Download Unit 3 4 Balancing Chemical Reaction Equations by Inspection

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NOTES: UNIT 3 (4): BALANCING EQUATIONS
I – IV) Mole & Stoichiometry
V) Balancing Equations by Trial and Error (a.k.a. Balancing by Inspection)
A) Predicated upon the Law of the Conservation of Matter
Take Home Message: The number of moles of each element (or species) on the reactant side
must equal the number of moles of the same element (species) on the product side.
1) During a chemical reaction, atoms can't just appear or disappear. We must account
for every atom or ion.
a) Balancing an equation gives you the * mole ratio
. You may change the
coefficients BUT, you may NEVER change * a subscript, in order to balance
b) Changing a coefficient changes the ratio between the substances, but changing
* subscript
changes the nature / identities of the * substances
themselves
Changing the nature of the substances is NOT the goal of balancing an equation.
The goal is to ensure the * conservation
of matter.
2) Coefficients must be in the simplest, whole number ratio (no fractions in Regents-land)
Take Home Message: The coefficient of a substance of a balanced equation = the number of
moles of the substance. (Which, of course can be converted to a mass)
So: * The coefficient = the number of moles of the substance
3) coefficient: the number written in front of a substance that indicates the total number of moles
of molecules, formula units or atoms of the entire substance.
4) subscript : gives the number of moles of a specific element (it does't apply to the entire substance
or formula)
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Before going on … I must introduce to you a little something called the polyatomic ions.
(Review: recall that the terms ion ≠ ionic)
B) Polyatomic ion: A chemically unstable group of bonded atoms, which end up with an
unequal number of protons and electrons.
Since most PAI are made of bonded nonmetal atoms, the bonding that holds the PAI together,
is often covalent bonds. The charge is shared by the entire polyatomic ion (sort of like a
like a group of passengers sharing a taxi … the taxi is like the charge)
A polyatomic ion tends to act chemically like a single species. (See your reference charts)
1) Emphasize that last statement … a PAI is a set of atoms which acts as a single species.
2)  Lastly, most inorganic compounds that contain a polyatomic ion are ALSO
classified as ionic compounds. The whole compound exists due to ionic bonding, yet it
also contains some covalent bonds because of the polyatomic ion’s structure.
Bonding is due to opposite charge attraction
Thus: NaNO3(s) Is an ionic compound. Na+1
NO3-1
But this is put together with covalent bonds
Remember: Just because something is an ION does NOT MEAN it is IONIC
Hence: NO3-1 is an ion... but made via covalent bonds and is not ionic, itself. NO3-1 is found
though in many ionic compounds ... such as NaNO3, Pb(NO3)2, Fe(NO3)3
http://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/Ions/ions.html
Cyanide PAI
Bicarbonate PAI
3) As far a subscripts go, you know already that: e.g. Ca(NO3)2
2 nitrate polyatomic groups
for every 1 calcium ion
Table of Common Polyatomic Ions
Name
Formula
Name
Formula
-1
C2H3O2
Acetate
Hypochlorite ClO-1
IO3-1
Ammonium NH4+1
Iodate
(not on Table E)
(not ammonia)
Arsenate
* or… 2 moles of nitrate ion for
every 1 mole of Ca2+ ion per 1
mole of the compound Ca(NO3)2
AsO4-3
Nitrate
NO3-1
CO3-2
ClO3-1
ClO2-1
CrO4-2
CN-1
Cr2O7-2
HCO3-1
Nitrite
Perchlorate
Peroxide
Phosphate
Sulfate
Sulfite
Thiocyanate
NO2-1
ClO4-1
OH-1
Thiosulfate
S2O3 -2
(not on Table E)
Carbonate
Chlorate
Chlorite
Chromate
Cyanide
Dichromate
Hydrogen
Carbonate
Hydroxide
O22PO4-3
SO4-2
SO3-2
SCN-1
Okay, now we can begin to balance reaction equations…
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C) PROCESS BALANCING BY INSPECTION: For each and every species....
On the reactant side: multiply: (coefficient )(by all appropriate subscripts for the species)
On the product side: multiply: (coefficient )(by all appropriate subscripts species)
Compare /Adjust the coefficients so that the results of the two multiplications are equal.
Assume a coefficient of 1 when there is none written
Assume a subscript of 1 when there is none written (hmm ! Is he trying to tell me something ?)
eg)
kJ
+ __ I2(g)

+ __ Cl2(g)
___ICl(s)
1) First year students often have 1 major misconception … they believe that the coefficients
of the reactant side and the coefficients of the product side need to add up to each other.
They are WRONG!! The number of moles of each element on the reactant side must be the
same as the number of moles of that element on the product side … but their coefficients may
be different.
2 Fe2O3(s) + 3 C(s) 
4 Fe(s) + 3 CO2(g)
for example: note that 2 +3 = 5
and that 4 + 3 = 7 …The sum of the coefficients is NOT the
same but matter is conserved…Analyze the
moles of atom & ions on both sides. They are
the same number on both sides
2) Helpful Hints for balancing by trial and error:
These first three
suggestions are
interchangeable as a
starting point.
These are just hints ☺
 Balance metal atoms or metal ions
 Balance nonmetal atoms or ions (but leave O until the very end)
 You can balance polyatomic ions which are INTACT on both sides
of the reaction arrow against each other (as 1 entity), if you choose.
 Use the H2O  MOH hint when possible
 With combustion reactions, or those with O2 as a reactant, think
about using the fraction rule
 Sometimes a reaction is balanced already, so be prepared, to leave
every coefficient as "1".
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D) Simple Examples:
1)
_____ Al(s) + _____ Cl2(g)  _____ AlCl3(s)
2) _____ PbCl2(aq) + _____ KI(aq)  _____ KCl(aq)
+ _____ PbI2(s)
3) _____ C2H4(g) + _____ O2(g)  _____ CO2(g) + _____ H2O(l)
4) _____ Na(s)
+ _____ Cl2(g) 
5) _____ HgO(s) 
_____ Hg(l)
6) _____ O2(g)

7) _____Al(s)
+ _____ O2(g) 
8) _____ KClO3(s) 
_____ NaCl(s)
+ _____ O2(g)
______ O3(g)
_____ Al2O3(s)
_____ O2(g) + _____ KCl(s)
9) _____ C6H12(l) + _____ O2(g)  _____ CO2(g) + _____ H2O(g)
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E) Simple Examples with Polyatomic Ions ....If it helps, put ( ) around the polyatomic ion if none exist.
e.g. On a test, you are given:
___Cu + ___AgNO3  ___Cu(NO3)2 + ___ Ag
you might find it helpful to put ( ): ___Cu
+ ___ Ag(NO3)  ___Cu(NO3)2 + ___ Ag

How many are represented here?
Note, how many of the polyatomic ion are represented?
Recall: A PAI is a group of atoms in which the #p  #e-, YET this group of bonded atoms functions as a single entity...
they “go around together” and tend to stay together.
Name
Formula Name
C2H3O2-1
Acetate*
Ammonium* NH4+1
Formula
(modified): LIST OF POLYATOMIC IONS
Hypochlorite ClO-1
IO3-1
Iodate
(not ammonia)
Arsenate
Carbonate*
Chlorate
Chlorite
Chromate
Cyanide
Dichromate
Hydrogen *
Carbonate
Hydroxide*
AsO4-3
CO3-2
ClO3-1
ClO2-1
CrO4-2
CN-1
Cr2O7-2
HCO3-1
Nitrate*
Nitrite
Perchlorate
Peroxide
Phosphate*
Sulfate*
Sulfite
Thiocyanate
NO3-1
NO2-1
ClO4-1
OH-1
Thiosulfate
S2O3 -2
O22PO4-3
SO4-2
SO3-2
SCN-1
1) _____Al(s) + _____ Cu(NO3)2(aq)
 _____ Al(NO3)3(aq) + _____ Cu(s)
2) ____Al2(SO4)3(aq) + ____ Ca(OH)2(aq)  ____ CaSO4(aq) + ____Al(OH)3(s)
3) _____Al(s) + _____H2SO4(aq)  _____H2(g) + _____ Al2(SO4)3(aq)
4) ____Ba(NO3)2(aq) + ____K2CrO4(aq)  ____BaCrO4(s) + ____ KNO3(aq)
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5) _____ Ca(s) + _____ HNO3(aq)  _____ Ca(NO3)2(aq) + _____H2(g)
6) _____ Ca(s) + _____ Zn(OH)2(aq)  _____ Ca(OH)2(s) + _____ Zn(s)
7) _____ Cu(s) + _____ AgNO3(aq)  _____ Ag(s)
+ _____ Cu(NO3)2(aq)
8) _____ Pb(NO3)2(aq) + _____ K3PO4(aq)  _____ Pb3(PO4)2(s) + _____ KNO3(aq)
9) _____BaBr2(aq) + _____Al2(S2O3)3(aq)  _____BaS2O3(s) + _____AlBr3(aq)
10) _____Pb(NO3)2(aq) + _____H3AsO4(aq)  _____PbHAsO4(s) + _____ HNO3(aq)
11) _____Na2Cr2O7(aq) + ____ AlCl3(aq)  ____ NaCl(aq)
+ ____ Al2(Cr2O7)3(aq)
12) ____K3PO4(aq) + ____MgCl2(aq)  ____Mg3(PO4)2(s) + ____ KCl(aq)
Challenge
13) __KI(s) + __H2SO4(aq) +__ MnO2(s)  __K2SO4(aq) + __MnSO4(aq) + __H2O(l) + ___I2(s)
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F) "Fraction Rule" Use a decimalized value (or fraction) to balance the reactant O2 in an equation.
You don't need to use this EVERY time … It's just a strategy for your bag of tricks...
LOOK FOR : * O2 as a reactant & use a decimalized coefficient (e.g. 1.5 , 4.5) to balance the O2
1) _____ C4H6(g) + _____ O2(g)  _____ CO2(g) + _____ H2O(l)
2) _____ C5H10(g)
+ _____ O2(g)  _____ CO2(g) + _____ H2O(l)
3) _____ C6H6(g) + _____ O2(g)  _____ CO2(g) + _____ H2O(l)
4) _____ C6H14(g) + _____ O2(g)  _____ CO2(g) + _____ H2O(l)
5) _____ C7H14
+ _____ O2  _____ CO2 + _____ H2O
6) _____ C8H14
+ _____ O2  _____ CO2 + _____ H2O
7) _____ C8H18 + _____ O2  _____ CO2 + _____ H2O
8) _____ FeS2(s) + _____ O2(g)  _____ Fe2O3(s) + _____ SO2(g)
9) _____ NH3(g) + _____ O2(g)  _____ NO (g) + _____ H2O(l)
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G) H2O  MOH Hint: When balancing an equation with H2O on one side of the reaction arrow and a
metal hydroxide on the other, re-write the water as H(OH) and balance the
lone hydrogen (H) separately from the (OH)
metal hydroxide : * metal cation (hydroxide polyatomic ion)
LOOK FOR :
e.g. NaOH, Fe(OH) 2
* a metal hydroxide on one side and water on the other side
1) _____ Na(s) + _____ H2O(l)  _____ NaOH(aq) + _____ H2(g)
2) _____Li(s)
+ _____ H2O(l)  _____ LiOH(aq)
+ _____ H2(g)
3) _____ HNO3(aq) + _____ Mg(OH)2(aq) _____ H2O(l) + _____Mg(NO3)2(aq)
4) ____ NH4Cl(s) + ____ Ca(OH)2(s)  ____CaCl2(s) + ____H2O(l) + ____ NH3(g)
5) ____NH4Br(aq)
+ ____KOH(aq)  ____H2O(l) + ____ NH3(g) + _____KBr(aq)
6) _____ Al(s) + _____ H2O(l)  _____ Al(OH)3(aq) + ______ H2(g)
7) _____ LiOH (aq) + _____ H2S(g)  _____ Li2S(s) + _____H2O(g)
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H) Other issues for coefficients and subscripts (…pretty much review)
1) Formula of a Hydrate
Reading:
Many ionic compounds when crystallized from aqueous solutions, retain a definite proportion of water as part of
their crystalline structure. This water is referred to as the water of hydration. These ionic crystals appear to be
perfectly "dry". When an ionic crystal is in possession of this "trapped water", the crystal is called a HYDRATE.
Upon heating a sample of the ionic crystal, this water may be driven off. As the water is driven off, the crystalline
structure is shattered and all that remains is a dry powder. This dry powder is called the ANHYDROUS salt.
(The prefix "AN-" simply means "WITHOUT" and "HYDROUS" refers to water, so, an "anhydrous salt" refers
to an ionic crystal WITHOUT WATER.)
a) MgSO4• 7 H2O
i) How many moles of water are in 0.50 moles of MgSO4• 7 H2O?
ii) How many moles of water are trapped in a 0.10 mole sample of
magnesium heptahydrate?
2) Empirical Formula: The symbolic representation of a substance using subscripts in
the simplest whole-number ratio.
a) The substance may not exist empirically in nature. The empirical formula
is just the basic ratio between the elements ... for instance, simple carbohydrates
have an empirical formula of CH2O (a 1:2:1 ratio). However, no such
compound exists in this ratio. The most basic carbohydrate exhibiting this
empirical formula is C6H12O6(s)
b) ionic compounds are written only in their respective empirical formula
3) Subscripts and Moles
a) Do you understand, that in 1 mole of H2O there are * 2
and *1 mole of O atom?
moles of H atom
b) How many moles of H atoms are in 1 mol of urea (NH2)2CO?
c) Use the formula: C12H14O6Ir3Au2Pt2AgSi5 which is a UV filter for OAKLEY®
Plutonite lenses for Oakley Razor Blades
Assuming 1 mol of the compound, identify the number of moles of each element.
270
HOMEWORK: BALANCING EQUATIONS
Directions: Begin to push yourself to remember the rules, procedures, processes, reference tables, facts…etc.
1. When balanced using the simplest whole number ratio, what is the molar ratio between carbon and
silicon?
_____SiO2(s)
_____C(s) 
+
a) 4:1
_____Si(s)
b) 1:2
+ _____CO(g)
c) 2:1
d) 5:3
2. When balanced, using the simplest whole number ratio, what is the coefficient of zinc metal?
_____Zn(s)
+
a) 1
b) 2

_____HCl(aq)
c) 3
_____ZnCl2(aq)
+
_____H2(g)
d) 4
3. When balanced, using the simplest whole number ratio what is the coefficient in front of oxygen gas?
_____O2(g) 
_____PH3(g)
+
a) 8
b) 6
_____P4O10(g)
c) 16
+
____ H2O(l)
d) 4
4. When balanced using the simplest whole number ratio, what is the coefficient of NaCl?
_____Na2Cr2O7(aq)
a) 1
+ _____AlCl3(aq)  _____NaCl(aq)
b) 12
c) 3
+ _____Al2(Cr2O7)3(aq)
d) 6
5. When balanced using the simplest whole number ratio, what are the molar ratios between the substances?
____C2H3Cl(g)
a) 2:5:4:2:2
+
____O2(g)  ____CO2(g) +
b) 1:3:4:2:6
____H2O(l)
c) 4:10:3:4:1
+
____HCl(g)
d) 1:1:3:4:8
6. When balanced using the simplest whole number ratio what is the coefficient in front of H2O? (Think about one
of the hints..)
_____Al4C3(s)
a) 24
+ _____H2O(l)
b) 2

c) 3
_____CH4(g)
+ _____Al(OH)3(g)
d) 12
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7. When balanced using the simplest whole number ratio, what is the coefficient in front of water?
_____Mg(OH)2(aq) 
_____HNO3(aq)
+
a) 1
b) 2
_____H2O(l)
c) 3
+
_____Mg(NO3)2(aq)
d) 4
8. When balanced using the simplest whole number ratio, what is the coefficient for barium bromide?
_____BaBr2(aq)
a) 1
+

_____Al2(S2O3)3(aq)
b) 7
c) 3
_____BaS2O3(aq) + ____AlBr3(aq)
d) 6
9. When balanced using the simplest whole number ratio, what is the number of moles of KNO3 produced?
_____Pb(NO3)2(aq)
a) 18
+
_____K3PO4(aq) 
b) 2
c) 6
_____Pb3(PO4)2(aq)
+ _____KNO3(aq)
d) 7
10 When balanced using the simplest whole number ratio, what is the coefficient for carbon dioxide?
_____C5H10(g)
a) 10
+
b) 2

______CO2(g)
c) 15
d) 5
_____O2(g)
+
______H2O(l)
11 When balanced using the simplest whole number ratio, what is the coefficient of oxygen gas?
_____C5H12(g)
a) 6
+
b) 5
_____O2(g)

c) 10
______CO2(g)
+
______H2O(l)
d) 8
ANSWERS:
1.
2.
3.
4.
5.
c
a remember, you can assume a coefficient of 1, if there is no other coefficient written in front of a symbol
a
d
a The molar ratios are represented by the coefficients of the balanced equation : this requires the use of the
"fraction" rule and then the doubling of the coefficients.
6. d Did you use the “Water-Hydroxide Hint“ ? It would help
7. b Did you use the “Water-Hydroxide Hint “
8. c
9. c
10 a
11 d
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MORE BALANCING EQUATIONS (only if you wish for more practice…)
1)
C6H6(g) + O2(g) 
2)
C2H6(g) +
O2(g) 
3)
C3H8(g) +
O2(g)
CO2(g)
+
CO2(g)

H2O(l)
+
H2O(l)
CO2(g) +
H2O(l)
CHALLENGE (a minor one)
4)
5)
CH3OH(g) +

O2(g)
O2(g) 
FeS2(s) +
CO2(g) +
Fe2O3(s)
+
+
C(s) 
H2O(l)
SO2(g)
CHALLENGE
6)
Ca3(PO4)2(s) +
SiO2(s)
7)
IBr(g)
8)
BCl3(s)
9)
Al(s) +
H2SO4(aq) 
Al2(SO4)3(aq) +
10)
Al(s) +
Cu(NO3)2(aq) 
Al(NO3)3(aq)
11)
PbCl2(aq) +
12)
Cu
13)
Fe3(PO4)2(aq) +

+ NH3(g)
+
+
P4(s)
+
KI(aq) 
HNO3 
NI3(s)
H2(g) 
PbI2(s) +
Cu(NO3)2 +
NaOH(aq) 
P4(g)
+
+
CaSiO3(l)
+ CO(g)
NH4Br(s)
BP(s)
+
HCl(g)
H2(g)
+
Cu(s)
KCl(aq)
H2O +
Na3PO4(aq)
+
NO2
Fe(OH)2(s)
273
CHALLENGE (a minor one)
SO2(g) 
14)
PbO(s) +
PbS(s) +
15)
Fe2O3(s) +
16)
Mg(s) +
O2(g) 
MgO(s)
17)
Na(s) +
I2(g) 
NaI(s)
18)
Al(s) +
19)
N2(g) +
H2(g) 
20)
C2H3SNH2
+
21)
CO +
H2 
22)
Re(s) +
Br2(l) 
23)
F2 +
24)
Fe +
25)
NaClO3 
26)
PF3(g) + H2O 
27)
PCl5(l) +
C(s) 
S(s) 
O2(g)
Fe(s) + CO2(g)
Al2S3(s)
NH3(g)
H2O 
H2S
(not nearly as tough as it looks)
ReBr3(s)
KF +
CuSO4 
Cu +
H2O(l) 
+
CH3OH
KCl 
NaCl +
C2H3O2NH4
Cl2
FeSO4
O2
H3PO3(aq) +
H3PO4(aq) +
HF(aq)
HCl(aq)
274
ANSWERS :
1)
2 C6H6(g) + 15 O2(g)  12 CO2(g) + 6 H2O(l)
 4 CO2(g) + 6 H2O(l)
2)
2 C2H6(g) + 7 O2(g)
3)
C3H8(g) + 5 O2(g)
4)
2 CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(l)
5)
4 FeS2(s) + 11 O2(g)  2 Fe2O3(s) + 8 SO2(g)
6)
2 Ca3(PO4)2(s)
7)
3 IBr(g) + 4 NH3(g)  NI3(s) + 3 NH4Br(s)
8)
4 BCl3 + P4 + 6 H2  4 BP + 12 HCl
9)
2 Al(s) + 3 H2SO4(aq)  Al2(SO4))3(aq) + 3 H2(g)
10)
2 Al(s) + 3 Cu(NO3)2(aq)  2 Al(NO3)3(aq)
11)
PbCl2(aq) + 2 KI(aq)  PbI2(s) + 2 KCl(aq)
12)
Cu + 4HNO3  Cu(NO3)2 + 2 H2O + 2 NO2
13)
Fe3(PO4)2(aq)
14)
2 PbO(s) + 2 SO2(g)  2 PbS(s)
15)
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
16)
2 Mg(s) + O2(g) 
17)
2 Na(s) + I2(g) 
18)
2 Al + 3 S 
19)
N2(g) + 3H2(g) 
20)
C2H3SNH2 + 2 H2O  C2H3O2NH4 + H2S
21)
CO + 2 H2  CH3OH
22)
2 Re(s) +
23)
F2 +
24)
Fe + CuSO4  Cu + FeSO4
25)
2 NaClO3 
26)
PF3(g)
+ 3 H2O(l)  H3PO3(aq)
+ 3 HF(aq)
27)
PCl5(l)
+ 4 H2O(l)  H3PO4(aq)
+ 5 HCl(aq)
 3 CO2(g) +
4 H2O(l)
+ 6 SiO2(s) + 10 C(s)  P4(g) + 6 CaSiO3(l) + 10 CO(g)
+ 3 Cu(s)
+ 6 NaOH(aq)  2 Na3PO4(aq)
+
3 Fe(OH)2(s)
+ 3O2(g)
2MgO(s)
2 NaI(s)
Al2S3
3 Br2(l)
2NH3(g)
 2 ReBr3(s)
2KCl  2KF
+ Cl2
2 NaCl + 3 O2
275