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

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1 Unit 8: Stoichiometry -- involves finding amts. of reactants & products in a reaction What can we do with stoichiometry? For generic equation: R A + R B P 1 + P Given the amount of R A (or R B ) one can find the amount of R B (or R A ) that is needed to react with it amount of R A or R B amount of P 1 or P that will be produced amount of P 1 or P you need to produce amount of R A and/or R B you must use patties + bread 1 Big Mac patties +? excess + 18 bread?? +? 5 Big Macs

2 TiO + Cl + C TiCl + CO + CO TiO + Cl + C TiCl + 1 CO + CO How many mol chlorine will react with.55 mol carbon? C Cl X mol Cl mol Cl mol C.55 mol C 6.07 mol Cl What mass titanium (IV) oxide will react with.55 mol carbon? C TiO X g TiO mol TiO 79.9 g TiO.55 mol C g TiO mol C 1mol TiO How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiO TiCl 6.0 x 10 X m' c TiCl 115 g TiO 1mol TiO 79.9 g TiO mol TiCl mol TiO m' c TiCl 1mol TiCl = 8.7 x 10 m cules TiCl

3 Island Diagram helpful reminders: 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.. 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.. The units on the islands at each end of the bridge being crossed appear in the conversion factor for that bridge.

4 Ir + Ni P Ni + IrP If 5. x 10 8 m cules nickel (II) phosphide react w /excess iridium, what mass iridium (III) phosphide are produced? Ni P IrP X girp 5. x 10 8 m' c Ni =.95 x 10 7 IrP P 1mol NiP 6.0 x 10 m' c Ni How many grams iridium will react with 65 grams nickel (II) phosphide? P mol IrP 1mol Ni P Ni P. girp 1mol IrP Ir X gir 65 gni P 1mol NiP 8.1gNi P mol Ir 1mol Ni P 19. girp 1mol Ir = 751 g Ir How many moles of nickel are produced if 8.7 x 10 5 atoms of iridium are consumed? Ir Ni 5 1mol Ir X mol Ni 8.7 x 10 at.ir 6.0 x 10 at.ir = 17 mol Ni mol Ni mol Ir

5 What volume hydrogen gas is liberated (at STP) if 50 g zinc react w /excess hydrochloric acid (HCl)? Zn + HCl ZnCl + H 1 Zn + HCl 1 ZnCl + 1 H 50 g excess X L X L H 1mol Zn 1mol H. L H 50 g Zn 17.1L H 65. g Zn 1mol Zn 1mol H At STP, how many m cules oxygen react w /6 dm butane (C H 10 )? C H 10 + O CO + H O C H O 8 CO + 10 H O 6 dm X m cules 6.0 x 10 X m' c O 6 dm C H 10 1mol C. dm H10 C H 10 1 mol O mol C H 10 m' c O 1mol O = 1.10 x 10 6 m cules O

6 Energy and Stoichiometry CH (g) + O (g) CO (g) + H O(g) kj How many kj of energy are released when 5 g methane are burned? X kj 5 g CH 1mol CH 16 g CH 891kJ 1mol CH 007 kj At STP, what volume oxygen is consumed in producing 50 kj of energy? X L O mol O. L O 50 kj 7 L O 891 kj 1mol O What mass of water is made if 10,50 kj are released? mol HO 18 gho X gho 10,50 kj 6 gho 891 kj 1mol H O

7 The Limiting Reactant A balanced equation for making a Big Mac might be: B + M + EE B M EE With 0 M 0 B 0 M and excess B and excess EE excess M and excess EE 0 B and excess EE one can make 15 B M EE 10 B M EE 10 B M EE A balanced equation for making a tricycle might be: W + P + S + H + F W P SHF With 50 P 50 S 50 P and excess of all other reactants excess of all other reactants 50 S and excess of all other reactants one can make 5 W P SHF 50 W P SHF 5 W P SHF

8 Solid aluminum reacts w /chlorine gas to yield solid aluminum chloride. Al(s) + Cl (g) AlCl (s) If 15 g aluminum react w /excess chlorine, how many g aluminum chloride are made? X g AlCl 15 g Al 1mol Al 7 g Al mol AlCl mol Al 1.5 g AlCl 1mol AlCl = 618 g AlCl If 15 g chlorine react w /excess aluminum, how many g aluminum chloride are made? X g AlCl 15 g Cl 1mol Cl 71g Cl mol AlCl mol Cl 1.5 g AlCl 1mol AlCl = 157 g AlCl If 15 g aluminum react w /15 g chlorine, how many g aluminum chloride are made? 157 g AlCl We re out of Cl.

9 limiting reactant (LR): the reactant that runs out first Amount of product is based on LR. Any reactant you don t run out of is an excess reactant (ER). Example Limiting Reactant Excess Reactant(s) Big Macs buns meat tricycles pedals W, S, H, F Al / Cl / AlCl Cl Al 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.. Divide by the respective coefficients in balanced equation.. Reactant having the smaller result is the LR.

10 For the Al / Cl / AlCl example: X mol Al 15 g Al 1mol Al 7 g Al.6 mol Al.1 1mol Cl X mol Cl 15 g Cl 1.76 mol Cl 71g Cl 0.58 Cl is LR Fe(s) + Cl (g) FeCl (s) g Fe 179 L Cl Which is the limiting reactant: Fe or Cl? X mol Fe g Fe 1mol Fe 55.8 gfe.0 mol Fe.00 1mol Cl X mol Cl 179L Cl 8.0 mol Cl. L Cl.66 Fe is LR How many g FeCl are produced? X gfecl mol FeCl 16. gfecl.0 mol Fe 69 gfecl mol Fe 1mol FeCl

11 H (g) + O (g) H O(g) 1 g H 80 g O Which is LR: H or O? 1mol H X mol H 1 gh 6.5 mol H gh.5 1mol O X mol O 80 g O.5 mol O g O 1.50 O is LR How many g H O are formed? mol HO 18 g HO X gh O.5 mol O 90 g HO 1mol O 1mol H O How many g O are left over? zero; O is all used up How many g H are left over? mol H g H X gh (USED).5 mol O 10 gh 1mol O 1mol H (USED) g H left over

12 Fe(g) + Br (l) FeBr (s) 181 g Fe 96.5 L Br Find LR. X mol Fe 181g Fe 1mol Fe 55.8 g Fe. mol Fe 1.6 1mol Br X mol Br 96.5 L Br.1mol Br. L Br 1. Br is LR How many g FeBr are formed? X gfebr mol FeBr 95.5 gfebr.1mol Br 89 gfebr mol Br 1mol FeBr How many g of the ER are left over? X gfe (USED).1mol Br mol Fe mol Br 55.8 gfe 1mol Fe 160 gfe (USED) 1 g Fe left over

13 Percent Yield molten + solid molten + solid sodium aluminum aluminum sodium oxide oxide Al + O Na 1+ O Na(l) + Al O (s) Al(l) + Na O(s) 6 Na(l) + 1 Al O (s) Al(l) + Na O(s) Find mass of aluminum produced if you start w /575 g sodium and 57 g aluminum oxide. X mol Na 575 gna 1mol Na g Na 5 mol Na mol X mol Al O 57 g Al O.5 mol Al O 10 g X g Al.5 mol Al O mol Al 1mol Al O 7 g Al 189 g Al 1mol Al 1.5 This amount of product is the theoretical yield. -- amt. we get if reaction is perfect -- found by calculation

14 Now suppose that we perform this reaction in the lab and get only 17 grams of aluminum. Why? -- couldn t collect all Al -- not all Na and Al O reacted -- some reactant or product spilled and was lost % yield actual yield theoretical yield x 100 % yield can never be > 100%. Find % yield for previous problem. % yield act. yld. x 100 theo. yld. 17 g Al 189 g Al x %

15 Reaction that powers space shuttle is: H (g) + O (g) H O(g) + 57 kj From 100 g hydrogen and 60 g oxygen, what amount of energy is possible? 1mol H X mol H 100 g H 50 mol H g H 5 1mol O X mol O 60 g O 0 mol O g O 1 0 X kj 0 mol O 57 kj 1mol O 11,0 kj What mass of excess reactant is left over? mol H g H X gh (USED) 0 mol O 80 gh 1mol O 1mol H (USED) 0 g H left over

16 On NASA spacecraft, lithium hydroxide scrubbers remove toxic CO from cabin. CO (g) + LiOH(s) Li CO (s) + H O(l) For a seven-day mission, each of four individuals exhales 880 g CO daily. If reaction is 75% efficient, how many g LiOH should be brought along? X g CO 880 g CO person- day ( p) (7 d),60 g CO X glioh,60 g CO 1mol CO g CO mol LiOH 1mol CO.9 glioh 1mol LiOH = 6,768 g LiOH (if reaction is perfect) Need more than this, so 6,768 glioh ,700 glioh ** Reality: Take 80,000 g 100,000 g, just in case.

17 Automobile air bags inflate with nitrogen via the decomposition of sodium azide: NaN (s) N (g) + Na(s) At STP and a % yield of 85%, what mass sodium azide is needed to yield 7 L nitrogen? 7 L N Shoot for 87.1L N 0.85 X gnan 87.1L N 1mol N. L N mol NaN mol N 65 gnan 1mol NaN = 169 g NaN B H 6 + O B O + H O 10 g 0 g X g 1mol X mol BH6 10 g BH6 0.6 mol BH6 7.6 g mol O X mol O 0 g O 0.98 mol O g O 0.1 X gb 1mol B O 69.6 gb O 0.98 mol O 1.8 gbo mol O 1mol BO O

18 C H 8 + O CO + H O + energy 00 g 00 g X kj 1 C H O CO + H O + 8 kj Strategy: 1. Find LR.. Use LR to calc. X kj. 1mol X mol CH8 00 g CH8.55 mol CH8 g mol O X mol O 00 g O 6.5 mol O g O X kj 6.5 mol O 8 kj 5 mol O 10 kj

19 ZnS + O ZnO + SO 100 g 100 g X g (assuming 81% yield) ZnS + O ZnO + SO Strategy: 1. Find LR.. Use LR to calc. X g ZnO (theo. yield). Actual yield is 81% of theo. yield. X mol ZnS 100 g ZnS 1mol 97.5 g 1.06 mol ZnS mol O X mol O 100 g O.15 mol O g O 1.0 X g ZnO 1.06 mol ZnS mol ZnO mol ZnS 81. g ZnO 1mol ZnO 8.5 g ZnO Actual yield is 8.5 g ZnO (0.81) 67.6 g ZnO

20 Al + Fe O Fe + Al O X g X g 800 g needed 80% yield Al + 1 Fe O Fe + 1 Al O Shoot for 800 gfe gfe X g Al 1000 g Fe 1mol Fe 55.8 g Fe mol Al mol Fe 7 g Al 1mol Al 8 g Al X gfe O 1000 gfe 1mol Fe 55.8 gfe 1mol Fe O mol Fe gfe O 1mol Fe O 10 gfe O

UNIT 9 - STOICHIOMETRY

UNIT 9 - STOICHIOMETRY General Stoichiometry Notes STOICHIOMETRY: tells relative amts of reactants & products in a chemical reaction Given an amount of a substance involved in a chemical reaction, we can figure out the amount