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
4g
1g
2g
** 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.
mass
vol.
Use coeff. from balanced
equation in crossing this bridge
MOL
mass
vol.
MOL
part.
part.
SUBSTANCE “B”
(unknown)
SUBSTANCE “A”
(known)
4 Na(s) + 1 O (g) 2 Na O(s)
2
2
4 Na(s) + O2(g)
2 Na2O(s)
Na
Na
2O
NaNa
O22O
How many moles oxygen will react with 16.8 moles sodium?
1 mol O2
16.8 mol Na
= 4.20 mol O2
4 mol Na
How many moles sodium oxide are produced from
87.2 moles sodium?
87.2 mol Na 2 mol Na2O
= 43.6 mol Na2O
4 mol Na
How many moles sodium are required
to produce 0.736 moles sodium oxide?
0.736 mol Na2O
4 mol Na
2 mol Na2O
= 1.47 mol Na
Unit 8: Stoichiometry
-- involves finding amts.
of reactants & products
in a reaction
What can we do with stoichiometry?
For generic equation:
Given the…
amount of RA
(or RB)
amount of RA
or RB
amount of P1
or P2 you need
to produce
RA + RB
P1 + P2
…one can find the…
amount of RB (or RA)
that is needed to
react with it
amount of P1 or
P2 that will be
produced
amount of RA
and/or RB
you must use
Governing Equation:
2 patties
+
3 bread
4 patties
+
?
6 bread
excess
?
50 patties
+ 18 bread
+
?
75 bread
1 Big Mac®
?
6 Big Macs®
25 Big Macs®
Stoichiometry Island Diagram
SUBSTANCE “A”
Mass
(g)
1 mol = molar mass
(in g)
1 mol = 22.4 L
Volume
3
(L or dm ) 1 mol = 22.4 dm3
Particle
(at. or
m’c)
SUBSTANCE “B”
Use coefficients
from balanced
equation
MOLE
(mol)
1 mol = 6.02 x 1023
particles
1 mol = molar mass
Mass
(in g)
(g)
MOLE
(mol)
1 mol = 22.4 L Volume
3
1 mol = 22.4 dm3 (L or dm )
1 mol = 6.02 x 1023
particles
Particle
(at. or
m’c)
2
4 2 + __C
3
__TiO
2 + __Cl
2
1
2
__TiCl
4 + __CO
2 + __CO
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
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
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.
middle
conversion
factor
is
only
onebeing
3.
onbridge
the from
islands
each
end
of the
the
bridge
1. The
Use units
coefficients
theatequation
only
when
crossing
thatmiddle
has appear
twobridge.
different
in
it. The
conversion
crossed
in the
conversion
factor
foralways
that bridge.
the
Thesubstances
other six bridges
have
factors
the other
six bridges
have the same
“1
mol” for
before
a substance’s
formula.
substance in both the numerator and denominator.
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?
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
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
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
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
= 17 L H2
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 O2…”
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
How many kJ of energy are released
when 54 g methane are burned?
54 g CH4
1 mol CH4
16 g CH4
891 kJ
1 mol CH4
= 3.0 x 103 kJ
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?
5430 kJ
2 mol O2
891 kJ
22.4 L O2
1 mol O2
What mass of water is made if
10,540 kJ are released?
10,540 kJ
2 mol H2O
18 g H2O
891 kJ
1 mol H2O
E
O2
= 273 L O2
E
H2O
= 425.9 g H2O
The Limiting Reactant
A balanced equation for making
a Big Mac® might be:
3 B + 2 M + EE
With…
30 M
30 B
30 M
B3M2EE
…and…
excess B and
excess EE
excess M and
excess EE
30 B and excess
EE
…one can
make…
15 B3M2EE
10 B3M2EE
10 B3M2EE
A balanced equation for making
a big wheel might be:
3W+2P+S+H+F
W3P2SHF
With…
…and…
…one can
make…
50 P
excess of all other
reactants
25 W3P2SHF
50 S
excess of all other
reactants
50 W3P2SHF
50 P
50 S + excess of
all other reactants
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?
125 g Al
1 mol Al
27 g Al
2 mol AlCl3
2 mol Al
Al
AlCl3
133.5 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
133.5 g AlCl3
71 g Cl2
1 mol AlCl3
3 mol Cl2
= 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?
(We’re out of Cl2…)
157 g AlCl3
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).
In a root beer float, the LR
is usually the ice cream.
(What’s the product?)
Deliciousness!
With…
30 M
50 P
125 g Al
…and…
…one can make…
30 B and excess
EE
50 S + excess of all
other reactants
25 W3P2SHF
125 g Cl2
157 g AlCl3
From Examples
Limiting Reactant
Above…
10 B3M2EE
Excess Reactant(s)
Big Macs
B
M, EE
tricycles
P
W, S, H, F
Al / Cl2 / AlCl3
Cl2
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)
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
27 g Al
2 mol AlCl3
2 mol Al
Al
AlCl3
133.5 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
133.5 g AlCl3
71 g Cl2
1 mol AlCl3
3 mol Cl2
= 157 g AlCl3
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
55.8 g Fe
2 mol FeCl3
2 mol Fe
162.3 g FeCl3
1 mol FeCl3
= 649 g FeCl3
179 L Cl2
1 mol Cl2
2 mol FeCl3
162.3 g FeCl3
22.4 L Cl2
3 mol Cl2
1 mol FeCl3
= 865 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
2.0 g H2
2 mol H2O
2 mol H2
18.0 g H2O
1 mol H2O
= 120 g H2O
80 g O2
1 mol O2
2 mol H2O
18.0 g H2O
32.0 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
2.0 g H2
32.0 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 FeBr3
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
= 160.4 g Fe
USED
1 mol Fe
20.6 g Fe left over
Al3+ O2– Percent Yield
Na+ O2–
solid
solid
molten
solid
aluminum
sodium
sodium
aluminum
oxide
oxide
6 Na(l) + 1 Al2O3(s)
2 Al(s) + 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
= 224.9 g Al
357 g Al2O3 1 mol Al2O3
102 g Al2O3
2 mol Al
26.98 g Al
1 mol Al2O3
1 mol Al
= 188.9 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
actual yield
% yield 
x 100
theoretica l yield
-- % yield can never be > 100%.
FG %
Batting
average
GPA
Find % yield for previous problem.
172 g Al
act. yld.
x 100 = 91.0%
% yield 
x 100 
189 g Al
theo. yld.
On NASA spacecraft, lithium hydroxide “scrubbers” remove
toxic CO2 from cabin.
CO2(g) + 2 LiOH(s)
Li2CO3(s) + H2O(l)
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
)
x (4 p) x (7 d) = 24,640 g CO2
CO2 Li2CO3
24, 640 g CO2 1 mol CO2 1 mol Li2CO3 73.9 g Li2CO3
= 41,374 g
1 mol CO2 1 mol Li2CO3
44 g CO2
Li2CO3
x
Act
100
% Y=
100 75 =
“theo” yield
41,373.9 g
Theo
x = 31030 g Li2CO3
On NASA spacecraft, lithium hydroxide “scrubbers” remove
toxic CO2 from cabin.
CO2(g) + 2 LiOH(s)
Li2CO3(s) + H2O(l)
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?
(
880 g CO2
person-day
24,640 g CO2
(
)
x (4 p) x (7 d) = 24,640 g CO2
)(
1 mol CO2
44 g CO2
)(
2 mol LiOH
1 mol CO2
= 26,768 g LiOH (act.)
= 0.75
theo.
CO2 LiOH
23.9 g LiOH
1 mol LiOH
)
theo. = 36,000 g LiOH
REALITY: TAKE 100,000 g
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
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
Act
% Y=
100
Theo
74 L
85 =
100 x = 87.1 L N2 → “Theo” yield
x
N2
87.1 L N2
1 mol N2
22.4 L N2
NaN3
2 mol NaN3 65 g NaN3
3 mol N2
1 mol NaN3
= 168.5 g NaN3
B2H6 + 3 O2
10 g
30 g
B2O3 + 3 H2O
X g?
B2O3
10 g B2H6 1 mol B2H6 1 mol B2O3 69.6 g B2O3
27.6 g B2H6 1 mol B2H6 1 mol B2O3
= 25.2 g B2O3
30 g O2
1 mol O2
1 mol B2O3
69.6 g B2O3
32 g O2
3 mol O2
1 mol B2O3
= 21.75 g B2O3
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
3 mol O2
1 mol ZnO
x
Act
100
% Y=
100 81=
83.5g ZnO
Theo
= 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
Act
% Y=
100
Theo
800 g
80 =
100
x
**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
= 483.1 g Al
Fe
Fe2O3
1000 g Fe 1 mol Fe 1 mol Fe2O3 159.7 g Fe2O3
= 1430 g Fe2O3
1 mol Fe2O3
55.85 g Fe 2 mol Fe