200 deda hsg tinh khoi 1112

Tp 200 thi n thi hc sinh gii

62 Cu 1: Hon thnh s p sau:+ O2 + Y1 + Y2 O C 4H 6O 2 C 4H 6O 4 C 7H 12O 4 C 10H 18O 4 + H2 X 2+Y1+Y 2 xt H 2SO 4 H 2SO 4

(X1) (X2) (X3) (X4) Cho X1 l anehit a chc mch thng, Y2 l ancol bc II. Cu 2: A, B, D l cc ng phn c cng cng thc phn t C6H9O4Cl, tha mn cc iu kin sau : 36,1g A + NaOH d 9,2g etanol + 0,4 mol mui A1 + NaCl. B + NaOH d mui B1 + hai ru (cng s nguyn t C) + NaCl D + NaOH d mui D1 + axeton + NaCl + H2O. Hy lp lun xc nh cng thc cu to ca A, B, D v vit cc phng trnh phn ng. Bit rng D lm qu tm. Cu 3: 1/ Vit p iu ch t lapsan t metan v cc cht v c cn thit 2/ Mt dung dch monoaxit HA nng 0,373% c khi lng ring bng 1,00 g/ml v pH = 1,70. Khi pha long gp i th pH = 1,89. a/ Tm Ka ca axit trn? b/ Tm M v cng thc ca axit ny bit n c %KL ca H = 1,46%; O = 46,72% v mt nguyn t X cha bit vi % khi lng cn li. Cu 4: 1/ Cho E0 ca Fe2+/Fe = -0,44 vn, Fe3+/Fe = -0,04 vn. a/ Tnh E0 ca Fe3+/Fe2+? b/ Tnh E ca Fe3+/Fe2+ trong dd c [OH-] = 1M bit T ca Fe(OH)2 = 10-14 v Fe(OH)3 = 10-36. 2/ Ha tan ht 7,33 gam hh kim loi M ha tr II v oxit ca n thu c 1 lt dd X c pH = 13. a/ Tm M? b/ Tnh th thch dd cha HCl v H2SO4 c pH = 0 cn thm vo 0,1 lt dd X thu c dd mi c pH = 1,699. Cu 5: Cho 13,36 gam hh A gm Cu, Fe3O4 vo dd H2SO4 c nng d thu c V1 lt SO2 v dd B. Cho B p vi NaOH d c kt ta C, nung kt ta ny n khi lng khng i c 15,2 gam cht rn D. Nu cng cho lng A nh trn vo 400 ml dd X cha HNO3 v H2SO4 thy c V2 lt NO duy nht thot ra v cn 0,64 gam kim loi cha tan ht. Cc p xy ra hon ton v cc kh o ktc. 1/ Tnh V1, V2? 2/ Tnh CM mi cht trong X bit dung dch sau p ca A vi X ch c 3 ion(khng k ion H+ v OH- do nc phn li ra)? Cu 6: Trn hirocacbon kh A vi oxi theo t l th tch A:O2 = 1:9 ri cho vo bnh kn thy p sut trong bnh l 1 atm 00C. Bt tia la in A chy ht, hh sau p c p sut l 1,575 atm 136,50C. 1/ Tm CTPT ca A? 2/ Vit CTCT c th c ca A bit tt c cc nguyn t cacbon trong A u c cng mt dng lai ha? 3/ Chn CTCT ca A trn vit s tng hp cht B(Anthracen) c CTCT nh sau:

Bit ta phi dng p ix-An c dng(R, R, X, Y c th l H, gc hirocacbon, nhm chc): R R X CH X HC HCHC CH HC Y Y

R' R' Cu 7: Thm t t 17,85 ml dung dch km clorua 17% (d =1,12g/ml) vo 25 ml dung dch kali cacbonat 3,0 mol/lt (d = 1,30 g/ml) to ra kt ta cacbonat baz. Sau phn ng lc b kt ta, tnh nng % cc cht trong nc lc. Cu 8: Cho hn hp gm 25,6 gam Cu v 23,2 gam Fe3O4 tc dng vi 400 ml dung dch HCl 2M cho n khi phn ng hon ton thu c dung dch A v cht rn B. Cho dung dch A phn ng vi dung dch AgNO3 d tch ra kt ta D. Tnh lng kt ta D. 63 Cu 1: 2,808 gam mt ieste quang hot A ch cha C, H, v O c x phng ha vi 30 ml dd NaOH 1M. Sau khi x phng ha cn 6 ml dd HCl 1M ch chun NaOH d. Sp x phng ha gm mui ca axit icacboxylic B khng quang hot, CH3OH v mt ancol quang hot C. Ancol C p vi I 2/NaOH cho kt ta vng. iaxit B p ch p vi Br 2/CCl4 theo t l 1:1 v ch cho mt sp D duy nht. Ozon phn B ch cho mt sn phm. Vit CTCT ca A, B, C khng cn vit cng thc lp th v vit p xy ra?Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

1


Cu 2: Hon thnh s p sau:Ai A2 B D C2H2 X1 Y X1 X2

A3

CH3CHO

X3

Cu 3: 1/ So snh v gii thch tnh baz ca metylamin, amoniac, imetylamin, etylamin, anilin, iphenylamin. 2/ Nu cch phn bit: etylamin, ietylamin v trietylamin v isopropylamin. 3/ So snh v gii thch tnh axit ca cc axit sau: axit axetic, axit lactic, axit acrylic, axit propionic. Cu 4: Cho 2,16 gam hn hp gm Al v Mg tan ht trong dung dch axit HNO 3 long, un nng nh to ra dung dch A v 448 ml ( o 354,9 K v 988 mmHg) hn hp kh B kh gm 2 kh khng mu, khng i mu trong khng kh. T khi ca B so vi oxi bng 0,716 ln t khi ca CO 2 so vi nit. Lm khan A mt cch cn thn thu c cht rn D, nung D n khi lng khng i thu c 3,84 gam cht rn E. Vit phng trnh phn ng, tnh lng cht D v % lng mi kim loi trong hn hp ban u. Cu 5: Hn hp cha km v km oxit c ha tan ht bng dung dch HNO 3 rt long nhn c dung dch A v khng c kh bay ra. C cn cn thn dung dch A ri nung khan 210 0C n khi khng cn thot ra th thu c 2,24 lt kh (o 191,1 K v 7,1. 104 Pa) v cn li 113,4 gam cht rn kh. Hy xc nh khi lng mi cht trong hn hp u(1 Pa = 9,87.10-6 atm). Cu 6: thy phn hon ton 0,74 gam mt hn hp este n chc cn 7,0 gam dung dch KOH 8% trong nc. Khi un nng hn hp este ni trn vi axit H2SO4 80% sinh ra kh X. Lm lnh X, a v iu kin thng v em cn, sau cho kh li t t qua dung dch brom d trong nc th thy khi lng kh gim 1/3, trong khi lng ring ca kh gn nh khng i. a/ Tnh khi lng mol ca hn hp este, xc nh thnh phn hn hp kh sau khi lm lnh v tnh khi lng ca chng. b/ Xc nh thnh phn hn hp este ban u. c/ Nu phn ng phn bit 2 este trn, vit phng trnh phn ng. Cu 7: a) Ti sao trong cc phn t H2O, NH3 cc gc lin kt HOH (104,290) v HNH (1070) li nh hn gc t din 0 (109 ,28) ? b) Xt 2 phn t H2O v H2S ti sao gc HSH (92015) li nh hn HOH (104029)

c) Xt 2 phn t H2O v F2O ti sao gc FOF (103015) li nh hn HOH (104029) Cu 8: Trn CuO vi mt oxit kim loi n ha tr II theo t l mol 1:2 c hn hp A. Dn mt lung kh H2 d i qua 3,6 gam A nung nng thu c hn hp B. ha tan ht B cn 60 ml dung dch HNO3 nng 2,5M v thu c V lt kh NO duy nht(ktc) v dung dch ch cha nitat kim loi. Xc nh kim loi ha tr II ni trn v tnh V? 64 Cu 1: Xc nh sc in ng E0, hng s cn bng ca phn ng: Hg 2+ Hg + Hg2+. 2Cho : E0 (Hg2+/ Hg 2+ ) = + 0,92; E0 (Hg2+/ Hg) = + 0,85V v K = 2nE

10 0,059

Cu 2: Cho V lt CO qua ng s ng 5,8 gam oxit st nung mt thi gian thu c hh kh A v cht rn B. Cho B p ht vi HNO3 long thu c dd C v 0,784 lt NO. C cn C thu c 18,15 gam mui st (III) khan. Nu ha tan B bng HCl d th thy thot ra 0,672 lt kh(cc kh o ktc) 1/ Tm cng thc ca oxit st ? 2/ Tnh %KL mi cht trong B ? Cu 3: Ngi ta d tnh ho tan 10-3 mol Mg(NO3)2 trong 1 lt dung dch NH3 0,5M; trnh s to thnh kt ta Mg(OH)2 phi thm vo dung dch ti thiu bao nhiu mol NH4Cl? Cho KNH3 = 1,8.10-5; T Mg(OH)2 = 1,0.10-11+ HCl +KOH H 2 O Cu 4: Hon thnh s p sau: (CH)3 C CH = CH 2 A B C Sau hy ngh mt c ch gii thch s to thnh C ? Cu 5: Hon thnh s p sau:

Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

2


C A n-Butan 550 - 600 C A10 0

D

Axeton

B G B1Mg

1,4 - dibrom - 2 - buten D1CH2

C11) CH2

GlyxerintriNitrat

O B2 C2 D2 2) H3O+ ete khan

IsoamylAxetat

+ Bit p ca dn xut halogen vi Mg trong ete khan to ra hp cht c magi nh sau: ete khan RX + Mg RMgX + Mt vi p ca hp cht c magi:+ HCHO 2 RMgX RCH2OMgX RCH2OH MgX 2 - Mg(OH) 21 H3O RMgX RR1CHOMgX R-CHOH-R1(ancol bc II)

H O/H ++

+ R CHO

1 2 H3O RMgX RR1 R2COMgX RR1-COH-R2(ancol bc III)

+ R -CO-R

+

2 2 H3O RMgX RCH2 CH2OMgX R CH2CH2OH

+ (CH ) O = etylen oxit + CO

+

H 2 O/H 2 RMgX RCOOMgX RCOOH Cu 6: X,Y l kim loi n ha tr II v III. Ha tan ht 14,0 gam hn hp X, Y bng axit HNO3 thot ra 14,784 lt (27,30C v 1,1atm) hn hp 2 kh oxit c mu nu v c t khi so vi He = 9,56 , dung dch nhn c ch cha nitrat kim loi. Cng lng hn hp 2 kim loi trn cho tc dng vi axit HCl d th cng thot ra 14,784 lt kh (27,30C v 1atm) v cn li 3,2 gam cht rn khng tan. Xc nh X, Y v tnh % lng mi kim loi trong hn hp u. Cu 7: 1/ Dung dch A gm cc cation: NH4+ ; Na+ ; Ba2+ v 1 anion X c th l mt trong cc anion sau: CH3COO ; NO3; SO42 ; CO32 ; PO43 . Hi X l anion no? Bit rng dung dch A c pH = 5 . 2/ Thm NaOH d vo dung dch CuSO4, thm tip NH4NO3 vo dung dch n d c hin tng g xy ra? Vit phng trnh phn ng. Cu 8: t chy hon ton 2,7 gam cht hu c A phi dng va ht 4,76 lt oxi ktc, sp thu c ch c CO2 v H2O c khi lng ca CO2 H2O = 5,9 gam. 1/ Tm CTPT ca A bit MA < Mglucoz? 2/ Tm CTCT ca A bit A khng p vi Na, NaOH. Khi A p vi nc brom th thu c 2 sp B v C c CTPT l C7H7OBr. Ch r CTCT ca B v C bit %B ln hn C? 3/ Vit p iu ch A t metan v cc cht v c khc? 65 Cu 1: Cho 45,24 gam mt oxit st p ht vi 1,5 lt dd HNO3 long thu c dd A v 0,896 lt hh kh B gm NO v N2O. Bit t khi ca B so vi H2 l 17,625. Thm vo A m gam Cu, sau p thy thot ra 0,448 lt NO duy nht v cn li 2,88 gam kim loi khng tan. Cc kh o ktc. a/ Tm cng thc ca oxit st? b/ Tnh m v nng mol/l ca dd HNO3 ban u? c/ Sau khi lc b kim loi khng tan ri em c cn dung dch th thu c bao nhiu gam mui khan? Cu 2: A, B, C, D l nhng hirocacbon c CTPT C9H12. Bit A ch cha 2 loi hiro. un nng vi KMnO4 th A cho C9H6O6 , B cho C8H6O4, un nng C8H6O4 vi anhirit axeitc cho sp l C8H4O3. C v D u p vi Cu2Cl2/NH3 u cho kt ta mu v p vi dd HgSO4 sinh ra C9H14O(C cho M v D cho N). Ozon phn M cho nona-2,3,8-trion cn N cho 2axetyl-3-metylhexaial. Tm CTCT ca A, B, C, D v vit p xy ra bit ank-1-in p vi Cu2Cl2/NH3 u cho kt ta mu theo p: R-C CH + Cu2Cl2 + NH3 R-C CCu + NH4Cl Cu 3: Hon thnh s p sau bit X l C6H8O4. (1): X + NaOH A + B + C (7): C + AgNO3 + NH3 + H2O L + E + Ag (2): A + H2SO4 A1 + Na2SO4. (8): L + NaOH L1 + N + H2O

+

(3): A1 + AgNO3 + NH3 + H2O D + E + Ag (4): D + HNO3 E + F + H2O 0

CaO ,t (9): L1 + NaOH P + I (10): B + H2SO4 Q + Na2SO4. 0

0

H 2 SO4 ,t CaO ,t (5): A + NaOH I + H (11): Q Z + H2O (6): F + NaOH I + H2O Cho Z l axit acrylic Cu 4: Cho 7 gam hn hp A gm Fe v Cu vo 500 ml dung dch AgNO3. Sau p c dung dch A v 21,8 gam cht rn B. Thm NaOH d vo A ri nung kt ta sinh ra trong khng kh n khi lng khng i c 7,6 gam cht rn.


3


1/ Tnh %m mi kim loi? 2/ Tnh V dung dch HNO3 2M min cn ho tan ht 7 gam A bit to ra NO? Cu 5: Mt hn hp rn A gm kim loi M v mt oxit ca kim loi . Ngi ta ly ra 3 phn, mi phn c 59,08g A. Phn th nht ho tan vo dung dch HCl thu c 4,48 lt kh H2. Phn th hai ho tan vo dung dch ca hn hp NaNO3 v H2SO4 thu c 4,48 lt kh NO. Phn th 3 em nung nng ri cho tc dng vi kh H2 d cho n khi c mt cht rn duy nht, ho tan ht cht rn bng nc cng toan th c 17,92 lt kh NO thot ra. Cc th tch o ktc. Hy tnh khi lng nguyn t, cho bit tn ca kim loi M v cng thc oxit trong hn hp A.Na2CO3(9 ) (4 ) (1 ) (2 ) (1 ) 0 (5 ) (6 )

A(3 ) (7 )

(8 ) Cu 6: Hon thnh s p sau: Cu 7: Mt monotecpenoit mch h A c cng thc phn t C10H18O (khung cacbon hai n v isopren ni vi nhau theo qui tc u-ui). Oxi ho A thu c hn hp ht A1, A2 v A3. Cht A1 (C3H6O) cho phn ng iodofom v khng lm mt mu brm. Cht A2 (C2H2O4) phn ng c vi Na2CO3 v vi CaCl2 cho kt ta trng tan trong axit axetic; A2 lm mt mu dung dch KMnO4 long. Cht A3 (C5H8O3) phn ng iodofom v phn ng c vi Na2CO3. a. Vit cng thc cu to ca A1, A2 v A3. b. V cng thc cc ng phn hnh hc ca A v gi tn theo danh php IUPAC. Cu 8: 1/ Ha tan 69 gam hh CuCl2, FeCl3 theo t l mol 1:2 vo nc c dung dch A. in phn A vi in cc tr, thi gian in phn ht cc mui l T. Tnh tng khi lng catot khi in phn trong thi gian 0,5T; 0,7T.(Cho th t p ln l Fe3+ > Cu2+ > Fe2+). 2/ Hn hp X gm NaCl, NaHCO3, Na2CO3 trong c mt mui ngm nc. 61,3 gam X p va ht vi 100 ml dd HCl 4,5M thu c V lt CO2 ktc, dd A. Cho A vo 100 ml dd AgNO3 6,5M th va thu c kt ta max. Nu cho dd NaOH d vo X th c dd Y, cho tip dd Ba(NO3)2 d vo Y th thu c lng kt ta ln nht l 68,95 gam. Tnh V v %KL mi cht trong X? 66 Cu 1: Axit axety salixilic(cn gi l aspirin) c dng HA l axit yu c pKa = 3,49. tan trong nc ca n nhit phng l 3,55 gam/dm3. a/ Tnh pH ca dung dch aspirin bo ho nhit phng? b/ Tnh KL ti thiu NaOH cn ho tan 0,1 mol aspirin vo nc thnh 1 lit dd ? Tnh pH ca dd ny? c/ Vit p iu ch aspirin t metan v cc cht v c cn thit? Cu 2: 1/ Cho E0 ca Sn4+/Sn2+ = 0,15V v Sn2+/Sn = -0,14V. Gii thch ti sao khi ha tan Sn trong dd HCl d khng to ra Sn4+ m ch to ra Sn2+? 2/ Cho E0 ca Fe3+/Fe2+ = 0,771 V v Ag+/Ag = 0,799 V. a/ Vit p xy ra khi cho 2 cp trn p vi nhau? b/ Trn 50 ml dd AgNO3 0,01M vi 25 ml dd Fe(NO3)2 0,02M; 25 ml dd Fe(NO3)3 0,05M v bt Ag d. Tnh G ca p sau: Fe2+ + Ag+ Fe3+ + Ag. T kt qu cho bit chiu p?

B

C

(Cho G = - nF E; K = 10n.E / 0,059 vi n = s e trao i; F = 96500; nu G y > 0 thay vo (*) ta c:146,6 > M > 130,6 M l Ba. b/ S mol ca OH- = 0,1.0,1 = 0,01 mol; Gi V l th tch cn tm s mol H+ = 1.V mol. V pH ca dd sau p < 7 nn sau p axit d tnh theo OH-. H+ + OH- H2O Mol b: V 0,01 Mol p: 0,01 0,01 Mol cn: V-0,01 0

V 0, 01 =10-1,699 V = 0,0122 lt V + 0,1


7


64 x + 232 y = 13,36 x = 0,1 Cu 5: 1/ Gi x, y l s mol Cu v Fe3O4 ta d dng lp c h sau: 3y y = 0, 03 80 x + 160. 2 = 15, 2 p dng LBT electron V1 = 22,4.(0,1.2+0,03.1)/2= 2,576 lt + Khi cho A vo dd X th c p: 3Fe3O4 + 28H+ + NO3- 9Fe3+ + NO + 14H2O mol: 0,03 0,09 0,01 3+ 2+ 2+ Cu + 2Fe Cu + 2Fe Mol: 0,045 0,09 0,045 0,09

phi c: 0,1-0,045 -

0, 64 = 0,045 mol Cu p vi H+ v NO3- theo p: 64

3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O mol: 0,045 0,045 0,03 V2 = 22,4.(0,01 + 0,03) = 0,896 lt 2/ Ta thy s mol HNO3 = NO = 0,04 mol. Dung dch sau p ca A vi X c: 0,09 mol Fe2+ + 0,09 mol Cu2+ v a mol SO42-. p dng LBT in tch a = 0,18 mol. + Vy trong X c HNO3 = 0,1M v H2SO4 = 0,45M Cu 6: 1/ Gi CxHy l cng thc ca A ta c: CxHy +(x+ Mol: a

y y ) O2 xCO2 + H2O 4 2 y y a(x+ ) ax a 4 2

+ S mol kh trc p = a + 9a = 10a mol

y y y + 9a a(x+ ) = 9a + a 2 4 4 n1RT1 n 2 RT2 10a.273 (9a + 0, 25ay ).409,5 = = V bnh kn = hay: y = 6. P1 P2 1 1,575+ S mol kh sau p = ax + a + V A l cht kh nn A c th l: C2H6; C3H6 v C4H6. 2/V tt c cc nguyn t C trong A u c cng mt dng lai ha nn CTCT ca A: CH3-CH3(sp3); (CH2)3(xiclopropan = sp3); CH2=CH-CH=CH2(sp2) vCH CH2 CH2 CH

(sp3)

3/ Ta chn A l buta-1,3-ien hon thnh s : buta-1,3-ien butan etilen.CH2 CH2

+CH2 CH2

+

HC HC

Cho sp

cui cng trong s p vi H2/Ni, t0 th thu c B Cu 7: + Ta c: ZnCl2 = 0,025 mol; K2CO3 = 0,075 mol. + Qu trnh xy ra: CO32- + H2O HCO3- + OH-. 2Zn2+ + 2OH- + CO32- [Zn2(OH)2]CO3 Do ta c p xy ra dng phn t l: 2ZnCl2 + 3K2CO3 [Zn2(OH)2]CO3 + 2KHCO3 + 4KCl mol b: 0,025 0,075 0 0 0 mol p: 0,025 0,0375 0,0125 0,025 0,05 mol cn: 0 0,0375 0,0125 0,025 0,05 Dung dch nc lc c: 0,0375 mol K2CO3 + 0,025 mol KHCO3 + 0,05 mol KCl + Khi lng dd nc lc = 17,85.1,12 + 25.1,3 0,0125.224 = 49,692 gam. + Vy: K2CO3 = 10,4%; KHCO3 = 5,03%; KCl = 7,5%Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

8


Fe3O4 + 8HCl 2FeCl3 + FeCl2 + 4H2O 0,1 0,8 0,2 0,1 3+ 2+ 2+ Sau : Cu + 2 Fe Cu + 2 Fe 0,1 0,2 0,1 0,2 Khi dung dch A cha CuCl2 (0,1 mol) v FeCl2 (0,3 mol) Khi cho dung dch A phn ng vi dung dch AgNO3 d c cc phn ng: + Ag + Cl AgCl 0,8 0,8 + 2+ 3+ Ag + Fe Ag + Fe 0,3 0,3 khi lng D = AgCl v Ag = (0,8 143,5) + (0,3 108) = 147,2 g p n 63 Cu 1: + S mol NaOH p vi A = 0,024 mol s mol A = 0,012 MA =234 vC CTPT l C13H14O4. + Da vo sp ca p x phng ha suy ra A c dng: CH3OOC-R-COOR C l ROH, m ROH quang hot v cho kt ta vng vi I2/NaOH nn R c dng: CH3-CHOH-CH2-... + Do Ozon phn B ch cho 1 sp nn B phi c cu to i xng, do B ch p c vi 1 brom nn A, B, C l: CH3-OOCCH=CH-COOCH(CH3)C6H5; HOOC-CH=CH-COOH, CH3-CHOH-C6H5. Cu 2: A1 l C2H6 ; A2 l C2H5Cl ; A3 l C2H5OH ; X1 l C2H4 ; X2 l C2H4Cl2 ; X3 l C2H4(OH)2 ; B, D, Y l CH3COO-CH = CH2 ; CH2 = CH Cl ; CH3CHCl2 Cu 3: 1/ imetylamin > etylamin > metylamin > amoniac > anilin > iphenylamin. 2/ Dng HNO2 th: trietylamin khng p; ietylamin cho kt ta vng; 2 cht cn li cho ni bay ra. nhn ra 2 amin c kh th ta da vo ancol tng ng c to ra l bc I v II. Nhn bit 2 ancol ny bng CuO ri AgNO3/NH3. 3/ axit lactic > axit acrylic > axit axetic > axit propionic Cu 4: Theo gi thit th B cha N2 v N2O vi s mol u l 0,01 mol s mol e nhn to ra 2 kh ny l : 0,01(10+8) = 0,18 mol (I) + 2+ 5 Mg + 12 H + 2 NO 3 5 Mg + N2 + 6 H2O + 2+ 4 Mg + 10 H + 2 NO 3 4 Mg + N2O + 5 H2O + 3+ 10 Al + 36 H + 6 NO 3 10 Al + 3 N2 + 18 H2O + 3+ 8 Al + 30 H + 6 NO 3 8 Al + 3 N2O + 15 H2O c th c p to NH4NO3 + + 2+ 4 Mg + 10 H + NO 3 4 Mg + NH 4 + 3 H2O + + 3+ 8 Al + 30 H +3 NO 3 8 Al + 3 NH 4 + 9 H2O D c Al(NO3)3, Mg(NO3)2 c th c NH4NO3. NH4NO3 N2O + 2H2O 2 NH4NO3 N2 + O2 + 4 H2O Cu 8: p xy ra: 4Al(NO3)3 2Al2O3 + 12 NO2 + 3O2 2Mg(NO3)2 2MgO + 4 NO2 + O2

E ch c Al2O3 v MgO.

27 x + 24 y = 2,16 + Gi x, y ln lt l s mol ca Al v Mg ta c h : x 102. 2 + 40 y = 3,84 x = Al = 0,04 mol v Mg = 0,045 mol s mol e cho = 0,21 mol (II) + T (I, II) suy ra phi c NH4NO3. T d dng tnh c kt qu sau: D gm: Al(NO3)3 (8,52 gam) ; Mg(NO3)2 (6,66 gam) ; NH4NO3 (2,4 gam) = 17,58 gam. Hn hp ban u c 50% lng mi kim loi. Cu 5: + S mol kh thot ra l 0,1 mol. P c th xy ra : ZnO + 2HNO3 Zn(NO3)2 + H2O (1)Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

9


4Zn + 10HNO3 4Zn(NO3)2 + NH4NO3 + 3H2O t 0 2100 NH4NO3 N2O + H2O0 0

(2) (3)

t 350 NH4NO3 N2 + 1/2O2 + H2O (3) Zn(NO3)2 ZnO + 2NO2 + O2 (4) + Ta thy NH4NO3 phn hy trc nn nu Zn(NO3)2 phn hy ht th s mol kh phi ln hn s mol cht rn l ZnO. + Xt trng hp Zn(NO3)2 phn hy ht cht rn l ZnO vi s mol ZnO = 113,4/81=1,4 mol > s mol kh l 0,1 mol Zn(NO3)2 cha b phn hy s mol Zn(NO3)2 = 113,4/189 = 0,6 mol. + Theo (3) th s mol kh = N2O = 0,1 mol s mol NH4NO3 = 0,1 mol Zn = 0,4 mol ZnO = 0,6-0,4 = 0,2 mol. Cu 6: T p thy phn R-COO-R + KOH R-COOK + ROH mol 2 este = 0,01 v M = 74 * C 2 kh nng xy ra : - C 2 este u c KL mol = 74 ( H-COO-C2H5 v CH3-COO-CH3) - Mt trong hai este c KL mol < 74 l H-COO-CH3. Nh vy c 2 kh nng u c 1 este Fomat, khi un nng vi H2SO4 b phn hy to ra CO (KL mol = 28), ngoi ra cn mt kh b hp th bi nc brom, kh phi l anken sinh ra khi phn ancol trong este b tch nc. Mt khc, khi lng ring hn hp kh khng i, tc l kh phi c KL mol = 28, l C 2H4. C 2H4 + Br2 C2H4Br2 . * Nu trong hn hp c H-COO-C2H5 CO + C2H4 + H2O th sau khi i qua nc brom khi lng kh phi gim i 1/2 (tri gi thit). Vy cc gc H-COO- v C2H5- phi thuc v 2 este khc nhau. * Hn hp cha H-COO-CH3 (x mol) v R-COO-C2H5 (y mol). Ta c : x + y = 0,01 ; x = 2y (do CO = 2 C2H4 ) y = 0,01/3 v x = 0,02/3

Ta c : 60

0,02 3

+ (R + 73)

0,01 3

= 0,74 R = 29 C2H5-COO-C2H5

- Khi lng hn hp kh sau phn ng vi H2SO4 = 28 ( 0,01 = 0,28 gam) 0,02/3 60 = 0,4 gam H-COO-CH3 54,1% v 0,34 gam C2H5-COO-C2H5 45,9% * Phn bit 2 este bng phn ng vi dung dch AgNO3 trong NH3 : H-COO-CH3 + 2 Ag(NH3) + + 2H2O (NH4)2CO3 + 2NH + + CH3OH + 2 Ag 2 4 Cu 7: a) Trong cc phn t trn, nguyn t trung tm trng thi lai ho sp3 c cc cp e cha lin kt y mnh hn cp e lin kt. b) Khi m in ca nguyn t trung tm gim (hoc khi m in ca phi t tng)th cc cp in t ca lin kt b y nhiu v pha cc nguyn t lin kt nn chng ch cn mt khong khng gian nh chung quanh nguyn t trung tm. m in ca S < O nn HSH < HOH .

c) m in ca F > H nn FOF < HOH Cu 8: Gi oxit kim loi phi tm l MO v a v 2a l s mol CuO v MO trong A. V hidro ch kh c nhng oxit kim loi ng sau nhm trong dy in ha nn c 2 kh nng xy ra: * Trng hp 1: M ng sau nhm trong dy in ha CuO + H2 Cu + H2O MO + H2 M + H2O 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O 3M + 8HNO3 3 M(NO3)2 + 2NO + 4H2O 80a + ( M + 16).2a = 3, 6 Ta c h pt: 8a a = 0,01875 v M = 40 Ca loi v Ca trc Al 16a = 0,15 3 + 3 * Trng hp 2: M ng trc nhm trong dy in ha M = 24 Mg tha mn. V=

% CuO = 41, 66% 0, 0225. 2 . 22,4 = 0,336 lt. 3 % MgO = 58,34%2+ 2

p n 64

Cu 1: Hg

c s oxi ha +1 do Hg

2+ 2

Hg .+


10


Hg

2+

+ 2e E1

Hg 2+ 2e E3

2+

+ 1e E2

Hg

2E3 = 2E1 + E2 2.0,85 = 2.0,92 + E2 E2 = -0,14 vn K = 4,26. 10 3 Cu 2: a) S mol Fe trong FexOy = s mol Fe trong Fe(NO3)3 = 0,075 s mol oxi trong FexOy = (5,8-0,075.56)/16 = 0,1 oxit l Fe3O4. b) B c th cha Fe, FeO (a mol) v Fe3O4 d (b mol) 3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + H2O 3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O Fe + 4 HNO3 Fe(NO3)3 + NO + 2H2O Fe + 2HCl FeCl2 + H2 , ta c :

nFe = nH2 = 0, 03(mol )

56.0,03 + 72a + 232 b = 5,16 a = 0 a b 0,03 + + = 0,035 b = 0,015 3 3 0,03.56 %m Fe = .100 % = 32,56% v % m Fe3O 4 = 100% 32,56% = 67,44% 5,16

2 4 2+ 0 2+ 3 Cu 3: khng c kt t Mg(OH)2 th [Mg ].[OH ] 10-11. Vi C (Mg ) = 10 th [OH ] 10 . + Cn bng NH3 + H2O NH 4 + OH Kb = 1,8.10-5. 4 4 4 [ ] 0,5 10 x + 10 10c

( x + 104 ) 104 0,5 104

4 = 1,8.10-5 (coi 10 38,24 nn kh cn li phi l NO = 30 < 38,24. V tnh c NO = 0,32 mol v NO2 = 0,34 molGio vin: Trn Hu Tuyn 0944478966 - 0393509744

11


3X + 8HNO3 3X(NO3)2 + 2NO + 4H2O Y + 4HNO3 Y(NO3)3 + NO + 2H2O X + 4HNO3 X(NO3)2 + 2NO2 + 2H2O Y + 6HNO3 Y(NO3)3 + 3NO2 + 3H2O X + 2HCl XCl2 + H2 hoc 2Y + 6HCl 2YCl3 + 3H2 Bin lun: * Nu kim loi Y khng tan trong axit HCl Theo pt: s mol X = 0,6 v lng X = 10,8 gam nn X = * Vy kim loi X khng tan trong axit HCl Theo pt: s mol Y = 0,4 v lng Y = 14- 3,2= 10,8 gam nn Y =

10,8 = 18 (khng c kim loi tha mn) 0, 6 10,8 = 27 Al 0, 4

Al 3e Al3+ t s mol X bng a: tng s e nhng = 0,4. 3 + 2a = 1,2 + 2a 2+ X 2 e X N +5 + 3e N +2 tng s e thu = 0,32. 3 + 0,34 = 1,30 +5 +4 N + 1e N 3, 2 1,2 + 2a = 1,3 a = 0,05. Vy X = = 64 Cu v % Al = 77,14% ; %Cu = 22,86% 0, 05Cu 7: 1/ X l NO3 v NH4NO3: mi trng axit pH< 7. 2/ + c kt ta mu xanh: Cu2+ + 2OH- Cu(OH)2 + c kh mi khai : NH4+ + OH- NH3 + H2O + kt ta tan to dung dch xanh thm: Cu(OH)2 + 4NH3 [Cu(NH3)4]2+ + 2OHCu 8: A l C7H8O vi CTCT l CH3-O-C6H5, B l p-Br-C6H4-O-CH3. iu ch A ta lm nh sau: + CH 3Cl Metan axetilen benzen clobenzen natri phenolat C6H5-O-CH3 + NaCl (khng dng p C6H5OH + CH3OH v phenol khng p vi ancol) p n 65 Cu 1: a/ NO = 0,025 mol v N2O = 0,015 mol s mol e cho = 0,195 mol s mol FexOy =

0,195 3x 2 y

45,24 =

0,195(56 x + 16 y ) x 3 124,8x = 93,6y = oxit st l Fe3O4. 3x 2 y y 4

b/ dd A c 0,585 mol Fe(NO3)3 v HNO3 d. Khi thm m gam Cu vo th c p sau: 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O mol: 0,03 0,08 0,02 0,02 2Fe3+ + Cu 2Fe2+ + Cu2+. Mol: 0,585 0,2925 m = 64.(0,03+0,2925) + 2,88 = 23,52 gam. + S mol HNO3 = 1,89 mol CM = 1,26M c/ khi lng mui = Fe(NO3)2 + Cu(NO3)2 = 165,93 gam. Cu 2: + Ta thy rng A, B phi c vng benzen + Ta bit rng mi nhnh vng benzen sau khi oxi ha cho 2 Oxi gn vi vng nn A phi c 3 nhnh v cho sp c 6 Oxi B c 2 nhnh v sp c 4 oxi. + Do A c 3 nhnh v ch cha 2 loi H nn A c CTCT l:CH3

H3C

CH3

Tht vy:


12

Tp 200 thi n thi hc sinh giiCH3 [O] + H 2O H3CCH3 C2H5 [O ] - CO2

COOH

CH3COOH

HOOC

COOHO CO

+ V sp ca B p vi anhirit axetic cho sp l C8H4O3 nn 2 nhnh ca B phi gn nhau. Do B l:CO+ (CH O 3C ) 2O

COOH - H 2O

- 2CH 3C O O H

+ C v D u p vi Cu2Cl2/NH3 cho kt ta nn phi l ank-1-in. Da vo sp ozon phn suy ra CTCT ca C v D l:C CH CH3 CH3 C CH

(C )

(D )

Tht vy:C CH CH3+ H2O/Hg2+

CO-CH3 CH3+ O3

CH3-CO-CO-CH2-CH2-CH2-CH2-CO-CH3.

vCH3 C CH+ H 2O/Hg2+

CH3 CO-CH3+O3

CH3 CH3-CO-CH-CH-CH2-CH2-CH=O CH=O

Cu 3: X l: HCOO-C2H4-COO-CH=CH2 vi 2 CTCT tha mn l: HCOO-CH2-CH2-COO-CH=CH2. v HCOO-CH(CH3)COO-CH=CH2; A l HCOONa; B l HO-CH2-CH2-COONa hoc HO-CH(CH3)-COONa; C l CH3-CHO Cu 4: 1/ P xy ra theo th t sau: Fe + 2AgNO3 Fe(NO3)2 + 2Ag (1) Cu + 2AgNO3 Cu(NO3)2 + 2Ag (2) C th c: Fe(NO3)2 +AgNO3 Fe(NO3)3 + 2Ag (3) + Gi s mol ca Fe v Cu ln lt l x v y ta c: 56x + 64y = 7 (I) + Ta phi xt cc trng hp sau: TH1: Ch c p (1) ch c Fe p. TH2: C p (1) v (2) Fe ht v Cu p 1 phn hoc va ht TH3: C p (1), (2) v (3) Fe v Cu ht v AgNO3 d sau (2) * TH 1: Ch c Fe p (1). Gi x l s mol Fe p, y l s mol Cu v z l s mol Fe d ta c: 56(x+z) + 64y = 7 (I) Fe + 2AgNO3 Fe(NO3)2 + 2Ag (1) Mol: x 2x x 2x A c Fe(NO3)2 = x mol. B c 2x mol Ag + Cu = y mol v c th c Fe d = z mol. + Theo gi thit ta c: 108.2x + 64y + 56z = 21,8 (II) + Khi A p vi NaOH ta c: Fe(NO3)2 Fe(OH)2 Fe(OH)3 Fe2O3. Mol: x x x 0,5x 0,5x.160 = 7,6 (III) + Thay (III) vo (I, II) ta c: 64y + 56z = 1,68 v 64y + 56z = 1,28 Loi trng hp ny. * TH2: C p (1, 2) Fe ht v Cu p 1 phn hoc va ht gi x l s mol Fe, y l s mol Cu p v z l s mol Cu d ta c: 56x + 64(y+z) = 7 (I) Fe + 2AgNO3 Fe(NO3)2 + 2Ag (1) Mol: x 2x x 2x Cu + 2AgNO3 Cu(NO3)2 + 2Ag Mol: y 2y y 2y A c Fe(NO3)2 = x mol v Cu(NO3)2 = y mol. B c (2x+2y) mol Ag + Cu d = z mol Gio vin: Trn Hu Tuyn 0944478966 - 0393509744 13


+ Theo gi thit ta c: 108.(2x+2y) + 64z = 21,8 (II) + Khi A p vi NaOH ta c: 0,5x.160 + 80y = 7,6 (III) + Gii (I, II, III) c: x = 0,045 mol; y = 0,05 mol v z = 0,02 mol Vy %Fe = 36%. *TH3: Xy ra p (3) khi B ch c Ag S mol e m Ag+ nhn s mol e m A cho 21,8/108 s mol e m A cho. Gi s A ch c Cu th s mol e cho l nh nht v = 2.7/64 21,8/108 14/64 iu ny v l 2/ 7 gam A c 0,045 mol Fe v 0,07 mol Cu. lng HNO3 min th xy ra p sau: Fe + 4HNO3 Fe(NO3)3 + NO + H2O Mol: 0,045 0,18 0,045 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O mol: a 8a/3 a v Cu + 2Fe(NO3)3 Cu(NO3)2 + 2Fe(NO3)2. Mol: 0,0225 0,045 a + 0,0225 = 0,07 a = 0,0475 mol HNO3 = 0,3067 mol V = 153,33 mol. S: 1/ Fe = 36% 2/ 153,3 ml Cu 5: t CT ca oxit l MxOy; gi s mol M v MxOy trong mt phn ln lt l a v b ta c: Ma + b(Mx+16y) = 59,08 (I) + Vi phn 1 ta c: 2M + 2nHCl 2MCln + nH2. Mol: a an/2 an = 0,4 (II) + Vi phn 2 ta c: 3M + 4mH+ + mNO3- 3Mm+ +mNO + 2mH2O 3MxOy + (4xm-2y)H+ +(mx-2y)NO3- 3xMm+ +(mx-2y)NO +(2mx-y)H2O am + b(mx-2y) = 0,2.3 (III) + Vi phn 3 ta c: MxOy + yH2 xM + yH2O Mol: b bx cht rn gm (a+bx) mol M. Do : 3M + mHNO3 + 3mHCl 3MClm + mNO + 2mH2O m(a+bx) = 0,8.3 (IV) + T (III v IV) ta c by = 0,9 mol thay vo (I) ta c: M(a+bx) = 44,68 (V) + Chia (V) cho (IV) c: M = 18,6 m m = 3 v M = Fe. T M l Fe v (II) n = 2 a = 0,2 mol bx = 0,6 mol v by = 0,9 mol x/y = 2/3 oxit cho l Fe2O3. Cu 6: A l CO2, B l CaCO3 v C l Ca(HCO3)2. Cu 7: A1 l CH3-CO-CH3; A2 l HOOC-COOH; A3 l CH3-CO-CH2-CH2-COOH. V A l tecpen nn khung cacbon ca A s c to thnh bng cch ghp 2 isopren vi nhau CTCT ph hp ca A l:

CH3 - C = CH - CH2 - CH2 - C = CH - CH2 - OH CH3Vit li A dng sau:CH3

CH33,7-imetylocta-2,6-ienol.

CH2OH H3C CH3

+ P xy ra: Cu 8: 1/ Ta c s mol CuCl2 = 0,15 mol v FeCl3 = 0,3 mol. tng KL catot bng KL kim loi sinh ra bm vo catot. + anot xy ra p: 2Cl- Cl2 + 2e mol: 1,2 1,2 Khi p ht th s mol e trao i l 1,2 mol khi p l 0,5T v 0,7T th s mol e trao i l 0,6 mol v 0,84 mol. + catot xy ra p theo th t:


14


Fe + 1e Fe Cu2+ + 2e Cu Fe2+ + 2e Fe + TH1: thi gian p l 0,5T ng vi 0,6 mol e trao i th c Fe3+ + 1e Fe2+ mol: 0,3 0,3 0,3 Cu2+ + 2e Cu mol: 0,15 0,3 0,15 tng KL catot = KL ca Cu = 0,15.64 = 9,6 gam + TH2: thi gian p l 0,7T ng vi 0,84 mol e trao i. Fe3+ + 1e Fe2+ mol: 0,3 0,3 0,3 Cu2+ + 2e Cu mol: 0,15 0,3 0,15 Fe2+ + 2e Fe mol: 0,12 0,24 0,12 tng KL catot = KL ca Cu + Fe = 16,32 gam. 2/ + t x, y, z ln lt l s mol Cl-; HCO3- v CO32- ta c: y + 2z = 0,1.4,5; x + 0,45 = 0,65 v y + z = 0,35 x =0,2 mol; y = 0,25 mol v z = 0,1 mol V = 22,4.(y+z)=7,84 lt S mol NaCl = 0,2 mol; NaHCO3 = 0,25 mol; Na2CO3 = 0,1 mol. Gi n l s mol nc ta c: 0,2.58,5 + 0,25.84 + 0,1.106 + 18.n = 61,3 n = 1 mol. c 3 kh nng l: NaCl.5H2O; NaHCO3.4H2O v Na2CO3.10H2O nhng ch c Na2CO3.10H2O l ph hp vi thc nghim. p n 66 Cu 1: a/Aspirin c cng thc l o-CH3COO-C6H4-COOH MAspirin = 180 vC

3+

2+

nng ban u ca Aspirin =

3,55 = 0,01972 M. V Aspirin c dng HA nn ta c: 180.1

HA H+ + A-. B: 0,01972 0 0 Cb: 0,01972-x x x

x.x = 10-3,49 x = 2,37.10-3M pH = 2,625. 0, 01972 x

b/ Khi cho 0,1 mol aspirin vo nc th c 0,01972 mol tan ra to ra 2,37.10-3 mol H+ s cn li tan ht th phi dng NaOH trung ha H+ v ha tan lng HA cn li theo p: HA + NaOH NaA + H2O S mol NaOH = H+ + HA = 2,37.10-3 + (0,1-0,01972) = 0,08265 mol NaOH = 3,306 gam * Tnh pH: Dung dch sau p trn c [HA] = 0,01972-2,37.10-3 = 0,01735 M v [A-] = x + (0,1-0,01972) = 0,08265M. Do ta c cn bng: HA H+ + A-. B: 0,01375 0 0,08265 Cb: 0,01375-x x 0,08265+ x

x.(0, 08265 + x) = 10-3,49 x = 5,383.10-5M pH = 4,27. 0, 01375 xCOOH

c/ CTCT ca aspirin nh sau:

OCOCH3

+ P iu ch: metan axetilen benzen clobenzen natri phenolat. Sau l cc p sau:


15

Tp 200 thi n thi hc sinh giiONa ONa OH COOH

OCOCH3COOH + (CH 3-CO) 2O -CH 3COOH

+ CO 2

Pcao

+ H 2SO 4-Na 2SO 4

COOH

Cu 2: 1/ Ta phi tnh E0 ca Sn4+/Sn theo 1 trong 2 cch sau: Cch 1: Sn4+ + 2e Sn2+ c G1 = -2F.0,15 Sn2+ + 2e Sn c G2 = -2F(-0,14) Sn4+ + 4e Sn c G = G1 + G2 hay -4.F.E0 = -2F(0,15-0,14) E0 = 0,005 vn Cch 2: Da vo s

Sn

4+

E1 + 2e

SnE3 + 4e

2+

E2 + 2e

Sn

4E3 = 2E1+ 2E2 4E3 = 2.0,15 + 2.(-0,14) E3 = 0,005v. + bit c khi Sn p vi HCl to ra Sn2+ hay Sn4+ th ta phi tnh E ca 2 p sau: Sn + 2H+ Sn2+ + H2. E1. (1) Sn + 4H+ Sn4+ + 2H2. E2. (2) 0 E1 = 0 (-0,14) = 0,14 v > 0 (E ca 2H+/H2 c qui c = 0) v E2 = 0 0,005) = -0,005 v < 0 E > 0 p mi xy ra nn t 2 gi tr ca E trn suy ra ch xy ra p (1) tc l ch to ra Sn2+ m khng to ra Sn4+. 2/ a/ V EAg+/Ag > EFe3+/Fe2+ nn p xy ra nh sau: Fe2+ + Ag+ Fe3+ + Ag b/ Ta c: Ep = EAg+/Ag - EFe3+/Fe2+ (*)

0, 059 [ox] lg nn n [kh] 0, 059 0,005 lg EAg+/Ag = 0, 799 + = 0,663 vn 1 1+ Mt khc ta c: E = E0 +

(V Ag l cht rn nn nng qui c l 1; [Ag+] sau khi trn c tnh = V EFe3+/Fe2+ = 0,771 +

0, 059 0, 0125 lg = 0,794 vn 1 0, 005

50.0, 01 = 0, 05 ) 50 + 25 + 25

Ep = 0,663 0,794 = - 0,131 vn < 0 p xy ra theo chiu ngc li. G = - nF E = -1.96500.(-0,131) = 12641,5 Jun >0 p xy ra theo chiu ngc li. Cu 3: 1/ C l propan-1ol 2/ 66,67% 3/ propyl axetat. Cu 4: + D thy A va c nhm axit va c nhm OH, khi un nng A th nhm OH b tch nc to ra B. Theo trn th C phi l axeton do A, B, C phi c cu to l: A: CH3-CHOH-CH2-COOH(A quang hot v c C bt i CH3-C*HOH-CH2-COOH) B: CH3-CH=CH-COOH C: CH3-CO-CH3. + P xy ra: t0 CH3-CHOH-CH2-COOH CH2=CH-CH2-COOH 4 CH2=CH-CH2-COOH + [O] + H2O CH2OH-CHOH-CH2-COOH. H 2CrO4 CH3-CHOH-CH2-COOH + [O] CH3-CO-CH3 + CO2 + H2O CH3-CO-CH3 + 3I2 +4NaOH CH3COONa + CHI3 + 3NaI + 3H2O Cu 5: 1/ H2SO4 = 0,08 mol v HCl = 0,12 mol H+ = 0,28 mol. + P cho dng: M + 2H+ M2+ + H2 (1) + Nu X ch c Zn th s mol kim loi trong X l nh nht = 10/65 = 0,154 mol S mol H+ t nht cn ha tan ht X = 2.0,154 = 0,308 mol > s mol H+ gi thit cho l 0,28 mol Trong p (1) H+ ht v kim loi d s mol H2 = 0,28/2 = 0,14 mol. + Khi cho 50% ng vi 0,07 mol hiro p vi CuO th: CuO + H2 Cu + H2O Mol: 0,07 0,07 0,07

KMnO

A c 0,07 mol Cu v (

a a - 0,07) mol CuO 0,07.64 + 80.( - 0,07) = 14,08 a = 15,2 gam 80 8016



2/ A c: 0,07 mol Cu v 0,12 mol CuO. Khi p vi AgNO3 ta c: Cu + 2AgNO3 Cu(NO3)2 + 2Ag Mol: x 2x 2x B c: 0,12 mol CuO + (0,07 - x) mol Cu + 2x mol Ag

mB = 14,08 + 152x

108.2 x = 0,2523 x = 0,02 mol. 14, 08 + 152 x

B c: 0,12 mol CuO + 0,05 mol Cu + 0,04 mol Ag + Khi B + HNO3 th: CuO + 2HNO3 Cu(NO3)2 + H2O Mol: 0,12 0,24 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O mol: 0,05 0,1333 3Ag + 4HNO3 3AgNO3 + NO + 2H2O mol: 0,04 0,0533 HNO3 = 0,42667 mol V =0,2133 lt Cu 6: 1/ 16 gam(Cu p c vi Fe3+ v H+ + NO3-). 2/+ Khi A p vi NaOH th s mol NaOH = 0,12 mol; s mol H2 = 0,12 mol. Suy ra NaOH dAl + NaOH + H2O NaAlO2 + 3/2 H2. Mol: 0,08 0,08 0,08 0,12 Sau p trn th hh c: FeCO3 + Fe + Cu + 0,04 mol NaOH d + 0,08 mol NaAlO2. + Khi thm vo 0,74 mol HCl vo th: NaOH + HCl NaCl + H2O Mol: 0,04 0,04 NaAlO2 + 4HCl + H2O NaCl + AlCl3 + 3H2O Mol: 0,08 0,32 S mol HCl cn li sau 2 p trn l 0,38 mol. B l hh kh nn B phi c CO2 + H2. C chc chn c Cu, c th c FeCO3 + Fe. Mt khc C + HNO3 NO2 l kh duy nht nn C khng th cha FeCO3 C c Cu v c th c Fe (FeCO3 b HCl ha tan ht). TH1: Fe d. Gi x l s mol FeCO3; y l s mol Fe b ha tan; z l s mol Fe d, t l s mol Cu ta c: 116x + 56(y + z) + 64t = 20 0,08.27 = 17,84 (I) FeCO3 + 2HCl FeCl2 + CO2 + H2O Mol: x 2x x x Fe + 2HCl FeCl2 + H2 Mol: y 2y y y S mol HCl = 2x + 2y = 0,38 (II) B c x mol CO2 + y mol hiro. Da vo p ca B vi nc vi trong x = 0,1 mol (III) C c z mol Fe d + t mol Cu 3z + 2t = 1,12/22,4 (IV) x = 0,1 mol; y = 0,09 mol; z = 0,01 mol v t = 0,01 mol. Vy A c: 0,1 mol FeCO3 + 0,1 mol Fe + 0,01 mol Cu + 0,08 mol Al %KL + Tnh tip ta c gi tr ca m = 1,6 gam. TH2: Fe ht C ch c Cu s molCu = NO2 = 0,025 mol. A c 0,1 mol FeCO3 + 0,08 mol Al + 0,01 mol Cu + a mol Fe = 20 gam a = 0,1 mol D dng tnh c m = 2 gam. Cu 7: 1/ Sc CO2 d vo cc cha sn V lt dd NaOH c V lt dd NaHCO3. Thm V lt dd NaOH vo V lt dd NaHCO3 trn c dd Na2CO3 nguyn cht. 2/ A l etan; B l etilen v C l axetilen. Cho tia la in qua A th: C2H6 2Crn + 3H2 kh nn lm tng th tch 3 ln. Cu 8: Na2CO3, (NH4)2CO3, BaCO3, PbCO3, FeCO3, ZnCO3. Cu 9: 16,8 v 4,48 p n 67 Cu 1: 1/T din u, thp tam gic(chp), lng thp tam gic 2/pH=0,52 3/pH=6,15 Cu 2: (H khi A 2006) 1/ X l CH3OH, p = 3,2 gam 2/ Y l axit propionic(39,14%); Z l etyl propionat(33,94%)Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

17


Cu 3: Ta thy X no => trong X ch c nhm ete hoc ancol hoc c hai. V X p c vi CuO nn X chc chn c nhm anol OH. Khi 1 nhm CH2 OH chuyn thnh CH=O hoc CH-OH thnh C=O th s H gim i 2 tc l KLPT s gim 2 vC. Theo gi thit th MY nh hn MX l 8 vC nn trong X phi c 4 nhm OH(X khng c nhm ete v X ch c 4 oxi)=> Y c CTPT l C5H4O4 hay MY = 128 gam. S mol Y = 2,56/128 = 0,02 mol; s mol Ag = 0,16 mol. Trong Y chc chn c nhm anehit CHO c th c nhm xeton C=O . t Y l R(CHO)n ta c R(CHO)n + 2nAgNO3 + 3nNH3 +nH2O R(COONH4)n + 2nAg + 2nNH4NO3 Mol: 0,02 0,16 n=4 X v Y c CTCT ln lt lCH2OH HOH2C C CH2OH

CH=O O=HC C CH=O

CH=O v Khi X + NaBr/H2SO4 c tng ng vi X p vi HBr v: t0 2NaBr + H2SO4 Na2SO4 + 2HBr Do ta c:CH2OH HOH2C C CH2OH +4 B H r BrH2C CH2Br C CH2Br +4 H2O

CH2OH

CH2OH

CH2BrCH2 +2 n Z H2C C CH2 +2 n r ZB2

Do Q c M < 90 nn Q

khng cn Br vy Q l sp ca p sau:CH2Br BrH2C C CH2Br

CH2Br

CH2

Cu 4: + Vi phn 1 ta c: 2FeCl3 + H2S 2FeCl2 + S + 2HCl mol: x 0,5x CuCl2 + H2S CuS + 2HCl Mol: y y 16x +96y = 1,28 (I) 2FeCl3 + Na2S 2FeCl2 + S + 2NaCl sau : FeCl2 + Na2S FeS + 2NaCl 2FeCl3 + 3Na2S 2FeS + S + 6NaCl mol: x x 0,5 x CuCl2 + Na2S CuS + 2NaCl Mol: y y 88x + 32.0,5x + 96y = 3,04 (II) + T (I, II) ta c: x = 0,02 mol v y = 0,01 mol m = 4,6 gam. 2/ Ta thy A c = 3. Mt khc: A (C9H14) + [O] + H2O Y (MY = 190) MY MA = 68 = 4.M-OH Trong p trn c 4 nhm OH c gn vo A tc l A c 2 lin kt i; m A c = 3 nn A cn c 1 vng. + V A + H2 propylxiclohexan nn A c mch cacbon tng t propylxiclohexan. Da vo p oxi ha ta suy ra A l 1propenylxiclohexen:CH2-CH=CH2

+ P xy ra: A + [O] CO2 + HOOC-CH2-CO-CH2-CH2-CH2-CH2-COOH. Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

18

Tp 200 thi n thi hc sinh giiOH OHCH2-CH=CH2 CH2-CH - CH2

HO2O

OH

+ [O] + H

+ Khi Y p vi CH3COOH th ch c 3 OH p; ring OH cacbon bc III khng p. Cu 5: 1/ t CTPT ca oxit l RxOy ta c: 2RxOy + (2nx-2y)H2SO4 xR2(SO4)n + (nx-2y)SO2 + (2nx-2y)H2O mol: 0,1x/(nx-2y) 0,1 0,1x(2R+96n) = 120(nx-2y) x(2R+96n) = 1200(nx-2y)

Rx = 552nx 1200y vi 2 n 4 v n > R = 552n -1200

2y x

y x

+ Vi n = 2 Rx = 1104x 1200y th ta thy khng c gi tr ph hp + Vi n = 3 Rx = 1656x 1200y x = 3; y = 4 v R = 56 ph hp Fe3O4. Cch 2: Ta c s mol e cho = nhn = 0,1.2 = 0,2 mol. Gi +x l s oxi ha ca R trong oxit ta c: R+x R+n + y.e Mol: 0,2/y 0,2 S mol R2(SO4)n = 0,1/y mol 0,1.(2R + 96n) = 120.y R = 600y 48n. + V n 4 nn nu y 1 th R > 408 v l nu y < 1 oxit cho l Fe3O4 vi y = 1/3. 2/ S mol HCl = 0,85 mol. P: Fe3O4 + 8HCl FeCl2 + 2FeCl3 + 4H2O Mol: x 8x x 2x 4FeCl2 + O2 + 4HCl 4FeCl3 + 2H2O mol: x1 x1 V A lm mt mu nc brom nn FeCl2 d trn v tham gia p: 6FeCl2 + 3Br2 4FeCl3 + 2FeBr3. Mol: 0,05 0,025 0,05 + x1 = x v 8x + x1 = 0,85 x = 0,1 mol v x1 = 0,05 mol KL Fe3O4 = 23,2 gam. Cu 6: 1/ Gi CTPT ca A l CxHyOz vi s mol l a ta c: a(12x + y + 16z) = 14,6 (I) + P chy: CxHyOz + (x+y/4-z/2) O2 xCO2 + y/2 H2O mol: a ax 0,5ay

44ax + 9ay = 35,4 (II) v

32ax + 8ay = 0,7684 (III) 35, 4

+ Gii (I, II, III) ta c: ax = 0,6 mol; ay = 1 mol v az = 0,4 mol x:y:z = 3:5:2 A c dng (C3H5O2)n. Da vo M < 160 vC n = 2. Vy A l C6H10O4. 2/ S mol A = 0,15 mol. A c dng RCOO-R-OOC-R hoc ROOC-R-COOR hoc RCOO-R(OH)2 hoc HOOC-R-COOR + TH1: A l RCOO-R-OOCR ta c: RCOO-R-OOCR + 2NaOH 2RCOONa + R(OH)2. Mol: 0,15 0,15 R(OH)2 = 13,8/0,15 = 92 R = 58 khng tha mn. + TH2: A c dng RCOO-R(OH)2 nn ta c: RCOO-R(OH)2 + NaOH RCOONa + R(OH)3. Mol: 0,15 0,15 R(OH)3 = 92 = C3H5(OH)3 = glixerol R l C2H3- A c 2CTCT tha mn l: HO-CH2-CHOH-CH2-OOC-CH=CH2 v HO-CH2-CH(OOC-CH=CH2) -CH2-OH + TH3: A c dng R(COOR)2 R = C2H5 R = 0 A l C2H5OOC-COOC2H5. + TH4: HOOC-R-COOR R = 75 khng tha mn. 3/ B c dng C6H8O4. Da vo s suy ra E l C2H5OH R l HO-CH2-CH2-COONa A l HCOO-CH2-CH2-COO-CH=CH2. Cu 7: 1/ Fe = 0,02 mol v Mg = 0,01 mol. P xy ra theo th t: Mg + CuSO4 MgSO4 + Cu (1)Gio vin: Trn Hu Tuyn 0944478966 - 0393509744

19


Fe + CuSO4 FeSO4 + Cu (2) + Gi s Mg v CuSO4 va ht (1): Mg + CuSO4 MgSO4 + Cu (1) Mol: 0,01 0,01 Cht rn ch c: 0,02 mol Fe + 0,01 mol Cu = 1,76 gam (*) + Gi s c Mg, Fe v CuSO4 u ht (1) v (2) ta c: Mg + CuSO4 MgSO4 + Cu (1) Mol: 0,01 0,01 Fe + CuSO4 FeSO4 + Cu (2) Mol: 0,02 0,02 Cht rn ch c 0,03 mol Cu ng vi 1,92 gam (*) + T (*); (*) v gi thit suy ra: Mg ht v Fe p mt phn, do ta c: Mg + CuSO4 MgSO4 + Cu (1) Mol: 0,01 0,01 0,01 Fe + CuSO4 FeSO4 + Cu (2) Mol: x x x Cht rn sau p c: (0,01+x) mol Cu + (0,02-x) mol Fe 64.(0,01+x) + 56(0,02-x) = 1,88 x = 0,015 mol CuSO4 = 0,01 + x = 0,025 mol CM = 0,1M. 2/ S mol NO = NO2 = 0,0175 mol s mol e m HNO3 nhn = 0,07 mol. p dng LBT e ta c: 3x =

5, 04 56 x .4 + 0,07 x = 0,07 mol. 32

3/ a/ S mol hiro = 0,045 mol Vit p xy ra ta s thy s mol HCl= 2 ln s mol hiro = 0,09 mol + p dng LBTKL ta c: m + 0,09.36,5 = 0,045.2 + 4,575 m = 1,38 gam. b/ +Ta c: M = 50,5 hai kh l NO2 v SO2. S mol 2 kh = 0,084 mol SO2 = 0,021 mol v NO2 = 0,063 mol S mol e nhn = 0,105 mol. Gi s mol Fe v M ln lt l x v y. p dng LBT e ta c:

3 x + ny = 0,105 56 x + My = 1,38 x = 0,015; ny = 0,06 v My = 0,054 M = 9n M l Al 2 x + ny = 0, 09 Cu 8: 1/ P xy ra: 2XSO4 + 2H2O 2X + 2H2SO4 + O2 (1) + Gi s khi p vi thi gian l 2t th XSO4 vn cn tc l ch c p (1) S mol kh thu c gp i s mol kh ng vi t giy v = 0,007.2 = 0,014 mol iu ny tri vi gi thit l 0,024 mol gi s l sai Khi thi gian p l 2t th XSO4 ht v c p p nc: dp 2H2O 2H2 + O2 + Vi thi gian l t giy ta c: 2XSO4 + 2H2O 2X + 2H2SO4 + O2 (1) mol: 0,014 0,007 + Vi thi gian l t giy tip theo th: 2XSO4 + 2H2O 2X + 2H2SO4 + O2 (1) mol: x 0,5x dp 2H2O 2H2 + O2 mol: y y 0,5y 0,5x + y + 0,5y = 0,024-0,007 x + 3y = 0,034 (I) + Khi thi gian l t giy th ch c oxi p (1) bay ra s mol O2 ng vi t giy = 0,007 mol s mol e trao i = 0,007.4 = 0,028 mol Trong thi gian t giy cn li s mol e trao i cng phi l 0,028 4.(0,5x+0,5y) = 0,028 x + y = 0,014 (II) + Gii (I , II) c: x = 0,004 mol v y = 0,01 mol S mol XSO4 = 0,004 + 0,014 = 0,018 mol 0,018(X + 186) = 4,5 X = 64 = Cu. 2/ a/ pHX = 2,7 b/ pHY = 4,75


20

200 deda hsg tinh khoi 1112

Documents