Huu co 12 cua BGD tu 2007 den 2019

ChinhChem H� thong Giao d1:1c Hoa HQc Online ChinhChem.com 0978.432.243 Lcringo 2007 - Nam BCD chuyen hinh th&c thi man H6a h9c tir ti! lugn sang trac nghi¢m Nam d6, Chfnh 11 tuoi Chfnh chira chu tam den dieu d6, Chfnh chira djnh hinh dir()'c Slf thay doi d6 dang ft nhieu anh hir&ng den minh L&n Jen, v&i chut dam me nho theo dirimg dgy h9c man h6a, nhieu dem thay minh thieu s6t, di dgy h9c tro nhirng thgt Slf chira hieu het ve h6a h9c, ve vi¢c suot mirM may nam qua m9i ngirM n6i gi v&i ve h6a h9c? V&i Slf phat trien cua internet, tai li¢u bai tg.p man h6a thlfc la kh6ng thieu, nhirng chfnh vi nhieu nen Chfnh cam thay minh bj roi, bj quan minh vao nhfrng bai toan giai nhanh xa rM thlfc te Tren dir&ng di tim lgi chfnh minh, qua nhieu Ian ir&c tfnh nang lir()'ng ban than, Chfnh th6i thuc minh xay d!fng b9 "Tuyen tgp cac cau hoi trac nghi¢m H6a h9c cua BCD tir nam 2007 - 2019" Nhin lgi chijng dir&ng thang xay d!fng ngan hang cau hoi nay, 11.}.c lgi cac de chfnh th&c cua b9 tir cac nguon : De thi tot nghi¢p, de thi cao dang, de thi dgi h9c khoi A,B chinh lgi cha dung font, djnh dgng roi bat tay vao giai chi tiet tirng cau bang Math type va sap xep theo dgng, theo th& ti! tir dl den kh6, l9c nhfrng cau hoi da giam tai D6 la m9t qua trinh thgt nhieu thir thach, dau tranh nhieu chfnh ban than minh Chfnh da xep lgi ngan hang cau hoi theo y chu quan cua minh, chira thgt vira y nhir mong muon vi thM gian hgn che, nhirng ft nhieu va phan nao ly giai dir()'c cau hoi cua ban than ve vi¢c "suot mirM may nam qua m9i ngirM n6i gi v&i ve h6a h9c?" Tir d6 c6 cai nhin tot hD'n de djnh hir&ng 6n tgp cha h9c tro cua minh De hoan tuyen tgp nay, Chfnh xin cam D'n quy thay co xira da chia se nhfrng ban word h6a de cua b9 Xin cam D'n Thay Nguyln Anh Phong vi nhfrng tir h6a h9c hay va xin phep thay cha minh dir()'c sir dl.}.ng tir cua thay m9t so bai giai Xin cam D'n thgt nhieu trang mgng da cha minh tham khao nhieu hD'n hir&ng giai quyet cac bai toan kh6, de minh hoan thi¢n ban than hD'n tirng I Va t6i biet, t6i phai n6i IM cam D'n Cho t6i song, nhfrng thang nhfrng rat xanh Chgm Jen trai tim thay CO'n mO' chay nong Nhieu dem trang xoa bay , long nhir c6 gi6 dau mu.a -DEN vAuBMT, 00hSp 10/12/2019 Quythay co muon chuyen giao ban word quyen sach (Banta.ch bai giai + boi dap an, ban bai giai du&i cau h6i, banthotheotrinhtl! cac nam) detien bien soan lai c6 the lien he v&i em qua cac dia chi sau a: SoDT: • 0978.432.243 Gmail: • trungchinhbolero@gmail.com FB ca nhan: • Trung Chinh Phan (ChinhChem) Fanpage: • H6a hoc ChinhChem H1;mh phuc cua ngU'ai thay la c1U'9'C song cu9c c1ai cua h9c sinh ! Trang MUCLUC ■ "0 :c { H ,O Lo nco2 ➔ t = ( amino axit no, don chuc); n 00_ = 0,7mol N Ta c6:nNH2 + n 0w = n Hci ➔ n Hci = 0,2.1+O,7 = 0,9mol BTKL:mMu6i = mamino axit +mNaOH +mKOH +mHCI -mH20 ➔ 0,2.Mx+0,3.40+0,4.56+0,9.36,5-0,7.18 = 75,25 ➔ Mx = 103 12,36 ➔ H NC H6 COOH; nx(i ,36 g) = = 0,12mol ➔ n H20 = 0,12.4,5 = 0,54mol � Cau 64 g9i nH2 N-C n H2 n -COOH = a Dap an D; Ta c6 : nH+ = nOw Cau 65 Cau 76 Cau 77 Cau 78 Cau 79 Cau 80 Cau 81 Cau 82 a.1+0,02.2 +0,04 = 0,09 ➔ a = 0,0 lmol BTKL : m Muoi = ffiamino axit +ffiHCJ +ffiNaOH + ill KOH - ffiH2 O I Cau 75 ➔ mol ; nOH_ =0,04+0,05=0,09 mol ➔ 0,01.M y+0,02.147 +0,04.36,5+0,04.40+0,05.56-0,09.18 = 8,21 ➔ 66C I I 67A 68C I 698 I 70D I Dap an C; C H 03 N2 = C n H 2n+4 03 N2 ➔C H 5N H N03 718 I 728 I 738 I M y= 103 I 74C C H 5NH 3N03 +NaOH➔C H 5NH +NaN0 +H 0➔Y:C H 5NH ➔Mv = 45 Dap an C; Dap an D; BT Na H >CH 3COONa:0,2➔mMu6i =1 6,4 gam NCH COONa:0,15 CH ,ffiOC, H , :a , , Dap an B; { N H CH COOC H :b ,- CIH , NCH , COOH : 0, 15 ua" • Muoi _ 25,5 gam ➔mMum { NaCl:0,15 He KL]:88a + 03b � 9, � 0, ➔{" f [NaOH]:a +b = 0,2 b = 0,1 , CH ,ffiONa: 0, ➔Muoi { ➔m mum =17,9 gam N H 2CH 2COONa:0,1 Dap an D; n C3 H702 N = 0,1 mol ; n N,OH =0,15 mol ;➔n NaOH du = 0,05 mol ➔0,1.RCOONa+0,05.40 = 1,7➔ R = 30( H 2N-CH -) ➔X:H 2N-CH -COOCH Dap an A ; m2 -m l =7,5➔22 -36,5 =7,5➔2 nh6m COOH va nh6m NH ➔C n H 2n -1 04N➔C H 904N , ' 82 = D'ap an D; nc3 H = -0,02 91 64 RCOONa = ' = 82 ( CH COONa) 0' 02 o N xc3 H o 2N >X:CH 3COONH 3CH 3: metylamoni axetat Dap an A; _!±_ = 0,1573➔M A= ➔ H 2N-CH -COOCH 3; MA n cH o H =n HcH o Il Ag = - = 003 mol ➔0,03 mol H N-CH -COOCH 3➔m =2 ,67g H1;mh phuc cua ngU'ai thay la c1U'9'C song cu9c c1ai cua h9c sinh ! Trang 128 Dap an B ; Ile co2 0,15 mol; n N 0, 025 mol ➔n N 0, 05 mol; n H 20 0,175➔n H 0,35 mol ➔CTDGN : C 3H7 O, N = Cau 83 = = = = T.;i,o H2 N-CH2 COONa ➔H2 N-CH2 -COOCH3 Dap Cau 84 = { an B ; nx = R I NH2 > 29 R-COONa+Br2 0,1 mol; X+NaOH➔Khi➔X: R-COONH3 -R ➔X :CH2=CH-COONH3 -CH3➔Muoi: ffi CH 2=eH -COONa 0,1.94 9, 4g , , 40,449 7,8 65 -Dap an C; ne :n H :n :n N = Cau 86 12 ➔nx , 45 = Dap an B; 25,6 gam { X{ 35,956 - - 0,05 mol; RCOONa= 89 ➔ X: H2 N-CH -COOCH3 = = CH-COONa = = Cau 85 H2 C 15, 73 : ➔CTPT:C 3H7 O2 N 16 4,85 - -➔R � 05 14 = 30 (H2 N-CH2-) y:NH4 OOC-COONH4 : a mol Z :H2 NCH2 CONHCH2COOH :b mol 124a+132b w ' ){ 2a 0, = Y:NH4 OOC-COONH4 : 0,1 mol = a 0,1 25, ➔{ b 0,1 0,2 mol ClH3 NCH2COOH Hei ➔m hiruca _ HCa:{ -�; 0,1 molHOOC-COOH Z:H2 NCH2 CONHCH2COOH :0,1 mol = = = 31' Cau 87 Cau 88 Dap C Cau 89 an B; = 2H7 O2 N ➔{ a molHCOONH3 CH3 a+b 0, 0H20+3O2 N➔{ b mol CH COONH ➔{31a+1 7b = C = ,._, 0,15 molHCOONa ➔Muoi { ➔m M6 u , 05 mol CH3 COONa b 0, a = 0,15 = = 0, 13, 75 14 , 3g HCOONH3 CH3 + NaOH➔HCOONa+CH NH2 +H2O , ,1 molHCOONa Ran { ➔m = 10,8g ,1 mol NaOH du Da p an B; C6 truang h9p, truang hqp dung la: H4 OOCCH2COONH4 N Cau 90 E { C 2H5 NH3 ) H4 N Muoi CO :b : a He · ; 138a+124 b { 2a+2 b = = 0,04 , 62 ➔{ a = , 01 b , 01 = , , 0, 03 NH3 (ch9n v1 Kh1 { 0, l C 2H5 NH2 NaOOCCH COONa :0, 01 ➔ffi Muoi - 2, gam ,, { Na CO : 0, 01 H1;mh phuc cua ngU'ai thay la c1U'9'C song cu9c c1ai cua h9c sinh ! Trang 129 X: C2H5H3NOOCR COONH 3C2H5 : 7a , , Dap anD; { ➔ C2H5NH2 : 7a+3 a=0,1 ➔ a=0, 01 Y: H2NR 2COONH3C2H5 :3 a Cau 91 ➔15, 09 gammu6i { NaOOCR H2NR COONa : 0, 07 2COONa : 0, 03 ➔ 7R +3R =3 22- ,{R1 Bien Juan - - R : 28 : 42 X: C2H5H3NOOCC2H4COONH3C2H5 : 0, 07 ➔%X=76,63 % ,._, , cung E{ Y: H2NC3H6COONH3C2H5 : 0, 03 C u01 Dap anB; 0, l molE{ Cau 92 + ,3 gammu6i { X: C n H2n+404N2 : a Y: C m H2m_4O N6 :6 NaOOCR a=0, 07 [mo l ]: a+6=0, l ➔{ { 6=0, 03 [ NaOH ]: 2a+ 66=0,3 ICOONa : 0, 07 + l 8R _ ➔ 7R H2NR 2COONa 0, 03 - 0,l8 CH3OOCC 2H4COOCH3 : 0, 07 + NenE { (Ala)6 : 0, 03 A He Bi�n lu�n , =700-��- { R R I =28 =28 ➔ %X=48,61 % X: C H H NOOCR COONH C H : a , , Dap anD; { ➔ C H NH : 2.3 a+5 a=0, 22 ➔ a=0, 02 Y: H NR COONH C H : a Cau 93 ➔ 21,66 gammu6i { NaOOCR H2NR ICOONa : 0, 2COONa : 0,1 ➔3R +5R = 2662 Bi�n lu�n , R - -{ , R I : 42 : 28 X: C2H5H3NOOCC3 H6COONH3C2H5 : 0, 06 ➔ %X=49,85 % ,._, , cung E{ Y: H2NC2H4COONH3C2H5 : 0,1 C u01 X: C n H 2n+4O4N2 : a Dap anA; , 26 molE{ Y: C m H2m_3O6N5 :6 Cau 94 + 2,9 gammu6i { _ ➔ 2R H2NR 2COONa 0, 06.5 - 0,3 C2H5OOCC 2H4COOC2H5 : 0, D'ap a'n A; aa : C n H2n+l O2N Cau 95 a=0, [mol]: a+6=0, 26 ➔{ { 6=0, 06 [ NaOH ]: 2a+ 66=0, NaOOCR COONa : 0, + NenE { (Ala \: 0, 06 A He' tripeptit +3R Bien Juan ', R =1 2-��{ R =28 =28 ➔%X=63, 42% > C3n H6n-l O4N3 0,3n mol CO2 D�t 0, mol c3n H6n-P4N3 ➔ { ( 0,3n- 0, 05) mol H20 ➔ 44 0,3n+ 18.( 0,3n- 0, 05)=5 4,9 ➔n=3 D ipeptit ➔ aa : C H O N > C6 H O3 N2; 0,2 mol C H O N2 ➔ 1, mol CO2 ➔ fficaco, =1 20 g Cau 96 Cau 97 _ 63,6 -60 = 0,2 mol; n =n 20 =0, mol;n C =2n Da,p a, n C., nH O dipeptit 0, mol H I di peptit H 18 dung _!_ X: ➔mmum, =_! (63,6+0, 4.36,5)=7,8 g 10 10 Dap an C ; Neu { a molAla-Gly-Ala-Val-Gly-Val molGly-Ala-Gly-Glu ➔{ a+ 26=0, (molGly) 2a+6=0,32 ( molAla) ➔{ a= 0, 12 6=0, 08 ➔m=472 a+3326=83,2 g H1;mh phuc cua ngU'ai thay la c1U'9'C song cu9c c1ai cua h9c sinh ! Trang 130 Dap an A; Thu duqc Gly; Ala ➔E { Cau 98 He Tri: C H y O4 N3 : a X Penta: C0H mO6 N : b Ja pghep a=0 ,03 - a+b=0 ,0 { ➔{ b=0 ,0 , 3a+5b=0 ,0 7+0 ,1 ➔ X:GlyAla { Y :Gly 2Ala mol Gly 2Ala +4H 2O+5HC1 ➔Mu6i; BTKL:mMuc,i =59,95 gam Dap an C; A+ B+ 3C gop chu&i � A -B -C-C-C+ 4H O ,1 M Cau 99 n Ala : n Val =16: ➔ [S6 MX] M = (16+ 7)k; T6ngLK Mu i {GlyNa: 0,1 ➔m =20 gam Muot ' AlaNa: 0,1 Cau Dap an B; 0,1 mol Gly-Ala Cau Dap an A; a mol Gly-Ala Cau ➔ ffi G!y-Ala =0,01 46=1 ,46 gam NH3:0 ,4 BTOXI:0,4 l , +3ncH,= - 2 ,625 , CH :1 ,55 at X D,a p ,an B; D"o n chA' BTCacbon ) COO:O ' 5=n NaOH ➔mNaOH =20 gam 101 102 103 Cau 104 ! Dap an A ; { Na oH ctll � , Gly-K:a 2,4 gam ffiUOi { ➔a=0,01 Ala-K:a a mol X4 +4NaOH ➔Mu6i + lH O 2a mol Y + 3NaOH ➔Mu6i + H 2O n Na oH =4a +6a=0 ,6 mol ➔a=0 ,0 mol ➔BTKL: mpepti t = 72,48+1 8(a + 2a)-0 ,6.40=51 , 72g Cau 105 Cau 106 4,06 nx =n =0 ,02 mol ➔Mx = =203=2Y +Z-1 8.2 ➔Y =75 ( Gly) ,0 89 , an , B; aa: C H O N ➔{ X: tripeptit: C 30H60_ O4 N3 Dap n 2n +I Y: tetrapept1t: C 40H 80_2 O5 N4 ,0 mol C 40H 80_ 2O N4 ➔ 44.0 ,0 5.4n + 8.0 ,0 5.(4n- ) = 6,3 ➔ n=3 D6t chay: ,01 mol X: C H l7 O4 N3 ➔ nco2 s inh = ,0 mol ➔mBaco, =0 ,0 9.1 97=1 7,33 g H1;mh phuc cua ngU'ai thay la c1U'9'C song cu9c c1ai cua h9c sinh ! Trang 131 Dap an A; a mol C nH2n-l ON Quy d61: { + a mol NaOH 0,16 mol H2O ? ➔a=0,6.1,5 =0,9 mol Cau 107 30,73gE { x mol C O H n , ON - y mol H2 O xn=1,16 14xn+29x+18y =30,73 X =0,45 ➔ 44xn+18(xn-0,5x+y)=69 ,31➔ y =0,08 0,9 a X 116 -= =-n=y 0,16 0,16 45 2u+3v 2u+3v 116 = = n= ; { u+v 0,9 45 +Quy ve 0,16 molE; { v mol H2N -C 2H4 -COONa u+v =0,9 u mol H2N-CH2 -COONa ➔{ u=0,38 u ➔-=0,73 V=0,52 V Dap an A ; G9i so matxich X ; Y tuong @g la m , n X Ta c6: T{ y: m+l+n+1 =13 ⇒ {m-1;>4 n-1 ::::: m+n=ll ⇒ {m;>5 ⇒ n:::::5 m=5 {n� x = 0, ,{x mol X {5x +6 y =3, (nN,ott) Co ➔ ➔{ ; y = 0,3 y molY6 x + y = 0, Cau 108 , +Lal co : Quy d6i : ? n= 0, n� ong X = 0,3.n�ong y { X,Y d�u chua Gly ( 2C); Ala(3C) ⇒ ex Cy 0,3 12 0, 4 16 2.5