# Quantitative Relationships in Chemical Equations 4 Na(s) + O 2 (g) 2 Na 2 O(s) Particles4 atoms1 m’cule2 m’cules Moles4 mol1 mol2 mol Grams4 g1 g2 g **

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Quantitative Relationships in Chemical Equations 4 Na(s) + O 2 (g) 2 Na 2 O(s) Particles4 atoms1 m’cule2 m’cules Moles4 mol1 mol2 mol Grams4 g1 g2 g ** Coefficients of a balanced equation represent # of particles OR # of moles, but NOT # of grams.

When going from moles of one substance to moles of another, use coefficients from balanced equation. part. vol. mass MOL mass vol. part. MOL SUBSTANCE “A” SUBSTANCE “B” (known) (unknown) Use coeff. from balanced equation in crossing this bridge 4 Na(s) + 1 O 2 (g) 2 Na 2 O(s)

= 4.20 mol O 2 4 mol Na 87.2 mol Na 4 Na(s) + O 2 (g) 2 Na 2 O(s) How many moles oxygen will react with 16.8 moles sodium? How many moles sodium oxide are produced from 87.2 moles sodium? How many moles sodium are required to produce 0.736 moles sodium oxide? 4 mol Na 1 mol O 2 16.8 mol Na = 43.6 mol Na 2 O 2 mol Na 2 O 4 mol Na 0.736 mol Na 2 O = 1.47 mol Na 2 mol Na 2 O O2O2 NaNa 2 O Na

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

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

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®

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)

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

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

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 () = 8.66 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

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

2 mol Ir How many moles of nickel are produced if 8.7 x 10 25 atoms of iridium are consumed? IrNi = 220 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)

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 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. gexcess? L

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

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. = 3.0 x 10 3 kJ 1 mol CH 4 891 kJ 54 g CH 4 16 g CH 4 EE CH 4

10,540 kJ 1 mol H 2 O 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 = 425.9 g H 2 O 18 g H 2 O 891 kJ E O2O2 EE H2OH2O

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

50 P A balanced equation for making a big wheel 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

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

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 “limited” by the LR In a root beer float, the LR is usually the ice cream. Deliciousness! (What’s the product?)

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

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. Using Stoichiometry, calculate the amount of product possible from both R A and R B (2 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 / Cl 2 / AlCl 3 example: 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 LR

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 ? 1 mol Fe 55.8 g Fe 223 g Fe 1 mol Cl 2 22.4 L Cl 2 179 L Cl 2 How many g FeCl 3 are produced? = 649 g FeCl 3 2 mol FeCl 3 2 mol Fe 162.3 g FeCl 3 1 mol FeCl 3 2 mol FeCl 3 3 mol Cl 2 162.3 g FeCl 3 1 mol FeCl 3 = 865 g FeCl 3 649 g FeCl 3 *Remember that the LR “limits” how much product can be made!

2 H 2 (g) + O 2 (g) 2 H 2 O(g) 13 g H 2 80 g O 2 Which is the limiting reactant: H 2 or O 2 ? 1 mol H 2 2.0 g H 2 13 g H 2 1 mol O 2 32.0 g O 2 80 g O 2 How many g H 2 O are produced? = 120 g H 2 O 2 mol H 2 O 2 mol H 2 18.0 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 18.0 g H 2 O 1 mol H 2 O = 90 g H 2 O 90 g H 2 O *Notice that the LR doesn’t always have the smaller amount (13 v. 80)

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 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. H2H2 O2O2 HAD 13 g, USED 10 g… Start with the LR and relate to the other… 1 mol O 2 32.0 g O 2 80 g O 2 2 mol H 2 1 mol O 2 2.0 g H 2 1 mol H 2 10 g H 2 USED = 3 g H 2 left over

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

How many g of the ER are left over? FeBr 2 181 g Fe 96.5 L Br 2 2 Fe(s) + 3 Br 2 (g) 2 FeBr 3 (s) HAD 181 g, USED 160.4 g… 1 mol Br 2 22.4 L Br 2 96.5 L Br 2 2 mol Fe 3 mol Br 2 55.85 g Fe 1 mol Fe 160.4 g Fe USED = 20.6 g Fe left over

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(s) 6 Na(l) + Na 2 O(s) 123 Na + O 2– Al 3+ O 2– Al 1 mol Na 22.99 g Na 575 g Na 1 mol Al 2 O 3 357 g Al 2 O 3 = 224.9 g Al 2 mol Al 6 mol Na 26.98 g Al 1 mol Al 2 mol Al 1 mol Al 2 O 3 26.98 g Al 1 mol Al = 188.9 g Al188.9 g Al 102 g Al 2 O 3

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 using “Stoich” 189 g couldn’t collect all Al not all Na and Al 2 O 3 reacted some reactant or product spilled and was lost Actual yield

= 91.0% Find % yield for previous problem. % yield can never be > 100%. -- Batting average FG % GPA

= 41,374 g Li 2 CO 3 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 Li 2 CO 3 will actually be produced? CO 2 (g) + 2 LiOH(s) Li 2 CO 3 (s) + H 2 O(l) CO 2 Li 2 CO 3 () 880 g CO 2 person-day x (4 p) x (7 d)= 24,640 g CO 2 percent yield 1 mol CO 2 24, 640 g CO 2 1 mol Li 2 CO 3 1 mol CO 2 73.9 g Li 2 CO 3 1 mol Li 2 CO 3 44 g CO 2 “theo” yield x = 31030 g Li 2 CO 3

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 () () theo. = 36,000 g LiOH REALITY: TAKE 100,000 g theo. (act.) = 0.75

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? EE 1 mol H 2 = 14300 kJ 2 mol H 2 572 kJ 100 g H 2 2 g H 2 1 mol O 2 = 11440 kJ 1 mol O 2 572 kJ 640 g O 2 32 g O 2 11440 kJ Review Questions

What mass of excess reactant is left over? 2 H 2 (g) + O 2 (g) 2 H 2 O(g) + 572 kJ H2H2 O2O2 Started with 100 g, used up 80 g… 20 g H 2 left over 1 mol O 2 640 g O 2 2 mol H 2 1 mol O 2 2 g H 2 1 mol H 2 = 80 g H 2 32 g O 2 100 g 640 g

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? N2N2 NaN 3 percent yield “act” yield x = 87.1 L N 2 → “Theo” yield 1 mol N 2 87.1 L N 2 2 mol NaN 3 3 mol N 2 65 g NaN 3 1 mol NaN 3 = 168.5 g NaN 3 22.4 L N 2

1 mol B 2 H 6 B 2 H 6 + 3 O 2 B 2 O 3 + 3 H 2 O 10 g 30 g X g? B2O3B2O3 1 mol B 2 H 6 27.6 g B 2 H 6 10 g B 2 H 6 1 mol O 2 30 g O 2 = 25.2 g B 2 O 3 1 mol B 2 O 3 69.6 g B 2 O 3 1 mol B 2 O 3 3 mol O 2 69.6 g B 2 O 3 1 mol B 2 O 3 = 21.75 g B 2 O 3 21.75 g B 2 O 3 32 g O 2

___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 ZnO 2 mol ZnS 1 mol ZnS 97.5 g ZnS 100 g ZnS 1 mol O 2 100 g O 2 = 83.5 g ZnO 2 mol ZnO 81.4 g ZnO 1 mol ZnO 2 mol ZnO 3 mol O 2 81.4 g ZnO 1 mol ZnO = 169.6 g ZnO 83.5 g ZnO 32 g O 2 x = 67.6 g ZnO

___Al + ___Fe 2 O 3 ___Fe + ___Al 2 O 3 X g? X g? 800 g needed **Rxn. has an 80% yield. 2112 FeAl FeFe 2 O 3 “act” yield “theo” = 1000 g Fe 2 mol Fe 1 mol Fe 55.85 g Fe 1000 g Fe = 483.1 g Al 2 mol Al 26.98 g Al 1 mol Al 2 mol Fe 1 mol Fe 55.85 g Fe 1000 g Fe = 1430 g Fe 2 O 3 1 mol Fe 2 O 3 159.7 g Fe 2 O 3 1 mol Fe 2 O 3

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