Unit 3C: Stoichiometry Review

The Mole 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 = That is, 1 mole of atoms = __________ atoms The mole is the SI unit for “amount of substance.” 602,213,673,600,000,000,000,000 atoms 6.02 x 10 23

If I had a mole of dollars, I could give every person on Earth… 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 x ?” BIG x marbles would cover the entire Earth (including the oceans) …to a height of 2 miles. There are ~ 6,880,900,000 people on Earth. $87.5 trillion or $87.5 x 10 12

Welcome to Mole Island 1 mole = 22.4 STP 1 mol = molar mass 1 mol = 6.02 x particles

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

Mole Island Diagram Mass Particles Volume Mole Mass Volume Particles Substance A Substance B 1 mole = molar mass (g) Use coefficients from balanced chemical equation 1 mole = 22.4 STP 1 mole = x particles (atoms or molecules) (gases) Mole 1 mole = molar mass (g) 1 mole = 22.4 STP 1 mole = x particles (atoms or molecules)

Mole Island Diagram Substance A Substance B

White Board Overlay Mass Particles Volume Mole Mass Volume Particles Known Unknown Substance A Substance B

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

3. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.66 x m’c TiCl g TiO 2 1 mol TiO g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol TiCl x m’c TiCl 4 Island Diagram helpful reminders: 1 mol coeff. 1 mol 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. 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

How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x m’c TiCl g TiO 2 1 mol TiO g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol TiCl x m’c TiCl 4 Island Diagram helpful reminders: 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. 1 mol coeff. 1 mol

How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x m’c TiCl g TiO 2 1 mol TiO g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol TiCl x 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 mol coeff. 1 mol

3. The units on the islands at each end of the bridge being crossed must 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 m’c TiCl g TiO 2 1 mol TiO g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol TiCl x m’c TiCl 4 Island Diagram helpful reminders: 1 mol coeff. 1 mol

2 Ir + Ni 3 P 2 3 Ni + 2 IrP 1. If 5.33 x 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 g IrP 1 mol Ni 3 P 2 2 mol IrP 1 mol Ni 3 P x m’c Ni 3 P x m’c Ni 3 P 2 2. How many grams iridium will react with 465 grams nickel (II) phosphide? = 751 g Ir 1 mol Ir g Ir 1 mol Ni 3 P 2 2 mol Ir 1 mol Ni 3 P g Ni 3 P g Ni 3 P 2 Ni 3 P 2 Ir

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

Zn 4. What volume hydrogen gas is liberated (at STP) if 50 g zinc react w/excess hydrochloric acid (HCl)? __ Zn + __ HCl __ H 2 + __ ZnCl H2H2 = 20 L H 2 1 mol H L H 2 1 mol Zn 1 mol H 2 1 mol Zn 65.4 g Zn 50 g Zn 50 gexcess? L

5. 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 m’c O 2 1 mol O 2 2 mol C 4 H mol O 2 1 mol C 4 H dm 3 C 4 H dm 3 C 4 H 10 __ C 4 H 10 + __ O 2 __ CO 2 + __ H 2 O x m’c O 2 Suppose the question had been,“How many ATOMS of oxygen…” 1.10 x m’c O 2 1 m’c O 2 2 atoms O = 2.20 x 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) kJ 1. 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 kJ 54 g CH g CH 4 EE CH 4

10,540 kJ 1 mol H 2 O 3. 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) kJ 2. At STP, what volume oxygen is consumed in producing 5430 kJ of energy? 2 mol O 2 = 273 L O 2 1 mol O L O kJ 891 kJ E 2 mol H 2 O = g H 2 O g H 2 O 891 kJ E O2O2 EE H2OH2O

Limiting Reactants A balanced equation for making a Big Mac® might be: 3 B + 2 M + EE BM 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 BM 10 BM 10 BM

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 Bw 3 W + 2 P + S + H + F Bw 50 Bw 25 Bw

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 g Al 125 g Al AlAlCl 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 g Cl g Cl 2 Cl 2 AlCl 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 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…

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 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 / 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 g Al 125 g Al AlAlCl 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 g Cl g Cl 2 Cl 2 AlCl g AlCl 3 2 mol AlCl 3 LR ER

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 g Fe 223 g Fe 1 mol Cl L Cl L Cl 2 How many g FeCl 3 are produced? = 648 g FeCl 3 2 mol FeCl 3 2 mol Fe g FeCl 3 1 mol FeCl 3 2 mol FeCl 3 3 mol Cl g FeCl 3 1 mol FeCl 3 = 864 g FeCl g FeCl 3 *Remember that the LR “limits” how much product can be made! Limiting Reactant Practice

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 g H 2 13 g H 2 1 mol O 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 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 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 g O 2 80 g O 2 2 mol H 2 1 mol O 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 g Fe 181 g Fe 1 mol Br L Br L Br 2 How many g FeBr 3 are produced? = 958 g FeBr 3 2 mol FeBr 3 2 mol Fe g FeBr 3 1 mol FeBr 3 2 mol FeBr 3 3 mol Br g FeBr 3 1 mol FeBr 3 = 849 g FeBr g FeBr 3 2 Fe(s) + 3 Br 2 (g) 2 FeBr 3 (s)

How many g of the ER are left over? FeBr g Fe 96.5 L Br 2 2 Fe(s) + 3 Br 2 (g) 2 FeBr 3 (s) HAD 181 g, USED g… 1 mol Br L Br L Br 2 2 mol Fe 3 mol Br g Fe 1 mol Fe 160. g Fe USED = 21 g Fe left over

g Al 2 O 3 Percent Yield molten sodium solid aluminum oxide molten 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 + O 2– Al 3+ O 2– Al 1 mol Na g Na 575 g Na 1 mol Al 2 O g Al 2 O 3 = 225 g Al 2 mol Al 6 mol Na g Al 1 mol Al 2 mol Al 1 mol Al 2 O g Al 1 mol Al = 189 g Al189 g Al

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. Note: % yield should never be > 100% Batting average FG % GPA

= 41,384 g Li 2 CO 3 On NASA spacecraft, lithium hydroxide “scrubbers” remove toxic CO 2 from cabin. 1. 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 g Li 2 CO 3 1 mol Li 2 CO g CO 2 “theo” yield A = 31,000 g Li 2 CO 3

2. 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 L N 2 2 mol NaN 3 3 mol N 2 65 g NaN 3 1 mol NaN 3 = 170 g NaN L N 2

___ZnS + ___O 2 ___ZnO + ___SO g100 g X g ? (assuming 81% yield) Strategy: 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 g O 2 = 83.5 g ZnO 2 mol ZnO 81.4 g ZnO 1 mol ZnO 2 mol ZnO 3 mol O g ZnO 1 mol ZnO = 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 FeAl FeFe 2 O 3 “act” yield “theo” = 1000 g Fe 2 mol Fe 1 mol Fe g Fe 1000 g Fe = 480 g Al 2 mol Al g Al 1 mol Al 2 mol Fe 1 mol Fe g Fe 1000 g Fe = 1400 g Fe 2 O 3 1 mol Fe 2 O g Fe 2 O 3 1 mol Fe 2 O 3

Reaction that powers space shuttle is: 2 H 2 (g) + O 2 (g) 2 H 2 O(g) kJ From 100 g hydrogen and 640 g oxygen, what amount of energy is possible? EE 1 mol H 2 = kJ 2 mol H kJ 100 g H 2 2 g H 2 1 mol O 2 = kJ 1 mol O kJ 640 g O 2 32 g O kJ Review Questions

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