35.-thpt-yen-phong-2-bac-ninh-nam-2017-lan-1-co-loi-giai

S GD & T TNH BC NINH THPT YấN PHONG 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: Polime cua loi vt liu nao sau õy co cha nguyờn tụ nit ? A Cao su buna B T nilon-6,6 C T visco D Nha poli (vinyl clorua) + HCl + NaOH Cõu 2: Cho chuụi phan ng sau: X Y X Chõt nao sau õy phu hp: A H2N-CH2-COOH B C6H5NH2 C Ala-Gly D HCOONH4 Cõu 3: Sụ ụng phõn este mch h ng vi cụng thc phõn t C3H6O2 la: A B C D Cõu 4: Chõt co phan ng vi dung dch Br2 la: A Phenyl clorua B Alanin C Metyl amin D Triolein Cõu 5: Trng hp nao sau õy to hp cht Fe(II) ? A Nhung st vao dung dch H2SO4 loang B ụt dõy st bỡnh ng kh Cl2 C Nhung st vao dung dch AgNO3 d D Cho bt Fe vao dung dch HNO3 d Cõu 6: Nhung mt la st (d) vao dung dch cha mt cac chõt sau: FeCl 3, AlCl3, CuSO4, Pb(NO3)2, H2SO4 c nong Sau phan ng lõy la st ra, co bao nhiờu trng hp to muụi st (II) ? A B C D Cõu 7: Dóy kim loi tan hoan toan H2O iờu kiờn thng la: A Fe, Na, K B Ca, Ba, K C Ca, Mg, Na D Al, Ba, K Cõu 8: Glyxin la tờn gi cua chõt nao sau õy? A C6H5NH2 B CH3NH2 C H2N-CH2-COOH D CH3-CH(NH2)-COOH Cõu 9: Cho 2,655 gam amin no, n chc, mch h X tac dng vi lng d dung dich HCl Sau phan ng xay hoan toan, thu c 4,8085 gam muụi Cụng thc phõn t cua X la: A C3H9N B C3H7N C CH5N D C2H7N Cõu 10: Phan ng nao sau õy la phan ng nhit nhụm ? A 3Al + 3CuSO4 Al2(SO4)3 + 3Cu o t B 8Al + 3Fe3O4 4Al2O3 + 9Fe D 2Al + 3H2SO4 Al2(SO4)3 + 3H2 pnc C 2Al2O3 4Al + 3O2 Cõu 11: Gluxit na o sau õy c goi la ng mớa? A Saccaroz B Tinh bụt C Glucoz D Fructoz Cõu 12: Cac kim loai Fe, Cr, Cu cung tan dung dich nao sau õy? A Dung dich HCl B Dung dich HNO3 c, nguụi C Dung dich HNO3 loóng D Dung dich H2SO4 c, nguụi Cõu 13: Kim loai X tỏc dung v i H2SO4 loóng cho kh H2 Mt khac oxit cua X bi kh H2 kh thnh kim loai nhiờt ụ cao X la kim loai na o ? A Fe B Al C Mg D Cu Cõu 14: Cho 10 gam hn hp gm Fe v Cu tỏc dng vi dung dch H 2SO4 loóng d Sau phan ng thu c 2,24 lớt H2 lớt khớ hidro ( kc) dung dic h X va m gam kim loai khụng tan Gia tri cu a m la A 6,4 gam B 3,4 gam C 4,4 gam D 5,6 gam Trang Cu(OH)2 /OH t Cõu 15: Cho s ụ sau: X dung dịch màu xanh lam kết tủa đỏ gạch X la dung dich nao sau õy: A Protein B Triolein C Glucoz D Vinyl fomat Cõu 16: Cho 33,9 gam hụn hp bt Zn v Mg (t l : 2) tan ht dung dch hn hp gm NaNO v NaHSO4 thu c dung dch A ch cha m gam hn hp cỏc mui trung hũa v 4,48 lớt (kc) hn hp khớ B gm N2O v H2 Hn hp khớ B cú t so vi He bng 8,375 Giỏ tr gn nht ca m l : A 240 B 300 C 312 D 308 Cõu 17: Hn hp X gụ m valin (co cụng thc C4H8NH2COOH)) va ipeptit Glyxylalanin Cho m gam X vo 100ml dung dich H2SO4 0,5M (loang), thu c dung dich Y Cho ton b Y phn ng va vi 100ml dung dch NaOH 1M v KOH 1,75M un núng thu c dung dch cha 30,725 gam mui Phn trm lng ca Valin X l : A 65,179% B 54,588% C 45,412% D 34,821% Cõu 18: Cho cỏc nhn nh sau: (1) Tt c cỏc ion kim loi ch b kh (2) Hp cht cacbohirat v hp cht amino axit u cha thnh phn nguyờn t ging (3) Dung dch mui mononatri ca axit glutamic lm qu tớm chuyn sang mu xanh (4) Cho kim loi Ag vo dung dch FeCl2 thỡ thu c kt ta AgCl (5) Tớnh cht vt lớ chung ca kim loi cỏc electron t gõy (6) Phn ng thy phõn este v protein mụi trng kim u l phn ng mt chiu S nhn nh ỳng l A B C D Cõu 19: Chn cp cht khụng xy phn ng? A dung dch AgNO3 v dung dch Fe(NO3)2 B dung dch Fe(NO3)2 v dung dch KHSO4 C dung dch H2NCH2COONa v dung dch KOH D dung dch C6H5NH3Cl v dung dch NaOH Cõu 20: Cho cỏc dung dch FeCl3, HCl, HNO3 loóng, AgNO3, ZnCl2 v dung dch cha (KNO3, H2SO4 loóng) S dung dch tỏc dng c vi kim loi Cu nhit thng l A B C D Cõu 21: Dóy cỏc polime t chỏy hon ton u thu c khớ N2 A t olon, t tm, t capron, cao su buna-N B t lapsan, t enng, t nilon-6, xenluloz C protein, nilon-6,6, poli(metyl metacrylat), PVC D amilopectin, cao su buna-S, t olon, t visco Cõu 22: Hũa tan hon ton hn hp H gm Mg (5a mol) v Fe3O4 (a mol) dung dch cha KNO3 v 0,725 mol HCl, cụ cn dung dch sau phn ng thỡ thu c lng mui khan nng hn lng hn hp H l 26,23g Bit kt thỳc phn ng thu c 0,08 mol hn khớ Z cha H2 v NO, t ca Z so vi H2 bng 11,5 % lng st cú mui khan cú giỏ tr gn nht vi A 17% B 18% C 26% D 6% Cõu 23: Cho cỏc s phn ng sau (theo ỳng t l mol): X1 + X2 + H2O X3 + NaCl C7H18O2N2 (X) + NaOH X1 + 2HCl X3 t nilon-6 + H2O X4 + HCl X4 Phỏt biu no sau õy ỳng A X2 lm qu tớm húa hng B Cỏc cht X, X4 u cú tớnh lng tớnh C Phõn t ca X ln hn so vi X3 D Nhit núng chy ca X1 nh hn X4 Trang Cõu 24: Trn 2,43 gam Al vi 9,28 gam Fe 3O4 ri nung núng cho phn ng xy mt thi gian, lm lnh c hn hp X gm Al, Fe, Al2O3, FeO v Fe3O4 Cho ton b X phn ng vi dung dch HCl d thu c 2,352 lớt H2 (ktc) v dung dch Y Cụ cn Y c a gam mui khan Xỏc nh giỏ tr ca a l A 27,965 B 16,605 C 18,325 D 28,326 Cõu 25: Cho 0,3 mol hn hp X gm este n chc tỏc dng va vi 200 ml dung dch NaOH 2M un núng, thu c hp cht hu c no mch h Y cú phn ng bc v 37,6 gam hn hp mui hu c t chỏy hon ton Y ri cho sn phm hp th ht vo bỡnh cha dung dch nc vụi d, thy lng bỡnh tng 24,8 gam Khi lng ca X l A 30,8 gam B 33.6 gam C 32,2 gam D 35,0 gam Cõu 26: Hn hp E gm chui peptit X, Y, Z u mch h (c to nờn t Gly v Lys) Chia hn hp lm hai phn khụng bng Phn 1: cú lng 14,88 gam c em thy phõn hon ton dung dch NaOH M thỡ dựng ht 180 ml, sau phn ng thu c hn hp F cha a gam mui Gly v b gam mui Lys Mt khỏc, t chỏy hon ton phn cũn li thỡ thu c t l th tớch gia CO v hi nc thu c l : T l a : b gn nht vi giỏ tr : A 1,57 B 1,67 C 1,40 D 2,71 Cõu 27: Cho hn hp X gm FexOy, Fe, MgO, Mg Cho m gam hn hp X trờn tỏc dng vi dung dch HNO3 d thu c 6,72 lớt hn hp khớ N 2O v NO (dktc) cú t so vi H l 15,933 v dung dch Y Cụ cn dung dch Y thu c 129,4 gam mui khan Cho m gam hn hp X tỏc dng vi dung dch H2SO4 c núng d thu c 15,68 lớt khớ SO (ktc, sn phm kh nht) v dung dch Z Cụ cn dung dch Z thu c 104 gam mui khan Giỏ tr gn nht ca m l A 22,0 B 28,5 C 27,5 D 29,0 Cõu 28: Phõn bit cỏc cht CaCl2, HCl, Ca(OH)2 cú th dựng dung dch A NaOH B NaHCO3 C Na2CO3 D NaNO3 Cõu 29: Nhn xột no di õy l ỳng A Tripeptit hũa tan Cu(OH)2 (phn ng mu biure) to dung dch xanh lam B Trong phõn t protein luụn cú nguyờn t nit C Tinh bt v xenluloz l ng phõn ca D Este l nhng cht hu c d tan nc Cõu 30: Trong kim loi sau: Fe, Na, Al, Cr Kim loi n tip xỳc vi axit v kim loi khỏ mm, d kộo si, d dỏt mng kim loi theo th t l A Na v Fe B Cr v Al C Na v Al D Cr v Fe Cõu 31: X phũng húa hn hp gm CH3COOCH3 v CH3COOC2H5 thu c sn phm gm A mui v ancol B mui v ancol C mui v ancol D mui v ancol Cõu 33: Nguyờn t húa hc no sau õy thuc nhúm kim loi kim th A Natri B Bari C Nhụm D Kali Cõu 34: Nhn xột no sau õy sai A Glucoz c dựng lm thuc tng lc cho ngi gi, tr em, ngi m B G c dựng ch bin thnh giy C Xenluloz cú phn ng mu vi iot D Tinh bt l mt s ngun cung cp nng lng cho c th Cõu 35: X, Y, Z, T l mt cỏc cht sau: glucoz, anilin (C6H5NH2), fructoz v phenol (C6H5OH) Tin hnh cỏc thớ nghim nhn bit chỳng v ta cú kt qu nh sau: Thuc th X T Z Y (+): phn ng Nc Br2 Kt ta Nht mu Kt ta (-) (-): khụng phn ng dd AgNO3/NH3, to (-) Kt ta (-) Kt ta dd NaOH (-) (-) (+) (-) Cỏc cht X, Y, Z, T ln lt l A glucoz, anilin, phenol, fructoz B anilin, fructoz, phenol, glucoz C phenol, fructoz, anilin, glucoz D fructoz, phenol, glucoz, anilin Trang Cõu 36: Cho 13,5 gam hn hp gm amin no, n chc, mch h tỏc dng va vi 300 ml dung dch HCl x M, thu c dung dch cha 24,45 gam hn hp mui Giỏ tr ca x l A 0,5 B 1,4 C 2,0 D 1,0 Cõu 37: 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 38: 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 39: 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 40: 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 36: Chn D - Phng trỡnh: RNH + HCl RNH 3Cl m m RNH BTKL n HCl = RNH3Cl = 0,3mol C M(HCl) = M 36,5 Cõu 37: 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 38: Chn A Cõu 39: Chn D Hn hp Na2O v Al2O3: Na2O + H2O 2NaOH + Al2O3 2NaOH 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 Trang 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 40: Chn B - Hng t 1: Fe + 2AgNO3 Fe + Cu(NO3)2 Fe(NO3)2 + 2Ag ; Fe(NO3)2 + Cu mol: x 2x y y (Y) Ag : 2x mol, Cu : y mol mol (x + y) mol 642x7mol 48 4t 48 } x + y mol t y mol 0,5x mol y mol } 64x7mol48 64y7mol48 78 } Fe + AgNO ,Cu(NO )2 } NaOH t 14 43 44 4 43 Fe + , Cu + , NO Fe(OH)3 ,Cu(OH) Fe O ,CuO 4 4 43 44 4 43 44 43 m (g) X 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: m m 45 mol n Fe 2O3 = mol m Fe(OH) = m gam 56 112 28 10 49b 70m BT: Cu M m Fe2 O3 + m CuO = b m CuO = b m ữgam m Cu(OH) = ữgam 40 45m 49b 70m + = a m = 8,575b 7a - Ta cú: m Fe(OH) + m Cu(OH) = a 28 40 BT: Fe n Fe = n Fe(OH) = HT Trang PHN TCH HNG DN GII THI TH THPT YấN PHONG BC NINH LN Cõu 1: Chn B Cõu 2: Chn B + HCl + NaOH C H 5NH (X) C H 5NH 3Cl(Y) C H 5NH (X) Cõu 3: Chn D Cú ng phõn este mch h ng vúi CTPT l C3H6O2 l HCOOC2H5, CH3COOCH3 Cõu 4: Chn D Cõu 5: Chn A B Fe + 3AgNO3(d) Fe(NO3)3 + 3Ag C Fe + Cl2 FeCl3 D Fe + 4HNO3(d) Fe(NO3)3 + NO + 2H2O Cõu 6: Chn C Cú trng hp to mui st (II) l : FeCl3, CuSO4, Pb(NO3)2 v H2SO4(c, núng) Cõu 7: Chn B Cõu 8: Chn C Cõu 9: Chn D m m amin BTKL n HCl = n amin = muối = 0,059 mol M amin = 45(C H N) 36,5 Cõu 10: Chn B Cõu 11: Chn A A Saccaroz B Tinh bụt C Glucoz D Fructoz ng mớa ng nho ng mt ong Cõu 12: Chn C Lu ý : Cr, Fe v Al b th ng húa bi HNO3, H2SO4 c nguụi Cõu 13: Chn A Cõu 14: Chn C BT:e n Fe = n H = 0,1 m Cu = 10 56n Fe = 4, (g) Cõu 15: Chn C Cõu 16: Chn D 0,3mol 0,6 mol 0,15mol 0,05mol 678 } } } } } 2+ 2+ + + Zn , Mg + NaNO , NaHSO Zn , Mg , Na , NH ,SO + N 2O , H + H 2O 43 44 4 43 4 4 4 4 4 44 43 0,3mol 0,6 mol hỗn hợp kim loại dung dịch hỗn hợp BT:e n NH + = dung dịch A 2n Zn + + 2n Mg + 8n N O 2n H hỗn hợp B BT:N = 0,0625 mol n NaNO3 = 2n N O + n NH + = 0,3625mol BT:H n NaHSO = 10n NH + + 10n N 2O + 2n H = 2,225mol n H 2O = n NaHSO 4n NH + 2n H 2 = 0,9375mol BTKL m A = m kim loại + 85n NaNO3 + 120n NaHSO m B 18n H 2O = 308,1375(g) Cõu 17: Chn B a mol b mol b mol 0,1mol 0,175mol 0,05mol } } } } } 678 } 64b7mol48 NaOH, KOH Val ,GlyAla + H SO dung dịch Y Val ,Gly , Ala , Na + , K + ,SO 4 22 34 4 4 44 4 4 4 43 a mol hỗn hợp X 0,05mol 30,725(g) muối Trang 116n Val + 74n Gly + 88n Ala = m muối 39n K + 23n Na + 96n SO 116a + 74b + 88b = 16,8 a = 0,075 BTDT a + b + b = 0,175 n +n +n = n + +n + 2n b = 0,05 Val Gly Ala K Na SO %m Val = 54,58 Cõu 18: Chn A (1) Sai, Fe2+, Cr2+ b oxi húa bi cỏc cht oxi húa mnh nh KMnO4, O2 (2) Sai, hp cht amino axit cha C, H, O v N cũn hp cht cacbohidrat khụng cha N (3) ỳng, HOOC-CH2-CH2-CH(NH2)-COONa (mono natri glutamate) lm qu tớm húa xanh (4) Sai, cho Ag tỏc dng vi dung dch FeCl2 khụng cú phn ng xy (5) ỳng, cỏc tớnh cht nh tớnh dn in, dn nhit, do, ỏnh kim u electron t gõy (6) ỳng, phn ng thy phõn este v protein mụi trng kim u l phn ng mt chiu Vy cú nhn nh ỳng l (3), (5), (6) Cõu 19: Chn C A AgNO3 + Fe(NO3)2 Fe(NO3)3 + Ag B 3Fe2+ + NO3- + 4H+ 3Fe3+ + NO + 2H2O D C6H5NH3Cl + NaOH C6H5NH2 + NaCl + H2O Cõu 20: Chn D Cú cht tỏc dng c vi kim loi Cu iu kin thng l FeCl 3, HNO3 loóng, AgNO3 v dung dch cha (KNO3, H2SO4 loóng) Cu + 2FeCl3 CuCl2 + 2FeCl2 Cu + HNO3(loóng) Cu(NO3)2 + NO + H2O 3Cu + 2NO + 8H+ 3Cu2+ + 2NO + 4H2O Cu + 2AgNO3 Cu(NO3)2 + 2Ag Cõu 21: Chn A Cỏc polime t chỏy hon ton u thu c khớ N2 chng t phõn t polime cú cha N Vy cỏc polime cú thnh phn cha N l : t olon, t tm, t capron, cao su buna-N Cõu 22: Chn A BT:N n NH + = n KNO3 n NO = (x 0,06) mol m muối m H = 39n K + + 18n NH + + 35,5n Cl 16n O(trong H) = 39x + 18(x 0,06) + 35,5.0,725 64a m muối m H = 26,23 57x 64a = 1,5725 x = 0,0725mol 10n NH + + 2n O(trong H) + 4n NO + 2n H = n HCl 10(x 0,06) + 8a + 0,28 = 0,725 a = 0,04 mol 0,04.3.56 m H = 24.5a + 232a = 14,08(g) m muối khan = 40,31(g) % m Fe = 100 = 16,67 40,31 Cõu 23: Chn B - Cỏc phn ng xy ra: to nH N[CH ]5 COOH (X ) ( HN [CH ]5 CO ) n + nH 2O nilon H N[CH ]5 COOH (X ) + HCl ClH N[CH ]5 COOH (X ) H N[CH ]5 COONa (X1 ) + 2HCl ClH N[CH ]5 COOH (X ) + NaCl H N [CH ]5 COO NH 3CH (X) + NaOH H N[CH ]5 COONa ( X1 ) + CH NH (X ) + H 2O A Sai, X2 lm qu tớm húa xanh B ỳng X v X4 u cú tớnh lng tớnh C Sai, Phõn t ca X l 162 phõn t ca X3 l 167,5 D Sai, Nhit núng chy ca X1 ln hn X4 Cõu 24: Chn A Trang 0,09 mol 0,04 7mol 0,105 mol } } to AlCl , FeCl , FeCl + H O + H2 - Quỏ trỡnh: Al , Fe3O hỗn hợp X + HCl dư 3 4 4 43 dung dịch sau pư BT: H + Ta cú: n O (Fe3O ) = n H 2O = 0,16 mol n HCl = 2(n H + n H 2O ) = 0,53 mol - Hng t 1: BTKL a = m Y + 36, 5HCl 2n H 18n H 2O = m X + 36, HCl 2n H 18n H 2O = 27,965 (g) - Hng t 2: + Ta cú: a = m KL + 35,5n Cl = 27n Al + 56n Fe + 35,5n HCl = 27,965 (g) Cõu 25: Chn C - Nhn thy: nX = 0,3 < nNaOH = 0,4 X cú cha este ca phenol (A) v este cũn li l (B) (A) : RCOOC H R ' n A + n B = 0,3 n A = 0,1 n H 2O = n A = 0,1 Vi (B) : R1COOCH = CHR 2n A + n B = 0, n B = 0, n Y = n B = 0, - Khi t chỏy cht Y no, n chc, mch h (R2CH2CHO: 0,2 mol) luụn thu c n CO = n H 2O n CO = : Y l CH 3CHO 44n CO + 18n H 2O = 24,8 n CO = 0, mol m C Y = 0, BTKL m X = mmui + mY + m H 2O mNaOH = 32, (g) Cõu 26: Chn A + Cỏc mt xớch to tng ng vi cỏc - amino axit: - Hng t 1: S dng CTTQ ca peptit + Gi s mc xớch ca Gly l x v Lys l y ta cú CTTQ ca E l: O2 ,t C2x + 6y H3x +12y + O x + y +1N x + 2y (2x + 6y)CO + (1,5x + 6y + 1)H 2O 3x + 12y + n CO2 = n H2O 2x + 6y = x = 2 n E = 0, 0658 mol + Ta cú: 14,88 y = 0, 7353 n NaOH = n E (x + y) (x + y) = 0,18 57x + 128y + n GlyNa = xn E = 0,1316 mol a BT: Gly Lys = 1,57 n LysNa = yn E = 0, 0484 mol b - Hng t 2: Quy i v hn hp cỏc - amino axit v H2O + H + Ta cú : Gly m Lys n (E) + H O mGly + nLys Gly(C H5O N):x mol BT: C n CO2 = 2x + 6y O ,t + E Lys (C6 H14 O N ):y mol n CO2 = n H 2O 0,5x + y = z (1) BT: H H O : z mol n H2O = 2,5x + 7y z Trang BTKL 75n Gly + 146n Lys = m E + 18n H 2O 75x + 146y = 14,88 + 18z (2) BT: Na x + y = 0,18 n + n = n GlyNa LysNa NaOH v a = 1,57 b - Hng t 2.1: Ta cú th quy i hn hp E v cỏc axyl v H2O nh sau: C H 3ON, C6 H12ON v H2O sau ú gii tng t nh trờn ta cng cú th tỡm c kt qu - Hng t 3: Tỏch cht + T (1), (2) ta tớnh c: x = 0,1316 mol ; y = 0, 0484 mol; z = 0,1141 mol Vy + Ta tách C H 5O N(Gly) C H 3ON + H O v cú: tách C H12O N (Lys) C H 3ON + (CH ) NH + H 2O C H 3ON :x mol BT:C n CO2 = 2x + 4z O2 ,t + E H O :y mol mà n H 2O = n CO2 0,5x 0,5z = y (1) BT:H2 n = 1,5x + 4,5z + y (CH ) NH :z mol H2 O BTKL + Ta cú : x = n NaOH = 0,18 mol 71n (CH )4 NH + 18n H 2O = m E 57n C2H3ON 71x + 18y = 4, 62 (2) y = 0,0658 mol n GlyNa = x z = 0,1316 mol a = 1,57 + T (1), (2) ta tớnh c: b z = 0, 0484 mol n LysNa = z = 0, 0484 mol Cõu 27: Chn D - Hng t 1: Fe3+ , Mg 2+ , NH 4+ , NO3 + { NO , N 2O (1) 4 44 4 43 0,26 mol { HNO Mg, O + - Túm tt quỏ trỡnh: Fe, 14 43 m gam X H 2SO4 Fe3+ , Mg 2+ ,SO24 + SO (2) 44 4 43 { 104 (g) dd Z BT: e cho (1) v (2) n NH NO3 = 0,04 mol 129,4(g) dd Y 2n SO2 3n NO 8n N2O 0,7 mol = 0, 0375 mol m KL + m NH +4 + m NO3 = 129, (1) BTKL BTDT cho (1) v (2) 2n SO 24 = n NO3 + n NH 4+ (2) m KL + m SO24 = 104 (1) (2) 62(2n SO 24 + n NH + ) 96n SO 42 = 24, 725 n SO 42 = 0,8 mol + Xột quỏ trỡnh (2): BT: e + BTDT 2n SO 24 = 2n SO + 2n O n O = 0,1 mol m = m KL + m O = 28,8 gam m = m m = 104 0,8.96 = 27, gam Y SO KL - Hng t 2: 2n SO2 3n NO 8n N 2O BT: e cho (1) v (2) n NH4 NO3 = = 0, 0375 mol + Gi T l hn hp mui cha Fe(NO3)2 v Mg(NO3)2 suy ra: m T = m Y m NH4 NO3 = 126, gam + p dng phng phỏp tng gim lng + bo ton in tớch cho hn hp T v Z ta cú: mT m Z 126, 104 2n NO = n SO = = = 0,8 mol 2.M NO M SO 2.62 96 BT: S + BT: H + Xột quỏ trỡnh (2): n H2O = n H2SO4 = n SO2 + n SO42 = 1,5 mol BTKL m X + m H 2SO4 = m Z + mSO2 + m H 2O m = 28,8gam Trang Cõu 28: Chn B Thuc th CaCl2 HCl Ca(OH)2 A NaOH Khụng hin tng B NaHCO3 Khụng hin tng Cú khớ CO2 thoỏt Cú kt ta trng C Na2CO3 Cú kt ta trng Cú khớ CO2 thoỏt Cú kt ta trng D NaNO3 Khụng hin tng Cõu 29: Chn B A Sai, Tripeptit hũa tan Cu(OH)2 (phn ng mu biure) to dung dch mu tớm B ỳng, Trong phõn t protein luụn cú nguyờn t nit C Sai, Tinh bt v xenluloz khụng phi l ng phõn ca D Sai, Este l nhng cht hu c ớt tan nc vỡ phõn t khụng cú liờn kt hiro Cõu 30: Chn C Cõu 31: Chn D - Phng trỡnh phn ng: CH 3COOCH + NaOH CH 3COONa + CH 3OH CH 3COOC H + NaOH CH 3COONa + C H 5OH mui v ancol Cõu 32: Chn B - Cỏc kim loi kim th thuc nhúm IIA gm cỏc nguyờn t: Be, Mg, Ca, Sr, Ba, Ra Cõu 34: Chn C - Xenluloz khụng cú phn ng vi I2 ch tinh bt mi cú phn ng ny to dung dch cú mu xanh tớm Cõu 35: Chn B T: C6H12O6 Y: C6H12O6 Thuc th X: C6H5NH2 Z: C6H5OH (glucoz) (fructoz) Nc Br2 Kt ta trng Nht mu Kt ta trng Khụng phn ng dd AgNO3/NH3, to Khụng phn ng Kt ta Ag Khụng phn ng Kt ta Ag dd NaOH Khụng phn ng Khụng phn ng Cú phn ng Khụng phn ng Cõu 36: Chn D - Phng trỡnh: RNH + HCl RNH 3Cl m m RNH BTKL n HCl = RNH3Cl = 0,3mol C M(HCl) = M 36,5 Cõu 37: 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 38: Chn A Cõu 39: Chn D Hn hp Na2O v Al2O3: Na2O + H2O 2NaOH + Al2O3 2NaOH 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 Trang 10 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 40: Chn B - Hng t 1: Fe + 2AgNO3 Fe + Cu(NO3)2 Fe(NO3)2 + 2Ag ; Fe(NO3)2 + Cu mol: x 2x y y (Y) Ag : 2x mol, Cu : y mol mol (x + y) mol 642x7mol 48 4t 48 } x + y mol t y mol 0,5x mol y mol } 64x7mol48 64y7mol48 78 } Fe + AgNO ,Cu(NO )2 } NaOH t + + 14 43 44 4 43 Fe , Cu , NO Fe(OH)3 ,Cu(OH) Fe O ,CuO 4 4 43 44 4 43 44 43 m (g) X 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: 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 45m 49b 70m + = a m = 8,575b 7a - Ta cú: m Fe(OH) + m Cu(OH) = a 28 40 BT: Fe n Fe = n Fe(OH) = Trang 11