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 ca loi vt liu nao sau õy co cha nguyờn t 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: S 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 chõt Fe(II) ? A Nhung st vao dung dch H2SO4 loang B t 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 lỏ st (d) vao dung dch cha mt cỏc chõt sau: FeCl 3, AlCl3, CuSO4, Pb(NO3)2, H2SO4 c nong Sau phan ng lõy lỏ st ra, co bao nhiờu trng hp to mui st (II) ? A B C D Cõu 7: Day 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 ca 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 tỏc dng vi lng d dung dich HCl Sau phan ng xay hoan toan, thu c 4,8085 gam mui Cụng thc phõn t ca 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 loa Fe, Cr, Cu cung tan dung dich nao sau õy? A Dung dich HCl B Dung dich HNO3 c, nguụi C Dung dich HNO3 loang D Dung dich H2SO4 c, nguụ Cõu 13: Kim loai X tỏc dung v i H2SO4 loang cho kh H2 Mt khac oxit cua X bi kh H2 kh kim loai nhiờt ụ cao X la kim loai na o ? A Fe B Al C Mg D Cu Cõu 14: Cho 10 gam hụn hp gụm Fe va Cu tỏc dng vi dung dch H 2SO4 loang d Sau phan ng thu c 2,24 lớt H2 lớt khớ hidro ( kc) dung dic h X va m gam kim loa khụng tan Gia tri c u 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 va Mg (t l : 2) tan ht dung dch hụn hp gụm NaNO va NaHSO4 thu c dung dch A ch cha m gam hụn hp cỏc mui trung hũa va 4,48 lớt (kc) hụn hp khớ B gụm N2O va H2 Hụn hp khớ B co t so vi He bng 8,375 Giỏ tr gn nhõt ca m la : A 240 B 300 C 312 D 308 Cõu 17: Hụn hp X gụ m valin (co cụng thc C4H8NH2COOH)) va ipeptit Glyxylalanin Cho m gam X vao 100ml dung dich H2SO4 0,5M (loang), thu c dung dich Y Cho toan b Y phan ng va vi 100ml dung dch NaOH 1M va KOH 1,75M un nong thu c dung dch cha 30,725 gam mui Phn trm lng ca Valin X la : A 65,179% B 54,588% C 45,412% D 34,821% Cõu 18: Cho cỏc nhn nh sau: (1) Tõt ca cỏc ion kim loi ch b kh (2) Hp chõt cacbohirat va hp chõt amino axit ờu cha phn nguyờn t ging (3) Dung dch mui mononatri ca axit glutamic lam qu tớm chuyn sang mau xanh (4) Cho kim loi Ag vao dung dch FeCl2 thỡ thu c kt ta AgCl (5) Tớnh chõt vt lớ chung ca kim loi cỏc electron t gõy (6) Phan ng thy phõn este va protein mụi trng kiờm ờu la phan ng mt chiờu S nhn nh ung la A B C D Cõu 19: Chn cp chõt khụng xay phan ng? A dung dch AgNO3 va dung dch Fe(NO3)2 B dung dch Fe(NO3)2 va dung dch KHSO4 C dung dch H2NCH2COONa va dung dch KOH D dung dch C6H5NH3Cl va dung dch NaOH Cõu 20: Cho cỏc dung dch FeCl3, HCl, HNO3 loang, AgNO3, ZnCl2 va dung dch cha (KNO3, H2SO4 loang) S dung dch tỏc dng c vi kim loi Cu nhit thng la A B C D Cõu 21: Day cỏc polime t chỏy hoan toan ờu 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 hoan toan hụn hp H gụm Mg (5a mol) va Fe 3O4 (a mol) dung dch cha KNO3 va 0,725 mol HCl, cụ cn dung dch sau phan ng thỡ thu c lng mui khan nng hn lng hụn hp H la 26,23g Bit kt thuc phan ng thu c 0,08 mol hụn khớ Z cha H2 va NO, t ca Z so vi H2 bng 11,5 % lng st co mui khan co giỏ tr gn nhõt vi A 17% B 18% C 26% D 6% Cõu 23: Cho cỏc s ụ phan ng sau (theo ung t l mol): X1 + X2 + H2O X3 + NaCl C7H18O2N2 (X) + NaOH X1 + 2HCl X3 t nilon-6 + H2O X4 + HCl X4 Phỏt biu nao sau õy ung A X2 lam qu tớm hoa hụng B Cỏc chõt X, X4 ờu co tớnh lng tớnh C Phõn t ca X ln hn so vi X3 D Nhit nong chay ca X1 nh hn X4 Trang Cõu 24: Trn 2,43 gam Al vi 9,28 gam Fe3O4 rụi nung nong cho phan ng xay mt thi gian, lam lnh c hụn hp X gụm Al, Fe, Al2O3, FeO va Fe3O4 Cho toan b X phan ng vi dung dch HCl d thu c 2,352 lớt H2 (ktc) va dung dch Y Cụ cn Y c a gam mui khan Xỏc nh giỏ tr ca a la A 27,965 B 16,605 C 18,325 D 28,326 Cõu 25: Cho 0,3 mol hụn hp X gụm este n chc tỏc dng va vi 200 ml dung dch NaOH 2M un nong, thu c hp chõt hu c no mch h Y co phan ng bc va 37,6 gam hụn hp mui hu c t chỏy hoan toan Y rụi cho san phm hõp th ht vao bỡnh cha dung dch nc vụi d, thõy lng bỡnh tng 24,8 gam Khi lng ca X la A 30,8 gam B 33.6 gam C 32,2 gam D 35,0 gam Cõu 26: Hụn hp E gm chuụi peptit X, Y, Z ờu mch h (c to nờn t Gly va Lys) Chia hụn hp lam hai phn khụng bng Phn 1: co lng 14,88 gam c em thy phõn hoan toan dung dch NaOH M thỡ dung ht 180 ml, sau phan ng thu c hụn hp F cha a gam mui Gly va b gam mui Lys Mt khỏc, t chỏy hoan toan phn cũn li thỡ thu c t l th tớch gia CO va hi nc thu c la : T l a : b gn nht vi giỏ tr : A 1,57 B 1,67 C 1,40 D 2,71 Cõu 27: Cho hụn hp X gụm FexOy, Fe, MgO, Mg Cho m gam hụn hp X trờn tỏc dng vi dung dch HNO3 d thu c 6,72 lớt hụn hp khớ N 2O va NO (dktc) co t so vi H la 15,933 va dung dch Y Cụ cn dung dch Y thu c 129,4 gam mui khan Cho m gam hụn hp X tỏc dng vi dung dch H2SO4 c nong d thu c 15,68 lớt khớ SO (ktc, san phm kh nhõt) va dung dch Z Cụ cn dung dch Z thu c 104 gam mui khan Giỏ tr gn nht ca m la A 22,0 B 28,5 C 27,5 D 29,0 Cõu 28: Phõn bit cỏc chõt CaCl2, HCl, Ca(OH)2 co th dung dung dch A NaOH B NaHCO3 C Na2CO3 D NaNO3 Cõu 29: Nhn xột nao di õy la ung A Tripeptit hũa tan Cu(OH)2 (phan ng mau biure) to dung dch xanh lam B Trong phõn t protein luụn co nguyờn t nit C Tinh bt va xenluloz la ụng phõn ca D Este la nhng chõt hu c d tan nc Cõu 30: Trong kim loi sau: Fe, Na, Al, Cr Kim loi n tip xuc vi axit va kim loi khỏ mờm, d kộo si, d dỏt mng kim loi theo th t la A Na va Fe B Cr va Al C Na va Al D Cr va Fe Cõu 31: Xa phũng hoa hụn hp gụm CH3COOCH3 va CH3COOC2H5 thu c san phm gụm A mui va ancol B mui va ancol C mui va ancol D mui va ancol Cõu 33: Nguyờn t hoa hc nao sau õy thuc nhom kim loi kiờm th A Natri B Bari C Nhụm D Kali Cõu 34: Nhn xột nao sau õy sai A Glucoz c dung lam thuc tng lc cho ngi gia, tr em, ngi m B Gụ c dung ch bin giõy C Xenluloz co phan ng mau vi iot D Tinh bt la mt s nguụn cung cõp nng lng cho c th Cõu 35: X, Y, Z, T la mt cỏc chõt sau: glucoz, anilin (C 6H5NH2), fructoz va phenol (C6H5OH) Tin hanh cỏc thớ nghim nhn bit chung va ta co kt qua nh sau: Thuc th X T Z Y (+): phan ng Nc Br2 Kt ta Nht mau Kt ta (-) o dd AgNO3/NH3, t (-) Kt ta (-) Kt ta (-): khụng phan ng dd NaOH (-) (-) (+) (-) Cỏc chõt X, Y, Z, T ln lt la 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 hụn hp gụm 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 hụn hp mui Giỏ tr ca x la A 0,5 B 1,4 C 2,0 D 1,0 Cõu 37: X, Y la hai hp chõt hu c n chc phõn t ch cha C, H, O Khi t chỏy X, Y vi s mol bng hoc lng bng ờu thu c vi t l mol tng ng : va vi t l mol tng ng : S cp chõt X, Y tha man la A B C D Cõu 38: Polime X dai, bờn vi nhit va gi nhit tt nờn dt vai, may qun ỏo õm , X la A Poliacrilonitrin B Poli (vinylclorua) C Polibutaien D Polietilen Cõu 39: Co hụn hp, mụi hụn hp gụm chõt rn co s mol bng nhau: Na 2O va Al2O3; Cu va Fe2(SO4)3; KHSO4 va KHCO3; BaCl2 va CuSO4; Fe(NO3)2 va AgNO3 S hụn hp co th tan hoan toan nc (d) ch to cỏc chõt tan tt nc la A B C D Cõu 40: Cho m gam bt st vao dung dch X cha AgNO3 va Cu(NO3)2 n cỏc phan ng kt thuc thu c chõt rn Y va dung dch Z Cho dung dch Z tỏc dng ht vi dung dch NaOH d, thu c a gam kt ta T gụm hai hidroxit kim loi Nung T n lng khụng i thu c b gam chõt rn Biu thc liờn h gia m, a, b co th la 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 co cung s mol, lng MX = MY n (X) n CO (X) : n CO (Y) = : X : C H 4O n C(X) 2 = = va - Ta co: n H(Y) n C(Y) Y : C H 8O n H O(X) : n H O(Y) = 1: + Co ụng phõn ca X C2H4O2 la: CH3COOH va HCOOCH3 + Co S ụng phõn ca Y C3H8O la: CH3CH2CH2OH; CH3CH(OH)CH3 va CH3OC2H5 Vy s cp (X, Y) tha la: 3.2 = Cõu 38: Chn A Cõu 39: Chn D Hụn hp Na2O va Al2O3: Na2O + H2O 2NaOH + Al2O3 2NaOH 2NaAlO2 + H2O mol mol mol mol - Dung dch sau phan ng cha NaAlO2 la chõt tan tt nc Hụn hp Cu va Fe2(SO4)3: 2Cu + Fe2(SO4)3 2CuSO4 + FeSO4 mol 0,5 mol - Dung dch sau phan ng cha CuSO4; FeSO4 va Fe2(SO4)3 d la cỏc chõt tan tt nc Hụn hp KHSO4 va KHCO3: KHSO4 + KHCO3 K2SO4 + CO2 + H2O mol mol - Sau phan ng thu c K2SO4 tan tt nc nhng khớ CO ớt tan H2O, vy hụn hp trờn khụng hoan toan tan nc Hụn hp BaCl2 va CuSO4: BaCl2 + CuSO4 BaSO4 + CuCl2 Trang mol mol - Sau phan ng thu c BaSO4 kt ta khụng tan nc Hụn hp Fe(NO3)2 va AgNO3: Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag mol mol - Sau phan ng thu c Ag kt ta khụng tan nc Vy co hụn hp co 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:2xmol, Cu: ymol (x+ y) mol 642xmol 48 4tmol 48 } x+ ymol tymol 0,5xmol ymol 64xmol 48 64ymol 48 78 } Fe + AgNO3 ,Cu(NO3)2 } 2+ } 2+ NaOH t 14 43 44 4 43 Fe , Cu ,NO3 Fe(OH)3 ,Cu(OH)2 Fe2O3 ,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 co h sau: Theo b gam t y = b m 80(x + y) + 80(t y) = b (3) 80 56 + Thay (x + y) va (t y) vao (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 Ma 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 co: 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 C6H 5NH2(X) C6H 5NH 3Cl(Y ) C 6H 5NH 2(X) Cõu 3: Chn D Co ụng phõn este mch h ng voi CTPT la C3H6O2 la 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 Co trng hp to mui st (II) la : FeCl3, CuSO4, Pb(NO3)2 va H2SO4(c, nong) Cõu 7: Chn B Cõu 8: Chn C Cõu 9: Chn D m mamin BTKL nHCl = namin = muối = 0,059mol M amin = 45(C2H7N) 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 va Al b th ng hoa bi HNO3, H2SO4 c nguụi Cõu 13: Chn A Cõu 14: Chn C BT:e nFe = nH2 = 0,1 mCu = 10 56nFe = 4,4(g) Cõu 15: Chn C Cõu 16: Chn D 0,3mol 0,6mol 0,15mol 0,05mol 678 } } } } } 2+ 2+ + + Zn , Mg + NaNO3,NaHSO4 Zn , Mg ,Na ,NH ,SO4 + N 2O , H + H 2O 43 44 4 43 4 4 4 4 4 44 43 0,3mol 0,6mol hỗn hợ p kim loại dung dịch hỗn hợ p BT:e nNH 4+ = dung dịch A 2nZn2+ + 2nMg2+ 8nN2O 2nH2 hỗn hợ p B BT:N = 0,0625mol nNaNO3 = 2nN2O + nNH 4+ = 0,3625mol BT:H nNaHSO4 = 10nNH 4+ + 10nN2O + 2nH2 = 2,225mol nH 2O = nNaHSO4 4nNH4+ 2nH2 = 0,9375mol BTKL mA = mkim loại + 85nNaNO3 + 120nNaHSO4 mB 18nH2O = 308,1375(g) Cõu 17: Chn B amol bmol bmol 0,1mol 0,175mol 0,05mol 678 } } } } } 48 } 64bmol NaOH,KOH + + Val ,GlyAla + H SO dung dịch Y Val ,Gly ,Ala , Na , K ,SO42 4 22 34 4 4 44 4 4 4 43 amol hỗn hợ p X 0,05mol 30,725(g) muối Trang 116nVal + 74nGly + 88nAla = mmuối 39nK + 23nNa+ 96nSO42 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 SO4 %mVal = 54,58 Cõu 18: Chn A (1) Sai, Fe2+, Cr2+ b oxi hoa bi cỏc chõt oxi hoa mnh nh KMnO4, O2 (2) Sai, hp chõt amino axit cha C, H, O va N cũn hp chõt cacbohidrat khụng cha N (3) ung, HOOC-CH2-CH2-CH(NH2)-COONa (mono natri glutamate) lam qu tớm hoa xanh (4) Sai, cho Ag tỏc dng vi dung dch FeCl2 khụng co phan ng xay (5) ung, cỏc tớnh chõt nh tớnh dn in, dn nhit, do, ỏnh kim ờu electron t gõy (6) ung, phan ng thy phõn este va protein mụi trng kiờm ờu la phan ng mt chiờu Vy co nhn nh ung la (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 Co chõt tỏc dng c vi kim loi Cu iờu kin thng la FeCl 3, HNO3 loang, AgNO3 va dung dch cha (KNO3, H2SO4 loang) Cu + 2FeCl3 CuCl2 + 2FeCl2 Cu + HNO3(loang) Cu(NO3)2 + NO + H2O 3Cu + 2NO3 + 8H+ 3Cu2+ + 2NO + 4H2O Cu + 2AgNO3 Cu(NO3)2 + 2Ag Cõu 21: Chn A Cỏc polime t chỏy hoan toan ờu thu c khớ N2 chng t phõn t polime co cha N Vy cỏc polime co phn cha N la : t olon, t tm, t capron, cao su buna-N Cõu 22: Chn A BT:N nNH4+ = nKNO3 nNO = (x 0,06)mol mmuối mH = 39nK + + 18nNH4+ + 35,5nCl 16nO(trong H) = 39x + 18(x 0,06) + 35,5.0,725 64a 57x 64a = 1,5725 x = 0,0725mol mmuối mH = 26,23 10nNH4+ + 2nO(trong H) + 4nNO + 2nH2 = nHCl 10(x 0,06) + 8a+ 0,28 = 0,725 a = 0,04mol 0,04.3.56 mH = 24.5a+ 232a = 14,08(g) mmuối khan = 40,31(g) %mFe = 100 = 16,67 40,31 Cõu 23: Chn B - Cỏc phan ng xay 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 lam qu tớm hoa xanh B ỳng X va X4 ờu co tớnh lng tớnh C Sai, Phõn t ca X la 162 phõn t ca X3 la 167,5 D Sai, Nhit nong chay ca X1 ln hn X4 Cõu 24: Chn A Trang 0,09mol 0,04 7mol 0,105mol } } to AlCl ,FeCl ,FeCl + H O + H2 - Quỏ trỡnh: Al ,Fe3O4 hỗn hợ p X + HCl d 3 4 4 43 dung dịch sau p BT: H + Ta co: 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 co: a = m KL + 35,5n Cl = 27n Al + 56n Fe + 35,5n HCl = 27,965 (g) Cõu 25: Chn C - Nhn thõy: nX = 0,3 < nNaOH = 0,4 X co cha este ca phenol (A) va este cũn li la (B) (A) : RCOOC H R ' n A + n B = 0,3 n A = 0,1 n H 2O = n A = 0,1 Vi (B) : R 1COOCH = CHR 2n A + n B = 0, n B = 0, n Y = n B = 0, - Khi t chỏy chõt Y no, n chc, mch h (R2CH2CHO: 0,2 mol) luụn thu c n CO = n H 2O 44n CO + 18n H 2O = 24,8 n CO = 0, mol ma C Y = n CO = : Y la CH 3CHO 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 la x va Lys la y ta co CTTQ ca E la: 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 H 2O 2x + 6y = x = 2 n E = 0, 0658 mol + Ta co: 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 co : Gly m Lys n (E) + H 2O 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 n = 2,5x + 7y z H O : z mol H2O Trang BTKL va 75n Gly + 146n Lys = m E + 18n H 2O 75x + 146y = 14,88 + 18z (2) BT: Na x + y = 0,18 n GlyNa + n LysNa = n NaOH a = 1,57 b - Hng t 2.1: Ta co th quy i hụn hp E vờ cỏc axyl va H2O nh sau: C H 3ON, C6 H12ON va H2O sau o giai tng t nh trờn ta cng co th tỡm c kt qua - 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 C2H5O2N(Gly) C2H3ON + H2O va co: tách C6H12O2N2(Lys) C2H3ON + (CH 2)4NH + H 2O C2H3ON :xmol BT:C nCO2 = 2x + 4z O2,t0 + E H2O :ymol mànH2O = nCO2 0,5x 0,5z = y (1) BT:H2 n = ,5x + 4,5z + y (CH ) NH :zmol H O 2 BTKL + Ta co : 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 O (1) 4 44 4 43 0,26 mol {2 HNO Mg, O + - Tom tt quỏ trỡnh: Fe, 14 43 m gam X H 2SO Fe3+ , Mg 2+ ,SO42 + SO (2) 44 4 43 { 104 (g) dd Z BT: e cho (1) va (2) n NH4 NO3 = 0,04 mol 129,4(g) dd Y 2n SO2 3n NO 8n N 2O 0,7 mol = 0, 0375 mol m KL + m NH +4 + m NO3 = 129, (1) BTKL BTDT cho (1) va (2) 2n SO 24 = n NO3 + n NH 4+ (2) m KL + m SO 24 = 104 (1) (2) 62(2n SO 24 + n NH + ) 96n SO 42 = 24, 725 n SO = 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 KL Y SO - Hng t 2: 2n SO2 3n NO 8n N2O BT: e cho (1) va (2) n NH NO3 = = 0, 0375 mol + Gi T la hụn hp mui cha Fe(NO3)2 va Mg(NO3)2 suy ra: mT = m Y m NH NO3 = 126, gam + p dng phng phỏp tng giam lng + bao toan in tớch cho hụn hp T va Z ta co: 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 Co khớ CO2 thoỏt Co kt ta trng C Na2CO3 Co kt ta trng Co khớ CO2 thoỏt Co kt ta trng D NaNO3 Khụng hin tng Cõu 29: Chn B A Sai, Tripeptit hũa tan Cu(OH)2 (phan ng mau biure) to dung dch mau tớm B ỳng, Trong phõn t protein luụn co nguyờn t nit C Sai, Tinh bt va xenluloz khụng phai la ụng phõn ca D Sai, Este la nhng chõt hu c ớt tan nc vỡ phõn t khụng co liờn kt hiro Cõu 30: Chn C Cõu 31: Chn D - Phng trỡnh phan ng: CH 3COOCH + NaOH CH 3COONa + CH 3OH CH 3COOC H + NaOH CH 3COONa + C H 5OH mui va ancol Cõu 32: Chn B - Cỏc kim loi kiờm th thuc nhom IIA gụm cỏc nguyờn t: Be, Mg, Ca, Sr, Ba, Ra Cõu 34: Chn C - Xenluloz khụng co phan ng vi I2 ch tinh bt mi co phan ng to dung dch co mau xanh tớm Cõu 35: Chn B T: C6H12O6 Y: C6H12O6 Thuc th X: C6H5NH2 Z: C6H5OH (glucoz) (fructoz) Nc Br2 Kt ta trng Nht mau Kt ta trng Khụng phan ng dd AgNO3/NH3, to Khụng phan ng Kt ta Ag Khụng phan ng Kt ta Ag dd NaOH Khụng phan ng Khụng phan ng Co phan ng Khụng phan 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 co cung s mol, lng MX = MY n (X) n CO (X) : n CO (Y) = : X : C H 4O n C(X) 2 = = va - Ta co: n H(Y) n C(Y) Y : C H 8O n H O(X) : n H O(Y) = 1: + Co ụng phõn ca X C2H4O2 la: CH3COOH va HCOOCH3 + Co S ụng phõn ca Y C3H8O la: CH3CH2CH2OH; CH3CH(OH)CH3 va CH3OC2H5 Vy s cp (X, Y) tha la: 3.2 = Cõu 38: Chn A Cõu 39: Chn D Hụn hp Na2O va Al2O3: Na2O + H2O 2NaOH + Al2O3 2NaOH 2NaAlO2 + H2O mol mol mol mol - Dung dch sau phan ng cha NaAlO2 la chõt tan tt nc Hụn hp Cu va Fe2(SO4)3: 2Cu + Fe2(SO4)3 2CuSO4 + FeSO4 Trang 10 mol 0,5 mol - Dung dch sau phan ng cha CuSO4; FeSO4 va Fe2(SO4)3 d la cỏc chõt tan tt nc Hụn hp KHSO4 va KHCO3: KHSO4 + KHCO3 K2SO4 + CO2 + H2O mol mol - Sau phan ng thu c K2SO4 tan tt nc nhng khớ CO ớt tan H2O, vy hụn hp trờn khụng hoan toan tan nc Hụn hp BaCl2 va CuSO4: BaCl2 + CuSO4 BaSO4 + CuCl2 mol mol - Sau phan ng thu c BaSO4 kt ta khụng tan nc Hụn hp Fe(NO3)2 va AgNO3: Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag mol mol - Sau phan ng thu c Ag kt ta khụng tan nc Vy co hụn hp co 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:2xmol, Cu: ymol (x+ y) mol 642xmol 48 4tmol 48 } x+ ymol tymol 0,5xmol ymol 64xmol 48 64ymol 48 78 } Fe + AgNO3 ,Cu(NO3)2 } 2+ } 2+ NaOH t 14 43 44 4 43 Fe , Cu ,NO3 Fe(OH)3 ,Cu(OH)2 Fe2O3 ,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 co h sau: Theo b gam t y = b m 80(x + y) + 80(t y) = b (3) 80 56 + Thay (x + y) va (t y) vao (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 Ma 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 co: m Fe(OH) + m Cu(OH) = a 28 40 BT: Fe n Fe = n Fe(OH) = Trang 11 ... dịch A 2nZn2+ + 2nMg2+ 8nN2O 2nH2 hỗn hợ p B BT:N = 0,0 625 mol nNaNO3 = 2nN2O + nNH 4+ = 0,3 625 mol BT:H nNaHSO4 = 10 nNH 4+ + 10 nN2O + 2nH2 = 2, 225 mol nH 2O = nNaHSO4 4nNH4+ 2nH2 = 0,9375mol... ta co CTTQ ca E la: O2 ,t C2x + 6y H3x +12 y + O x + y +1N x + 2y (2x + 6y )CO + (1, 5x + 6y + 1) H 2O 3x + 12 y + n CO2 = n H 2O 2x + 6y = x = 2 n E = 0, 0658 mol + Ta co: 14 ,88 y = 0, 7353 ... cng co th tỡm c kt qua - Hng t 3: Tỏch cht + T (1) , (2) ta tớnh c: x = 0 ,13 16 mol ; y = 0, 0484 mol; z = 0 ,11 41 mol Vy + Ta tách C2H5O2N(Gly) C2H3ON + H2O va co: tách C6H12O2N2(Lys) C2H3ON