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chapter7.1c
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à 7.1èThe 3-Step Mole Method (Mass-Mass Problems)
äèPlease fïd ê number moles ç reactants or ç products ï ê followïg reactions.
âèFïd ê number ç moles ç CO that will react with 8.0 mole
ç Fe╖O╕.èThe reaction is Fe╖O╕(s) + 3CO(g) ──¥ 2Fe(l) + 3CO╖(g).
The chemical equation tells us that three moles ç CO will react with one
mole ç Fe╖O╕.èWe can treat this problem as a unit conversion problem.
è 3 mol CO
? mol CO = 8.0 mol Fe╖O╕ x ─────────── = 24 mol CO.
è 1 mol Fe╖O╕
éSèWe frequently need ë know ê relationship between ê amounts
ç reactants å products ï chemical reactions.èWhen we synêsize a
compound, we need ë know how much ç ê reactants are needed ë produce
ê desired amount ç product.
Consider ê reaction between ammonia å oxygen, one ç ê steps ï ê
manufacture ç nitric acid.
4NH╕(g) + 5O╖(g) ──¥ 4NO(g) + 6H╖O(g).
One way ç readïg this balanced equation is ë say four molecules ç
ammonia react with five molecules ç oxygen ë produce four molecules ç
nitric oxide å six molecules ç water.èCan we weigh such a small num-
ber ç molecules?èOf course not.èOn a larger scale, we can say four
moles ç ammonia react with five moles ç oxygen ë produce four moles ç
nitric oxide å six moles ç water.
How many moles ç O╖ will react with 12 moles ç NH╕?èWe can use ê
unit conversion process ë solve this problem.èWe want ë convert moles
ç NH╕ ïë moles ç O╖.èThe balanced equation provides ê conversion
facër because it says that 4 moles ç NH╕ react with 5 moles ç O╖.
Settïg-up ê problem as a unit conversion, we fïd
5 mol O╖
? mol O╖ = 12 mol NH╕ x ─────────è= 15 mol O╖.
4 mol NH╕
How many moles ç water would form along with 20 moles ç NO?èAgaï,
this is a unit conversion.èThe balanced equation tells us that when
4 moles ç NO form, 6 moles ç H╖O will also form.
6 mol H╖O
? mol H╖O = 20 mol NO xè───────── = 30 mol H╖O
4 mol NO
1èGiven ê reaction: 2Al(s) + Fe╖O╕(s) ──¥ 2Fe(l) + Al╖O╕(s),
how many moles ç Fe╖O╕ are required ë react 5.0 mol Al completely?
A) 1.0 mol B) 2.5 mol
C) 5.0 mol D) 10. mol
ü The balanced equation shows that 1 mol Fe╖O╕ reacts with 2 mol Al.
This mole ratio provides ê conversion facër between Fe╖O╕ å Al.
è 1 mol Fe╖O╕
? mol Fe╖O╕ = 5.0 mol Al x ─────────── = 2.5 mol Fe╖O╕
è 2 mol Al
Ç B
2èPhosphorous is obtaïed via ê reaction:
è2Ca╕(PO╣)╖(s) + 6SiO╖(s) + 10C(s) ──¥ 6CaSiO╕(s) + 10CO(g) + P╣(s).
How many moles ç P╣ are formed when 42 mol ç CaSiO╕ are formed?
èA) 0.17 mol P╣èè B) 1.0 mol P╣èè C) 7.0 mol P╣èè D) 252 mol P╣
üèFrom ê balanced equation, we know that 6 mol ç CaSiO╕ form
when 1 mole ç phosphorous forms.èPerformïg ê unit conversion from
ê given number ç moles ç CaSiO╕ ë ê desired moles ç P╣, we get
èè1 mol P╣
? mol P╣ = 42 mol CaSiO╕ x ──────────── = 7.0 mol P╣
è 6 mol CaSiO╕
Ç C
3èConsider ê reaction,
èèèCa╕P╖(s) + 6H╖O(l) ──¥ 3Ca(OH)╖(s) + 2PH╕(g).
How many mol ç PH╕ would be produced by ê reaction ç 0.24 mol H╖O?
A) 0.080 mol B) 0.48 mol C) 0.72 mol D) 1.4 mol
üèFrom ê balanced equation, you know that 6 mol ç water makes
2 mol ç phosphïe, PH╕.èYou need ë convert ê 0.24 mol ç water ïë
ê number ç moles ç PH╕.
è 2 mol PH╕
? mol PH╕ = 0.24 mol H╖O x ───────── =è0.080 mol PH╕
è 6 mol H╖O
Ç A
äèPlease fïd ê number ç grams or moles ç reactants or ç products ï ê followïg
reactions.
âèHow many grams ç Fe can be obtaïed from 15.0 g ç Fe╖O╕ via
ê reaction, 2Al(s) + Fe╖O╕(s) ──¥ 2Fe(l) + Al╖O╕(s).
1 mol Fe╖O╕èèè 2 mol Feèèè55.85 g Fe
? g Fe = 15.0 g Fe╖O╕ x ─────────────── x ─────────── x ──────────
èèèèèèèèèèèè159.70 g Fe╖O╕èè1 mol Fe╖O╕è 1 mol Fe
? g Fe = 10.5 g Fe
éSèLet's consider ê oxidation ç ammonia, NH╕, agaï.
4NH╕(g) + 5O╖(g) ──¥ 4NO(g) + 6H╖O(g).
We know that ê balanced chemical equation relates ê number ç moles
ç reactants å ç products ë each oêr.èIn Chapter 4, we solved
problems relatïg ê masses å moles ç substances.èConsequently, ê
balanced chemical equation also relates ê masses ç reactants å prod-
ucts ë each oêr.èIn sëichiometric problems, we perform ê same
types ç manipulations agaï å agaï å agaï.
With ê oxidation ç ammonia as a model, we could start with a number
grams ç ammonia å calculate how many grams ç oxygen are needed ë
react with ê ammonia, how many grams ç nitric oxide would be produced,
å how many grams ç water also would be produced.èThe normal sequence
ç calculations is:
┌───────┐ step 1 ┌─────────┐ step 2 ┌────────────┐ step 3 ┌───────────┐
│ g NH╕ |───────¥│ mol NH╕ │───────¥│ mol O╖, NO,│───────¥│ g O╖, NO, │
└───────┘èèèè└─────────┘èèèè|èor H╖Oèè│èèèè│èor H╖Oè │
èèèèèèèèèèèèèèèèèè └────────────┘èèèè└───────────┘
In step 1, we divide ê mass ç ê ammonia by ê molar mass ç ammonia
That operation gives us ê moles ç ammonia ï ê reaction.
In step 2, we look at ê balanced chemical equation ë fïd ê mole
relationship between ê ammonia å ê compound ç ïterest (O╖, NO, or
H╖O).
In step 3, we multiply by ê molar mass ç ê compound ç ïterest (O╖,
NO, or H╖O) ë convert from moles ë grams.
We might begï å fïish ê sequence at different places dependïg upon
what ïformation is given å is desired.èFor example, we could convert
from grams ç product ë moles ç reactant, ïstead ç grams ç product
ë grams ç reactant.
How many grams ç O╖ are required ë react 75.0 g ç NH╕ completely?
This problem is ê same as oêr unit conversion problems.
èèèèèèèè step 1èèèèstep 2èèèstep 3
èè 1 mol NH╕èè 5 mol O╖èè32.00 g O╖
?g O╖ = 75.0 g NH╕ x ─────────── x ───────── x ────────── =è176 g O╖
èè 17.03 g NH╕è 4 mol NH╕è 1 mol O╖
The molar mass ç NH╕ is 14.01 + 3(1.008) = 17.03 g/mol.èThe 5:4 mole
ratio comes from ê balanced equation.èThe molar mass ç O╖ is 2(16.00)
which is 32.00 g/mol.
How many grams ç NO would be produced by ê complete reaction ç 500. g
ç O╖?èThe conversion pathway is g O╖ ─¥ mol O╖ ─¥ mol NO ─¥ g NO.
èèèèèèèèstep 1èèè step 2èèèstep 3
èè 1 mol O╖èè4 mol NOèè30.01 g NO
?g NO = 500. g O╖ x ────────── x ───────── x ────────── =è375 g NO
èè32.00 g O╖è 5 mol O╖èè1 mol NO
The error that many student make is not lookïg at ê balanced reaction
ë get ê mole ratio for step 2.
For one last example, how many moles ç NH╕ are required ë prepare
600. g ç NO?èHere, ê conversion path is g NO ─¥ mol NO ─¥ mol NH╕.
1 mol NOèè 4èmol NH╕
? mol NH╕ = 600. g NO x ────────── x ────────── = 20.0èmol NH╕
30.01 g NOè 4 mol NO
We sëpped at ê end ç step 2 ï ê general scheme because we were
only asked ë fïd ê number ç moles ç NH╕ å did not need ë obtaï
ê number ç grams ç NH╕.
4èThe complete combustion ç propane, C╕H╜, is
C╕H╜(g) + 5O╖(g) ──¥ 3CO╖(g) + 4H╖O(l).
How many grams ç O╖ are required for ê complete combustion ç 35.0 g
ç propane?
A) 127 g O╖ B) 635 g O╖
C) 25.4 g O╖ D) 9.64 g O╖
ü The path for ê conversion is g C╕H╜ ¥ mol C╕H╜ ¥ mol O╖ ¥ g O╖.
We need ê molar masses ï order ë convert between grams å moles:
C╕H╜: 3(12.01) + 8(1.008) = 44.09 g/mol, å O╖: 2(16.00) = 32.00 g/mol.
The balanced equation reveals that 5 mol O╖ reacts with 1 mol C╕H╜.
èèè1 mol C╕H╜èè 5 mol O╖èè 32.00 g O╖
?g O╖ = 35.0 g C╕H╜ x ──────────── x ────────── x ────────── = 127 g O╖.
èèèèèèèèèèè44.09 g C╕H╜è 1 mol C╕H╜è 1 mol O╖
Ç A
5èOne ïdustrial method for makïg acetylene, C╖H╖, is
2CH╣(g) ──¥ C╖H╖(g) + 3H╖(g).
How many grams ç CH╣ are required ë make 750. g C╖H╖?
A) 1.85x10Äg CH╣ B) 609 g CH╣
C) 304 g CH╣ D) 925 g CH╣
ü The path for ê conversion is g C╖H╖ ─¥ mol C╖H╖ ─¥ mol CH╣ ─¥
g CH╣.èWe need ê molar masses ï order ë convert between grams å
moles: C╖H╖: 2(12.01) + 2(1.008) = 26.04 g/mol, å
èèè CH╣:èè12.01 + 4(1.008)è= 16.04 g/mol.
For ê second step ï ê conversion, ê equation shows that 2 mol ç
CH╣ produces 1 mol ç C╖H╖.
èèè 1 mol C╖H╖èè 2 mol CH╣è 16.04 g CH╣
?g CH╣ = 750. g C╖H╖ x ──────────── x ────────── x ────────── = 925 g CH╣.
èèèèèèèèèèè 26.04 g C╖H╖è 1 mol C╖H╖è 1 mol CH╣
Ç D
6èBreathalyzers use ê followïg reaction ë determïe ê
amount ç ethyl alcohol, C╖H║OH, ï a person's breath (å blood).
3C╖H║OH + 2K╖Cr╖O╝ + 8H╖SO╣ ──¥ 3HC╖H╕O╖ + 2Cr╖(SO╣)╕ + 2K╖SO╣ + 11H╖O.
How many grams ç ethyl alcohol react with 0.188 g K╖Cr╖O╝?
A) 0.0767 g B) 0.0227 g C) 0.0511 g D) 0.691 g
ü The path for ê conversion is g K╖Cr╖O╝ ─¥ mol K╖Cr╖O╝ ─¥
mol C╖H║OH ─¥ g C╖H║OH.èWe need ê molar masses ï order ë convert
between grams å moles: K╖Cr╖O╝, 294.20 g/mol; C╖H║OH, 46.07 g/mol.
For ê second step ï ê conversion, ê equation shows that 3 mol ç
C╖H║OH reacts with 2 mol K╖Cr╖O╝.
?g C╖H║OH =
èèè1 mol K╖Cr╖O╝èè 3 mol C╖H║OHèè46.07 g C╖H║OH
èè0.188 g K╖Cr╖O╝ x ─────────────── x ───────────── x ──────────────
èèè254.2 g K╖Cr╖O╝è 2 mol K╖Cr╖O╝è 1 mol C╖H║OH
?g C╖H║OH = 0.0511 g C╖H║OH
Ç C
7 How many grams ç water are required for ê oxidation ç
5.47 g ç Fe(OH)╖?èThe reaction is:
èè 4Fe(OH)╖(s) + O╖(g) + 2H╖O(l) ──¥è4Fe(OH)╕(s).
A) 0.548 g H╖O B) 0.274 g H╖O
C) 2.19 g H╖O D) 54.6 g H╖O
üèThe balanced equation shows that 2 mol ç water are required for
ê oxidation ç 4 mol ç Fe(OH)╖.èSïce we are workïg with grams, we
need ê molar masses ë convert between moles å grams.èThe molar
masses are: Fe(OH)╖, 89.87 g/mol; å H╖O, 18.02 g/mol.èThe scheme ë go
from 5.47 g ç Fe(OH)╖ ë ê number ç grams ç water isè5.47 g Fe(OH)╖
─¥ mol Fe(OH)╖ ─¥ mol H╖O ─¥ g H╖O.
è 1 mol Fe(OH)╖èè2 mol H╖Oèèè 18.02 g H╖O
?g H╖O = 5.47 g Fe(OH)╖ x ─────────────── x ───────────── x ───────────
è89.87 g Fe(OH)╖è 4 mol Fe(OH)╖èè1 mol H╖O
?g H╖O = 0.548 g H╖O
Ç A
8èWhat mass ç NaCl is required ë produce 1500. g ç Cl╖ via
ê reaction,è2NaCl(aq) + 2H╖O ──¥ 2NaOH(aq) + H╖(g) + Cl╖(g)?
A) 1820. g NaCl B) 2473 g NaCl
C) 4946 g NaCl D) 909.9 g NaCl
üèThe balanced equation shows that 2 mol ç NaCl are required for
each mol ç Cl╖.èTo convert between moles å grams, we need ê molar
masses.èThe molar masses are: NaCl, 58.44 g/mol; å Cl╖, 70.90 g/mol.
The pathway from grams ç chlorïe ë grams ç NaCl is g Cl╖ ─¥ mol Cl╖
─¥ mol NaCl ─¥ g NaCl.
1 mol Cl╖èè 2 mol NaClè 58.44 g NaCl
?g NaCl = 1500. g Cl╖ x ─────────── x ────────── x ───────────
70.90 g Cl╖è 1 mol Cl╖èè1 mol NaCl
?g H╖O = 2473 g H╖O
Ç B
9èThe reaction, 2N╖H╣(l) + N╖O╣(l) ──¥ 3N╖(g) + 4H╖O(g), is used
ï some small rocket moërs.èHow many grams ç N╖O╣ will react with
50.0 g ç N╖H╣?
A) 144 g N╖O╣ B) 71.8 g N╖O╣
C) 34.8 g N╖O╣ D) 29.5 g N╖O╣
üèThe pathway for ê solution follows ê normal three step se-
quence: g N╖H╣ ─¥ mol N╖H╣ ─¥ mol N╖O╣ ─¥ g N╖O╣.èWe need ê molar
masses ë convert between grams å moles.èThe molar masses are:è
N╖H╣, 32.05 g/mol; N╖O╣, 92.02 g/mol.èThe balanced reaction provides ê
conversion facër between moles ç N╖H╣ å moles ç N╖O╣.
1 mol N╖H╣èè 1 mol N╖O╣è 92.02 g N╖O╣
? g N╖O╣ = 50.0 g N╖H╣ x ──────────── x ────────── x ────────────
32.05 g N╖H╣è 2 mol N╖H╣è 1 mol N╖O╣
? g N╖O╣ = 71.8 g N╖O╣
Ç B
10èSelf-contaïed breathïg apparatus replenish oxygen via ê
reaction: 4KO╖(s) + 4CO╖(g) + 2H╖O(l) ──¥ 4KHCO╕(s) + 3O╖(g).èHow many
grams ç KO╖ are needed ë supply 15.0 g ç O╖?
A) 33.3 g KO╖ B) 25.0 g KO╖
C) 5.06 g KO╖ D) 44.4 g KO╖
üèFïdïg ê needed grams ç KO╖ follows ê usual three step path
g O╖ ─¥ mol O╖ ─¥ mol KO╖ ─¥ g KO╖.èWe need ê molar masses ë convert
between grams å moles.èThe molar masses are: O╖, 32.00 g/mol;
KO╖, 71.10 g/mol.èThe balanced reaction provides ê conversion facër
between moles ç O╖ å moles ç KO╖: 4 mol KO╖ produce 3 mol O╖.
èèè1 mol O╖èèè 4 mol KO╖è 71.10 g KO╖
? g KO╖ = 15.0 g O╖ x ──────────── x ───────── x ───────────
èèè32.00 g O╖èè 3 mol O╖èè1 mol KO╖
? g KO╖ = 44.4 g KO╖
Ç D