 left over 2 Fe(s) + 3 Br2(g) 181 g Fe 96.5 L Br2 2 FeBr3(s) Which is the limiting reactant: Fe or Br2? 181 g Fe 1 mol Fe 55.85 g Fe 2 mol FeBr3 2 mol Fe 295.55 g FeBr3 1 mol FeBr3 = 958 g FeBr3 96.5 L Br2 1 mol Br2 22.4 L Br2 2 mol FeBr3 295.55 g FeBr3 3 mol Br2 1 mol FeBr3 = 849 g FeBr33 How many g FeBr3 are produced? 2 Fe(s) + 3 Br2(g) 181 g Fe 96.5 L Br2 2 FeBr3(s) How many g of the ER are left over? 96.5 L Br2 1 mol Br2 2 mol Fe 22.4 L Br2 3 mol Br2 HAD 181 g, USED 160.4 g… Br2 Fe 55.85 g Fe = 160. g Fe USED 1 mol Fe 21 g Fe left over Al3+ O2– Percent Yield Na+ O2– solid solid molten molten aluminum sodium sodium aluminum oxide oxide 6 Na(l) + 1 Al2O3(s) 2 Al(l) + 3 Na2O(s) Find mass of aluminum produced if you start w/575 g sodium and 357 g aluminum oxide. 575 g Na 1 mol Na 22.99 g Na 2 mol Al 6 mol Na 26.98 g Al 1 mol Al = 225 g Al 357 g Al2O3 1 mol Al2O3 2 mol Al 101.96 g Al2O3 1 mol Al2O3 26.98 g Al 1 mol Al = 189 g Al Al 189 g This amt. of product (______) is the theoretical yield. • amt. we get if reaction is perfect • found by calculation using “Stoich” Actual yield Now suppose that we perform this reaction and get only 172 grams of aluminum. Why? • couldn’t collect all Al • not all Na and Al2O3 reacted • some reactant or product spilled and was lost % yield actual yield theoretica l yield x 100 Note: % yield should never be > 100% FG % Batting average GPA Find % yield for previous problem. % yield act. yld. theo. yld. x 100 172 g Al 189 g Al x 100 = 91.0% On NASA spacecraft, lithium hydroxide “scrubbers” remove toxic CO2 from cabin. CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l) 1. For a seven-day mission, each of four individuals exhales 880 g CO2 daily. If reaction is 75% efficient, how many g Li2CO3 will actually be produced? percent yield ( 880 g CO2 person-day 24, 640 g CO2 % Yield Act Theo ) x (4 p) x (7 d) = 24,640 g CO2 CO2 Li2CO3 1 mol CO2 1 mol Li2CO3 73.9 g Li2CO3 = 41,384 g 44.0 g CO2 1 mol CO2 1 mol Li2CO3 Li2CO3 x 100 0.75 A 41,384 g “theo” yield A = 31,000 g Li2CO3 2. Automobile air bags inflate with nitrogen via the decomposition of sodium azide: 2 NaN3(s) 3 N2(g) + 2 Na(s) At STP and a % yield of 85%, what mass sodium azide is needed to yield 74 L nitrogen? percent yield “act” yield % Yield Act Theo 87.1 L N2 x 100 0.85 75 L x 1 mol N2 22.4 L N2 x = 87.1 L N2 → “Theo” yield N2 2 mol NaN3 65 g NaN3 3 mol N2 NaN3 1 mol NaN3 = 170 g NaN3 2 3 2 ___ZnS + ___O 2 2 ___ZnO + ___SO 2 100 g 100 g X g ? (assuming 81% yield) Strategy: 1. Balance and find LR 2. Use LR to calc. X g ZnO (theo. yield) 3. Actual yield is 81% of theo. yield ZnO 100 g ZnS 1 mol ZnS 97.5 g ZnS 2 mol ZnO 81.4 g ZnO = 83.5 g ZnO 2 mol ZnS 1 mol ZnO 100 g O2 1 mol O2 2 mol ZnO 81.4 g ZnO 32 g O2 % Yield Act Theo x 100 3 mol O2 0.81 1 mol ZnO x 83.5 g ZnO = 169.6 g ZnO x = 67.6 g ZnO 2 1 ___Al + ___Fe 2O3 X g? X g? 2 1 ___Fe + ___Al 2O3 800. g needed “act” yield % Yield Act x 100 0.80 Theo 800. g **Rxn. has an 80.% yield. “theo” = 1000 g Fe x Fe 1000 g Fe 1 mol Fe 2 mol Al 55.85 g Fe 2 mol Fe 26.98 g Al 1 mol Al Al = 480 g Al Fe Fe2O3 1000 g Fe 1 mol Fe 1 mol Fe2O3 159.7 g Fe2O3 = 1400 g Fe2O3 1 mol Fe2O3 55.85 g Fe 2 mol Fe Review Questions 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 of energy is possible? E 100 g H2 1 mol H2 2 g H2 640 g O2 1 mol O2 32 g O2 572 kJ 2 mol H2 = 14300 kJ 572 kJ 1 mol O2 = 11440 kJ 100 g 640 g 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ What mass of excess reactant is left over? O2 640 g O2 1 mol O2 2 mol H2 2 g H2 32 g O2 1 mol O2 1 mol H2 Started with 100 g, used up 80 g… H2 = 80 g H2 20 g H2 left over