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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 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 O2(g) 2 Na2O(s) 
= 4.20 mol O2
4 mol Na
87.2 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 from
87.2 moles sodium?
How many moles sodium are required
to produce 0.736 moles sodium oxide?
4 mol Na
1 mol O2
16.8 mol Na
= 43.6 mol Na2O
2 mol Na2O
4 mol Na
0.736 mol Na2O
= 1.47 mol Na
2 mol Na2O
O2
Na
Na2O
Na2O
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 need
to produce
amount of P1 or
P2 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 bread
50 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 L
1 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 L
1 mol = 22.4 dm3
1 mol = 6.02 x 1023 particles
SUBSTANCE “B”
 Particle
(at. or m’c)
Use coefficients from balanced equation
Mass
(g)
 Particle
(at. or m’c)
Volume
(L or dm3)
Mass
(g)
 Particle
(at. or m’c)
Volume
(L or dm3)
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 dm3
1 mol = 22.4 L
1 mol = 22.4 dm3
1 mol = 6.02 x 1023 particles
1 mol = 6.02 x 1023 particles
2
__TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO 
How many mol chlorine will react with 4.55 mol carbon? 
3 mol C
4 mol Cl2
4.55 mol C
= 6.07 mol Cl2
What mass titanium (IV) oxide will react with 4.55 mol carbon?
= 242 g TiO2
4
3
2
2
1
C
Cl2
C
TiO2
3 mol C
2 mol TiO2
4.55 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 made
		from 115 g titanium (IV) oxide?
TiCl4
TiO2
(
)
= 8.66 x 1023 m’c TiCl4
115 g TiO2
1 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 mol
1 mol
1 mol
1 mol
coeff.
1 mol
1 mol
2 Ir + Ni3P2 3 Ni + 2 IrP 
If 5.33 x 1028 m’cules nickel (II) phosphide
react w/excess iridium, what mass iridium (III)
phosphide is produced?
Ni3P2
IrP
= 3.95 x 107 g IrP
1 mol IrP
223.2 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 with
465 grams nickel (II) phosphide?
= 751 g Ir
1 mol Ir
192.2 g Ir
1 mol Ni3P2
2 mol Ir
1 mol Ni3P2
238.1 g Ni3P2
465 g Ni3P2
Ni3P2
Ir
2 mol Ir
How many moles of nickel are produced
if 8.7 x 1025 atoms of iridium are consumed?
Ir
Ni
= 220 mol Ni
3 mol Ni
1 mol Ir
6.02 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/excess
hydrochloric acid (HCl)? 
__ Zn + __ HCl
__ H2
+ __ ZnCl2 
1
2
1
1
H2
= 17 L H2
1 mol H2
22.4 L H2
1 mol Zn
1 mol H2
1 mol Zn
65.4 g Zn
50. g Zn
50. g
excess
? L
At STP, how many m’cules oxygen react
with 632 dm3 butane (C4H10)? 
C4H10
O2
= 1.10 x 1026 m’c O2
1 mol O2
2 mol C4H10
13 mol O2
1 mol C4H10
632 dm3 C4H10
22.4 dm3 C4H10
__ C4H10 + __ O2
__ CO2 + __ H2O 
1
4
5
2
8
10
13
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 released
when 54 g methane are burned? 
AND moles-to-energy.
= 3.0 x 103 kJ
1 mol CH4
891 kJ
54 g CH4
16 g CH4
E
E
CH4
10,540 kJ
1 mol H2O
What mass of water is made if
10,540 kJ are released?
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) + 891 kJ 
At STP, what volume oxygen is consumed
in producing 5430 kJ of energy? 
2 mol O2
= 273 L O2
1 mol O2
22.4 L O2
5430 kJ
891 kJ
E
2 mol H2O
= 425.9 g H2O
18 g H2O
891 kJ
E
O2
E
E
H2O
The Limiting Reactant 
A balanced equation for making
a Big Mac® might be:
3 B + 2 M + EE B3M2EE
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 B3M2EE
10 B3M2EE
10 B3M2EE
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 W3P2SHF
3 W + 2 P + S + H + F W3P2SHF
50 W3P2SHF
25 W3P2SHF
Solid aluminum reacts w/chlorine gas to yield solid
aluminum 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 Al
27 g Al
125 g Al
Al
AlCl3
133.5 g AlCl3
2 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 Cl2
71 g Cl2
125 g Cl2
Cl2
AlCl3
133.5 g AlCl3
2 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 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 / Cl2 / AlCl3
tricycles
Big Macs
Excess Reactant(s)
Limiting Reactant
From Examples Above…
157 g AlCl3
125 g Cl2
125 g Al
25 W3P2SHF
50 S + excess of all other reactants
50 P
10 B3M2EE
30 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 Reactant
For the generic reaction
 RA + RB 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 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 / 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 Al
27 g Al
125 g Al
Al
AlCl3
133.5 g AlCl3
2 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 Cl2
71 g Cl2
125 g Cl2
Cl2
AlCl3
133.5 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 Fe
55.8 g Fe
223 g Fe
1 mol Cl2
22.4 L Cl2
179 L Cl2
How many g FeCl3 are produced?
= 649 g FeCl3
2 mol FeCl3
2 mol Fe
162.3 g FeCl3
1 mol FeCl3
2 mol FeCl3
3 mol Cl2
162.3 g FeCl3
1 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 H2O
2 mol H2
18.0 g H2O
1 mol H2O
2 mol H2O
1 mol O2
18.0 g H2O
1 mol H2O
= 90 g H2O
90 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.
H2
O2
 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 Fe
55.85 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 Fe
295.55 g FeBr3
1 mol FeBr3
2 mol FeBr3
3 mol Br2
295.55 g FeBr3
1 mol FeBr3
= 849 g FeBr3
849 g FeBr3
2 Fe(s) + 3 Br2(g) 2 FeBr3(s) 
How many g of the ER are left over?
Fe
Br2
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 
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. 
+ Al2O3(s)
Al(s)
6
Na(l)
+ Na2O(s)
1
2
3
Na+
O2–
Al3+
O2–
Al
1 mol Na
22.99 g Na
575 g Na
1 mol Al2O3
357 g Al2O3
= 224.9 g Al
2 mol Al
6 mol Na
26.98 g Al
1 mol Al
2 mol Al
1 mol Al2O3
26.98 g Al
1 mol Al
= 188.9 g Al
188.9 g Al
102 g Al2O3
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 Al2O3 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 
Li2CO3
On NASA spacecraft, lithium hydroxide “scrubbers” remove
toxic CO2 from cabin. 
For a seven-day mission, each of four individuals exhales
880 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-day
x (4 p) x (7 d)
= 24,640 g CO2
percent yield
1 mol CO2
24, 640 g CO2
1 mol Li2CO3
1 mol CO2
73.9 g Li2CO3
1 mol Li2CO3
44 g CO2
“theo” yield
x = 31030 g Li2CO3
On NASA spacecraft, lithium hydroxide “scrubbers” remove
toxic CO2 from cabin. 
For a seven-day mission, each of four individuals exhales
880 g CO2 daily. If reaction is 75% efficient, how many g LiOH
should be brought along? 
CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l)
= 26,768 g LiOH
1 mol LiOH
23.9 g LiOH
(
)
1 mol CO2
2 mol LiOH
1 mol CO2
44 g CO2
24,640 g CO2
CO2
LiOH
(
)
880 g CO2
person-day
x (4 p) x (7 d)
= 24,640 g CO2
(
)
(
)
theo. = 36,000 g LiOH
REALITY: TAKE 100,000 g
theo.
(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 of
energy is possible?
E
E
1 mol H2
= 14300 kJ
2 mol H2
572 kJ
100 g H2
2 g H2
1 mol O2
= 11440 kJ
1 mol O2
572 kJ
640 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 
H2
O2
Started with 100 g, used up 80 g…
20 g H2 left over
1 mol O2
640 g O2
2 mol H2
1 mol O2
2 g H2
1 mol H2
= 80 g H2
32 g O2
100 g
640 g
Automobile air bags inflate with nitrogen via the decomposition
of sodium azide: 
2 NaN3(s) 3 N2(g) + 2 Na(s)
At STP and a % yield of 85%, what mass sodium azide
is needed to yield 74 L nitrogen?
N2
NaN3
percent yield
“act” yield
x = 87.1 L N2 → “Theo” yield
1 mol N2
87.1 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 H2O
10 g	 30 g	 X g? 
B2O3
1 mol B2H6
27.6 g B2H6
10 g B2H6
1 mol O2
30 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 B2O3
21.75 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
2
2
ZnO
2 mol ZnS
1 mol ZnS
97.5 g ZnS
100 g ZnS
1 mol O2
100 g O2
= 83.5 g ZnO
2 mol ZnO
81.4 g ZnO
1 mol ZnO
2 mol ZnO
3 mol O2
81.4 g ZnO
1 mol ZnO
= 169.6 g ZnO
83.5 g ZnO
32 g O2
x = 67.6 g ZnO
___Al + ___Fe2O3 ___Fe + ___Al2O3 
 X g?		 X g?	 800 g needed **Rxn. has an
							80% yield. 
2
1
1
2
Fe
Al
Fe
Fe2O3
“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 Fe2O3
1 mol Fe2O3
159.7 g Fe2O3
1 mol Fe2O3
100
 x 
 yield
l
theoretica
 yield
actual
 
 
 yield
%
=
100
 x 
 yld.
theo.
 yld.
act.
 
 
 yield
%
=
100
 x 
 Al
g
 
189
 Al
g
 
172
 
®
100
 x 
 yield
l
theoretica
 yield
actual
 
 
 yield
%
=
100
 x 
 yld.
theo.
 yld.
act.
 
 
 yield
%
=
100
 x 
 Al
g
 
189
 Al
g
 
172
 
®
Act
%Y=100
Theo
g
75=100
41,373.9
x
g
g
Act
%Y=100
Theo
g
75=100
41,373.9
x
g
g
Act
%Y=100
Theo
g
74L
85=100
x
g
Act
%Y=100
Theo
g
74L
85=100
x
g
Act
%Y=100
Theo
g
x
81=100
83.5gZnO
g
Act
%Y=100
Theo
g
x
81=100
83.5gZnO
g
Act
%Y=100
Theo
g
800g
80=100
x
g
Act
%Y=100
Theo
g
800g
80=100
x
g