Unit 7 Stoichiometry Review The Mole The mole

Unit 7: 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 = 602, 213, 673, 600, 000, 000 atoms 6. 02 x 1023 atoms That is, 1 mole of atoms = _____

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 6. 022 x 1023? ” BIG. 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

Welcome to Mole Island 1 mol = molar mass 1 mole = 22. 4 L @ STP 1 mol = 6. 02 x 1023 particles

Mole Island Diagram Substance A Substance B Mass Volume (gases) m 1 m ola o r m le as = s( g) 1 mole = 22. 4 L @ STP Particles Mass Use coefficients from balanced chemical equation Mole = 3 2 e l o 0 1 s) 1 m 2 x les cule c 6. 0 arti ole p rm o s tom (a Mole = (g) e l o ass m 1 rm la mo 1 mole = 22. 4 L @ STP 1 6. mo 02 le (a to m par x 1 = s o tic 0 23 r m les ole cu les ) Volume (gases) Particles

Stoichiometry Practice Problems 2 4 2 + __C 3 __Ti. O 2 + __Cl 2 1 2 __Ti. Cl 4 + __CO 2 + __CO 1. How many mol chlorine will react with 4. 55 mol carbon? 4. 55 mol C C Cl 2 4 mol Cl 2 3 mol C = 6. 07 mol Cl 2 2. What mass titanium (IV) oxide will react with 4. 55 mol carbon? 4. 55 mol C 2 mol Ti. O 2 C Ti. O 2 3 mol C 79. 9 g Ti. O 2 1 mol Ti. O 2 = 242 g Ti. O 2

3. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? 1 m ol 1 mol ol m 1 ( ol m 1 1 mol 1 m ol coeff. Ti. O 2 )( 115 g Ti. O 2 1 mol Ti. O 2 79. 9 g Ti. O 2 Ti. Cl 4 )( ) 2 mol Ti. Cl 4 6. 02 x 1023 m’c Ti. Cl 4 2 mol Ti. O 2 1 mol Ti. Cl 4 = 8. 66 x 1023 m’c Ti. Cl 4 Island Diagram helpful reminders: 2. The middle bridge conversion factorend is the onlybridge one 3. The units on the islands at each of the 1. Use coefficients from the equation only when crossing that has two different substances inconversion it. The conversion being crossed must. The appear the factor the middle bridge. otherinsix bridges always havefor factors for the other six bridges have the same that bridge “ 1 mol = “ as a part of the conversion. substance in both the numerator and denominator.

2 Ir + Ni 3 P 2 3 Ni + 2 Ir. P 1. If 5. 33 x 1028 m’cules nickel (II) phosphide react w/excess iridium, what mass iridium (III) phosphide is produced? 5. 33 x 1028 m’c Ni 3 P 2 1 mol Ni 3 P 2 6. 02 x 1023 m’c Ni 3 P 2 2 mol Ir. P 1 mol Ni 3 P 2 238. 1 g Ni 3 P 2 Ir. P 223. 2 g Ir. P 1 mol Ir. P = 3. 95 x 107 g Ir. P 2. How many grams iridium will react with 465 grams nickel (II) phosphide? 465 g Ni 3 P 2 2 mol Ir 1 mol Ni 3 P 2 = 751 g Ir Ni 3 P 2 Ir 192. 2 g Ir 1 mol Ir

2 Ir + Ni 3 P 2 3 Ni + 2 Ir. P 3. How many moles of nickel are produced if 8. 7 x 1025 atoms of iridium are consumed? 8. 7 x 1025 at. Ir 1 mol Ir 6. 02 x 1023 at. Ir Ir 3 mol Ni 2 mol Ir = 220 mol Ni iridium (Ir) nickel (Ni) Ni

4. What volume hydrogen gas is liberated (at STP) if 50 g zinc react w/excess hydrochloric acid (HCl)? Zn H 2 1 Zn + __ 2 HCl __ 50 g excess 50 g Zn 1 mol Zn 65. 4 g Zn 1 H 2 + __ 1 Zn. Cl 2 __ ? L 1 mol H 2 1 mol Zn 22. 4 L H 2 1 mol H 2 = 20 L H 2

5. At STP, how many m’cules oxygen react with 632 dm 3 butane (C 4 H 10)? C 4 H 10 1 C 4 H 10 + __ 2 13 O 2 __ 632 dm 3 C 4 H 10 O 2 4 CO 2 + __ 5 H 2 O 8 10 __ 1 mol C 4 H 10 22. 4 dm 3 C 4 H 10 13 mol O 2 2 mol C 4 H 10 6. 02 x 1023 m’c O 2 1 mol O 2 = 1. 10 x 1026 m’c O 2 Suppose the question had been, “How many ATOMS of oxygen…” 1. 10 x 1026 m’c O 2 2 atoms O 1 m’c O 2 = 2. 20 x 1026 at. O

Energy and Stoichiometry CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) + 891 k. J A balanced eq. gives the ratios of moles-to-moles AND moles-to-energy. E CH 4 1. How many k. J of energy are released when 54 g methane are burned? 54 g CH 4 1 mol CH 4 891 k. J 16. 04 g CH 4 1 mol CH 4 = 3. 0 x 103 k. J

CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) + 891 k. J 2. At STP, what volume oxygen is consumed in producing 5430 k. J of energy? 5430 k. J 2 mol O 2 891 k. J 22. 4 L O 2 1 mol O 2 3. What mass of water is made if 10, 540 k. J are released? 10, 540 k. J 2 mol H 2 O 891 k. J 18. 02 g H 2 O 1 mol H 2 O E O 2 = 273 L O 2 E H 2 O = 426. 2 g H 2 O

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

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

Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. 2 Al(s) + 3 Cl 2(g) 2 Al. Cl 3(s) If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? Al 125 g Al 1 mol Al 26. 98 g Al 2 mol Al. Cl 3 2 mol Al Al. Cl 3 133. 34 g Al. Cl 3 1 mol Al. Cl 3 = 618 g Al. Cl 3 If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? Cl 2 Al. Cl 3 125 g Cl 2 1 mol Cl 2 2 mol Al. Cl 3 133. 34 g Al. Cl 3 70. 91 g Cl 2 3 mol Cl 2 1 mol Al. Cl 3 = 157 g Al. Cl 3

2 Al(s) + 3 Cl 2(g) 2 Al. Cl 3(s) If 125 g aluminum react w/125 g chlorine, how many g aluminum chloride are made? 157 g Al. Cl 3 (We’re out of Cl 2…) 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). Think of this analogy…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.

How to Find the Limiting Reactant For the generic reaction RA + R B P, assume that the amounts of RA and RB are given. Should you use RA or RB in your calculations? 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 si tcu de

For the Al / Cl 2 / Al. Cl 3 example: ER 2 Al(s) + 3 Cl 2(g) LR 2 Al. Cl 3(s) If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? 125 g Al 1 mol Al 26. 98 g Al 2 mol Al. Cl 3 2 mol Al Al Al. Cl 3 133. 34 g Al. Cl 3 1 mol Al. Cl 3 = 618 g Al. Cl 3 If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? Cl 2 Al. Cl 3 125 g Cl 2 1 mol Cl 2 2 mol Al. Cl 3 133. 34 g Al. Cl 3 70. 91 g Cl 2 3 mol Cl 2 1 mol Al. Cl 3 = 157 g Al. Cl 3

Limiting Reactant Practice 2 Fe(s) + 3 Cl 2(g) 223 g Fe 179 L Cl 2 2 Fe. Cl 3(s) Which is the limiting reactant: Fe or Cl 2? 223 g Fe 1 mol Fe 55. 85 g Fe 2 mol Fe. Cl 3 2 mol Fe 162. 21 g Fe. Cl 3 1 mol Fe. Cl 3 = 648 g Fe. Cl 3 179 L Cl 2 1 mol Cl 2 22. 4 L Cl 2 2 mol Fe. Cl 3 162. 21 g Fe. Cl 3 3 mol Cl 2 1 mol Fe. Cl 3 = 864 g Fe. Cl 3 How many g Fe. Cl 3 are produced? *Remember that the LR “limits” how much product can be made!

2 H 2(g) + O 2(g) 13 g H 2 2 H 2 O(g) 80 g O 2 Which is the limiting reactant: H 2 or O 2? 13 g H 2 1 mol H 2 2. 02 g H 2 2 mol H 2 O 2 mol H 2 18. 02 g H 2 O 1 mol H 2 O = 120 g H 2 O 80 g O 2 1 mol O 2 2 mol H 2 O 18. 02 g H 2 O 32. 00 g O 2 1 mol H 2 O = 90 g H 22 O How many g H 2 O are produced? *Notice that the LR doesn’t always have the smaller amount (13 v. 80)

How many g O 2 are left over? zero; O 2 is the LR and therefore is all used up How many g H 2 are left over? 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. Start with the LR and relate to the other… 80 g O 2 1 mol O 2 2 mol H 2 2. 02 g H 2 32. 00 g O 2 1 mol H 2 HAD 13 g, USED 10 g… H 2 = 10 g H 2 USED 3 g H 2 left over

2 Fe(s) + 3 Br 2(g) 181 g Fe 96. 5 L Br 2 2 Fe. Br 3(s) Which is the limiting reactant: Fe or Br 2? 181 g Fe 1 mol Fe 55. 85 g Fe 2 mol Fe. Br 3 2 mol Fe 295. 55 g Fe. Br 3 1 mol Fe. Br 3 = 958 g Fe. Br 3 96. 5 L Br 2 1 mol Br 2 22. 4 L Br 2 2 mol Fe. Br 3 295. 55 g Fe. Br 3 3 mol Br 2 1 mol Fe. Br 3 = 849 g Fe. Br 33 How many g Fe. Br 3 are produced?

2 Fe(s) + 3 Br 2(g) 181 g Fe 96. 5 L Br 2 2 Fe. Br 3(s) How many g of the ER are left over? 96. 5 L Br 2 1 mol Br 2 2 mol Fe 22. 4 L Br 2 3 mol Br 2 HAD 181 g, USED 160. 4 g… Br 2 Fe 55. 85 g Fe = 160. g Fe USED 1 mol Fe 21 g Fe left over

Al 3+ O 2– Percent Yield Na+ O 2– solid molten aluminum sodium aluminum oxide 6 Na(l) + 1 Al 2 O 3(s) 2 Al(l) + 3 Na 2 O(s) Find mass of aluminum produced if you start w/575 g sodium and 357 g aluminum oxide. 575 g Na 1 mol Na 22. 99 g Na 2 mol Al 6 mol Na 26. 98 g Al 1 mol Al = 225 g Al 357 g Al 2 O 3 1 mol Al 2 O 3 2 mol Al 101. 96 g Al 2 O 3 1 mol Al 2 O 3 26. 98 g Al 1 mol Al = 189 g Al Al

189 g This amt. of product (______) is theoretical yield. • 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 Al 2 O 3 reacted • some reactant or product spilled and was lost

Note: % yield should never be > 100% Batting average FG % GPA Find % yield for previous problem. = 91. 0%

On NASA spacecraft, lithium hydroxide “scrubbers” remove toxic CO 2 from cabin. CO 2(g) + 2 Li. OH(s) Li 2 CO 3(s) + H 2 O(l) 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? percent yield ( 880 g CO 2 person-day 24, 640 g CO 2 ) x (4 p) x (7 d) = 24, 640 g CO 2 Li 2 CO 3 1 mol CO 2 1 mol Li 2 CO 3 73. 9 g Li 2 CO 3 = 41, 384 g 44. 0 g CO 2 1 mol Li 2 CO 3 “theo” yield A = 31, 000 g Li 2 CO 3

2. Automobile air bags inflate with nitrogen via the decomposition of sodium azide: 2 Na. N 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? percent yield “act” yield x = 87. 1 L N 2 → “Theo” yield 87. 1 L N 2 1 mol N 2 22. 4 L N 2 2 mol Na. N 3 65 g Na. N 3 3 mol N 2 Na. N 3 1 mol Na. N 3 = 170 g Na. N 3

2 3 2 ___Zn. S + ___O 2 2 ___Zn. O + ___SO 2 100 g X g ? (assuming 81% yield) Strategy: 1. Balance and find LR 2. Use LR to calc. X g Zn. O (theo. yield) 3. Actual yield is 81% of theo. yield Zn. O 100 g Zn. S 1 mol Zn. S 97. 5 g Zn. S 2 mol Zn. O 81. 4 g Zn. O = 83. 5 g Zn. O 2 mol Zn. S 1 mol Zn. O 100 g O 2 1 mol O 2 2 mol Zn. O 81. 4 g Zn. O 32 g O 2 3 mol O 2 1 mol Zn. O = 169. 6 g Zn. O x = 67. 6 g Zn. O

2 1 ___Al + ___Fe 2 O 3 X g? 2 1 ___Fe + ___Al 2 O 3 X g? 800. g needed “act” yield **Rxn. has an 80. % yield. “theo” = 1000 g Fe Fe 1000 g Fe 1 mol Fe 2 mol Al 55. 85 g Fe 2 mol Fe 26. 98 g Al 1 mol Al Al = 480 g Al Fe Fe 2 O 3 1000 g Fe 1 mol Fe 2 O 3 159. 7 g Fe 2 O 3 = 1400 g Fe 2 O 3 1 mol Fe 2 O 3 55. 85 g Fe 2 mol Fe

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

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