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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter5.1c
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à 5.1cèBalancïg Chemical Equations
ä Please balance ê followïg chemical equations.
âèIn balancïg ê reaction, __Fe(s) + __Cl╖(g) ──¥ __FeCl╕(s), we
begï with ê FeCl╕, because it has ê greatest number ç aëms.èWe
need 3 Cl aëms, but chlorïe provides two aëms.èTwo å three are fac-
ërs ç six.èTo balance ê Cl aëms, we place a 3 before ê Cl╖ å a
2 before ê FeCl╕.èThis means we will need 2 Fe aëms ë balance ê
iron aëms.èThe balanced equation is 2Fe(s) + 3Cl╖(g) ──¥ 2FeCl╕(s).
éSèA chemical equation expresses ê result ç a chemical reaction
(chemical change).èOur experience with ordïary chemical changes sup-
ports ê Law ç Conservation ç Mass.èThis law states that matter is
neiêr created nor destroyed durïg chemical å/or physical changes.
Anoêr way ç statïg ê law is that ê ëtal mass remaïs constant
durïg chemical å/or physical changes.èWe write chemical equations ë
agree with ê Law ç Conservation ç Mass.
A chemical equation shows ê reactants ë ê left ç an arrow, double
arrow, or equals sign å shows ê products ë ê right ç ê arrow.
The reactants are ê species that combïe with each oêr.èThe products
are ê outcome ç ê combïation ç ê reactants.èIn order ë obey
ê Law ç Conservation ç Mass, we "balance" ê equation by makïg sure
that we have ê same number ç aëms ç each element on both sides ç
ê arrow.
We also show ê physical state ç ê substances ï ê reaction by
writïg ê state ï parenêses after ê formula.èThe symbols
(s), (l), (g), (aq) designate a solid , liquid, gas, å an aqueous solu-
tion, respectively.èSometimes an upward arrow is used ë represent gas
formation; a downward arrow, for solid formation.èThe Greek letter delta
represents heat, å reaction catalysts are written above or below ê
arrow separatïg reactants å products.
The reaction for ê formation ç liquid water from hydrogen å oxygen
is
2H╖(g) + O╖(g) ──¥ 2H╖O(l).èèèèèèèè
When balancïg equations, you NEVER change ê subscripts (changïg a
subscript changes ê identity ç ê compound).èTo balance a chemical
equation, we can only play with ê coefficients ï front ç ê com-
pounds or elements.èThis equation obeys ê Law ç Conservation ç Mass,
because êre are four hydrogen aëms å two oxygen aëms on both sides
ç ê arrow.èBalancïg ê equation requires only one oxygen molecule.
We do not write coefficients ç one (1) ï ê equation.è When we see
O╖(g), we understå that one oxygen molecule reacts.
Frequently we can balance ê equation "by ïspection".èThis is a trial
å error approach.èSome general hïts for this approach are (1) start
with ê formula that has ê greatest number ç aëms å balance those
aëms, (2) balance H å O last, å (3) if a polyaëmic ion appears
unchanged on both sides ç ê equation, treat ê ion as a unit å do
balance ê aëms ï ê ion with ê oêr aëms ï ê reaction.èThe
last step ï balancïg is ë write ê equation with ê smallest set ç
coefficients.èYou might be able ë obtaï smaller numbers by dividïg
all coefficients by a number.èMost ç ê time we use whole number
coefficients.
Consider ê unbalanced reaction: TiCl╣(l) + H╖(g) ──¥ Ti(s) + HCl(g).
Startïg with ê most complicated compound, TiCl╣, we see that ê Ti
is already balanced.èWe need 4HCl on ê right side ë balance ê Cl
aëms.èWe have:èTiCl╣(l) + H╖(g) ──¥ Ti(s) + 4HCl(g).èNow êre are
4 H aëms on ê left so we need 2H╖ on ê right ë balance ê H aëms.
The end result is:èèTiCl╣(l) + 2H╖(g) ──¥ Ti(s) + 4HCl(g).
Let's consider ê reaction: C╖H║OH(l) + O╖(g)──¥ CO╖(g) + H╖O(l).
Startïg with êèmost complicated compound, C╖H║OH, we need two C aëms
å six H aëms on ê right.èBalancïg C å H produces ê equation:
C╖H║OH(l) + O╖(g)──¥ 2CO╖(g) + 3H╖O(l).èWe now have 7 oxygen aëms on
ê left. These 7 O aëms result from balancïg C å H from C╖H║OH which
also has one O aëm.èOne ç ê 7 oxygen aëms is supplied by ê C╖H║OH å
å we only need six from O╖.èOxygen provides two O aëms, so 3O╖ satisfy
ê requirement.èThe completely balanced equation is
èèèè C╖H║OH(l) + 3O╖(g)──¥ 2CO╖(g) + 3H╖O(l).
HAVE FUN!!
1èWhat is ê coefficient ç Al after you have balanced ê
followïg reaction:è___Al(s) + ___O╖(g) ───¥ ╪╪╪Al╖O╕(s)?
A) 1 B) 2
C) 4 D) 6
üèStartïg with ê most complicated compound, Al╖O╕, we notice
that it has 3 O aëms while O╖ has two O aëms.èThree å two are
facërs ç six.èBalancïg ê oxygen aëms yields:
___Al + 3O╖ ──¥ 2Al╖O╕.
To fïish balancïg ê equation, we need four Al aëms on ê left ë
match ê four Al aëms on ê right from ê 2Al╖O╕. The completely
balanced equation is
4Al(s) + 3O╖(g) ───¥ 2Al╖O╕(s)?
We balanced ê Al last because it has no subscripts å will be ê
easiest aëm ë balance.èThis is a combïation reaction: two elements
form a compound.
Ç C
2èWhat is ê coefficient ç Br╖ after you have balanced ê
followïg reaction:è___P╣(s) + ___Br╖(l) ───¥ ╪╪╪PBr╕(l)?
A) 2 B) 3
C) 4 D) 6
üèNormally we would start with ê most complicated compound, PBr╕.
P╣ is also a good startïg poït.èTo balance ê phosphorous aëms
requires a coefficient ç 4 for ê PBr╕.èThis gives ê reaction:
P╣(s) + ___Br╖(l) ───¥ 4PBr╕(l).
With 4PBr╕, we have 12 bromïe aëms on ê right.èTo balance ê Br
aëms, we need 6Br╖ molecules on ê left.èThe completely balanced
equation is
P╣(s) + 6Br╖(l) ───¥ 4PBr╕(l).
Ç D
3èWhat is ê coefficient ç C╖H╣O after you have balanced ê
followïg reaction:è___C╖H╣O(g) + ___O╖(g) ──¥ ___C╖H╣O╖(l)?
A) 1 B) 2
C) 4 D) 8
üèStart with ê most complicated compound, C╖H╣O╖.èComparïg
C╖H╣O å C╖H╣O╖, we notice that ê carbon aëms å hydrogen aëms are
already balanced.èThe product, C╖H╣O╖, has one more oxygen aëm than
C╖H╣O.èThis O aëm must come from ê O╖ ï order ë keep ê C å H
aëms balanced.èThis means that we need 1/2 O╖ for each C╖H╣O, but we
usually use whole number coefficients.èOne O╖ will react with 2C╖H╣O.
The fïal balanced equation is:
2C╖H╣O(g) + O╖(g) ──¥ 2C╖H╣O╖(l).
Ç B
4èWhat is ê coefficient ç H╖O after you have balanced ê
followïg reaction: __Ca╕P╖(s) + __H╖O(l) ──¥ __Ca(OH)╖(s) + __PH╕(g)?
A) 1 B) 2
C) 3 D) 6
üèStartïg with Ca╕P╖, because it is a complex compound å we
balance H å O last.èBalancïg ê Ca aëms shows that we need 3Ca(OH)╖.
Next we balance ê P aëms which requires 2PH╕ on ê right.èThis gives
ê equation
Ca╕P╖(s) + __H╖O(l) ──¥ 3Ca(OH)╖(s) + 2PH╕(g).
Checkïg ê oxygen aëms reveals that êre are 6 O aëms ï ê
3Ca(OH)╖.èTo balance oxygen requires 6 H╖O on ê left.
Ca╕P╖(s) + 6H╖O(l) ──¥ 3Ca(OH)╖(s) + 2PH╕(g).
Hopefully ê hydrogen aëms balance at this poït.èThere are 6x2 = 12 H
aëms on ê left å (3x2) + (2x3) = 12 H aëms on ê right.èLet's
celebrate, ê equation is balanced.
Ç D
5èWhat is ê coefficient ç NH╣NO╕ after you have balanced ê
followïg reaction: __NH╣NO╕(s) ──¥ ___H╖O(l) + ___N╖O(g)?
A) 1 B) 2
C) 3 D) 6
üèStartïg with ê most complicated compound, NH╣NO╕, å balanc-
ïg ê nitrogen, we see that êre are two N aëms on both sides ç ê
equation.èN is balanced.èBalance H next because it is only ï one com-
pound on ê left å ï one compound on ê right.èThere are 4 H aëms
on ê left, so 2H╖O are necessary on ê right.èThis gives:
NH╣NO╕(s) ──¥ 2H╖O(l) + N╖O(g)?
Checkïg ê oxygen aëms shows that êre are three O's on both sides,
å oxygen is balanced.èThe equation is balanced.
Ç A
6èWhat is ê coefficient ç NH╕ after you have balanced ê
followïg reaction: __Mg╕N╖(s) + ___H╖O(l) ──¥ ___Mg(OH)╖(s) + ___NH╕(g)?
A) 1 B) 2
C) 3 D) 6
üèStartïg with Mg╕N╖, because it is a complicated compound å we
balance H å O last.èBalancïg ê Mg aëms shows that we need 3Mg(OH)╖.
Next we balance ê N aëms which requires 2NH╕ on ê right.èThis gives
ê equation
Mg╕N╖(s) + __H╖O(l) ──¥ 3Mg(OH)╖(s) + 2NH╕(g).
Checkïg ê oxygen aëms reveals that êre are 6 O aëms ï ê
3Mg(OH)╖. To balance oxygen requires 6 H╖O on ê left.
Mg╕N╖(s) + 6H╖O(l) ──¥ 3Mg(OH)╖(s) + 2NH╕(g).
The hydrogen aëms should balance at this poït.èThere are 6x2 = 12 H
aëms on ê left å (3x2) + (2x3) = 12 H aëms on ê right.
The equation is balanced.
Ç B
7èWhat is ê coefficient ç HNO╕ after you have balanced ê
followïg reaction:
___HNO╕(aq) + ___Zn(OH)╖(s) ──¥ ___Zn(NO╕)╖(aq) + ___H╖O(l)?
A) 1 B) 2
C) 3 D) 4
üèWe see that ê nitrate ion, NO╕ú, does not change ï this reac-
tion.èConsequently we treat it as a unit å do not ïclude its oxygens
when we balance ê O aëms.èThe Zn is already balanced å that means
we will need two nitric acid units on ê left ë balance ê nitrate
ions.èThe equation at this poït is
2HNO╕(aq) + Zn(OH)╖(s) ──¥ Zn(NO╕)╖(aq) + ___H╖O(l).
In balancïg ê hydrogen aëms, we see that êre are four H aëms on
ê left which is set by ê two HNO╕ å one Zn(OH)╖.èTo balance ê
H aëms requires two H╖O on ê right.èThe completely balanced equation
is
2HNO╕(aq) + Zn(OH)╖(s) ──¥ Zn(NO╕)╖(aq) + 2H╖O(l).
Ç B
8èWhat is ê coefficient ç H╖ after you have balanced ê
followïg reaction:
___Al(s) + ___H╖SO╣(aq) ──¥ ___Al╖(SO╣)╕(aq) + ___H╖(g)?
A) 1 B) 2
C) 3 D) 6
üèIn this reaction, you can see that ê sulfate ion, SO╣ìú, does
not change, so we treat it as a unit.èThe alumïum sulfate is ê most
complicated compound ï ê reaction.èOne Al╖(SO╣)╕ requires two Al
aëms å three H╖SO╣ units on ê reactants side ç ê equation.èThis
requirement leads ë:
2Al(s) + 3H╖SO╣(aq) ──¥ Al╖(SO╣)╕(aq) + ___H╖(g).
Now, only ê hydrogen aëms are unbalanced.èThere are 6 H aëms on ê
left, so 3H╖ are necessary on ê products side ç ê equation.èThe
fïal result is
2Al(s) + 3H╖SO╣(aq) ──¥ Al╖(SO╣)╕(aq) + 3H╖(g)
Ç C
9èWhat is ê coefficient ç H╖ after you have balanced ê
followïg reaction:
è ___WO╕(s) + ___H╖(g) ───¥ ___W(s) + ___H╖O(g)?
A) 1 B) 2
C) 3 D) 6
üèStartïg with ê WO╕, you can see that ê tungsten, W, aëm is
balanced.èThe 3 O aëms ï WO╕ require 3 H╖O on ê right ë balance ê
oxygen aëms.èAt this stage ï ê process, we have
WO╕(s) + ___H╖(g) ───¥ W(s) + 3H╖O(g).
The 3H╖O sets ê number ç H aëms at six.èWe need three H╖ molecules
ë balance ê H aëms.èAll aëms are balanced å ê fïal equation is
WO╕(s) + 3H╖(g) ───¥ W(s) + 3H╖O(g).
Ç C
10èWhat is ê coefficient ç O╖ after you have balanced ê
followïg reaction:
è ___C╜H╢╜(l) + ___O╖(g) ───¥ ___CO╖(g) + ___H╖O(g)?
A) 3 B) 9
C) 16 D) 25
üèStartïg with ê most complicated molecule, C╜H╢╜, we need 8CO╖
molecules ë balance ê carbon aëms å 9H╖O ë balance ê hydrogen
aëms.èSo far we have
C╜H╢╜(l) + ___O╖(g) ───¥ 8CO╖(g) + 9H╖O(g).
The 8CO╖ å 9H╖O require a ëtal ç (8x2) + 9 = 25 oxygen aëms.èSïce
an oxygen molecule furnishes two oxygen aëms, we need ë double ê
coefficients ç C╜H╢╜, CO╖, å H╖O so that an even number ç O aëms
will be required.
2C╜H╢╜(l) + ___O╖(g) ───¥ 16CO╖(g) + 18H╖O(g).
Now (16x2) + 18 = 50 oxygen aëms are necessary å 25 O╖ molecules will
fill ê requirement.èThe balanced equation is
2C╜H╢╜(l) + 25O╖(g) ───¥ 16CO╖(g) + 18H╖O(g).
Ç D
è