Đáp án chi tiết 20 đề thi thử thpt năm 2017 - môn hóa học

TNG HP P N CHI TIT 20 THI TH THPT QUC GIA 2017 MễN HểA HC - PHN S GD & T THI TH THPT QUC GIA LN NM 2017 TNH HI PHềNG Mụn: KHOA HC T NHIấN HểA HC THPT TIấN LNG Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt Mó : 132 ( thi cú 40 cõu / trang) 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: Ngõm mt inh st dung dch HCl, phn ng xy chm phn ng xy nhan hn, ngi ta thờm tip vo dung dch axit mt vi git dung dch no sau õy A NaCl B FeCl3 C H2SO4 D Cu(NO3)2 Cõu 2: nhit cao, khớ CO kh c cỏc oxit no sau õy A Fe2O3 v CuO B Al2O3 v CuO C MgO v Fe2O3 D CaO v MgO Cõu 3: t chỏy 4,56 gam hn hp E cha metylamin, imetylamin, trimetylamin cn dựng 0,36 mol O Mt khỏc ly 4,56 gam E tỏc dng vi dung dch HCl loóng, d thu c lng mui l A 9,67 gam B 8,94 gam C 8,21 gam D 8,82 gam Cõu 4: Hũa tan hon ton hn hp Mg, Al, Fe v Cu dung dch HNO3 (loóng d) thu c dung dch X Cho dung dch NaOH d vo dung dch X c kt ta Y Nung kt ta Y n phn ng nhit phõn kt thỳc thu c ti a bao nhiờu oxit A B C D Cõu 5: Hũa tan hon ton 7,5 gam hn hp gm Mg v Al bng lng va V lớt dung dch HNO3 1M Sau cỏc phn ng kt thỳc, thu c 0,672 lớt N ( ktc) nht v dung dch cha 54,9 gam mui Giỏ tr ca V l A 0,72 B 0,65 C 0,70 D 0,86 Cõu 6: Este n chc X cú t hi so vi CH4 l 6,25 Cho 20 gam X tỏc dng vi 300 ml dung dch KOH 1M (un núng) Cụ cn dung dch sau phn ng thu c 28 gam cht rn khan Cụng thc cu to ca X l A CH3-CH2-COO-CH=CH2 B CH2=CH-COO-CH2-CH3 C CH2=CH-CH2- COO -CH3 D CH3-COO-CH=CH-CH3 Cõu 7: Cho cỏc phỏt biu sau: (a) Cỏc cht CH3NH2, C2H5OH, NaHCO3 u cú kh nng phn ng vi HCOOH (b) Thnh phn chớnh ca tinh bt l amilopectin (c) Cỏc peptit u tỏc dng vi Cu(OH)2 cho hp cht cú mu tớm c trng (d) Anilin (C6H5NH2) tan ớt nc Trong cỏc phỏt biu trờn, s phỏt biu ỳng l A B C D Cõu 8: Dung dch X gm 0,01 mol Cu(NO3)2 v 0,1 mol NaHSO4 Khi lng Fe ti a phn ng c vi dung dch X l (bit NO l sn phm kh nht ca NO3-) A 3,36 gam B 5,60 gam C 2,80 gam D 2,24 gam Cõu 9: un núng dung dch cha 0,2 mol hn hp gm glyxin v axit glutamic cn dựng 320 ml dung dch NaOH 1M, cụ cn dung dch sau phn ng, thu c lng mui khan l A 36,32 gam B 30,68 gam C 35,68 gam D 41,44 gam Cõu 10: Cho bit th t t trỏi sang phi ca cỏc cp oxi húa kh dóy in húa (dóy th in cc chun) nh sau: Zn2+/Zn; Fe2+/Fe; Cu2+/ Cu; Fe3+/Fe2+; Ag+/Ag Cỏc kim loi v ion u phn ng c vi ion Fe2+trong dung dch l A Ag, Fe3+ B Zn, Ag+ C Ag, Cu2+ D Zn, Cu2+ Cõu 11: X, Y, Z, T l mt cỏc dung dch sau: (NH 4)2SO4, K2SO4, NH4NO3, KOH Thc hin thớ nghim nhn xột chỳng v cú c kt qu nh sau: Cht X Z T Y dd Ba(OH)2, t0 Cú kt ta xut hin Khụng hin tng Kt ta v khớ thoỏt Cú khớ thoỏt Cỏc dung dch X, Y, Z, T ln lt l A K2SO4, (NH4)2SO4, KOH, NH4NO3 B (NH4)2SO4, KOH, NH4NO3, K2SO4 C KOH, NH4NO3, K2SO4, (NH4)2SO4 D K2SO4, NH4NO3, KOH, (NH4)2SO4 Cõu 12: Trung ho 9,0 gam mt amin n chc Y cn 200 ml dung dch HCl 1M CTPT ca Y l A C4H11N B CH5N C C3H9N D C2H7N Cõu 13: Phỏt biu no sau õy l sai A Metyl amin l cht khớ, lm xanh qu tớm m B Cỏc ipeptit hũa tan c Cu(OH)2 nhit thng C Cỏc amino axit cú nhit núng chy cao D Cỏc cht bộo cú gc axit bộo khụng no thng l cht lng Cõu 14: Kim loi cú tớnh kh mnh nht l A Fe B Sn C Ag D Au Cõu 15: Trong cỏc kim loi Na, Fe, Cu, Ag, Al Cú bao nhiờu kim loi ch iu ch c bng phng phỏp in phõn A B C D Cõu 16: Khi cho HNO3 c vo ng nghim cha anbumin thy cú kt ta mu A xanh thm B tớm C en D vng Cõu 17: Hai dung dch u phn ng c vi kim loi Fe l: A AgNO3 v H2SO4 loóng B ZnCl2 v FeCl3 C HCl v AlCl3 D CuSO4 v HNO3 c ngui Cõu 18: Este X cú cụng thc phõn t C 2H4O2 un núng 9,0 gam X dung dch NaOH va n phn ng xy hon ton thu c m gam mui Giỏ tr ca m l A 8,2 B 10,2 C 12,3 D 15,0 Cõu 19: Cho hn hp Cu v Fe hũa tan vo dung dch H 2SO4 c núng ti phn ng hon ton, thu c dung dch X v mt phn Cu khụng tan Cho dung dch NH d vo dung dch X thu c kt ta Y Thnh phn ca kt ta Y gm A Fe(OH)2 B Fe(OH)2, Cu(OH)2 C Fe(OH)3, Cu(OH)2 D Fe(OH)3 Cõu 20: Este X cú CTPT CH3COOC6H5 Phỏt biu no sau õy v X l ỳng A Tờn gi ca X l benzyl axetat B X cú phn ng gng C Khi cho X tỏc dng vi NaOH (va ) thỡ thu c mui D X c iu ch bng phn ng ca axit axetic vi phenol Cõu 21: Cho 6,88 gam hn hp cha Mg v Cu vi t l mol tng ng l : vo dung dch cha 0,12 mol Fe(NO3)3 Sau cỏc phn ng hon ton thu c m gam kim loi Giỏ tr ca m l: A 5,12 B 3,84 C 2,56 D 6,96 Cõu 22: Cho cỏc s chuyn húa sau: C + H 2O O2 X 1500 Y Z + T ; HgSO , H 2SO o + H2 ,t KMnO +T Y P + Q o E Pd/PbCO Bit phõn t E ch cha mt loi nhúm chc Phõn t ca E l H 2SO , t A 132 B 118 C 104 D 146 Cõu 23: in phõn dung dch X gm FeCl2 v NaCl (t l mol tng ng l : 2) vi in cc tr mng ngn xp thu c dung dch Y cha hai cht tan, bit lng dung dch X ln hn lng dch Y l 4,54 gam Dung dch Y hũa tan ti a 0,54 gam Al Mt khỏc dung dch X tỏc dng vi lng d dung dch AgNO3 thu c m gam kt ta Giỏ tr m l A 14,35 B 17,59 C 17,22 D 20,46 Cõu 24: Hũa tan ht m gam hn hp X gm Mg, MgO, Mg(HCO 3)2, MgSO3 bng mt lng va dung dch H2SO4 30%, thu c 11,2 lớt (ktc) hn hp khớ Y v dung dch Z cú nng 36% T ca Y so vi He bng Cụ cn Z c 72 gam mui khan Giỏ tr ca m l A 20 B 10 C 15 D 25 Cõu 25: Phỏt biu no sau õy l sai A Theo chiu tng dn in tớch ht nhõn, cỏc kim loi kim cú nhit núng chy gim dn B cựng mt chu k, bỏn kớnh kim loi kim ln hn bỏn kinh kim loi kim th C KAl(SO4)2.12H2O, NaAl(SO4)2.12H2O v (NH4)Al(SO4)2.12H2O c gi l phốn nhụm D Cỏc kim loi kim, kim th u tỏc dng vi nc iu kin thng Cõu 26: Khi b m, mt sc hoc sau cỏc ca phu thut, nhiu ngi bnh thng c truyn dch m c th sm hi phc Cht dch truyn cú tỏc dng trờn l A glucoz B saccaroz C amino axit D amin Cõu 27: Phỏt biu no sau õy l ỳng: A Tt c cỏc polime tng hp u c iu ch bng phn ng trựng ngng B Tt c cỏc polime u khụng tỏc dng vi axit hay baz C Protein l mt loi polime thiờn nhiờn D Cao su buna-S cú cha lu hunh phõn t Cõu 28: Hai cht hu c X, Y cú thnh phn phõn t gm C, H, O (M X < MY < 70) C X v Y u cú kh nng tham gia phn ng bc v u phn ng c vi dung dch KOH sinh mui T hi ca Y so vi X cú giỏ tr l: A 1,403 B 1,333 C 1,304 D 1,3 Cõu 29: Cp cht khụng xy phn ng l A dung dch NaNO3 v dung dch MgCl2 B dung dch NaOH v Al2O3 C K2O v H2O D Na v dung dch KCl Cõu 30: Cỏc amino axit u cú A kh nng lm i mu qu tớm B ỳng mt nhúm amino C ớt nht nhúm COOH D ớt nht hai nhúm chc Cõu 31: 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 32: 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 A B C D Cõu 33: 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 34: Thc hin s phn ng (ỳng vi t l mol cỏc cht) sau: (2) X1 + H 2SO Na 2SO + X t (4) nX + nX T lapsan + 2nH 2O t (1) X + NaOH X1 + X + 2H 2O t (3) nX + nX Nilon 6, + 2nH 2O Nhn nh no sau õy l sai? A X cú cụng thc phõn t l C14H22O4N2 C X3 v X4 cú cựng s nguyờn t cacbon B X2 cú tờn thay th l hexan-1,6-iamin D X2, X4 v X5 cú mch cacbon khụng phõn nhỏnh Cõu 35: Hũa tan ht m gam kim loi M cn dựng 136 gam dung dch HNO 31,5% Sau kt thỳc phn ng thu c dung dch X v 0,12 mol khớ NO nht Cụ cn dung dch X thu c (2,5m + 8,49) gam mui khan Kim loi M l: A Mg B Cu C Ca D Zn Cõu 36: t chỏy hon ton mt este no chc mch h X Sc ton b sn phm chỏy vo dung dch Ca(OH)2 d, sau phn ng thu c 5,0 gam kt ta v lng dung dch gim 2,08 gam Bit x phũng húa X ch thu c mui ca axit cacboxylic v ancol S ng phõn ca X l: A B C D Cõu 37: Hn hp A gm mt peptit X v mt peptit Y (mi cht c cu to t loi aminoaxit, tng s nhúm CO-NH- loi phõn t l 5) vi t l s mol n X : nY = : Khi thy phõn hon ton m gam hn hp A thu c 5,625 gam glyxin v 10,86 gam tyrosin Gớa tr ca m l: A 14,865 gam B 14,775 gam C 14,665 gam D 14,885 gam Cõu 38: Cho 1,752 gam amin n chc X tỏc dng vi dung dch HCl loóng d, thu c 2,628 gam mui S ng phõn cu to ca X l A B C D Cõu 39: Cht no sau õy khụng phn ng vi dung dch HCl A H2N-CH2-COOH B CH3COOH C C2H5NH2 D C6H5NH2 Cõu 40: Ngi ta thng dựng cỏc vt dng bng bc co giú cho ngi b trỳng giú (khi ngi b mt mi, chúng mtdo c th tớch t cỏc khớ c nh H 2S) Khi ú vt bng bc b en phn ng: 4Ag + O2 + 2H2S 2Ag2S + 2H2O Cht kh phn ng trờn l A O2 B H2S C Ag D H2S v Ag PHN TCH HNG DN GII THI TH THPT TIấN LNG HI PHềNG LN Cõu 1: Chn D - Khi ngõm mt inh st vo dung dch HCl thỡ: Fe + 2HCl FeCl2 + H2 + Khớ H2 sinh mt phn bỏm li trờn inh st lm gim kh nng tip xỳc vi ion H + nờn phn ng xy chm v khớ H2 sinh s ớt - Khi nh thờm dung dch Cu(NO3)2 vo thỡ: Fe + Cu(NO3)2 Fe(NO3)2 + Cu + Trong dung dch lỳc ny hỡnh thnh mt pin in in cc Fe Cu cú s chuyn dch cỏc electron v ion H+ dung dch s nhn electron vỡ vy lm cho phn ng xy nhanh v khớ H thoỏt nhiu hn Cõu 2: Chn A - nhit cao, khớ CO, H2 cú th kh c cỏc oxit kim loi ng sau Al dóy in húa Cỏc cht tha l: Fe O3 v CuO Cõu 3: Chn B - Quy i hn hp E: CH NH , (CH ) NH, (CH ) N thnh CnH2n+3N: a mol - t chỏy E: CnH2n+3N + (1,5n + 0,75)O2 nCO2 + (n + 1,5)H2O + 0,5N2 n O = (1,5n + 0, 75)a = 0,36 (1) v m E = (14n + 17) a = 4,56 (2) T (1), (2) ta tớnh c: a = 0,12 mol BTKL - Cho E tỏc dng vi HCl thỡ n HCl = n E = 0,12 mol mmui = mE + 36,5nHCl = 8,94 (g) Cõu 4: Chn D Mg, Cu + HNO Mg(NO ) , Cu(NO ) + NaOH Mg(OH) , Cu(OH) t o MgO, CuO + H 2O Fe, Al Fe(NO ) , Al(NO ) Fe(OH) Fe 2O3 - Lu ý: + Cho mt lng d NaOH vo Al3+, ban u cú kt ta trng keo khụng tan sau ú tan dn v to dung dch sut + H2O cng l mt oxit vỡ theo nh ngha oxit l hp cht ca nguyờn t ú cú nguyờn t l oxi Vy cú ti a l oxit Cõu 5: Chn D - Hng t 1: Bo ton nguyờn t N Mg, Al + HNO Mg(NO ) , Al(NO ) 3, NH NO + N {2 + H 2O 4 4 44 4 4 43 - Quỏ trỡnh: 3 7,5 (g) V (l) 54,9 (g) hỗn hợp muối 0,03 mol 24n Mg + 27n Al = 7,5 n Mg = 0, + Ta cú: 148n Mg(NO ) + 213n Al(NO ) + 80n NH NO = 54,9 n Al = 0,1 BT: e n NH NO3 = 0, 05 2n Mg + 3n Al = 8n NH NO3 + 10n N BT: N VHNO3 = 2n Mg(NO3 ) + 3n Al(NO )3 + 2n NH NO + 2n N = 0,86 (l) - Hng t 2: Tớnh theo s mol HNO3 m Al3+ + m Mg 2+ + 18n NH + + 62n NO3 = 54,9 n NH + = 0, 05mol +Ta cú: BT e + BTDT n NO3 = n NH + + (8n NH 4+ + 10n N ) n HNO3 = 12n N + 10n NH + = 0,86 mol Cõu 6: Chn A - T t ta suy M X = 100 : C5H O v nX = 0,2 mol ; nKOH = 0,3 mol t CTTQ ca X l: RCOOR - Hng t 1: Tỡm gc R + Ta cú: nKOH p = nRCOOK = nX = 0,2 mol nKOH d = 0,1 mol m M RCOOK = m rắn 56n KOH dư = 112 0,2 R l C2H5 Vy cụng thc cu to ca X l C H 5COOCH = CH - Hng t 2: Tỡm gc R BTKL M R'OH = m X + m KOH m rắn = 44 : CH 3CHO R l C2H3 Vy X l C H 5COOCH = CH 0,2 Cõu 7: Chn A (a) ỳng, Phng trỡnh phn ng: CH3NH2 + HCOOH HCOONH3CH3 H 2SO HCOOC2H5 + H2O C2H5OH + HCOOH o t NaHCO3 + HCOOH HCOONa + CO2 + H2O (b) ỳng, Amilopectin chim t 70 80% lng tinh bt amiloz chim t 20 30% lng tinh bt (c) Sai, Cỏc tripeptit tr lờn (ipeptit khụng cú phn ng ny) u tỏc dng vi Cu(OH) cho hp cht cú mu tớm c trng (d) ỳng, Anilin (C6H5NH2) l cht lng, sụi 184 oC, khụng mu, rt c, tan ớt nc nhng tan nhiu benzen v etanol Vy cú phỏt biu ỳng l (a), (b), (d) Cõu 8: Chn C S oxi húa S kh Fe Fe2+ + 2e 4H+ + NO3- + 3e NO + 2H2O (vỡ lng Fe phn ng ti a nờn Fe chuyn lờn Fe2+) 0,08 0,02 0,06 0,01 Cu2+ + 2e Cu 0,01 0,02 2H+(d) + 0,02 BT:e n Fe = 2e H2 0,02 0,01 3n NO + 2n Cu 2+ + 2n H = 0, 05 mol m Fe = 2,8 (g) Cõu 9: Chn B - Hng t 1: Xỏc nh cỏc cht cú mui 0,32 mol 678 H NCH 2COOH : x mol H NCH 2COONa : x mol + NaOH + H 2O HOOC(CH ) CH(NH ) COOH : y mol NaOOC(CH ) CH(NH ) COONa : y mol x + y = 0, x = 0, 08 mmui = 97x + 191y = 30, 68 (g) + Ta cú: x + 2y = 0,32 y = 0,12 - Hng t 2: Bo ton lng BTKL + Ta cú: n H 2O = n NaOH = 0,32 mol mmui = m gly + m glu + 40n NaOH m H 2O = 30, 68 (g) Cõu 10: Chn B - Da vo quy tc ta xỏc nh c cỏc cp cht cú phn ng vi Fe2+ l Zn, Ag + Phn ng: 2+ Zn + Fe2+ Zn + Fe 3+ Fe2+ + Ag+ Fe + Ag Cõu 11: Chn D Cht X: K2SO4 Z: KOH T: (NH4)2SO4 Y: NH4NO3 dd Ba(OH)2, t0 Cú kt ta xut hin Khụng hin tng Kt ta v khớ thoỏt Cú khớ thoỏt - Phng trỡnh phn ng: Ba(OH)2 + K2SO4 BaSO4 trng + 2KOH Ba(OH)2 + NH4NO3: khụng xy Ba(OH)2 + (NH4)2SO4 BaSO4 trng + 2NH3 + 2H2O Ba(OH)2 + 2NH4NO3 Ba(NO3)2 + 2NH3 + 2H2O Cõu 12: Chn D - Ta cú: M Y = mY = = 45 : Y l C H N n Y 0, Cõu 13: Chn B Cõu 14: Chn A Cõu 15: Chn B - Na, Al c iu ch bng phng phỏp in phõn núng chy - Fe, Cu c iu ch bng c phng phỏp l in phõn dung dch, nhit luyn v thy luyn - Ag c iu ch bng phng phỏp l in phõn dung dch v thy luyn Vy ch cú kim loi Na v Al c bng mt phng phỏp in phõn Cõu 16: Chn D - Nh vi git dung dch HNO c vo ng nghim cha anbumin thy cú kt ta mu vng Cõu 17: Chn A A Fe + 2AgNO Fe(NO ) + 2Ag B Fe + 2FeCl3 3FeCl2 Fe + H 2SO loóng FeSO + H C Fe + 2HCl FeCl2 + H2 D Fe + CuSO4 FeSO4 + Cu Cõu 18: Chn B - Phng trỡnh: mol: HCOOCH + NaOH HCOONa + CH 3OH 0,15 0,15 m HCOONa = 10, (g) Cõu 19: Chn A + NH Fe 2+ , Cu 2+ ,SO 42 Fe(OH) + H 2SO 4 4 4 - Quỏ trỡnh: Fe, Cu Cu v d dd X - Lu ý: Cỏc hiroxit hay mui ca cỏc kim loi Cu, Ag, Zn, Ni to phc tan dung dch NH3 d Cõu 20: Chn C A Sai, Tờn gi ca X l phenyl axetat B Sai, X khụng cú phn ng gng C ỳng, Phng trỡnh: CH3COOC6H5 + NaOH CH3COONa + C6H5ONa + H2O D Sai, X c iu ch bng phn ng ca anhirit axetic vi phenol : C6H5OH + (CH3CO)2O CH3COOC6H5 + CH3COOH Cõu 21: Chn B - Hng t 1: Cõn bng phng trỡnh 2+ 2+ Mg + 2Fe3+ Mg + 2Fe mol: 0,02 0,04 0,04 n Fe3+ cũn li = 0,08 mol (a) 3Ag + 4HNO3 (loóng) 3AgNO3 + NO + 2H2O (b) Cr + 2HCl(loóng) CrCl2 + H2 (c) Fe + H2SO4(loóng, ngui) FeSO4 + H2 (d) H2S + FeCl2 : khụng xy phn ng (e) Na2O + H2O 2NaOH (f) Al2O3 + KHSO4 Al2(SO4)3 + K2SO4 + H2O Cõu 31: Chn D Nhiờn liu húa thch l cỏc loi nhiờn liu c to thnh bi quỏ trỡnh phn hy k khớ ca cỏc sinh vt cht b chụn vựi cỏch õy hn 300 triu nm - Cỏc nhiờn liu húa thch l ti nguyờn khụng tỏi to bi phi mt hang triu nm to chỳng - Vic t nhiờn liu húa thch to khong 21,3 t tn CO mi nm, m CO l mt nhng khớ gõy hiu ng nh kớnh, lm cho nhit trung bỡnh ca b mt ca trỏi t tng Cõu 32: Chn B ( C6 H10O5 ) n (X) + H+ nH 2O nC6 H12O6 (Y) CH 2OH[CHOH]4 CHO(Y) + Br2 + H 2O CH 2OH[CHOH]4 COOH(axit gluconic) + 2HBr CH 2OH[CHOH]4 COOH + NaHCO3 CH 2OH[CHOH]4 COONa (natri gluconat) + CO (Z) + H 2O 6nCO (Z) + 5nH 2O ( C6 H10O5 ) n (X) clorophin,as Vy X, Y ln lt l tinh bt v glucoz Cõu 33: Chn B 0,02 mol } Fe(NO )3 , Al,Cu,MgCO NO, N 2O, N , H + Fe n + , Al 3+ ,Mg + ,Cu + , NH + ,SO + H O 4 42 4 43 44 4 43 4 4 44 4 4 4 43 H 2SO 28,4 (g) A hỗn hợp khí Z 65,48(g)B BT:SO - Khi cho B tỏc dng vi BaCl2 thỡ n H 2SO = n BaSO = 0, 53 BT:Na - Khi cho B tỏc dng vi NaOH thỡ n NaOH = 2n Na 2SO = 1,06 v n NH = n H 2O BTKL 18n H 2O + 17n NH = m B + 40n NaOH m 142n Na 2SO n NH + = n NH = n H 2O = 0,02 - Khi cho A tỏc dng vi H2SO4 thỡ : BT:H n H 2O = nZ = 2n H 2SO4 4n NH + 2n H2 BTKL = 0, 47 m Z = m A + 98n H 2SO m B 18n H 2O = 6, m Z 6, = = 0,2 VZ = 4, 48(l) M Z 16.2 Cõu 34: Chn C BTKL n CO2 = m A + 32n O2 18n H 2O m 12n CO2 2n H 2O = 1,3 n O(A) = A = 0,5 44 32 - Gi s hn hp ch cha este n chc (ch c to bi axit propionic) n A max = n O(A) = 0,25 quan hệ n A (k A 1) = n CO n H 2O k A = 1,6 n Br2 (pư max) = n A (k A 1) = 0, CO H O 2 - Gi s hn hp ch cha este hai chc (ch c to bi axit oxalic) n A = n O(A) = 0,125 quan hệ n A (k A 1) = n CO n H 2O k A = 3,2 n Br2 (pư min) = n A (k A 1) = 0,275 CO H O 2 Cõu 35: Chn C 0,1mol x mol 3x mol } 678 } } + + 2 H NC x H y (COOH) (X) + H SO XH , H ,SO X ,SO , Na + , K + 42 43 4 44 4 43 4 44 4 43 22 34 0,1mol NaOH, KOH 0,1mol 0,1mol dung dịch X 36,7(g) Y BTDT n Na + + n K + = 2n SO + 2n X 4x = 0, x = 0,1 (Y) m X = m Y 96n SO 39n K + 23n Na + = 13,1 M X = 13,1 = 131 M X = 133 %N = 10,52 0,1 Cõu 36: Chn C - Xột phn ta cú : n mantôzơ = n Ag = 0,015mol - Xột phn ta cú : n Ag = (4n mantôzơ + 2n tinh bột ).H + 2n mantôzơ (dư) (4.0,015 + 2x).0,6 + 0,015.0, 4.2 = 0,192 x = 0,12 Vy m = 2(342n mantôzơ + 162n tinh bột ) = 49,14 (g) Cõu 37: Chn B - Khi cho 3,83 gam X tỏc dng vi HCl thỡ : n NH = n HCl = 0,03 mol - Theo bi ta cú : m O 80 80m N 80.0,03.14 = nO = = = 0,1mol m N 21 21.16 21.16 - Khi t chỏy hon ton 3,83 gam hn hp X thỡ : BT:O 2n CO + n H 2O = 0,385 n CO2 = 0,13mol 2n CO2 + n H 2O = n O + 2n O2 12n CO + 2n H 2O = m X 16n O m N 12n CO + 2n H 2O = 1,81 n H 2O = 0,125mol - Dn sn phm chỏy vo nc vụi d thỡ : n CaCO3 = n CO = 0,13mol m CaCO = 13(g) Cõu 38: Chn D - Xột ton b quỏ trỡnh phn ng ta cú : n H 2O = n HCl = 0,04 mol BTKL m X = m muối + m ancol + 18n H 2O 36,5n HCl m KOH = 13,76 (g) Cõu 39: Chn D - Quy i 0,4 mol hn hp E thnh C2H3ON, -CH2 v H2O, ú : n C H3ON = n Gly + n Ala + n Val = 1,1 mol , n CH = n Ala + 3n Val = mol v n H 2O = n E = 0, mol Vy lng ca 0,4 mol E l : m E = 57n C H3ON + 14n CH + 18n H 2O = 83,9(g) - Khi t chỏy 0,4 mol E thỡ : 44n CO + 18n H 2O = 195,7(g) n CO = 2n C H 3ON + n CH = 3,2 m CO ,H O(khi đốt m gam E) 78,28 m = = m = E = 33,56 (g) n H 2O = 1,5n C H 3ON + n CH + n H 2O = 3,05 m 2,5 CO ,H 2O (khi đốt 0,4 mol E) 195,7 Cõu 40: Chn D Al, Al O + NaNO , H 2SO Al 3+ , Na + , NH + ,SO + N x O y , H + H 2O 14 43 4 43 4 44 4 43 14 43 7,65(g) hỗn hợp X dung dịch Y dung dịch Z m (g) khí T - Khi cho dung dch Z tỏc dng vi BaCl2 d thỡ : n H 2SO = n BaSO = 0, mol - Xột quỏ trỡnh hn hp X tỏc dng vi dung dch Y ta cú : + n Al (trong X) = 0,17 mol n Al2O3 = m X 27n Al BT:Al = 0, 03mol n Al3+ (trong Z) = n Al + 2n Al2O3 = 0, 23mol 102 + Khi dung dch Z tỏc dng vi 0,935 mol NaOH thỡ : BT:H n NH + = n NaOH 4n Al3+ = 0, 015 mol n H 2O = 2n H 2SO 2n H 4n NH + = 0,355 mol BTDT n Na + = 2n SO n NH + 3n Al3+ = 0, 095 mol m Z = 23n Na + + 27n Al3+ + 18n NH 4+ + 96n SO = 47, 065 (g) (Z) BTKL m T = m X + 98n H 2SO + 85n NaNO3 18n H 2O m Z = 1, 47 (g) S GD & T THI TH THPT QUC GIA LN NM 2017 TNH BèNH PHC Mụn: KHOA HC T NHIấN HểA HC THPT PHC LONG Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt Mó : 132 ( thi cú 40 cõu / trang) 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: Khi un núng cht X cú cụng thc phõn t C 5H10O2 vi dung dch NaOH thu c C2H5COONa v ancol Y Y cú tờn l A Ancol Etylic B Ancol Propyolic C Ancol isopropyolic D Ancol Metylic Cõu 2: Cho dung dch lng trng trng tỏc dng vi dung dch axit nitric c,cú hin tng A Kt ta mu tớm C Kt ta mu vng B Dung dch mu xanh D Kt ta mu trng Cõu 3: Cho dóy cỏc cation kim loi :Ca2+, Cu2+, Na+, Zn2+ Cation kim loi no cú tớnh oxi húa mnh nht dóy A Ca2+ B Cu2+ C Na+ D Zn2+ Cõu 4: Phỏt biu no sau õy l sai ? A Theo chiu tng dn ca in tớch ht nhõn,nhit núng chy ca kim loi kim gim dn B nhit thng, tt c cỏc kim loi kim th u tỏc dng c vi nc C Na2CO3 l nguyờn liu quan trng cụng nghip sn xut thy tinh D Nhụm bn mụi trng khụng khớ v nc l mng oxi Al 2O3 bn vng bo v Cõu 5: Cho s sau : HCl (dư) CH 3OH, HCl (khan) KOH NaOH,t X1 X2 X3 H2N-CH2COOK X (C4H9O2N) Vy X2 l : A ClH3N-CH2COOH B H2N-CH2-COOH C H2N-CH2-COONa D H2N-CH2COOC2H5 Cõu 6: Cho hn X gm Zn, Fe vo dung dch cha AgNO v Cu(NO3)2, sau phn ng thu c hn hp Y gm kim loi v dung dch Z Cho NaOH d vo dung dch Z thu c kt ta gm hidroxit kim loi.Dung dch Z cha A Zn(NO3)2, AgNO3,Fe(NO3)3 C Zn(NO3)2, Fe(NO3)2, Cu(NO3)2 B Zn(NO3)2 ,Fe(NO3)2 D Zn(NO3)2,Cu(NO3)2, Fe(NO3)3 Cõu 7: Oxit no sau õy l lng tớnh ? A Fe2O3 B CrO C Cr2O3 D CrO3 Cõu 8: in phõn dung dch no sau õy, thỡ cú khớ thoỏt c in cc (ngay t lỳc mi u bt u in phõn) A Cu(NO3)2 B FeCl2 C K2SO4 D FeSO4 Cõu 9: Hp cht H2N-CH2-COOH phn ng c vi : (1) NaOH, (2) HCl, (3) C 2H5OH, (4) HNO2 A (1), (2), (3), (4) B (2), (3), (4) C (1), (2), (4) D (1), (2), (3) Cõu 10: Amin X cú phõn t nh hn 80 Trong phõn t X, nit chim 19,18% v lng.Cho X tỏc dng vi dung dch hn hp gm KNO v HCl thu c ancol Y Oxi húa khụng hon ton Y thu c xeton Z Phỏt biu no sau õy ỳng ? A Tỏch nc Y ch thu c anken nht B.Tờn thay th ca Y l propan-2-ol C.Phõn t X cú mch cacbon khụng phõn nhỏnh D.Trong phõn t X cú liờn kt bi Cõu 11: Dóy kim loi u cú th iu ch bng phng phỏp in phõn dung dch mui ca chỳng l A Na, Cu B Ca, Zn Cõu 12: Phỏt biu no sau õy khụng ỳng ? A Enzin l nhng cht hu cht cú bn cht protein B Cho glyxin tỏc dng vi HNO cú khớ bay C Fe, Ag D K, Al C Phc ng saccarozo cú cụng thc l (C 12H21O11)2Cu D Tetrapeptit thuc loi polipeptit Cõu 13: Cht no sau õy trng thỏi rn iu kin thng ? A Glyxin B Triolein C Etyl aminoaxetat D Anilin Cõu 14: Hũa tan hon ton hn hp X (gm x mol Fe, y mol Cu, z mol Fe 2O3, v t mol Fe3O4) dung dch HCl khụng thy khớ cú khớ bay khi bỡnh, dung dch thu c ch cha mui Mi quan h gia s mol cỏc cht cú hn hp X l : A x + y = 2z + 2t B x + y = z + t C x + y = 2z + 2t D x + y = 2z + 3t Cõu 15: Cho t t tng git ca dung dch cha b mol HCl vo dung dch cha a mol Na 2CO3 thu c V lớt khớ CO2 Ngc li cho t t tng git ca dung dch cha a mol Na 2CO3 vo dung dch cha b mol HCl thu c 2V lớt khớ CO (cỏc th tớch khớ o cựng iu kin) Mi quan h gia a v b l : A a = 0,75b B a = 0,8b C a = 0,35b D a = 0,5b Cõu 16: Dung dch CuSO4 loóng c dựng lm thuc dit nm cho hoa iu ch 800 gam dung dch CuSO4 5%, ngi ta hũa tan CuSO 4.5H2O vo nc Khi lng CuSO 4.5H2O cn dựng l ? A 32,0 gam B 40,0 gam C 62,5 gam D 25,6 gam Cõu 17: Thy phõn 14,6 gam Gly-Ala dung dch NaOH d thu c m gam mui Giỏ tr ca m l : A 20,8 B 18,6 C 22,6 D 20,6 Cõu 18: Thy phõn 44 gam hn hp T gm este cựng cụng thc phõn t C 4H8O2 bng dung dch KOH d Chng ct dung dch sau phn ng thu c hn hp ancol Y v cht rn khan Z un núng Y vi H2SO4 c 1400C, thu c 14,3 gam hn hp cỏc ete Bit cỏc phn ng xy hon ton Khi lng mui Z l A 53,2 gam B 50,0 gam C 34,2 gam D 42,2 gam Cõu 19: Cho hn hp M gm hai cht hu c X, Y Trong ú X l mt axớt hu c hai chc, mch h, khụng phõn nhỏnh (trong phõn t cú mt liờn kt ụi C=C) v Y l ancol no, n chc, mch h t chỏy hon ton 22,32 gam M thu c 14,40 gam H 2O Nu cho 22,32 gam M tỏc dng vi K d thu c 4,256 lớt H2 (ktc) Phn trm lng ca Y M gn nht vi giỏ tr no sau õy? A 27,25% B 62,40% C 72,70% D 37,50% Cõu 20: Cht X cú cụng thc phõn t C 3H4O2, tỏc dng vi dung dch NaOH thu c CHO 2Na Cụng thc cu to ca X l A HCOO-C2H5 B CH3-COOH C CH3-COO-CH3 D HCOO-C2H3 Cõu 21: Dóy gm cỏc cht c sp xp theo chiu tng dn nhit sụi t trỏi sang phi l A HCOOCH3, C2H5OH, HCOOH, CH3COOH B.CH3COOH, HCOOH, C2H5OH, HCOOCH3 C CH3COOH, C2H5OH, HCOOH, HCOOCH3 D.HCOOH, CH3COOH, C2H5OH, HCOOCH3 Cõu 22: Cho vo ng nghim ml dung dch lũng trng trng 10%, thờm tip ml dung dch NaOH 30% v git dung dch CuSO 2% Lc nh ng nghim, hin tng quan sỏt c l A Cú kt ta xanh lam, sau ú tan to dung dch xanh lam B Cú kt ta xanh lam, sau ú kt ta chuyn sang mu gch C Cú kt ta xanh lam, sau ú tan to dung dch mu tớm D Cú kt ta xanh lam, kt ta khụng b tan Cõu 23: Cho cỏc cht sau: HCl, AgNO 3, Cl2, KMnO4/H2SO4 loóng, Cu S cht tỏc dng c vi dung dch Fe(NO3)2 l A B C D Cõu 24: Cho m gam hn hp X gm glyxin v axit glutamic tỏc dng vi 0,4 mol HCl thu c dung dch Y, Y phn ng ti a vi 0,8 mol NaOH thu c 61,9 gam hn hp mui % Khi lng glyxin cú X l A 50,51% B 25,25% C 43,26% D 37,42% Cõu 25: X, Y l hai hp cht hu c n chc phõn t ch cha C, H, O Khi t chỏy X, Y vi s mol bng hoc lng bng u thu c vi t l mol tng ng : v vi t l mol tng ng : S cp cht X, Y tha l A B C D Cõu 26: Polime X dai, bn vi nhit v gi nhit tt nờn dt vi, may qun ỏo m , X l A Poliacrilonitrin B Poli (vinylclorua) C Polibutaien D Polietilen Cõu 27: Cú hn hp, mi hn hp gm cht rn cú s mol bng nhau: Na 2O v Al2O3; Cu v Fe2(SO4)3; KHSO4 v KHCO3; BaCl2 v CuSO4; Fe(NO3)2 v AgNO3 S hn hp cú th tan hon ton nc (d) ch to cỏc cht tan tt nc l A B C D Cõu 28: Cho m gam bt st vo dung dch X cha AgNO3 v Cu(NO3)2 n cỏc phn ng kt thỳc thu c cht rn Y v dung dch Z Cho dung dch Z tỏc dng ht vi dung dch NaOH d, thu c a gam kt ta T gm hai hidroxit kim loi Nung T n lng khụng i thu c b gam cht rn Biu thc liờn h gia m, a, b cú th l A m = 8,225b 7a B m = 8,575b 7a C m = 8,4 3a D m = 9b 6,5a Cõu 29: Thc hin cỏc thớ nghim sau: (1) Nung hn hp Fe v KNO3 khớ tr (2) Cho lung khớ H2 i qua bt CuO nung núng (3) t dõy Mg bỡnh kớn cha y SO2 (4) Nhỳng dõy Ag vo dung dch HNO3 S thớ nghim xy phn ng oxi húa kim loi: A B C D Cõu 30: Hn hp X gm a mol Al v b mol Na Hn hp Y gm b mol Al v a mol Na Thc hin thớ nghim sau Thớ nghim 1: Hũa tan hn hp X vo nc d thu c 5,376 lớt khớ H2, dung dch X1 v m gam cht rn khụng tan Thớ nghim 2: Hũa tan hn hp Y vo nc d thu c dung dch Y1 ú lng NaOH l 1,2 gam Bit th tớch khớ o ktc Tng lng Al hn hp X v Y l A 6,75 gam B 7,02 gam C 7,29 gam D 7,56 gam Cõu 31: Chia dung dch hn hp X gm Al2(SO4)3 v Fe2(SO4)3 thnh hai phn bng Phn mt hũa tan va ỳng 2,56 gam bt Cu Phn hai tỏc dng vi 200 ml dung dch Ba(OH) 1M, thu c 50,5 gam kt ta Cỏc phn ng xy hon ton T l mol gia Al 2(SO4)3 v Fe2(SO4)3 dung dch hn hp X l A : B : C : D : Cõu 32: Nhit phõn mui amoni icromat: (NH4)2Cr2O7 thu c sn phm l A Cr2O3, N2, H2O B Cr2O3, NH3, H2O C CrO3, N2, H2O D CrO3, NH3, H2O Cõu 33: un núng triglyxerit X vi dung dch NaOH va thu c dung dch Y cha mui natri ca axit stearic v oleic em cụ cn dung dch Y thu c 54,84 gam mui Bit X lm mt mu va dung dch cha 0,12 mol Br2 Phn Khi lng phõn t ca X l A 886 B 888 C 884 D 890 Cõu 34: Aminoaxit X (CnH2n+1O2N), ú phn trm lng cacbon chim 51,28% Giỏ tr ca n l A B C D C Dung dch Br2 D Cu(OH)2 Cõu 35: Cht no sau õy tỏc dng vi tripanmitin A H2 B Dung dch NaOH Cõu 36: Cho 5,6 gam Fe vo 200 ml dung dch Cu(NO 3)2 0,5M v HCl 1,2 M thu c khớ NO v m gam kt ta Xỏc nh m Bit rng NO l sn phm kh nht ca NO3- v khụng cú khớ H2 bay A 0,64 B 2,4 C 0,32 D 1,6 Cõu 37: phõn bit cỏc dung dch glucoz, saccaroz v h tinh bt cú th dựng dóy cht no sau õy lm thuc th ? A AgNO3/NH3 v NaOH B Cu(OH) v AgNO3/NH3 C HNO3 v AgNO3/NH3 D Nc brom v NaOH Cõu 38: Cho 35 gam hn hp cỏc amin gm anilin, metylamin, imetylamin, imetylmetylamin tỏc dng va vi 300ml dung dch HCl 1M Khi lng mui khan thu c sau phn ng l: A 45,65 gam B 45,95 gam C 36,095 gam D 56,3 gam Cõu 39: Nhng phn ng húa hc ln lt chng minh rng phõn t glucoz cú nhúm chc CHO v cú nhiu nhúm OH lin k l: A Phn ng gng v phn ng lờn men ru B Phn ng gng v phn ng vi Cu(OH) nhit phũng cho dung dch mu xanh lam C Phn ng to phc vi Cu(OH) v phn ng lờn men ru D Phn ng lờn men ru v phn ng thy phõn Cõu 40: Ly 0,3 mol hn hp X gm H2NC3H5(COOH)2 v H2NCH2COOH cho vo 400 ml dung dch HCl 1M thỡ thu c dung dch Y Y tỏc dng va vi 800 ml dung dch NaOH 1M thu c dung dch Z Lm bay hi Z thu c m gam cht rn khan Giỏ tr ca m l: A 61,9 B 28,8 C 52,2 HT D 55,2 PHN TCH - HNG DN GII THI TH THPT PHC LONG LN Cõu 1: Chn A t C H 5COOC H + NaOH C H 5COONa + C H 5OH etylpropylat Natripropylat etanol Cõu 2: Chn C Cõu 3: Chn B Tớnh oxi húa gim dn theo dóy : Cu2+ > Zn2+ > Ca2+ > Na+ Cõu 4: Chn B A ỳng, Theo chiu tng dn in tớch ht nhõn,nhit núng chy ca kim loi kim gim dn B Sai, Mg tỏc dng vi nc nhit cao, Be khụng tỏc dng vi nc mi iu kin nhit C ỳng, Na2CO3 l nguyờn liu quan trng cụng nghip sn xut thy tinh D ỳng, Thc t, cỏc son ni lm bng nhụm cú lp oxit Al 2O3 bn ngn cn khụng cho Al tip xỳc vi H2O iu ny cng xy tng t vi Cr v Zn Cõu 5: Chn A NH2CH2COOC2H5 (X) + NaOH H2N-CH2-COONa (X1) + C2H5OH H2N-CH2-COONa (X1) + HCl ClH3N-CH2-COOH (X2) + NaCl ClH3N-CH2-COOH (X2) + CH3OH ClH3N-CH2-COOCH3 (X3) + H2O ClH3N-CH2-COOCH3 (X3) + 2KOH H2N-CH2-COOK + KCl + CH3OH + H2O Cõu 6: Chn C Ag,Cu :hỗn hợp rắn Y NaOH Zn, Fe + AgNO ,Cu(NO ) Zn 2+ , Fe 2+ ,Cu + , NO Fe(OH) ,Cu(OH) + Na ZnO , NaNO 123 44 4 43 4 44 4 43 44 4 43 44 4 43 hỗn hợp X dung dịch hỗn hợp dung dịch Z hỗn hợp kết tủa - Vy dung dch Z chc Zn(NO3)2, Fe(NO3)2 v Cu(NO3)2 Cõu 7: Chn C A Fe2O3 B CrO C Cr2O3 D CrO3 Oxit baz Oxit baz Oxit lng tớnh Oxit axit Cõu 8: Chn C Ti catot H 2O + Ti anot 2e 2OH - + H2 H2O 4H+ + O2 + 4e dung dịch sản phẩm Bn cht ca in phõn dung dch K 2SO4 l cụ cn dung dch Cõu 9: Chn A H2N-CH2-COOH + NaOH H 2N-CH2-COONa + H2O H2N-CH2-COOH + HCl ClH3N-CH2-COOH + H H2N-CH2-COOH + C2H5OH H2N-CH2-COOC2H5 + H2O 0 C H2N-CH2-COOH + HONO HO -CH2-COOH + N2 + H2O Cõu 11: Chn D Dóy kim loi u cú th iu ch bng phng phỏp in phõn dung dch mui ca chỳng l nhng kim loi hot ng mnh nh kim loi kim, kim th v nhụm Cõu 12: Chn D A ỳng, bn cht ca enzim l nhng cht hu cht cú bn cht protein B ỳng, Cho glyxin tỏc dng vi HNO cú khớ bay 0 C H2N-CH2-COOH + HONO HO -CH2-COOH + N2 + H2O C ỳng, Phc ng saccarozo cú cụng thc l (C 12H21O11)2Cu 2C 12H22O11 + Cu(OH)2 (C12H21O11)2Cu + 2H2O D Sai, peptit c chia thnh hai loi : * Oligopeptit gm cỏc peptit gm cỏc peptit cú t 10 gc aminoaxit * Polipeptit gm cỏc peptit cú t 11 n 50 gc aminoaxit Cõu 13: Chn A - Triolein, Etyl aminoaxetat, Anilin trng thỏi lng Glyxin trng thỏi rn Cõu 14: Chn B BT:e 2n Fe + 2n Cu = 2n Fe3O + 2n Fe 2O x + y = z + t Cõu 15: Chn A - Cho t t a mol HCl vo b mol Na 2CO3 thỡ : n CO2 (1) = n HCl n Na 2CO3 n CO = b a - Cho t t b mol Na 2CO3 vo a mol HCl thỡ : n CO (2) = - Theo bi ta cú : n HCl = 0,5b n CO2 (1) V b a = = = a = 0,75b n CO2 (2) 2V 0,5b Cõu 16: Chn C - Ta cú: n CuSO = 800.0,05 = 0,25mol m CuSO4 5H 2O = 62,5(g) 160 Cõu 17: Chn A t - Phn ng : Gly Ala + 2NaOH GlyNa + AlaNa + H 2O m muối = 97n GlyNa + 111n AlaNa = 20,8(g) Cõu 18: Chn A BTKL - un núng hn hp Y vi H2SO4 thỡ : n H 2O = 0,5n T = 0,25mol m Y = m ete + 18n H 2O = 18,8(g) BTKL - T tỏc dng vi NaOH thỡ m Z = m T + 40n NaOH m Y = 53,2 (g) (với n NaOH = n T = 0,5mol) Cõu 19: Chn D - Khi t 22,32 gam M thỡ : n CO2 = m M 2n H 2O 16n O(trong M) 22,32 2.0,8 16(4n X + n Y ) = 12 12 - p dng bt bóo hũa ta c : n CO n H 2O = 2n X n Y 22,32 2.0,8 16(4n X + n Y ) 0,8 = 2n X n Y 88n X + 4n Y = 11,12(1) 12 - Khi cho lng M trờn tỏc dng vi K d thỡ : 2n X + n Y = 2n H = 0,38(2) - T ta gii h (1) v (2) c : n X = 0,12 mol n Y = 0,14 mol , suy n CO2 = 0,9 mol BT:C - Xột hn hp M ta cú : an X + bn Y = n CO 0,12a + 0,14b = 0,9 a = b = - Vy X v Y ln lt l : HOOC CH = CH COOH (0,12 mol) v C 3H 7OH (0,14 mol) %m C 3H 7OH(Y) = 0,14.60 100 = 37,63 22,32 Cõu 20: Chn D t HCOOCH = CH (C 3H 4O ) + NaOH HCOONa + CH 3CHO Cõu 21: Chn A Cõu 22: Chn C Cõu 23: Chn D Cú cht tỏc dng c vi Fe(NO3)2 l HCl, AgNO3, Cl2 v KMnO4/H2SO4 loóng PT phn ng : 3Fe + + 4H + + NO 3Fe 3+ + NO + 2H 2O Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag 6Fe(NO3)2 + 3Cl2 4Fe(NO3)3 + 2FeCl3 10Fe(NO3)2 + 2KMnO4 + H2SO4 Fe(NO3)3 + Fe2(SO4)3 + K2SO4 + 2MnSO4 + H2O Cõu 24: Chn A - Xột ton quỏ trỡnh phn ng ta cú h sau : a + 2b = 0, a = 0,2 n Gly + 2n Glu = n NaOH n HCl 97a + 191b = 38,5 b = 0,1 97n GlyNa + 191n GluNa = m muối 58,5n NaCl %m Gly = 75n Gly 100 = 50,51 75n Gly + 147n Glu Cõu 25: Chn B - Khi t chỏy X, Y cú cựng s mol, lng MX = MY n (X) n CO (X) : n CO (Y) = : X : C H 4O n C(X) 2 = = v - Ta cú: n H(Y) n C(Y) Y : C H 8O n H O(X) : n H O(Y) = 1: + Cú ng phõn ca X C2H4O2 l: CH3COOH v HCOOCH3 + Cú S ng phõn ca Y C3H8O l: CH3CH2CH2OH; CH3CH(OH)CH3 v CH3OC2H5 Vy s cp (X, Y) tha l: 3.2 = Cõu 26: Chn A Cõu 27: Chn D Hn hp Na2O v Al2O3: Na2O + H2O 2NaOH 2NaOH + Al2O3 2NaAlO2 + H2O mol mol mol mol - Dung dch sau phn ng cha NaAlO2 l cht tan tt nc Hn hp Cu v Fe2(SO4)3: 2Cu + Fe2(SO4)3 2CuSO4 + FeSO4 mol 0,5 mol - Dung dch sau phn ng cha CuSO4; FeSO4 v Fe2(SO4)3 d l cỏc cht tan tt nc Hn hp KHSO4 v KHCO3: KHSO4 + KHCO3 K2SO4 + CO2 + H2O mol mol - Sau phn ng thu c K 2SO4 tan tt nc nhng khớ CO ớt tan H2O, vy hn hp trờn khụng hon ton tan nc Hn hp BaCl2 v CuSO4: BaCl2 + CuSO4 BaSO4 + CuCl2 mol mol - Sau phn ng thu c BaSO4 kt ta khụng tan nc Hn hp Fe(NO3)2 v AgNO3: Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag mol mol - Sau phn ng thu c Ag kt ta khụng tan nc Vy cú hn hp cú th tan tt nc d Cõu 28: Chn B - Hng t 1: Fe + 2AgNO3 Fe(NO3)2 + 2Ag mol: x ; 2x y y (Y) Ag : 2x mol, Cu : y mol mol 642x7mol 48 4t 48 } Fe + AgNO ,Cu(NO )2 14 43 44 4 43 (x + y) mol m (g) X Fe + Cu(NO3)2 Fe(NO3)2 + Cu x + y mol t y mol 0,5x mol y mol } } 64x7mol48 64y7mol48 78 } NaOH t Fe + , Cu + , NO Fe(OH)3 ,Cu(OH) Fe O ,CuO 4 4 43 44 4 43 44 43 dung dịch X dung dịch Z a (g) b(g) Theo m gam m 56(x + y) = m (1) x + y = Theo a gam 56 (1), (3) 90(x + y) + 98(t y) = a (2) + Ta cú h sau: Theo b gam t y = b m 80(x + y) + 80(t y) = b (3) 80 56 + Thay (x + y) v (t y) vo (2) ta c biu thc: m = 8,575b 7a - Hng t 2: BT: Fe n Fe = n Fe(OH) = m m 45 mol n Fe 2O3 = mol m Fe(OH) = m gam 56 112 28 10 49b 70m BT: Cu M m Fe 2O3 + m CuO = b m CuO = b m ữgam m Cu(OH) = ữ gam 40 - Ta cú: m Fe(OH) + m Cu(OH) = a 45m 49b 70m + = a m = 8,575b 7a 28 40 Cõu 29: Chn D - Cỏc phn ng xy ra: (1) Nỳng núng KNO3: to 2KNO 2KNO + O sau ú Fe tỏc dng to 3Fe + 2O Fe 3O o t (2) H + CuO Cu + H 2O : phn ng kh oxit kim loi (3) 2Mg + SO 2MgO + S (4) 3Ag + 4HNO3 3AgNO3 + NO + 2H 2O Vy cú phn ng oxi húa kim loi l (1), (3) v (4) Cõu 30: Chn C a mol b mol } } n H2 Al + H 2O dư { Al , Na X1 + H - Thớ nghim 1: 14 {2 + m gam Ta cú: n Na = b = = 0,12 mol 43 0,24 mol hỗn hợp X vi O 2: 0,03mol 0,12 mol 64 48 } } + H 2O BT: Na Al , Na NaOH , NaAlO - Thớ nghim 2: 4 a = 0,15 mol 4 4 32 Khi ú: 0,12 mol a mol dung dịch Y dung dịch Y2 m Al = 27.(a + b) = 7, 29 gam Cõu 31: Chn B - Gi x l s mol ca Al2(SO4)3 n Al3+ = 2x mol - Phn 1: hũa tan va ỳng vi 0,04 mol Cu n Cu = n Fe (SO )3 = 0, 04 mol - Phn 2: tỏc dng vi 0,2 mol dung dch Ba(OH)2 thu c kt ta gm: Fe(OH)3: Nhn thy 3n Fe3+ < n OH n Fe(OH)3 = 0, 08 mol m Fe(OH)3 = 8,56 (g) BaSO4: Nu n BaSO = n Ba 2+ = 0, mol m BaSO + m Fe(OH)3 > 50,5 gam n SO 24 = 3(n Al2 (SO )3 + n Fe (SO )3 ) = 3x + 0,12 m BaSO = 699x + 27,96 (g) Al(OH)3: Xột trng hp to kt ta ca Al(OH)3 (vi n OH cũn li = 0,16 mol) - Trng hp 1: Al(OH)3 khụng b hũa tan + Khi ú 3n Al3+ = 6x n OH = 0,16 x > 0, 0267 m n SO 24 = 3x + 0,12 < 0, x < 0, 0267 (vụ lớ) - Trng hp 2: Al(OH)3 b hũa tan mt phn + Khi ú: n Al(OH)3 = 4n Al3+ n OH = 8x 0,16 m Al(OH)3 = 624x 12, 48 (g) m m Fe(OH)3 + m BaSO + m Al(OH) = 50,85 x = 0, 02 mol Vy n Al (SO )3 = n Fe (SO ) Cõu 32: Chn A o t - Nhit phõn mui amoni icromat: (NH4)2Cr2O7 Cr2O3 + N2 + 4H2O Cõu 33: Chn A - Gi s triglyxerit X c to t gc axit stearic v gc axit oleic Trong phõn t ca X lỳc ny cha liờn kt C=C - Khi cho X tỏc dng vi dung dch Br2 thỡ: n X = n Br2 = 0, 06 mol n C17 H 33COONa = 2n X = 0,12 mol + NaOH - Khi cho X mmui = 54,84 gam (tha yờu cõu bi toỏn) n C17 H 35COONa = n X = 0, 06 mol - Nu nh trng hp trờn khụng tha thi ta suy trng hp cũn li l X c to t gc axit stearic v gc axit oleic Khi ú MX = 888 Vy M X = 886 Cõu 34: Chn C - Ta cú: %C = 12n = 0, 5128 n = 14n + 47 Cõu 35: Chn B - Tripanmitin cú cụng thc (C15H31COO)3C3H5 tham gia phn ng thy phõn mụi trng axit v kim (C15H31COO)3C3H5 + 3NaOH 3C15H31COONa + C3H5(OH)3 Cõu 36: Chn A S oxi húa S kh Fe Fe2+ + 2e 0,1 3e + 4H+ + NO3- NO + 2H2O ; 0,18 0,24 0,2 0,01 Cu2+ + 2e Cu 0,1 0,2 (vỡ sau phn ng cú cht rn nờn Fe chuyn lờn Fe2+) - Nhn thy: ne nhn > ne cho nCu p = 2n Fe 3n NO = 0,01 mol mrn = mCu p = 0, 64 (g) Cõu 37: Chn B Thuc th Glucoz Saccaroz H tỡnh bt Cu(OH)2 Phc xanh lam Phc xanh lam Khụng hin tng AgNO3/NH3 Kt ta bc Khụng hin tng Khụng hin tng Cõu 38: Chn B BTKL m muối khan = m amin + 36,5n HCl = 45,95(g) Cõu 39: Chn B Tớnh cht ca ancol a chc (poliancol hay poliol): - Dung dch glucoz hũa tan Cu(OH)2 cho dung dch phc ng glucoz cú mu xanh lam 2C6H12O6 + Cu(OH)2 (C6H11O6)2Cu + 2H2O Tớnh cht ca anehit: - Vi dung dch AgNO3 NH3, un núng (thuc th Tollens) cho phn ng bc o t CH2OH[CHOH]4CHO + 2[Ag(NH3)2]OH CH2OH[CHOH]4COONH4 + 2Ag + 3NH3 + H2O Cõu 40: Chn A n A + n B = n X n A + n B = 0,3 n A = 0,1mol - Xột ton b quỏ trỡnh phn ng ta cú 2n A + n B = n NaOH n HCl 2n A + n B = 0, n B = 0,2 mol - Khi cho dung dch Y tỏc dng vi NaOH thỡ n H 2O = n NaOH = 0,8 mol BTKL m rắn khan = 147n A + 75n B + 36,5n HCl + 40n NaOH 18n H O = 61,9 (g) ... kh HT S GD & T THI TH THPT QUC GIA LN NM 201 7 TNH THI BèNH Mụn: KHOA HC T NHIấN HểA HC THPT PH DC Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt Mó : 132 ( thi cú 40 cõu / trang) Cho... 40: Chn C S GD & T KIM TRA CHT LNG HC Kè I NM 201 7 TNH LM NG Mụn: KHOA HC T NHIấN HểA HC Thi gian lm bi: 50 phỳt, khụng k thi gian phỏt Mó : 132 ( thi cú 40 cõu / trang) Cho bit nguyờn t ca cỏc... C2H5OH + HCOOH o t NaHCO3 + HCOOH HCOONa + CO2 + H2O (b) ỳng, Amilopectin chim t 70 80% lng tinh bt amiloz chim t 20 30% lng tinh bt (c) Sai, Cỏc tripeptit tr lờn (ipeptit khụng cú phn ng