43.-thpt-nguyen-dinh-chieu-ben-tre-nam-2017-lan-1-co-loi-giai

S GD & T TNH BN TRE THPT NGUYN èNH CHIU THI TH THPT QUC GIA LN NM 2017 Mụn: KHOA HC T NHIấN HểA HC Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt ( thi cú 40 cõu / trang) Mó : 132 Cho bit nguyờn t ca cỏc nguyờn t : H =1; C = 12; N = 14; O = 16; Na = 23; Mg = 24; Al = 27; S =32; Cl = 35,5; K = 39; Ca = 40; Cr = 52; Fe = 56; Cu = 64; Zn = 65; Ag = 108; Ba=137 Cõu 1: Oxit no sau õy l oxit axit A CrO B Al2O3 C CrO3 D Fe2O3 Cõu 2: Kim loi no sau õy l kim loi kim th A Ca B Na C Al D Fe Cõu 3: Phn trm lng ca nguyờn t oxi glucoz l A 44,44% B 53,33% C 51,46% D 49,38% Cõu 4: T no sau õy thuc t nhõn to A t olon B t tm C t visco D t nilon-6,6 Cõu 5: Nhit phõn hiroxit Fe (II) khụng khớ n khi lng khụng i thu c A Fe3O4 B FeO C Fe2O3 D Fe Cõu 6: Phn ng no sau õy l sai A Cr(OH)3 + NaOH NaCrO2 + 2H2O B 3Zn + 2CrCl3 3ZnCl2 + 2Cr C 2Cr + 3Cl2 2CrCl3 D 2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O Cõu 7: Dung dch anilin (C6H5NH2) khụng phn ng c vi cht no sau õy A NaOH B Br2 C HCl D HCOOH Cõu 8: Trng hp no sau õy to sn phm l ancol v mui natri ca axit cacboxylic t0 t0 A HCOOCH=CH-CH3 + NaOH B HCOOC(CH3)=CH2 + NaOH 0 t t C CH2=C(CH3)COOH + NaOH D HCOOCH2CH=CH2 + NaOH Cõu 9: Trong cụng nghip, Al c iu ch bng cỏch no di õy A in phõn núng chy AlCl3 B in phõn dung dch AlCl3 C Cho kim loi Na vo dung dch AlCl3 D in phõn núng chy Al2O3 Cõu 10: Nhn nh no sau õy l ỳng A Nc cng l nc cha nhiu ion HCO3- v SO42B lm mm tớnh cng ca nc cng vnh cu bng cỏch un núng C Nc t nhiờn thng cú c tớnh cng tm thi v tớnh cng vnh cu D Nc cng l tỏc nhõn gõy ụ nhim ngun nc hin Cõu 11: Mt phõn t xenluloz cú phõn t l 15.10 6, bit rng chiu di mi mc xớch C6H10O5 khong 5.10-7 (mm) Chiu di ca mch xenluloz ny gn ỳng l A 3,0.10-2 (mm) B 4,5.10-2 (mm) C 4,5.10-1 (mm) D 3,0.10-1 (mm) Cõu 12: t chỏy 0,01 mol este X n chc bng lng oxi va , thu c 448 ml khớ CO (ktc) Mt khỏc un núng 6,0 gam X vi dung dch KOH va , thu c lng mui l : A 10,0 gam B 6,8 gam C 9,8 gam D 8,4 gam Cõu 13: Cho 2a mol bt Fe vo dung dch cha 5a mol AgNO3, sau phn ng xy hon ton, thu c dung dch gm cỏc cht Trang A Fe(NO3)3 B Fe(NO3)2 v Fe(NO3)3 C Fe(NO3)2, AgNO3 D Fe(NO3)3 v AgNO3 Cõu 14: iu khng nh no sau õy l sai A Al(OH)3 phn ng c vi dung dch HCl v dung dch KOH B Trong cỏc phn ng húa hc, kim loi Al ch úng vai trũ l cht kh? C Kim loi Al tan c dung dch H2SO4 loóng, ngui D Kim loi Al cú tớnh dn in tt hn kim loi Cu Cõu 15: Hũa tan ht 3,24 gam bt Al dung dch HNO3 d, thu c 0,02 mol khớ X nht v dung dch Y cha 27,56 gam mui Khớ X l A NO2 B N2O C N2 D NO Cõu 16: X phũng húa hon ton triglyxerit X dung dch NaOH d, thu c glyxerol, natri oleat, natri stearat v natri panmitat Phõn t ca X l A 860 B 862 C 884 D 886 Cõu 17: phõn bit cỏc dung dch riờng bit: anilin, glucoz v alanin, ta dựng dung dch no sau õy A NaOH B AgNO3/NH3 C HCl D Br2 Cõu 18: Hn hp X gm FeCl2 v KCl cú t l mol tng ng l : Hũa tan hon ton 16,56 gam X vo nc d thu c dung dch Y Cho dung dch AgNO3 d vo Y, kt thỳc phn ng thu c m gam kt ta Giỏ tr m l A 40,92 gam B 37,80 gam C 49,53 gam D 47,40 gam Cõu 19: S ng phõn cu to ng vi amin cú cụng thc phõn t C4H11N l A B C D Cõu 20: Phỏt biu no sau õy l ỳng A Xenluloz tan tt imetyl ete B Saccaroz cú kh nng tham gia phn ng gng C Glucoz th hin tớnh oxi húa tỏc dng vi H2 (xỳc tỏc Ni, to) D Amiloz v amilopectin l ng phõn ca Cõu 21: Cho V1 ml dung dch NaOH 0,4M vo V2 ml dung dch H2SO4 0,6M Sau kt thỳc phn ng thu c dung dch ch cha mt cht tan nht T l V1 : V2 l A : B : C : D : Cõu 22: Hũa tan ht m gam hn hp X gm Cu v Fe dung dch HCl loóng d, thu c 0,09 mol khớ H2 Nu cho m gam X trờn vo dung dch HNO3 loóng d, thu c 0,15 mol khớ NO (sn phm kh nht ca N+5) Giỏ tr ca m l A 12,48 gam B 10,80 gam C 13,68 gam D 13,92 gam Cõu 23: Thớ nghim no sau õy cú phn ng húa hc xy A Cho dung dch Fe(NO3)3 vo dung dch AgNO3 B Cho Cr2O3 vo dung dch NaOH loóng C Nh dung dch Br2 vo dung dch cha NaCrO2 v NaOH D Cho bt Fe vo dung dch HNO3 c, ngui Cõu 24: Nhn nh no sau õy l sai A iu kin thng, cỏc amino axit l cht rn, d tan nc B trng thỏi tinh th, cỏc amino axit tn ti dng ion lng cc (mui ni phõn t) C Axit glutamic l thuc h tr thn kinh D Trựng ngng axit 6-aminohexanoic thu c nilon-6 cú cha liờn kt peptit Cõu 25: Hiro húa hon ton a mol cht hu c X mch h cn dựng a mol H2 (xỳc tỏc Ni, t0) thu c este Y cú cụng thc phõn t C4H8O2 S ng phõn tha ca X l A B C D Trang Cõu 26: Cho hn hp gm Na, Ba v Al vo lng nc d, thy thoỏt V lớt khớ H2 (ktc); ng thi thu c dung dch X v cũn li 1,08 gam rn khụng tan Sc khớ CO d vo X, thu c 12,48 gam kt ta Giỏ tr ca V l A 1,792 lớt B 3,584 lớt C 7,168 lớt D 8,960 lớt Cõu 27: Cho 15,94 gam hn hp gm alanin v axit glutamic tỏc dng va vi 200 ml dung dch NaOH 1M, thu c dung dch X Cho 450 ml dung dch HCl 0,8M vo dung dch X, cụ cn dung dch sau phn ng thu c m gam rn khan Giỏ tr m l A 32,75 gam B 33,48 gam C 27,64 gam D 33,91 gam Cõu 28: un núng 0,2 mol hn hp X cha etyl fomat v etyl axetat vi dung dch AgNO 3/NH3 (dựng d) thu c 17,28 gam Ag Nu thy phõn hon ton 28,84 gam X vi dung dch KOH va , thu c m gam mui Giỏ tr ca m l A 37,24 gam B 26,74 gam C 31,64 gam D 32,34 gam Cõu 29: Thc hin cỏc phn ng sau: (1) Cho Na vo dung dch CuSO4 (2) in phõn dung dch CuSO4 bng in cc tr (3) Thi lung khớ H2 n d qua ng nghim cha CuO (4) Nung núng hn hp gm Al v CuO khớ tr (5) Cho bt Fe vo dung dch CuCl2 Sau kt thỳc thớ nghim, s trng hp thu c Cu n cht l A B C D Cõu 30: t chỏy hon ton 14,24 gam hn hp X cha este u no, n chc, mch h thu c CO v H2O cú tng lng l 34,72 gam Mt khỏc un núng 14,24 gam X vi dung dch NaOH va , thu c hn hp Y cha ancol k tip v hn hp Z cha mui ca axit cacboxylic k tip, ú cú a gam mui A v b gam mui B (MA < MB) T l gn nht ca a : b l A 0,6 B 1,25 C 1,20 D 1,50 Cõu 31: Cho hn hp gm Fe2O3 v Cu vo lng d dung dch H2SO4 loóng thu c dung dch X v cũn li mt phn rn khụng tan Dung dch X tỏc dng c vi bao nhiờu cht s cỏc cht sau: Fe, NaNO3, Cl2 v KMnO4 A B C D Cõu 32: Hn hp X cha hai hp cht hu c gm cht Y (C2H7O2N) v cht Z (C4H12O2N2) un núng 9,42 gam X vi dung dch NaOH d, thu c hn hp T gm hai amin k tip cú t so vi He bng 9,15 Nu cho 9,42 gam X tỏc dng vi dung dch HCl loóng d, thu c dung dch cú cha m gam mui ca cỏc hp cht hu c Giỏ tr ca m l A 10,31 gam B 11,77 gam C 14,53 gam D 7,31 gam Cõu 33: Cho 17,82 gam hn hp gm Na, Na 2O, Ba, BaO (trong ú oxi chim 12,57% v lng) vo nc d, thu c a mol khớ H2 v dung dch X Cho dung dch CuSO4 d vo X, thu c 35,54 gam kt ta Giỏ tr ca a l A 0,08 B 0,12 C 0,10 D 0,06 Cõu 34: Cho dóy cỏc cht: triolein; saccaroz; nilon-6,6; t lapsan; xenluloz v glyxylglyxin S cht dóy cho c phn ng thy phõn mụi trng kim l A B C D Cõu 35: Thc hin cỏc thớ nghim sau (1) Cho bt Al vo dung dch NaOH (d) (2) in phõn dung dch NaCl bng in cc tr, khụng mng ngn xp (3) Cho dung dch KI vo dung dch cha Na2Cr2O7 v H2SO4 (4) Dn lung khớ NH3 qua ng s cha CrO3 (5) Cho bt Fe vo lng d dung dch FeCl3 S thớ nghim thu c n cht l Trang A B C D Cõu 36: Hn hp X cha cỏc este u mch h gm hai este n chc v mt este a chc, khụng no cha mt liờn kt ụi C=C t chỏy hon ton 0,24 mol X cn dựng 1,04 mol O2, thu c 0,93 mol CO2 v 0,8 mol H2O Nu thy phõn X NaOH, thu c hn hp Y cha ancol cú cựng s nguyờn t cacbon v hn hp Z cha mui Phn trm lng ca este n chc cú lng phõn t ln X l A 22,7% B 15,5% C 25,7% D 13,6% Cõu 37: Thc hin s phn ng (ỳng vi t l mol cỏc cht) sau: t (1) X + NaOH X1 + X + 2H 2O (2) X1 + H 2SO Na 2SO + X 0 t t (3) nX + nX (4) nX + nX Nilon 6, + 2nH 2O T lapsan + 2nH 2O Nhn nh no sau õy l sai? A X cú cụng thc phõn t l C14H22O4N2 B X2 cú tờn thay th l hexan-1,6-iamin C X3 v X4 cú cựng s nguyờn t cacbon D X2, X4 v X5 cú mch cacbon khụng phõn nhỏnh Cõu 38: Tin hnh in phõn dung dch CuSO4 v NaCl bng in cc tr, mng ngn xp vi cng dũng in khụng i, ta cú kt qu ghi theo bng sau: Thi gian (s) Khi lng catot tng Anot Khi lng dung dch gim 3088 m (gam) Thu c khớ Cl2 nht 10,80 (gam) 6176 2m (gam) Khớ thoỏt 18,30 (gam) T 2,5m (gam) Khớ thoỏt 22,04 (gam) Giỏ tr ca t l : A 8878 giõy B 8299 giõy C 7720 giõy D 8685 giõy Cõu 39: Hn hp E cha ba peptit u mch h gm peptit X (C4H8O3N2), peptit Y (C7HxOyNz) v peptit Z (C11HnOmNt) un núng 28,42 gam E vi dung dch NaOH va , thu c hn hp T gm mui ca glyxin, alanin v valin t chỏy ton b T cn dựng 1,155 mol O2, thu c CO2, H2O, N2 v 23,32 gam Na2CO3 Phn trm lng ca X hn hp E l A 4,64% B 6,97% C 9,29% D 13,93% Cõu 40: Hũa tan ht 12,06 gam hn hp gm Mg v Al 2O3 dung dch cha HCl 0,5M v H 2SO4 0,1M Sau kt thỳc phn ng thu c dung dch X Cho t t dung dch NaOH 1M vo dung dch X, phn ng c biu din theo th sau: Nu cho t t V ml dung dch NaOH 0,3M v Ba(OH) 0,1M vo dung dch X, thu c kt ta ln nht, ly kt ta nung ngoi khụng khớ n lng khụng i thu c m gam rn khan Giỏ tr ca m l A 37,860 gam B 41,940 gam C 48,152 gam D 53,124 gam -HT - Trang PHN TCH - HNG DN GII THI TH THPT NGUYN èNH CHIU LN Cõu 1: Chn C - CrO, Fe2O3 l oxit baz, CrO3 l oxit axit, Al2O3 l oxit lng tớnh Cõu 2: Chn A Cõu 3: Chn B - CTPT ca glucoz l C6H12O6 cú %O = 53,33 Cõu 4: Chn C - T nhõn to (t bỏn tng hp) l t cú ngun gc t thiờn nhiờn c pha ch thờm bng phng phỏp húa hc nh t visco, t xenluloz axetat Cõu 5: Chn C - Phng trỡnh: 4Fe(OH)2 + O2(khụng khớ) 2Fe2O3 + H2O Cõu 6: Chn B - Triong mụi trng axit, mui Cr(III) th hin tớnh oxi húa v d b nhng cht kh nh Zn kh thnh mui Cr(II): Zn + 2CrCl3 ZnCl2 + 2CrCl2 Cõu 7: Chn A Cõu 8: Chn D t A HCOOCH=CH-CH3 + NaOH HCOONa + CH3CH2CHO t B HCOOC(CH3)=CH2 + NaOH HCOONa + CH3COCH3 t C CH2=C(CH3)COOH + NaOH CH2=C(CH3)COONa + H2O t D HCOOCH2CH=CH2 + NaOH HCOONa + CH2=CH-CH2-OH Cõu 9: Chn D Cõu 10: Chn C A Sai, Nc cng l nc cha nhiu ion Ca2+ v Mg2+ B Sai, Phng phỏp un núng ch cú th lm mm c nc cng tm thi vỡ nc cng tm thi cú cha cỏc ion Mg2+, Ca2+ v HCO un núng: 2+ 2+ t MgCO , CaCO (kt ta) 2HCO3 CO 32 + CO + H 2O ú Mg , Ca + CO o Lc b kt ta ta s thu c nc mm C ỳng, Trong nc t nhiờn cha nhiu cỏc ion Ca2+, Mg2+ v HCO3 , Cl , SO 24 nờn cú c tớnh cng tm thi v tớnh cng vnh cu D Sai, Nc cng gõy nhiu tr ngi vi i sng hng ngy v cho cỏc ngnh sn xut nh: lm qun ỏo mc nỏt, lm gim mựi v thc n, úng cn lm tc ng dn nc, lm hng nhiu dung dch pha ch nhng khụng phi l tỏc nhõn gõy ụ nhim ngun nc Cõu 11: Chn B Ta cú: M mạch xenlulôzơ = 92592 l mạch xenlulôzơ = n mắc xích d mắc xích = 0,046 (mm) ; 4,5.10 (mm) 162 Cõu 12: Chn D n CO = : X l C2H4O2 cú cụng thc cu to: HCOOCH3 - Ta cú: C X = nX n mắc xích = Trang - Khi cho 0,1 mol X tỏc dng vi KOH thu c mui l HCOOK: 0,1 mol mmui = 8, (g) Cõu 13: Chn B - Vỡ 2n Fe < n AgNO3 < 3n Fe nờn dung dch sau phn ng ch cha Fe(NO3)2 v Fe(NO3)3 Cõu 14: Chn D Al(OH)3 + 3HCl AlCl3 + 3H 2O A ỳng, Cỏc phn ng ca Al(OH)3: Na[Al(OH) ] Al(OH)3 + NaOH B ỳng, Al l mt kim loi cú tớnh kh mnh, d b nhng e v chuyn thnh Al3+ C ỳng, Phng trỡnh: 2Al + 3H2SO4 Al2(SO4)3 + 3H2 D Sai, Chiu gim dn dn in ca kim loi: Ag > Cu > Au > Al > Fe Cõu 15: Chn B m Y 213n Al(NO3 )3 BT:Al n Al(NO3 )3 = n Al = 0,12 mol n NH NO3 = = 0, 025 mol 80 BT:e - Go a l s e nhn ca X ỏp dng: 0,02a = 3n Al 8n NH NO3 = 0,16 a = Vy X l N 2O Cõu 16: Chn A + NaOH C17 H 33COONa + C17 H 35COONa + C15H 31COONa + C3 H (OH) - Cho: (RCOO)3 C3H triglyxerit X natri oleat natri stearat natri panmitat glyxerol Vy cu to ca X l: Cõu 17: Chn D Thuc th Br2 Anilin: C6H5NH2 Mt mu dung dch brom v cú kt ta trng Glucoz: C6H12O6 Mt mu dung dch brom Alanin: C3H7O2N Khụng hin tng Cõu 18: Chn A 0,06 mol 0,12 mol } } + H 2O + AgNO3 FeCl , KCl dd Y AgCl, Ag - Quỏ trỡnh: 14 43 44 43 16,56 (g ) X m (g) BT: Cl n AgCl = 2n FeCl + n KCl = 0, 24 mol m = 40,92 (g) - Kt ta gm: BT: e n = n = 0, 06 mol Ag FeCl Cõu 19: Chn C ng phõn amin bc n- C4H9NH2, CH3CH(CH3)CH2NH2, CH3CH2CH(CH3)NH2 v (CH3)3CNH2 ng phõn amin bc C2H5NHC2H5, n-C3H7NHCH3 v (CH3)2CHNHCH3 ng phõn amin bc C2H5NH(CH3)2 Vy amin cú CTPT l C4H11N cú ng phõn cu to Cõu 20: Chn C A Sai, Xenluloz khụng tan nc c un núng v khụng tan dung mụi hu c nh benzene, ete (trong ú cú imetyl ete) Trang B Sai, Saccaroz khụng cú kh nng tham gia phn ng gng C ỳng, Glucoz (C6H12O6) th hin tớnh oxi húa tỏc dng vi H2 (xt Ni, t0) to sorbitol (C6H14O6) D Sai, Amiloz v amilopectin l thnh chớnh ca tinh bt cú cụng thc l (C 6H10O5)n ú phõn t ca amilopectin ln hn so vi amiloz, vỡ vy cht trờn khụng phi ng phõn ca Cõu 21: Chn D - Phng trỡnh: 2NaOH + H 2SO Na 2SO + H 2O mol: 0,4V1 0,6V2 - Vỡ sau phn ng ch thu c cht tan nờn phn ng trờn va 0, 4V1 V = 0, 6V2 = V2 Cõu 22: Chn B - Khi cho X tỏc dng vi HCl loóng ta cú: n H = n Fe = 0, 09 mol 3n Fe 3n NO BT: e = 0, 09 mol m X = 10,8 (g) - Khi cho X tỏc dng vi HNO3 loóng ta cú: n Cu = Cõu 23: Chn C A Fe(NO3)2 + AgNO3 : khụng phn ng B Cr2O3 ch phn ng vi NaOH c núng C 3Br2 + 2NaCrO2 + 8NaOH 2Na2CrO4 + 6NaBr + 4H2O D Cr, Fe v Al b th ng húa vi HNO3 v H2SO4 c ngui Cõu 24: Chn D A ỳng, iu kin thng, cỏc amino axit l cht rn, d tan nc B ỳng, Trong phõn t amino axit, nhúm NH2 v nhúm COOH tng tỏc vi to ion lng cc Vỡ vy amino axit kt tinh tn ti dng ion lng cc Trong dung dch, dng ion lng cc chuyn mt phn nh thnh dng phõn t: C ỳng Axit glutamic l thuc h tr thn kinh D Sai, Trựng ngng axit 6-aminohexanoic thu c nilon-6 cú cha liờn kt peptit to nH N [CH ]5 COOH n ( HN [CH ]5 CO ) n + nH 2O Liờn kt CO-NH trng hp ny khụng phi l liờn kt peptit vỡ khụng dc to t cỏc - amino axit Cõu 25: Chn D Ni,t - Phng trỡnh: C H 6O (X) + H C H 8O (Y) mol: a a a Vy X l este mch h, khụng no (cú mt liờn kt C = C) v n chc X cú ng phõn l: HCOOCH=CHCH3 (cis trans), HCOOC(CH3)=CH2, HCOOCH2CH=CH2 CH3COOCH=CH2 v CH2=CHCOOCH3 Cõu 26: Chn C +H O + CO + 2+ 2 Na,Ba, Al Al { d Na ,Ba 4,AlO 32 Al(OH) 14 433 - Quỏ trỡnh: 1,08 (g) dung dịch X 0,16 mol BTDT (X) n Na + + 2n Ba + = 0,16 - Ta cú: n AlO = n Al(OH)3 = 0,16 mol m 2n Na + 4n Ba = n H n H = 2(n Na + + 2n Ba + ) = 0,32 mol VH = 7,168 (l) Trang Cõu 27: Chn A 89n Ala + 147n Glu = 15,94 n Ala = 0, 08 mol - Ta cú: n Ala + 2n Glu = n NaOH = 0, n Glu = 0, 06 mol BT:Na n NaCl = n NaOH = 0, mol m X = 125,5n AlaHCl + 183,5n GluHCl + 58,5n NaCl = 32, 75(g) Cõu 28: Chn D - Cho 0,2 HCOOC H + AgNO /NH X CH 3COOC H mol n Ag = 0,08 mol n CH 3COOC 2H = 0,12 mol HCOOC H : 0,14 mol HCOOK : 0,14 mol + KOH mmui = 32,34 (g) - Cho 28,84 gam X CH 3COOC H : 0, 21 mol CH 3COOK : 0, 21 mol n HCOOC 2H = Cõu 29: Chn C - C trng hp thu c Cu n cht l (2), (3), (4), (5) Cỏc phn ng xy ra: (1) 2Na + CuSO4 + 2H2O Cu(OH)2 + Na2SO4 + H2 o dpdd t (2) 2CuSO4 + 2H2O 2Cu + 2H2SO4 + O2 (3) H2 + CuO Cu + H2O o t (4) 2Al + 3CuO Al2O3 + 3Cu Cõu 30: Chn B (5) Fe + CuCl2 FeCl2 + Cu - Khi t X cha este no, n chc, mch h thỡ ta luụn cú: n CO = n H 2O = 0,56 mol m mX 2n + n H 2O 2n O BTKL BT: O n O = CO + H 2O = 0, 64 mol n X = CO = 0, mol 32 0,56 = 2,8 Vỡ cho X tỏc dng vi NaOH thu c ancol k tip v mui ca axit - Ta cú: C X = 0, cacboxylic k tip nờn este X ln lt l: HCOOCH : x mol x + y = 0, x = 0,12 CH 3COOC H : y mol 2x + 4y = 0,56 y = 0, 08 - Hn hp mui Z gm HCOONa (A): 0,12 mol v CH3COONa (B): 0,08 mol a : b = 1, 243 Cõu 31: Chn D H 2SO (d ) Fe O ,Cu Fe + ,Cu + , H + (d ) ,SO + Cu (rắn không tan) - Phn ng: 142 343 4 44 4 43 hỗn hợp dung dịch X - Dung dch X tỏc dng c vi cht sau: Fe + CuSO Cu + FeSO Fe + H 2SO FeSO + H 3Fe 2+ + NO + 4H + 3Fe 3+ + NO + 2H 2O 6FeSO + 3Cl 2Fe (SO ) + 2FeCl 10FeSO + 2KMnO + 8H 2SO 5Fe (SO ) + K 2SO + 2MnSO + 8H 2O Cõu 32: Chn B Trang - Cho Y : HCOONH 4 43CH 33 x mol 9, 42 (g)X COONH 2H Z : H 24NCH 424 4 34C43 y mol 64x7mol48 4y7mol48 T (M T = 36, 6) : CH NH v C H NH + NaOH 4x7mol4 44y7mol4 48 44y7mol4 48 CH Cl,4Cl4H43NCH Cl 43 NH 43 44 24COOH 4 4, C42H 54NH 343 + HCl m (g ) - Hng t 1: Tớnh theo lng ca tng mui 77x + 120y = 9, 42 x = 0, 06 + Ta cú: mmui = 11, 77 (g) 31x + 45y = 36, 6.(x + y) y = 0, 04 - Hng t 2: Bo ton lng BTKL + Ta cú: nHCl = x + 2y = 0,14 mol mmui = mX + mHCl - mHCOOH = 11, 77 (g) Cõu 33: Chn A - Quy i hn hp thnh Na, Ba v O Ta cú : 23n Na + 137n Ba = 15,58 23n Na + 137n Ba = 17,82 16n O n Na = 0,32 mol n Na + + 2n Ba 2+ + 233n Ba + = 35,54 n Ba = 0, 06 mol 98n Cu(OH) + 233n BaSO = 35,54 98 n + 2n Ba 2n O BT:e n H = Na = 0, 08 mol Cõu 34: Chn B - Cú cht thy phõn mụi trng kim l: triolein; nilon-6,6; t lapsan; glyxylglyxin Cõu 35: Chn B - Cú thớ nghim thu c n cht l (1), (2), (3) v (4) Phng trỡnh phn ng: (1) 2Al + 2NaOH + 2H2O 2NaAlO2 + 3H2 dpdd (2) 2NaCl + 2H2O 2NaOH + Cl2 + H2 (3) 6KI + Na2Cr2O7 + 7H2SO4 4K2SO4 + Cr2(SO4)3 + Na2SO4 + 3I2 + 7H2O (4) 2NH3 + 2CrO3 N2 + Cr2O3 + 3H2O Cõu 36: Chn D 0,93 = 3,875 Vỡ thy phõn X NaOH thu c ancol cú cựng cacbon v - Ta cú: C X = 0, 24 mui Trong X cú cht l HCOOC2H5 (A), este n chc cũn li l RCOOC2H5 (B), este chc (C) c to t etylenglicol C2H4(OH)2 v hai axit cacboxylic HCOOH; RCOOH (trong gc R cú liờn kt ụi C=C) quan hệ + Lỳc ny kA = ; kB = ; kC = p dng n B + 2n C = n CO n H 2O = 0,13 (1) CO H O BT: O 2n A + 2n B + 4n C = 2n CO + n H 2O 2n O = 0,58 (2) v nA + nB + nC = 0,24 (3) + T (1), (2), (3) ta tớnh c: nA = 0,16 mol ; nB = 0,03 mol ; nC = 0,05 mol BT: C 3.0,16 + 0, 03.C B + 0, 05.C C = 0,93 (4) (vi CB > 4, CC > 5) + Nu CB = thay vo (4) ta cú: CC = Tha (nu CB cng tng thỡ CC < nờn ta khụng cn xột na) 7, 25 = 13, 61 Vy (B) l CH2=CH-COOC2H5: 0,03 mol %m C = 22, 04 Cõu 37: Chn C Trang t0 (1) C H (COONH ) (CH ) + 2NaOH C 6H (COONa) (X1) + NH (CH ) NH (X ) (2) C H (COONa) (X1 ) + H 2SO C H (COOH) (X ) + Na 2SO t (3) nHOOC(CH ) COOH(X ) + nNH (CH ) NH (X ) ( OC(CH ) CONH(CH ) NH ) n + 2nH 2O nilon-6,6 t0 (4) nC H (COOH) (X ) + nC H (OH)2 (X ) ( OCC H COOC H O ) n + nH O Tơ lapsan - Cht X3 l C H (COOH) cú nguyờn t C v X4 l HOOC(CH ) COOH cú nguyờn t C Cõu 38: Chn A - Gi a l s mol ca NaCl Quỏ trỡnh in phõn xy nh sau: Thi gian (s) Ti catot Ti anot 2+ t = 3088 Cu + 2e Cu 2Cl 2e + Cl2 ne = 2x x 2x x 2x 2x x 2+ Cu + 2e Cu 2Cl 2e + Cl2 t = 6176 0,16 0,32 0,16 a 0,5a ne = 4x = 0,32 2H2O 4e + 4H+ + O2 0,32 - a 0,08 - 0,25a 2+ Cu + 2e Cu 2Cl 2e + Cl2 0,2 0,4 0,2 0,2 0,2 0,1 t=? 2H2O + 2e 2OH- + H2 2H2O 4e + 4H+ + O2 2y y 2a (0,4 + 2y) - 0,2 0,5y + 0,05 - Ti thi im t = 3088 cú: m dd giảm = 64n Cu + 71n Cl 135x = 10,8 x = 0,06 mol - Ti thi im t = 6176 (s) cú: m dd giảm = 64n Cu + 71n Cl + 32n O = 18,3 a = 0,2 mol - Ti thi im t (s) cú: m dd giảm = 64n Cu + 71n Cl + 32n O + 2n H = 22,04 y = 0,03 mol 0, 46 n e(trao đổi) = 2n Cu + 2n H = 2.0,2 + 2.0,03 = 0, 46 mol t = 3088 = 8878(s) 0,16 Cõu 39: Chn A m BT:Na n Ala, Gly,Val = (1,5n CO n O ) n CO = 0, 99 mol n AlaNa, GlyNa, ValNa = 2n Na 2CO3 = 0, 44 mol - Quy i hn hp E thnh C2H3ON, CH2 v H2O Lp h sau: 57n C 2H3ON + 14n CH + 18n H 2O = 28, 42 n C 2H 3ON = 0, 44 BT:C 2n C 2H 3ON + n CH = n CO = 0,99 n CH = 3n Val + n Ala = 0,11 2, 25n n C H3ON + 1,5n CH = n O = 1,155 H 2O = n X + n Y + n Z = 0,1 n mắc xích = n Gly,Ala,Val = 4, nX + nY + nZ Trong E cú cha peptit cú s mc xớch ln 4,4 (5, hoc 7) Vy Z l pentapeptit (Gly)4Ala, X l ipeptit (Gly)2 v Y l ipeptit AlaVal (khụng th l tripeptit (Gly) 2Ala vỡ ú thy phõn hn hp E s khụng thu c mui ca Val) BT:C 4n X + 7n Y + 11n Z = n CO = 0,99 n X = 0,01 0, 01.132 n Y = 0, 01 %m X = 100% = 4, 64 - Ta cú: 2n X + 2n Y + 5n Z = 2n NaOH = 0, 44 28, 42 132n + 174n + 317n = 28, 42 n = 0, 08 X Y Z Z Cõu 40: Chn B - Ti v trớ kt ta khụng i thỡ: n Mg(OH) = 2a Trang 10 BT:Al - Ti v trớ kt ta t cc i thỡ: n Al(OH)3 = 5a n Mg(OH) = 3a n Al 2O3 = n Al(OH)3 = 1,5a m 102n Al2O3 + 24n Mg = 12, 06 102a + 3a.24 = 12, 06 a = 0, 06 mol n H + (d ) = n NaOH 4n Al(OH)3 2n Mg(OH)2 = 17a 4.3a 2.2a = 0,06 mol + Khi ú: n HCl + 2n H 2SO = 2n Mg + 6n Al 2O3 + n H + (d ) x = 1, n H 2SO = 0,12 mol; n HCl = 0, mol - Xột TH1 : Al(OH)3 t cc i Gi V lớt dung dch KOH, Ba(OH)2 0,12 mol 0,18mol 0,06 mol 0,12 7mol 0,6 mol 0,3V mol 0,1V mol 0,12 mol 0,18mol 0,1V mol 0,09 mol 0,12 mol } } } } 64 48 64 48 64 48 78 } t Mg 2+ , Al 3+ , H + ,SO , Cl + NaOH , Ba(OH) Mg(OH) , Al(OH) ,BaSO MgO , Al O ,BaSO 4 4 4 4 4 43 4 4 4 44 4 4 43 4 44 4 43 dung dịch X dung dịch hỗn hợp hỗn hợp kết tủa hỗn hợp rắn + Ta cú: n OH = 2n Ba(OH)2 + n KOH = 3n Al(OH)3 + 2n Mg(OH) + n H + V = 1, 68 lớt + Nhn thy: n Ba + > n SO n BaSO (max) = n SO = 0,12 mol mrn = 102n Al2O3 + 40n MgO + 233n BaSO = 41, 94 gam - Vỡ trng hp thỡ lng Al(OH)3 v BaSO4 ó kt ta cc i nờn ta khụng xột trng hp tip theo HT Trang 11