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Unit 3C: Stoichiometry Review

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 = .

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Unit 3C: Stoichiometry Review

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  1. Unit 3C: Stoichiometry Review

  2. 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 That is, 1 mole of atoms = __________ atoms

  3. 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). BIG. I Meant, “How Big is 6.022 x 1023?” 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

  4. 1 mol = molar mass 1 mole = 22.4 L @ STP 1 mol = 6.02 x 1023 particles Welcome to Mole Island

  5. Mole Island Diagram Substance ASubstance B Mass Mass Use coefficients from balanced chemical equation 1 mole = molar mass (g) 1 mole = molar mass (g) Volume Mole Mole Volume 1 mole = 22.4 L @ STP 1 mole = 22.4 L @ STP (gases) (gases) 1 mole = 6.02 x 1023 particles (atoms or molecules) 1 mole = 6.02 x 1023 particles (atoms or molecules) Particles Particles

  6. Stoichiometry Practice Problems 2 4 3 2 1 2 __TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO 1. How many mol chlorine will react with 4.55 mol carbon? 4 mol Cl2 4.55 mol C = 6.07 mol Cl2 C Cl2 3 mol C 2. What mass titanium (IV) oxide will react with 4.55 mol carbon? 79.9 g TiO2 2 mol TiO2 4.55 mol C C TiO2 1 mol TiO2 3 mol C = 242 g TiO2

  7. ( ) ( ( ) ) 3. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? 1 mol 1 mol coeff. 1 mol 1 mol 1 mol 1 mol TiO2 TiCl4 2 mol TiCl4 1 mol TiO2 6.02 x 1023 m’c TiCl4 115 g TiO2 79.9 g TiO2 2 mol TiO2 1 mol TiCl4 = 8.66 x 1023 m’c TiCl4 Island 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. 3. The units on the islands at each end of the bridge being crossed must appear in the conversion factor for that bridge 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.

  8. 2 Ir + Ni3P23 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? Ni3P2 IrP 1 mol Ni3P2 2 mol IrP 223.2 g IrP 5.33 x 1028 m’c Ni3P2 1 mol IrP 1 mol Ni3P2 6.02 x 1023 m’c Ni3P2 = 3.95 x 107 g IrP 2. How many grams iridium will react with 465 grams nickel (II) phosphide? Ni3P2 Ir 1 mol Ni3P2 2 mol Ir 465 g Ni3P2 192.2 g Ir 1 mol Ir 1 mol Ni3P2 238.1 g Ni3P2 = 751 g Ir

  9. 2 Ir + Ni3P23 Ni + 2 IrP iridium (Ir) nickel (Ni) 3. How many moles of nickel are produced if 8.7 x 1025 atoms of iridium are consumed? Ir Ni 8.7 x 1025 at. Ir 1 mol Ir 3 mol Ni 2 mol Ir 6.02 x 1023 at. Ir = 220 mol Ni

  10. 4. What volume hydrogen gas is liberated (at STP) if 50 g zinc react w/excess hydrochloric acid (HCl)? Zn H2 1 2 1 1 __ Zn + __ HCl __ H2 + __ ZnCl2 50 g excess ? L 1 mol Zn 1 mol H2 22.4 L H2 50 g Zn 1 mol H2 1 mol Zn 65.4 g Zn = 20 L H2

  11. 5. At STP, how many m’cules oxygen react with 632 dm3 butane (C4H10)? C4H10 O2 1 4 5 2 13 8 10 __ CO2 + __ H2O __ C4H10 + __ O2 632 dm3 C4H10 1 mol C4H10 13 mol O2 6.02 x 1023 m’c O2 22.4 dm3 C4H10 2 mol C4H10 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 = 2.20 x 1026 at. O 1 m’c O2

  12. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ E E Energy and Stoichiometry A balanced eq. gives the ratios of moles-to-moles AND moles-to-energy. CH4 1. How many kJ of energy are released when 54 g methane are burned? 54 g CH4 1 mol CH4 891 kJ = 3.0 x 103 kJ 16.04 g CH4 1 mol CH4

  13. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ E E E E H2O O2 2. At STP, what volume oxygen is consumed in producing 5430 kJ of energy? 2 mol O2 22.4 L O2 5430 kJ = 273 L O2 891 kJ 1 mol O2 3. What mass of water is made if 10,540 kJ are released? 10,540 kJ 2 mol H2O 18.02 g H2O = 426.2 g H2O 891 kJ 1 mol H2O

  14. 3 B + 2 M + EE BM Limiting Reactants A balanced equation for making a Big Mac® might be: With… …and… …one can make… 30 M excess B and excess EE 15 BM 30 B excess M and excess EE 10 BM 30 M 30 B and excess EE 10 BM

  15. 3 W + 2 P + S + H + F Bw A balanced equation for making a big wheel might be: …and… With… …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

  16. 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? Al AlCl3 1 mol Al 2 mol AlCl3 125 g Al 133.34 g AlCl3 26.98 g Al 1 mol AlCl3 2 mol Al = 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 1 mol AlCl3 3 mol Cl2 = 157 g AlCl3

  17. 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? (We’re out of Cl2…) 157 g AlCl3 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). 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. Think of this analogy…

  18. For the generic reaction RA + RBP, assume that the amounts of RA and RB are given. Should you use RA or RB in your calculations? How to Find the Limiting Reactant 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

  19. For the Al / Cl2 / AlCl3 example: LR ER 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 AlCl3 1 mol Al 2 mol AlCl3 133.34 g AlCl3 125 g Al 26.98 g Al 1 mol AlCl3 2 mol Al = 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 1 mol AlCl3 3 mol Cl2 = 157 g AlCl3

  20. 2 Fe(s) + 3 Cl2(g) 2 FeCl3(s) Limiting Reactant Practice 223 g Fe 179 L Cl2 Which is the limiting reactant: Fe or Cl2? 223 g Fe 1 mol Fe 2 mol FeCl3 162.21 g FeCl3 1 mol FeCl3 55.85 g Fe 2 mol Fe 648 g FeCl3 = 648 g FeCl3 179 L Cl2 1 mol Cl2 2 mol FeCl3 162.21 g FeCl3 22.4 L Cl2 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!

  21. 2 H2(g) + O2(g) 2 H2O(g) 13 g H2 80 g O2 Which is the limiting reactant: H2 or O2? 13 g H2 1 mol H2 2 mol H2O 18.02 g H2O 1 mol H2O 2.02 g H2 2 mol H2 = 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 H2O = 90 g H2O How many g H2O are produced? *Notice that the LR doesn’t always have the smaller amount (13 v. 80)

  22. HAD 13 g, USED 10 g… 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… O2 H2 80 g O2 1 mol O2 2 mol H2 2.02 g H2 10 g H2 USED = 32.00 g O2 1 mol O2 1 mol H2 3 g H2 left over

  23. 2 Fe(s) + 3 Br2(g)2 FeBr3(s) 181 g Fe 96.5 L Br2 Which is the limiting reactant: Fe or Br2? 181 g Fe 1 mol Fe 2 mol FeBr3 295.55 g FeBr3 1 mol FeBr3 55.85 g Fe 2 mol Fe = 958 g FeBr3 96.5 L Br2 1 mol Br2 2 mol FeBr3 295.55 g FeBr3 22.4 L Br2 3 mol Br2 1 mol FeBr3 = 849 g FeBr3 849 g FeBr3 How many g FeBr3 are produced?

  24. 2 Fe(s) + 3 Br2(g)2 FeBr3(s) HAD 181 g, USED 160.4 g… 181 g Fe 96.5 L Br2 How many g of the ER are left over? Br2 Fe 96.5 L Br2 1 mol Br2 2 mol Fe 55.85 g Fe 160. g Fe USED = 22.4 L Br2 3 mol Br2 1 mol Fe 21 g Fe left over

  25. solid aluminum oxide solid sodium oxide molten sodium molten aluminum Percent Yield Al3+ O2– Na+ O2– 6 + Al2O3(s) 1 2 + Na2O(s) 3 Na(l) Al(l) Find mass of aluminum produced if you start w/575 g sodium and 357 g aluminum oxide. Al 575 g Na 1 mol Na 2 mol Al 26.98 g Al 1 mol Al 22.99 g Na 6 mol Na = 225 g Al 357 g Al2O3 1 mol Al2O3 2 mol Al 26.98 g Al 1 mol Al2O3 1 mol Al 101.96 g Al2O3 = 189 g Al 189 g Al

  26. This amt. of product (______) is the theoretical yield. 189 g • 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

  27. Note: % yield should never be > 100% FG % Batting average GPA Find % yield for previous problem. = 91.0%

  28. CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l) ( ) On NASA spacecraft, lithium hydroxide “scrubbers” remove toxic CO2 from cabin. 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 x (4 p) x (7 d) = 24,640 g CO2 person-day CO2 Li2CO3 24, 640 g CO2 1 mol CO2 1 mol Li2CO3 73.9 g Li2CO3 = 41,384 g Li2CO3 1 mol CO2 1 mol Li2CO3 44.0 g CO2 “theo” yield A = 31,000 g Li2CO3

  29. 2 NaN3(s) 3 N2(g) + 2 Na(s) 2. Automobile air bags inflate with nitrogen via the decomposition of sodium azide: At STP and a % yield of 85%, what mass sodium azide is needed to yield 74 L nitrogen? percent yield “act” yield x = 87.1 L N2→ “Theo” yield N2 NaN3 1 mol N2 87.1 L N2 65 g NaN3 2 mol NaN3 3 mol N2 1 mol NaN3 22.4 L N2 = 170 g NaN3

  30. 2 3 2 2 ___ZnS + ___O2___ZnO + ___SO2 100 g 100 g X g ? (assuming 81% yield) Balance and find LR Strategy: 1. 2. 3. Use LR to calc. X g ZnO (theo. yield) Actual yield is 81% of theo. yield ZnO 1 mol ZnS 2 mol ZnO 81.4 g ZnO 100 g ZnS = 83.5 g ZnO 83.5 g ZnO 1 mol ZnO 97.5 g ZnS 2 mol ZnS 100 g O2 1 mol O2 2 mol ZnO 81.4 g ZnO = 169.6 g ZnO 3 mol O2 1 mol ZnO 32 g O2 x = 67.6 g ZnO

  31. 2 1 2 1 ___Al + ___Fe2O3___Fe + ___Al2O3 X g? X g? 800. g needed **Rxn. has an 80.% yield. “act” yield “theo” = 1000 g Fe Fe Al 1000 g Fe 1 mol Fe 26.98 g Al 2 mol Al = 480 g Al 1 mol Al 55.85 g Fe 2 mol Fe Fe Fe2O3 1 mol Fe 1 mol Fe2O3 159.7 g Fe2O3 1000 g Fe = 1400 g Fe2O3 1 mol Fe2O3 55.85 g Fe 2 mol Fe

  32. 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ E E Review Questions Reaction that powers space shuttle is: From 100 g hydrogen and 640 g oxygen, what amount of energy is possible? 100 g H2 1 mol H2 572 kJ = 14300 kJ 2 mol H2 2 g H2 1 mol O2 572 kJ 640 g O2 = 11440 kJ 11440 kJ 1 mol O2 32 g O2

  33. 100 g 640 g 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ What mass of excess reactant is left over? O2 H2 640 g O2 1 mol O2 2 mol H2 2 g H2 = 80 g H2 1 mol O2 1 mol H2 32 g O2 20 g H2 left over Started with 100 g, used up 80 g…

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