PPT: Stoichiometry Review

Report
Unit 3C:
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 = 602,213,673,600,000,000,000,000 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
Use coefficients
from balanced
chemical equation
Volume
(gases)
1 mole =
22.4 L @ STP
Particles
Mole
Mole
Mass
1 mole =
22.4 L @ STP
Volume
(gases)
Particles
Stoichiometry Practice Problems
2
4 2 + __C
3
__TiO
2 + __Cl
2
1
2
__TiCl
4 + __CO
2 + __CO
1. How many mol chlorine will react with 4.55 mol carbon?
4.55 mol C
C
Cl2
4 mol Cl2
= 6.07 mol Cl2
3 mol C
2. What mass titanium (IV) oxide will react with 4.55 mol carbon?
4.55 mol C 2 mol TiO2
C
TiO2
3 mol C
79.9 g TiO2
1 mol TiO2
= 242 g TiO2
3. How many molecules titanium (IV) chloride can be made
from 115 g titanium (IV) oxide?
coeff.
1 mol
TiO2
(
115 g TiO2
)(
1 mol TiO2
79.9 g TiO2
1 mol
TiCl4
)(
)
2 mol TiCl4
6.02 x 1023 m’c TiCl4
2 mol TiO2
1 mol TiCl4
= 8.66 x 1023 m’c TiCl4
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
mustThe
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 + Ni3P2
3 Ni + 2 IrP
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 Ni3P2
1 mol Ni3P2
6.02 x 1023 m’c Ni3P2
2 mol IrP
1 mol Ni3P2
Ni3P2
IrP
223.2 g IrP
1 mol IrP
= 3.95 x 107 g IrP
2. How many grams iridium will react with
465 grams nickel (II) phosphide?
465 g Ni3P2
1 mol Ni3P2
238.1 g Ni3P2
2 mol Ir
1 mol Ni3P2
= 751 g Ir
Ni3P2
Ir
192.2 g Ir
1 mol Ir
2 Ir + Ni3P2
3 Ni + 2 IrP
3. How many moles of nickel are produced
if 8.7 x 1025 atoms of iridium are consumed?
Ir
8.7 x 1025 at. Ir
1 mol Ir
6.02 x 1023 at. 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
H2
1 Zn + __
2 HCl
__
50 g excess
50 g Zn
1 mol Zn
65.4 g Zn
1 H2 + __
1 ZnCl2
__
?L
1 mol H2
1 mol Zn
22.4 L H2
1 mol H2
= 20 L H2
5. At STP, how many m’cules oxygen react
with 632 dm3 butane (C4H10)?
C4H10
1 C4H10 + __
2
13 O2
__
632 dm3 C4H10
O2
4 CO2 + __
5 H2O
8
10
__
1 mol C4H10
22.4 dm3 C4H10
13 mol O2
2 mol C4H10
6.02 x 1023 m’c O2
1 mol O2
= 1.10 x 1026 m’c O2
Suppose the question had been,“How many ATOMS of oxygen…”
1.10 x 1026 m’c O2
2 atoms O
1 m’c O2
= 2.20 x 1026 at. O
Energy and Stoichiometry
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g) + 891 kJ
A balanced eq. gives the ratios of moles-to-moles
AND moles-to-energy.
E
CH4
1. How many kJ of energy are released
when 54 g methane are burned?
54 g CH4
1 mol CH4
891 kJ
16.04 g CH4 1 mol CH4
= 3.0 x 103 kJ
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g) + 891 kJ
2. At STP, what volume oxygen is consumed
in producing 5430 kJ of energy?
5430 kJ
2 mol O2
891 kJ
22.4 L O2
1 mol O2
3. What mass of water is made if
10,540 kJ are released?
10,540 kJ
2 mol H2O
891 kJ
18.02 g H2O
1 mol H2O
E
O2
= 273 L O2
E
H2O
= 426.2 g H2O
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
10 BM
A balanced equation for making
a big wheel might be:
3W+2P+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 Cl2(g)
2 AlCl3(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 AlCl3
2 mol Al
AlCl3
133.34 g AlCl3
1 mol AlCl3
= 618 g AlCl3
If 125 g chlorine react w/excess aluminum,
how many g aluminum chloride are made?
Cl2 AlCl3
125 g Cl2 1 mol Cl2
2 mol AlCl3 133.34 g AlCl3
70.91 g Cl2 3 mol Cl2
1 mol AlCl3
= 157 g 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
• 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 + 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 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 / Cl2 / AlCl3 example:
ER
2 Al(s) + 3 Cl2(g)
LR
2 AlCl3(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 AlCl3
2 mol Al
Al
AlCl3
133.34 g AlCl3
1 mol AlCl3
= 618 g AlCl3
If 125 g chlorine react w/excess aluminum,
how many g aluminum chloride are made?
Cl2 AlCl3
125 g Cl2 1 mol Cl2
2 mol AlCl3 133.34 g AlCl3
70.91 g Cl2 3 mol Cl2
1 mol AlCl3
= 157 g AlCl3
Limiting Reactant Practice
2 Fe(s) +
3 Cl2(g)
223 g Fe
179 L Cl2
2 FeCl3(s)
Which is the limiting reactant: Fe or Cl2?
223 g Fe
1 mol Fe
55.85 g Fe
2 mol FeCl3
2 mol Fe
162.21 g FeCl3
1 mol FeCl3
= 648 g FeCl3
179 L Cl2
1 mol Cl2
22.4 L Cl2
2 mol FeCl3 162.21 g FeCl3
3 mol Cl2
1 mol FeCl3
= 864 g FeCl3
How many g FeCl3 are produced?
*Remember that the LR “limits” how much product can be made!
2 H2(g) + O2(g)
13 g H2
2 H2O(g)
80 g O2
Which is the limiting reactant: H2 or O2?
13 g H2
1 mol H2
2.02 g H2
2 mol H2O
2 mol H2
18.02 g H2O
1 mol H2O
= 120 g H2O
80 g O2
1 mol O2
2 mol H2O
18.02 g H2O
32.00 g O2
1 mol O2
1 mol H2O
= 90 g H22O
How many g H2O are produced?
*Notice that the LR doesn’t always have the smaller amount (13 v. 80)
How many g O2 are left over?
zero; O2 is the LR and
therefore is all used up
How many g H2 are left over?
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.
Start with the LR and relate to the other…
80 g O2
O2
1 mol O2
2 mol H2
2.02 g H2
32.00 g O2
1 mol O2
1 mol H2
HAD 13 g, USED 10 g…
H2
= 10 g H2
USED
3 g H2 left over
2 Fe(s) +
3 Br2(g)
181 g Fe
96.5 L Br2
2 FeBr3(s)
Which is the limiting reactant: Fe or Br2?
181 g Fe
1 mol Fe
55.85 g Fe
2 mol FeBr3
2 mol Fe
295.55 g FeBr3
1 mol FeBr3
= 958 g FeBr3
96.5 L Br2
1 mol Br2
22.4 L Br2
2 mol FeBr3 295.55 g FeBr3
3 mol Br2
1 mol FeBr3
= 849 g FeBr33
How many g FeBr3 are produced?
2 Fe(s) +
3 Br2(g)
181 g Fe
96.5 L Br2
2 FeBr3(s)
How many g of the ER are left over?
96.5 L Br2
1 mol Br2
2 mol Fe
22.4 L Br2
3 mol Br2
HAD 181 g, USED 160.4 g…
Br2
Fe
55.85 g Fe
= 160. g Fe
USED
1 mol Fe
21 g Fe left over
Al3+ O2– Percent Yield
Na+ O2–
solid
solid
molten
molten
aluminum
sodium
sodium
aluminum
oxide
oxide
6 Na(l) + 1 Al2O3(s)
2 Al(l) + 3 Na2O(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 Al2O3 1 mol Al2O3
2 mol Al
101.96 g Al2O3 1 mol Al2O3
26.98 g Al
1 mol Al
= 189 g Al
Al
189 g
This amt. of product (______)
is the 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 Al2O3 reacted
• some reactant or product spilled and was lost
% yield 
actual yield
theoretica l yield
x 100
Note: % yield should never be > 100%
FG %
Batting
average
GPA
Find % yield for previous problem.
% yield 
act. yld.
theo. yld.
x 100

172 g Al
189 g Al
x 100
= 91.0%
On NASA spacecraft, lithium hydroxide “scrubbers” remove
toxic CO2 from cabin.
CO2(g) + 2 LiOH(s)
Li2CO3(s) + H2O(l)
1. 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?
percent yield
(
880 g CO2
person-day
24, 640 g CO2
% Yield 
Act
Theo
)
x (4 p) x (7 d) = 24,640 g CO2
CO2 Li2CO3
1 mol CO2 1 mol Li2CO3 73.9 g Li2CO3
= 41,384 g
44.0 g CO2 1 mol CO2 1 mol Li2CO3
Li2CO3
x 100
0.75 
A
41,384 g
“theo” yield
A = 31,000 g Li2CO3
2. 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?
percent yield
“act” yield
% Yield 
Act
Theo
87.1 L N2
x 100
0.85 
75 L
x
1 mol N2
22.4 L N2
x = 87.1 L N2 → “Theo” yield
N2
2 mol NaN3 65 g NaN3
3 mol N2
NaN3
1 mol NaN3
= 170 g NaN3
2
3 2
___ZnS
+ ___O
2
2
___ZnO
+ ___SO
2
100 g
100 g
X g ? (assuming 81% yield)
Strategy:
1. Balance and find LR
2. Use LR to calc. X g ZnO (theo. yield)
3. Actual yield is 81% of theo. yield
ZnO
100 g ZnS 1 mol ZnS
97.5 g ZnS
2 mol ZnO 81.4 g ZnO
= 83.5 g ZnO
2 mol ZnS 1 mol ZnO
100 g O2 1 mol O2 2 mol ZnO 81.4 g ZnO
32 g O2
% Yield 
Act
Theo
x 100
3 mol O2
0.81 
1 mol ZnO
x
83.5 g ZnO
= 169.6 g ZnO
x = 67.6 g ZnO
2
1
___Al
+ ___Fe
2O3
X g?
X g?
2
1
___Fe
+ ___Al
2O3
800. g needed
“act” yield
% Yield 
Act
x 100
0.80 
Theo
800. g
**Rxn. has an
80.% yield.
“theo” = 1000 g Fe
x
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
Fe2O3
1000 g Fe 1 mol Fe 1 mol Fe2O3 159.7 g Fe2O3
= 1400 g Fe2O3
1 mol Fe2O3
55.85 g Fe 2 mol Fe
Review Questions
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
100 g H2
1 mol H2
2 g H2
640 g O2
1 mol O2
32 g O2
572 kJ
2 mol H2
= 14300 kJ
572 kJ
1 mol O2
= 11440 kJ
100 g
640 g
2 H2(g) + O2(g)
2 H2O(g) + 572 kJ
What mass of excess reactant is left over?
O2
640 g O2
1 mol O2
2 mol H2
2 g H2
32 g O2
1 mol O2
1 mol H2
Started with 100 g, used up 80 g…
H2
= 80 g H2
20 g H2 left over

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