CHUYÊN ĐỀ BỒI DƯỠNG HÓA HỌC 11 - NGUYỄN ĐÌNH ĐỘ
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Transcript of CHUYÊN ĐỀ BỒI DƯỠNG HÓA HỌC 11 - NGUYỄN ĐÌNH ĐỘ
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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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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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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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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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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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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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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 .
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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 lng- ring 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)/
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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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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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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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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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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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. 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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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
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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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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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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
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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
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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
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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
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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)
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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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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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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
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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
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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)
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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
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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 :
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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.
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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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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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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
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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
96
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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 cng- thc 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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= 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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+ 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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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 ng- phn, 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
118
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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.
121
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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 :
122
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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
123
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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
124
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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
125
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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
128
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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 - propyl- clorua. 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
132
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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 :
134
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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
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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 mono- brom. Bit kh nng phn ng ca nguyn t H c bc I l 1; c bc II l82; c bc IH l 1600.
138
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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