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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12 CNG N THI HC SINH GII MN HO 12 NM HC 2012-2013

A. K HOCH N THITun 1. n tp cu to nguyn t- bng tun hon- lin kt ho hcTun 2. n tp cc phn ng trong ho v c.Tun 3. n tp cc dng bi tp v l thuyt phn ng trong ho v c.Tun 4. n tp cc dng bi tp v l thuyt phn ng trong ho v c(tip)Tun 5. n tp mt s dng bi ton v c: phn ng ca mt s cht oxh mnh: HNO 3, H2SO4 c.

Tun 6. n tp mt s dng bi ton v c: kim loi kim, kim th, nhm, bi ton in phn.Tun 7. n tp cc phn ng trong ho hu cTun 8. Cc dng bi tp v l thuyt phn ng trong ho hu c.Tun 9,10,11.Bi tp v cc hp cht hu c: hirocacbon, ancol, phenol, ax cacboxylic, este,cacbohirat,amin, aminoax, peptit.Tun 12,13,14. Hng dn hc sinh lm bi tp trc nghim v luyn mt s thi trc nghim.Tun 15. Hng dn hc sinh lm bi thi t lun v luyn tng hp.

B. NI DUNG

Tun 1.n tp cu to nguyn t- bng tun hon- lin kt ho hcI. Tm tt l thuyt

II. Bi tp vn dng

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Chuyn cu to nguyn t- bng tun hon lin kt ha hc

Bi 1. Hp cht A c cng thc l MXx, trong M chim 46,67% v khi lng. M l kim loi, X l phi kimchu k 3. Bit trong ht nhn nguyn t ca M c: n p = 4, ca X c n = p (trong n, n, p, p l s ntron vproton). Tng s proton trong MXx l 58.

1. Xc nh MXx ?2. Ho tan 1,2 gam A hon ton va trong dung dch HNO 3 0,36M th thu c V lt kh mu nu

(ktc) v dung dch B lm qu tm ho .Hy xc nh gi tr V v th tch dung dch HNO3 cn dng.

Bi 2. Hp cht A c cng thc l MXx, trong M chim 46,67% v khi lng. M l kim loi, X l phi kimchu k 3. Bit trong ht nhn nguyn t ca M c: n p = 4, ca X c n = p (trong n, n, p, p l s ntron vproton). Tng s proton trong MXx l 58.

1. Xc nh MXx ?2. Ho tan 1,2 gam A hon ton va trong dung dch HNO 3 0,36M th thu c V lt kh mu nu

(ktc) v dung dch B lm qu tm ho .Hy xc nh gi tr V v th tch dung dch HNO3 cn dng.

1. Xc nh MXx ?- Trong M c: n p =4 n = p + 4

- Trong X c: n = p- Do electron c khi lng khng ng k nn: M = 2p + 4 (1)

X = x.2p (2)2p 4 46,67 7

(1), (2) 7p ' x 8p 16 (3)x.2p ' 53,33 8

+ = = =

- Theo bi: px + p = 58 (4)- Gii (3), (4) px = 32, p = 26, n = 30

p = 26 nn M l Fe.- Do x thuc s nguyn dng:Bin lun:

x 1 2 3 4 . . .p 32 16 10,7 8Kt lun Loi Nhn Loi Loi

X = 2, p = 16 nn X l S.Vy cng thc ca A l FeS22. Hy xc nh gi tr V v th tch dung dch HNO3 cn dng:Phng trnh phn ng:

FeS2 + 18HNO3 Fe(NO3)3 + 15NO2 + 2H2SO4 + 7H2O0,01(mol) 0,18 0,15

A

1,2n 0,01(mol)

120= =

V = 0,15.22,4 = 3,36(mol)

3HNO

0,18V 0,5(lt)

0,36= =

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12

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Tun 2,3. n tp cc dng bi tp v l thuyt phn ng trong ho v c.

Bi 1. Ch dng thm phenolphtalein. Hy phn bit cc dung dch ng ring bit sau: NaCl, NaHSO 4, CaCAlCl3, FeCl3, Na2CO3. (Vit phn ng xy ra dng ion)Bi 2. Nu hin tng xy ra v vit cc phng trnh phn ng (nu c) khi:

a) Cho Fe3O4 tc dng vi dung dch HI d.

b) Cho kim loi Al vo dung dch hn hp gm KNO3 v KOH.c) Cho dung dch H2SO4 long vo dung dch Fe(NO3)2.d) Cho mui natri axetat vo dung dch K2Cr2O7.

Bi 3. Cnbng phn ng oxi ho- kh sau theo phng php thng bng electron:a.FexSy + 3NO

+ H+ 3Fe + +24SO

+ NO + H2O

b. FeCl2 + KMnO4 + KHSO4 c. AlCl3 + KMnO4 + KHSO4 d. FeCO3 + KMnO4 + KHSO4

e. 3 2 3 2 4 3 2 2FeCO + FeS + HNO Fe (SO ) + CO + NO + H O .

g.Cu2FeSx + O2 Cu2O + Fe3O4 +.h. CH3-C6H4- C2H5 + KMnO4 + H2SO4

Bi 4. Vit phng trnh phn ng xy ra trong cc trng hp sau:(1) Dn kh O3 vo dung dch KI. (2) Dn kh H2S vo dung dch FeCl3.(3) Trn dung dch KI vi dung dch FeBr3. (4) Dn kh Cl2 vo dung dch NaOH.(5) Dn kh SO2 vo dung dch KMnO4. (6) Dn kh F2 vo nc nng.(7) Trn dung dch FeCl2 vi dung dch AgNO3 d. (8) Dn kh SO2 v dung dch H2S.(9) Dn kh CO2 v dung dch NaAlO2 ( Na[Al(OH)4).(10) Cho dung dch AlCl3 vo dung dch Na2CO3(11) Ho tan hon ton Fe2O3 trong dung dch HI(12) Sc kh CO2 vo dung dch BaCl2

(13) Cho dung dch NH3 d vo dung dch FeCl3(14) Nh dung dch NaHSO4 vo dung dch Ba(HCO3)2(15) Ha tan hon ton hn hp Cu, Fe2O3 trong dung dch gm NaNO3 v KHSO4.(16) Nhit phn cc cht sau: NH4NO3, NH4NO2, Fe(NO3)2, hn

Bi 5. a)Hon thnh cc phng trnh phn ng sau:MnO2 + HCl Kh A; FeS + HCl Kh BNa2SO3 + HCl Kh C; NH4HCO3 + NaOH Kh D

b) Cho kh A tc dng vi kh D; cho kh B tc dng vi kh C; cho kh B tc dng vi kh A trong nc. Vicc phng trnh phn ng xy ra.Bi 6. Cho cc dung dch ring bit mt nhn sau: Na2SO4, AlCl3, FeSO4, NaHSO4, FeCl3. Ch dng dung dcK2S nhn bit cc dung dch trn ngay ln th u tin. Vit cc phng trnh ho hc minh ha.Bi 7. a) Hon thnh cc phng trnh phn ng sau:

KMnO4 + HCl Kh A FeS + HCl Kh BNa2SO3 + H2SO4 Kh C NaCl + H2O dd D(n phn mng ngn

b) Cho kh A tc dng vi dung dch D, kh B tc dng vi kh C.Cho kh C tc dng vi dung dch D vi t l mol 1:1.Vit phng trnh phn ng xy ra.

Bi 8: Hp cht A c dng M3X2. Khi cho A vo nc, thu c kt ta trng B v kh C l mt cht c. Kt tB tan c trong dung dch NaOH v dung dch NH3. t chy hon ton kh C ri cho sn phm vo nc dthu c dung dch axit D. Cho D t t vo dung dch KOH, phn ng xong thu c dung dch E cha 2 muDung dch E phn ng vi dung dch AgNO3 cho kt ta mu vng F tan trong axit mnh.

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 121/ Lp lun chn cng thc ha hc ng cho cht A. Vit cc phng trnh phn ng xy ra theo th t t An F. Bit M v X u l nhng n cht ph bin.Bi 9: Vit phn ng xy ra trong cc th nghim sau:1/ Cho t t dung dch AlCl3 vo dung dch NaOH cho n d2/ Cho dung dch FeCl3 ln lt tc dung vi Na2CO3; HI; H2S; K2S.3/ Cho As2S3 tc dng vi HNO3 c nng.4/ Cho NH4Cl tc dng vi dung dch NaAlO2; phenol tc dng vi natri cacbonat

1/+ u tin khng c kt ta: AlCl3 + 4NaOH NaAlO2 + 3NaCl + 2H2O+ Khi d AlCl3 s xut hin kt ta: 3NaAlO2 + AlCl3 + 6H2O 4Al(OH)3 + 3NaCl2/ 2FeCl3 + 3Na2CO3 + 3H2O 2Fe(OH)3 +3CO2 + 6NaCl

FeCl3 + HI FeCl2 + HCl + I2.2FeCl3 + H2S 2FeCl2 + S + 2HCl

2FeCl3 + 3Na2S 2FeS + S + 6NaCl3/ As2S3 + 28HNO3 2H3AsO4 + 3H2SO4 + 28NO2 + 8H2O4/ NH4Cl + NaAlO2 + H2O NH3 + Al(OH)3 + NaClC6H5OH + Na2CO3 C6H5ONa + NaHCO3.

Bi 10. Hy nhn bit cc l mt nhn cha cc dung dch sau : Na2CO3 , Na2CO3, Na2SO4, NaHCO3, Ba(HCO3, Pb(NO3)2.

Bi 11. C cc mui A,B,C ng vi cc gc axit khc nhau, cho bit :A + dung dch HCl c kh thot raA + dung dch NaOH c kh thot raB + dung dch HCl c kh thot raB + dung dch NaOH c kt ta. dng dung dch C + A c kh thot ra dng dung dch C + B c kt ta v kh thot ra

Xc nh cng thc phn t ca 3 mui, vit phng trnh phn ng.

Bi 12. Trong phng th nghim, b dng c v di y c th dng iu ch nhng cht kh no trong scc kh sau: Cl2, O2, NO, NH3, SO2, CO2, H2, C2H4, gii thch. Mi kh iu ch c, hy chn mt cp cht

v B thch hp v vit phn ng iu ch cht kh ?

Bi 13. Xc nh cc cht A, B, C v hon thnh 3 phn ng sau:NaBr + H2SO4 (c)

0t Kh A + ........ (1)

NaI + H2SO4 (c)0t Kh B + ........ (2)

A + B C (rn) +.... (3)

Bi 14.Sc kh A vo dung dch cha cht B ta c rn C mu vng v dung dch D.

Kh X c mu vng lc tc dng vi kh A to ra C v F. Nu X tc dng vi kh A trong nc to ra Y v F, r

thm BaCl2 vo dung dch th c kt ta trng. A tc dng vi dung dch cht G l mui nitrat kim loi to ra k

ta H mu en. t chy H bi oxi ta c cht lng I mu trng bc.

Xc nh A, B, C, F, G, H, I, X, Y v vit phng trnh ha hc ca cc phn ng.

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Bi 15. Ch dng cht ch th phenolphtalein, hy phn bit cc dung dch ng trong cc l ring bit b mt nhnNaHSO4, Na2CO3, AlCl3, Fe(NO3)3, NaCl, Ca(NO3)2. Vit cc phng trnh ho hc di dng ion thu gn. Bi 16. Xc nh cc cht v hon thnh s bin ha:

A X + D

X B Y +Z

C A + G

A l H2S v X l S ; B l SO2 ; C l FeS ; D l H2O ; Y l HBr ; Z l H2SO4 ;G l FeBr2 hoc FeSO4.S + H2

0t H2S ;

S + O20

t SO2 ;

S+ Fe0

t FeS ;

2 H2S + SO2 3S + H2O ;SO2 + 2 H2O + Br2 H2SO4 + 2 HBr ;

FeS +2 HBr FeBr2 + H2S ;FeS + H2SO4 FeSO4 + H2S ;

Tun 4.n tp mt s dng bi ton v c: phn ng ca mt s cht oxh mnh: HNO 3, H2SO4 c.

Bi 1: Cho m gam hn hp X gm Na v kim loi M vo nc d.Phn ng xy ra hon ton thu c 8,96 ltkh (ktc), dung dch Y v mt phn cht rn khng tan.Cho ton b lng cht rn ny tc dng vi 1,628 ltdung dch HNO3 0,5 M (ly d 10% so vi lng cn thit)sau phn ng thu c 0,448 lt N2( ktc) v dungdch Z. C cn Z c 46,6 gam cht rn khan.Vit phng trnh phn ng v xc nh m,M? .s: Al; 15,4 gam

Bi 2: Ha tan hon ton 1,62 gam Al vo 280ml dung dch HNO3 1M c dung dch A v kh NO l sn phmkh duy nht. Mt khc cho 7,35 gam 2 kim loi kim thuc 2 chu k lien tip vo 150ml dung dch HCl c

+Br2+D

H2, tO + B

+O2

+Fe +Y hoc Z

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12dung dch B v 2,8 lt H2 ktc.Trn dung dch A v B c 1,56 gam kt ta.Xc nh 2 kim loi kim v tnh CMca dung dch HCl. .s Na,K v 0,3M

Bi 3: Cho hn hp A gm kimloaij R ha tr 1 v kim loi X ha tr 2.Ha tan 3 gam A vo dung dch chaHNO3 v H2SO4 thu c 2,94 gma hn hp kh B gm NO2 v kh D, c V= 1,344 lt ktc.

a. Tnh khi lng mui khan thu c khi c cn dung dch.b. Nu t l NO2/D thay i th khi lng mui thay i trong khong no?

c. Nu cho cng mt lng Clo ln lt tc dng vi R v X th mR=3,375mX, mRCl=2,126mXClTnh %m mi kim loi trong hn hp..s:a.7,06; b.6,367,34; c. 64%

Bi 4. Cho 3,28 gam hn hp A gm Fe v Mg vo 400ml dung dch CuSO4 . Sau phn ng hon ton thu c4,24 gam cht rn B v dd C.Thm dd NaOH d vo dung dch C lc ly kt ta nung ngoi khng kh n khilng khng i c 2,4 gam cht rn D.

a. Tnh CM dung dch CuSO4.b. % mFe trong hn hpc. Ha tan B bng dung dch H2SO4 c nng d thu c V lt SO2 ktc.Tnh V?

.a: a.0,1; b. 85,366%Fe; c. 1,904 lt.

Bi 5. 1) Cho hn hp gm a mol FeS2 v b mol Cu2S tc dng va vi dung dch HNO3 th thu c dundch A (ch cha 2 mui sunfat) v 26,88 lt hn hp kh Y gm NO 2 v NO iu kin tiu chun (khng csn phm kh no khc), t khi ca Y so vi H 2 l 19. Cho dung dch A tc dng vi Ba(OH)2 d th thu kt ta E. Nung E n khi lng khng i th thu c m gam cht rn.

a. Tnh % theo th tch cc kh.

b. Tnh gi tr m.

2) Cho BaO tc dng vi dung dch H2SO4 thu c kt ta A v dung dch B. Cho B tc dng vi kiloi Al d thu c dung dch D v kh E. Thm K2CO3 vo dung dch D thy to kt ta F. Xc nh cc cht AB, D, E, F v vit cc phng trnh phn ng xy ra.

Bi 6.Ha tan hon ton mt lng hn hp A gm Fe 3O4 v FeS2 trong 25 gam dung dch HNO3 to kh dunht mu nu c th tch 1,6128 lt (ktc). Dung dch thu c cho tc dng va vi 200 ml dung dcNaOH 1M, lc kt ta em nung n khi lng khng i, c 3,2 gam cht rn. Tnh khi lng cc chtrong A v nng % ca dung dch HNO3 (gi thit HNO3 khng b mt do bay hi trong qu trnh phn ng).Bi 7:a. Cho hn hp A gm Fe2O3 v Cu vo dung dch HCl d, thu c dung dch B v cn li 1 gam Cu khng tanSc NH3 d vo B, lc ly kt ta v nung ngoi khng kh n khi lng khng i thu c 1,6 gam cht rnTnh khi lng ca hn hp A ban u.b. Cho 48 gam Fe2O3 vo m gam dung dch H2SO4 long nng 9,8%, sau khi kt thc phn ng thu c dundch c khi lng 474 gam( dung dch A). Tnh C% cc cht tan trong dung dch A.

Cho 48 gam Fe2O3 vo m gam dung dch H2SO4 9,8%, sau sc kh SO2 vo cho n d. Tnh C% ca c

cht tan trong dung dch thu c, bit rng phn ng xy ra hon ton.Bi 8.Nung 8,08 gam mt mui A, thu c cc sn phm kh v 1,6 gam mt hp cht th rn khng tan tronnc. Ton b sn phm kh c hp th ht bi 200 gam dung dch NaOH nng 1,2%, sau phn ng thc dung dch ch cha mt mui B duy nht c nng 2,47%. Tm cng thc phn t A, bit khi nung s oha ca kim loi khng thay i.Bi 9.

1.Cho 20,8 gam hn hp Fe, FeS, FeS2, S tc dng vi dung dch HNO3 c nng d thu c V lt kh NO

(l sn phm kh duy nht, o ktc) v dung dch A. Cho A tc dng vi dung dch Ba(OH) 2 d thu c 91

gam kt ta. Tnh V?

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 122. Cho m gam hn hp hai kim loi Fe, Cu (trong Fe chim 30% v khi lng) vo 50 ml dung dc

HNO3 nng 63% (d = 1,38 g/ml) un nng, khuy u hn hp ti cc phn ng hon ton thu c rn cn nng 0,75 m gam, dung dch B v 6,72 lt hn hp kh NO2 v NO ( ktc). Hi c cn dung dch B th thc bao nhiu gam mui khan ? (Gi s trong qu trnh un nng HNO3 bay hi khng ng k)

Bi 10. Cho 3,9 gam hn hp M gm hai kim loi X, Y c ho tr khng i ln lt l II v III vo dung dchH2SO4 long (d), sau khi cc phn ng xy ra hon ton thu c dung dch A v 4,48 lt kh H 2 (ktc).

a) Tnh khi lng mui trong A.

b) Cho 3,9 gam hn hp M tc dng va vi V lt dung dch HNO 3 1M, sau phn ng thu c 0,84 lkh B duy nht (ktc) v dung dch C. C cn cn thn dung dch C c 29,7 gam mui khan. Tm cng thphn t ca B v tnh gi tr ca V?Bi 11.Cho 3,64 gam hn hp A gm oxit, hiroxit v mui cacbonat trung ha ca mt kim loi M c ha tr tc dng va vi 117,6 gam dung dch H2SO4 10%. Sau phn ng thu c 448 ml kh CO2 (ktc) v dundch X cha mt mui duy nht. Dung dch X c c nng phn trm v nng mol ln lt l 10,876% v0,545M. Khi lng ring ca dung dch X l 1,093 g/ml. a) Xc nh tn kim loi M. b) Tnh % khi lng ca cc cht c trong hn hp A.Bi 12. Dung dch X cha HCl 4M v HNO3 aM.Cho t t Mg vo 100 ml dung dch X cho ti khi kh ngn

thot ra thy tn ht b gam Mg, thu c dung dch B ch cha cc mui ca Mg v thot ra 17,92 lt hn hp khY gm 3 kh. Cho Y qua dung dch NaOH d thy cn li 5,6 lt hn hp kh Z thot ra c

2Z Hd / =3,8. Cc ph

ng xy ra hon ton. Th tch cc kh u o ktc. Tnh a, b?Bi 13. Cho 5,8 gam FeCO3 tc dng vi dung dch HNO3 va thu c dung dch X v hn hp Y gm CONO. Cho dung dch HCl d vo dung dch X c dung dch Y. Dung dch Y ho tan ti a m gam Cu to ra sphm kh NO duy nht. Tnh m?

Cu I : (5 im)1. nY = 0,8 mol; nZ = 0,25 mol 2 0 55NOn mol,= (0,5 )

V khi qua dung dch NaOH ch c kh NO2 hp th nn Z phi cha kh H2 v kh A 7 6ZM( , )= .

Ta c2

10 2

2H HCln n ,= = mol nA = 0,05 mol.

0 2 2 0 057 6

0 25A

Z

MM

, . , .,

,

+= = MA = 30 A l NO. (0,5 )

Gi nMg phn ng l x mol.Qu trnh oxi ha: Qu trnh kh:Mg Mg+2 + 2e 2H+ + 2e H2x 2x 0,4 mol 0,2 mol

N+5 + 1e N+40,55 mol 0,55 mol

N+5 + 3e N+20,15 mol 0,05 mol (0,5)

p dng bo ton electron ta c: 2x = 0,4 + 0,55 + 0,15 x = 0,55 mol. b = 0,55.24 = 13,2 gam. (0,5)

3 3 3HNO pu NO NO

n n pu n muoi( ) ( ) ( ) = + = 0,55 + 0,05 + 2 (0,55 0,2) = 1,3 mol. (0,5)

[ ]31 3

130 1

HNO M,

,= = a = 13M. (0,5)

2. (2 im): 3 33 30 05 3 0 15FeCOFe NO Fen n mol n n mol, ; ,+ += = = = (0,5)

3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O (0,5)0 1 5 3

2mol

, .0,15 mol

Cu + 2Fe3+ Cu2+ + 2Fe2+ (0,5)

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 120,025 mol 0,05 mol

Vy m = 64 (0 1 5 3

2

, .+0,025) = 16 gam.

Bi 14. t chy hon ton 4.4g sunfua ca kim loi M (cng thc MS) trong oxi d. Cht rn sau phn ng emho tan trong 1 lng va dung dch HNO3 37.8% thy nng phn trm ca mui trong dung dch thu l 41.72%. Khi lm lnh dung dch ny th thot ra 8.08g mui rn. Lc tch mui rn thy nng phn trmca mui trong dung dch l 34.7%. Xc nh cng thc ca mui rn.

V O2 d nn M c ho tr cao nht trong oxit2MS + (2 + n:2)O2M2On + 2SO2 (0,25 )a 0,5aM2On + 2nHNO3 2M(NO3)n + n H2O (0,25 )0,5a an a

Khi lng dung dch HNO3m = an 63 100 : 37,8 = 500an : 3 (g)

Khi lng dung dch sau phn ngm = aM + 8an + 500an : 3 (g)

Ta c (aM + 62an) : (aM + 524an: 3) = 0,4172Nn M = 18,65n (0,50 )Chn n = 3 Suy ra M = 56 (Fe)Ta c: a(M+32)= 4,4 Suy ra a = 0,05

khi lng Fe(NO3)3 lm= 0,05 242 = 12,1(g)

Khi lng dung dch sau khi mui kt tinh :mdd = aM + 524an: 3 8,08 =20,92 (g)

Khi lng Fe(NO3)3 cn li trong dung dch l :m = 20,92 34,7 : 100 = 7,25924 (g)

Khi lng Fe(NO3)3 kt tinhm = 12,1 - 7,25924 = 4,84 (g) (0,50 )

t cng thc Fe(NO3)3 . nH2OSuy ra 4,84:242 (242 + 18n) = 8,08 Suy ra n = 9CT Fe(NO3)3 . 9H2OBi 15. Cho hn hp gm a mol FeS2 v b mol Cu2S tc dng va vi dung dch HNO3 th thu c dundch A (ch cha 2 mui sunfat) v 26,88 lt hn hp kh Y gm NO 2 v NO iu kin tiu chun ( l nhng sphm kh duy nht), t khi ca Y so vi H2 l 19. Cho dung dch A tc dng vi Ba(OH)2 d th thu c kta E. Nung E n khi lng khng i th thu c m gam cht rn.

1.Tnh % theo th tch cc kh ?2.Tnh gi tr m?

Bi 16. 1. Hp cht X to bi 2 nguyn t A, B v c khi lng phn t l 76.A, B c s oxi ha cao nht trong cc oxit l + no v + mo v c s oxi ha m

trong cc hp cht vi hiro l - nH v - mH tha mn iu hin: | no|= | nH| v| m0| = 3| mH|Tm cng thc phn t ca X. Bit A c s oxi ha cao nht trong X

Bi 17. Ha tan hon ton kim loi M vo dung dch HNO 3 aM (long) thu c dung dch X v 0,2 mol N(sn phm kh duy nht). Ha tan hon ton kim loi M vo dung dch HNO 3 aM ch thu c dung dch YTrn X v Y c dung dch Z. Cho dung dch NaOH d vo Z thu c 0,1 mol kh v mt kt ta E. Nung n khi lng khng i c 40 gam cht rn F. Hy xc nh M, M. Bit:M, M u l cc kim loi ha tr II.M, M c t l nguyn t khi l 3:8. Nguyn t khi ca M, M u ln hn 23 v nh hn 70.

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12V M vo dung dch HNO3 ch thu c dung dch Y, nn dd Y phi cha 4NH

+, v kh thu c l NH3

3 4NH NH

n n 0,1mol+

= = . Theo bo ton electron, ta c:

' '

2. 0,2.3 0,3

2. 0,1.8 0,4

= =

= =M M

M M

n n

n n

* Trng hp 1: Cht rn F l hn hp oxit MO, MO(Kim loi Hg hoc oxit khng lng tnh)nF = 0,3 + 0,4 = 0,7 molnOxi trong F= 0,7 molmOxi = 0,7.16 = 11,2 gam

2KLm = 40 - 11,2 = 28,8 gam.

+ Nu3

' 8=

M

M, th ta c: 0,3.M + 0,4.M = 28,8

0,3.M + 0,4.3

8.M = 28,8

M = 64 (Cu) M = 24 (Mg) (nhn)

+ Nu' 3

8=

M

M, th ta c: 0,3.M + 0,4.M = 28,8

M; 56,2 ; M ; 21,1 (loi)

* Trng hp 2: F ch c 1 oxit MO hay MO

NOM =

40

0,4= 100 M' = 84 (loi)

MOM =

40

0,3= 133,33 M = 117,3 (loi)

Bi 18. X l hn hp Cu, Fe. Ho tan ht m gam X bng V lt dung dch H2SO4 98%, t0 d (d=1,84 g/ml) dung dch A. Pha long dung dch A ri in phn vi in cc tr dng in I= 9,65A n ht Cu 2+ th mtpht 20 giy (H = 100%). Dung dch B nhn c sau phn ng va ht 100mldung dch KMnO4 0,04M.1/ Tnh phn trm khi lng 2 kim loi trong X.2/ Tnh V, bit lng axit dng ch ht 10% so vi lng c.1/ t x, y ln lt l s mol Fe v Cu, ta c:

2Fe + 6H2SO4 Fe2(SO4)3 + 3SO2 + 6H2Omol: x 3x 0,5x

CuSO4 + 2H2SO4 CuSO4 + SO2 + 2H2OMol: y 2y y A c Fe2(SO4)3; CuSO4 v H2SO4 d. Khi p ht Cu2+ th c 2 p sau(Fe3+ p trc Cu2+):

2Fe2(SO4)3 + 2H2O in phn 4FeSO4 + 2H2SO4 + O2mol: 0,5x x

CuSO4 + H2O in phn Cu + H2SO4+O2Mol: y y

+ Da vo thi gian suy ra s mol e trao i trong qu trnh p =It

F= 0,056 mol

0,5x.2 + 2y = 0,056 hay x + 2y = 0,056 (I) Dung dch B c: FeSO4 v H2SO4. Khi p vi thuc tm th:

10FeSO4 + 2KMnO4 + 8H2SO4 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2Omol: x 0,2x 0,2x = 0,004 (II)+ T (I v II) ta c: x = 0,02 mol v y = 0,018 mol%mFe = 49,3% Cu = 50,7%.2/ S mol H2SO4 p = 3x + 2y = 0,096 mol s mol axit ban u = 0,96 mol V = 52,17 mlBi 19.Cho 20,0 gam hn hp gm mt kim loi M v Al vo 200 ml dung dch hn hp gm H 2SO4 v HCl (Smol HCl gp 3 ln s mol H2SO4) th thu c 11,2 lt kh H 2 (ktc) v vn cn d 3,4 gam kim loi. Lc lphn dung dch ri em c cn th thu c mt lng mui khan. Bit M c ho tr II trong cc mui ny.

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 121. Hy vit phng trnh phn ng di dng ion rt gn.2. Tnh tng khi lng mui khan thu c v nng mol/l ca mi axit trong dung dch.3. Xc nh kim loi M, bit s mol mi kim loi tham gia phn ng bng nhau.

1. Hy vit phng trnh phn ng di dng ion rt gn:M + 2H+ M2+ + H2 (1)2Al + 6H+ 2Al3+ + 3H2 (2)H2SO4 2H+ + SO42

a(mol) 2a aHCl H+ + Cl

3a(mol) 3a 3a

2. Tnh tng khi lng mui khan thu c v nng mol/l ca mi axit trong dung dch:Gi a l s mol ca H2SO4 v 3a l s mol ca HCl

+

=

=

=

24

H 5a(mol)

SO a(mol)

Cl 3a(mol)

Gt: Kim loi cn d nn axit phn ng ht.

(1), (2) + = = = = =2HH

11,2n 5a 2n 2. 1 a 0,2(mol)

22,4

= =M(HCl)0,6

C 3M0,2

= =2 4M(H SO )

0,2C 1M

0,2

3. Xc nh kim loi M:Theo nh lut bo ton khi lng:

= + + =muoi5a

m (20 3,4) (98a 36,5.3a) 2 57,1(gam)2

Gi x l s mol ca kim loi M v Al:

(1), (2) + = =3

x x 0,5 x 0,2(mol)2

Mx + 27x = 20 3,4 = 16,6 M = 56 (Fe)Bi 20. Ha tan hn hp rn (gm Zn, FeCO3, Ag) bng dd HNO3 (long, d) thu c hn hp kh A gm kh khng mu c t khi so vi hiro l 19,2 v dung dch B. Cho B phn ng vi dung dch NaOH d, lc kta to thnh v nung n khi lng khng i c 2,82 gam cht rn. Bit rng mi cht trong hn hp chkh HNO3 to thnh mt cht.

1. Lp lun tm kh cho.2. Tnh khi lng mi cht trong hn hp ban u (bit trong hn hp s mol Zn = s mol FeCO3).

Bi 21. 1. Cho 10,40 gam hn hp X (gm Fe, FeS, FeS 2, S) tc dng vi dung dch HNO3 c nng d thc V lt kh NO2 (l sn phm kh duy nht, o ktc) v dung dch A. Cho A tc dng vi dung dcBa(OH)2 d thu c 45,65 gam kt ta.a) Vit cc phng trnh ho hc ca cc phn ng xy ra di dng phng trnh ion.b) Tnh V v s mol HNO3 trong dung dch cn dng oxi ha hon ton hn hp X.a) Trong hai kh chc chn c CO2 = 44 vC. V AM = 38,4 < MCO2 nn kh cn li c M < 38,4 vC. V l khkhng mu nn l NO hoc N2

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12+ Do Ag l kim loi yu nn khng th kh HNO3 xung sn phm ng vi s oxi ha thp nh nit, amoninitrat nn kh cn li ch c th l NO.+ V mi cht trong hh ch kh HNO3 n mt cht kh nht nh nn Zn s kh HNO3 xung NO hoc NH4NOb) Gi x l s mol Zn s mol FeCO3 = x, gi l s mol Ag= y.+ Nu ch c Zn cng kh HNO3 to ra kh NO th ta c:

3Zn + 8HNO3 Zn(NO3)2 + 2NO + 4H2Omol: x 2x/3

3Ag + 4HNO3 3AgNO3 + NO + 2H2Omol: y y/3

3FeCO3 + 10HNO3 3Fe(NO3)3 + 3CO2 + NO + 5H2Omol: x x x/3

Kh to thnh c: x mol CO2 v3x y

3

+mol NO.

+ V hh kh c t khi so vi hiro l 19,2 nn s mol CO2 = 1,5.nNO

x =3x y

1,5.3

+ y = -x (loi) (1,0 im)

sm phm kh phi c NH4NO3 l sp kh ng vi Zn do ta c:

4Zn + 10HNO3 4Zn(NO3)2 + NH4NO3 + 3H2Omol: x x x/43Ag + 4HNO3 3AgNO3 + NO + 2H2O

mol: y y y/33FeCO3 + 10HNO3 3Fe(NO3)3 + 3CO2 + NO + 5H2O

mol: x x x x/3

kh to thnh c x mol CO2 vx y

3

+mol NO. V s mol CO2 = 1,5. nNO

x = y+ Khi B + NaOH d v nung th cht rn ch c:

Fe(NO3)3 NaOH Fe(OH)30t 0,5 Fe2O3

AgNO3 NaOH 0,5Ag2O0t Ag

0,5x mol Fe2O3 + y mol Ag. V x = y nn ta c:80x + 108x = 2,82 x = 0,015 mol.

Vy c 3 cht trong hh cho u c s mol l 0,015 mol.Do : mZn = 0,975 gam; mFeCO3 = 1,74 gam v mAg = 1,62 gam. (1,5 im)a)Cc phng trnh phn ng: (1,0 im)Fe + 6H+ + 3NO3- Fe3+ + 3NO2+ 3H2O (1)FeS + 10 H+ + 9NO3- Fe3+ + SO42- + 9NO2 + 5H2O (2)FeS2 + 14H+ + 15NO3- Fe3+ + 2SO42- + 15NO2 + 7H2O (3)S + 4H+ + NO3- SO42- + 6NO2 + 2H2O(4) (4)

Dung dch sau phn ng c: Fe3+, SO42-, H+H+ + OH- H2OFe3+ + 3OH- Fe(OH)3Ba2+ + SO42- BaSO4b) Coi hn hp gm Fe v S ta c s : (2,0 im)

3 2

3

3

( )

24

4

( )

aSOdHNO Ba OH

Fe OHFe Fexmol xmolxmolS BSOymol ymolymol

+

+ +

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Theo bi ra ta c h:56 32 10,4 0,1

107 233 45,65 0,15

+ = =

+ = =

x y x mol

x y y mol

p dng nh lut bo ton eletron ta c:Fe Fe+3 + 3e0,1mol 3.0,1molS S+6 + 6e

0,15mol 6.0,15molN+5 + 1e N+4

a.1mol a molp dng nh lut bo ton e ta c:a = 0,3 + 0,9 = 1,2 mol V = 1,2.22,4 = 26,88 ltTheo (1) v (4):

3

6. 4 1, 2+= = + =HNO Fe SHn n n n mol

Bi 22.Ho tan hon ton hn hp MgCl2, FeCl3, CuCl2 vo nc ta c dung dch A. Cho t t dng kh Hvo A cho n d th thu c lng kt ta nh hn 2,51 ln lng kt ta to ra khi cho dung dch Na 2S d v

dung dch A.Tng t, nu thay FeCl3 trong A bng FeCl2 vi khi lng nh nhau (c dung dch B) th lng k

ta thu c khi cho H2S vo B nh hn 3,36 ln lng kt ta to ra khi cho dung dch Na2S vo B.Vit cc phng trnh phn ng v xc nh thnh phn phn trm khi lng mi mui trong hn h

ban u.p n

Gi x, y, z ln lt l s mol CuCl2 , MgCl2 , FeCl3.* Tc dng vi dung dch Na2S

CuCl2 + Na2S CuS + 2NaClMgCl2 + Na2S + 2H2O Mg(OH)2+ H2S + 2NaCl2FeCl3 + 3Na2S 2FeS + S + 6NaCl (0,25 )

* Tc dng vi dung dch H2SCuCl2 + H2S CuS + 2HCl2FeCl3 + H2S 2FeCl2 + 2HCl + S (0,25 )MgCl2 + H2S khng xy ra

-Nu thay FeCl3 bng FeCl2 cng khi lng :* Tc dng vi dung dch Na2S

CuCl2 + Na2S CuS + 2NaClMgCl2 + Na2S + 2H2O Mg(OH)2 + H2S + 2NaCl

FeCl2 + Na2S FeS + 2NaCl (0,25 )* Tc dng vi dung dch H2S

CuCl2 + H2S CuS + 2HCl

96x + 88z + 32.z

2+ 58y = 2,51

z96x + 32.

2

(1) (0,25 )

S mol FeCl2 =127

5,162 z(0,25 )

96x + 58y +127

5,162 z.88 = 3,36.96x (2) (0,25 )

Ta c: y = 0,664x v z = 1,67x (0,25 )%MgCl2 = 13,45 ; %FeCl3 = 57,80 v %CuCl2 = 28,75 (0,25 )

Bi 23. Trong bnh kn dung tch 2,112 lt cha kh CO v mt lng hn hp bt A gm Fe3O4 v FeCO3

27,30

C p sut trong bnh l 1,4atm (th tch cht rn khng ng k). Nung nng bnh nhit cao cc ph

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ng xy ra hon ton. Hn hp sau phn ng c t khi so vi H2 l27

554. Ha tan hon ton hn hp A trong d

HNO3 long, thu c3

792,1lt hn hp kh gm NO v CO2 ktc. Tnh th tch dd HCl 2M ha tan ht

hn hp A.

1. )mol(12,0)3,27273(x082,0

4,1x112,2nCO =

+

=

Gi x, y l s mol Fe3O4, FeCO3 trong hn hp ACc ptp:

Fe3O4 + 4CO = 3Fe + 4 CO2 (1)

x 4x 4x

FeCO3 + CO = Fe + 2 CO2 (2)y y 2y

Hn hp sau phn ng (1) v (2): n hn hp = n CO2+ n CO d = 4x + 2y + 0,12 - (4x + y) = 0,12 + y

4127

554x2M NO,2CO =

Ha tan A trong HNO3 long: 3Fe3O4 + 28HNO3 = 9Fe(NO3)3 + NO + 14H2O (3)x x/3

3FeCO3 + 10HNO3 = 3Fe(NO3)3 + NO + 3CO2 + 5H2O (4)y y/3 y

T (3) v (4):3

08,0y

3

y

3

xn NO,2hhCO =++=

T ta c h phng trnh

+=++

=+

)y12,0(41)yx412,0(28)y2x4(44

08,0yx

Gii h, ta c:

==

015,0y

02,0x

Ptp ha tan A trong dd HCl l:Fe3O4 + 8HCl = 2FeCl3 + FeCl2 + 4H2O (5)0,02 mol 0,16 molFeCO3 + 2HCl = FeCl2 + CO2 + H2O (6)0,015 mol 0,03 mol

mol19,016,003,0nHCl =+=

mol095,0219,0VddHCl ==

Bi 24.Cho 4,93 gam hn hp gm Mg v Zn vo cc cha 215 ml dung dch H 2SO4 1M (long). Sau khi phnng hon ton thm tip vo cc 0,6 lt dung dch hn hp gm Ba(OH)2 0,05M v NaOH 0,7M. Khuy u chophn ng hon ton, lc ly kt ta ri nung n khi lng khng i th thu c 13,04gam cht rn.

a. Vit cc phng trnh phn ng xy ra (i vi cc phn ng xy ra trong dung dch cn vit phngtrnh dng ion thu gn).

b. Tnh khi lng mi kim loi trong hn hp ban u.Cc phn ng xy ra

2H+ + Mg Mg2+ + H2 (1)2H+ + Zn Zn2+ + H2 (2)

H+

d + OH-

H2O (3)

t0

t0

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Ba2+ + SO42- BaSO4 (4)Mg2+ + 2OH-Mg(OH)2 (5)Zn2+ + 2OH- Zn(OH)2 (6)

nu OH- d Zn(OH)2 + 2OH- ZnO22- +2 H2O (7)Mg(OH)2

0t MgO + 2 H2O (8)

Zn(OH)20

t ZnO + 2 H2O (9)

* S mol H2SO4 : 0,215x1 =0,215 mol H+: 0,43 mol ; SO42- : 0,215 mol

4,93/65 < nkim loi < 4,93/240,0758 < nkim loi < 0,2054

(1) v (2) s mol H+ phn ng 2. 0,2054 = 0,4108 < s mol H+ban u H+ cn d, 2 kim loi ht.* Dung dch baz c: OH-: 0,48 mol ; Ba2+: 0,03 mol ; Na+:0,42 mol ; SO42- : 0,215 molt x: s mol MgMg2+ : x molMgO : x mol

y: s mol Zn Zn2+ : y molBaSO4 : 0,03 mol

s mol OH- p = s mol H+ = 0,43 mol s mol OH- d = 0,48 0,43 = 0,05 mol p (7) xy raRn thu c sau phn ng: BaSO4 , MgO, c th ZnO nu Zn(OH)2 khng ha tan ht.Xt 2 trng hpTH1 : Rn thu c BaSO4 , MgOmBaSO4 = 0,03. 233=6,99gmMgO = 13,04 6,99=6,05g mMg = 0,15125. 24 = 3,63g

mZn = 4,93 3,63 = 1,3 gTH2: Rn thu c BaSO4 , MgO, ZnO0,03. 233+ 40x + (y 0,025)81 =13,04 (10)24x + 65y = 4,93 (11)

x = 0,191 ; y = 0,00518Theo (6) nu s mol Zn(OH)2 = y duOHn lm tan ht Zn(OH)2 = 2y = 2. 0,00518=0,01036 mol < duOHn =0,05 mol v l.Vy trng hp 1 chp nhn.Bi 25. Cho 5,15 gam hn hp A gm Zn v Cu vo 140 ml dung dch AgNO3 1M. Sau khi phn ng xong thuc 15,76 gam hn hp kim loi v dung dch B. Chia B thnh hai phn bng nhau, thm KOH d vo phn 1,thu c kt ta. Lc kt ta, em nung n khi lng khng i, c m gam cht rn.

a. Tnh m?b. Cho bt Zn ti d vo phn 2, thu c dung dch D. Cho t t V lt dung dch NaOH 2M vo dung

dch D thu c 2,97 gam kt ta. Tnh V, cc phn ng xy ra hon ton.a. V sau phn ng thu c hn hp kim loi nn c th gm 2 trng hp sau:

+ Trng hp 1: AgNO3 ht, Zn cn d, Cu cha phn ng ( hn hp KL gm: Zn d, Cu, Ag ).Gi nZn, n Cu (hhA) l x v y, nZn phn ng l a ( mol ).Zn + 2AgNO3 Zn(NO3)2 + 2Ag (1)a 2a a 2a

mA = 65x + 64y = 5,15 (I); mKL = 65(x-a) + 64y + 108. 2a = 15,76 (II)nAgNO3 = 2a = 0,14 (III). H phng trnh I, II, III v nghim (loi).+ Trng hp 2: Zn ht, Cu phn ng mt phn, AgNO3 ht. gi n Cu phn ng l b (mol).

Zn + 2AgNO3 Zn(NO3)2 + 2Ag (1)x 2x x 2x

Cu + 2AgNO3 Cu(NO3)2 + 2Ag (2)b 2b b 2b

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12mA = 65x + 64y = 5,15 (I); mKL = 64(y-b) + 108( 2x + 2b ) = 15,76 (II)nAgNO3 = ( 2x + 2b ) = 0,14 (III). Gii h phng trnh I, II, III ta c: x = 0,03, y = 0,05, b = 0,04.+ Trong mi phn c: 0,015 mol Zn(NO3)2 v 0,02 mol Cu(NO3)2 .

Zn(NO3)2 Zn(OH)2 K2ZnO2.Cu(NO3)2 Cu(OH)2 CuO.0,02 0,02 m = 0,02.80 = 1,6 gam.

b. Zn + Cu(NO3)2 Zn(NO3)2 + Cu (1)0,02 0,02+ nZn(NO3)2 (dd D) = 0,015 + 0,02 = 0,035. C th gm 2 trng hp sau:+ Trng hp 1: Zn(NO3)2 d.

Zn(NO3)2 + 2NaOH Zn(OH)2 + 2Na(NO3) (2)0,06 0,03

V = 0,06/2 = 0,03 lt.+ Trng hp 2: Zn(NO3)2 ht.

Zn(NO3)2 + 2NaOH Zn(OH)2 + 2Na(NO3) (2)0,035 0,07 0,035

Zn(OH)2 + 2NaOH Na2ZnO2 + 2H2O (3)

0,005 0,01+ nNaOH = 0,07 + 0,01 = 0,08 V = 0,08/2 = 0,04 lt.

Tun 6. n tp mt s dng bi ton v c: kim loi kim, kim th, nhm, bi ton in phn.

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Tun 7. n tp cc phn ng trong ho hu c

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Tun 8

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Tun 9

Tun 8. Cc dng bi tp v l thuyt phn ng trong ho hu c.

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Hai ng phn X, Y ch cha cc nguyn t C, H, O trong H chim 2,439% v khi lng. Khi t chy Xhoc Y u thu c s mol H2O bng s mol mi cht. Hp cht Z c khi lng phn t bng khi lng pht ca X v cng cha C, H, O. Bit 1 mol X hoc Z tc dng va vi 3 mol AgNO3 trong dung dch NH3,mol Y tc dng va vi 4 mol AgNO3 trong dung dch NH3.Tm cng thc phn t ca X, Y, Z, bit rng chng u c mch cacbon khng phn nhnh.Vit phng trnh phn ng ha hc xy ra.(HSG H TNH 2012)Cu 9:

a. Hp cht A cha vng benzen, c cng thc phn t l C9H11NO2. A phn ng c vi dung dch NaOH vdung dch HCl. Bit rng:A tc dng c vi HNO2 sinh ra B ( C9H10O3). un nng B vi axit H2SO4 c c cht C (C9H8O2), C ph

ng vi dung dch KMnO4 trong mi trng H2SO4 long, un nng to ra cht D (C8H6O4), D c tnh i xncao.Xc nh cng thc cu to ca A, B, C, D.b. Hp cht A c cng thc C2H8N2O3. Cho 2,16 gam A tc dng vi dung dch NaOH d, un nng c dundch B ch cha cc cht v c v kh D c mi c trng. Vit cng thc cu to c th c ca A, tnh khlng mui c trong B. .(HSG H TNH 2012)

1) Cho dy phn ng sau:

A

X

B

Y

C

Z

D E

M + Cl2, as

1 : 1 (mol)

+ dd NaOH

+ dd NaOH

+ O2,

Cu, t0 + dd AgNO3/NH

3, t0

+ H2SO4, t

0

+ H2SO4, t0

- H2OPolistiren ?

xt, t0, p

(1)(2) (3) (4) (5)

(6)

(7) (8) (9)

t0cao

(10)

Cho bit cng thc cu to thu gn ca cc cht ng vi cc ch ci M, A, B, X, Y trong dy phn nVit phng trnh phn ng (4), (5), (9), (10)?

Cu 1. (3,0 im)

t chy hon ton 1,60 gam mt este n chc E thu c 3,52 gam CO2 v 1,152 gam nc

a. Tm cng thc phn t ca E.b. Cho 10 gam E tc dng vi NaOH va , c cn dung dch sau phn ng thu c 14 gam cht rn kha

G. Cho G tc dng vi dung dch H2SO4 long thu c G1 khng phn nhnh. Tm cng thc cu to ca E , vi

cc phng trnh phn ng

c. X l mt ng phn ca E, X tc dng vi NaOH to ra mt ancol m khi t chy hon ton mt th tc

hi ancol ny cn 3 th tch kh O2 o cng iu kin (nhit v p sut). Xc nh cng thc cu to v g

tn ca X

Cu 3 . (4,0 im)

1. Cht X c cng thc phn t C8H15O4N. T X c hai bin ha sau:

C8H15O4N 0t,OHdungdichNa C5H7O4NNa2 + CH4O + C2H6O

C5H7O4NNa2 ldungdichHC C5H10O4NCl + NaCl

Bit : C5H7O4NNa2 c mch cacbon khng phn nhnh v c nhm NH2 v tr . Xc nh cng thc cu to c th

c ca X v vit phng trnh ha hc ca cc phn ng theo hai bin ha trn di dng cng thc cu to.

2. Hp cht A c cng thc C9H8 c kh nng kt ta vi dung dch AgNO 3 trong NH3 v phn ng vi bro

trong CCl4 theo t l mol 1:2. un nng A vi dung dch KMnO 4 ti khi ht mu tm, ri thm lng d dun

dch HCl c vo hn hp sau phn ng thy c kt ta trng l axit benzoic ng thi gii phng kh CO 2 v C

Xc nh cng thc cu to ca A v vit phng trnh ha hc ca cc phn ng xy ra.

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Cu 5 . (3,0 im)

1. T kh thin nhin v cc cht v c cn thit, thit b phn ng y . Hy vit phng trnh iu ch c

cht sau : mH2NC6H4COONa v pH2NC6H4COONa

2. Hai hp cht thm A v B l ng phn c cng thc phn t C nH2n-8O2. Hi B c khi lng ring 5,44

gam/lt ( ktc). A c kh nng phn ng vi Na gii phng H 2 v c phn ng trng gng. B phn ng

vi NaHCO3 gii phng kh CO2.

a) Vit cng thc cu to ca A v B.

b) Trong cc cu to ca A, cht A1 c nhit si nh nht. Hy xc nh cng thc cu to ng ca A 1.c) Vit phng trnh phn ng chuyn ha ocrezol thnh A1.

Cu IV (5,25 im)1.Hai cht A, B c cng cng thc phn t C5H12, tc dng vi Cl2 theo t l mol 1:1 c chiu sng th

ch to ra 1 dn xut monoclo duy nht, B to ra 4 dn xut monoclo. Vit cng thc cu to ca A, B v dxut clo.

2. Mt hp cht A c cng thc phn t C6H6 khi tc dng vi dung dch AgNO3 trong NH3 to ra hcht B. Khi lng mol phn t ca B ln hn ca A l 214 vc. Vit cng thc cu to v gi tn A theo danphp IUPAC.

3. Cho s phn ng sau

Mi ch ci ng vi mt cht hu c, mi mi tn 1 phn ng, ch c dng thm cc cht v c; xtc cn thit vit phng trnh phn ng thc hin s trn.

4.Thy phn hon ton 0,5 mol peptit (A) th thu c cc - amino axit l: 1,5 mol Glyxin, 0,5 mAlanin, 0,5 mol Valin. Khi thy phn khng hon ton (A), ngoi thu c cc amino axit th cn thy c ipeptit: Ala-Gly; Gly- Ala v 1 tripeptit Gly-Gly-Val.

a) Hy vit cng thc cu to cc

- amino axit.b) Hy xc nh trnh t cc -amino axit trong A.

Bi 8: 1/ Vit cng thc cu to ca 5 cht hu c m trong mi phn t ch c 2 nguyn t hiro u phn ngc vi dung dch AgNO3/NH3? Vit phn ng xy ra?2/ Bn cht hu c A, B, C, D c cng CTPT l C4H4O4 cha hai nhm chc u phn ng c vi dung dchNaOH trong :

+ A, B to ra mui v nc, B c ng phn hnh hc+ C to ra mui v ancol+ D to ra mui, anehit v nc.Tm CTCT ca 4 cht trn v vit phn ng xy ra?

/ CTCT ca 5 cht l:CHCH; CHC-CCH; HCHO; HCOOH; O=HC-CH=O.+ P xy ra:CHCH + 2AgNO3 + 2NH3 CAgCAg + 2NH4NO3.CHC-CCH + 2AgNO3 + 2NH3 CAgC-CCAg + 2NH4NO3.HCHO + 4AgNO3 + 6NH3 + 2H2O (NH4)2CO3 + 4NH4NO3 + 4AgHCOOH + 2AgNO3 + 3NH3 + H2O (NH4)2CO3 + 2NH4NO3 + 2AgO=HC-CH=O + 4AgNO3 + 6NH3 + 2H2O NH4OOC-COONH4 + 4NH4NO3 + 4Ag2/ + A v B l axit c CTCT: CH2=C(COOH)2 v HOOC-CH=CH-COOH+ C l este vng:

CH4 A BC D

D E

F CH4

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12COO CH2

COO CH2

C p to ra: NaOOC-COONa + HO-CH2-CH2OH.+ B l HOOC-COO-CH=CH2.

+ P xy ra:.

Tun 9,10,11.Bi tp v cc hp cht hu c: hirocacbon, ancol, phenol, ax cacboxylic, este,cacbohirat,amin, aminoax, peptit.

. Khi tin hnh este ha mt 1 mol axit CH3COOH vi 1 mol ancol C2H5OH th hiu sut phn ng este ha cc i 66,67 %. hiu sut este ha t cc i 80% th cn tin hnh phn ng este ha 1 mol axit vi banhiu mol ancol.

Cu 4: (2,0 im)-V.P 2012

Cho 2,760 gam cht hu c A (cha C, H, O v c 100 < MA< 150) tc dng vi dung dch NaOH va sau lm kh, phn bay hi ch c nc, phn cht rn khan cn li cha hai mui ca natri c khi ln4,440 gam. Nung nng 2 mui trong oxi d, sau khi phn ng xy ra hon ton thu c 3,180 gam Na 2CO2,464 lt CO2 ( ktc) v 0,900 gam nc.

Xc nh cng thc phn t v cng thc cu to ca A.

Cu 5: (1,0 im) V.P 2012

Khi thy phn khng hon ton mt peptit A c khi lng phn t 293 thu c 2 peptit B v C. M0,472 gam peptit B phn ng va vi 18 ml dung dch HCl 0,222 M khi un nng v mu 0,666 gam peptit Cphn ng va vi 14,7 ml dung dch NaOH 1,6% (khi lng ring l 1,022 g/ml) khi un nng. Xc nhcng thc cu to v gi tn A, bit rng khi thy phn hon ton A thu c hn hp 3 amino axit l glyxin,

alanin v phenyl alanin. - n HCl = 0,018 0,222 0,004 (mol) ; nNaOH = 1, 6 1, 022 14, 7100 40

(mol)

- m N (A) = 29314,3

100= 42 => trong (A) c 3 nguyn t N

=> 2 peptit B v C l 2 ipeptit (0,5 )* Xt phn ng B + dung dch HCl :H2N-R-CO-NH-R-COOH + 2HCl + H2O ClH3N-R-COOH + ClH3N-R-COOH

=> nB =1

2nHCl = 0,002 (mol) => MB =

0,472

0,002= 236 (g/mol)

=> R + R = 132+ Nu R = 14 (CH2) => R = 118+ Nu R = 28 (CH3CH R = 104 (C6H5CH2CH).

** Xt phn ng C + dung dch NaOHH2N-R1-CO-NH-R1-COOH + 2NaOH H2N-R1-COONa + H2N-R1-COONa + H2O

=> nC =1

2nNaOH = 0,003 (mol) => MC =

0,666

0,003= 222 (g/mol)

=> R1 + R1 = 118+ Nu R1 = 14 (CH2) => R1 = 104 (trng vi kt qu ca B )+ Nu R1 = 28 (CH3CH R1 = 90 (loi)

=> B l CH3 CH(NH2)CONH CH(CH2-C6H5)COOH

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12=> C l NH2 CH2CONH CH(CH2-C6H5)COOH (1,0 )Vy A c 2 cu to:

NH2 CH2CONH CH(CH2-C6H5) CONHCH(CH3)COOHGLY-PHE ALA

CH3 CH(NH2)CONH CH(CH2-C6H5) CONHCH2COOHALA PHE GLY

2. S mol O2 v N2 trong khng kh:

)mol(15,010020x

4,228,16n

2O==

)mol(6,0100

80x

4,22

8,16n

2N==

Gi n l s nguyn t C trung bnh trong 2 phn t aminoaxt CTPT chung l: NOHC 21n2n +

Phn ng chy : 222221n2n N21

OH2

1n2COnO

4

3n6NOHC +

++

+

+(1)

x xnGi x l s mol 2 aminoaxt, ta c: xnn

2CO=

Hn hp kh B gm: CO2, N2 cho B tc dng vi Ca(OH)2 d:

CO2 + Ca(OH)2 = CaCO3 + H2O (2) )mol(095,0

100

5,9nn

2)OH(Ca2CO===

Ta c h phng trnh:

=

=+

095,0xn

21,3x)47n14(

Gii h ta c x = 0,04 375,2n =CTPT ca 2 aminoaxt: C2H5O2N CTCT : H2N-CH2-COOH

C3H7O2N CTCT: H2N-C2H4-COOH

Gi a, b ln lt l s mol 2 aminoaxt

=+

=+

21,3b89a75

04,0ba

=

=

015,0b

025,0a

m C2H5O2N = 75 x 0,025 = 1,875 (g)

m C3H7O2N = 89 x 0,015 = 1,335 (g)

Hn hp B sau phn ng: nO2d = )mol(0375,0x4

3n615,0 =

nN2 = )mol(62,0

2

x6,0 =+

nCO2 = 0,095(mol)

nB = 0,037 + 0,62 + 0,095 = 0,7525 (mol)

p sut trong bnh: )atm(505,18,16x273

4,22x)5,136273(x7525,0P =

+=

2. Dng 16,8 lt khng kh ktc (O2 chin 20% v N2 chim 80% th tch) t chy hon ton 3,21 gam hnhp A gm 2 aminoaxt k tip nhau c cng thc tng qut CnH2n+1O2N. Hn hp thu c sau phn ng emlm kh c hn hp kh B, cho B qua dd Ca(OH)2 d thu 9,5gam kt ta. Tm CTCT v khi lng ca 2aminoaxt.

Nu cho kh B vo bnh dung tch 16,8 lt, nhit 136,50C th p sut trong bnh l bao nhiu? Cho bitaminoaxt khi t chy to kh N2.

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Cu 6: (1,0 im) V.P 2012

un nng hn hp gm 1 mol HCOOH; 1 mol CH 3COOH v 2 mol C2H5OH c H2SO4 c xc tc toC (tronbnh kn dung tch khng i) n trng thi cn bng th thu c 0,6 mol HCOOC2H5 v 0,4 mCH3COOC2H5. Nu un nng hn hp gm 1 mol HCOOH, 3 mol CH 3COOH v a mol C2H5OH iu kin nhtrn n trng thi cn bng th thu c 0,8 mol HCOOC2H5. Tnh a.

Cu 1 ( 2,5 ). Tin hnh ln men gim 200ml dung dch ancoletylic 5,750 thu c 200ml dung dch Y. L100ml dung dch Y cho tc dng vi Na d th thu c 60,648 lt H 2 (ktc). Tnh hiu sut ca phn ng l

men gim. (Bit2 5C H OH

d = 0,8 g/ml)

Cu 4 ( 3,5 ). un hn hp gm ancol A v axit B (u l cht c cu to mch h, khng phn nhnh) thu este X. t chy hon ton m gam X thu c 1,344 lt kh CO 2 (ktc) v 0,72 gam nc. Lng oxi cn dng 1,344 lt (ktc).a/ Tm cng thc phn t ca X, bit t khi hi ca X so vi khng kh nh hn 6.b/ Xc nh cng thc cu to ca A, B, X bit gia A, B v X c mi quan h qua s sau:

CxHy Q A M B XCu 4 (4,0 im).

Cho 2,76 gam cht hu c A (ch cha C, H, O v c cng thc phn t trng vi vi cng thc n gin nht) tdng vi dung dch NaOH (va ), sn phm thu c em lm bay ht hi nc, phn cht rn khan cn li l hmui ca natri c khi lng 4,44 gam. Nung nng hai mui ny trong oxi (d), sau khi phn ng xy ra hon tothu c 3,18 gam Na2CO3; 2,464 lt kh CO2 (ktc) v 0,9 gam H2O.

a) Tm cng thc phn t v vit cng thc cu to ca A tha mn cc tnh cht trn.b) Cht B l mt ng phn ca A, khi cho B tc dng vi lng d dung dch NaOH hoc vi lng d dun

dch NaHCO3 to ra sn phm khc nhau ln lt l C 7H4Na2O3 v C7H5NaO3. Vit cng thc cu to ca B vphng trnh ha hc ca cc phn ng xy ra.

Cu 7 ( 2,5 ). Khi thy phn hon ton 43,40 gam mt peptit X (mch h) thu c 35,60 gam alanin v 15,0gam glixin. Vit cng thc cu to c th c ca peptit X.

Bi 9: t chy ht 0,02 mol cht hu c A cn 21,84 lt khng kh (ktc). Sau phn ng, cho sn phm chgm CO2, H2O, N2 hp th ht vo bnh ng dung dch Ba(OH)2 d thy khi lng bnh tng ln 9,02 gam vc 31,52 gam kt ta. Kh thot ra khi bnh c th tch 17,696 lt (ktc).a/ Xc nh cng thc phn t ca A. Bit rng khng kh gm 20% oxi v 80% nit theo th tch?b/ Xc nh cng thc phn t ca A bit rng A khng lm mt mu brom trong CCl4 v A c hnh thnh cht hu c X v cht hu c Y, phn t khi ca X v Y u ln hn 50; khi X tc dng vi nc brom to kt ta trng. Mi quan h gia A v X, Y th hin trong cc s phn ng di y:

A + NaOH X + B + H2O (1)A + HCl Y + D (2)D + NaOH X + NaCl + H2O (3)B + HCl Y + NaCl (4)

Bi 10: un hn hp ru A vi axit B (u l cht c cu to mch h, khng phn nhnh) thu c este X. chy hon ton m gam X thu c 1,344 lt kh CO2 (ktc) v 0,72 gam nc. Lng oxi cn dng l

Bi 9: t chy ht 0,02 mol cht hu c A cn dng 21,84 lt khng kh (ktc). Sau phn ng, cho tonb sn phm chy gm CO 2, H2O, N2 hp th ht vo bnh ng dung dch Ba(OH)2 d thy khi lngbnh tng ln 9,02 gam v c 31,52 gam kt ta. Kh thot ra khi bnh c th tch 17,696 lt (ktc).a/ Xc nh cng thc phn t ca A. Bit rng khng kh gm 20% oxi v 80% nit theo th tch v coinh nit khng b nc hp th.b/ Xc nh cng thc phn t ca A bit rng A khng lm mt mu brom trong CCl4 v A c hnhthnh t cht hu c X v cht hu c Y, phn t khi ca X v Y u ln hn 50; khi X tc dng vinc brom to ra kt ta trng. Mi quan h gia A v X, Y th hin trong cc s phn ng di y:

A + NaOH X + B + H2O (1)

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12A + HCl Y + D (2)D + NaOH X + NaCl + H2O (3)B + HCl Y + NaCl (4)

Gii1/ Ta c KK = 0,975 mol; C = BaCO3 = 0,16 mol; H = 2H2O = 0,22 mol. O = 0,16.2 + 0,11 0,975/5 =0,04 mol; N = 2.(17,696/22,4-0,975.4/5) = 0,02 mol C:H:O:N = 8:11:2:1 A c dng (C8H11O2N)n.

+ p dng LBTKL ta tnh c mA = 3,06 gamMA = 3,06/0,02 = 153 vC n = 1+ Vy CTPT ca A l: C8H11O2N.2/ Lp lun suy ra A l CH3COONH3C6H5; X l C6H5NH2(anilin); Y l CH3COOH. Cc p xy ra:

Bi 10: un hn hp ru A vi axit B (u l cht c cu to mch h, khng phn nhnh) thu c esteX. t chy hon ton m gam X thu c 1,344 lt kh CO 2 (ktc) v 0,72 gam nc. Lng oxi cn dngl 1,344 lt (ktc).a/ Tm cng thc phn t ca X, bit t khi hi ca X so vi khng kh nh hn 6.b/ Xc nh cng thc cu to ca A, B, X bit gia A, B v X c mi quan h qua s sau:

CxHy Q A M B XGii

CTGN l C3H4O2 CTPT l (C3H4O2)n. V dX/kk< 6 nn n = 1 hoc 2.+ Vi n = 1 Ch c 1 CTCT l HCOO-CH=CH2 loi v ancol tng ng l CH2=CH-OH khngbn+ Vi n = 2 th CTPT l C6H8O2.2/ Da vo s trn th CxHy l xiclo-C3H6; A l CH2OH-CH2-CH2-OH; B lHOOC-CH2-COOH X l este vng c CTCT:

CH2

COO

COO

CH2

CH2

CH2

+ Tht vy p xy ra nh sau:

Thm NH3 d vo dd c 0,5 mol AgNO3 ta c dd A. Cho t t 3 gam kh X vo A n phn ng hon toc dung dch B v cht rn C. Thm t t HI n d vo B thu c 23,5 gam kt ta vng v V lt kh Y ktc thot ra. Bin lun tm X, khi lng cht rn C v th tch kh Y.+ V X p vi AgNO3/NH3 c cht rn C nn X l anehit hoc ank-1-in hoc HCOOH. Nu l ank-1-in th khicho HI vo B khng c kh thot ra X l anehit hoc HCOOH+ Khi cho HI vo B th ta c:

Ag+ + I- AgI

V s nAgI =23,5

235=0,1 mol s mol Ag+ cn li trong B l 0,1 mol; v c kh thot ra nn phi c 23CO

. Do

s mol Ag+ p vi kh X l 0,4 mol s mol X l 0,2 mol (HCOOH) hoc 0,1 mol (HCHO)

MX tng ng l 15 vC (3

0,2); 30 vC (

3

0,1). Ta thy ch c HCHO ph hp.

HCHO 3 3AgNO /NH (NH4)2CO3 + 4Ag0,1 0,1 0,4

23CO + 2H+ H2O + CO2

0,1 0,1+ Khi lng ca C= mAg = 43,2 gam;

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12th tch Y = 2,24 lt. Cu 5: (4,0 im).

Cho 2,64 gam mt este A vo mt bnh kn c dung tch 500 ml ri em nung bnh n 2730C, ton b esho hi th p sut bng 1,792 atm.

1. Xc nh cng thc phn t ca A v tnh nng mol/l ca dung dch NaOH cn thit thu phn hlng este ni trn, bit rng th tch dung dch NaOH l 50 ml.

2. Xc nh cng thc cu to ca A v tnh khi lng mui thu c sau phn ng (vi hiu sut phng l 100%).

1. Xc nh cng thc phn t ca A v tnh nng mol/l ca dung dch NaOH cn thit thu phn htlng este ni trn:Gi cng thc ca este A l CxHyOz

= = = =+ x y z

A C H O

1,792.0,5.273 2,64n 0,02(mol) M 132(gam/mol)

22,4(273 273) 0,02

Hay: 12x + y + 16z = 132V A l este nn l s chn.*Nu z = 2 12x + y = 100

x 1 2 3 4 5 6 7y 88 76 64 52 40 28 16Chn x = 7, y = 16.Vy cng thc phn t ca A l: C7H16O2

+

= =2.7 16 2

0(loai)2

*Nu z = 4 12x + y = 68

x 1 2 3 4 5y 56 44 32 20 8

Chn x = 5, y = 8.

Vy cng thc phn t ca A l: C5H8O4

+

= =2.5 8 2

2(nhan)2

nNaOH = 2.0,02 = 0,04(mol)

CM(NaOH) = = =M(NaOH)0,04

C 0,8M0,05

*Nu z = 6 12x + y = 36

x 1 2 3 4y 24 12 0 -12

Kt lun: loi

2. Xc nh cng thc cu to ca A v tnh khi lng mui thu c sau phn ng:*Trng hp 1:Este A c to bi t 2 axit n chc v 1 ancol hai chc:Cng thc ca A c dng:

R3

R1COO

R2COO

Ta c: C1 + C22 = 52 = 3 (vi 2 chc COO)V R1 R2 v ancol 2 chc c ti thiu 2C nn ancol hai chc l: CH2OHCH2OH

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 12Nn R1 l H (HCOOH) v R2 l CH3(CH3COOH)Vy cng thc cu to ca este l:

HCOO

CH3COO

CH2

CH2

HCOO

CH3COO

CH2

CH2

+2NaOH

t0HCOONa

CH3COONa

0,02

0,020,02

+ CH2OHCH2OH

= + =muoim 0,02.68 0,02.82 3(gam)

*Trng hp 2: Este A c to bi t 1 axit hai chc v 2 ancol n chc:R1 COOR2R3OOC

Ta c: R1 + R2 + R3 = 5 2 = 3V hai ancon ng ng k tip nhau v ancol c ti thiu 1CNn R2 l CH3 (CH3OH) v R3 l CH3CH2 (CH3CH2OH) v R1 = 0Vy cng thc cu to ca A l:

CH3OOC COOCH2CH3COOCH3

COOCH2CH3

+ 2NaOHt0 COONa

COONa

+

CH3OH

CH3CH2OH0,02 0,02

muoim 0,02.134 2,68(gam)= =

1.+ V s H gp i s C nn c A v B u c dng: CnH2nOx. Mt khc A, B p vi Na u cho lng hiro nhnhau nn A, B c cng s nhm OH.+ Ta thy A, B u c 1lin kt trong phn t nn 1 mol A hoc B ch p c vi 1 mol hiro theo gi thisuy ra khi 1 mol A hoc B p vi Na ch cho 0,5 mol hiro c A, B ch c 1 nhm OH. Vy A, B c CTPT ph hp vi mt cc trng hp sau: TH1: A l CnH2n-1OH (a mol); B l HO-CmH2m-CHO (b mol) TH2: A l HO-CnH2n-CHO (a mol); B l HO-CmH2m-CHO (b mol)+ ng vi trng hp 1 ta c h:

a(16 14n) b(14m 46) 33,8

5,60,5a 0,5b

22,4

13, 442b

22,4

+ + + =

+ =

= a = 0,2; b = 0,3 v 2n + 3m = 12 n = 3 v m = 2 tha mn+ ng vi trng hp 2 ta c h:

a(46 14n) b(14m 46) 33,8

5,60,5a 0,5b

22,4

13,442a 2b

22,4

+ + + =

+ =

+ =

a + b = 0,5 v a + b= 0,3 loi.

+ Vy A l: CH2=CH-CH2-OH v B l HO-CH2-CH2-CHO

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 122. phn ng vi thuc tm m sn phm thu c ancol a chc l cht A:3CH2=CH-CH2-OH + 4H2O+2KMnO4 3CH2OH-CHOH-CH2OH + 2MnO2 + 2KOHmol: 0,2 0,4/3 th tch dd KMnO4 = 1,33 lt.

CuV. (4,0 im)Hn hp X gm hai cht hu c A, B ch cha chc ancol v anehit. Trong c A, B s nguyn t H u g

i s nguyn t C, gc hirocacbon c th no hoc c mt lin kt i. Nu ly cng s mol A hoc B cho phng ht vi Na th u thu c V lt hiro cn nu ly s mol nh th cho phn ng ht vi hiro th cn 2V lCho 33,8 gam X phn ng ht vi Na thu c 5,6 lt hiro ktc. Nu ly 33,8 gam X phn ng ht vi AgNOtrong NH3 sau ly Ag sinh ra phn ng ht vi HNO3 c th thu c 13,44 lt NO2 ktc.

1. Tm CTPT, CTCT ca A, B?2. Cn ly A hay B khi phn ng vi dung dch thuc tm ta thu c ancol a chc? Nu ly lng A ho

B c trong 33,8 gam X th cn bao nhiu ml dung dch thuc tm 0,1M phn ng va vi X to ra ancolchc? CuV. (4,0 im)

Hn hp X gm hai cht hu c A, B ch cha chc ancol v anehit. Trong c A, B s nguyn t H u gi s nguyn t C, gc hirocacbon c th no hoc c mt lin kt i. Nu ly cng s mol A hoc B cho phng ht vi Na th u thu c V lt hiro cn nu ly s mol nh th cho phn ng ht vi hiro th cn 2V l

Cho 33,8 gam X phn ng ht vi Na thu c 5,6 lt hiro ktc. Nu ly 33,8 gam X phn ng ht vi AgNOtrong NH3 sau ly Ag sinh ra phn ng ht vi HNO3 c th thu c 13,44 lt NO2 ktc.

1. Tm CTPT, CTCT ca A, B?2. Cn ly A hay B khi phn ng vi dung dch thuc tm ta thu c ancol a chc? Nu ly lng A ho

B c trong 33,8 gam X th cn bao nhiu ml dung dch thuc tm 0,1M phn ng va vi X to ra ancol chc?1.+ V s H gp i s C nn c A v B u c dng: CnH2nOx. Mt khc A, B p vi Na u cho lng hiro nhnhau nn A, B c cng s nhm OH.+ Ta thy A, B u c 1lin kt trong phn t nn 1 mol A hoc B ch p c vi 1 mol hiro theo gi thisuy ra khi 1 mol A hoc B p vi Na ch cho 0,5 mol hiro

c A, B ch c 1 nhm OH. Vy A, B c CTPT ph hp vi mt cc trng hp sau: TH1: A l CnH2n-1OH (a mol); B l HO-CmH2m-CHO (b mol) TH2: A l HO-CnH2n-CHO (a mol); B l HO-CmH2m-CHO (b mol)+ ng vi trng hp 1 ta c h:

a(16 14n) b(14m 46) 33,8

5,60,5a 0, 5b

22,4

13,442b

22,4

+ + + =

+ =

=

a = 0,2; b = 0,3 v 2n + 3m = 12 n = 3 v m = 2 tha mn+ ng vi trng hp 2 ta c h:

a(46 14n) b(14m 46) 33,8

5,60,5a 0, 5b

22,4

13,442a 2b

22,4

+ + + =

+ =

+ =

a + b = 0,5 v a + b= 0,3 loi.

+ Vy A l: CH2=CH-CH2-OH v B l HO-CH2-CH2-CHO

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Trng THPT Nguyn Du Chuyn n thi hc sinh gii ha 122. phn ng vi thuc tm m sn phm thu c ancol a chc l cht A:3CH2=CH-CH2-OH + 4H2O+2KMnO4 3CH2OH-CHOH-CH2OH + 2MnO2 + 2KOHmol: 0,2 0,4/3 th tch dd KMnO4 = 1,33 lt.Cu 8 (4,0 im)

Cht hu c X c cng thc phn t trng vi cng thc n gin nht. t chy 5,2 gam X cn 5,04 ltoxi (kc), hn hp kh CO2 v hi H2O thu c c t khi so vi H2 bng 15,5. X tc dng c vi natri. Khi

un nng 5,2 gam X vi dung dch NaOH va , thu c 3,4 gam mui v cht hu c Y khng c kh nngha tan Cu(OH)2.

a. Tm cng thc phn t, cu to ca X, Y.b. Hy ngh s iu ch X v Y t cc hidrocacbon n gin nht tng ng (Khng qu 5

phn ng).

t CTPT ca X CxHyOzs mol oxi : 5,04/22,4 = 0,225 molmC02 + mH2O = 5,2 + 0,225. 32 = 12,4gKLPTTB ca (CO2 +H2O) = 15,5 .2 = 31.nCO2= nH2O = 0,2 mol.

nC = 0,2 mol ; nH = 0,4 mol ; nO2 = 0,15 molx:y:z = 0,2:0,4:0,15 = 4:8:3CTPT l C4H8O3X tc dng vi natri, chng t trong X c nhm OH ca axit hoc ancol.X tc dng vi NaOH to ra mui v cht hu c Y. Vy phn t X c 1 nhm chc este.t CTTQ l RCOOR

nRCOONa = nx = 0,05 mol nn MRCOONa = 3,4/0,05=68 gam.Vy RCOONa l HCOONaX c dng: HCOOC3H6OH v Y l C3H6(OH)2 thuc loi ancol no hai chc.V Y khng ha tan Cu(OH)2 nn Y phi c hai nhm OH cch xa nhau hay cu to Y l: CH2OH CH2 CH2OH

X l: HCOO CH2 CH2 CH2 OHiu ch

CH4 HCHO HCOOH

CH2 CH2CH2 BrCH2 CH2 CH2 Br HOCH2 CH2 CH2OH (Y)

HCOOH + HOCH2 CH2 CH2OH X

Tun 12,13,14. Hng dn hc sinh lm bi tp trc nghim v luyn mt s thi trc nghim.Tun 15. Hng dn hc sinh lm bi thi t lun v luyn tng hp.

Br2 dd NaOHdd

1:1

t0 ,xt

CD HSG 12 day du

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