chuyÊn ĐỀ bỒi dƯỠng hÓa hỌc 11 - nguyỄn ĐÌnh ĐỘ

8/9/2019 CHUYN BI DNG HA HC 11 - NGUYN NH

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NGUYN NH

CAO HC HA HC

CHUYM D BI DII

HA HC 11Gm: 37 CH PHNG PHP GII TON BI TP TNG CH

BI TP NNG CAO 228 BI TON CHN LC

Nh xut bn Nng

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ng gp PDF bi GV. Nguyn Thanh T


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CHUYN BI DNG

H0HC11



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CHUYN

S IN LI UNG DCH*

CH81 CCH VIT PHUNG TRNH IN LI

S in i ca mt cht c biu din bng phng tr in i

- Phg inh in i ca axt X l :

y i r ' = n* +x?-V d HJ304 = 2H*+ SO f - Phvtmg trnh in i cabaz M(H)n l

\ M(oma = Mat +nOH' V d: B(0H)2 = Ba2*--20H'

- Phng trnh in i ca mu MIKm

- ' M +mX?~

V d: Al S O / = 2AP* +3SOf * Ch - Vi cht in i yn th du "c thay bng " =

V d : CHgCOOH = CH3C + H*I - Cc a axit in li theo tng nc, V d H$S04 ta c :

Nc : H04 - H++HS~.

Nc 2: ESOl - H++sof-.Tngqut: HpSO = 2H+ +S O

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3 P1 Vit phng irnh iu li ca

a) CaC2 b) A1(N03 ) 3

c) H3P04 (theo tng nc v tng qut) d) NaHS4

GIIa) o o I I C a 2+ + 2 C1-

b ) A1(N0 - Al3+ + N O

e) h p o - H + + t t j P O ;

H P 0 ; = H + + H P O -

H P O - H* + P O f

h p o = 3H*' + PO,"

d; N a H S O , =4 N a* +H++ s

CHI) B 2

D nh gi m vo in li a )

_ S' pi in a = ""

Nu : a > 0,3a < 0,03

0,03 < a < 0,3

D IN L ( a )

ih, yu ca cht in i A gi ta da

hn t A (hay s'mo A) in i )hn t A(ha mo A) bn u

/ A cht i mnh ' A cht in li yu """ A cht in l trung bmh

Bft] TP 2 T r o n g 0 ,5

t n g c n g 3 , 1 3 . 1G2 i n l c a C

C h o h n g s A v o

l t d u n g d c h C H C O O H , M c p h n t c h a p h n l i v io n . T n h

C O O H n n g .

adro N = 6 , 02 .1 02:.

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GI

S moi CHgCOOH ban u = 0,5 . 0,01 = 0,005mol Suy r s"phn t CHgCOOH ban u = 0,005.6,02.1023

= 3,01. O21Gi s d ;Ck phn t CHgCOOH din li tkeo phng trnh:

C3COOH = CHoCOO' + H+: c t k k

=> (3,01.1021 k) + k + k = 3,13 . 1021Rt ra k = 0,12 . 1021

Sphn t CH3COOH in liVy in li a = 7-------- -------------7 s phn t CHjCOOH ban u0 1 2 1 0 2 1

= ^ ^ 1 7 = 0,0398 hay 3,98%3,01.1021

BI TP3 Nng ion H+ rong dung dch CH3COOH

0>1M O,O013mol/l. Tnh in li ca CH3 CQOH nng .

GII

C 1 lt dung dch CHgCOOH c cha 0,1 mol CH3COOH, trong c 0,0013 mol CH3COOH in li theo phn ng :

CH3COOH = CH3COO* + H*0,0013 moi 0,0013 moi

=> in li a = 9P = 0,013 hay 1,3%0,1 J

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BI TAP4

a) Chng minh rang nu mt dung dch monoaxt yu c nng ban u l C(mo/), v hng s axt K th in lia ca axit nng trn c

tnh theo cng thc

b) T cngthc trn cho bit trong dung dch cng long, s phn li ca cht in li yn cng t hay cng gim ?

GIIa) Xt ung- dch monoaxit yu HA c nng ban

C(znoi/I},hng s axit l K v in li l a, ta c phtrinh in li :

HA > H+ + /ALc du c 0 0PMnli ca c aLc cn bng (C - (a) c* ca

[H+]m** [HA].

K

CK.Ca' c - c a

K

C(1 - ctK

V HA .l cit in li yu nn a io (i-ct)~

_ c2a2 rr Rt ra :

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c c ^ j v h a y a W

b) Vi dung chcng long th c cng nh, m a v c t l nghch vi nhau nn a cng ln. Vy dung dch cng long

s phn E ca cht in li yu cng mnhBI P 5 Bit axit hpoclor HCO c hng s axit K =

5.10'8. Tnh in l ct ca axit trn trng dung dch axit hipoclor0 ,1 M.

GII

in li a

= V . P = 7.10-4 hay 0,07%

BI TPs Cho hng s axt ca CH3COOH K = l^.io*5. Tnh in li a ca CHgCOOH trong dung dch CH3COH 0,4M.

GII

inli a = v c 0.4= V4 5 . 10-6- = 6,7.10- hay 0,67%

\ _________________

CH3TNH PH CA MT DUNG DCH\

- Vit phng trinh in i tnh c nng ion H* trong dung dch

- p dng cng thc pH = - g[H+

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BI P 7 Cho mt dung dch A ca 2 axit trong nc : H2S04 0,3M v HCI 0,4M. Tnh pH ca dung dch A.

GII

Trong 1 lt dung dch A c 0,3 mol H2S04 v 0,4 mol HC1.Cc phng trinh in li :

H2S04 = 2H+ . + S0 24-0;3 moi 0,6 molHC1 H+ + Ci-0,4 moi 0,4 moi

=> Trong mt lt dung dch A c 0.6 + 0,4-1 moi H+=> [H+] = 1(mol/1)=> pH = -g[H+J= -Igl = 0

81TP3 Ha tan 4 gamNaOH rn vo nc c 1 0 lt dng chA. Tnh pH.ca dung dch A thu .

GIISmoi NaOH = 4 : 40 = 0,1 moiPhng trnhin li :

NaOH = Na+ + OH:0,1 moi 0,1 moi

Vy [OH'j= ' = 0,011 0

do pOH = -lg[OH'j = -IglO'2 = 2Suy ra pH = 14 - pOH - 12* Ch : Vi dung dch nc ca cc cht ta lun c

pH + pOH = 14

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BI TP9 Pha trn 40cm3 nc v 10cm3 dung dch HCI c pH = 2. Tnh pH ca dung dch thu dc.

GII

Dung dch HC1 c pH = 2 -lgfH+j = 2 [H+] - 1Q'2(mol/1)

Suy ra trong 10cm3 hay 0,01 lit dung dich HC1 trn c cha0,0.10*2 = 10*4 moi H+:

ung dch sau pha long c th tch = 40 + 10 = 50cm3 hay0,051 .

10-4 Nng' H+ trong dung dch sau pha long =3- 2.10'3Vy pH = -lg[H+J = *1 g 2.10-3 = 2,7

BI TP10 Ha tan3g axit axetic vo nc c 0,5 lt dung dcha) Tm nng mo/I ca ion H trng dung dch,

sp ra in l a ca axit axetic/ Cho rng s a x it -----:c CH3COOH l1 ,8 .1 0 '5.

b) Thm vo dung dch trn 0,3 moi CH3COONa. Tnh pH ca dung dch thu c. Gi thit CH3COONa l cht in li hon ton c th tch dung dch khng thay i*

GII

a) S' mol CHoCOOH dng = = 0,05 3 60

Nng dung dich CH3COOH = - = 0-.1M0,5Xt phng trnh in l ca axit axetic

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Ul3(JUUii CHgCOO- + H+Lc u 0,IM 0 0Phn li xM xM xMLc cn bng- (0,1 -x)M : xM xM

M[CH3C00-]. [H+1 :

[CH3C0CH] "1,8k )-5

Nn X2 1.8.10'50 , 1 - X

gii phng.trnh trn, ta gi s X < 0,1, tc - 0,1.. Kh'y rt ra X2 - 1,8.lG"6.hay X= 1,34,10"3

R rng 1,34.10"3 0 ,1 .nn gi tri X gii trn chnin c.

Vy [H+] = X= 1,34.1 O'3 = mol/lTrong 1 lt dung1dch axit axetic ni trn c cia 0,1

CH3COOH, nrng ch c l^ . io*3 mol CHgCOOH . in li n1,34.10" 3 in li a as ----- = 1,34. O' 2 hay 1,34%0,1

. ,* GIl : Cng c th tnh in li atheo cng thc

- V I " - V o f 1 =

b) Khi thm vo dung dch trn 0,3 mo CHgCOONa th* mt dung dch mi, trong nng CH3CH vn

v nng CH3COONa = = Q,6M.' 0,5CHgCOONa l cht in i hon ton nn :

CHgCOONa = CH3COO- 4- Na+. 0,6M 0,6M ,

Xt phng trnh in i ca CH3COOH :

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V-zXlgV/V/ KSJ.J. -----Lc u 0 ,1 M 0,6M 0Phn li xM xM xM Lc cn bng (0,1 - x)M ' (0,6 + x)M xM

(0 , 6 + x)x . -=> - =1 ,8 .1 0 * 50,1 - X gii phng- trnh trn ta gi sX 0,1, fcc (0,1 - x)

0,1. Khi y rt ra X2 + 0,6x - 1,8.106 = 0Phng trnh trn cho nghim otogX = 3.10' 6 0,1 nn

gi tr X gii chp nhn dc.=> pH = -lg[H+] = -lg3.10' 6 = 5,56

CHB A TN VTAN

tan (a) ca mt cht ( mt nhit xc nh) l gam ti a ca cht tan c trong 100 gam nc t thnh dung dch b ha.

Biu thc quan h gia nng phn trm ca ung d

bo ha cht tan X v tan (a) ca cht X : c% = ~ ~ ~100-ha

* Ch : Mt dung dch mui bo ha nhit t n h xung nhit t2 th s xut hin mt ng mui tinh, trng phn dung dch cn i l dung dch bo ha nhit t2

BI TP2 2 tan ca NaCl 20cl 36g

a) Tnh nng phn trm dng dch NaCI bo ha nhit ni trn.

b) Tnh ng kt ta NaCl thu ckhi m lnh 1360 gam dung dch NaCl bo ha 20cxung cn 10c.Cho tan ca NaCl 10c l35,8g.

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GII

a) Nng dung dich NaCl bo ha = -ft-' = 26,47 (%)100 + 36

b) C 136g dug- dch NaC bo ha 20c c 36 gam NaCl, do 1360g dung dch ny c cha 360 gam NaClGi s thu c Xgam NaCl kt ta khi h nhit t 20c

xung 10c, nh vy phn dung dch cn li c khi lng l(1360 - x) gam v c cha (360 - x) gam NaCl tan trong .y l dung ch bo ha 10c. Suy ra :

C 135,8g dung dch NaCl bo ha 10c c cha 35,3 gam NaCl

(1360 - x)g dung dch NaCl .bo ha 10c c cha (360 -x) gam NaCl

Rt ra : (1360 - x)35,8 = 135,8 (360 - x)X = 2

Vy d c 2 gam NaCl kt ta.

EIP12 C 29 gam dung dch Fe(N Q)g 41,72%, Lmlnh dung dch thy thot ra 8,08g mui rn. Lc tch mui rn thy nng Fe(N03 ) 3 trong nc lc l

I 34,7%I Tn cng thc mui rn thot ra trn, .

GII

S gam Fe(N03)3 trong ung dch u = ~ ~ ~ ~ ~ - 12,1 g

Id'ic: rn tch ra mu ngm nc. Gi s c XmolFe(NOo)s tch ra di dng mui ngm nc theo phng trnh:

Fe(NO*)5 + nH20 = FeN03) . nH20 X X

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Nh vy phn nc lc c khi-lng'l (29 - 8,08) = 20,92g, v c cha (12,1 - 242x) gam Fe(N03)3' tan trong . Suy ra tac h :

*(242 + ln) = 3,08'(12,1 - 242x)100= 34,720,92

Gii ra cX as ,02 v n = 9 Vy mui tch ra l Fe(NOs)g . 9H20

BI TP13 C nhn xt g v tan ca cc cht di y da vo th biu din tan ca cung

tah

Ctt-So

yM T~*~----------------r : *-* ** X --- : -to 40 6 400 ic 0 O/ 40 got fooizo Vc

GII

Khi tng nhit t0cn 100c, tan ca KN03 v NH4CI u tng, tan ca KN03 tng mnh khi nhit tng.

Tri li, tan ca CaS04 tng ch trong gii hn t0c n 40c. Trn 40c, tan ca CaS04 gim dn.



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_________ - ~ Y * v n M.C.. w\w ury iKOrvcjMUl

DNGDCH

X T S TN G TC GIA DNG gii c dng bi-tp ny, cn bit r tnh tan

cc ch% s tn ti ca cc cht, cng nh kh

hy cc cht trong ung dch.V d:+' Khng tn. ti ion Ba2+v SO ~trong dung dch v Ba

mui kt ta.+ Khng ton ti cht Hg(OE)2 vi Hg(OH)2 km b

hy theo phn ng Eg(OH)2 ~ Hg + H20+ Khng tn ti dung dch AI$33 v mui ny b th

mnh theo phng trnh : Ifi3 + 6H20 - 2A(OH)3 + 3H

Bng sau y gip ta nm r dc cc vn nu

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B n g s a u

y g i

p t n m

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bv o is

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ocu

2-ca._ HH 1117

: K h

n g

l a n

____________________

_____________

:

t l a n

___

:

K h n g

l n

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BI P H C 3 dung dch, mi dung dch cha 2 ion Ckhng trng lp nhau) trong s6 ion sau : Na+, Mg2%P b h N O, CO y' , S O ^ 'Cho bit cc ion cha trong mi dung dch.

GII

HNG DN : Mui ch rt d kt ta, ch c Pb(NO)2 ltan trong nc. Vy phi c dung dch Pb(N0,)9

Mui MgC03 kt ta, cn MgS04 l mui tan nn phi cdung dch MgS04.

VY : Cc ion trong ba dung dch cho lDung dch 1 : Pb2+, NO;Dung dch 2 : Mg2*, SO]?~

Dung ch 3 : Na+, CO3"

B1P15 C 5 dung d ch : Na2C03, FeCl3, NaOH, A12(S04 ) 3 v AgNO^. Vit cc phn ng xy ra (nu c)

khi ln lt cho mt dung dch ny phn ng vi cc dung dch cn, li. ______ _____ ____________________ I

GII

Kt qu th nghim c cho bi bng sau : Na2CO, FCI3 NaOH A12(SG4;3 AgN03

Na2COv (1) - . (2) (3)FCI3 4) - (5)

NaOH (6);(7) (8)A17{S04)3 (9)AgNOa



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Cc phn ng xy ra : 1): 2FeCl3 + SNajCOj + 3f20 = 2Fe(OH)3/ +3C02 + 6 NaC 2):A12(S04) 3 + 3Na2C03 + 3H20 = 2A1(0H)3 >u 3C02 +3Na2S04

(3): 2AgN0g + NagCOg = 2NaN03 1 Ag20 i +C02

(4): FeClg + 3NaOH = FeiOHJjj I +3NaCl(5): FeCls + 3AgN03 = 3AgCI + Fe{N0 3>3

(6): A y s o ^ + 6NaOH = 2A1(0H)3 + SNagSO*(7): A1(H)3 + NaOH =NaAI02 + 2H20

(xy ra nu NaOH dng d). (8 ): 2AN03 -h 2NaOH =2NaN03 +Ag20 i + H20

(9): s0 4)z + 6AgN03 = 3Ag2S04 i + 2AI(N03 ) 3 Lu (1) l tng hp ca 2 phn ng :

2 FeCI3 + SNagCOg - Fe2(C0 3) 3 +6 NaCl

Fe(C03 ) 3 + 3H20 = 2 Fe(OH)3 h3C02

(2) l tng hp ca 2 phn ngA12(S04) 3 + NagCOg = A12(C03 ) 3 + 3Na2S04A12(C03)3 + 3H20 = 2A(OH)3 +3C02

(3) l tng hp ca 3 pMn ng :2AgN03 + NagCOg = Ag2C03 + 2NaNOsAg2C03 + H20 = 2AgOH*+ C02

2gOH = Ag20 + HgO (8) l tng hp ca 2 phn ng :

AgN03 + NaOH = AgOH + NaNOg2AgOH ... = Ag20 + H20

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mt nhit khng i v trong mt dung mi I tch nng' vi y thsi thch hp cc ion ca mt I tan trng dung ch Mo ha mui mt gi t j nh v bng tch -s tan T, .I V d: TCj= l,56.10-w. . iu ny. c ngha trong dung dch AgG ho ha t C-) = ,56.0 -. Nh vy xt nng, thc s ca cc ion Ag* vmt dung dch kho st, nu ta c :

[g- [Cl < 1,56.1 10: Dung dch AgC cha ho hch thy kt t AgC xut b

[Ag*] [C'] ~ ,56 .l : Ding dch AgC bo ha - [fJ [C '] > L .l 10: Dung dch AgC qu bo ha n )

xut kin, kt ta AgC.! Tm li iu ki:i c kt ta AgCL xt hinc [Ag+l ic-] > ,56.10O Da v qu i tc ny ta. c th tn hton d on xut hin kt ta ca cc cht.

Vic sc si tch s' tan gia h ai mu i cn c th gip ta bit c mui no d kt t

Mui t tan no c tch s tan cng -nh th cngta V d TAsC =1 ,56 .1010 v TAt7 l 16 ch thy mu

Ag d kt tua hn mui gCL

I B1TP16 Trh 2ml ung dch HC1 0,01M vi 8ml dcli AgNOg 0,0125M. Don xem c Xttt hin kta khng ? Cho T ^ , = 1,56.1050

I ________________ x _______________ _ ________

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GIIDung dch sau. pha trn c

V = 2 +8 = 10n =iH S mol HC1 baa u =2 .1 0 -3 .0 , 0 1 = 2 .1 0 ' 5

S' mol AgNOjban u = 8.10"5.0,0125 = 10'42 10-5Nng HC1 sau pha trn = ~ - 2 .1 0 '3M10

X0-4 '10

Cc phn ng iD li :

XuNng gNOg sau pha trn = ~ - 10'2M

HC1 H+ + Cl-2 .1 0 -3M 2 .KrMAgNQs - Ag+ + NO;1 0 -%! O^M

Nh vy [Ag+][C1'J = 2.10*3.10-2 = 2.10' 5 > TAgC = 1,56-10 do d c kt ta xut hin.

-10

8I TP17 Cho TAgCI =1,56.10*10 = T.(

T ^ I O - ^ T *Mt dung dch mui nari c [Cr] =1 0 ' 1 v [] = 10~ 3

! Dng lng thch hp dung dch AgNOg tc dng vi dung dch trn.

a) Kt ta no s xut hin trurc ? Gii thchb) nh gi khnng tch hon ton ion ' ra khi

ion Cl" t dung dch trn. Bit rng khi nng mt ion t tr xung th c th coiion c tch ht.

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GII

Cc phnng to kt tua : g+ + Cl' = AgCl Ag* + r = Agl

Theo qui tc v tch s tan thi- AgCl kt ta ta phi c [Ag*] Cl'] >1,56.10'10

156 1 0 _lRt ra [Ag+j > ~ = 1,56.lO'3]^& 1 0 - 1

- Agl k ta t =10"3

Nh vy Ag kt ta th nng ion Ag* b hn nhiuso vi vic d AgCl kt ta, tc g d .kt ta hn gC.

b) Theo trn th Ag kt ta trc, v iu kin1 Agl kt

ta l [Ag ] > = 103M[IITrong- khi bt u kt ta thi nng-d ion I" trong-dng'

dich. gim dn, khin [g+] tng dn ln v n.mt lc noT,

nu [Ag+ > - l = 1,56.1Q'3M th-AgCl bt u kt ta. Nh [C1 ]vy thi im AgC bt u kt ta th :

T7'S>' in =IC1-]

Do [IJ < 10'6 nn c th coi tch c ht I* ra khi dung-dch.



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CH7PH T RN C C D UN G D C H c NN G P H A N

TRM KHC NHAU (CHAT TAN GION G NHAU)

Loi toi ny c cch gii hah theo phng php ng cho ' Gi mx v C ln lut khi ng v nong phn trm

ca dngdch Gi v C2 n t kh ng v nng' phn trm

ca dung dch I Khi trn dung dch vi dung dch c dung dch

mi c nong- phn trm c, ta p ng cho sauCr C2-C

>X X c2 !Cr CI

(Ly tr tuyt i cc hiu trn . c s dng)

Khi y ta s c : * =\C1~C\

\Cl ~C\

BI ? 18 Cn phi pha bao nhiu gam dung dch mui n nng 20% vo 40Qg dung dch mui n nng 15% c dung dch mui n c nng 16%.

GII

p ng phng php ng- cho ;20 1\ /

16 / \15 4

^ _ 1tnv 4

haym,400



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23/177

Rutramj = 1 0 0 gam Vy phi dng' Ogam dung dch mui n nng- 2

BfP19Gn thm bao nhiu nc vo 600 gm dundch NaQH 18% c dung dch NaOH 15%

'GII

Xem nc l- dung dch NaH 0%, p ng phngng cho ta c :

0 315

/ \18 15

h m i _ 1m2 ~ 15 ~~5 600 ~ 5

Rt ra m, = 120Vy phi dng- 120 gam nc

TP20 Cn ha tan thm bo nhiu gam st rn vo

800 gam dung ddb. NaOH1 0

% c dung dch NaOH 2 0 %

GII

Xem st rn l dung dch NaOH 00#, p dng phngng cho :

' 100 . 10

2010 80

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10 1 h 1^ m2 = 80 =8 . ay 800 ~8

Rt ra m. = 1 0 0 g Vy phi dng- 100 ga NaOH rn.

CH B 8 QUN H NNG PH A Nt r m

V NNG M OL/i

chuyn i t nng phn trm qua nng- mo (hay ngc ) nht thit phi bit khi ng ring' c dng dch

Nhc li cng thc tnh khi lng ring d (g/m)

. mdd d = vvdd

Ngoi ra c th s dng cng thc quan h gia 2 'loi nng :

c - 7c* * M - *

B TP21 Ha tan 2,3 gam natri kim loi vo 197,8 gam nca) Tnh nng %dung dch NaOH thucb) Tnh nng mo/l dung dch trn, cho, khi lng

ring dung dch l d = l,08g/ml.

GII

a) S mol Na d dng = 2,3 : 23 = 0,1 Ta c phn ng: :

25



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25/177

Na + Ho0 NaOH +

0,1 mo 0,1 mo 0,05 molKhi lng- dung dch thu c =2,3 -f 197,8 - 0,05.2 = 200g-

E1.J 900 b) Th tch dung dch thu c = = = 185ml

BI P 22

*) Khng dng cng thc, tnh nng mol/1 ca dung dch H2S04 19,6% (d = l,25g/ml)

b) Th li cng thcCy. = . c%1 M

GIIa) Xt 1 t hay 100ml dung dch H2S04 19,6% d cho

Khi lng 1 lt dung dch ny = 1000 . 1,25 = 1250g _ 19 6Khi lng H.?SO, cha trong 1 iit dung dch - 120 .

= 245g

2453 mo H2S04 cha trong- 1 lt dung dch = * = 2,5 mo98.3 uy ra CVi = 2s5iM

1,08

Vy Cy = = 0,54MM 0,185

b) CM=

o c-



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26/177

CH D 9: PH TRN C C DNG D CH C NN G

M OL /L KHC NH (CH AT TAN G VG NHAU)Tng t nh nng- phn trm t cng c phng php

ng cho khi pha trn dung dch (th tch V] ; nng Cx vi dung dch (th tch V2 nng C2) nb sau

C C2-C

c2 lCr C (Ly tr tuyt i cc hiu trn l s dng)

v; \c2~c\ Khi y ta s c

BI TP23 V n dng bao nhiu ml dung dch HC1 2M phatrn vi 500ml dung dch HC1 IM c dung dch c nng 1,2M.

GIIp dng phng" php ng cho :........

2 0,2\ /

1 0 8

X = 0 1 _ 1 j Z l =

* v2 0,8 4 m 500 4Rt ra Vj = 12ml

Vy phi^^ng 125ml dung dch HC1 2M ____________

SI TPk Cn dng bao nhiu ml dung dch H2S 04 2t5M pha vi bao nhiu ml dung dch H2SQ4 IM c 600m dung dch H2S04 1,5M.



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27/177

GIAI

p dng phng php dng cho : '2 ,5 0,5

_ 1,5 1 ^ N 1

! m v2= 1 = 2

M Vj + V2 = Nn d dng rt- ra V1= 20Gml v V2 =40GznlVy phi dng 200ml dung ch H2S04 2 }5Vv 4G0mi un

dch IM.-

! CH B 10 PHA T RN H AI DNG CH c

I RING KH C NHAU (CHA T TN G I Ta cng c phng- php ng cho cho loi ton n pha dung dch (th tch V ; khi, lng ring d) vi d I ch (th tch V2 ; khi ng ring d2

' d, ld2-dl d

d2 fdr d!I (Ly tr tuyt i cc hiu trn s' dng) _ V. \ - d \

Khi y ta c = 7 7 J 2 K !.A-rf

BI ?P 25 Cn pha bao nhiu ml dng dch NaOH (d l,26g/ml) vi bao nu mi duiig dch NaOH (d l,06g/ml) c dung dch NaOH c d =1.16g/ml .

28



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28/177

u m i

p dng phng php ng cho1,26 0,1

1,16 1,06 0,1

V, 01v 2 0,1

= 1tc Vj - V2

Vy phi pha cc dung- dch trn theo t th tch 1:1

81 TP26 Khi lng ring ca mt dung dch thay i ra sao khi ta p ha long ung dch ny bng nc ? Xem trc c khi lngring d = Xg/m.

.Xt mt ung dch, bt k, trc khi pha ong- n c khml

hxng ring dj =r r 1

Gi s ta pha long n bngX m ncV X m ncn n g X g a m nn d u n g dch p h a long c

m t + Xkhi lngring l d2 m2V2 =

GII

m. m,+X Nh v dj - d2

V, V, + X

Ij (Vj > x) ~ V j (n i j4- x )

V j . ( V j + X )

x(m, - v t)V,(Vt +x)/

29



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29/177

x(m.-V)o : d, - dy > 0 ~ ------ >0

i i V \ T VI(V1+ X)

xfmj - Vj) > 0 v mu s un dng)m1- Vj > 0 ( vX lun ng) . 1V,dj > 1 %

Nh vy vi mt dung dch c khi lng ring d > Ig/mthi dj - 9 > 0, hay d > d2> ngha l s nha ong bng nc

s lm khi lng ring dung dch gim.Chng minh tng- t ta c :d1- d2< 0 dj < 1

Nh vy vi mt dung dch c khi lng ring' d < 1 g/mlth dj - < 0 hay dj < 2, ngh l s pha long bng ncs l m khi lng ring dung dch tng.

LYN TPU Vit phng trnh in li ca

a) (NH4)2 S04 b) CH3COOH

c) K,Cr,07 ) HCO

L2 _ a) Vit cng thc cu to ca axit photphoror H3P03,

bit y l mt axit.b) Vit phng trnh in li ca Na^HPOj

30



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30/177

HNG DNa) H - 0 0

H - 0 Hb) Na2HP03 = 2Na+ + HPO|-

L3 Trn Sn dung ch NaCl2 M vi 150ml dung ch BaCl2 0,4M. Tinh nng mol/1 cc ion trong dung dich thu c

HNG DN

S mol NaCl = 0,05 . 2 = 0,1' S moi BaCl2 = 0,4 . 0,15 = 0,06

Cn c cc. phn ng in li thy c s" mol cc iontrong- dung dch sau pha trn l

Na* = 0,1

Do : c

nBa2-..= 0,06

ncr = 0,1 + 0,06.2 = 0,22

M0,2

0,06 0,2

0,22

'Ba

0,2

= 0,5M

= 0,3M

= 1 ,1M

U Tnh lng kt ta NaCl thuc khi m lnh 1370g dung ch NaCl bo ha 50cxung cn20c. cho tan ca NaC 50Cv20C ln lt l 37g v 36g

31



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31/177

HUNUDAV

50c, c lOOg nc ha tan ti a 37g-.NaC cho dnng' 137g. Suy ra dung.dch NaCi bo ha . 0c ch NaCL

Gi s klii lm lnh th c Xgam NaC kt ta. phn nc lc cn li nng- (1370 - x)g v cha (370 xg NaCtan trong . y dung dch bo ha 20c, Bgha.l :C 136g dung dch NaCl bo ha 20c cha36% NaCl

(1370 - x)g dug dch NCl bo ha 20c cha 370x)gSuy ra : 136 370 - x).= (1370 - x)36Vy thu c10g kt ta NaCl

U ^ Tnh lng bc nitrat kt ta c khi lm l2500g ung dch AgNOg 60c xung cn X0c. C tan ca AgNOg 60c v 1 0 c ln lt l525g v170g

HNG NL lun nh bi 1.4, ta c phng1trnh

- (2100 - x)280 . = {2500 - x)17x = 1420g

Vy c 1420 g AgNO: kt ta . .

K5 ^Ha tan 3,86g tinh th BaCl2 . nH20 vo nc 300ml dung dch Bal2 0,O5M. Tm cng the mu dng.

HNG DNS moi BaCl2trong dung dch = 0,3 . 0,05 = 0,015Suy ra 5mo 3aCL2. nH20 d dng = 0,015

32



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32/177

Phn t lng BaCl2. nH20 = = 2440,015208 + 18n = 244n = 2Mui trn. l BaCl2. 2H20

[J Xc nh lng tinh th MgS04 . 6H20 .tch ra khi dung dch khi lm lnh lS42g dung dch MgS04 bo ha t 80c xung cn20c. Cho tan ca MgS04 80cv 20c ln lt l 64,2g v 44,5g.

HNG DN

1642 642

Lng MgS04 c trong dung dch u = ' = 642 g

Gi s c Xgam MgS04 . 6H20 tch ra khi dung dch.120x

Ch rng X gam mui ngm nc ny c cha TT1 =0,526x228

gam MgS04.'Nh vy (642 - x)g dung dch MgS04 bo ha 20c c

cha (642 - 0,526x) gam MgSOp cn 144,5g dung dch MgS04 bo ha 20c c cha 44,5 g MgS04Suy ra : (1642 - x)44,5 = (642 - 0,526x) 144,5

=3- X = 6 2 5 g

Vy c625g MgS04 . 6H20 tch r

18 un nng 1600g dung dch KA1(S04 ) 2 nng 2%

lm bay hi bt1 2Og nc, ri lm lnh dung dch cn li xung cn 20C th c bao nhiu gam tinh th KA1(S04 ) 2 . 1 2 HzO kt tinh ? Cho tan caKA1(S04)2 20c l 6 gam.

3-C... HH 11 33



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33/177

HNG DN

Lng K1(S04)2 c trong dung dc u = -PQ-- - S 32 g

Gi s c X gam tinh th KAI(S04) 2

. 12H20 kt tnh. Ch258x rng X jam tinh th ny c cha = 0,544x gam474

KA1(S04 ) 2Sau khi un, dung dch cn li nng 1600 - 1200 =QOg ri

lm lnii thc X gam phn kt tinh. Nh vy :(400 - x)g vug ch 1CAI(S04 )2 bao ha 20c cha (32 -

0,544x) gam KAI(S04)2, cn 106 gam dungdchKAlSO^ boha t trn c cha6 gam KAl(SO-4)2. Suy ra :

(400-x)6 * (32 - 0,544x)106X = 19,2g

1.9

a) Tnh pH ca dung dch CE3COOH 0,4M, bit hng s axit cda CHgCOOH l 1,8.105b) Tnh in li a cda CHgCbOH trong dungch

trn. Th li vi cng thc a = u

HNG DN

a) CHgCOOH CHgCO +. H+Trc phnli 0,4M 0 0Cn bng (0,4 - x)M xM xM

= 1?8.0-50.4 -X



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34/177

. Gi s X 0,4 thi(0,4 -x) - 0,4. Khi y ta c :X2 V l l -5 . 0,4 *7,2,10^X = 2,7.10'3 0,4 (nhn c),

pH = 4g 2,7.10'3 = ^ s a -fe)

s mol CHgCOOH phn li (xt trong1 lt dung dch)Ta ca = s moi CHgCOOH ban u (xt trong1 lt dung dch)

2,7. 10- 30,4 =*0,67%

/ k * *V 8 T l "5Nu p dungCng' thc th a = ---*

U0 Tnh pH ca dung dch CHgCOOH 0,5M

S : pH = -Ig 3.10*3 = 2,56

1.11 C 3 dung dch, mi dung dch cha 2 pation v 2 anion (khng trng lp) trong s cc ion sau : NH ;Na+ ; A g+ ; Ba2 ; Mg2+ ; A l3+ ; CI' ; B r ;NOj;C Of;SO- vPOf Cho bit cc ion trong mi dung dch.

HNG DN, ' " ...... *t '

Cc cp cation v anion phi khng to kt ta mi dngdch khi kt hp tng i mt vi nhau. Vy chng l 3 dung dch :

-Na+,NH ;;C02-,P f --Ar,Al3 +;NO;,.SOf - Ba2+; Mg2+; Cl\ Bt

3*



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35/177

u 2 Ch ra mnh sai a) S in l no cng to ra cc ion m v on [ b) Nari kim loi cng l cht in li v n tan j nc to ra dung dch NaOH dn in c c) Mt axit mnh cng l mt cht in li mn d) ng saccaroz khng phi l cht in,li

HNG DN

fe) l mnh sai v ki cho natri vo nc xyng ha hc to thnh dung dch NaOH nh sau :

2Na + 2H20 = 2NaOH +. %

M3 iu khng nh no sau y lun ng

a) ung dch mui axit nil t thit phi c mi trI a xt j b) Dung dch mui trung ha nht th i t c pH = c) Bng dch c pH < 7 nht thit lm qu tm

d) Dung ch c pH > 7 nht thi t lm phenolphtalha hng

e) Nc ct c pH = 7 ! ____________________ ___ ___ _ -________

HNG DN

Mnh d e) lun ng. Tht vy....a) Dung .dch mui NaHC03 c mi trng ba z

b) Dung- dch mui CH3COONa c mi trng bazc) Dung ch c pH < 5 mi lm qui tm ha d) Dung dch c pH > 8 mi lm phenolphtalein ha

36



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36/177

1.14 Trn0 ,n] dung dch H2S04 Oj0 0 2 M vi l,2 nal dung dch BaCl2 0,005M th c xut hin kt ta BaS04khng ? Cho tch s tan ca BaS04 l T = 10'10.

i. - - - ------------------------------------------------------3

GIIpnc

2 NaCl = 2Na + , 2 Na + 2HjO = 2NaOH + H22Fe + 3C2 . = 2FeCl3FeClj + 3NaOH = Fe(OH)3 v3NaCl



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63/177

SHD15 VIT CHUI PHN NG

I Cn nm vng tnh cht ha hc cc cht trong vit ng phng- trnh vhn ng. Chng hn cc cho c phn ng' vi sxit n'vi chui

C uSO .^ ...th phn ng ng

Cu(OH)2+H2SOA = CuSO-r2Kp' cn phxi ng : Cu(OH)2 +Na^SO = CaS04 +2NaOH

Ngoi ra phi bit chn ha cht thch 'hp ph xy ra c. V d vi chui

KC * KNOz ...

thi phn ng : KC + NN03 = KN03 + NaC (sai)I m phn ng ng :I KC +AgNOr = AgC + KN03

j 8I TP44 Vit cc phn ng biu din chui san

a) Zn ZnO*- ZnCl2 Zn(N03)2 Zn(OH)2 9*ZnS04

b) Fe FeCL, FeClg Pe(0f>3 Fe(N03)3- Fe20 3

a) 2Zn T- 0 2 = 2ZnOZnO + 2HC1 = _ ZnCl2 + H20ZnCio + 2AgNOg = Zn{N03)2 + 2AgCl *>Zii{N03)2 + 2NaOH = Zn(OH)2 + 2NaN03Mg- + ZiiS04 "= MgS4 +' Zn

b) Fs + 2HC1 . = FeCl2 ..+ H22FeCl2'+ Cl2 ' = . 2FeC3

GII

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64/177

r cv /ig *r M i a u i i wv - ------

Fe(OH), + HNO = Fe(N0 + 3H20t

4Fe{N03)3 2Pe20g + 2N02+ 302Fe20g +. 3H2 = 2Fe + 3H20

B TP45 Hon th n i chui bin ha sau :( ' - u ( sU L

_ +A H ! nc NH, da _ +HC1Cu ~ CuCl, ----- D CuCl,

0> ( I


65/177

GII

a) l kh NH3. Tht vyA -- nc * dung dich NH3

NH3 '+ HC = NH4Cl',NH4C1+ . NaOH = Na'GI+-NH3 + H20 NHg + HN03 = NH4NOg.

. . tNH4N03 > = N20 + 2H20

b) X, Y l NH


66/177

BI TP47 Ch s bin h

A - c - E - CaCO;

CaCOo < .)%/['+x V +Y - +z " . --Srp - Q ^ R ~ CaCOg

Xc nh A, B,c . . . bit chng l nhng cht khc nhau

Cc cht cn tm l

A : CaO B : H20 c : Ca(GH}2 : HC1 E : CaC2F:Na2C03

Tht vy :

GII

P:C02 X : Ba(OH)2 Q: Ba(HC03)2Y : KOH RrKCOgZ:Ca(NG3)2 .

CaCOa CaO + H20CaOH)2-t- 2HC1 = CaC2 + Na2CO = 2C02 + Ba(OH)2 =

Ba(HC03)2 + 2K0H K 2CO3 + Ca(N03) 2 =

CaO + C02Ca(OH)2 CaC2^ 2 l fi CaCOg + 2NaCl

Ba(HC03)2

BaCOg + i COg + 2H20 C a C 0 + K N O

67



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67/177

aHUiZn + Clj = ZnCV(2): ZnClj +2NH3 + 2H20 = Zn(OH)2 + 2NH4C1(3): ZnCLj +6 NH3 1 2&;0 = [Zn(NH3)4](OH), + 2NH4C1

(i ) : [Zd(NH3)4](OH)2 + 6HC1 = ZnClj + 4NH4Ci + 2,05): ZnCl2 + 2NaOH = Zn(OH)2 + 2NCl(6): Zn(OH)2 + 2NaOH- = Na2Zn02 +2H.fi(7): NajZnO, + 4HC1 = ZnCl2 + 2NaCl + 2H20

b) 2N H 3 + C02 = (NH2)2CO + H20 (NH2)2CO + 2H20 ; = (NH4)2C03(NH4)2G03 + 2HC1=' ' 2NHC11- C02 + H20( N H j2 C O , + 2 K O H = C C ^ + 2 N H , + 2 H j,0

59



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68/177

CH g 17CHI BIN HA DNG CC PHN NGCA GC CHT A, B, d . CHA R CNG TH

Vi dng ny; ta cn m ron's cc pin ng- cho mt phn ng c trng, tc phn ng gip ta xc nh

A (hoc B, c...) cht c th no, t d ng S iyra cng thc cc cht cn i.

3 P 49 Hon thnh s sau :

' / 7 4 - I s. *n + 0 2

+ o 2 +n 2o

B + HjO

r HNOa

GIIHNG DN

Trong 4 -phn ng nu thi phn ng cui gip ta xc nh c E l N02. Do D l NO ;c l NHS ; A, B l N2 v H2

Cc phn ng xy ra N 2 + 3 H2

p . t* . xt 2 N H 3

4NHg + 502 4 N 0 + 6 H 202 N O +0 2 ----- 2N02

4 N 0 2 + 0 2 + 2 H 20 ----- 4HN03

3TP5C Hon i&nh s sau ;A + B cC + B^H2

- D fh ~ X (mi trng thi)

69



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69/177

d +h 2oF + Ec + x

E Ag2S04+C + H20 A + H20

GIIHNG DAN

Trong" 6 phn ng nu th phn ng th 3 gip ta nhdc A l lu hunh. Do .c S02 ; D l S03 ; B l 0 2 ; El H2S04 ; F l A g v X l H2S

Cc phnng xy ra :

s + 0 2 = S 0 2

2 S 0 2 + 0 2 J L 2 S O 3

S + H2 a - H 2S

s o 3+ h 2o = E2S042Ag +2 H 2 S 0 4 = Ag-2 S 0 4 + S 0 2 + H 20S 0 2 + 2 H 2S = 3 S 4 - 2 H 20

51Tp 51 Hon thnh s .5t % H c

FeS2 + 02 A (kh) + Ba V o 2 c +T) E

Ag + E F + A + D A + D G G + KOH H+D

KMn04 + H + E + K + D



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70/177

HNG DAN

Trong s' cc phn ng nu th phn ng u tin cho

thy A l S02. Do c l S03 ; D H20 ; E la H2S04 ; F lAg2S04 ; G l H2S03 ; H .qSOg ; I l KjSO< ; K l MnSO* Cc phn ng xy ra :

4FeS2 + U 02 = 8S02 + 2Fe2Os

2SO, + 02 =^= 2 SO3S03 HO . = SO*

2Ag + 2H2S04 = Ag2S04 + S02 + 2H20S 0 2 + H 2 = H2 S 0 3

HjSOj + 2 KOH = K SOg + 2H20 2 K M n0 4 + K S O g + 3 H2 S 0 4 = 6 K S 04 + 2 M n S 04 + 3 H 20

s.

BI TP 5Z N 4 d t . . .v A + N'aOH X

Xp + H,

0 2

* - G + NaH

fib X

MD

GII H N G D N

Trong s cc phn ng trn th phn ng u tin cho thy A phi l mt mui clorua. Mui ny tc dng vi NaOH gii phng ra mt kh nn ch cth l mui amoni hoc mui ca



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71/177

amin hu c (s hc chng trnh 12). Tuy nhin numui ca amin hu cth s to ra B l amn hu c.Aminhu c khi. chy khng th to ra2 sn pLm nh phn nth 2 , do A phi l NH4C1

Bit c A,. d .g suy ra cc cht cn li theo cc.ng : -

t .NH4 G1 + NaOH ===== NH3 + H20 + NaCltc

4NH3 + 302 ===== 2N2 + 6H20x t , t , p

N2 + 3H2 = 2NH3NH3 +HCI- = NH4CI

N2 + 02 = 2N02 NO + 02 = 2N02

2N02 + 2NaOH = NaN03 + NaN


72/177

U T Cu(N )2, trnh by 3 phng php ha hc khc nhau iu ch ra CuO

J " 1 *

/ , . |V ^ ^ (J ( V ^

+ 4NOj+2 ^ sl

HNG DN* Phng php I

2 C u ( N 03 ) 2 = 2 C u O +4N02 2

* Phng php II / Zn + Cu(N03) 2 = Zn(N3 ) 2 + Cu

2 Cu + 02 = 2CO* Phng php III

Cu(NOs ) 2 + 2NaOH = Cu(OH) 2 1 + 2NaN03t

Cu{OH) 2 = CuO + H20

111.3 H l mt mui v c c s dng rt ph bin trong cng nghip v sng. S phn ng iu ch mui H nh sau :

'

l 1 a n u S 3 U * - ' r r \ ~ \

vo 1*. A + B (kh)C u V. ; --

G . _L H + B+H.0NH,Cl + GK'~iMH + B + H2 0

SNG DNMui H cho Na2C03

73



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U Vi cc ha ch t : H230 47 NaOH , NH,NOa , CaC03, Fe v FeS c th iu ch c bao nhiu kh ? Vit phcmg trnh phn ng

HNG DNC th iu ch c 10 kh sau : H2 , NH3 , C02 , H2S ,

S02 , 0 2 , N2 , NO , N02 , N20Cc phn ng :

Fe + H2S04 NH4N03 + NaOH

CaC03 + H2S04

FeS04 + H2 Na N03+ NH3+ H20.

H2S + FeS04

i 2NaNOs2H2S + 3024NH3 + 302

4NH3 + 5 02xt

2NaN02 + 0 22S02 + 2H202N2*. 6H20

4NO + 6H20

NH4N03,2 N 0 + 0 2 =

N20 + 2H202NO,

rnu Vit cc phn ng biu din chui sau :

a) N, NH3 NO N02HN0, AgNO, -Wig AgSO,

-

b) Fe(NO;i):i'3 FeClo

B D



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J u A( l

c)

A T

A1(N03)3

'

B ^ G

Al(OH)o

tll.6 B tc chui phn ng

B = = D + E (k)D + C = F

E + dung^dch trong nc ca F = B + G + E^oBit rngc l kh cn cho s chy. D l kim loi

chy vi ngn la mu tm

HNG DNA phi l mt mui kali. Nhit phn cho ra Og- nn l KG103.

Cc phn ng2 KCO3 = 2KC1 + 302

,KC1 = K + C124K +0 2 = - 2 K , 0 -V .

Cl2 + 2K0H - KC1 + KC10 + H20

111.7 Hy vit cc phn ng iu ch phn m S.A t cc nguyn liu : than, nc, khng kh v lu hunh

HNG DNPhn S,A c cng thc (NH4)2S04



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Cc phn ng': Chng-ct phn on khng- kh lng c N2 v 02

C + H20 . = CO + H2N2 + 3H2 ' ' = 2 NH3s + 0 2 = S022S02 + 02 2SO3 S03 + H20 = H2S042 NH3 + H2804 = (NH4)2S04 r

111.8 Trnh by cch iu ch btn NH4HC03 t cancacbonat, amonntrat v nc

HNG DN

CaC03 s CaO + C02CaO + H20 = Ca(OH)2

2NH4N03 + Ca(OH)2 = Ca(N03)2 + 2NHS+ 2H2

NH3 + C02 + H20 = NH4HC03

!---------------------------------;--------------------------------------- HU) Khi t nng mnh trong h quang in mI hp (X) gm th tch bng nhau ca hai k k mu thu c mt kh khng mu. Khi ny nh chng ha nu ngoi khng khi. Ha tan ht kh ny vo mic (vi s hin in ca kh oxi) j 2 0 0 g dung dch axit (Y)

I a) Vit cc phn ng xy raI b) Tnh nng phn trm dung dch (Y). Choban u c th tch l 8,96 t (kc)

76



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HNG DAM

Smol x = = 0,422,4Hai kh trong X phi N2 (0 , 2 mclj v 02 (0,2 moi)

Cc phn ng : Oo N 2 + . .0,2 moi 0>2mol2NO + 020,4 moi4N02 + Oo + 2H0 s 0,4 mo

2 NO 0,4 moi 2N02 0,4 mol 4HNO 0,4 mol

HNO. = =1 2 ,8 %200

U.10 Cho cc cht su y : N2 ; NO ; N02 ; NH4N02 ; HN02. Hy t cc cht trn vo cc thch hp trong S sau v vit cc phng trnh phn ng

(1 )A

L- t - J (7)-


77/177

B : NO D:N0 2Cc phn ng :

'AOX

(li Na + Os ----- 2NO(2 ) : 2NO + 02 = 2 NOz(3) 2N02 + H = h n o2 +.KNO3(4) h n o 2 + n h3 nh 4no2(5) : NH4N02 + NaOH = NaN02 + NH3 + H20(6 ) : 4NH3 + 502 = 4NO + 6H20

(7) n h4n o 2 ----- N2 + 2H20

(8 ) : ' N2 + 3H2 = 2 NH3

!!!.

11 C 2 ng nghim ng dung dch

AgNOj v AI(N0 3)3I a) Cho mt lng d nc amonic vo2 ng nghim,

thy mt ng c kt ta trng, mt ng c kt ta xut hin nhng li tan ngay. Gii thch v vit phng trnh phn ng. b) Thm vo ng nghim m kt ta xut hin li tan I mt t dung dch KI th thy kt ta xut hin tr I li. Vit phng trnh phn ng gii thch hin

tng trn.

HNG DNa) Ong A1(N03 ) 3 c kt ta xut hin :

A1(NO;)3 + 3 NH3 + 3H20 = l(OH)31 + 3NH4NOaOng AgN03 c kt ta xut hin nhng sau li tan :

78



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AgNOj + NHS+ H2Q = AgOH + NH4N03AgOH + 2NH3 = [Ag(NH3)2]OH (tan)

b) C s xut hin kt ta Agl theo phn ng

[Ag(NH3)2]OH +-KI * Agl i +-KOH +2 NHS* Ch Phc cht cng in li trong dung dch. S in li ca phc

ch 'dinra qua hai giai on. Giai on u gi giai on s cp, din ra hu nh hon ton tng t nL vi cht inli mnh. Chng hn xt phc eht [Ag(NH3)2OH, ta c giai

on phn li s cp :[A g(N H 3)2 ]OH = [A g ( N H 3)2]+ + OH*

Sau xy ra s phn i vi mc nh ion phc. y l giai on phn li th cp :

[Ag(NH3+ =Ag* + 2N3

Nh vy trong ung dch phc cht [Ag(NH3)2]OH c. s hin

din rt nh mt lng ion Ag+. kim, chng' iu ny, ngita cho mt dung dch mui clorua b't ki vo dung dch[Ag(NH3)2]OH th khng thy c kt ta xut hin. Tuy nhinnu thy ung ch clonia bng dung dch iodua th thy cxut hin kt ta Ag. iu ny c th gii thch c da votch s" tan ca AgC n hn nhiu so vi tch s" tan ca Agl{1,8.10-10 v 10'16), nh khi da vo dung -dch [Ag(NH3)2JOH

mt lng d ion Cl' Cling- khng t dc iu kin [Ag+1CI']> l,8.1Cr10, cn n dng mui iodua thi iu kin [Ag+l [I*] > 1 0 lfi xy ra.

79



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CHUYN 17

TON V AXIT NITRIC

Ni nh axit ohidric, axit sulfuric... ch tc dng vkim loi ng' trc hidro trong, dy din ha v ch giH.J, th axit nitric cn tc ng' c vi nhng kim o5au hidr.o trong dy itn ha tr (Au, Pt, ng thito ra nhiu khkhc nhau nli NO; NO.>7'N. N.,0 hoch co ra mt hn hp kh trong s' cc kh nu trn.

CH B 18 PHN NG GI PHNG MT KH - t a, b... l 50 mo cc kim oi dng - Vit phn ng xy ra, t s mo a., b,.. vo phn

Tnh s' mc cc cht in quan.- Lp h phng trnh. Gii h

1. - ____ I - ' - - . : ------ -------- ' - - --------- - - -

r ~---; T 7 7 ~ r ~ ^ 81p 53 Cho 605g hn hp C, A g, Au tc dng vi HNy m c, d. Sau phn ng thu c 0,896 N 09 (0c ; 2 atm) v l,97g cht rn. Tnh% khi lnmi kim loi trong hn hp-

GIIAu khng tc dng vi HNO^. Vy lng Au = l,97g v

Cu, Ag l (6,05 - 1,97 = 4,08gGi a, b l 50' mo Cu, Ag- ng. Ta c phn ngCu + 4HNO3 = Cu(NCL)., + 2NO, + 2H,0



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A g + a t i N . = A # N U . . r N U , +

b b

Suy rar:64a + lsb = 4,08n 1 PV 2.0,896 , rt2 a + b = = - 0,08

RT 0,082 . 273

Gii ra c a = 0,03 v b = 0,02197Vy 9c u = = 32,07c6,05 64.0,03%Cu = - = 31,79;

6,05% g = 35,8%

SI TP5fr Chia 2,2g hn hkhi lcmg mi kbn loi trong hn hp

GIIa) Gi a, b l s mol Fe v M mi phn

Cc phn ng phn 1Fe 4- 2HC = FeCl, + H2

a a2M + 2nHCl = 2MC1, + nH,, bn b --

2

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Cc phnng. phn 2 :Fe + 6HN0;j = Fe(N03)5 + 3N02 - 3H20

a 3a

M + 2n HN03 = M(N03)a + n N02 + n H20 b bn

2 26a T- bM = = 1,11 2

Suy ra :I a bn _ 0,896 : 22,4 = 0,04 23'a + bn - 2,0 16 : 22 ,4 = 0,09

Rt ra a = 0,0.1 ; bn = 0,06 ; bM - 0,54 _ bM 0,54 ,Do : = 9 tc M = 9n bn 0,06

*Xt bng n 1 2 3

M 9 18 27Ch c n - 3 : M - 27 l ph hp fAI)

n 56 . 0,01 b) %Fe = = 50,9%1,1

% Ai = 49,1#

B1 TP55 Ha tan hon ton 0?368g hn hp Ai, Zn cn va 25 lit dung ch HNOg 0,00IM. Sau phn ng thu c dung dch cha 3 mui. Tnh s gam mi

kim loai ban u.GII

Trong 3 mu(Ti to t h n h ch c chn ph c m t(N03)ov Zn(N03)2. Mi cn li ch c th l NH4NO;3

C-ia, b l s moi A, Zn dng. Ta c phn ng :



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82/177

8A1 + 30HNO,_=8 A](NOj) 3 + 3NH^N03 + 911,0

4Zn+ I O H N O3 4Zn(NOg}2-r N H4 N O2 + 3 H2O2

Gii ra c a = 0,0040. v b = 0,0040-Vy lng A - 27 . 0,004 s 0,108g

lng Zn = 65 . 0,004 - 0,26g

BI TP56 Ha tan honton 9,74g hn hp Cu v Ag (mi cht c thnh phn thay i t0 n1 0 0 %) bng HN03 long thu rc V t NO (kc)a) Tim khong xc nh ca V

b) Khi V = 1,121, xc nh khi lng mi kim loi trong hn hp.

I GII.........a) * Gi s. 9.74g hn hp trn ch cha C

S moi Cu = = 0,15364Phn ng khi y

3Cu + 8HNO3 = 3 Cu (NO 3 }2 + 2NO + 4H200,153 mol . 0,102 mi:=> V = 22,4 . 0,102 =2,2S{ .

* gi s 9,74g hn hp trn ch cha Ag

a 30a8

10b4

27a +-6 b = 0,368



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83/177

S* mol Ag = = 0,09* 108

Phn ng kh y :3 A g + 4 H N 0 3> 3A g-N O + N O .+ 2 H 20

0,09 ml 0,03 mol

=> V = 0,03 . 22,4 = 0,672^ Nh vy : 0,6721 V 2,28^

])) Gi a5 b l s moi Cu v Ag ng Ta c phn ng:

3Cu + 8HNO3 = 3Cu(N03)2 4- 2N0.+ 4H20

2

a 33Ag-+HNO = 3A gN 03 + NO +H20 -

b *' '3

64a + lOSb = 9,742 a b 1 , 1 2 *'

.+ = 77=0,05Suy-ra m = a(A + 186> +bB + 186)

92



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+ . 0 0 . ,

b) Theo, bi th :

0,148 ^ a + M 0,307Do vi biu thc m = aA + bB + 186(a + b) th

8,3 + 186. 0,148 < m ^ 8,3 + 186 . 0,307 38,52g H 1 68,12g

' c) V hn hp c thnh phn thay i t 0 n 100% nn :* NO = 3,32 => hn hp ch gm mt kim loi uy nh't

vi phn t lng nng hnGi s kim oi c phn t hxng nng hn l A, suy ra :

* ^NO 6,881 = hn hp ch gm mt kim oi duy nht vi phn t lng nh hom

=> B = 27( {) => Hn hp bn du cha st v nhm

B (ha tr 2). Ha tan 3 gam (X) vo dung ch c cha HN03 v H2S04 thu 1,344 lt (kc) hu hp (Y) gm N09v SO,, Bit t khi ca (Y) so vi hiro la 2 4 5

a) Tnh lng mui khan thu c.b) Nu khng bit t khi ca () so vi hidro th ch

xc inh c khong giihn ca lng mui khan trn. Tim khong gii hn ny.

ff.7 Hn hp(X) gm kim loi A (ba tr 1) v kim loi



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S : a) 7,06g; b)6,72g< ng-mui< 8,76 g

I IO_ Ha tan hon ton 9g nhm bng dung dch UNO, ' I thu c hn h


94/177

CHUYN B V

THIT LP CNG THC PHN Iff CHT

CHB20 THIT LP CNG THC PHN T DA VOPHNG T RNH PHN NG. (HOC DA VO

VC GI H PHNG TRiNH TON HC)

. Phng php, ny gii ch ht cc bi ton thit p cng thc phn t. Cc bc tin. hnh :- t cng thc tng qut ca cht hu c cho- Gi a s'moi ch hu c dng. Vit phn ng t s mo a vo phng trnh. Ti smol cc cht lin

quan.- Lp h phng trnh. Gii h(Ch : Nu bi ton khng- c phn ng, tt nhin khng'

cn t s mo cho cht hu c).

BI TP80 tchy hon ton 7,3g mt hp cht hu c thu c 13,2g C2 v 4,5g H20. Mt khc ha hi hon ton 29,2 gam cht hu . c trn thu c th tch hoi bngvi th tch ca 6,4 gam oxi (trong cng iu kin).Tm cng thc phn t cht huc trn.

_ _ _ _ _ v- c 1 0 ^

t chy cht hu c thu c C02 v H20 => cht hu cchcc, H v c th c 0

95



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Gi a s mo cht hu c ng, ta c phn ng

.+x+ - f>* - xC2 + 2 **2

a ax 2

Mt khc gi M l phn t'lng cht hu c 7 theo nh29 2 "6 4 lut Avogaro ta c : =,M 32

V, 32.29,2 Rt ra M = ~ 5S1466,4

Suy ra:

a(12x + y + 6 z) = 7,3 (1)

az = = 0,3(2)44 .

^ = 7 7 = 0,25(3)2 181 2x + y + 18z = 146 (4)

(1 ), (4).=* a = 0,t)5. Th no (2) => X=6(3) => y= 10

(4) => z = 4Vy CTPT cn tim l CgH]0O4

SI P 3 t ciy hon ton 23,2g hp cbt hu c thu c 24,84 lit C02 (kc); 9g H20 v1 0 ,6 g NajCQgTim CTPT ca (X), bit ch cha mt ngn t

oxi trong phn -t.

GII

t cng thc ca (X) l C^EyONa.Gi a l 5"mol (X) d dng, ta c phn ng

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96/177

4x y + z _ 2CxHy0Naz+ ( )02 (2x,~ z) C02 + yH20+ zNa2CO: A c cng thc nguyn CH)D=> 13n < 110 hay n < 8,4=> c th c CTPT l C22, CH p CfHfi) CgHg

: t chy hon tn hp cht hu c A (chac ,! H, O) bng lng oxi va nhn thy s moi C02I s moi H20 = s moi 0 2. Bit 40 < MA < 58. Tmi CTPT ca A.

1 - 0



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HNG DN

xC02 r * i )

a a(x + -^) ' ax1 . ay2

=> y = 2x v z =X

=> A c cng thc C Hgx x^ 40 < 30x < 58. Rt raX = 1=> A c CTPT l CH20

V.7 t chy ht l,8 8 g cht A chac , H, o cn 1,904 lt 0 2 (kc), thu c C02 v hi nc theo t l th tch tcmg ng l 4 : 3. Bit MA< 200 cLv.c, hy tm

CTPT ca A.HUNGDN

Xt phn ng bi V.6 , ta 'C: a{1 2 x + y + 16z) =1 , 8 8

2ax _ 4 ay = 3Gii ra c ax = 0,08 ; ay = 0,12-; a2 = 0,05

=> ax : ay raz = 0,08 : 0,12 : 0,05

Vy c cngthc nguyn (CgH^OgJjjV 188n < 200, r t ra n = i. Do d c CTPT l CgH120 5

X : y : z = 8 :12 :0



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y, Phn tch nh lng mt hp cht hu c A kt qu : %c = 40% ;% H = 6 ,6 6 %; %0 = 53,34%

Bit rng cha 2 nguyn t oxi trong phn t Tm CTPT cua A. _______________

S:C2H402

VJ) t chy 14,4g cht hu c A c 28,6g CO.?; 4,5 H20 v 5,3g Na9COy Tm CTPT ca A, bit A cha2nguyn t0X1 trong phn t.

S : C7H502iNa

IY.13 t chy hon ton mt lng cht hu c A cl ,6 g oxi v thu c 4,4g C02 ; l ,8 g H20. Tn CTPT ca A, bit Ma < 60.

BS : CH202

y.n Cho 400cm3 hn hp gm hrocacbon A v N2 vo 900cm3 oxi (d) ri t. Th tch hn hp sau ph ng l 1400cm3. Sau khi eho nc ngng t cn SOOcm3, sau cho qua dung dch KOH d ch cn 400c3n3. (Cc th tch o cng iu kin).

Tm CTPT ca A. Gi s trong iu kin phn n cho N2 cha tc dng c vi 0 2.

HNG DN

S bi ton 400cm3CxHy + N2J'

1900cm3 Oo

iC02H20 (hi -HnO

02 cn d1400cm3

co 20 2 cn d

N* ________ .800cm2

-co, 400cm3(N2 + 02 cn d



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112/177

= 1 4 0 0 - 8 0 0 = 6 0 0 c n rVco 3S800 - 400 = 400cm3

Gi s dng a cm3 CxHy v b cm3 N2, ta c phn ng

Gii ra c a= 200 ; b = 200 ;X = 2 ; y = 6 Vy A c CTPT l C2Hg

A2 t chy hon ton 3,61g cht hc (X) thu c hn hp kh ch gm C02, hi nc v HC. Dn hn

hp ny qua bnh cha dung dch AgNOg d (trong HN03) nhit thp thy c2,87g kt ta v bnh cha tng thm 2,17g. Kh thot ra di vo dung d ch nc vi trong d thy c 12g kt ta. Tm CTPT ca X, bit M < 200.

HNG DN

X chy c C02, H20, HCi= 3 %cha c, H, CI v c th c 0t cng thc X l v a l s mol X dng, tac phn ng

C*v(x +J)02 *CQ2+ 2h 20a.ax2

^ = 6002

113



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113/177

+ 0 .

a mol

xC02 +

ax

2 + ( )H20 + tHCl2

y - 1 a( ) at 2

AgN03 + HC . AgCI I HN03at at

3nh cha tng thm 2,17 g l do hp th H20 v HC1 nn

C02 + C(OH)2 s CaC03 -i + H20

Gii ra ' x : a y : az :at = 0,12: 0,18 : 0,08 : 0,02

hay : : y : z : t = 6 : 9 : 4 : 1=> X c cng thc nguyn (CgH904Cl)n= 180,5n < 200

=> n =1 .Vy X c CTPT l CfiH904Cl

V.13 Hp ^ht A cha cc nguyn t Gj H, o, N. cphn t lng l 89 .v.c. t chy hon ton 1moiA c 3 moi CO; ; hi nc v 0,5 moi N2. Tm CTPTca .



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115/177

CHUYN D VI

NS PHN - 0NB m

ng' phn l nhng- cht c cng cng thc phn tkhc nhau cng thc cu to. Chang1hn CH3 - 0 - CCH, - CH2 - OH hai ng phn ca nhau, vi ty c ccu to khc shall nhng- ging- nhau cng thc phn C2HgO

ng: dng nhng cht c cu to ha hc v t

ha hc tng fc nhau nlnEg- thnh phn plin t hnhau mt hay nhiu nhm CH2Khi ta tp hp cc cht cimh cht tng t nh vy

thnh mt dy, ta c mt dy ng ng cc cht hChng hn cc cht CH , C2HS, C3Hg... thuc dy nankan

Cn lu rng trong mt y ng ng, s ng p

nhanh khi 50 cacbon tng. Chng: hn trong dy ng.nkan, nu nh C4HW, ch c 2 ng' phn, thi C7H16 ngphn, cn CqEL^ n ti 159 ng phn, ri C13H2Slng phn, C20H42 l 366.319 ng phn v C304.111.846.763 dng phn.

CH B 24 CCH VIT NG PHN

Gi s ta phi vit cc ng phn ca hp cht CxH- Tnh s' ni i ca CJyONt theo cng thc :

^ ^ *2 j t - y-r t +2 so ni i - -------------72

116



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/

- Lp ccdng mch c c th.c. Nu hp cht c ni cn phi di chuyn v tr ni i trn mch c ' to ra cc mch c c v tr i i khc nhau.

- in H vo cc nguyn t c, 0, N cho ph hp vi ha tr ca chng.

Bi TP81 Vit CTCT cc ng phn c CTPT ;

a) C,BU b) C4H8 c) C3H4

GIa) CH3-CH2-CH2-CH2-CH3

CH,1 3c h 3 - CH -c h 2 -c h 3 c h 3- C -c h 3

c h 3 c h 3

b) CH2 = CH - CH2 - CHg CH3 - CH = CH - CH

CH, - CH,

CH? - CH,

CH2 = C- C H3 c h 2- c h 2

h 3 - L z - L ^c h 2 - c h - c h 3\ / g h 2

c)CH 2 = C = CH. CHoC-CH3HC = CH

\_ /CHrt

* Cb :- Qua cu b ta thy 1 n i i tcmg ng mt vng- Qua cu c ta thy ni i tng cmg1 ni ba

117



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B P 92 Vit CTCT cc ng phn mch h c CTPT:

a) CjHgO b) -C3H7N

GII

a) CH3 - CH2 - CH2 - OH CH3 - CH - OH CH3- 0 - GH2 - CH3ch 3

b)CH2 = CH- CH2 -NH2 CH3 - CH = CHr NHgCH2 = C-NH2 c h2 = c h - n h - c h3

c h 3ch3 - CH = N - c h3 c h3 - C2 - N =s c h2

B1TP93 Nhn ni 3 k u :

: C H 3 c - C H gB : CH2 = c = CH2c : CH = CH

\ __ /c h ;

u no sau y ng ? Nu sai ch r im sai a) A, B,c l 3 ng phah) , B, c l 3 ng ngc) , B,c c Cling cng thc, cu tod) , B, c u thuc dy ng ng ankin

GIIa) ng. Ba cht A, B,c l ng phn ca nhaub) Sai. Ba cht A, B,c khng th l ng ng v chng c

cng CTPT.c) Sai. Ba cht A, B,c u c cu to khc nhaud) Sai. Chi c cht thuc dy ng ng ankin

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=* Cng thc chung ca dy dng ng metan l CmH2m+2

LYN TP

n Vit tt c cc ng phn c CTPT l C4HI0 ;

C4H9CI; C4Hi00 ; C4HnN. :11. 2 i u k h n g n h n o d c r y l u n n g ?

a) Cc cht l ng ng ca nhau khng th cha cng s nguyn t hiro trong phn t.

b) Cc eht c cng cng thc phn t nht thit phi l cc ng phn ca nhan.

c) Cc cht c phn t lng hn km nhan 14 vc n h t t h i t p h i l h a ing ng lint ip.

j ) Bt c chth uc no c cha nguyn tnit I u c phn t lng l mt s l.

HNG Na) ng

b) Khng ng. Cc cht c cng CTPT nhng CTCT khc nhaumi l ng phn ca nhau

c) Khngng. V d naptalen C10Hg v unecan C0K22 cphn t lng hn km nhau 14 QVC nhng khng phi l 2 ng n g lin tip.

d) Khng ng. V d C2HgN2 c M = 60 (chn)

L3 Vit cng t h c c u t a o c c c h t d n x u t c a benzen c CTPT l C?HO.

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HNG DNc 5 cng thc cu to

Vi.4 iu kin olefin c ng phn Cis-trans l g ?

Cho v d.HNG DN

iu dn mt oein c ng phn cis-trans l n -phc dng baC = Cab. Trong :

a a a _b

b b b acis . trans

Vi d olefin CH3 - CH = CH - CH3 c 2 ngphn cis vtrans sau :

H H H CH,c = e " c - c ^

c h 3 c 3 c h 3 htis-buten-2 trans-butec-2

Kl.5 Khi cho axetilen phn ng cng vi clo theo t l moi 1:1 nhn thy trong sn phm phn ng c 2 cht ng phn ca nhau. Vit GTCT 2 ng phn . c ti.

HNG DNPhn ng xy ra :

CHeCH + Cljj CHCUCHC1 Hai ng phn trn l :

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H Clw .

Cl *X Htrans-,2-icloetyen

y Bit rng cng thc tng qut ca mt idro- eacbon mch h bt ki c dng CnH2n +2 - 2k & ^ lin kt Jt),

Kt lun g v dy ng ng ca hidrocacbon A bit rng khi t chy hon ton A thu c

c) nH0 < nco^

HNG DNXt phn ng

< v w * + - n C O j + ( c + 1 - k lH 20

a an a(n + 1 - k)a) Ta c (n + 1 - k)a > na

=> 1 - k> 0 hay 1 > k Vy k - 0. Hidrocacbon ny l ankan

b) Ta c (n + 1 - k)a =s na=> k = 1

Hidrocacbon ny l ankenc) (n T 1 - k)a< na1 - k < 0 hay 1 < k

Hirocacbon ny c t 2 lin ktti tr ln (ankin ; ankaien; ankatrien...)

Kx Hc = c ^

C1 CIs-l;2-icoetyen

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dxig ca benzen theo 2 phng php ;a) a vo s electron ha tr ca nguyn t b) Da vo ih ngha v ng ng.

HUNGN

b) Benzen c CTPT/ CgHg=* Cc ng dng: benzen c cng thc :

C6Hg(CH2)ahayCg+nHg+2nt 8 + n s m th in -m - 6, do d :

6 + 2n = 6 + 2(m - 6) = 2m - 6 Vy cc ng ng benzenc cng thc CraH2m. g

Yl8 Hidrocacbcn A c CTPT l C5H,9. A te dng vi clo theo t l mo 1:1 trong iu kiii chiu sng ch to r1 sn phm th duy nht. Xc nh cu to ca A

HNG DNC5H12 c th c 3 -cu to sau (nh50 ln-lt l , II, III)

CH,CH3 -C H2 -CH 2-CK2 -CH3 CHg-C-CHg

CH,CH3-CH-CH2-CH3

CHgTa c ohn g :

C-H12 + Cl2~^~ C5Hj jC1+H C 1 .Nh vy xt tng cu to c th ca ankan C5Hl2 th : ' u to (l cho 3 sn Dhih th eo nh sau

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CCH2 - CH2 - CH2 - CH2 - CHS CHg - GHC1. CH2 - CH2 - CBgCH3 - CH2 - CH - CH2 - GHg

C1Vy (I) khng ph hp

* Cu to (II) cho 4 sn phm th cl nh sauC H 2C - C H - C H 2 - C H 3 C H 3 - C C 1 - C H j - G H gCH3 CH3 .

CHS-CH-C HC1 - CH3 CE, - CH - CE - CH2CICH3 CH3

Yy (II) khng pi hrp* Cu to (III) cho 1 sn phm th clo duy nht :

CH3CH2CI-C-CH3

CHoVy (III) l cu to ph bp

Yi.9 Hidrocacbon c6 CTPT CgH14. Khi phn ag vi clo theo t l moi1 : 1 trong iu kn chiu sng'dh to2 sn phm duy nht. Sc nh CTCT aiikaii trn.

S: CH3 -CI-CH-CK 3CH, CH,

V.1Q Clo ha mt ankan theo t moi1 : 1 c mt sn phm th trong clo chim 33,33% v khi lng

a) Tm CTPT ankan.b) Nu sn phm h trn l duy nht, nh cu to ankan trn.

S : a) C5E 2

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I|IU1 Phn ti ch nh ng mt hp cht huc a,I mch h, cha mt nguyn t oxi trong phn t c % c = 54,54 ; %H 5 9,09 v% o = 3d6,37

a) Tm CTPT cua A.b) Vit CTCT ca A. Bit hp cht c nhm OH gn vo Qguyn tc mang ni i th khng bn.

S aj C2H40

A l mt hp cht cha 24,24%c ; 4,04% H v 71,72% CI v khi lnga) Tim cng thc nguyn ca A.l) nh CTPT ca A.c) Vit t c CTCTc th c ca .

S: bXa*r~~ ---- ------------~ ~ ;------------------------------------ Vit tt c CTCT c th c ca hp cht hi c c CTPT l ( HgCLj

b) Nu ch bit hp cht hu c trn c cng thcI nguyn l (C2H4Cl)nth c bin untm c CTPT

HNG DN

fc) S ni i trong (C2H4Cl)n hay C2nH4nCln = -r 5n + 22 - n

2 -2 - n

S ni di (T&am t cht s* 0 ------- 0 2 5=n2 Nhng nX 1 cho cng thc C>H4C1 khng' ph hp ha tr.Vy n.= 2



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CHUYN VII

Off GH PHN NG

Hu ht nhng phn ng'hvi c u xy r qua nhiu bc trung gian. Ton b cc bc trung gin ny c gi l c ch phn ng.

Vie vn dng c ch phn ng gip gii tMcH c s hnh

thnh cc sn phm ph bn cnh' sn phm chng trong qu trnh phn ng. .

CH C CH Gc TRONG PHN NG THE

CLOVOANKAN n gin ta xt c ch phn ng th co vo metan* Bc khi mo phn ng Khi hp th nng ng nh sng, phn t C2 b ct thnh

2 nguyn t C

C L -C ~^C + * Bc pht trin .phn ng Nguyn t Co rt hot ng s tc dng1vi CH to ra

gc t Co CH3 v HCL Sau gc CH3 tc dng vi phn't C2khc to ra CHZCL v nguyn t Co

H-CH3+ . CHZ+ HC

CH2 +C1-C1 CH3C +C Qu trnh ny p i lp i nhiu n



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* -Bc tt mch ph ng Rc pht tin phn ng s b dng li dc xy r ci kt hp sa :

Cl+ Clz

CBZ+ CH2C h3 + H3 . C z H q .

* Ch brm cng' cho p h n ng t h tng t nhCQ nhng xy ra kh khn hn. Io cn cho phn ng khn, nhiu hn na. -

Si P 38 Gii thch v sao trong phn ng th eo v- mtan c s tao thnhC3L ?

L _ ZZL _________GII

Theo c ch phnng th xt ch 26, ta thy-c to thnh C,H6 trong qn trnh phn ng

* Ch : C2H6 sinhra c th tip -tc phn ng th Clo theo c ch gc xt trn to ra

c h 3-c h 2-c h 2- c h 3

SH|j27 TNH HIU ST TO THNH CC

SN PHM DNG PHN TH

Vi cc snhan t propan tr i, phn ng' th u tin rs nguyn t H gn vo nguyn t c bc cao hn (l ng t c Hn kt vi t hdro hn).

Do bit kh nng phn. ng tng i ca tng n t hirg ( tng nguyn t c bc , bc , bc ) ta c tnh c hiu sut to thnh cc sn phm ng phn

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BI TP97 Khi cho propan phn ng vi co trong iu kin chi sng v t l moi 1:1 thu c hai sn phm th monoclo l n - propyl corua v so - propylclorua. Tnh hiu sut to thnh cc sn phm. Cho kh nng phn ng ca nguyn t H c bc I l 1;

b c n l 4 ,3 .

GII

CH3-CH2.CH3cn n[2 nguyn t H bc IITng kh nng phn ng ca co nguyn t H ny l :

6 .1 + 2.4,3 = 14,3 .Cc phn ng c th xy ra :

c h 3 - c h 2 - C H 3 + C Lj c h 3 - c h 2 - c h 2 c i

C&J - CH2- CH? + Cl2 . CH3 - CHC1 - CHg + HC16 1=>Hiu sut to n - propylcorua = =41,1%14,3

2 4 3Hiu sut to iso - propylclorua - ~ ~ = 58,9%14,3

BI TP98 Trong iu kin c chiu sng v 127c,brom phn ng vi iso - pentan theo t l mol 1:1 to ra 4 sn phm th monobrom. Tnh hiu sut to thnh mi sn phm. Cho bit trong iu kin nu, kh nng phn ng ca nguyn t H cacbon bc l 1; bcn l 82 v bc III l 1600.

GII

r so - pentn CH.J - CH *CH2 -CH3C

CH,,

1 nguyn tH bc II2 nguyn tH bc II9 nguyn tH bc I

9-CB...HH 11 129



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r Tng' khnngphn ng ca chng = 9.1 + 2.82 + 1.1600 = 1773 S phn ng to 4 sn phm th :

' ' - BrCHg-CH-CHj-CH3 '(A)

ch3+ Br2 cH3-CBr-CH2-CH8 (B)CK3 - CH - CH2 - CH3------ -----1 CH3

CH3 \ CHr CH -CHB r~CH 3 (C)\ c h 3

CH j - CH CH j - GHjBr (D)CH3

=? Hiu sut to A = " = 0,34%1773

Hiu sut tao B = = 90,24%17732 32Hiu sut to c = = 9,25%1773

_ 3 1Hiu sut tao D = -- = 0,17%1773

H I T X 7 7 T ' ;mn co CH PHN NG CNG ELECTROPHIN! YO ANKEN ! n gin ta xt phn ng cng ca HBr vo CH, =CH .I Phn n? cng din ra qua 2 giai on k tip sau :

I CHt = Ch\ + B r C H j -CH2+ Br~ 1)CH;i - +H2 + B r CH3 - CH2 B (2) -

Giai on { xy ra chm qavt nh tc chung ca phn ng, trong c s tn cng ca H~ to thnh ion dng CH., - 'CR2 nn gi l phn ng cng eectrophin (hay cng cc in t; 4

1 Giai on (2) xy ra nhanh.



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E1P 89 V it CO ch phn ng cng xay ra kh co propen tc ng vi dung dch brm.

GIIPhn ng: din ra qua 2 giai on* Giai on 1 (chm)

CH3-c h - &, + Br-5- Br- C ,-*f- CLBr + Br* Giai on 2 (nhantu

CH3- +CH - CH2 Br + Brx O.-CHBr-CH^r

Bi TP100 Khi dn ety en vo dung dch Br2 (in ch bng cch cho CI2 va tc ng vi dang dch NaBr) th thu c hai sn phm hu c. Gii thch v vit cc phn ng.

GIIDung dch Br2 trn c s hin din NaCl do phn ng :

2NaBr + Cl2 = 2NaCl + Br2 Xt c ch phn ng cng Br2 vo etyln :* Giai on 1 (chm)

CH2 = CH2 + B r+s - Br - 5 C H ^ r +C2 + B r

* Giai on 2 (nhanh)CH2 Br - +CH2 +' Br CH2Br - CH^r

Ngoi ra dung dch c s hin din cn C1 nn cng xy ra: CH2Br - +CH2 + cr CH2Br - CH2C1

Vy thu c 2 sn phm lCH2Br - CH.zBr v CH2Br - CH2C1



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GHHfl 29 D OAN KHA NANG PHAN NG CNG

ELECTROPHNTheo c ch cng eectrophin xt ch 28

on tn 'cng c tiu phn mang in tch dngi to ra mt ion dng trung gian l gii o

nh vn tc chung- ca phn ng. T c th d nu mt electron ni dpi cng ln th phn ng xy ra.

1;7----------- : ------- -------- -------- ----8MP1S1 Sp xp theo chiu tng dn tc ng cng bronL khi cho cc cht sau y tc dvi dung dch brom (Bitrng CH3 l nhm electron)a) CHy - CH = CH2 b) CH2= CH2

I c) CH3 -c = CH - CH3 d) CH,' c s" C - C,

o c mt 4 nhm - CH3 nn mt d electron ni iCcht d ln nht, phn ng c tc ln nht.Cn vi cht b, mt electron ni i thp

phn ng- c tc nh nht.Vy chiu tng dn tc phn ng i :

CH. CH? CH3

GIX

b < a < c < d

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CHB 30: C CH PHN NG TH E EL ECTR OPH

VNG BENZEN n gin ta xt ph ng' thbom vo benzen kh ch

benzen tc ig vi brm ng nguyn cht v c s hin din cda bt st.

- Trc ht Fe tc dng vi Br2 to ra xc tc FeBr32Fe + 3Br2 = 2FeBr2

- Ri Br2 tc dung tip vi FBr3 to ra cation B* ; F e B 3 + B r - Bt ~ [ F e B r + B i*

- Sau. ction Bi* kt hp vi mt trong su nguyn t c ca benzen to phc c khng bn :

+ B* * J Br

(phc )- Phc ( nhanh chng phn hy theo phn ng :

r ^ j f +[FeBr4 + t FeBt\ + HBr

8I TP 102

a) Trnh by c ch phn ng nitro ha benzen khi cho benzen tc dng vi hn hp HNOg m c v H2S04 m c.

b) C phi vai tr ca H2S04 c l ht nc trong phn ng phn ng nitro ha xy ra chiu thun {tc chiu tng hiu sut phi ng) khng?



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G I I

a) Trc ht, cation "NO, c to ra nh dat xc tc H^so^ HO - NO, + e-= H - 0* - NQ2 = H20 + *N02HO - NO, + B^H -O *- N02 = H20 + *N02

H- Sau cation +N02 kt hp vi 1 trong-6 nguyn tc

ca benzen to phc khng bn :H

^ + +n o 2 - n o 2+

- Phc nhanh chng phn hy theo phn ng :

NO2 ---- * r fA - . N02H .

C r+ H+

bi Vai tr ca H2S04 khng- phi ht nc ca phn ng HoSOo

c 6h 6+h n o 3 - c .h 5n o 2+h 2o

m l to cation *N09. Tht vy :- Phn ngnitro ha. benzen khng phi phn ng' tliun

nghch nn c ht nc cng khng nh hng g n phn ng. .

- Nu thay H2S04 dc bng- mt axit mnh khc khng c* kh nng ht nc th phn ng nitro. ha benzen vn

xy ra :

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CHO31 D ON KH NNG THE NO VNG 5ENZEN

S c mt ca mt nhm th trong vng benzen m cho kh nng th tip theo vo vng' benzen hoc d hn, hoc

kh hn. Do phn ng' th xy ra theo c ch electrophin (in t)

nn cc nhm th c hiu ng tng qut y electron s lm tng mt. -electro trong vng, gip pkn ng th xy ra d hn. Ngc li nhm th c hiu, ng tng qut ht electron s m gim mt electron trong vng, lm phn ng th xy ra kh hn.

BI TP 103 S p x p t h e o c h i u t n g d n t c p h n n g

mononitro ha ca cc hp cht sau ya) C6H5 - H b)CKH5-CH3c ) C 6H 5 - N 0 2

GII

Do -CHy l nhm y electron nn mt electron trongvng benzen.tng, phn ng xy ra d hn

Do -NO.> l nhm ht electron nn mt electron trongvng benzen gim, phn ng xy ra kh hn

Vy chiutng dn tc phnng l : b > a > c

CHB32 QU LUT TH VO VNG BENZEN

* Khi vng benzen c sn cc gc - CHV -C2Hr ... (hy cc nhm 'OE; -OR; -NH2; -C; -Br..J phn ng' th sy ra d dng hn (tr trng' hp -C, -B) v s u tin xy ra cc v tr ortho; para.



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.... .Wit-nnom nguyn t cha in k i hoc ba), phn ng th xy ra kh hn v s u tin xy ra v tri met. '

V d : T l phn trm cc sn phm khi nitro ha .

t ohien( J y CHg v axit benzon


136/177

slTP104 T henzen vit s iu ch (ghi r iu kin) cc cht sau :a) orthobromonitrobenzenb) metabromonitrobenzenc) orthobromotohien

d) metanirotoluen

GII

HNG DNCh qui lut th vo vng benzen gn c cc nhm

th vo v tr thch hp.Br

+HNO,a) 0

b) 0

c) 0

d) 0

Fe

+HNO,H2S 0 4

+CH3C1

AC U

+HNOa

0

NO,hT'2s o 4

CH,

NO.

+Br2

Fe

+Br2

Fe

+CH;:CI

A1CL

Br

m ' n o 2

n o 2

Br

0 Br

CH,

NO.,

0 CH,

LUYN TP



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HNG DNas

b) - Bc khi mo : Cl2 2C'l- Bc pht trin : CHg + c I CgH- + HC1

C-H, + Cl2 - C2H5CI + 01.

- Bc tt mch : C1 + C'l * Cl,C2H5 + CjHj CHj - CH, - GH2 - CHj

C2h 5 + c-l C2H5C1I-------------------------------------------------------------------------------I V1T2 Cho ha n - butan theo t l mol 1:1 thu dc hai I sn^phm th monoco. Xc nh t l cc sn phmI trn bit rng kh nng phn ng ca nguyn t HI c bc I l 1; c bc n l4.

HNG DN

n - batan CH3 - CH,. CH, - CH3 C| 6 ngU3f * i f l 14 nguyn t H bc II

Tng kh nng phn ng ca chng = 6 . 1 + 4 . 4 = 22

=9 Hiusut tao CH,C1 -CH, - CH, - CH, = = 2 2 2 3 22 22

= Hiusut to CH,- CHC1 - CH, -CH. = r r ^ = 3 2 3 22 22 6 16=> 1 l gia 2 sn phm trn = : =3:8

22 22

I VIL3 Cho iso - butan phn ng vi brom theo t l moi 1:1 v c chiu sng thu c nhng sn phm mono* brom no ? Cho bit hiu sut to thnh mi monobrom. Bit kh nng phn ng ca nguyn t H c bc I l 1; c bc II l82; c bc IH l 1600.

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H U N G D N

Iso - butan CH., - CH - C, c I t_ ^ * J 3 \ l nguyn t H bc IIICH3

Tng- kh nng phn ng ca cc nguyn t H = 9-1 + 1600.1 = 1609

Ta thu c 2 monobrom sau :

CGH2Br-CH-CH3 (A)

CH

8

CH3- CBr-CH, (B)h i ,

9 1=> Hiu sut tao A = = 0,6%1609

Tr. 1. 1600 Hitt sut tao B - =2 99,4%1

chuyÊn ĐỀ bỒi dƯỠng hÓa hỌc 11 - nguyỄn ĐÌnh ĐỘ

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