Quantitative Relationships in Chemical Equations

4 Na(s) + O2(g) 2 Na2O(s)

Particles 4 atoms 1 m’cule 2 m’cules

Moles 4 mol 1 mol 2 mol

Grams 4 g 1 g 2 g

** Coefficients of a balanced equation represent# of particles OR # of moles,

but NOT # of grams.

When going from moles of one substance to molesof another, use coefficients from balanced equation.

part.

vol.

mass

MOL

mass

vol.

part.

MOL

SUBSTANCE “A” SUBSTANCE “B”(known) (unknown)

Use coeff. from balancedequation in crossing this bridge

4 Na(s) + 1 O2(g) 2 Na2O(s)

= 4.20 mol O2

4 mol Na87.2 mol Na

4 Na(s) + O2(g) 2 Na2O(s)

How many moles oxygen will react with 16.8 moles sodium?

How many moles sodium oxide are produced from87.2 moles sodium?

How many moles sodium are requiredto produce 0.736 moles sodium oxide?

4 mol Na

1 mol O216.8 mol Na

= 43.6 mol Na2O

2 mol Na2O4 mol Na0.736 mol Na2O = 1.47 mol Na

2 mol Na2O

O2Na Na2ONa2O Na

Unit 8: Stoichiometry

-- involves finding amts. of reactants & products in a reaction

amount of RA

and/or RB

you must use

amount of P1

or P2 you needto produce

amount of P1 orP2 that will be

produced

amount of RA

or RB

amount of RB (or RA) that is needed to

react with it

amount of RA

(or RB)

…one can find the…Given the…

What can we do with stoichiometry?

For generic equation: RA + RB P1 + P2

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 dm3)

1 mol = 22.4 L1 mol = 22.4 dm3

1 mol = 6.02 x 1023 particles

SUBSTANCE “A”

Stoichiometry Island Diagram

MOLE(mol)

Mass(g)

1 mol = molar mass (in g)

Volume(L or dm3)

1 mol = 22.4 L1 mol = 22.4 dm3

1 mol = 6.02 x 1023 particles

SUBSTANCE “B”

Particle(at. or m’c)

2__TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO

How many mol chlorine will react with 4.55 mol carbon?

3 mol C

4 mol Cl24.55 mol C = 6.07 mol Cl2

What mass titanium (IV) oxide will react with 4.55 mol carbon?

= 242 g TiO2

4 3 2 21

C Cl2

C TiO2 3 mol C2 mol TiO24.55 mol C

1 mol TiO2

79.9 g TiO2

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 madefrom 115 g titanium (IV) oxide?

TiCl4TiO2

( )= 8.66 x 1023 m’c TiCl4

115 g TiO21 mol TiO2

79.9 g TiO2( )2 mol TiO2

2 mol TiCl4 ( )1 mol TiCl4

6.02 x 1023 m’c TiCl4

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 mol1 mol

1 mol1 molcoeff. 1 mol

1 mol

2 Ir + Ni3P2 3 Ni + 2 IrP If 5.33 x 1028 m’cules nickel (II) phosphidereact w/excess iridium, what mass iridium (III)phosphide is produced?

Ni3P2 IrP

= 3.95 x 107 g IrP

1 mol IrP223.2 g IrP

1 mol Ni3P2

2 mol IrP1 mol Ni3P2

6.02 x 1023 m’c Ni3P2

5.33 x 1028 m’c Ni3P2

How many grams iridium will react with465 grams nickel (II) phosphide?

= 751 g Ir

1 mol Ir192.2 g Ir

1 mol Ni3P2

2 mol Ir1 mol Ni3P2

238.1 g Ni3P2

465 g Ni3P2

Ni3P2 Ir

2 mol Ir

How many moles of nickel are producedif 8.7 x 1025 atoms of iridium are consumed?

Ir Ni

= 220 mol Ni

3 mol Ni1 mol Ir6.02 x 1023 at. Ir

8.7 x 1025 at. Ir

2 Ir + Ni3P2 3 Ni + 2 IrP

iridium (Ir) nickel (Ni)

Zn

What volume hydrogen gas is liberated(at STP) if 50. g zinc react w/excesshydrochloric acid (HCl)?

__ Zn + __ HCl __ H2 + __ ZnCl2 1 2 1 1

H2

= 17 L H2

1 mol H2

22.4 L H2

1 mol Zn1 mol H21 mol Zn

65.4 g Zn50. g Zn

50. g excess ? L

At STP, how many m’cules oxygen reactwith 632 dm3 butane (C4H10)?

C4H10 O2

= 1.10 x 1026 m’c O2

1 mol O22 mol C4H10

13 mol O21 mol C4H10632 dm3 C4H10

22.4 dm3 C4H10

__ C4H10 + __ O2 __ CO2 + __ H2O 1 4 52 8 1013

6.02 x 1023 m’c O2

Suppose the question had been “how many ATOMS of O2…”

1.10 x 1026 m’c O2

1 m’c O2

2 atoms O = 2.20 x 1026 at. O

1 mol CH4

A balanced eq. gives the ratios of moles-to-moles

Energy and Stoichiometry

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ

How many kJ of energy are releasedwhen 54 g methane are burned?

AND moles-to-energy.

= 3.0 x 103 kJ1 mol CH4

891 kJ54 g CH4

16 g CH4

EECH4

10,540 kJ

1 mol H2O

What mass of water is made if10,540 kJ are released?

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ

At STP, what volume oxygen is consumedin producing 5430 kJ of energy?

2 mol O2 = 273 L O21 mol O2

22.4 L O25430 kJ891 kJ

E

2 mol H2O = 425.9 g H2O18 g H2O

891 kJ

E O2

EE H2O

The Limiting Reactant

A balanced equation for makinga Big Mac® might be:

3 B + 2 M + EE B3M2EE

30 B and excess EE30 M

excess M and excess EE30 B

excess B and excess EE30 M

…one can make…

…and…With…

15 B3M2EE

10 B3M2EE

10 B3M2EE

50 P

A balanced equation for makinga big wheel might be:

…one can make…

…and…With…

50 S + excess of all other reactants

excess of all other reactants50 S

excess of all other reactants50 P 25 W3P2SHF

3 W + 2 P + S + H + F W3P2SHF

50 W3P2SHF

25 W3P2SHF

Solid aluminum reacts w/chlorine gas to yield solidaluminum chloride.

2 Al(s) + 3 Cl2(g) 2 AlCl3(s) If 125 g aluminum react w/excess chlorine,how many g aluminum chloride are made?

= 618 g AlCl3

2 mol Al 1 mol AlCl3

1 mol Al27 g Al

125 g AlAl AlCl3

133.5 g AlCl32 mol AlCl3

If 125 g chlorine react w/excess aluminum,how many g aluminum chloride are made?

= 157 g AlCl3

3 mol Cl2 1 mol AlCl3

1 mol Cl271 g Cl2

125 g Cl2

Cl2 AlCl3133.5 g AlCl32 mol AlCl3

2 Al(s) + 3 Cl2(g) 2 AlCl3(s)

If 125 g aluminum react w/125 g chlorine,how many g aluminum chloride are made?

157 g AlCl3 (We’re out of Cl2…)

limiting reactant (LR): the reactant that runs out first

--

Any reactant you don’t run outof is an excess reactant (ER).

amount of product is “limited” by the LR

In a root beer float, the LRis usually the ice cream.

Deliciousness!(What’s the product?)

Al / Cl2 / AlCl3

tricycles

Big Macs

Excess Reactant(s)Limiting ReactantFrom Examples Above…

157 g AlCl3125 g Cl2125 g Al

25 W3P2SHF50 S + excess of all other reactants50 P

10 B3M2EE30 M

…one can make……and…With…

30 B and excess EE

B

P

Cl2

M, EE

W, S, H, F

Al

How to Find the Limiting ReactantFor the generic reaction

RA + RB P,assume that the amountsof 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 calculations)

2. Whichever reactant produces the smaller amount of product is the LR3. The smaller amount of product is the maximum amount produced

For the Al / Cl2 / AlCl3 example:

2 Al(s) + 3 Cl2(g) 2 AlCl3(s)

If 125 g aluminum react w/excess chlorine,how many g aluminum chloride are made?

= 618 g AlCl3

2 mol Al 1 mol AlCl3

1 mol Al27 g Al

125 g AlAl AlCl3

133.5 g AlCl32 mol AlCl3

If 125 g chlorine react w/excess aluminum,how many g aluminum chloride are made?

= 157 g AlCl3

3 mol Cl2 1 mol AlCl3

1 mol Cl271 g Cl2

125 g Cl2

Cl2 AlCl3133.5 g AlCl32 mol AlCl3

LR

2 Fe(s) + 3 Cl2(g) 2 FeCl3(s) 223 g Fe 179 L Cl2

Which is the limiting reactant: Fe or Cl2?

1 mol Fe55.8 g Fe

223 g Fe

1 mol Cl222.4 L Cl2

179 L Cl2

How many g FeCl3 are produced?

= 649 g FeCl3

2 mol FeCl32 mol Fe

162.3 g FeCl31 mol FeCl3

2 mol FeCl33 mol Cl2

162.3 g FeCl31 mol FeCl3

= 865 g FeCl3

649 g FeCl3

*Remember that the LR “limits” how much product can be made!

2 H2(g) + O2(g) 2 H2O(g) 13 g H2 80 g O2

Which is the limiting reactant: H2 or O2?

1 mol H2

2.0 g H2

13 g H2

1 mol O2

32.0 g O2

80 g O2

How many g H2O are produced?

= 120 g H2O

2 mol H2O2 mol H2

18.0 g H2O1 mol H2O

2 mol H2O

1 mol O2

18.0 g H2O1 mol H2O

= 90 g H2O90 g H2O

*Notice that the LR doesn’t always have the smaller amount (13 v. 80)

How many g O2 are left over?

How many g H2 are left over?

zero; O2 is the LR and therefore is all used up

We know how much H2 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.

H2O2

HAD 13 g, USED 10 g…

Start with the LR and relate to the other…

1 mol O2

32.0 g O2

80 g O2 2 mol H2

1 mol O2

2.0 g H2

1 mol H2

10 g H2

USED=

3 g H2 left over

181 g Fe 96.5 L Br2

Which is the limiting reactant: Fe or Br2?

1 mol Fe55.85 g Fe

181 g Fe

1 mol Br2

22.4 L Br2

96.5 L Br2

How many g FeBr3 are produced?

= 958 g FeBr3

2 mol FeBr3

2 mol Fe295.55 g FeBr3

1 mol FeBr3

2 mol FeBr3

3 mol Br2

295.55 g FeBr3

1 mol FeBr3

= 849 g FeBr3849 g FeBr3

2 Fe(s) + 3 Br2(g) 2 FeBr3(s)

How many g of the ER are left over?

FeBr2

181 g Fe 96.5 L Br2

2 Fe(s) + 3 Br2(g) 2 FeBr3(s)

HAD 181 g, USED 160.4 g…

1 mol Br2

22.4 L Br2

96.5 L Br2 2 mol Fe

3 mol Br2

55.85 g Fe

1 mol Fe 160.4 g Fe USED=

20.6 g Fe left over

Percent Yield

moltensodium

solidaluminum

oxide

solidaluminum

solidsodiumoxide

Find mass of aluminum produced if you start w/575 g sodiumand 357 g aluminum oxide.

+ Al2O3(s) Al(s)6 Na(l) + Na2O(s)1 2 3

Na+ O2–Al3+ O2–

Al1 mol Na22.99 g Na

575 g Na

1 mol Al2O3357 g Al2O3

= 224.9 g Al

2 mol Al6 mol Na

26.98 g Al1 mol Al

2 mol Al

1 mol Al2O3

26.98 g Al1 mol Al

= 188.9 g Al188.9 g Al

102 g Al2O3

Now suppose that we performthis reaction and get only 172grams 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 Al2O3 reactedsome reactant or product spilled and was lost

Actual yield

= 91.0%

Find % yield for previous problem.

% yield can never be > 100%.

100 x yieldltheoretica

yieldactual yield%

--

100 x yld.theo.

yld.act. yield% 100 x Alg 189 Alg 172

Batting average

FG %

GPA

= 41,374 g Li2CO3

On NASA spacecraft, lithium hydroxide “scrubbers” removetoxic CO2 from cabin.

For a seven-day mission, each of four individuals exhales880 g CO2 daily. If reaction is 75% efficient, how many g Li2CO3 will actually be produced?

CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l)

CO2 Li2CO3

( )880 g CO2

person-dayx (4 p) x (7 d) = 24,640 g CO2

percent yield

1 mol CO224, 640 g CO2 1 mol Li2CO3

1 mol CO2

73.9 g Li2CO3

1 mol Li2CO344 g CO2

“theo” yieldAct%Y= 100Theo

75= 10041,373.9

xg

x = 31030 g Li2CO3

On NASA spacecraft, lithium hydroxide “scrubbers” removetoxic CO2 from cabin.

For a seven-day mission, each of four individuals exhales880 g CO2 daily. If reaction is 75% efficient, how many g LiOHshould be brought along?

CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l)

= 26,768 g LiOH

1 mol LiOH23.9 g LiOH( )1 mol CO2

2 mol LiOH1 mol CO2

44 g CO2

24,640 g CO2

CO2 LiOH

( )880 g CO2

person-dayx (4 p) x (7 d) = 24,640 g CO2

( ) ( )theo. = 36,000 g LiOH

REALITY: TAKE 100,000 gtheo.

(act.) = 0.75

Reaction that powers space shuttle is: 2 H2(g) + O2(g) 2 H2O(g) + 572 kJ

From 100 g hydrogen and 640 g oxygen, what amount ofenergy is possible?

EE

1 mol H2 = 14300 kJ2 mol H2

572 kJ100 g H2

2 g H2

1 mol O2 = 11440 kJ1 mol O2

572 kJ640 g O2

32 g O2

11440 kJ

Review Questions

What mass of excess reactant is left over?

2 H2(g) + O2(g) 2 H2O(g) + 572 kJ

H2O2

Started with 100 g, used up 80 g… 20 g H2 left over

1 mol O2640 g O2 2 mol H2

1 mol O2

2 g H2

1 mol H2

= 80 g H232 g O2

100 g 640 g

Automobile air bags inflate with nitrogen via the decompositionof sodium azide: 2 NaN3(s) 3 N2(g) + 2 Na(s)At STP and a % yield of 85%, what mass sodium azideis needed to yield 74 L nitrogen?

N2 NaN3

percent yield“act” yield

Act%Y= 100Theo

74L85= 100x

x = 87.1 L N2 → “Theo” yield

1 mol N287.1 L N2 2 mol NaN3

3 mol N2

65 g NaN3

1 mol NaN3

= 168.5 g NaN3

22.4 L N2

1 mol B2H6

B2H6 + 3 O2 B2O3 + 3 H2O10 g 30 g X g?

B2O31 mol B2H6

27.6 g B2H6

10 g B2H6

1 mol O230 g O2

= 25.2 g B2O3

1 mol B2O3 69.6 g B2O3

1 mol B2O3

1 mol B2O3

3 mol O2

69.6 g B2O3

1 mol B2O3

= 21.75 g B2O321.75 g B2O3

32 g O2

___ZnS + ___O2 ___ZnO + ___SO2

100 g 100 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

2 3 22

ZnO

2 mol ZnS1 mol ZnS97.5 g ZnS

100 g ZnS

1 mol O2100 g O2

= 83.5 g ZnO2 mol ZnO 81.4 g ZnO

1 mol ZnO

2 mol ZnO

3 mol O2

81.4 g ZnO1 mol ZnO

= 169.6 g ZnO

83.5 g ZnO

32 g O2

Act%Y= 100Theo

x81= 100

83.5g ZnO x = 67.6 g ZnO

___Al + ___Fe2O3 ___Fe + ___Al2O3 X g? X g? 800 g needed **Rxn. has an

80% yield.

2 1 12

Fe Al

Fe Fe2O3

“act” yieldAct%Y= 100Theo

800g80= 100x

“theo” = 1000 g Fe

2 mol Fe1 mol Fe55.85 g Fe

1000 g Fe= 483.1 g Al2 mol Al 26.98 g Al

1 mol Al

2 mol Fe1 mol Fe55.85 g Fe

1000 g Fe= 1430 g Fe2O3

1 mol Fe2O3 159.7 g Fe2O3

1 mol Fe2O3