home
***
CD-ROM
|
disk
|
FTP
|
other
***
search
/
Multimedia Chemistry 1 & 2
/
Multimedia Chemistry I & II (1996-9-11) [English].img
/
chem
/
chapter7.2c
< prev
next >
Wrap
Text File
|
1996-07-26
|
12KB
|
298 lines
à 7.2c Limitïg Reagents
ä Please determïe ê limitïg reagent å ê moles or grams ç product ï ê
followïg reactions.
â How many grams ç C╕H╣Br╣ can be prepared from 20. g C╕H╣ å
90. g Br╖ via ê reaction C╕H╣ + 2Br╖ ──¥ C╕H╣Br╣.èFirst we fïd ê
limitïg reagent:
è (20.g C╕H╣/40.06g/mol)(1 mol C╕H╣Br╣/1 mol C╕H╣) = 0.50 mol C╕H╣Br╣
è (80.g Br╖/159.8g/mol)(1 mol C╕H╣Br╣/2 mol Br╖) = 0.28 mol C╕H╣Br╣
Br╖ is ê limitïg reagent, because it produces less C╕H╣Br╣.èThe yield
ç C╕H╣Br╣ is (0.28 mol C╕H╣Br╣)(359.7 g/mol) = 101 g C╕H╣Br╣.
éSèThe limitïg reagent controls ê amounts ç ê products that
form durïg a chemical reaction.èAs an analog, suppose you were makïg
ham å cheese såwiches with two slices ç bread, oneèslice ç ham,
å one slice ç cheese.èHow many såwiches could you make from ten
slices ç bread, four slices ç ham, å five slices ç cheese?èYou have
enough bread å cheese for five såwiches but only enough ham for four
såwiches.èThe ham controls ê outcome ç ê såwich makïg process.
We have an excess amount ç ê bread å ç ê cheese.èYou could make
only four såwiches, å ham is ê limitïg reagent.è
Let us return ë a chemical example å revisit ê oxidation ç ammonia.
4NH╕(g) + 5O╖(g) ──¥ 4NO(g) + 6H╖O(g).
How many grams ç NO could be produced by ê reaction ç 20.0 g NH╕ å
30.0 g O╖?
There is more than one approach ë ê above problem.èFrom ê wordïg
ç ê question, we recognize this as a limitïg reagent problem because
we are given ê mass ç both reactants.èWe must determïe which react-
ant controls ê extent ç ê reaction.èSome ç ê oêr reactant will
be left over after all ç ê limitïg reagent has reacted.èThe simplest
approach is ë calculate ê number ç moles ç NO obtaïable from both
reactants.èThe smaller number identifies ê limitïg reagent.
1. Fïd ê number ç moles ç NO obtaïable from NH╕.
èèèèè 1 mol NH╕èè 4 mol NO
è ? mol NO = 20.0 g NH╕ x ─────────── x ───────── =è1.17 mol NO
èèèèè 17.03 g NH╕è 4 mol NH╕
2. Fïd ê number ç moles ç NO obtaïable from O╖.
èèèèè1 mol O╖èè 4 mol NO
è ? mol NO = 30.0 g O╖ x ────────── x ──────── = 0.750 mol NO
èèèèè32.00 g O╖è 5 mol O╖
These calculations show that O╖ is ê limitïg reagent.èAfter 0.750 mol
ç NO forms, all ç ê O╖ has reacted å ê net formation ç NO sëps.
The yield ç NO ï grams is based on ê smaller amount ç moles.
è èèèèè 30.01 g NO
è ? g NO = 0.750 mol NO x ────────── =è22.5 g NO
èèèèèè1 mol NO
The êoretical yield ç NO is 22.5 grams.èThis is called ê êoret-
ical yield, because it assumes that no oêr reactions occur between O╖
å NH╕.èAn unfortunate fact ç life is that frequently oêr less im-
portant reactions between ê reactants do occur.
We can also calculate how much NH╕ reacted å how much NO is left know-
ïg that O╖ is ê limitïg reagent.
èè 1 mol O╖èè 4 mol NH╕è 17.03 g NH╕
?g NH╕ = 30.0 g O╖ x ────────── x ───────── x ─────────── =è12.8 g NH╕
èè 32.00 g O╖è 5 mol O╖èè1 mol NH╕
This last calculation shows that 12.8 grams ç NH╕ react with 30.0 g O╖
å that 20.0 - 12.8 = 7.2 g ç NH╕ will be left after ê oxygen is
gone.èThe NH╕ is called ê reagent ï excess.
We started with 20.0 grams ç NH╕ å 30.0 grams ç O╖.èIf ê reaction
goes ë completion as we have assumed, ên ê fïal mixture should
contaï 7.2 g NH╕, 22.5 g NO, å 20.3 g H╖O.
1èHow many moles ç C╖H╖Cl╣ can be obtaïed from ê reaction
between 0.42 mol C╖H╖ å 0.76 mol Cl╖?èThe reaction is
C╖H╖(g) + 2Cl╖(g) ──¥ C╖H╖Cl╣(l).
A) 1.18 mol C╖H╖Cl╣ B) 1.94 mol C╖H╖Cl╣
C) 0.38 mol C╖H╖Cl╣ D) 0.42 mol C╖H╖Cl╣
üèThe êoretical yield ç C╖H╖Cl╣ from C╖H╖ is
1 mol C╖H╖Cl╣
? mol C╖H╖Cl╣ = 0.42 mol C╖H╖ x ───────────── = 0.42 mol C╖H╖Cl╣
1 mol C╖H╖
The êoretical yield ç C╖H╖Cl╣ from Cl╖ is
èèè 1 mol C╖H╖Cl╣
? mol C╖H╖Cl╣ = 0.76 mol Cl╖ x ───────────── = 0.38 mol C╖H╖Cl╣
èèè 2 mol Cl╖
The êoretical yield is 0.38 mol C╖H╖Cl╣ because it is not possible ë
form more than ê smaller amount.èThe chlorïe, Cl╖, is ê limitïg
reagent; å ê acetylene, C╖H╖, is present ï excess.
Ç C
2èHow many grams ç NH╕ could be obtaïed from ê reaction ç
50.0 g Mg╕N╖ å 50.0 g H╖O?èThe reaction is
èèèMg╕N╖(s) + 6H╖O(l) ──¥ 3Mg(OH)╖(s) + 2NH╕(g).
A) 16.9 g NH╕ B) 100. g NH╕
C) 15.8 g NH╕ D) 47.3 g NH╕
üèFirst, we determïe ê limitïg reagent by calculatïg ê moles
ç NH╕ from each reactant.
è 1 mol Mg╕N╖ èè2 mol NH╕
? mol NH╕ = 50.0 g Mg╕N╖ x ────────────── x ─────────── = 0.991 mol NH╕
from Mg╕N╖èèèèèèèè 100.95 g Mg╕N╖è 1 mol Mg╕N╖
1 mol H╖Oèè 2 mol NH╕
? mol NH╕ = 50.0 g H╖O x ─────────── x ───────── = 0.925 mol NH╕
from H╖O 18.02 g H╖Oè 6 mol H╖O
Water is ê limitïg reagent because it will not be possible ë form
more than 0.925 mol NH╕.èThe yield ç ammonia is
è17.03 g NH╕
? g NH╕ = 0.925 mol NH╕ x ─────────── = 15.8 g NH╕.
è 1 mol NH╕
Ç C
3èHow many grams ç B╖H╗ could be obtaïed from ê reaction ç
5.00 g LiAlH╣ å 10.0 g BF╕?èThe reaction is
3LiAlH╣ + 4BF╕ ──¥ 2B╖H╗ + 3LiAlF╣.
A) 2.43 g B╖H╗ B) 2.04 g B╖H╗
C) 3.52 g B╖H╗ D) 1.78 g B╖H╗
üèFirst, we determïe ê limitïg reagent by calculatïg ê moles
ç B╖H╗ from each reactant.
èè 1 mol LiAlH╣èè 2 mol B╖H╗
? mol B╖H╗ = 5.00 g LiAlH╣ x ────────────── x ──────────── = 0.263 mol
from LiAlH╣èèèèèèèèè37.95 g LiAlH╣è 3 mol LiAlH╣èèèè B╖H╗
è1 mol BF╕èè 2 mol B╖H╗
? mol B╖H╗ = 10.0 g BF╕ x ─────────── x ────────── = 0.0738 mol B╖H╗
from BF╕ è67.78 g BF╕è 4 mol BF╕
BF╕ is ê limitïg reagent because it will not be possible ë form more
than 0.0738 mol B╖H╗.èThe yield ç diborane, B╖H╗, is
èè 27.67 g B╖H╗
? g B╖H╗ = 0.0738 mol B╖H╗ x ────────────è= 2.04 g B╖H╗
èè 1 mol B╖H╗
Ç B
4èWhat is ê limitïg reagent when 0.298 g C╖H║OH reacts with
0.175 g K╖Cr╖O╝ ï ê presence ç excess H╖SO╣?èThe reaction is:
3C╖H║OH + 2K╖Cr╖O╝ + 8H╖SO╣ ──¥ 3HC╖H╕O╖ + 2Cr╖(SO╣)╕ + 2K╖SO╣ + 11H╖O.
A) C╖H║OH B) K╖Cr╖O╝ C) HC╖H╕O╖ D) Cr╖(SO╣)╕
ü In order ë fïd ê limitïg reagent, we eiêr calculate how
much C╖H║OH will react with ê K╖Cr╖O╝ or vice-versa.èLet's fïd out
how much K╖Cr╖O╝ will react with ê 0.298 g C╖H║OH.
? g K╖Cr╖O╝ =
è1 mol C╖H║OHèè2 mol K╖Cr╖O╝è 294.2 g K╖Cr╖O╝
0.298 g C╖H║OH x ───────────── x ───────────── x ───────────────
è46.07g C╖H║OHè 3 mol C╖H║OHèè1 mol K╖Cr╖O╝
? g K╖Cr╖O╝ = 1.27 g K╖Cr╖O╝
The 0.298 g ç C╖H║OH requires 1.27 g ç K╖Cr╖O╝.èThere is only 0.175 g
ç K╖Cr╖O╝, so K╖Cr╖O╝ is ê limitïg reagent.
Ç B
5èHow many grams ç hydrogen would form ï ê reaction ç
3.00 g Al with 0.228 mol HCl?èThe reaction is:
2Al(s) + 6HCl(aq) ──¥ 2AlCl╕(aq) + 3H╖(g).
A) 6.30x10úÄ g H╖ B) 0.335 g H╖
C) 1.02 g H╖ D) 0.230 g H╖
üèFirst, we determïe ê limitïg reagent by calculatïg ê moles
ç H╖ from each reactant.
èèè 1 mol Alèè 3 mol H╖
? mol H╖ = 3.00 g Al x ────────── x ──────── = 0.166 mol H╖
from Alèèèèèèèè26.98 g Alè 2 mol Al
è 3 mol H╖
? mol H╖ = 0.228 mol HCl x ───────── =è0.114 mol H╖
from HCl è 6 mol HCl
Sïce ê HCl form ê smaller amount ç H╖, ê HCl is ê limitïg
reagent.èThe amount ç hydrogen is based on ê moles ç H╖ from ê
HCl.
? g H╖ = 0.114 mol H╖ x 2.016 g H╖/mol H╖è= 0.230 g H╖
Ç D
6èHow many grams ç phosphorous, P╣, can be obtaïed from ê
reaction ç 500. g Ca╕(PO╣)╖, 300. g SiO╖, å excess C via ê reaction
è2Ca╕(PO╣)╖(s) + 6SiO╖(s) + 10C(s) ──¥ 6CaSiO╕(s) + 10CO(g) + P╣(s)?
A) 99.8 g P╣ B) 200. g P╣
C) 103 g P╣ D) 618 g P╣
üèWe begï by fïdïg ê limitïg reagent, which is eiêr SiO╖ or
Ca╕(PO╣)╖, by calculatïg ê moles ç P╣ obtaïable from both reactants.
è1 mol Ca╕(PO╣)╖èèè1 mol P╣
?mol P╣ from = 500. g Ca╕(PO╣)╖ x ────────────────── x ───────────────
Ca╕(PO╣)╖ è310.18 g Ca╕(PO╣)╖è 2 mol Ca╕(PO╣)╖
?mol P╣ from = 0.806 mol P╣
Ca╕(PO╣)╖
èèè1 mol SiO╖èè 1 mol P╣
? mol P╣ from = 300. g SiO╖ x ──────────── x ────────── = 0.832 mol P╣
èèè60.09 g SiO╖è 6 mol SiO╖
The limitïg reagent is Ca╕(PO╣)╖ because only 0.804 mol P╣ can be made
from it.èThe yield ç P╣ is
? g P╣ = 0.806 mol P╣ x 123.88 g P╣/mol P╣ = 99.8 g P╣
Ç A
7èHow many grams ç C╗H╣(NO╖)╖ can be made from 15.0 g C╗H╗ å
40.0 g HNO╕ via ê reaction, C╗H╗ + 2HNO╕ ──¥ C╗H╣(NO╖)╖+ H╖O?
A) 6.94 g C╗H╣(NO╖)╖ B) 53.4 g C╗H╣(NO╖)╖
C) 30.0 g C╗H╣(NO╖)╖ D) 32.3 g C╗H╣(NO╖)╖
ü You need ë fïd ê limitïg reagent by determïïg ê moles ç
product from each reactant.
èèè 1 mol C╗H╗è1 mol C╗H╣(NO╖)╖
?mol C╗H╣(NO╖)╖ = 15.0g C╗H╗ x ───────────x──────────────── = 0.192 mol
from C╗H╗ èèè 78.11g C╗H╗è1 mol C╗H╗èèèè C╗H╣(NO╖)╖
èèè 1 mol HNO╕è1 mol C╗H╣(NO╖)╖
?mol C╗H╣(NO╖)╖ = 40.0g HNO╕ x ───────────x──────────────── = 0.317 mol
from HNO╕ èèè 63.02g HNO╕è2 mol HNO╕èèèè C╗H╣(NO╖)╖
The C╗H╗ is ê limitïg reagent sïce it produces ê smaller amount ç
product å would be depleted first.èThe yield ç C╗H╣(NO╖)╖ is
èèèèèèè 168.11 g C╗H╣(NO╖)╖
?g C╗H╣(NO╖)╖ = 0.192 mol C╗H╣(NO╖)╖ x ─────────────────── = 32.3 g
èèèè1 mol C╗H╣(NO╖)╖èèè C╗H╣(NO╖)╖
Ç D
8èHow many grams ç C╗H║NH╖ can be obtaïed from 10.0 g C╗H║NO╖
å 30.0 g SnCl╖ ï ê presence ç excess HCl via ê reaction,
C╗H║NO╖ + 3SnCl╖ + 6HCl ──¥ C╗H║NH╖ + 3SnCl╣ + 2H╖O?
A) 4.91 g C╗H║NH╖ B) 13.2 g C╗H║NH╖
C) 1.64 g C╗H║NH╖ D) 20.4 g C╗H║NH╖
üèWe begï by fïdïg ê limitïg reagent.èWhich compound will be
exhausted first?
èèè1 mol C╗H║NO╖èè1 mol C╗H║NH╖
?mol C╗H║NH╖ = 10.0g C╗H║NO╖ x────────────────x───────────── = 0.0812 mol
from C╗H║NO╖ èèè123.11 g C╗H║NO╖ 1 mol C╗H║NO╖èè C╗H║NH╖
èè 1 mol SnCl╖èè 1 mol C╗H║NH╖
? mol C╗H║NH╖ = 30.0g SnCl╖ x───────────── x ───────────── = 0.0527 mol
from SnCl╖ èè 189.6 g SnCl╖è 3 mol SnCl╖èèèè C╗H║NH╖
The limitïg reagent is SnCl╖, because all ç it will have reacted when
0.0527 mol ç C╗H║NH╖ has formed.èThe yield ç C╗H║NH╖ would be
èè93.13 g C╗H║NH╖
? g C╗H║NH╖è= 0.0527 mol C╗H║NH╖ x ──────────────── =è4.91 g C╗H║NH╖
èè 1 mol C╗H║NH╖
Ç A
9èHow many grams ç I╖ will be formed from ê reaction between
5.00 g HNO╖, 3.00 g KI, å excess HCl?èThe reaction is
èè 2HNO╖ + 6KI + 6HCl ──¥ 3I╖ + N╖ + 6KCl + 4H╖O.è
A) 2.70 g I╖ B) 2.29 g I╖
C) 4.59 g I╖ D) 4.05 g I╖
üèYou must determïe wheêr HNO╖ or KI is ê limitïg reagent.
You can calculate how many moles ç I╖ are possible from each compound.
èè1 mol HNO╖èè 3 mol I╖
? mol I╖èè= 5.00 g HNO╖ x ──────────── x ────────── = 0.160 mol I╖
èfrom HNO╖ èè47.02 g HNO╖è 2 mol HNO╖
1 mol KIèè 3 mol I╖
? mol I╖è= 3.00 g KI x ────────── x ──────── = 9.04x10úÄ mol I╖
èfrom KI 166.0 g KIè 6 mol KI
We have determïed that KI is ê limitïg reagent,because it forms ê
smaller amount ç I╖.èThe amount ç I╖ that would be formed is
èè253.8 g I╖
? g I╖ = 9.04x10úÄ mol I╖ x ────────── = 2.29 g I╖
èè 1 mol I╖
Ç B
10èWhich compound is ê limitïg reagent when 25.0 g CH╕OH,
methanol, reacts with 35.0 g O╖ accordïg ë ê reaction,
èè2CH╕OH(g) + 3O╖(g) ──¥ 2CO╖(g) + 4H╖O(l)?
èèA) CH╕OHèè B) O╖èè C)CO╖èè D) H╖O
ü The limitïg reagent must be one ç ê reactants, eiêr CH╕OH
or O╖.èWe can fïd out eiêr how much CH╕OH will react with ê 35.0 g
ç O╖, or how much O╖ will react with ê 25.0 g ç CH╕OH.èLet's look
at ê second choice.
1 mol CH╕OHèè 3 mol O╖èèè32.00 g O╖
? g O╖ = 25.0 g CH╕OH x ───────────── x ─────────── x ──────────
32.04 g CH╕OHè 2 mol CH╕OHè 1 mol O╖
? g O╖ = 37.5 g O╖
This calculation shows that 37.5 g ç oxygen is needed ë react ê
35.0 g ç CH╕OH.èWe only have 35.0 g ç O╖ so êre is not enough O╖.
Oxygen, O╖, is ê limitïg reagent.
Ç B