S GD&T NGH AN K THI CHN HC SINH GII CP TNH LP 12 NM HC 2010 2011Mn thi: HO HC LP 12 THPT - BNG AThi gian lm bi: 180 phtCu 1 (6,0 im)1. a) C 5 kh A, B, C, D, E. Kh A c iu ch bng cch nung KMnO4 nhit cao,kh B c iu ch bng cch cho FeCl2 tc dng vi dung dch hn hp KMnO4 trongH2SO4long d, kh C c iu ch bng cch cho st II sunfua tc dng viH2SO4 c nng, kh D c iu ch bng cch cho st pirit vo dung dch HCl trong iu kin thch hp, kh E c iu ch bng cch cho magie nitrua tc dng vi nc. Hy vitphng trnh ha hc ca cc phn ng xy ra.b) Cho cc kh A, B, C, D, E ln lt tc dng vi nhau tng i mt, trng hp no c phn ng xy ra ? Vit phng trnh ha hc ca cc phn ng v ghi r iu kin (nu c).2. a) Cho t t 100 ml dung dch hn hp Na2CO3 1M v KHCO3 aM vo 200 ml dung dch HCl 1M, sau khi phn ng xy ra hon ton th thu c 2,688 lt CO2 ( ktc). Tnh a ?b) Tnh pH ca dung dch to thnh khi ha tan 0,1 mol PCl3 vo 450 ml dung dch NaOH 1M. Cho hng s axit ca H3PO3 l :210 . 6 , 11aK, 710 . 0 , 72aK Cu 2 (4,0 im)1. ThmV(ml)dungdchBa(OH)20,1M vo100ml dung dchKAl(SO4)20,1Mthuc 2,1375 gam kt ta. Tnh V ?2. Cho 5,22 gam mt mui cacbonat kim loi tc dng hon ton vi dung dch HNO3 long, d thu c dung dch X v hn hp kh Y gm 0,336 lt NO v CO2, cc kh o ( ktc). Xc nh mui cacbonat v th tch kh CO2 thu c.Cu 3 (4,0 im)1. Cng thc n gin nht ca cht M l C3H4O3 v cht N l C2H3O3. Hy tm cng thc phn t ca M v N. Bit M l mt axit no a chc, N l mt axit no cha ng thi nhm chc OH; M v N u l mch h. Vit cng thc cu to c th c ca M v N.2. Hp cht A c cng thc phn t C8H14O5Bit : + A 2, H O H++ C2H5OH + B (bit s mol A phn ng = s mol C2H5OH = s mol B)+ Glucoz m e n B trng ngng polimeXc nh cng thc cu to ca A, B v vit phng trnh ha hc ca cc phn ng.Cu 4 (6,0 im)1. nhit khng i, hng s phn ly Ka ca cc cht : phenol; p-crezol; p-nitro phenol; 2,4,6 trinitro phenol (axit picric); glixerol l 7,0.10-5 ; 6,7.10-11 ; 1,28.10-10 ; 7,0.10-8 ; 4,2.10-1. Hy gn Ka vo cc cht trn theo th t tng dn. Gii thch ? 2. Ch dng nc v brom hy trnh by phng php nhn bit 6 cht lng ring bit sau :Benzen, anilin, xiclo hexen, axit acrilic, axit fomic, axit propionic.3. Cho 2,76 gam cht hu c X (ch cha C, H, O v c cng thc phn t trng vi cng thc n gin nht) tc dng vi dung dch NaOH va , sau chng kh th thu c hi nc, phn cht rn cha hai mui ca natri c khi lng 4,44 gam. t chy hon ton 4,44 gam hn hphai mui ny trong oxi th thu c 3,18 gam Na2CO3 ; 2,464 lt CO2 ( ktc) v 0,9 gam nc. Tm cng thc phn t, vit cng thc cu to c th c ca X.(Cho : H=1, C=12, N=14, O=16, Na=23, Mg=24, Al=27, S=32, Fe=56; Cu=64, Zn=65, Ba =137)- - - Ht - - - chnh thc H v tn th sinh:...................................................................S bo danh:.......................S GD& T NGH AN K THI CHN HC SINH GII CP TNH LP 12NM HC 2010 - 2011HNG DN V BIU IM CHM THI CHNH THCMn thi: HO HC THPT BNG A(Hng dn v biu im gm 05 trang)Cu Ni dung imCu 1 6,01.a)1.5Tm c 5 kh v vit ng 5 phng trnh hoc (nu vit ng 5 phng trnh vn cho im ti a 1,5 im) : A l O2; B : Cl2; C: SO2; D : H2S; E : NH32 KMnO4 0t K2MnO4 + MnO2 + O2 10 FeCl2 + 2 KMnO4 + 18 H2SO4 5 Fe2(SO4)3 + 2MnSO4 + K2SO4 + 10Cl2 + 18H2O2FeS + 10 H2SO4c nng Fe2(SO4)3 + 9SO2+ 10H2O2FeS2 + 4 HCl 2FeCl2 + 2S + 2H2SMg3N2 + 6 H2O 3Mg(OH)2 + 2NH3 0,30,30,30,30,3 1.b)2,02SO2 + O2 02 5450 , C V O2SO32H2S + 3O2 0t 2SO2+ 2H2OHoc : 2H2S + O2 (thiu) 2S+ 2H2O4NH3 + 5O2 0850 , C Pt 4NO + 6H2O Hoc : 4NH3 + 3O2(thiu) 0t 2N2 + 6H2OCl2 + SO2 0t SO2Cl2 Cl2+ H2S S + 2HCl3Cl2 + 2NH3 N2+ 6HClHoc : 3Cl2 + 8NH3 6NH4Cl + N2 2H2S + SO2 3S+ 2H2OH2S + NH3 NH4HS0,250,250,250,250,250,250,250,25 2.a)1,25Ta c 23230,1 ; 0, 20,1 ; 0,12CO HCOHCOn mol n moln amol n mol + ' H+ ht ,CO32- v HCO3- d Cc phng trnh phn ng xy ra ng thi: CO32- + 2H+ CO2+ H2O (1) x2x x HCO3- + H+CO2+ H2O (2)yyy0,250,250,25Goi x v y l s mol ca CO32- v HCO3- tham gia phn ng (1) v (2) Ta c h phng trnh 2 0, 2 0, 080,12 0, 04x y xx y y+ ' '+ 0,25Ta c : 0,1 0, 080, 50,1 0, 04xa My a 0,25 2.b)1,25S mol NaOH = 0,45.1 = 0,45 molPCl3 + 3H2O H3PO3 + 3HCl 0,100,10 0,300,25HCl + NaOH NaCl + H2O0,30 0,30 H3PO3 + NaOH NaH2PO3 + H2OBan u0,100,15 0,00P 0,100,10 0,10Sau p0,000,05 0,10NaOH + NaH2PO3 Na2HPO3 + H2OBan u0,050,10 0,00P 0,050,05 0,05Sau p0,000,05 0,050,250,25Vy sau phn ng thu c dung dch c s mol H2PO3- = s mol HPO32-

= 0,05 mol 0,25Ta c cn bng :H2PO3- H+ + HPO32-2232 3aH HPOK HH PO+ + 11 ] ] 1 ] 1 ] pH = lg 27)lg(7, 0.10 6,155aKH p+ 1 ]0,25Cu 2 4,012,0Ba(OH)2 Ba2+ +2OH-

(a l s mol Ba(OH)2 ) a a 2aKAl(SO4)2 K+ + Al3+ + 2SO42- 0,01 0,010,02Ba2+ + SO42- BaSO4 (1)Al3+ + 3OH- Al(OH)3 (2)Al(OH)3 + OH- Al(OH)4-

(3)Nu SO42- kt ta ht th : 4aS B Om 0,02.233 = 4,66 (gam) > 2,1375 (gam) SO42- d0,250,25Trng hp 1 : Al3+ tham gia va hoc d ch xy ra phn ng(1) v (2) khi 20, 01 0, 0153aa . Khi lng kt ta m c tnh : 4 32aS ( )dd ( )2233 .78 2,1375 0, 007530, 00750, 075( ) 75( )0,1 0,1B O Al OHBa OHam m m a aaV l hay ml + + 0,50,5Trng hp 2 : Xy ra phn ng (1), (2), (3) th : 20, 01 0, 0153aa > >Al3+ + 4OH- Al(OH)4- phn ng va khi a = 0,02 Vy 0, 015 0, 02 a < Khi a = 0,015 nu kt ta tnh theo BaSO4 l : 0,015.233 = 3,495 > 2,1375 (gam) loi 0,522,0Gi cng thc mui M2(CO3)n c s mol l x3M2(CO3)n + (8m 2n) HNO3 6M(NO3)m + 2(m - n)NO + 3nCO2+ (4m n)H2O (*)Theo phng trnh (*) ta c :2( ) 0, 015 2( ) 0, 0453NOn m n x m n x (1)Ta li c : (2M + 60n)x = 5,22(2)0,50,25 Gii (1) v (2) ta c :M = 116 m 146 nTa c bng sau :m 2 3 3n 1 1 2M 86 202 56Nghim loi loi Fe0,250,250,25Vy cng thc ca mui cacbonat l FeCO323 3.2. .22, 4 .0, 015.22, 4 1, 0082( ) 2(3 2)CO NOnV nm n (l)0,5Cu 3 4,01 2,0Cng thc phn t M c dng (C3H4O3)n c bi lin kt :a = n + 1Hay : (C6/3H 8/3O2)3n/2axit c 3n/2 nhm chc do : n + 1 = 322nn Cng thc phn t ca M l C6H8O6 hay C3H5(COOH)3. Vy cng thc cu to ca M l : HOOC CH2CH2CH COOH COOHHOOC CH2 CH CH2COOHCOOHHOOCCH3CH CH2COOHCOOHHOOC CHCH3CHCOOHCOOHHOOC CCOOHCOOHCH2CH30,250,250,5Cng thc phn t N c dng (C2H3O3)m (1) Hay C2m-yH3m-(x+y)(OH)x(COOH)y (2)T (1) bi lin kta = 2 22 2m my+ + (a)T (1) v (2) ta c : 3m = x + 2y (theo O)(b)Do N l ancol nn x2m y(c)T (a), (b), (c) 2 y 0,250,25Khi y =1theo (a) m = 0 v nghimKhi y = 2 theo (a) m =2 suy ra x = 2Vy cng thc phn t N l : C4H6O6 hay C2H2(OH)2(COOH)2Ta c cng thc cu to ca N l : HOOC CHOHCHOHCOOH HOOCCCOOHOHCH2OHv0,52 2,0Do bi lin kt a= 2 v 2 52B C H OHn n = 2nAVy A l este no hai chc cha hai gc axit ging nhau v 1 gc C2H5B c to ra t glucoz, c phn ng trng ngng Do cng thc cu to ca B l :0,250,25

CH3CH COOHOHA c cng thc cu to l :0,5Cc phng trnh phn ng :0,5C6H12O6 menlactic 2CH3 CH(OH) COOH 0,250,25Cu 46,01 1,0Sp xp cc cht trn theo th t Ka tng dn :

CH2CH CH2OH OH OHOHCH3OHOHNO2NO2O2NOHNO2 Ka

6,7.10-111,28.10-10

7,0.10-87,0.10-5 4,2.10-1 0,5Gii thch : Glixerol c tnh axit yu hn cc phenol l do phenol c vng benzen ht electron lm tng phn cc ca lin kt OH. Tnh axit ca phenol mnh hay yu ph thuc vo nhm th lin kt vi vng benzen. Nu nhm th y electron (nhm CH3 ) lm gim phn cc lin kt OH. Nn tnh axit gim. Nu nhm th ht electron (nhm NO2) lm tng phn cc lin kt OH. Axit picric c tnh axit mnh nht v c 3 nhm (NO2) ht electron; p- cresol c tnh axit yu hn cc phenol cn li v c nhm CH3 y electron.0,52 2,0Trch mu th, sau cho nc vo 6 cht lng ta c hai nhm :+ Nhm tan : axit acrilic, axit fomic, axit propionic+ Nhm khng tan : benzen, anilin, xiclo hexen0,250,25Cho dung dch brom vo 3 cht tanAxit acrilic lm mt mu dung dch brom, axit fomic lm mt mu dung dch brom v c kh thot 0,25 ra. Axit propionic khng lm mt mu dung dch brom.CH2 = CH COOH + Br2 CH2Br CHBr COOHHCOOH + Br2 CO2+ H2O0,250,25Cho dung dch brom vo 3 cht khng tan : bezen to thnh hai cht lng phn lp. Anilin to kt ta trng C6H5-NH2 + 3Br2C6H2(Br3)NH2 + 3HBrXiclo hexen lm mt mu dung dch brom BrBrBr20,250,250,253 3,02 3Na COn= 3,180, 03106mol ; 2COn= 2, 4640,1122, 4 molX + NaOH hai mui ca natri + H2O (1)Hai mui ca natri + O2 0t Na2CO3 + CO2 + H2O (2)S mol Na = 0,06 mol; S mol C = 0,03 + 0,11 = 0,14 mol 0,250,25p dng nh lut bo ton khi lng (1) ta c :2 2 2i0, 72(2, 76 2, 4) 4, 44 0, 72 0, 0418X NaOH mu H O H O H Om m m m m gam n mol + + + Tng s mol H trong nc = 2 s mol H2O = 2.(0,04 +0,05) = 0,18 molS mol H trong 0,06 mol NaOH = 0,06 molS mol H trong X l : 0,18 0,06 = 0,12 molKhi lng O trong X l : 2,76 (0,14.12 +0,12) = 0,96 (gam) hay nO = 0,06 mol0,250,25 0,250,25Ta c t l : nC : nH : nO = 0,14 : 0,12 : 0,06 = 7 : 6 :3Vy cng thc phn t ca X l : C7H6O3 Do : nX = 2, 760, 02138mol ; 0, 0630, 02NaOHXnn V X c bi lin kt a = 5Nn cng thc cu to ca X l :

OOCHOHOOCHOHOOOHCH0,750,75Ghi ch : - Th sinh lm cch khc nhng ng kt qu vn cho im ti a- Phng trnh ha hc ghi thiu iu kin tr i s im.S GD& T NGH AN K THI CHN HC SINH GII TNH LP 12 NM HC 2009 - 2010

Mn thi: HO HC - THPT BNG AThi gian: 180 pht (khng k thi gian giao ) thi chnh thc ( thi gm 02 trang)Cu 1 (2,5 im).Phenol v anilin u phn ng vi dung dch nc brom, nhng toluen th khng.1. T kt qu thc nghim c th rt ra kt lun g?2. Anisol (metylphenyl ete) c phn ng vi dung dch nc brom khng ? Gii thch.3. Nu cho dung dch nc brom ln lt vo tng cht ptoludin (paminotoluen), pcresol (pmetylphenol) theo t l mol 1 : 2 th thu c sn phmchnh l g? Cu 2 (2,5 im).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, ri 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 kt ta H mu en. t chy H bi oxita 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.Cu 3 (4,0 im)1. Cht X c cng thc phn t C8H15O4N. T X c hai bin ha sau :C8H15O4N 0t , OH dungdichNa C5H7O4NNa2 + CH4O + C2H6OC5H7O4NNa2 l dungdichHC C5H10O4NCl + NaClBit: 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 didng cng thc cu to.2. Hp cht A c cng thc C9H8 c kh nng kt ta vi dung dch AgNO3 trong NH3 v phn ng vi brom trong CCl4 theo t l mol 1 : 2. un nng A vi dung dch KMnO4 ti khi ht mu tm, ri thm lng d dung dch HCl c vo hn hp sau phn ng thy c kt ta trng l axit benzoic ng thi gii phng kh CO2v Cl2. Xc nh cng thc cu to ca A v vit phng trnh ha hc ca cc phn ng xy ra.Cu 4 (3,0 im). Cho hn hp Y gm ba kim loi K, Zn, Fe vo nc d thu c 6,72 lt kh (ktc) v cn li cht rn B khng tan c khi lng 14,45 gam. Cho B vo 100 ml CuSO4 3M, thu c cht rn C c khi lng 16,00 gam. Xc nh khi lng mi kim loi trong Y? Cu 5 (4,0 im)1. T kh thin nhin v cc cht v c cn thit, thit b phn ng y . Hy vit phng trnh iu ch cc cht sau : mH2NC6H4COONa v pH2NC6H4COONa2. Hai hp cht thm A v B l ng phn c cng thc phn t CnH2n-8O2. Hi B c khi lng ring 5,447 gam/lt ( ktc). A c kh nng phn ng vi Na gii phng H2v c phn ng trng gng. B phn ng c vi NaHCO3 gii phng kh CO2.a) Vit cng thc cu to ca A v B.b) Trong cc cu to ca A c cht A1 c nhit si nh nht. Hy xc nh cng thc cu to ng ca A1.c) Vit cc phng trnh phn ng chuyn ha ocrezol thnh A1.Cu 6 (4,0 im).1. Cho 20,80 gam hn hp Fe, FeS, FeS2, S tc dng vi dung dch HNO3 c nng d thu c V lt kh NO2(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 91,30 gam kt ta. Tnh V? 2. Trong mt bnh kn A dung tch 1 lt 500 0C, hng s cn bng ca phn ng tng hp HI t H2 v I2 bng 46. a)Tnh nng mol cc cht trng thi cn bng? Bit ban u trong bnh A c 1mol H2 v 1mol I2b) Nu ban u cho 2 mol HI vo bnh A nhit 500 0C th nng cc cht lc cn bng l bao nhiu?c) Nu h ang trng thi cn bng cu a, ta thm vo h 1,5 mol H2 v 2,0 mol HI th cn bng dch chuyn theo chiu no?(Cho H=1, C=12, N=14, O=16, Na=23, S=32, K =39, Fe=56; Zn=65, Ba =137)- - - Ht - - H v tn th sinh:.................................................................S bo danh:.......................S GD& T NGH AN K THI CHN HC SINH GII TNH LP 12NM HC 2009 - 2010HNG DN V BIU IM CHM THI CHNH THCMn thi: HO HC - THPT BNG A(Hng dn v biu im gm 06 trang) Cu Ni dung imCu 1 2,5Khi cho phenol, anilin, toluen vo nc brom, cc cht phn ng vi nc brom l :NH2+ 3Br2NH2Br BrBr+ 3HBr + 3Br2Br BrBr+ 3HBrOHOHT kt qu c th rt ra kt lun : Kh nng phn ng ca phenol v anilin mnh hn nhiu so vi toluen. T suy ra cc nhm NH2, -OH c tc dng hot ha nhn thm mnh hn nhm CH30,250,250,25Cu to ca phenol, anisol l:O H O CH3So snh cu to ca phenol v anisol ta thy anisol c nhm CH3 y electron nn nhm OCH3 y electron mnh hn nhm OH, lm mt electron trong vng benzen ca anisol ln hn ca phenol. V vy anisol phn ng vi nc brom.+ 3Br2Br BrBr+ 3HBrOCH3OCH30,250,250,253CH3+2Br2+2HBrNH2CH3NH2Br BrCH3+2Br2+2HBrCH3Br BrOHOH0,50,5Cu 2 2,5A : H2S; B : FeCl3; C : S ;F : HCl ; G : Hg(NO3)2 ; H : HgS ; I : Hg ; X : Cl2 ; Y : H2SO4Th sinh c th khng lp lun ch cn xc nh ng cc cht v vit phng trnh cho im ti a (B ngoi FeCl3 c th ly cc cht khc) Phng trnh ha hc ca cc phn ng :H2S + 2FeCl3 2FeCl2 + S + 2HCl (1)Cl2 + H2S S + 2HCl(2)2,5Ghi ch : - Th sinh lm cch khc nhng ng vn cho im ti a, - Phng trnh ha hc ghi thiu iu kin tr i s im.S GD&T NGH AN K THI CHN HC SINH GII TNH LP 12 NM HC 2008 - 2009Mn thi: HO HC 12THPT- BNG BThi gian lm bi: 180 phtCu1 (2,5 im):Vit tt c cc ng phn ca C3H5Br3. Cho cc ng phn ln lt tc dng vi dd NaOH un nng. Hy vit cc phng trnh phn ng.Cu 2 (1,5 im): Cho cc dd: NaOH, HCl, CH3COONa, H2NCH2COOH, CH3NH2, NH4Cl. Hy vit cc phng trnh phn ng khi cho cc cht tc dng vi nhau tng i mtCu 3 (2,5 im). T nguyn liu ban u l than, vi, nc, ta iu ch c kh A. T A c s chuyn ha sau:A B DEF GH IBit cht E khng cha oxi, khi t chy hon ton E cn 3,808 dm3 O2 (ktc), sn phm sinh ra c 0,73 g HCl, cn CO2 v hi nc to ra theo t l th tch V(CO2) : V(H2O) =6:5 (o cng iu kin nhit , p sut). Tm cng thc cu to cc cht hu c ng vi ch ci c trong s v vit cc phng trnh phn ng.Cu 4 (1,5 im). Polime A c to ra do phn ng ng trng hp gia but-1,3-dien v stiren. Bit 6,234 g A phn ng va ht vi 3,807 g Br2. Tnh t l s mt xch but-1,3-dien v stiren trong polime trn. Vit cng thc cu to mt on mch bt k ca A tha mn t l trn.Cu 5 (2,0 im). Cht A c cng thc phn t C5H6O4l este hai chc, cht B c cng thc phn t C4H6O2 l este n chc. Cho A v B ln lt tc dng vi dd NaOH d, sau c cn cc dung dch ri ly cht rn thu c tng ng nung vi NaOH (c mt ca CaO) th trong mi trng hp ch thu c mt kh duy nht l CH4. Hy tm cng thc cu to ca A, B v vit cc phng trnh phn ng xy ra.Cu 6 (2,0 im). C th dng dd nc Br2 phn bit cc kh sau y: NH3, H2S, C2H4, SO2 ng trong cc bnh ring bit c khng? Nu c hy nu hin tng quan st, vit phng trnh phn ng gii thch.Cu 7 (2,5 im).FeO l oxit baz, va c tnh oxi ha, va c tnh kh. Al(OH)3l cht lng tnh. CaCO3va c tnh baz va khng bn nhit. HCl l axit c tnh kh. NH3c tnh baz yu hn KOH. Hy vit phng trnh phn ng (dng phn t) minh ha.Cu 8 (1,5 im). Ly mt si dy in gt b v nha bng PVC ri t li ng trn ngn la n cn th thy ngn la c mu xanh l m. Sau mt lc, ngn la mt mu xanh. Nu p li dy ng ang nng vo v nha trn ri t th ngn la li c mu xanh l m. Gii thch hin tng. Vit cc phng trnh phn ng xy ra.Cu 9 (1,0 im). Kt qu xc nh s mol ca cc ion trong dung dch X nh sau: Na+ c 0,1 mol; Ba2+ c 0,2 mol; HCO3- c 0,05 mol; Cl- c 0,36 mol. Hi kt qu trn ng hay sai? Gii thch.Cu 10 (3,0 im). Ha tan hon ton hn hp A gm FeS v FeCO3 bng dung dch HNO3 c, nng thu c hn hp B mu nu nht gm hai kh X v Y c t khi i vi H2 l 22,8 v dung dch C. Bit FeS phn ng vi dung dch HNO3 xy ra nh sau:FeS + HNO3 Fe(NO3)3 + H2SO4 + NO2 + H2Oa. Tnh t l % theo khi lng cc mui trong Ab. Lm lnh hn hp kh B nhit thp hn c hn hp D gm ba kh X, Y, Zc t khi so vi H2 l 28,5. Tnh thnh phn % theo th tch cc kh trong Dc. -110C hn hp D chuyn thnh hn hp E gm hai kh. Tnh t khi ca E so vi H2 Bit: C=12; H=1;O=16; N=14; Fe=56; Br= 80; S = 32;-------------Ht------------- chnh thc6000cThan-HCl+dd Cl2 +NaOHH2SO4c1700CH v tn th sinh:....................................................................S bo danh:.....................S GD&T NGH ANK THI CHN HC SINH GII TNH LP 12 Nm hc 2008 - 2009HNG DN V BIU IM CHM CHNH THC(Hng dn v biu im chm gm 05 trang)Mn: HO HC 12 THPT - BNG BCu Ni dungBiu im1 2,5Cc ng phn ca C3H5Br3 l:CH2Br-CHBr-CH2Br (1)CH2Br-CH2-CHBr2 (2)CHBr2-CHBr-CH3(3)CH2Br-CBr2-CH3(4)CH3-CH2-CBr3(5)5CT x 0,25 =1,25CH2Br-CHBr-CH2Br+ 3NaOH CH2OH-CHOH-CH2OH +3NaBrCH2Br-CH2-CHBr2 + 3NaOHCH2OH-CH2-CHO + 3NaBr + H2OCHBr2-CHBr-CH3+ 3NaOHCH3-CHOH-CHO+3NaBr + H2OCH2Br-CBr2-CH3+ 3NaOH CH2OH-CO-CH3 + 3NaBr + H2OCH3-CH2-CBr3 + 4NaOHCH3COONa + 3NaBr + 2H2O5p x 0,25 =1,252 1,5NaOH + HClNaCl + H2O(1)NaOH + H2NCH2COOH H2NCH2COONa + H2O(2)NaOH + NH4Cl NaCl + NH3 + H2O(3)HCl + CH3COONa CH3COOH + NaCl(4)HCl + CH3NH2CH3NH3Cl(5)HCl + H2NCH2COOH NH3ClCH2COOH(6)6p x 0,25 =1,53 2,5Xc nh cht E:23, 8080,1722, 4On mol; 0,730,0236,5HCln molTheo gi thit, cht E cha 3 nguyn t C, H, Cl nn oxi c trong CO2, H2O bng lng oxi tham gia phn ng (theo nh lut bo ton khi lng)Nu coi s mol CO2 = 6a th s mol H2O = 5aTa c 6a. 2 + 5a = 0,17.2 a=0,02 Suy ra s mol CO2 = 6a = 0,12 s mol C = 0,12S mol H2O = 5a = 0,1 s mol H = 0,2Xc nh c cng thc phn t cht E cho 0,5 S mol HCl = 0,02 s mol H = s mol HCl = 0,02Tng s mol H = 0,2 + 0,02 = 0,22 T l C:H:Cl = 0,12:0,22:0,02 = 6 : 11 : 1Cng thc n gin ca E l: C6H11Cl. Theo s cho, cng thc ca E phi l C6H11ClTm cng thc cc cht hu c nu trong s v vit phng trnh phn ngCaCO3CaO + CO2(1)CaO + 3C CaC2 + CO(2)CaC2 + 2H2O C2H2 + Ca(OH)2(3) (A)3C2H2 (4)(B)+ 3H2(5) (D) + Cl2+ HCl (6) (E)

+ + HCl (7) (F)+ Cl2 (8)(G)+ 2NaOH+ 2NaCl(9) (H) + 2H2O(10)hoc+ 2H2O

hoc+ 2H2OimXc nh 8 cng thc cu to cho 1 im. Vit c 10 phn ng cho 1 im.Nu khng xc nh E m vn hon thnh s th tr 0,5 im t0l inthan6000Nit0asClCl kimtrong ancolClClClClt0OHOHOHOHH2SO4 c1800c (I) 4 1,5Gi A l (C4H6)n-(C8H8)m. Phng trnh phn ng vi Br2(-CH2-CH=CH-CH2)n(CH2-CH-)m + nBr20,5Theo phng trnh c (54n + 104m) g cn 160n g Br2 Theo d kin 6,324 g cn 3,807 g Br2 Ta c 54 104 1606,324 3,807n m n + 12nm 0,5Cng thc cu to 1 on mch ca A:-CH2-CH=CH-CH2-CH-CH2-CH-CH2-0,55 2Cc cu to tho mn ca A v cc phng trnh phn ng:CH2-COO-CH2hoc COO-CH2Cu to ca B: CH3-COOCH=CH2Vit c 3 cng thc cu to cho 0,75 imCc phn ng:CH2-COO-CH2+ 2NaOHCH2-COONa + C2H4(OH)2(1)COO-CH2COONa + 2NaOH CH2-COONa + CH3CHO + H2O (2) COONaCH3COOCH=CH2 + NaOH CH3COONa + CH3CHO (3)CH2-COONa +2NaOHCH4 + 2Na2CO3 (4)COONaCH3COONa + NaOH CH4 + Na2CO3 (5)Vit c 5 phng trnh phn ng cho 1,25 im C6H5 C6H5CH2COOCOOCH-CH3CH2COOCOOCH-CH3CaOt0CaOt0(-CH2-CHBr-CHBr-CH2)n (CH2-CH-)m 6 2C th dng dd nc Br2 nhn bit cc kh , c th:. NH3: dd Br2 mt mu, c kh khng mu khng mi thot ra2NH3 + 3Br2N2+ 6HBrHoc 8NH3 + 3Br2N2+ 6NH4Br0,5. H2S: dd Br2 mt mu, c kt ta mu vngH2S + Br22HBr + S0,5. C2H4: dd brom mt mu, to cht lng phn lpC2H4 + Br2C2H4Br20,5. SO2: dd brom mt mu, to dd trong sut ng nhtSO2 + Br2 + 2H2O2HBr + H2SO40,57 2,5FeO + 2HCl FeCl2 + H2OFeO + CO Fe + CO23FeO + 10HNO33Fe(NO3)3 + NO + 5H2OAl(OH)3 + 3HCl AlCl3 + 3H2OAl(OH)3 + NaOH Na[Al(OH)4]CaCO3 + 2HCl CO2 + H2O + CaCl2CaCO3CaO + CO2KOH + NH4Cl KCl + NH3 + H2OHCl + NH3NH4Cl 4HCl + MnO2MnCl2 + Cl2 + 2H2O10 p x 0,25 = 2,58 1,5Khi gt b v PVC, li ng t nhiu vn cn PVC nn khi t s c qu trnh sinh ra CuCl2, CuCl2 phn tn vo ngn la, ion Cu2+ to mu xanh l m c trng. Khi ht CuCl2 (ht PVC) ngn la li khng mu. Nu cho dy ng p vo PVC th hin tng lp li0,5Cc phn ng: PVC chy: (-CH2-CHCl-)n + 2,5nO22nCO2 + nH2O + nHCl0,52Cu + O22CuO 0,252HCl + CuO CuCl2 + H2O (Nu hc sinh vit Cu + HCl th khng cho im)0,25 t0t0t0t0t09 1Trong dd X tng in tch dng: 0,1 + 0,2.2 = 0,5Trong dd X tng in tch m: 0,05 + 0,36 = 0,410,5Kt qu trn l sai v tng in tch dng khng bng tng intch m0,510 3a Theo ra th hn hp kh B phi l NO2 v CO2 theo phn ng sau 0,25FeS + 12HNO3Fe(NO3)3 + H2SO4 +9NO2 + 5H2OFeCO3 + 4HNO3 Fe(NO3)3 + CO2 +NO2 + 2H2O0.5t nFeS=a (mol), nFeCO3 = b (mol) suy ra nNO2=9a + b, nCO2 = b 0,25Ta c: 46(9 ) 4422,82(9 2 )a b ba b+ ++a:b=1:3nFeS : nFeCO3 = 1:30,25T l khi lng:388 20,18%348 79,82%FeSFeCO 0,25b Lm lnh B c phn ng sau: 2NO2 N2O4 khi 2 492N OM , lm M = 570,25Gi x l s mol N2O4 c trong hn hp D Trong D gm: (9a + b) - 2x = 4b -2x mol NO2, x mol N2O4 , b mol CO20,25Suy ra 46(4b-2x)+92x+44b=57(4b-2x+x+b) x=b0,25Tng s mol trong D =4b gm NO2 = 2b chim 50%, N2O4=b chim 25%, CO2=b chim 25%0,25C -110c phn ng: 2NO2N2O4 xy ra hon tonHn hp E gmN2O4 v CO2 trong nN2O4 =2b; nCO2=b0,25T khi i vi H2: 92.2 44382( 2 )b bb b++0,25----------------------------------------------Ch : Hc sinh gii theo cch khc nu ng vn cho im ti a.S GD&T NGH AN K THI CHN HC SINH GII TNHNM HC 2007-2008Mn thi: HO HC LP 12 THPT - BNG AThi gian: 180 pht (khng k thi gian giao )Cu 1 (5,0 im): chnh thc 1. Nu hin tng, vit phng trnh phn ng xy ra trong cc trng hp sau:a) Trn dung dch Na2CO3 vi dung dch FeCl3.b) Sc kh H2S n bo ho vo dung dch FeCl3 .c) Cho ur vo dung dch Ba(OH)2 .2. Vit cc qu trnh xy ra cc in cc v phng trnh in phn cc dung dch sau: a) BaCl2 (c mng ngn)b) CuSO4 c) K2SO43. Gii thch hin tng st ty, tn b n mn trong khng kh m. Cu 2 (7,0 im).1. Cc cht hu c A, B, C, D c cng cng thc phn t C4H6O4 u phn ng vidung dch NaOH theo t l mol 1:2. Trong : - A, B u to ra mt mui, mt ru.- C, D u to ra mt mui, mt ru v nc.Bit rng khi t chy mui do A, C to ra th trong sn phm chy khng c nc. Xc nh A, B, C, D v vit phng trnh phn ng vi NaOH.2. C th tn ti bao nhiu loi lin kt hyr trong ru etylic c ho tan phenol. Vit cng thc biu din cc mi lin kt ny v cho bit lin kt no bn nht, lin kt no km bn nht? Gii thch.3. Cho 9,2g mt hp cht hu c n chc X tc dng va vi dung dch cha 0,2mol Ag2O trong NH3thu c 21,6 g Ag. Xc nh cng thc cu to ca X. Vit phng trnh phn ng ho hc xy ra.Cu 3 (5,0 im). Cho 39,84g hn hp F gm Fe3O4 v kim loi M vo dung dch HNO3 un nng, khuy u hn hp phn ng hon ton thu c 4,48 ltkh NO2 l sn phm kh duy nht ( ktc), dung dch G v 3,84gkim loi M. Cho 3,84g kim loi M vo 200ml dung dch H2SO4 0,5M v KNO3 0,5M khuy u th thu c dung dch H, kh NO duy nht. Cho dung dch NH3d vo dung dch G thu c kt ta K. Nung K trong khng kh n khi lng khng i thu c 24g cht rn R. a) Tm kim loi M (bit M c ho tr khng i trong cc phn ng trn).b) C cn cn thn dung dchH thu c bao nhiu gam mui khan?Cu 4 (3,0 im). 1. xi ho mt ru X bi xi c bt ng lm xc tc, c cht kh Y. xi ho Y vi xc tc Pt thu c axt Z. Cho Z tc dng vi xt c mui T. Cho T tc dng vi dung dch Ag2O/ NH3 c Ag kim loi. Tm cng thc cu to ca X, Y, Z, T. Vit cc phng trnh phn ng.2. Trn Y vi mt anht P c 5,9g ri un nng nh vi mt lng d dung dch Ag2O trong NH3 thu c 2,24 lt CO2 ( ktc) v 64,8g Ag. Xc nh cng thc cu to ca P.( Cho Fe = 56; Al = 27; Cu = 64; N = 14; O = 16; H = 1; Ag = 108, Zn = 65,Mg=24, C = 12, Pb = 207, K = 39)Th sinh khng c s dng bt c ti liu g.--------------Ht -------------S GD& T NGH AN K THI CHN HC SINH GII TNH LP 12NM HC 2009 - 2010

Mn thi: HO HC - THPT BNG AThi gian: 180 pht (khng k thi gian giao )Cu 1 (4 im). thi d b ( thi gm 02trang)Ha tan 48,8 gam hn hp A gm Cu v mt oxit st trong HNO3 d thu c dung dch B v 6,72 lt kh NO (ktc) l sn phm kh duy nht. C cn dung dch B thu c 147,8 gam hn hp hai mui. 1. Xc nh cng thc oxit st 2. Cho 48,8 gam hn hp A trn vo 400 ml dung dch HCl 2M cho n khi cc phn ng xy ra hon ton thu c dung dch C v cht rn D. Cho C tc dng vi dung dch AgNO3 d thu c cht rn E. Tnh khi lng ca E3. Tnh th tch NO duy nht (ktc) thu c khi cho D tc dng vi HNO3 dCu 2 ( 4 im).t chy hon ton 9,0 gam mt axit cacboxylic A thu c 6,72 lt CO2(ktc) v 5,4 gam H2O1. Xc nh cng thc cu to v gi tn A2. Cho A tc dng vi hn hp 2 ancol n chc, mch h l ng ng k tip nhau thu c 8,0 gam hn hp este. Cho ton b este ny tc dng vi NaOH va thu c 8,15gam mui. Xc nh cng thc phn t ca 2 ancolCu 3 ( 3 im. un nng 28,75 gam etanol vi H2SO4 c 1700C. Cho ton b sn phm thu c qua bnh 1 ng CuSO4 khan, sau qua bnh 2 ng dung dch KOH c d, cui cng qua bnh 3 ng dung dch Br2d trong dung mi CCl4. Sau th nghim khi lng bnh 3 tng 10,5 gam. Li cho dung dch H2SO4 long d tc dng vi dung dch bnh 2 thu c hn hp kh A, A c kh nng lm mt mu dung dch Br2, KMnO4 long1. Vit cc phng trnh phn ng xy ra2. Tnh hiu sut ca phn ng to thnh cht hp th bnh 3Cu 4 ( 3 im). Trong phng th nghim c 2 dung dch cha bit nng l NaHSO4 v Na2CO3. Ngi ta tin hnh cc th nghim sau:Th nghim 1: Cho t t 100 gam dung dch NaHSO4 vo 100 gam dung dch Na2CO3 thu c 198,9 gam dung dch ATh nghim 2: Cho t t 100 gam dung dch Na2CO3 vo 100 gam dung dch NaHSO4 thu c 197,8 gam dung dch BTh nghim 3: Cho t t 50 gam dung dch Na2CO3vo 100 gam dung dch NaHSO4thu c 150 gam dung dch C1. Hy gii thch kt qu cc th nghim trn2. Tnh C% ca 2 dung dch ban uCu 5 ( 3 im). Hp cht hu c X no mch h, phn t cha mt chc OH v chc COOH, c cng thc (C4H6O5)n 1. Xc nh cng thc phn t v vit cc ng phn cu to c th c ca X. 2.Xc nh cng thc cu to ng ca X, bit X tch nc cho hai sn phmngphn Y, Z. Vit cng thc cu to ca Y, Z. 3.So snh nhit nng chy ca Y, Z gii thch. 4.So snh tnh axit ca Y v Z. Gii thch Cu 6 ( 3 im). Dung dch A l dung dch hn hp CH3COOH 0,1M;CH3COONa 0,1M1. Tnh pH ca dung dch A2. Tnh pH ca dung dch thu c khi a) Cho 0,01 mol HCl vo 1 lt dung dch A b) Cho 0,01 mol NaOH vo 1 lt dung dch AGi s th tch dung dch thay i khng ng k (Cho H=1, C=12, N=14, O=16, Na=23, S=32, K =39, Fe=56; Zn=65, Ba =137)- - - Ht - - -H v tn th sinh:...................................................................S bo danh:...................S GD&T NGH AN K THI CHN HC VIN GII TNH LP 12 NM HC 2008 - 2009Mn thi: HO HC 12B TC THPTThi gian lm bi: 180 phtCu 1 (1,5 im). Cho cc dd: NaOH, HCl, CH3COONa, H2NCH2COOH, CH3NH2, NH4Cl. Hy vit cc phng trnh phn ng khi cho cc cht tc dng vi nhau tng i mtCu2 (3,75 im).Vit tt c cc ng phn ca C3H5Br3. Cho cc ng phn ln lt tc dng vi dd NaOH un nng. Hy vit cc phng trnh phn ng.Cu 3 (3,25 im). Cc cht sau: Na, H2S, CO2, Cl2, CH4, HCHO, C6H12O6 (glucoz) c tnh xi ho hay kh? Mi tnh cht vit mt phng trnh phn ng minh ho.Cu 4 (2,0 im). Cho cc hp cht sau: NH3, C6H5NH2, CH3NH2, C6H5OH, (CH3)2NH. Hy sp xp tnh baz theo th t tng dn. Gii thch ngn gn. Cu 5 (1,5 im). Kt qu xc nh s mol ca cc ion trong dung dch X nh sau: Na+ c 0,1mol, Ba2+ c 0,2mol, HCO3- c 0,05mol, Cl- c 0,36mol. Hi kt qu trn ng hay sai? Gii thch.Cu 6 (4,0 im). t chy hon ton 1,06g hn hp 2 ancol no, n chc, mch h l ng ng k tip nhau ri cho ton b sn phm vo 1 lt dung dch NaOH 0,2M. Sau th nghim nng dung dch NaOH cn li 0,1M (gi s th tch dung dch khng thay i).a. Xc nh cng thc phn t, cng thc cu to v gi tn 2 ancol.b. Xc nh % theo khi lng mi ancol.Cu 7 (4,0 im). Cho dung dch HNO3 tc dng vi 2 kim loi nhm v magi, sau phn ng thu c 0,896 lt (ktc) hn hp 2 kh l NO v N2O (l sn phm kh duy nht) c t khi so vi hyr l 16,75.a. Tnh % theo th tch ca mi kh.b. Tnh s mol ion NO3- to thnh trong mui.Bit: C=12; H=1;O=16; Al = 27; Mg = 24; Na = 23.-------------Ht------------- chnh thcS Gd&t Ngh an K thi chn hc vin gii tnh lp 12 Nm hc 2008 - 2009hng dn v biu im Chm chnh thc(Hng dn v biu im chm gm 03 trang)Mn: Ha Hc - b tc thpt ---------------------------------------Cu Ni dung im1 1,5NaOH + HClNaCl + H2O(1)NaOH + H2NCH2COOH H2NCH2COONa + H2O (2)NaOH + NH4Cl NaCl + NH3 + H2O(3)HCL + CH3COONa CH3COOH + NaCl (4)HCL + CH3NH2CH3NH3Cl (5)HCl + H2NCH2COOH NH3ClCH2COOH(6)6 p x 0,25 = 1,252 3,75Cc ng phn ca C3H5Br3 l:CH2Br-CHBr-CH2Br (1)CH2Br-CH2-CHBr2 (2)CHBr2-CHBr-CH3(3)CH2Br-CBr2-CH3(4)CH3-CH2-CBr3(5)5p x 0,5 =2,5CH2Br-CHBr-CH2Br+ 3NaOH CH2OH-CHOH-CH2OH +3NaBrCH2Br-CH2-CHBr2 + 3NaOHCH2OH-CH2-CHO + 3NaBr + H2OCHBr2-CHBr-CH3+ 3NaOHCH3-CHOH-CHO+3NaBr + H2OCH2Br-CBr2-CH3+ 3NaOH CH2OH-CO-CH3 + 3NaBr + H2OCH3-CH2-CBr3 + 4NaOHCH3COONa + 3NaBr + 2H2O5p x 0,25 =1,253 3,25Cht c tnh kh: Na, H2S, CH40,25Cht c tnh oxi ho: CO20,25Cht va c tnh oxi ho va c tnh kh: Cl2, HCHO, C6H12O6, (CH4)0,52Na + Cl22NaCl (1)2H2S + O22S + 2H2O (2)CH4 + 2O2 CO2 + 2H2O (3)CO2 + 2Mg 2MgO + C (4)Cl2+ 2NaOHNaCl + NaClO + H2O (5)HCHO + H2CH3OH 9 p x 0,25 =2,25 (6)HCHO + 4AgNO3 + 6NH3 + 2H2O(NH4)2CO3 + 4Ag + 4NH4NO3 (7)CH2OH-(CHOH)4-CHO + H2CH2OH-(CHOH)4-CH2OH (8)CH2OH-(CHOH)4-CHO+ 2AgNO3 + 3NH3 + H2OCH2OH-(CHOH)4-COONH4 + 2Ag + 2NH4NO3 (9)4 2C6H5OH < C6H5NH2 < NH3