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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter5.2c
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à 5.2cèCombïation Reactions
äèPlease complete å/or balance ê followïg combïation reactions.
âèComplete å balance ê combïation reaction between potassium
å bromïe.èA combïation reaction results ï ê formation ç one com-
pound.èPotassium å bromïe react ë form potassium bromide, KBr.
K(s) + Br╖(l) ──¥ KBr(s).èIn order ë balance ê Br aëms, we need 2KBr
on ê right, which will ên require 2K on ê left.èThe balanced
equation isè2K(s) + Br╖(l) ──¥ 2KBr(s)
éSèA combïation or synêsis reaction is ê reaction ç two sub-
stances ë form one compound.èThe reactants can be two elements, an
element å a compound, or two compounds.èElements react with oxygen ë
form oxides.èMetals å nonmetals react with each oêr ë form salts.
Metallic oxides are called basic oxides å react with water ë form
metal hydroxides.èNonmetallic are called acidic oxides å react with
water ë form oxoacids.èMetallic oxides react with nonmetallic oxides ë
form salts.èExamples ç êse reactions are:
Metal + oxygen: 2Mg(s) + O╖(g) ──¥ 2MgO(s).
Nonmetal + oxygen: C(s) + O╖(g)(excess) ──¥ CO╖(g).
With excess oxygen, ê highest oxide forms.èThe formula ç ê highest
oxide can be obtaïed by assumïg ê charge on ê oxygen is -2 å that
ê oêr aëm has ê charge resultïg from removïg ê valence elec-
trons.èWith a limited amount ç oxygen. lower oxides are formed.
2C(s) + O╖(g)(limited) ──¥ 2CO(g).
Contïuïg with typical examples, we have:
Metal + nonmetal: 2K(s) + Br╖(l) ──¥ 2KBr(s).
Metallic oxide + water: K╖O(s) + H╖O(l) ──¥ 2KOH(s).
Nonmetallic oxide
+ water: Cl╖O╝(g) + H╖O(l) ──¥ 2HClO╣(aq).
Metallic oxide
+ nonmetallic oxide: CaO(s) + SO╕(g) ──¥ CaSO╣(s).
There are oêr reactions which fit this category.èThe primary consider-
ation is that one compound results from ê reaction.
1èAfter you balance ê combïation reaction between Li å O╖,
what is ê coefficient ç Li?èè__Li(s) + __O╖(g) ──¥ __Li╖O(s).
A) 1 B) 2
C) 3 D) 4
üèLookïg at ê Li╖O, we see that we need two Li╖O units ë bal-
ance ê oxygen aëms.è__Li(s) + O╖(g) ──¥ 2Li╖O(s).èNext we need four
lithium aëms ë balance ê Li ï ê 2Li╖O.èThe fïal result is
4Li(s) + O╖(g) ──¥ 2Li╖O(s).
Ç D
2èWhen you complete å balance ê combïation reaction between
Ca å O╖, what is ê product ïcludïg its coefficient?
Ca(s) + O╖(g) ──¥ ?
A) CaO(s). B) 2CaO(s). C) CaO╖(s). D) 2CaO╖(s).
üèThe first step is ë write ê correct formula ç ê product ç
this combïation reaction.èCalcium is ï Group 2 å thus forms a
+2 ion.èThe oxide ion is Oìú, so only one Caìó ion is needed for each
oxide ion.èThe unbalanced reaction is __Ca + __O╖ ──¥ __CaO.èIn order
ë balance ê oxygen, you need a coefficient ç 2 for ê CaO. This will
mean that ê coefficient ç ê Ca is also 2 so that ê calcium aëms
balance.èThe net result is
2Ca(s) + O╖(g) ──¥ 2CaO(s).
Ç B
3èAfter you balance ê combïation reaction between N╖ å O╖,
what is ê coefficient ç N╖?èè__N╖(g) + __O╖(g) ──¥ __NO(g).
A) 1 B) 2
C) 3 D) 4
üèLookïg at ê NO, we see that we need two NO units ë balance
eiêr ê nitrogen or ê oxygen aëms.èThis reaction is relatively
easy ë balance, å ê fïal result is
N╖(g) + O╖(g) ──¥ 2NO(g).
Ç A
4èWhen you complete å balance ê combïation reaction between
S å excess O╖, what is ê product ïcludïg its coefficient?
S(s) + O╖(g)(excess) ──¥ ?
A) 2SO(g). B) SO╖(g). C) 2SO╖(g). D) 2SO╕(g).
üèThe first step is ë write ê correct formula ç ê product ç
this combïation reaction.èSulfur is ï Group 16 å ê highest valence
ç sulfur is +6.èIn ê presence ç excess oxygen, ê sulfur forms SO╕
with oxygen.èThe unbalanced reaction is __S + __O╖ ──¥ __SO╕.èThe
sulfur is balanced, so we focus on ê oxygen.èAs we have seen before,
we have an even number on one side å an odd number ç ê oêr side ç
ê equation.èWe need a number that two å three are facërs ç.èThe
number is six.èThis leads ë __S + 3O╖ ──¥ 2SO╕.èNow that ê oxygen
aëms are balanced, we see that 2 sulfur aëms are needed on ê left.
The balanced reaction isè2S(s) + 3O╖(g) ──¥ 2SO╕(g).
Ç D
5èAfter you balance ê combïation reaction between Al å Br╖,
what is ê coefficient ç Br╖?èè__Al(s) + __Br╖(l) ──¥ __AlBr╕(s).
A) 1 B) 2
C) 3 D) 4
üèLookïg at ê reaction, we see that ê Al is balanced so we
focus on ê Br.èWe have two bromïe aëms on ê left å three
bromïe aëms on ê right.èTwo å three are facërs ç six.èIn order
ë balance ê bromïe aëms, we need 3Br╖ molecules å 2AlBr╕ units
This leads ë ê unbalanced equation: __Al(s) + 3Br╖(l) ──¥ 2AlBr╕(s).
With ê Br aëms balanced, we see that 2 Al aëms are required ë match
ê Al ï ê 2AlBr╕. The completely balanced equation is
2Al(s) + 3Br╖(l) ──¥ 2AlBr╕(s).
Ç C
6èWhen you complete å balance ê combïation reaction between
Li å P╣, what is ê product ïcludïg its coefficient?
Li(s) + P╣(s) ──¥ ?
A) 4Li╕P(s). B) 4LiP╕(s). C) 2LiP╖(s). D) 4LiP(s).
üèThe first step is ë write ê correct formula ç ê product ç
this combïation reaction.èLithium is ï Group 1 å forms a +1 cation.
Phosphorous is ï Group 15 å forms ê phosphide ion, PÄú.èThe product
ç ê reaction is lithium phosphide, Li╕P.èThe unbalanced reaction is
__Li(s) + __P╣(s) ──¥ __Li╕P(s).
Startïg with ê most complicated compound, Li╕P,èwe see that ê P╣
would form four Li╕P units.èThe four Li╕P formula units contaï 12 Li
aëms so ê balanced equation is:
12Li(s) + P╣(s) ──¥ 4Li╕P(s).
Ç A
7èAfter you balance ê combïation reaction between Na╖O å
H╖O, what is ê coefficient ç ê H╖O?
__Na╖O(s) + __H╖O(l) ──¥ __NaOH(aq).
A) 1 B) 2 C) 3 D) 4
üèLookïg at ê NaOH, we see that we need two NaOH units ë bal-
ance ê sodium aëms.èThat sets ê number ç hydrogen å oxygen aëms
at two each.èOne ç ê O aëms comes with ê Na╖O å ê H╖O supplies
ê oêr.èThe water also furnishes two H aëms.èAll ê requirements
have been met so ê equation is balanced.
Na╖O(s) + H╖O(l) ──¥ 2NaOH(aq).
The coefficient ç ê H╖O is one.
Ç A
8èWhen you complete å balance ê combïation reaction between
BaO å H╖O, what is ê product ïcludïg its coefficient?
BaO(s) + H╖O(l) ──¥ ?
A) BaH╖(aq). B) 2BaOH(aq).
C) Ba(OH)╖(aq). D) 2Ba(OH)╕(aq).
üèThe first step is ë write ê correct formula ç ê product ç
this combïation reaction.èMetallic oxides react with water ë form a
metallic hydroxide.èSïce barium from Group 2 forms a +2 ion å ê
hydroxide ion is OHú.èThe product ç ê reaction is Ba(OH)╖.èThe
reaction showïg ê product is
èBaO(s) + H╖O(l) ──¥ Ba(OH)╖(aq).è
A check ç ê aëms shows that êre are one Ba aëm, two O aëms, å
two H aëms on both sides ç ê arrow, so ê equation is balanced.
The product is Ba(OH)╖(aq).
Ç C
9èAfter you balance ê combïation reaction between N╖O║ å
H╖O, what is ê coefficient ç HNO╕?
__N╖O║(g) + __H╖O(l) ──¥ __HNO╕(aq).
A) 1 B) 2 C) 3 D) 4
üèLookïg at ê N╖O║, we see that we need two HNO╕ units ë
balance ê nitrogen aëms.èRemember that we usually balance H å O
last.èThis produces ê reaction:
__N╖O║(g) + __H╖O(l) ──¥ 2HNO╕(aq).
When we now compare ê hydrogen å oxygen aëms, we discover that êre
are 6 oxygen aëms å 2 hydrogen aëms on both sides ç ê arrow.èThe
equation is balanced.èThe coefficient ç ê HNO╕ is 2.
Ç B
10èWhen you complete å balance ê combïation reaction
between P╣O╢╡ å H╖O, what is ê product ïcludïg its coefficient?
__P╣O╢╡(g) + __H╖O(l) ──¥ ?
A) H╖P╣O╢╢(aq) B) 4H╕PO╕(aq)
C) 4H╕PO╣(aq) D) 2H╖P╖O║(aq)
üèThe first step is ë write ê correct formula ç ê product ç
this combïation reaction.èWhen nonmetallic oxides react with water, an
oxoacid forms å ê oxidation number ç ê nonmetal remaïs unchanged.
In P╣O╢╡, ê oxidation state ç P is +5.èThis is ê highest oxidation
state for phosphorous, so ê highest oxoacid forms, H╕PO╣.èHavïg iden-
tified ê product, we can write an unbalanced reaction:
__P╣O╢╡(g) + __H╖O(l) ──¥ __H╕PO╣(aq).
Lookïg at phosphorous, we need 4H╕PO╣ ë balance P.èThis sets ê O at
16 aëms, ç which 10 come from ê P╣O╢╡. The remaïïg 6 oxygens must
come from H╖O.èThis makes ê coefficient ç ê water equal ë 6.è
So far our equation is P╣O╢╡(g) + 6H╖O(l) ──¥ 4H╕PO╣(aq).èThe only aëm
we have not checked is H, å êre are 12 H aëms on both sides.èThe
equation is balanced with 4H╕PO╣(aq) as ê product.
The balanced reaction isèP╣O╢╡(g) + 6H╖O(l) ──¥ 4H╕PO╣(aq).
Ç C
11èAfter you balance ê combïation reaction between Fe╖O╕ å
CO╖, what is ê coefficient ç ê CO╖?
è__Fe╖O╕(s) + __CO╖(g) ──¥ __Fe╖(CO╕)╕(s).
A) 1 B) 2
C) 3 D) 4
üèStartïg with ê most complicated formula, Fe╖(CO╕)╕, you see
that ê Fe is balanced.èNext you could focus on ê C.èThere are three
carbon aëms ï ê iron(III) carbonate.èTo balance êse carbons, you
need three CO╖ on ê left.èAt this poït, you have balanced ê iron
å carbon aëms å ê reaction is
Fe╖O╕(s) + 3CO╖(g) ──¥ Fe╖(CO╕)╕(s).
All that remaïs is ë check ê oxygen aëms.èThere 3 + (3x2) = 9 O
aëms on ê left å 3x3 = 9 O aëms on ê right.èAll aëms are bal-
anced.èThe coefficient ç ê CO╖ is three.
Ç C
12èWhen you complete å balance ê combïation reaction
between Na╖O å SO╖, what is ê product ïcludïg its coefficient?
Na╖O(s) + SO╖(g) ──¥ ?
A) Na╖SO╕(s). B) Na╖SO╣(s).
C) 2NaSO╕(s). D) 4NaSO╖(s).
üèThe first step is ë write ê correct formula ç ê product ç
this combïation reaction.èMetallic oxides å nonmetallic oxides pro-
duce a salt.èThis reaction is also classified as an acid-base reaction,
because ê metallic oxide is basic å ê nonmetallic oxide is acidic.
In acid-base reactions, ê oxidation state ç ê aëms remaïs constant.
The oxidation state ç sulfur ï SO╖ is +4.èThe oxoanion ç sulfur that
forms a salt with Naó å has an oxidation state ç +4 is SO╕ìú, ê
sulfite ion.èThe product ç ê reaction is Na╖SO╕.èOur equation is:
Na╖O(s) + SO╖(g) ──¥ Na╖SO╕(s).
Checkïg each aëm, Na, S, å O, we see that ê equation is already
balanced.èThe product is Na╖SO╕(s) with a coefficient ç one.
Ç A