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

Feb 23, 2016

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Unit 8: Stoichiometry . -- involves finding amts. of reactants & products in a reaction. For generic equation: R A + R B P 1 + P 2. …one can find the…. Given the…. What can we do with stoichiometry?. amount of R B (or R A ) that is needed to react with it. amount of R A - PowerPoint PPT Presentation
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Page 1: Unit 8: Stoichiometry

Unit 8: Stoichiometry

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

Page 2: Unit 8: Stoichiometry

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

Page 3: Unit 8: Stoichiometry

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®

Page 4: Unit 8: Stoichiometry

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)

Page 5: Unit 8: Stoichiometry

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

Page 6: Unit 8: Stoichiometry

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

( )= 8.7 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

Page 7: Unit 8: Stoichiometry

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 IrP( )1 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 Ir( )1 mol Ni3P2

238.1 g Ni3P2

465 g Ni3P2

Ni3P2 Ir

Page 8: Unit 8: Stoichiometry

2 mol Ir

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

Ir Ni

= 217 mol Ni

( )3 mol Ni( )1 mol Ir6.02 x 1023 at. Ir

8.7 x 1025 at. Ir

2 Ir + Ni3P2 3 Ni + 2 IrP

iridium (Ir) nickel (Ni)

Page 9: Unit 8: Stoichiometry

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

( )= 17.1 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 X L

Page 10: Unit 8: Stoichiometry

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

C4H10 O2

= 1.10 x 1026 m’c O2

1 mol O2( )2 mol C4H10

13 mol O2( )1 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

Page 11: Unit 8: Stoichiometry

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.

= 3007 kJ( )1 mol CH4

891 kJ54 g CH4

16 g CH4( )

EECH4

Page 12: Unit 8: Stoichiometry

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 O2( )1 mol O2

22.4 L O25430 kJ891 kJ( )

E

2 mol H2O = 426 g H2O( )1 mol H2O18 g H2O10,540 kJ

891 kJ( )

E O2

EE H2O

Page 13: Unit 8: Stoichiometry

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

Page 14: Unit 8: Stoichiometry

50 P

A balanced equation for makinga tricycle 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

Page 15: Unit 8: Stoichiometry

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 AlCl3

( ) 133.5 g AlCl32 mol AlCl3

Page 16: Unit 8: Stoichiometry

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 based on LR

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

Your enjoyment!(What’s the product?)

Page 17: Unit 8: Stoichiometry

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

Page 18: Unit 8: Stoichiometry

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. Calc. # of mol of RA and RB you have.

2. Divide by the respective coefficients in balanced equation. 3. Reactant having the smaller result is the LR.

Page 19: Unit 8: Stoichiometry

2

For the Al / Cl2 / AlCl3 example:

1 mol Al27 g Al

125 g Al( ) = 4.63 mol Al (HAVE)

1 mol Cl271 g Cl2

125 g Cl2( ) = 1.76 mol Cl2 (HAVE)

Step #1

._.

._. 3 = 0.58

= 2.31

Step #2

LR

Step #3

“Oh, bee- HAVE !”

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

(And start every calc. with the LR.)

Page 20: Unit 8: Stoichiometry

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

Which is the limiting reactant: Fe or Cl2?

21 mol Fe55.8 g Fe

223 g Fe( ) = 4.0 mol Fe (HAVE)

1 mol Cl222.4 L Cl2

179 L Cl2( ) = 8.0 mol Cl2 (HAVE)

._.

._. 3 = 2.66

= 2.0 LR= Fe

How many g FeCl3 are produced?

= 649 g FeCl3( )2 mol Fe 1 mol FeCl3( )4.0 mol Fe 162.3 g FeCl32 mol FeCl3

FeCl3Fe(“Oh, bee-HAVE!”)

Page 21: Unit 8: Stoichiometry

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

Which is LR: H2 or O2?

21 mol H2

2 g H2

13 g H2( ) = 6.5 mol H2

(HAVE)

1 mol O2

32 g O2

80 g O2( ) = 2.5 mol O2

(HAVE)

._.

._. 1 = 2.50

= 3.25

LR= O2

How many g H2O are formed?

= 90 g H2O( )1 mol O2 1 mol H2O( )2.5 mol O2 18 g H2O2 mol H2O

H2OO2(“Oh, bee-HAVE!”)

Page 22: Unit 8: Stoichiometry

2 g H2

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 started with (i.e., 13 g). To find how much is left over, we first need to figure out how much was USED UP in the reaction.

= 10 g H2 used up( )1 mol O2 1 mol H2( )2.5 mol O2 2 mol H2

H2O2

Started with 13 g, used up 10 g… 3 g H2 left over

Page 23: Unit 8: Stoichiometry

2 Fe(s) + 3 Br2(g) 2 FeBr3(s) 181 g Fe 96.5 L Br2

Find LR.

21 mol Fe55.8 g Fe

181 g Fe( ) = 3.24 mol Fe (HAVE)

1 mol Br2

22.4 L Br2

96.5 L Br2( ) = 4.31 mol Br2

(HAVE)

._.

._. 3 = 1.44

= 1.62

LR= Br2

How many g FeBr3 are formed?

= 849 g FeBr3( )3 mol Br2 1 mol FeBr3( )4.31 mol Br2 295.5 g FeBr32 mol FeBr3

FeBr3Br2(“Oh, bee-HAVE!”)

Page 24: Unit 8: Stoichiometry

How many g of the ER are left over?

55.8 g Fe= 160 g Fe used up

( )3 mol Br2 1 mol Fe( )4.31 mol Br2 2 mol FeFeBr2

Started with 181 g, used up 160 g… 21 g Fe left over

2 Fe(s) + 3 Br2(g) 2 FeBr3(s) 181 g Fe 96.5 L Br2

Page 25: Unit 8: Stoichiometry

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

Na1+ O2–Al3+ O2–

61 mol23 g

575 g Na( ) = 25 mol Na (HAVE)

1 mol102 g

357 g Al2O3( ) = 3.5 mol Al2O3

(HAVE)

._.

._. 1 = 3.5

= 4.17

LR

= 189 g Al1 mol Al2O3 1 mol Al

3.5 mol Al2O3 27 g Al2 mol AlAlAl2O3

( )( )

Page 26: Unit 8: Stoichiometry

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

189 g

couldn’t collect all Al not all Na and Al2O3 reactedsome reactant or product spilled and was lost

Page 27: Unit 8: Stoichiometry

= 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

Page 28: Unit 8: Stoichiometry

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?

21 mol H2

2 g H2

100 g H2( ) = 50 mol H2

(HAVE)

1 mol O2

32 g O2

640 g O2( ) = 20 mol O2

(HAVE)

._.

._. 1 = 20

= 25

LR

= 11,440 kJ( )1 mol O2

20 mol O2 572 kJ

EEO2

Page 29: Unit 8: Stoichiometry

What mass of excess reactant is left over?

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

2 g H2 = 80 g H2

used up( )1 mol O2 1 mol H2

( )20 mol O2 2 mol H2

H2O2

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

Page 30: Unit 8: Stoichiometry

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

( ) ( )(if reactionis perfect) (Need morethan this!) ( 0.75) ._.

= 35,700 g LiOH

REALITY: TAKE 100,000 g

Page 31: Unit 8: Stoichiometry

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?

Shoot for… 74 L ( 0.85) ._.

= 87.1 L N2 N2 NaN3

1 mol NaN3

65 g NaN3( )3 mol N2

2 mol NaN31 mol N2

22.4 L N2

87.1 L N2( ) = 169 g NaN3

( )__C3H8 + __O2 __CO2 + __H2O + __kJ

Strategy: 1. 2.

X g Y g ? kJ

Conceptual question (How would you do it?)

Find LR. Calc. ? kJ. EELR

Page 32: Unit 8: Stoichiometry

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

11 mol27.6 g

10 g B2H6 ( ) = 0.362 mol B2H6

(HAVE)

1 mol32 g

30 g O2 ( ) = 0.938 mol O2

(HAVE)

._.

._. 3 = 0.313

= 0.362

LR

= 21.8 g B2O3( )3 mol O2 1 mol B2O3( )0.938 mol O2 69.6 g B2O31 mol B2O3

B2O3O2

Page 33: Unit 8: Stoichiometry

___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

21 mol97.5 g

100 g ZnS( )= 1.026 mol ZnS (HAVE)

1 mol32 g

100 g O2( ) = 3.125 mol O2

(HAVE)

._.

._. 3 = 1.042

= 0.513 LR

= 83.5 g ZnO( )2 mol ZnS 1 mol ZnO( )1.026 mol ZnS 81.4 g ZnO2 mol ZnOZnOZnS

Actual g ZnO = 83.5 (0.81) = 68 g ZnO

Page 34: Unit 8: Stoichiometry

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

80% yield.

2 1 12

Shoot for… 800 g Fe ( 0.80) ._.

= 1000 g Fe

Fe Al

1 mol Al27 g Al( )2 mol Fe

2 mol Al1 mol Fe55.8 g Fe

1000 g Fe( ) = 484 g Al( )

Fe Fe2O3

1 mol Fe2O3

159.6 g Fe2O3( )2 mol Fe1 mol Fe2O31 mol Fe

55.8 g Fe1000 g Fe( ) = 1430 g

Fe2O3( )

Page 35: Unit 8: Stoichiometry
Page 36: Unit 8: Stoichiometry

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