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TUYN TP CC BI TP HA HC HAYV LI GII CA DIN N BOXMATH

(sa ln 1, ngy 17/04/2012)

BoxMath, thng 1 2012

MC LC

LI NG ......................................................................................................................................................... 1

DANH SCH THNH VIN THAM GIA BIN SON........................................................................... 2

PHN 1. CC BI TON V C................................................................................................................. 3

Bi 1 10 ...................................................................................................................................................... 3

Bi 11 20 .................................................................................................................................................... 9

Bi 21 30 .................................................................................................................................................. 13

Bi 31 40 .................................................................................................................................................. 17

Bi 41 47 .................................................................................................................................................. 22

PHN 2. CC BI TON HU C............................................................................................................ 25

Bi 1 10 .................................................................................................................................................... 25

Bi 11 20 .................................................................................................................................................. 29

Bi 21 30 .................................................................................................................................................. 33

Bi 31 40 .................................................................................................................................................. 37

Bi 41 50 .................................................................................................................................................. 41

Bi 51 54 .................................................................................................................................................. 46

1 http://boxmath.vn 17/04/2012

LI NG

C nhiu ngi nhn xt Ha hc l b mn ca tr nh, hc ha phi nh rt nhiu th iu c thng, nhng cha phi l tt c. Ha hc giu trong lp o nng n ca nhng cng thc, nhng phn ng phc tp l mt v p tinh t ca s t duy logic. V p y c bit c hin din qua nhng bi ton ha, nhng bi ton cho bn rt nhiu con ng, nhng s ch c mt con ng p nht, ngn nht i n c chn l. Nhim v ca ngi hc ha l phi vn dng cc thao tc nh gi, phn on tm ra c con ng y, con ng khng ch a n kt qu m con a n mt nim vui, nim hng khi nh mt cht keo gn tri tim bn vi Ha hc.Tng c nhng giy pht v a trong sung sng tng c nhng khonh khc chi vi, b tc gia nhng s liu, nhng phng trnh phn ng Chng ti tin hnh lm tuyn tp ny vi mc ch u tin l nim mong mun ng cm, l kht khao chia s tnh yu Ha hc ca chng ti vi cc bn thng qua nhng bi tp ha mi l, nhng li gii hay.

Mi bi tp khng ch n gin l tnh ton, ng sau l nhng tng.

Mi li gii khng ch l p dng phng php m thc s l mt qu trnh phn tch v sng to.

Ha hc ang c tin hnh thi theo phng php trc nghim, th nhng li gii u tin m chng ti a ra vn lun lun l mt li gii y , i t nhng d kin ca bi ton n kt qu m hon ton khng ph thuc vo cc p n cho trc. Bi chng ti hiu v mong cc bn s hiu, ch c nh vy, chng ta mi c th i ht c v p ca ha hc, rn luyn c t duy suy lun logic cho bn thn, hc tp mt cch thc cht v sng to.

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Mi kin xin gi v [email protected]

Thay mt ban bin tp ti xin chn thnh cm n.

H Ni, ngy 18/1/2012

i din nhm bin son

Ch bin

F7T7


DANH SCH THNH VIN THAM GIA BIN SON

1. Ph Tin Cng THPT chuyn i hc S phm H Ni

2. Thi Mnh Cng THPT chuyn Phan Bi Chu, Ngh An

3. Trn Bo Dng THPT Ng Quyn, B Ra Vng Tu

4. Nguyn Th Thu Hi THPT Trn Ph, H Tnh

5. Nguyn Quc Oanh THPT So Nam, Qung Nam

6. Nguyn Hu Ph THPT Ty Sn, Lm ng

7. Phan Qunh Nga THPT Hng Kh, H Tnh


PHN 1. CC BI TON V C

Bi 1 10

Bi 1. _________________________________________________________________________________Hn hp A gm mui sunfit, hidrosunfit v sunfat ca cng mt kim loi kim M. Cho 17,775 gam hn hp A vo dung dch 2( )Ba OH d, to thnh 24,5275 gam hn hp kt ta. Lc kt ta, ra sch v cho kt ta

tc dng vi dung dch HCl d, thy cn 2,33 g cht rn. Kim loi kim M lA. Li B. K C. Rb D. Na

Li gii.Phng trnh phn ng

2 24 4Ba SO BaSO

+ -+ 2

3 2 3OH HSO H O SO- -+ +

2 23 3Ba SO BaSO

+ -+

Cht rn cn li l 4BaSO : 42,33

0,01( )233BaSO

n mol= =

3

(24,5275 2,33)0,1023( )

217BaSOn mol

-= =

17,775158

0,1023 0,01M = =

+trung binh

Ta c 80 158 2 96TBM M M+ < < += . Ch c M = 39 tha mn iu kin ny. Chn p n B.

Bi tp tng t. Hn hp X gm 2 3 3,M CO MHCO v MCl (M l kim loi kim). Cho 32,65 gam X tc

dng va vi dung dch HCl thu c dung dch Y v c 17,6 gam 2CO thot ra. Dung dch T tc dng

vi dung dch 3AgNO d c 100,45 gam kt ta. Kim loi M l

A. Na B. Li C. K D. Rb

Bi 2. ______________________________________________________________________________Cho mt hp cht ca st tc dng vi 2 4H SO c nng, to ra 2SO (sn phm kh duy nht). Nu t l

2 4H SO em dng v 2SO to ra l 2 4 2: 4 :1H SO SOn n = th cng thc phn t ca X l:

A. Fe B. FeS C. FeO D. Fe3O4Li gii.

t ( )2SO

n a mol= Ta c:

Nu X l oxit ca st th ta c qu trnh kh l

( ) ( )6 4

2a2

mol a molS e S+ ++

( )2 4

4 3H SOn a a a mol == -tao muoi

( ) ( )32 4 3( )

2Fe SO Fen a mol n a mol+ = =

( ) ( )

3

2 22 2(3 ) 2x

a mol a molFe x e Fe+ ++ -

2x =Ngoi ra khng cn nguyn t no nhng hay nhn eVy cng thc X l FeO


Nu X l mui sunfua ca st th X c CTPT dng x yFe S (a mol)

Ta c cc qu trnh oxi ha kh

( ) ( )6 4

22

mo bb l molS e S+ ++

( (

2

) ) ( )

3 1ax mol ax mol ax molFe eFe+ + +

24

4

( ) ( )

2 ( )

24

ay mol ay mol

ay ax mo

y

l

xx

S S ey

++

-

+ -

Vy 2

( )SOn ay b mol= + , ( )6 22 4 432H SO S SO taomuoiax

n n n b mol+ -= + = +

Suy ra ( )3 4( ) 12ax

b ay b+ = +

Theo nh lut bo ton electron ta li c ( )2 (4 2 ) 2b ax ay ax= + -

Gii h phng trnh (1) v (2) suy ra 10 3

, : 10 : 3b b

x y x ya a

= = = (khng tn ti hp cht no

tha mn iu ny). Vy trng hp th hai khng tha mn.Kt lun: Ch c FeO tha mn bi. Chn p n C.Bnh lun: Vi trng hp X l oxit ca st, d nhn thy 2 3Fe O phn ng khng to ra 2SO , ch xt FeO

v 3 4Fe O . Ta thy 1 mol ca hai cht ny khi phn ng u nhng 1 mol electron. Vy nu t

( )2SO

n a mol= th s mol ca FeO v 3 4Fe O u l 2 ( )a mol . Nhng ch c FeO mi to ra c mui

cha 243 ( )a mol SO- tha mn yu cu bi ra. Chn ngay FeO, p n B.

Bi tp tng t. X l mt hp cht ca Fe. Cho X tc dng vi 2 4H SO c nng thy thot ra kh 2SO vi

t l mol X v 2SO l 2:9. X l:

A. 3 4Fe O B. FeS C. 2FeS D. FeO

p s: B. FeS

Bi 3. _____________________________________________________________________________Cho t t a gam st vo V ml dung dch 3HNO 1M khuy u cho n khi tan ht thy thot ra 0,448 lt kh

NO (ktc) ng thi thu c dung dch A . Dung dch A c kh nng lm my mu hon ton 10 ml dung

dch 4KMnO 0,3M trong mi trng axit. Gi tr ca a v V l:

A. a =1,4 gam; V = 80 ml B. a = 1,12 gam; V = 80 mlC. a = 0,56 gam; V = 56 ml D. a = 0,84 gam; V = 60 ml

Li gii.Ta c cc phng trnh th hin qu trnh oxi ha kh:

( )( )5 20,06 0,023 1N e N NO+ ++

( )333 2

x x xFe Fe e+ +

( )222 3

y y yFe Fe e+ +

7 2

0,003 0,0155 (4)Mn e M+ ++

2 3 1 (5)y y

Fe Fe e+ + +

Theo phn ng kh 4KMnO , t (4) v (5) ta c 0,015y =

Theo phn ng kh 3HNO , t (1), (2) & (3), ta c 3 2 0,06 0,01( )x y x mol+ = =


Vy 0,025 1,4Fen m g= =

3 3 33 2 0,02 0,08( )HNO NO NOn n n x y mol- -= + = + + = tao muoi bi khu

Suy ra ( )80V ml=

Bi 4. _____________________________________________________________________________Hn hp X gm c , x yAl Fe O . Tin hnh nhit nhm hon ton ( )m g hn hp X trong iu kin khng c

khng kh thu c hn hp Y. Chia Y thnh hai phn.Phn 1. Cho tc dng vi NaOH d thu c 1,008 lt 2H (ktc) v cn li 5,04 gam cht rn khng tan.

Phn 2 c khi lng 29,79 gam, cho tc dng vi dung dch 3HNO long d thu c 8,064 lt NO (ktc,

l sn phm kh duy nht).Gi tr ca m v cng thc ca oxit st l

A. 39,72 gam & FeO B. 39,72 gam & 3 4Fe O

C. 38,91 gam & FeO D. 36,48 gam & 3 4Fe O

Li gii.Cch 1

2 32 3 3 (1)x yyAl Fe O yAl O xFe+ +

Phn ng nhit nhm hon ton m phn 1 to kh 2H nn hn hp Y gm 2 3, ,Al Fe Al Od Phn 1

( )Aln a mol= ; ( )Fen b mol= ; ( )2 3Al On c mol=Phng trnh phn ng

2 2 20,03 0,0452 2 2 2 3Al NaOH H O NaAlO H+ + +

5,04Fem m g= =ran . Suy ra ( )0,09Fen mol= Phn 2

( )Aln ka mol= ; ( )Fen kb mol= ; ( )2 3Al On kc mol=Theo nh lut bo ton electron ta c: ( )3 3 .0,123. 0,36.3 3NOk a b k n k+ = = =

29,79 (27.0,09 56.0,27)0,04( )

3.102c mol

- + = =

29,79.439,72

3m g = =

T (1) ta c 0,04.3 0,09 4 3x y x y= =Chn p n B

Cch 2. Phn 1

Sau khi tc dng vi NaOH to ra kh nn sau khi nhit nhm th Al cn d 0,03Aln =

Cht rn cn li l Fe 0,09Fen mol =13

Al

Fe

nn

=

Phn 2Gi s mol ca Al l a th s mol ca Fe l 3a. Bo ton electron ta c:12 1,08 0,09a a= =

2 3 2 312,24 0,12Al O Al Om n = =


Suy ra phn 1 c 2 3

0,04Al On =39,72( )m gam =

Da vo s mol ca O v Fe suy ra CTPT ca oxit l 3 4Fe O .

Bi 5. ______________________________________________________________________________in phn in cc tr dung dch cha 0,2 mol 3AgNO vi cng dng in 2,68 A, trong thi gian t

(gi) thu c dung dch X (hiu sut ca qu trnh in phn l 100%). Cho 16,8 gam bt Fe vo X thy thot ra kh NO (sn phm kh duy nht) v sau cc phn ng hon ton thu c 22,7 gam cht rn. Gi trca t l:

A. 2,00 B. 1,00 C. 0,50 D. 0,25Li gii.

3 2 3 24 2 4 4AgNO H O Ag HNO O+ + +

Gi x l s mol 3AgNO d v y l s mol 3HNO to ra. Tng s mol ca 3AgNO d v 3HNO chnh bng s

mol 3NO- khng i. Do , ta c

30, 2

NOx y n -+ = =

Ag ti a to ra cng ch 0,2 mol tc l khi lng cht rn sau phn ng nh hn 21,6 gam. Vy trong

cht rn cn Fe d.22 2Fe Ag Fe Ag+ ++ +

3 3 3 24 ( ) 2Fe HNO Fe NO NO H O+ + +

Suy ra s mol Fe phn ng l3

0.58y

x +

S mol Ag to ra l x, vy c 3

108 (0.5 .56) 22,7 16,88y

x x- + = -

T tm c 0,1x y= = hay 3600( ) 1( ).t s h= =

Bi 6. _______________________________________________________________________________Cho 8,64 gam Al vo dung dch X (X c to thnh bng 74,7 gam hn hp Y gm 2CuCl v 3FeCl vo

nc). Kt thc phn ng thu c 17,76 gam cht rn gm hai kim loi. T l mol ca 3 2:FeCl CuCl trong

hn hp Y l:A. 2:1 B. 3:2 C. 3:1 D. 5:3

Li gii.- Nu d Al th chc chn hn hp kim loi phi cha c 3 kim loi Al, Cu v Fe. Do Al phi ht

sau phn ng v hai kim loi cn li l Cu v Fe .- Do Fe nn chc chn c 3 phn ng sau xy ra theo th t:

3 3 2

3

3 e 3 ex xx

Al F Cl AlCl F Cl+ +

2 323

2 3 2A 3yyy

Al CuCl lCl Cu+ +

2 3 32

2 3 3 3z x z

Al FeCl AlCl Fe+ + ( y 32

x z , do Al phn ng ht.

Ta c 2

0,32( ) 0,323 3Alx

n mol y z= + + =

Khi lng hai mui ban u l 74,7 gam, suy ra .162,5 .135 74,7x y+ =

Khi lng hai kim loi thu c l 17,76, suy ra 3

64 56. 17,762

x z+ =


Gii h ba phng trnh trn suy ra: 0,36; 0,12; 0,12x y z= = =Vy : 3 :1x y = . Chn p n C.

Bi 7. _______________________________________________________________________________ Cho 240 ml dung dch 2( )Ba OH 1M vo 200 ml dung dch hn hp 3AlCl a mol/lt v 2 4 3( )Al SO 2a mol/lt

thu c 51,3 gam kt ta. Gi tr ca a l:A. 0,12 B. 0,15 C. 0,16 D. 0,2

Li gii.

Trong cc dung dch c 0,24 mol 2 ;Ba + 0,48 mol ,OH - a mol 3 ;Al + 0,6a mol ;Cl - 1,2a mol 24SO- .

Xt bng sau da theo cc gi tr ca aa 0,12 0,16 0,2

Nhn xt 4 0,48;a < 1,2 0,24a (loi)

TH2: to 3CH OH trong phn ng vi KOH

Khi lng cht rn khan thu c l:74 0,7.0,2.56 32 12,88 0,12; 8,88m a a a m g= + = + = =

p n 8,88 gam.Bnh lun. Bi ton t ra hai thao tc, trong vic xc nh c CTPT ca este l thao tc u tin v mu cht. tm c CTPT ca este ta phi gii mt bi ton v phn ng chy m hon ton c th ng ring ra lm thnh mt bi ton c lp. Bi tp trn thc ra khng kh, nhng li c t vo hai bi ton khc nhau, tnh ton li kh di dng, nn li tr thnh bi tp kh.

Bi 26. ______________________________________________________________________________________Cho m gam tinh bt ln men thnh ancol (ru) etylic vi hiu sut 81% . Ton b lng 2CO hp th hon

ton vo dung dch nuc vi trong, thu c 550 g kt ta v dung dich Y . un k dung dch Y thu thm 100 gam kt ta. Khi lng m l bao nhiu ?

Li giiCc phng trnh phn ng:

2 2 3 25,5 5,5

( )Ca OH CO CaCO H O+ +

1 12 2 2 3 2( ) ( )Ca OH CO H O Ca HCO+ +

21

3 3 2 21

( )Ca HCO CaCO H O CO + +

Tng s mol 2CO phn ng l


( )

( ) ( )2

5,5 1.2 7,5

1257,5 : 2 :81% 750

27

CO

tinh bot tinh bot

n mol

n mol m gam- -

= + =

= = =

Bi 27.______________________________________________________________________________________Cho 200 gam mt loi cht bo c ch s axit bng 7 tc dng va vi mt lng NaOH, thu c 207,55gam hn hp mui khan. Khi lng NaOH tham gia phn ng l:

A. 31 gam B. 32,36 gam C. 30 gam D. 31,45 gamLi gii

1 g cht bo cn trung ha bi 7 mg KOHSuy ra 200 g s cn 1400 mg KOH tc l cn 0,025 mol KOH

20,025 0,025R COOH NaOH R COONa H O- + - +

x3 3 5 3 5 33( ) 3 3 ( )

xRCOO C H NaOH RCOONa C H OH+ +

Ta c 1400

0,025( )56.1000KOH NaOH

n n mol= = =

p dng LBT KL: 200 0,025.40 120 207,55 18.0,025 92 0,025( )x x x mol+ + = + + =

3.0,25.40 0,025.40 31NaOHm g= + = .Chn p n C.

Bi 28. ______________________________________________________________________________________Khi thu phn kim 265,2 g cht bo to bi mt axitcacboxylic thu uc 288 gam mui kali. Cht bo ny c tn gi l:

A. glixerol tristearat B. glixerol trioleat C. glixerol trilinoleat D. glixerol tripanmitat

Li giiDng phng php tng gim khi lng vi X l s mol ca este.(393 41) 288 265,2 22,8 0,3x x mol- = - = =

17 33884 .axitM C H COOH= Chn p n B.

Bi 29. _____________________________________________________________________________________A l mt hn hp gm hai cht thuc dy ng ng ca stiren c khi lng phn t hn km nhau 14

vC. t chy hon ton m gam A bng 2O d. Cho sn phm chy hp th vo 300ml dung dch NaOH

2M. Khi lng bnh ng dung dch tng 22,44 gam v thu c dung dch D. Cho 2BaCl d vo dung

dch D thu c 35,46 gam kt ta. cc phn ng Xy ra hon ton. Tm CTPT ca 2 hidrocacbon trong A.Li gii

Gi CTPT trung bnh ca A l 2 8n nC H -

3 2

2 2 2

35,460,18( ) 0,18 (0,3.0,2 0,182) 0,42( )

1970,42 0,22

44 18 22,44 0,22( ) 0,054

0,428,4

0,05

BaCO CO

CO H O H O A

n mol n mol

n n n mol n

n

= = = + - =

-+ = = = =

= =

Suy ra hai hidrocacbon cn tm l 8 8 9 10,C H C H

Bi 30. _____________________________________________________________________________________


Hp cht X mch h c CTPT l 4 9 2C H NO . Cho 10,3 g X phn ng va vi dung dch NaOH sinh ra 1

cht kh Y v dung dch Z. Kh Y nng hn khng kh, lm qu tm m chuyn mu Xanh. dung dch Z c kh nng lm mt mu nc brom. C cn dung dch Z thu c m g mui khan. Gi tr m?

Li gii

Kh Y nng hn khng kh, lm qu tm m chuyn mu Xanh suy ra: 3 2CH NH

Dung dch Z c kh nng lm mt mu nc brom suy ra phi c lin kt i C C= . Suy ra X l

2 3 2CH CHCOONH CH= .

Do 10,3

94 .94 9,4103muoi muoi muoi X

m n M n g= = = =

Bi 31 40

Bi 31. ____________________________________________________________________________Aminoaxit Y cng thc c dng ( )x y mNC H COOH . Ly mt lng axit aminoaxetic ( X ) v 3,104 gam Y. Bit X v Y c cng s mol. t chy hon ton lng X v Y trn, th tch 2O cn dng t chy Y

nhiu hn X l 1,344l (ktc). CTCT thu gn ca Y l:

3 2.A CH NHCH COOH 2 2 2.B H NCH CH COOH

( )3 3.C N CH COOH ( )4 8 2.D NC H COOHLi gii.

t:

1: : ( )x y zY C H O N k mol

2 5 2: : ( )X C H O N k mol

1 2 2 2 2

1( )

4 2 2 2x y zy z y

C H O N x O xCO H O N+ + - + +

2 5 2 2 2 2 2

5 2 5 1(2 ) 2

4 2 2 2C H O N O CO H O N+ + - + +

5 2 1,344(2 ). ( ). (1)

4 2 22,4 4 2y z

k x k + - + = + - vi 3,104 3,10412 16 14 Y

kx y z M

= =+ + +

Thay vo ta c:

( ) ( )3,104 9 4x 2z 0,06 15. 12x 16z 14 194 9 4x+ 2z12x 16z 14 4

596x 179 628z 1956 0

yy y

y

y

- - + = - + + + = - + - + + + + - - =

T y suy ra 4y , ch c p n C tha mn. Th li p n ny, ta chn C l ph hp.

Bi 32. _____________________________________________________________________________Cho 0,02 mol mt este X phn ng va ht vi 200 ml dung dch NaOH 0,2M, sn phm to ra ch 1 mui v mt ancol u c s mol bng s mol este, u c cu to mch thng. Mt khc khi x phng ho hon ton 2,58 gam este bng 20 ml dung dch KOH 1,5M va thu c 3,33 gam mui. Vy X l:

A. etilenglicol oxalat B. etilenglicol adipatC. imetyl adipat D. ietyl oxalat

Li gii.


TN1. Do 12este NaOH

n n= , nn este phi l este hai chc hoc l este ca phenol.

Li c muoi ancol esten n n= = , nn este phi l este ca ancol hai chc v axit hai chc.

Suy ra CTPT ca este l 2( )R COO R .

TN2.

Ta c 1

0,0152este muoi KOH

n n n mol= = =

T tnh ra c 4 8R C H= v 2 4R C H = . Suy ra este l C. imetyl adipat.

Bi 33. ____________________________________________________________________________Mt este A ( khng cha chc khc) c to nn t 1 axit hu c B v 1 ancol C. Ly m gam A cho tc dng vi KOH d thu c m1 gam mui. Ly m gam A cho tc dng vi 2( )Ca OH d thu c 2m gam

mui. Bit 2 1m m m< < . CT thu gn ca C l:

A. 2 5C H OH B. 3CH OH

C. 3 7C H OH D. 4 8C H OH

Li gii.

1m gam m gamRCOOR RCOOK

Ta c 1m m> nn 39R nn 2 40R >

Vy 20 39R< =

-

Bi 35. ____________________________________________________________________________un 0,4 mol hn hp 2 ancol no, n chc, mch h k tip trong dy ng ng bng dung dch 2 4H SO

140o C thu c 7,704 g hn hp 3 ete. Tham gia phn ngs ete ho c 50% s mol ancol c khi lng phn t ln v 40% ancol c khi lng phn t nh. Xc nh cng thc hai ancol.


A.Metylic v etylic B. etylic v n-propylicC. n-propylic v n-butylic D. propan-2-ol v butan-2-ol

Li gii.

Gi hn hp 2 ancol l ROH vi 14 1R n= + (do hai ancol no, n chc, h)Hai ancol l ROH (x mol), R OH (y mol). Vi R R >Ta c h phng trnh h phng trnh

0,4

0,5 0,4(2 16)( ) 7,704

2

x y

x yR

+ = +

+ =

12,848 0,320,2 1,6

Rx

R-

=+

T iu kin 0 0,4 30,52 40,15x R< < < > nn khi t A th phi c 2 2H O COn n> .

t CTPT ca A l 2 3 2 2 2n n x y xC H O N+ - - , vi y l bt bo ho ca gc R ( 0;y x 1)

Nu 2x y+ th khi t A c 2 2H O CO

n n< .

Do vy 0, 1y x= = , khi A c CTPT: 2 1 2n nC H O N+Khi t chy th

2 22 2H O CO An n n- =

Vi B th 2 2H O CO

n n= .

Do vi 0,25 mol ban u c: ( )2.0,675 2.0,6 0,15 mol- = cht A.Nn 0,2 mol X s c 0,12 mol cht AX do vy s phn ng va vi ( )0,12a mol= HCl.


Bi 43. _____________________________________________________________________________Cho hn hp gm 2 este mch h l ng phn ca nhau. Ly 0,2 mol X cho phn ng hon ton vi 300 ml dung dch NaOH 1M, ri tin hnh chng ct c 8,5 gam hn hp hai ancol ng ng k tip nhau. C cn phn dung dch cn li sau chng ct c cht rn A. Nung A trong oxi d n phn ng hon ton c 22 gam 2CO v 7,2 gam 2H O cng 1 lng 2 3Na CO . Cng thc phn t ca hai este l:

A. 2 3 2 5C H COOC H & 3 5 3C H COOCH

B. 2 5 2 5C H COOC H & 3 7 3C H COOCH

C. 2 3 3 7C H COOC H & 3 5 2 5C H COOC H

D. 3 5 3 7C H COOC H & 4 6 2 5C H COOC H

Li gii. bi cho mt gi thit quan trng l hai ancol ng ng k tip nhau, ta ngh ngay ti vic li dng iu ny tm ra cng thc hai ancol thng qua khi lng mol trung bnh. Tuyt vi hn na, ta cng

c khi lng v s mol hai ancol ny. Vy d dng suy ra 8,5

42,50,2ancol

M = =

Suy ra hai ancol l 3CH OH v 2 5C H OH , loi p n C, D

Ta c s ng cho

( )

3

2 5

(32) 3,51

42,53

46 10,5

CH OH

C H OH

=

T l hai ancol cng chnh l t l hai este nn:Suy ra

3 2 50,05 0,15esteRCOOCH esteR COOC Hn n = =

Ch cn hai p n A, B. V chng ta cn phn ng th 2 cha s dng. Ta quan st hai mui to ra t cc este ca A, B ch khc nhau s nguyn t H. Ta s da vo s H chn p n ng. phn ng (2) ta bo ton nguyn t H.

22 0,7H trong muoi H O NaOH dun n n= - =

Vi cc s mol, ta th hai p n A, B th ch thy A to ra c 0,7. Chn A.Bnh lun. Ta c th gii nhanh hn da vo cc p n A, B, C, D nh sau:

Ta c: 0,2 ( )

:: 0,1 ( )

RCOONa molA

NaOH mol

d

S mol H trong mui 2

2 0,7 ( )H O NaOH duRCOONa n n mol= - =

S H trung bnh trong 0,7

3,50,2

RCOONa = = . Vy loi cc p n B, D

Th p n C xem ng khng nh.

t: 2 3 3 7 3 5 2 5

0, 2; (*)

60 46 8,5C H COOC H C H COOC Hx y

n x yx y

n+ =

= = + =Gii h (*) cho ta nghim m. Vy chn p n A.

Bi 44. _____________________________________________________________________________Ly 15,66g amin n chc, mch h X ( X c khng qu 4 lin kt p ) trn vi 168 lt khng kh (ktc). Bt tia la in t chy hon ton X , hn hp sau phn ng c a v 0 C, 1atm ngng t ht hi nc th c th tch l 156,912 lt. Xc nh s ng phn ca X.

A.2 B.17 C.16 D.8Li gii


Gi amin l 2 3 2 2 2 2 2323 3 1( 4)

2 2 2n n kn k

C H N k O nCO n k H O N+ -- +

+ + - + +

32 53 1 15,66. 0, 495

02 2 14 17 2

nn kn

kn k

=- + - - =

=+ -

5 13C H N

Cc ng phn lAmin bc 1: 5 11 2C H NH- c 8 ng phn (ging ancol)

Amin bc 2: 3 4 9CH NH C H- - c 1.4 4= ng phn (do 3CH c 1 ng phn v 4 9C H c 4 ng phn) ;

2 5 3 7C H NH C H- - c 1.2 2= ng phn

Amin bc 3: 3 2 3 7( )CH N C H- c 1.1.2 2= ng phn ; 3 2 5 2( )CH N C H- c 1.1.1 1= ng phn

Chn p n B.

Bi 45. _____________________________________________________________________________Oxi ha 38 gam hn hp propanal, ancol A no n chc bc 1 v este B (to bi mt axit ng ng ca axit acrylic v ancol A) c hn hp X gm axit v este. Mt khc, cho lng X phn ng vi 0,5 lt dung dch NaOH 1,5M th sau phn ng trung ha ht NaOH d cn 0,15 mol HCl c dung dch D. C cn D

c hi cht hu c E, cn li 62,775 gam hn hp mui. Cho E tch nc 140o C ( 2 4H SO c xc tc)

c F c t khi vi E l 1,61. A v B ln lt l:A. 2 5C H OH v 23 55C H COOC H C. 3CH OH v 4 7 3C H COOCH

B. 3CH OH v 3 5 3C H COOCH D. 2 5C H OH v 4 7 2 5C H COOC H

Li gii.

Ta nhn thy E chnh l ancol to este B. t CTPT ca E l ROH. Do 1,61 1FE

MM

= > nn F l ete, do

( )2 52 16

1,61 2917

RR C H

R+

= = -+

. Vy E hay ancol A chnh l 2 5C H OH . Loi p n B v C

Gi x, y, z ln lt l s mol ca 2 5 ,C H CHO 2 5 ,C H OH 2 1 2 5m mC H COOC H-(Lu , do axit to thnh este l ng ng ca axit acrylic nn 3m )Ta c 58 46 (72 14 ) 38x y z m+ + + =

Oxi ha hn hp s to ra hn hp X gm x mol 2 5 ,C H COOH y mol 3 ,CH COOH z mol este

S mol NaOH phn ng vi hn hp sn phm: 0,5.1,5 0,15 0,6( ) 0,6NaOHn mol x y z= - = + + = .

C cn D s to ra x mol 2 5 ,C H COONa y mol 3 ,CH COONa z mol 2 1m mC H COONa- v 0,15 mol

NaCl. Suy ra 96 82 (66 14 ) 0,15.78,5 64,775x y z m+ + + + =Nh vy ta c h 3 phng trnh:

38 5858 46 (72 14 ) 38 (1) (1')14

0,6 (2) 38 (2 ')

96 82 (66 14 ) 0,15.5

x 46 72zz

x 36 6z 16

08,5 62,775 (3) (3'),6

my

y

x y z

x y z m

x y z

x y z m

- -

+ - =+ + =

- =+ + + = + + =

+ + + + =

T (2) v (3) suy ra 2 17

36( ) 16 6 38( ) 36(0,6 ) 16 6 38(0,6 ) (*)15 110

x y z x y z z z z+ < + < + - < + < -

Tuyển tập một số bài toán hóa học hay và khó

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