S GD & T TNH THANH HểA THPT CHUYấN LAM SN ( thi cú 40 cõu / trang) THI TH LN THPT QUC GIA NM 2017 Mụn: KHOA HC T NHIấN HểA HC Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt 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: Phn ng thy phõn este mụi trng kim, un núng c gi l phn ng ? A X phũng húa B Trỏng gng C Este húa D Hidro húa Cõu 2: Kim loi iu ch c bng phng phỏp thy luyn, nhit luyn v in phõn : A Mg B Na C Al D Cu Cõu 3: Cho 7,68 gam hn hp Fe 2O3 v Cu tỏc dng vi HCl d, sau phn ng cũn li 3,2 gam Cu Khi lng ca Fe2O3 ban u l: A 2,3 gam B 3,2 gam C 4,48 gam D 4,42 gam Cõu 4: Thy phõn este X cú CTPT C4H6O2 Khi X tỏc dng vi dung dch NaOH thu c hn hp cht hu c Y v Z ú Z cú t hi so vi H l 16 Tờn ca X l: A Etyl axetat B Metyl propionat C Metyl axetat D Metyl acrylat Cõu 5: Khi in phõn NaCl núng chy (in cc tr), ti catot xy ? A s kh ion Na + B s kh ion Cl C s oxi húa ion ClD s oxi húa ion Na + Cõu 6: Nung núng mt ng s cha 36,1 gam hn hp gm MgO, CuO, ZnO v Fe 2O3 ri dn hn hp khớ X gm CO v H d i qua n phn ng xy hon ton, thu c 28,1 gam cht rn Tng th khớ X (ktc) ó tham gia phn ng kh l : A 5,6 lớt B 11,2 lớt C 22,4 lớt D 8,4 lớt Cõu 7: Dóy kim loi no di õy iu ch c bng cỏch in phõn núng chy mui clorua ? A Al, Mg, Fe B Al, Mg, Na C Na, Ba, Mg D Al, Ba, Na Cõu 8: Dung dch cha cht no sau õy lm qu tớm i thnh mu xanh ? A Glutamic B Anilin C Glyxin D Lysin Cõu 9: t chỏy hon ton hp cht hu c no sau õy thỡ thnh phn sn phm thu c khỏc vi cht cũn li ? A Protein B Cao su thiờn nhiờn C Cht bộo D Tinh bt Cõu 10: T c sn xut t xenluloz ? A to tm B t capron C t nilon-6,6 D t visco Cõu 11: Cho cỏc hp kim sau: Cu-Fe (I); Zn-Fe (II); Fe-C (III); Sn-Fe (IV) Khi tip xỳc vi dung dch cht in li thỡ cỏc hp kim m ú Fe u b n mũn trc l: A I, III v IV B II, III v IV C I, II v IV D I, II v III Cõu 12: Cho dóy cỏc cht sau: glucoz, saccaroz, isoamyl axetat, phenylamoni clorua, poli(vinyl axetat), glyxylvalin (Gly-Val), etilenglicol, triolein S cht tỏc dng vi dung dch NaOH un núng l: A B C D Cõu 13: Cho s phn ng sau: X + NaOH CH3COONa + cht hu c Y ; Y + O2 Y1 + NaOH Y1 ; CH3COONa + H2O S cht X tha s trờn l: A B C D Cõu 14: Cht no sau õy va phn ng c vi NaOH va phn ng c vi HCl? A C2H5OH B C6H5NH2 C NH2-CH2-COOH D CH3COOH Cõu 15: Cho tng cht : NH2-CH2-COOH; CH3COOH; CH3COOCH3 ln lt tỏc dng vi dung dch NaOH (un núng) v vi dung dch HCl (un núng) S trng hp xy phn ng l: Trang A B C D Cõu 16: Cho dóy cỏc polime gm: t tm, t capron, nilon 6,6, t nitron, poli(metyl metacrylat), poli(vinyl clorua), cao su buna, t axetat, poli(etylen terephtalat) S polime c tng hp t phn ng trựng hp l: A B C D Cõu 17: Cht X cú CTPT C2H7NO2 tỏc dng vi dung dch HCl v dung dch NaOH Cht X thuc loi hp cht no sau õy ? A Mui amoni hoc mui ca amin vi axit cacboxylic B Aminoaxit hoc mui ca amin vi axit cacboxylic C Aminoaxit hoc este ca aminoaxit D Este ca aminoaxit hoc mui amoni Cõu 18: Metylamin khụng phn ng c vi dng dch no sau õy ? A CH3COOH B FeCl C HCl D NaOH Cõu 19: Tin hnh cỏc thớ nghim sau: - Thớ nghim 1: Cho Fe vo dung dch H 2SO4 loóng - Thớ nghim 2: Nhỳng Fe vo dung dch H 2SO4 loóng cú thờm vi git dung dch CuSO - Thớ nghim 3: Nhỳng Cu vo dung dch FeCl - Thớ nghim 4: Nhỳng Fe vo dung dch FeCl S trng hp n mũn in húa l: A B C D VUI LềNG T MUA XEM Y NI DUNG Cõu 29: Axit no sau õy l axit bộo? A Axit glutamic B Axit stearic C Axit axetic D Axit aipic Cõu 30: Cho lung khớ H d qua hn hp cỏc oxit CuO, Fe 2O3, Al2O3, MgO nung núng nhit cao Sau phn ng, hn hp cht rn thu c gm ? A Cu, Fe, Al, Mg B Cu, FeO, Al 2O3, MgO C Cu, Fe, Al 2O3, MgO D Cu, Fe, Al, MgO Cõu 31: Cho 0,01 mol mt este tỏc dng va vi 100ml dung dch NaOH 0,2M, un núng Sn phm to thnh gm mt ancol v mt mui cú s mol bng v bng s mol este Mt khỏc, x phũng húa hon ton 1,29 gam este ú bng mt lng va 60ml dung dch KOH 0,25M, cụ cn dung dch sau phn ng thu c 1,665 gam mui khan Cụng thc ca este ú l: A C2H4(COO)2C4H8 B C4H8(COO)2C2H4 C CH2(COO)2C4H8 D C4H8(COO)C3H6 Cõu 32: Hn hp X cha cht A (C5H16O3N2) v cht B (C4H12O4N2) tỏc dng vi dung dch NaOH va , un núng cho n phn ng xy hon ton ri cụ cn thu c m gam hn hp Y gm mui D v E (MD < ME) v 4,48 lớt (ktc) hn hp Z gm amin no, n chc, ng ng k tip cú t so vi H2 bng 18,3 Khi lng ca mui E hn hp Y l: A 4,24 B 3,18 C 5,36 D 8,04 Cõu 33: Hn hp gm 1,3 mol Mg v x mol Zn vo dung dch cha mol Cu 2+ v mol Ag+ sau phn ng hon ton, lc b phn dung dch thu c cht rn gm kim loi Giỏ tr ca x cú th l: A 1,8 B C 2,2 D 1,5 Cõu 34: Cho cỏc s phn ng sau: (a) X + O2 Y; (b) Z + H2O G (c) Z + Y T (d) T + H2O Y + G Bit X, Y, Z, T, G u cú phn ng vi dung dch AgNO NH3 to kt ta v G cú nguyờn t cacbon phõn t Phn trm lng ca nguyờn t oxi phõn t T cú giỏ tr xp x bng ? A 37,21% B 44,44% C 53,33% D 43,24% Cõu 35: Cho cỏc phỏt biu sau v cacbohidrat : (a) Glucoz v saccaroz u l cht rn cú v ngt, d tan nc (b) Tinh bt v xenluloz u l polisaccarit (c) Trong dung dch glucoz v saccarozo u hũa tan Cu(OH) to phc mu xanh lam Trang (d) Khi thy phõn hon ton hn hp gm tinh bt v saccaroz mụi trng axit ch thu c mt loi monosaccarit nht (e) Khi un núng glucoz vi dung dch AgNO 3/NH3 thu c Ag (g) Glucoz v saccaroz u tỏc dng vi H (xỳc tỏc Ni, un núng) to sorbitol S phỏt biu ỳng l: A B C D Cõu 36: Tin hnh in phõn vi in cc tr v mng ngn xp mt dung dch cha m gam hn hp CuSO4 v NaCl cho n nc bt u b in phõn c hai in cc thỡ dng li anot thu c 0,896 lớt khớ (kc) Dung dch sau in phõn cú th hũa tan ti a 3,2 gam CuO Giỏ tr ca m l: A 11,94 B 9,60 C 5,97 D 6,40 Cõu 37: Hũa tan hon ton 25,3 gam hn hp X gm Mg, Al v Zn bng dung dch HNO Sau phn ng kt thỳc thu c dung dch Y v 4,48 lớt (kc) khớ Z (gm hai hp cht khớ khụng mu) cú lng 7,4 gam Cụ cn dung dch Y thu c 122,3 gam hn hp mui S mol HNO ó tham gia phn ng gn nht vi giỏ tr no sau õy ? A 1,81 mol B 1,95 mol C 1,8 mol D 1,91 mol Cõu 38: Cho 5,2 gam hn hp gm Al, Mg v Zn tỏc dng va vi dung dch H 2SO4 10% thu c dung dch Y v 3,36 lớt khớ H (kc) Khi lng ca dung dch Y l: A 152 gam B 146,7 gam C 175,2 gam D 151,9 gam Cõu 39: Hn hp M gm LysGlyAla, LysAlaLysLysLysGly v AlaGly ú oxi chim 21,3018% v lng Cho 0,16 mol M tỏc dng va vi dung dch HCl thu c m gam mui Giỏ tr ca m l: A 86,16 B 90,48 C .83,28 D 93,26 Cõu 40: Hn hp E gm X, Y v Z l peptit u mch h (M X > MY > MZ) t chỏy 0,16 mol X hoc Y hoc Z u thu c s mol CO ln hn s mol H 2O l 0,16 mol Nu un núng 69,8 gam hn hp cha X, Y v 0,16 mol Z vi dung dch NaOH va ự thu c dung dch cha 101,04 gam hai mui ca alanin v valin Bit n X < nY Phn trm lng ca X E gn nht vi : A 12 B 95 C 54 D 10 HT Trang PHN TCH HNG DN GII CHI TIT THI TH LN CHUYấN LAM SN THANH HểA LN Cõu 1: Chn A Cõu 2: Chn D - Cỏc kim loi Na, Mg, Al u c c iu ch bng phng phỏp in phõn núng chy - Kim loi Cu c iu ch bng c phng phỏp: Phng phỏp thy luyn: Zn + CuSO4 ZnSO4 + Cu o t Phng phỏp nhit luyn: CO + CuO Cu + CO2 pdd Phng phỏp in phõn: 2CuSO4 + 2H2O 2Cu + 2H2SO4 + O2 Cõu 3: Chn B - Phng trỡnh phn ng : Fe2O3 + 6HCl Cu + 2FeCl 2FeCl3 + 3H2O ; CuCl2 + 2FeCl mol: x 2x mol: x 2x - Ta cú: m Fe 2O3 + mCu p = 7,68 3,2 160x + 64x = 4,48 x = 0,02 mol m Fe 2O3 = 3, (g) Cõu 4: Chn D - Ta cú: M Z = 32 Z l CH3OH Vy este X cú cụng thc cu to l: CH = CH COOCH cú tờn gi l metyl acrylat Cõu 5: Chn A - Cỏc phn ng xy cỏc in cc: catot (cc õm) xy s kh ion Na+ thnh Na: Na+ + e Na anot (cc dng) xy s oxi húa ion Cl- thnh Cl2: 2Cl- Cl2 + 2e pnc - Phng trỡnh in phõn: 2NaCl 2Na + Cl2 Cõu 6: Chn B 36,1 28,1 - Ta cú: VX = (n CO + n H ).22, = ữ.22, = 11, (l) 16 Cõu 7: Chn C - Trong cụng nghip, kim loi kim v kim loi kim th c iu ch bng phng phỏp in phõn núng chy mui halogenua tng ng: pnc 2MCln 2M + nCl2 (M l kim loi) - Tuy nhiờn, Al c iu ch bng cỏch in phõn núng chy Al2O3: pnc 2Al2O3 4Al + 3O2 Cõu 8: Chn D - Dung dch anilin v cỏc amin thm ca chỳng khụng lm i mu qu tớm - Tỏc dng lờn thuc th mu ca cỏc aminoaxit: (H2N)x R (COOH)y Khi: x = y thỡ amino axit trung tớnh, qu tớm khụng i mu x > y thỡ amino axit cú tớnh baz, qu tớm húa xanh x < y thỡ amino axit cú tớnh axit, qu tớm húa Cõu 9: Chn A - t chỏy hon ton cao su thiờn nhiờn, cht bộo, tinh bt thỡ sn phm thu c l CO2 v H2O - Khi t chỏy hon ton protein thỡ ngoi sn phm thu c l CO2 v H2O cũn cú khớ N2 thoỏt Cõu 10: Chn D - Cỏc loi t c sn xut t xenluloz: T visco: hũa tan xenluloz NaOH loóng v CS2 thu c dung dch keo rt nht l t visco T axetat: hũa tan xenluloz vi anhirit axetic (cú H2SO4 c) thu c xenluloz iaxetat v xenluloz triaxetat Trang Cõu 11: Chn A - Trong cp in cc: kim loi kim loi thỡ kim loi no cú tớnh kh mnh hn úng vai trũ l cc õm (anot) thỡ kim loi ú b n mũn tip xỳc vi dung dch cht in li - Trong cp in cc: kim loi phi kim thng hay gp nht l Fe C thỡ kim loi úng vai trũ l cc õm v b n mũn tip xỳc vi dung dch cht in li - Do vy cỏc cp hp kim tha l: I, III, IV Cõu 12: Chn A - Cỏc cht hu c tỏc dng vi NaOH thng gp l : o t Dn xut halogen: R-X + NaOH ROH + NaX (Chỳ ý: C6H5Cl khụng tỏc dng NaOH un núng, phn ng ch xy cú y cỏc iu kin xỳc tỏc, nhit d v ỏp sut) Phenol: C6H5OH + NaOH C6H5ONa + H2O Axit cacboxylic (-COOH): -COOH + NaOH -COONa + H2O o t Este (-COO-): RCOOR + NaOH RCOONa + ROH Mui ca amin: RNH3Cl + NaOH RNH2 + NaCl + H2O Aminoaxit: H2NRCOOH + NaOH H2NRCOONa + H2O Mui ca aminoaxit: HOOCRNH3Cl + 2NaOH NaOOCRNH2 + NaCl + 2H2O Mui amoni ca axit hu c: RCOONH3R+ NaOH RCOONa + RNH2 + H2O Mui amoni ca axit vụ c: RNH3NO3, (RNH3)2CO3, RNH3HCO 3, RNH3HSO 4, (RNH3)2SO4 - Cu to ca cỏc cht dóy: Glucoz Saccaroz Isoamyl axetat Phenylamoni clorua C6H12O6 C12H22O11 CH3COOCH2CH2CH(CH3)2 C6H5NH3Cl Poli(vinyl axetat) glyxylvalin Etilenglicol Triolein ( CH[OOCCH ] CH ) Gly - Val C2H4(OH)2 (C17H33COO)3C3H5 Vy cú cht tha Cõu 13: Chn B - T cỏc d kin ca bi ta suy c cu to ca X l: CH3COOR v Y1 l: CH3COOH - Cỏc cụng thc cu to ca Y tha phn ng: Y + O2 Y1 2+ men giaỏm Mn C2H5OH (Y) + O2 CH3COOH + H2O ; 2CH3CHO (Y) + O2 2CH3COOH Vy cú cht X tng ng l: CH 3COOC2H5 ; CH3COOCH=CH2 ; CH3COOCH(OH)-CH v CH3COOCH(Cl)-CH Cõu 14: Chn C A C2H5OH khụng tỏc dng c vi NaOH v HCl B C6H5NH2 + HCl C6H5NH3Cl C H NCH 2COOH + HCl ClH NCH 2COOH H NCH 2COOH + NaOH H NCH 2COONa + H 2O D CH3COOH + NaOH CH3COONa + H2O Cõu 15: Chn C - Cỏc phn ng xy ra: to H NCH 2COOH + NaOH H NCH 2COONa + H 2O to H NCH 2COOH + HCl ClH NCH 2COOH o t CH3COOH + NaOH CH3COONa + H2O Trang to CH 3COOCH + NaOH CH 3COONa + CH 3OH HCl,t o CH 3COOH + CH 3OH CH 3COOCH + H 2O Cõu 16: Chn C - T tm thuc loi t thiờn nhiờn - T axetat c iu ch t phn ng ca xenluloz vi anhirit axetic (xỳc tỏc H2SO4 c) H SO ủaởc [C6H7O2(OH)3]n + 3n(CH3CO)2O [C6H7O2(OCOCH3)3]n + 3nCH3COOH Xenluloz anhirit axetic xenluloz triaxetat axit axetic - T capron Trựng hp caprolactam: Caprolactam Nilon (t capron) - T nitron: Trựng hp acrilonitrin: - Cao su buna: Trựng hp buta-1,3-ien: Buta-1,3-ien Polibutaien hay cao su buna - Poli(metyl metacrylat): Trựng hp metyl metacrylat: - T Nilon6,6: ng trựng ngng axit aipic v hexametyleniamin: o t nH OOC[CH ]4 COOH (X ) + nNH 2[CH ]6 NH (X ) ( NH[CH ]6 NHCO[CH ]4 CO ) n + 2nH O axit aipic hexametyleniamin poli(hexametylen-aipamit) hay t nilon 6,6 - T lapsan: ng trựng ngng axit terephtalic v etylen glycol: to n(p HOOCC H 4COOH)+ n(HOCH 2CH 2OH) ( OC C6 H CO OCH CH O ) n + 2nH 2O Axit terephtalic Etylen glicol Poli (etylen terephtalat) hay tụ lapsan Vy cú polime c iu ch bng phn ng trựng hp l t capron, t nitron, poli(metyl metacrylat), poli (vinyl clorua), cao su buna VUI LềNG T MUA XEM Y NI DUNG Cõu 32: Chn D - Khi cho hn hp X tỏc dng vi NaOH va thỡ : t0 (C H NH ) CO (A) + 2NaOH Na 2CO (D) + 2C 2H NH + 2H 2O t0 (COONH 3CH ) (B) + 2NaOH (COONa) (E) + CH 3NH + 2H 2O - Xột hn hp khớ Z ta cú : Trang n C 2H5 NH + n CH3NH = 0, n C 2H NH = 0, 08 mol n E = 0,5n CH NH = 0, 06 mol 45n C 2H 5NH + 31n CH NH = 0, 2.18,3.2 n CH 3NH = 0,12 mol m E = 0, 06.134 = 8, 04 (g) Cõu 33: Chn D - Hn hp kim loi gm Ag v Cu, gi s hn hp gm Mg v Zn tỏc dng va vi dung dch cha 2n + + n Ag + 2n Mg mol Cu2+ v mol Ag+ thỡ : n Zn = Cu = 1,7 mol (Khụng cú ỏp ỏn) - Chng t dung dch sau phn ng cha Mg2+, Zn2+ v Cu2+ Vỡ vy n Zn < 1,7 mol Cõu 34: Chn B - Cỏc phn ng xy ra: xt Hg + (b) C2 H (Z) + H 2O (a) HCHO(X) + O HCOOH(Y) CH 3CHO(G) (c) HCOOH(Y) + C H (Z) HCOOC 2H (T) + H (d) HCOOC2 H (T) + H 2O HCOOH(Y) + CH 3CHO (G) Vy %O(T) = 44, 44 Cõu 35: Chn B - Cú nhn nh ỳng l (a), (b), (c) (e) (d) Sai, Khi thy phõn tinh bt ch thu c glucoz cũn thy phõn saccaroz thỡ thu c c glucoz v fructoz (g) Sai, Ch cú glucoz phn ng vi H (Ni t0) thu c sorbitol, saccaroz thỡ khụng tham gia phn ng hiro húa Cõu 36: Chn A - Vỡ dung dch hũa tan c CuO nờn dung dch sau in phõn cú cha H + (tc l ti anot nc ó in phõn) Ta cú : n H + = 2n CuO = 0,08 mol Cu2+ x mol + Ti catot 2e 2x mol Ti anot 2Cl Cl2 + 2e 2y mol y mol 2y mol + H2O 4H + O2 + 4e 0,08 mol 0,02 mol 0,08 mol Cu x mol BT:e 2n Cu + = 2n Cl2 + 4n O Xột hn hp khớ ta cú: n Cl2 = n khí n O 2x 2y = 0,08 x = 0,06 mol y = 0,02 y = 0,02 mol m = 160n CuSO + 58,5n NaCl = 11,94 (g) Cõu 37: Chn D n NO + n N O = 0,2 n NO = 0,1mol - Xột hn hp khớ Z ta cú : 30n NO + 44n N 2O = 7, n N 2O = 0,1mol n NO3 (trong muối) = 3n NO + 8n N 2O + 9n NH + = 1,1 + 9x - Ta cú: m muối = m kim loại + 18n NH + + 62n NO 122,3 = 25,3 + 18x + 62(1,1 + 9x) x = 0,05mol n HNO3 = 10n NH + + 4n NO + 10n N 2O = 1,9 mol Cõu 38: Chn D - Ta cú: n H 2SO = n H = 0,15mol m dd H 2SO4 = 98n H 2SO 100 = 147(g) C% BTKL m Y = m kim loại + m dd H 2SO 2n H = 151,9(g) Cõu 39: Chn B - Nhn thy rng hn hp M cú dng GlyAla(Lys) x (CTPT ca M l C 5+6x H10+12x O 3+ x N 2+ 2x ) Trang - Theo ta cú: %m O = 16(3 + x) = 0,213018 x = 1,5 12(5 + 6x) + 16(3 + x) + 10 + 12x + 14(2 + 2x) t0 5HCl + 2,5H 2O GlyHCl + AlaHCl + Lys(HCl) - Khi cho M tỏc dng HCl thỡ: GlyAla(Lys)1,5 + 0,8mol 0,16 mol 0,4 mol BTKL m muối = m M + 36,5n HCl + 18n H 2O = 90, 48(g) Cõu 40: Chn A n CO n H 2O 0,16 = nX = 0,16 k X = 0,5k X 0, 5k X - Tng t t ln lt 0,16 mol Y v Z thỡ ta c kY = kZ = - Gi x l s mol ca hn hp E Khi un núng 69,8 gam E vi NaOH va thỡ : - Khi t 0,16 mol X thỡ : BTKL m muối = m E + 18n H 2O 40n NaOH 101,04 = 69,8 + 40.4t 18t t = 0, 22 mol + Xột hn hp mui ta cú: 111n AlaNa + 139n ValNa = m muối 111n AlaNa + 139n ValNa = 101,04 n AlaNa = 0,76 n AlaNa + n ValNa = 4n E n AlaNa + n ValNa = 0,88 n ValNa = 0,12 - Ta nhn thy rng nZ > nValNa, nờn peptit Z E l (Ala)4 (0,16 mol) - Gi x, y ln lt l s mol ca X v Y Theo ta cú X l (Val)a(Ala)4 a v b l (Val)b(Ala)4 b BT:Val xa + yb = 0,12 xa + yb = 0,12 x = 0,02 y = 0,04 BT:Ala x ... %m O = 16 (3 + x) = 0, 213 018 x = 1 ,5 12 (5 + 6x) + 16 (3 + x) + 10 + 12 x + 14 (2 + 2x) t0 5HCl + 2,5H 2O GlyHCl + AlaHCl + Lys(HCl) - Khi cho M tỏc dng HCl thỡ: GlyAla(Lys )1 ,5 + 0,8mol 0 ,16 mol... t 0 ,16 mol X thỡ : BTKL m muối = m E + 18 n H 2O 40n NaOH 10 1,04 = 69,8 + 40.4t 18 t t = 0, 22 mol + Xột hn hp mui ta cú: 11 1n AlaNa + 13 9n ValNa = m muối 11 1n AlaNa + 13 9n ValNa = 10 1,04... + 8n N 2O + 9n NH + = 1, 1 + 9x - Ta cú: m muối = m kim loại + 18 n NH + + 62n NO 12 2,3 = 25, 3 + 18 x + 62 (1, 1 + 9x) x = 0,05mol n HNO3 = 10 n NH + + 4n NO + 10 n N 2O = 1, 9 mol Cõu 38: Chn D -