1. Unit 8: Stoichiometry -- involves finding amts. of reactants & products in a reaction

2. amount of R A and/or R B you must use amount of P 1 or P 2 you need to produce amount of P 1 or P 2 that will be produced amount of R A or R B amount of R B (or R A ) that is needed to react with it amount of R A (or R B ) …one can find the… Given the… What can we do with stoichiometry? For generic equation: R A + R B P 1 + P 2

3. 4 patties + ? Governing Equation: 2 patties + 3 bread 1 Big Mac® excess + 18 bread ? ? + ? 25 Big Macs® 6 bread 75 bread50 patties 6 Big Macs®

4. Use coefficients from balanced equation MOLE (mol) Mass (g) Particle (at. or m’c) 1 mol = molar mass (in g) Volume (L or dm 3 ) 1 mol = 22.4 L 1 mol = 22.4 dm 3 1 mol = 6.02 x 10 23 particles SUBSTANCE “A” Stoichiometry Island Diagram MOLE (mol) Mass (g) 1 mol = molar mass (in g) Volume (L or dm 3 ) 1 mol = 22.4 L 1 mol = 22.4 dm 3 1 mol = 6.02 x 10 23 particles SUBSTANCE “B” Particle (at. or m’c)

5. 2 __TiO 2 + __Cl 2 + __C __TiCl 4 + __CO 2 + __CO How many mol chlorine will react with 4.55 mol carbon? 3 mol C 4 mol Cl 2 4.55 mol C = 6.07 mol Cl 2 What mass titanium (IV) oxide will react with 4.55 mol carbon? () = 242 g TiO 2 43221 () CCl 2 CTiO 2 3 mol C 2 mol TiO 2 4.55 mol C () 1 mol TiO 2 79.9 g TiO 2

6. 3. The units on the islands at each end of the bridge being crossed appear in the conversion factor for that bridge. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x 10 23 m’c TiCl 4 115 g TiO 2 1 mol TiO 2 79.9 g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol TiCl 4 6.02 x 10 23 m’c TiCl 4 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. 1. Use coefficients from the equation only when crossing the middle bridge. The other six bridges always have “1 mol” before a substance’s formula. 1 mol coeff. 1 mol

7. 2 Ir + Ni 3 P 2 3 Ni + 2 IrP If 5.33 x 10 28 m’cules nickel (II) phosphide react w/excess iridium, what mass iridium (III) phosphide is produced? Ni 3 P 2 IrP () = 3.95 x 10 7 g IrP 1 mol IrP 223.2 g IrP () 1 mol Ni 3 P 2 2 mol IrP () 1 mol Ni 3 P 2 6.02 x 10 23 m’c Ni 3 P 2 5.33 x 10 28 m’c Ni 3 P 2 How many grams iridium will react with 465 grams nickel (II) phosphide? () = 751 g Ir 1 mol Ir 192.2 g Ir () 1 mol Ni 3 P 2 2 mol Ir () 1 mol Ni 3 P 2 238.1 g Ni 3 P 2 465 g Ni 3 P 2 Ni 3 P 2 Ir

8. 2 mol Ir How many moles of nickel are produced if 8.7 x 10 25 atoms of iridium are consumed? IrNi = 217 mol Ni () 3 mol Ni () 1 mol Ir 6.02 x 10 23 at. Ir 8.7 x 10 25 at. Ir 2 Ir + Ni 3 P 2 3 Ni + 2 IrP iridium (Ir)nickel (Ni)

9. Zn What volume hydrogen gas is liberated (at STP) if 50 g zinc react w/excess hydrochloric acid (HCl)? __ Zn + __ HCl __ H 2 + __ ZnCl 2 1211 H2H2 () = 17.1 L H 2 1 mol H 2 22.4 L H 2 1 mol Zn 1 mol H 2 1 mol Zn 65.4 g Zn 50 g Zn ()() 50 gexcessX L

10. At STP, how many m’cules oxygen react with 632 dm 3 butane (C 4 H 10 )? C 4 H 10 O2O2 = 1.10 x 10 26 m’c O 2 1 mol O 2 () 2 mol C 4 H 10 13 mol O 2 () 1 mol C 4 H 10 632 dm 3 C 4 H 10 22.4 dm 3 C 4 H 10 __ C 4 H 10 + __ O 2 __ CO 2 + __ H 2 O 145 2810 13 6.02 x 10 23 m’c O 2 () Suppose the question had been “how many ATOMS of O 2 …” 1.10 x 10 26 m’c O 2 () 1 m’c O 2 2 atoms O = 2.20 x 10 26 at. O

11. 1 mol CH 4 A balanced eq. gives the ratios of moles-to-moles Energy and Stoichiometry CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) + 891 kJ How many kJ of energy are released when 54 g methane are burned? AND moles-to-energy. = 3007 kJ () 1 mol CH 4 891 kJ 54 g CH 4 16 g CH 4 () EE CH 4

12. What mass of water is made if 10,540 kJ are released? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) + 891 kJ At STP, what volume oxygen is consumed in producing 5430 kJ of energy? 2 mol O 2 = 273 L O 2 () 1 mol O 2 22.4 L O 2 5430 kJ 891 kJ () E 2 mol H 2 O = 426 g H 2 O () 1 mol H 2 O 18 g H 2 O 10,540 kJ 891 kJ () E O2O2 EE H2OH2O

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

14. 50 P A balanced equation for making a tricycle might be: …one can make… …and… With… 50 S + excess of all other reactants excess of all other reactants 50 S excess of all other reactants 50 P 25 W 3 P 2 SHF 3 W + 2 P + S + H + F W 3 P 2 SHF 50 W 3 P 2 SHF 25 W 3 P 2 SHF

15. Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 (s) If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? = 618 g AlCl 3 () 2 mol Al 1 mol AlCl 3 () 1 mol Al 27 g Al 125 g Al AlAlCl 3 () 133.5 g AlCl 3 2 mol AlCl 3 If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? = 157 g AlCl 3 () 3 mol Cl 2 1 mol AlCl 3 () 1 mol Cl 2 71 g Cl 2 125 g Cl 2 Cl 2 AlCl 3 () 133.5 g AlCl 3 2 mol AlCl 3

16. 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 (s) If 125 g aluminum react w/125 g chlorine, how many g aluminum chloride are made? 157 g AlCl 3 (We’re out of Cl 2.) limiting reactant (LR): the reactant that runs out first -- Any reactant you don’t run out of is an excess reactant (ER). amount of product is based on LR In a root beer float, the LR is usually the ice cream. Your enjoyment! (What’s the product?)

17. Al / Cl 2 / AlCl 3 tricycles Big Macs Excess Reactant(s)Limiting Reactant From Examples Above… 157 g AlCl 3 125 g Cl 2 125 g Al 25 W 3 P 2 SHF 50 S + excess of all other reactants 50 P 10 B 3 M 2 EE30 M …one can make……and…With… 30 B and excess EE B P Cl 2 M, EE W, S, H, F Al

18. How to Find the Limiting Reactant For the generic reaction R A + R B P, assume that the amounts of R A and R B are given. Should you use R A or R B in your calculations? 1. Calc. # of mol of R A and R B you have. 2. Divide by the respective coefficients in balanced equation. 3. Reactant having the smaller result is the LR.

19. 2 For the Al / Cl 2 / AlCl 3 example: 1 mol Al 27 g Al 125 g Al () = 4.63 mol Al (HAVE) 1 mol Cl 2 71 g Cl 2 125 g Cl 2 () = 1.76 mol Cl 2 (HAVE) Step #1. _.. _. 3= 0.58 = 2.31 Step #2 LR Step #3 “Oh, bee- HAVE !” 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 (s) (And start every calc. with the LR.)

20. 2 Fe(s) + 3 Cl 2 (g) 2 FeCl 3 (s) 223 g Fe 179 L Cl 2 Which is the limiting reactant: Fe or Cl 2 ? 2 1 mol Fe 55.8 g Fe 223 g Fe () = 4.0 mol Fe (HAVE) 1 mol Cl 2 22.4 L Cl 2 179 L Cl 2 () = 8.0 mol Cl 2 (HAVE). _.. _. 3= 2.66 = 2.0 LR = Fe How many g FeCl 3 are produced? = 649 g FeCl 3 () 2 mol Fe 1 mol FeCl 3 () 4.0 mol Fe 162.3 g FeCl 3 2 mol FeCl 3 FeCl 3 Fe (“Oh, bee-HAVE!”)

21. 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 13 g H 2 80 g O 2 Which is LR: H 2 or O 2 ? 2 1 mol H 2 2 g H 2 13 g H 2 () = 6.5 mol H 2 (HAVE) 1 mol O 2 32 g O 2 80 g O 2 () = 2.5 mol O 2 (HAVE). _.. _. 1= 2.50 = 3.25 LR = O 2 How many g H 2 O are formed? = 90 g H 2 O () 1 mol O 2 1 mol H 2 O () 2.5 mol O 2 18 g H 2 O 2 mol H 2 O H2OH2OO2O2 (“Oh, bee-HAVE!”)

22. 2 g H 2 How many g O 2 are left over? How many g H 2 are left over? zero; O 2 is the LR and therefore is all used up We know how much H 2 we started with (i.e., 13 g). To find how much is left over, we first need to figure out how much was USED UP in the reaction. = 10 g H 2 used up () 1 mol O 2 1 mol H 2 () 2.5 mol O 2 2 mol H 2 H2H2 O2O2 Started with 13 g, used up 10 g… 3 g H 2 left over

23. 2 Fe(s) + 3 Br 2 (g) 2 FeBr 3 (s) 181 g Fe 96.5 L Br 2 Find LR. 2 1 mol Fe 55.8 g Fe 181 g Fe () = 3.24 mol Fe (HAVE) 1 mol Br 2 22.4 L Br 2 96.5 L Br 2 () = 4.31 mol Br 2 (HAVE). _.. _. 3= 1.44 = 1.62 LR = Br 2 How many g FeBr 3 are formed? = 849 g FeBr 3 () 3 mol Br 2 1 mol FeBr 3 () 4.31 mol Br 2 295.5 g FeBr 3 2 mol FeBr 3 FeBr 3 Br 2 (“Oh, bee-HAVE!”)

24. How many g of the ER are left over? 55.8 g Fe = 160 g Fe used up () 3 mol Br 2 1 mol Fe () 4.31 mol Br 2 2 mol Fe FeBr 2 Started with 181 g, used up 160 g… 21 g Fe left over 2 Fe(s) + 3 Br 2 (g) 2 FeBr 3 (s) 181 g Fe 96.5 L Br 2

25. Percent Yield molten sodium solid aluminum oxide solid aluminum solid sodium oxide Find mass of aluminum produced if you start w /575 g sodium and 357 g aluminum oxide. + Al 2 O 3 (s) Al(l) 6 Na(l) + Na 2 O(s) 123 Na 1+ O 2– Al 3+ O 2– 6 1 mol 23 g 575 g Na () = 25 mol Na (HAVE) 1 mol 102 g 357 g Al 2 O 3 () = 3.5 mol Al 2 O 3 (HAVE). _.. _. 1= 3.5 = 4.17 LR = 189 g Al 1 mol Al 2 O 3 1 mol Al 3.5 mol Al 2 O 3 27 g Al 2 mol Al AlAl 2 O 3 () ()

26. Now suppose that we perform this reaction and get only 172 grams of aluminum. Why? -- This amt. of product (______) is the theoretical yield. -- amt. we get if reaction is perfect found by calculation 189 g couldn’t collect all Al not all Na and Al 2 O 3 reacted some reactant or product spilled and was lost

27. = 91.0% Find % yield for previous problem. % yield can never be > 100%. --

28. Reaction that powers space shuttle is: 2 H 2 (g) + O 2 (g) 2 H 2 O(g) + 572 kJ From 100 g hydrogen and 640 g oxygen, what amount of energy is possible? 2 1 mol H 2 2 g H 2 100 g H 2 () = 50 mol H 2 (HAVE) 1 mol O 2 32 g O 2 640 g O 2 () = 20 mol O 2 (HAVE). _.. _. 1= 20 = 25 LR = 11,440 kJ () 1 mol O 2 20 mol O 2 572 kJ EE O2O2

29. What mass of excess reactant is left over? 2 H 2 (g) + O 2 (g) 2 H 2 O(g) + 572 kJ 2 g H 2 = 80 g H 2 used up () 1 mol O 2 1 mol H 2 () 20 mol O 2 2 mol H 2 H2H2 O2O2 Started with 100 g, used up 80 g… 20 g H 2 left over

30. On NASA spacecraft, lithium hydroxide “scrubbers” remove toxic CO 2 from cabin. For a seven-day mission, each of four individuals exhales 880 g CO 2 daily. If reaction is 75% efficient, how many g LiOH should be brought along? CO 2 (g) + 2 LiOH(s) Li 2 CO 3 (s) + H 2 O(l) = 26,768 g LiOH 1 mol LiOH 23.9 g LiOH () 1 mol CO 2 2 mol LiOH 1 mol CO 2 44 g CO 2 24,640 g CO 2 CO 2 LiOH () 880 g CO 2 person-day x (4 p) x (7 d)= 24,640 g CO 2 () () (if reaction is perfect) (Need more than this!) ( 0.75). _. = 35,700 g LiOH REALITY: TAKE 100,000 g

31. Automobile air bags inflate with nitrogen via the decomposition of sodium azide: 2 NaN 3 (s) 3 N 2 (g) + 2 Na(s) At STP and a % yield of 85%, what mass sodium azide is needed to yield 74 L nitrogen? Shoot for…74 L ( 0.85). _. = 87.1 L N 2 N2N2 NaN 3 1 mol NaN 3 65 g NaN 3 () 3 mol N 2 2 mol NaN 3 1 mol N 2 22.4 L N 2 87.1 L N 2 () = 169 g NaN 3 () __C 3 H 8 + __O 2 __CO 2 + __H 2 O + __kJ Strategy:1. 2. X g Y g ? kJ Conceptual question (How would you do it?) Find LR. Calc. ? kJ. EE LR

32. B 2 H 6 + 3 O 2 B 2 O 3 + 3 H 2 O 10 g 30 g X g 1 1 mol 27.6 g 10 g B 2 H 6 () = 0.362 mol B 2 H 6 (HAVE) 1 mol 32 g 30 g O 2 () = 0.938 mol O 2 (HAVE). _.. _. 3= 0.313 = 0.362 LR = 21.8 g B 2 O 3 () 3 mol O 2 1 mol B 2 O 3 () 0.938 mol O 2 69.6 g B 2 O 3 1 mol B 2 O 3 B2O3B2O3 O2O2

33. ___ZnS + ___O 2 ___ZnO + ___SO 2 100 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. 2322 2 1 mol 97.5 g 100 g ZnS () = 1.026 mol ZnS (HAVE) 1 mol 32 g 100 g O 2 () = 3.125 mol O 2 (HAVE). _.. _. 3= 1.042 = 0.513 LR = 83.5 g ZnO () 2 mol ZnS 1 mol ZnO () 1.026 mol ZnS 81.4 g ZnO 2 mol ZnO ZnOZnS Actual g ZnO = 83.5 (0.81) = 68 g ZnO

34. ___Al + ___Fe 2 O 3 ___Fe + ___Al 2 O 3 X g X g 800 g needed **Rxn. has an 80% yield. 2112 Shoot for…800 g Fe ( 0.80). _. = 1000 g Fe FeAl 1 mol Al 27 g Al () 2 mol Fe 2 mol Al 1 mol Fe 55.8 g Fe 1000 g Fe () = 484 g Al () FeFe 2 O 3 1 mol Fe 2 O 3 159.6 g Fe 2 O 3 () 2 mol Fe 1 mol Fe 2 O 3 1 mol Fe 55.8 g Fe 1000 g Fe () = 1430 g Fe 2 O 3 ()

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