Nối tiếp nội dung phần 1 tài liệu Cẩm nang ôn luyện thi Đại học - Rèn luyện giải nhanh các đề thi ba miền Bắc - Trung - Nam Hóa học, phần 2 giới thiệu tới người đọc 13 bộ đề thi thử Đại học và hướng dẫn giải chi tiết từ các trường THPT chuyên trên cả nước. Mời các bạn tham khảo.
dm nang On luyQn thi BH mign B&c - Trung - Nam mOn H6a hpc - Cti Thanh Toan Cty TNHH MTV DVVH Khang Vi^t PTHH cac phan ting: HCOOH + NaHCOj HCOONa + CO2 t +H2O C H C O O H + NaHC03 HCOOH + Ag20 " rftuj.-' Cac PTHH cac phan umg xay ra: HCOOCH(CH3)2 + NaOH HCOONa + (CH3)2CHOH ' (X) CH^COONa + CO2 T +H2O ^"^ > 2Ag i +CO2 t +H2O (CH3 )2 CHOH + CuO "^^'^^^ ' ' ' Dap an dung la B Cau 25: Theo bai ra: n^^^^o^(0^^2)3 = 297/297 - (kmol) C6H702(OH)3+3HN03^C6Hv02(ON02)3+3H20 ^ Vi H = 80% nen: m t n , , , H „ i , r i c 63% C^HjCOONa + CH3CHO Z, + AgN03 /NH3 -> Khdng xay ' Dap an diing la D (2) CH2 = C H C O O C H + NaOH -> CH2 = CHCOONa + CH3OH EsteX (C4H8O2) + NaOH - + Dan chiJc + Tac dung duoc vdi dung dich NaOH + Khdng CO phan ling trang guong 1) C H C H C O O H ; Dap an diing la C Cau 28: So do: V- => X la axit cacboxylic hoac este (kh6ng phai este ciia axit fomic) Cac CTCr ciia X: C6H7O2 (OH)^ ;C6H702 (ONO2)3 (bo he s6' n) ddH2S04l i ' >m 100 162/80-202,5 (kg) ^CS2.NaOH ^ ^ ^ ^ ^ •I tham gia phan iJng trang guong vi khong nhom OH hemiaxetal ndn khdng CO kha nang ma vong tao nhom chiic andehit Vay phat bieu A khong chinh xac Dap an diing la A f-^;, qy,,: Cau 30: Hop cMt X c6 cac tfnh ch^: ,^ , ^^ ^ ' : Dap an diing la D Chu y; De dan gian ta vie't c6ng thiic cua xenluloza, xenlulozo trinitrat Mn la Xenluloza (Z,) V C^u 29: San ph^m cua phan ung giua mantoza va CH30H(xt HCl) khong the = (3 63 100 100)/(63.80) = 375 (kg) (1) CH3COOC2H5 + NaOH ' • '' , , " (Z) )CH3 - CO - CH3 + Cu + H2O (Z) CH3CHO + A g — ^ ^ ^ ^ A g i + C H C O O H (Y) •4% dm Cty TMHH MTV DWH Khang Vi^t nang fln luy$n thi DH mjgn B^c - Trung - Nam mfln H6a hgc - Cii Thanh Toan BO D £ THI SO M N HOA HOC K H O I A, B (Trich de thi thu Idn 2- Tru&ng THPT Chuyen Le Hong Phong ) Cau 1: Thirc hien phan urng ete hoa h6n hop g6m ancol don chiifc thu dirge h6ii horp cac ete La'y mOt s6' cac ete dem dO't chay hoan toan thu duoc CO, va HjO CO so' mol bang va deu ga'p ISn s6' mol ete da dot chay C I FT cu;, ancol la C^O 9: ^'^ phenol trung ngung v6\5 mol HCHO (xiic tac H ^ t°), thi thu duge bao nhieu gam poli(phenol-fomandehit) co ca'u true maeh khOng pha nhanh? Bid't h\tu sua't phan ling la 100% A 15,9 gam B 21,2 gam C 26,5 gam D 10,6 gam 10: X la a-amino axit chiia nhom -COOH Cho 10,3 gam X tac dung het vol 200ml dung djch HCl 1,5M thu duge dung djch Y De phan ung het vdi cae chat Y thi cSn dung vCra dij 400 ml dung djch NaOH I M CTCT diing cua X la: A C H O H va C H ^ O H B C H O H va CjH^OH A C H - C H - C H ( N H ) - COOH C C H O H va C H O H D C H S O H va C H O H B C H - C H ( N H ) - C H - COOH Cau 2: Mot hop chat huii co X c c6ng thvJc phan tu CHjO.N La'y 0,2 mol X tac dung vd'i dung djch co chtra 0,25 mol NaOH va dun nong, thu dugc chat lam xanh quy tim dm va dung dich Y Co can dung djch Y thu dirge m gam chat rin khan Tri so' cua m la: A 13,6 B 15,6 C 14,6 D 19,2 Cau 3: Cho cac polime: cao su buna; polistiren; amilozo; amilojjectin; xenlulozo; to capron; nhira bakelit So' polime co ca'u true maeh khOng phan nhanh la: A , B.4 C D Cau 4: Oxi hoa gam mot ancol don chuc X bang O2 (co mat xiic tac Cu) thu dugt 5,6 gam h6n hop Y gom andehit, ancol va nu6c N6'u cho toan b6 h6n hgp Y tac dung vdi lugng du dung dich AgNO^/NH,, dun nong thi khd'i lugng Ag thu duge la: A 10,8 gam B 21,6 gam C 32,4 gam D 43,2 gam Cau 5: Hay xac djnh co bao nhieu loai to co nguon goc tir xenlulozo so cac loai to sau: to tam; to viseo; to axetat; to nilon-6,6 A B.4 C D Cau 6: S6' dong phan amin bae ciia C4H,|N la A B C.9 D Cau 7: Thuy phan hoan toan h6n hgp X g6m este don chiJc A, B cdn dung vira du 100 ml dung dich NaOH I M thu duge 6,8 gam muO'i nha't va 4,04 gam h6n hgp ancol la d6ng dang lien tia'p C n g thiic ca'u tao ciia este la: A C H , - C O O C H B C H - C O O C H va C H va -COOC2H5 CH3-COOC2H, C H - C O O C H , va H - C O O C H D H - C O O C H va H-COOC2H5 Cau 8: Etanol la chat trung gian de san xua't cao su nhan tao, to t6ng hgp, la chfl' thay the cho cac nhien lieu co nguon gde hoa thach Co th^ di^u chd'etanol baiit^ each lan men cac nguydn li6u eo ehufa tinh b6t Ti'nh lugng ngu cd'c chiia 65'/' tinh bgt de san xua't duge 2,3 ta'n etanol Biet hao hut qua trinh san xua't 25% A 8,3 ta'n '*B 1,013 ta'n C 5,4 ta'n D 1,56 X&n ' C CH3-CH(CH3)-CH(NH2)-COOH " • D C H - C H ( N H ) - COOH 'Hm^m ! cau 11: DCing thudc thir nha't nao sau day &i phan biat cac dung djch: mantozo, C H O H , C2H5(OH)3,CH3COOH,HCOOCH3 va long trang triing? A Phenolphtalein B Na C Cu(0H)2 low D Quy tim ,^ ^ ^ , Cau 12: Phat bieu khong diing la: A Hgp chat H N - C H - C O O H N - C H la este cua glyxin ^ »(>i B Amino axit la nhOng chat rdn, ket tinh, tan t6't nu6c va co vi ngot C Trong dung djch, H N - C H - COOH t6n tai dang ion ludng cue H3N^-CH2-COO" D Amino axit la hgp cha't huu co tap chiie, phan tiJr ehiia d6ng thori nhom amino va nhom cacboxyl Cau 13: Chat hiru co X don chu-c co c6ng thiic phan tir la C H O X tac dung vdi NaOH dun nong thu duge mud'i Y co phan tir kh6'i nho hon phan tu kh6'i ciia X Ten ggi diing ciia X la: A Axit propionic B Metyl fomat C Metyl axetat D Etyl fomat Cau 14: Thuy phan hoan toan m6t pentapeptit X maeh ho thu duge cac amino axit A, B, C, D, E Khi thuy phan khong hoan toan peptit X thi thu duge cac peptit ED, AE, CB va EDC Peptit X phai co thii tu cac amino axit phSn la: A CBAED B AEDCB C EDCAB D CDEAB 15: Chon phat bieu khong chinh xac: , A Tinh b6t bj thuy phan hoan toan m6i tru6ng axit va dun nong thi thu duge san ph^m cu6'i cung la glucozo i , B- Depolime hoa polistiren didu kien thi'ch hgp thu duge stiren C Thiiy phan poli(vinyl axetat) m6i trucmg ki6m san phdm thu duge la poli(vinyl ancol) va muoi axetat f > •P'S i * C>- Xenlulozo bj thiiy phan hoan toan mdi trucmg axit va dun nong thi thu duge san phdm cu6'i cung la fruetozo C^m nang On luyen thi DH mign Bic - Trung - Nam mOn H6a hqc - Cu Thanh Toan Cau 16: Phan ting nao gSy sir phan cat mach polime? A Cao su isopren tac dung vdi HCl B Poll (vinyl clorua) tac dung v6i Cl-, C Thuy phan to capron D Thiiy phan poli(vinyl axetat) Cau 17: Cho 100ml dung dich ancol etylic 57,5° (bid't khO'i lugng ridng ciia ancol nguySn chat la 0,8 g/ml va cua nu6c la 1,0 g/ml) tac dung v6i Na dir, sau phan ling thu duoc V lit (dktc) Gia trj ciia V la: A 25,432 lit B 37,632 lit C 11,200 lit D 75,264 lit Cau 18: Cho cac hoa cMt sau: (1) AgNOj/ddNH,, t°; (2) H , (Ni, t*^; (3) phenol ( H \ (4) CuCOH):/ NaOH, t"; (5) Na; (6) dung dich Br,; (7) CH3COOH Hay cho biet andehit fomic tac dung vdi chat nao so' cac hoa chat tren? A (1); (2); (4); (5); (6) B (1); (2); (3); (4); (5); (6) C (1); (2); (3); (4); (5); (6); (7) D ( I ) ; (2); (3); (4); (6) Cau 19: M6t a-amino axit X (chi chiia m6t nhom -NH, va mot nhom -COOH) Cho 0,89 gam X phan urng vCra du v6i HCl tao 1,255 gam muO'i C6ng thiic ca'u tao ciia X la: A H2N - CH2 - CH2 - CH2 - COOH B H2N - CH2 - COOH C H2N - CH2 - CH2 - COOH D CH2 - CH(NH2) - COOH Cau 20: Cho cac dung dich khOng mau: HCOOH, H^N-CHrCOOH (glyxin), HOOC-CH:-CH2-CH(NH.)COOH (axit glutamic), H2N-(CH2)4-CH(NH2)-COOH (lysin) ThuO'c thir c6 th^ phan biet cac chat trdn bao g6m: A Quy tim, dung dich NaOH B Quy tim, dung djch AgNOj/NHj du C Quy tim, dung djch CUSO4 D Dung dich HCl, dung djch Ba(OH)2 Cau 21: H6n hop X gom axit axetic va ancol etylic Cho 14,52 gam hOn hop X tac dung vori Na du thu diroc 3,024 lit H , (dktc) Tie'p tuc cho H.SO4 dac vao h6n hop X va dun nong thi thu duoc 8,8 gam este Hidu sua't ciia phan umg este hoa la: A 83,33% B 75,00% C 66,67% D 80,00% Cau 22: De dot chay hoan toan 0,1 mol mOt axit cacboxylic don cMc thi c5n vira dii V lit oxi (dktc), thu duoc 0,3 mol CO, va 0,2 mol nude Gia trj ciia V la: A 0,86 B 11,2 C.4,48 D 6,72 Cau 23: H6n hop X gom axit axetic v^ etyl axetat (ti le mol 1:1) D6't chay hoan toan 0,1 mol X, sau cho toan bo san ph^m chay vao binh dung 500ml dung djch Ba(OH), 0,4M thu duoc ket tua Y va dung djch Z Khi khO'i luong dung djch Z so v6i khoi luong dung djch Ba(0H)2 se thay doi nhu the nao? A Tang 13,2 gam B Tang 25,1 gam ' C Giam 1,1 gam D Giam 10,95 gam Cau 24: H6n hop g6m 0,02 mol etanol va 0,02 mol ancol no X tac dung he't v6i N^' sau phan utig thu duoc 0,672 lit or dktc va 3,76 gam muO'i natri X la: A propanl,2-diol B etylen glicol C metanol D glixerol Cau 25: Chat huu co X (chura C, H, O) don chiic, mach ho va c6 phan tir kh6'i la 60 SO'CTCT ciia X thoa man dieu kien tren la: A B C.4 D.5 ^ Cty TNHH MTV DWH Khang Vi$t ^^U 26: Di dieu che' 100 gam thuy tinh hOu co thi cSn bao nhifiu gam ancol metylic va bao nhieu gam axit metacrylic, biet hieu sua't qua trinh phan utig dat 80% A Axit 68,8 gam; ancol 25,6 gam B Axit 107,5 gam; ancol 40 gam C Axit 107,5 gam; ancol 32 gam D Axit 86,0 gam; ancol 32 gam Q^a 27: H6n hop X g6m 0,25 mol hai andehit d6ng dang ke' tie'p tac dung vdi luong du AgNOV ddNH,, dun nong Sau phan ihig xay hoan toan, thu duoc 67,5 gam Ag PMn tram so' mol ciia hai andehit h6n hop X la: A.50%va50% B 25%va75% C 35% va 65% D 30% va 70% cau 28: Day gdm cac chat d^u co the' tao true tie'p duoc ancol etylic la: A CH3CH2Cl,CH3COOC2H„C6H,206 (glucozo), C2H4 B CH3CH2Cl,CH,COOC2H„C6H,206 (glucozo), C2H2 4,mm t>h r C CH3CH2CI,CH3COOC2H5,C6H,206 (glucozo), C2H6 ' D CH3CH2Cl,CH3COOC2H5,C6H,206 (glucozo), C H f cau 29: Dieu kien de' mot monome tham gia phan ting trung hop la: A Co lien ket boi • fHOrJiW: B Co lien ket bOi hoac vong kem b^n C Co lien ket d6i, ba hoac vong 3, canh D Co mot lien ket doi hoac vong kem b^n ' ' ' ' ' J f I- cau 30: Thiiy phan hoan toan 10 gam mOt loai bOng thien nhien dung djch H,S04 loang va dun nong Toan b6 luong glucozo tao dem thirc hien phan ihig trang guong thi thu duoc 12,96 gam Ag Ham luong xenlulozo co loai bdng thien nhien la: A 97,2% B.95,4% C 93,6% D 98,1% HUONG DAN GIAI Cau 1: Vi cac ancol don chiic => ete tao don chirc E>6't chay ete tao so mol CO, bang s6' mol H.O ' => Ete dang C„H,„0 (don cMc) C n H „ - ^ 2 _ , n C + nH20 l(mol) ' n(mol) Theo b^i z:> n = (C4H8O;CH3 - O - CH2 - CH = CH2 ) Vay ancol co th^ \k CH3OH va C3H5OH E>ap an dung la B * ^ , 2: X CO cac tinh chA: CTPT: CH5O2N , ' X + NaOH - ) khf lam xanh quy ^m + dd Y ==> X phai c6 ca'u tao: HCOONH4 ^'"^ ' Ca'm nang 6n luygn thi DH mign B^c - Trung - Nam mOn H6a hgc - Cii Thanh Toan w PTHH: HCOONH4 + NaOH Ban ddu: 0,2 Phaniing: 0,2 0,2 Con: 0,05 Cty TNHH MTV DWH Khang HCOONa + N H , t + H O RCOOR + N a O H 0,25 -> 0,1 0,2 ''J'' 0,2 "''Chat ran gom: H C O O N a ( , m o l ) ; N a O H ( , m o l ) '*' ' " ' V a y m =0,2.68 + 0,05.40 = 15,6(gam) il,;'' Dap an dung la B Cau 3: - Cac ,a'v¥l ; polime mach kh6ng phan nhanh: Cao su buna; polistiren; amilozo; Polime mach phan nhanh: amilopectin Polime mach khdng gian: nhira bakelit Dap an diing la D Cau 4: Dat ancol X la R C H ^ O H = , l = ^ C H O H va C H , O H Vay este la H C O O C H > R - CHO + H O Theo PTPIT: n„,d,hi, = 2.no = 0,1 ( m o l ) ; '31 Jig n„„„| (ban d^u) > 0,1 (mol) 92 ta'n ^ 2,3.162 - = 4,05 (ta'n) 'm = 92 = my n„,„„, (phan ling) = n,„d,hi, = 0,1 (mol) m„ = 4,05.100.100 65.75 Dap an dung la A Chui: (C6H,oO,)_^ 2,3 ta'n i = 8,3 (ta'n) > C H H + 2C02 Cau 9: So phan irng: OH >4Ag + 0,1 -> 0,4 =>mAg =0,4.108 = 43,2(gam) Dap an dung la D Cau 5: Cac to c6 nguon g6c tir xenluloza: to visco, to axetat (2 to) Dap an dung la A Cau 6: C T C T cac amin bac CO CTPT C4H,|N la: 1) C H - N H - C H - C H - C H CH3-NH-CH(CH3)2 - NH - +nH20-.nC6H,20 C6H,20, => andehit la H C H O - CH2 + 2CO2 Vi ngu c6'c chi chiia 65% tinh b6t va H = 75% nen: => M,n^„, < / 0,1 = 40 => ancol la CH3OH 3) CH3 !V c6 so do: 162 tan m =:>4 + mo^ =5,6=>mQ2 = l , g = > n Q = , ( m o l ) 2) HCOOC2H3 Dap an diing la D cau 8: Ta rlO^!'' Theo dinh luat bao toan khoi luong, ta c6: m ^ + HCHO -> O , l ( m o l ) CeH.oO., — ^ ^ ^ C H , ^ = ^ C H , O H - ^ 0,1 Suyra: M^cooNa = , / , = ( H C O O N a ) xenlulozo tcfcapron (5 chat) RCH O H + - O , RCOONa + ROH CH2 - Ban d^u: 0,3 Phan ling: Sau P/ir: 0,25 0,25 0,25 0,05 Vay kh6'i luomg polime thu duoc: m = 0,25.106 = 26,5(gam) Dap an diing la C Cau 10: Tlieobai ra: n^^^^ = 0,3(mol);nN^oH = , ( m o l ) "x CH3 Dap an diing la D Cau 7: Theo bai ra: np^^oH ~ 0,1 ( m o l ) Dat c6ng thiic este la RCOOR CH2O =nNaOH - I H C I = , - , = 0,1 (mol) = > M x =10,3/0,1 = 103 Vay CTCT c u a X : CH,-CH.-CH(NH2)-C00H (a -aminoaxit) £5ap an diing la A Cin\g On luy^n thi DH mign B^c - Trung - Nam mOn H6a hpc - Cii Thanh ToAn Cty TNHH MTV DWH Khang Vi$t Cku 11: Chon thud'c thix la CuCOH)^ / O H " A E D C B + H2O A E D C B + -> A E+ 2H2O D C B A + E D C + B Pap an diing la B CaU 15: Xet cac phuang an: Cu(OH),/OH- A Dung: (C^H.oOs)^^ + n H - H+,tO tinh hot kh6ng ht dd xanh lam dd t i m CH,COOH CHsOH; HCOOCH, Mantozo; C,H,(OH)3 Cu(OH)2/OH-,t'' glucoza • V OH'./i; ! I dd xanh nhat B Dung: [-CH2 Long t r l n g trirng tiC>^B:,.•I - Oh )'!• ^nC6H,206 - C H ( C H , ) - polistiren - ^ n C H - C H = CH2 stiren C Dung: Cu(OH),/OH -CH2 - C H ( O O C C H , ) + nNaOH - > [-CH2 poli (vinyl axetat) - C H ( O H ) -]\a poli (vinyl ancol) + D Khdng chinh xac: (CgHioOj)^^ + n H j O CO i khong i ihdng i C,H5(OH)3 C,H,OH Mantozof CO xenluloza Dap an dung la D HCOOCH, A Khdng dung vi hop cha^t H N - C H -COOH3N-CH3 la mu6'i cua glyxin / • ; L ; cao su isopren vdi metylamin B [ - C H - C H C l -I B Diing: amino axit la mu6'i n6i phan ttr D Dung: (H2N),R(COOH)b polime + nCl2 [-CHC1 - C H C l poli (vinyl clorua) C Dung: H ^ N - C H , - C O O H ^ H3N+ - C H , - C O O " , + nHCl 5« ;, polime C [ - H N - ( C H ) , - CO ' I + n H ^ n H N - ( C H )^ - C O O H ' hap cha't tap chirc tacapron Dap an dung la A monome D [ - C H - C H ( 0 C C H ) - ] „ + nNaOH - > [ - C H c a u 13: Sa 66: X (CjHeO,, don chirc) + N a O H - > mud'i Y + poli (vinyl axetat) => X la axit hoSc este RCOONa + H2O ^RCOONa > ^ + nCH3C00Na - C H ( O H ) poli (vinyl ancol) Dap an diing la C V i M Y < M x => X kh6ng th^ la axit => X la este V I M y < , A [ - C H - C ( C H ) = C H - C H ] _ ^ +nHa ^ [ - C H - C a ( C H ) - C H - C H - C a u 12: Xet cac phuang an: j Cau 17: Trong 100 m l dung djch c6: + 57,5 m l C H , O H ^ + 100 - 57,5 = 41,5 m l H O mc^HjOH = 57,5.0,8 = 46(g) nc^HjOH = / = l ( m o l ) RCOOH ) => g6c hidrocacbon cua ancol (trong X ) 16n han 23 => g6'c hidrocacbon cua ancol la C H - ( ) PTHH: mH20 2C2H3OH + 2Na ^ Vay este la HCOOC2H, (etyl fomat) = 42,5.1 = 42,5(g) => n H = 2,36(mol) C H , O N a + H2 T -> 0,5 (mol) i 2H2O + 2Na - > N a O H + Hj t Dap an dung la D C a u 14: Peptit X la A E D C B That vay: A E D C B + 4H2O ^ glucoza Cau 16: Xet cac phuang an: Dap an dung la C (RCOOH + NaOH •>nC6H,206 ^g) 4- A+E+D+C+B A E D C B + 2H2O ^ A + E D + CB 2,36 l,18(mol) =>n^^^ = , +1,18 = l,68(mol) iz> E)ap an dung la B I, =1,68.22,4 = 37,632(0 if' ,j„;t ' f Cty TNHH MTV DWH Khang Vi$t dm nang On luy^n thi DH mign B^c - Trung - Nam mOn H6a hpc - Cu Thanh Toan C a u 18: Cac hoa chat tac dung vori HCHO: 2CH3COOH + 2Na (1) H C H O + l A g N O , + 3NH3 + H2O -> 2Ag + HCOONH4 + ZNH^NOj (2) HCHO+H2 Ni,t 2CH3COONa + H j ^ X 0,5x 2C2H5OH + 2Na -^CHjOH y (3) H C H O + C^HsOH (4) HCHO + 2Cu (OH) + NaOH (6) HCHO + Br2 + H O ^ CU2O + HCOONa + 3H2O CO Cau 19: Dat X la R - CHCNHj) - COOH CH(NH3C1) - COOH l,255(g) Theo dinh luat bao to^n khd'i luofng: Bandau: 0,15 Phan ling: 0,1 vay H = 0,89(g) +"^HC\ ': > ' • ' O2 -> V(l) , CO2 + H2O 0,3 mol i-10,0; 0,2 mol ' > ^ v Ap dung djnh luat bao toan nguydn t6' cho nguydn to oxi, ta c6: "O(axit) + "o(oxi) = " ( C ) '* • 'l'"^^ "o(H20) i" V.2 = 0,3.2 + 0,2.1 =^V = 6,72 > 0,1.2 + 22,4 vay X la CH3 - C H ( N H ) - COOH (alanin) Dap an diing la D Cau 20: Chon thudc thir gom: Cau 23: Theo bai ra: n Ba(OH)2 =0.2(mol); nc,H402 =nc4H802 = 0,05(mol) axit fomic, glyxin, axit glutamic; lysin C2H4O2 Quy tim 0,05 xanh , , axit fomic; axit glutamic lysin >2C02 + 2H2O ^ dd Agl IO3/NH3 Con: Cau 21: Theobai ra: n^^^ =0,135(mol);ncH3cooc2H5 =0,l(mol) Xac djnh sd'mol CHjCOOH.CjHsOH 14,52 gam X: ^ )4C02 + 4H2O ^> 0,2^0,2 0,2 (mol) CO2 Phan ling: Dap an dimg la B 0,05 +0- CO2 + Ba(OH)2 ^ BaCOj i +H2O khdng c6 kt Ag axit glutamic 0,1-> 0,1 C4H8O2 =^mco2 + m H o - ( + 0,2).44 + (0,l + 0,2).18 = 18,6(g) 0,2 C H O H + C H j C O O N a A,0 C H O H + 2Na -> 2C2H50Na + H2 0,02 C6H|20(, ^ C H S O H + C O ^0,01 R(OH)^ +aNa ^ R ( O N a ) ^ + ^ H -a Cdu 28: Xet cac phuong an: C a u 24: Theo bai ra: n^^^ =0,03(mol) ^ y = 0,1875 -• 0,0625.100% v a y % s6 m o l m6i andehit X la: ^-^^ = 25% m ddZ - m' d,d B a ( O H ) = , - , = - l , l ( g ) 0,02 fx =0,0625 C H + H2O C2H5OH B Loai, vl C H t C Loai, V I CjHft 0,02 Ta c6: • -> 0,02 0,01a D Loai, V I CH4 Dap an diing la A 0,01+0,0 l a - , 0,02.68 + 0,02.(R + 39a) = 3,76 •a = 2;R = ( - C H - ) Chuj- C2H50Na + H C l ^ C H O H + NaCl v a y X C O the' la C H - C H ( O H ) - C H j O H (propan - , - diol) H30 X chiia C, H , O c6: M = 60 ^ Cac CTPT cua X c6 the' c6: C H ; C H - Cac 1)HC00CH3; 2)CH3COOH; 4) ( C H ) C H O H ; 5) C H - O - CH2 - C H 3) C H C H C H O H CH3COOH 86g 32g lOOg mi m, lOOg 100-32.100 m, = - — ^ — ^ - ( g ) ; 100.80 Cau 29: D i ^ u kien monome tham gia phan ling trung hop: + C6 lien ket b6i Thf du CH2 = C H ; C H - C H - C H j ; CH2 = C H - C H = CH2 ;CH2 = C H - CI; 100.86.100 ^/ ^ = — — - — = 107,5(g) 100.80 + Hoac C O vong kem b^n: T h i du: caprolactam; etilen oxit; Dap an dung la B Cau 30: Theo bai ra: n^g =0,12(mol) C.U^^O, C a u 27: Theo bai ra: n^g - , ( m o l ) 0,06 V I 2.nx = 0,5 < n^g = 0,625 < n x = 1,0 C6H,oO, + H - C , H , -> A g 0,06 C H , H + C j H j B r Dap an diing la B X )C2H50H C H C O O H + C H N a -> C H O H + C H C 0 N a CH2 - C ( C H ) C O O H + C H H - ^ [ C H = C ( C H ) ( C O O C H ) ] + H H C H O ->• A g i »h> vA : \-n, H3O"' C a u 26: So phan ung: xrH «nr/ ^ V i H = 80%nen ^ C H H + CH30H -j^C2H50H C H C H O + H2 (dcfn chufc, mach ho): Dap an dung la D ^ LiAlH CH3COOCH3 C a u 25: - C2H5OH + N2+H2O C H , N H + HNO2 Dap an dung la A Crcr cua X : if: M t so phan umg khac sinh C H O H : C H C H O -> A g i y ^ 2y Vay % xenlulozo = ^ ' = , % Dap an dung la A itJ A Cty TNHH MTV DWH Khang Vigt Cafm nang On luy^n thi DH mign B&c - Trung - Nam mOn H6a hqc - Cu Thanh ToAn BO D ETHI SO MON 9: Thdi Hj du qua hdn hop gdm cac oxit MgO,Al203,Fe203,CuO va PbO d nhiet dd cao thi hdn hop chat ran thu ducic sau phan ling gdm: H O A H O C K H O I A, B (Trich de thi thii Idn 3- Truong THPT Chuyen Le Hong Phong ) Cau 1: Dot chay hoan toan 0,1 mol mot axit cacboxylic don chiic, can vira du 6,72 lit O (dktc), thu duoc 0,3 mol CO, va a mol H,0 Gia tri ciia a la: A 0,2 B.0,4 C 0,1 D.0,3 Cau 2: Hoa tan gam mudi ngslm nude CuSO^.nH^O vao nude, dien phan dung taokettuakhi: C a / b < D a/b = r phan ling tach nudre tir X thu dugc mot hidrocacbon Y, trung hop Y tao polistireii (PS) Sd cdng thiirc ca'u tao ciia X thoa man dieu kien tren la: B C D thi luong ket tua thu duoc la: B 1,17 gam C 0,234 gam D Mg,Cu,Fe,Al203,Pb ' , / ^ j, , (1): H O C H - C H O H (2): H O C H - C H - C H O H (3): C H - C H - O - C H (4): H O C H - CHOH - C H O H (5): C6H,206(glucozo) (6): C H C O O H vrf^ '' A.(l);(3);(5);(6) B (1); (2); (4); (5) C.(l);(4);(5);(6) D (2); (3); (4); (5) aFeS2 + bHN03 0,672 lit (dktc) khf H, Cho dung dich A tac dung vdi dung dich chiia 0,016 mol A 0,312 gam C Mg,Cu,Fe,Al,Pb Cau 12: Xac djnh cac ht sd a, b, c phuong trinh hoa hoc sau: Cau 5: Hoa tan hdn hop X gdm Na, K nude du, tao dung dich A va AICI3 ' Cac cha't (trong nude) tac dung dugc vdi C U ( H ) la: Cau 4: Hop chat h&u co thorn X c6 cdng thiic phan tu CgHjoO Khi thuc hien A B MgO,Cu,Fe,Al,Pb A Khir CaO bang lugng du CO a nhiet dd eao B Khir CaO bang lucmg du cacbon d nhiet dd cao C Tir CaCO, dieu che CaCl, khan rdi dien phan CaCl, D Nhiet phan CaO nhi6t dd cao Cau 11: Cho cac chat sau: Cau 3: Cho a mol C O , ha'p thu het vao dung dich chiia b mol Ca(OH)2 Phan ung B l < a / b < A MgO,Cu,Fe,Al203,Pb Cau 10: Cd the didu ehe kim loai canxi bang phuong phap nao sau day? dich thu duoc cho A6n c6 thoat tai catot thi dCrng lai, thu duoc dung djch A Trung hoa dung dich A cin mot luong vira dii dung dich chiia 1,6 gam NaOH Vay n c6 gia trj la: A B C D A a / b < l / •f D 0,78 gam cau 6: Tinh the tich dung dich FeS04 1,5M cSn dung d^ phan ling vira du vdi - '' Fe2 ( S O + H2SO4 + cNO + H O A a = 2,b = 5,c = B a = 2,b = 10,c = 10 C a = 2,b = 4,c = D a = 2,b = 8,c = cau 13: Dun ndng hdn hgp ancol don chiic mach hd vdi H2SO4 dac 140°C thu 100ml dung djch chiia KMn04 0,2M va K2Cr3O7 , l M mdi trudng axit? dugc hdn hgp ete Ddt chay hoan toan mot ba ete thi thu dugc A 0,32 lit caebonie va hoi nude cd ti le mol la n^Q^ : n^^^Q = : Cdng thuc phan tir ciia B 0,64 lit C 0,106 lit D 0,08 lit Cau 7: Este E cd cdng thiic phan tu la C5Hg02 Thiiy phan E mdi trudiig ancol la: axit thu dugc dimetyl xeton Cdng thiic ca'u tao thu gon ciia E la: ' A C H C O O C H = CHCH3 B CH3COOC(CH3) = C H A C H O H va C H , H B C H S O H va C HCOOCH = C H C H C H D HC00C(CH3) = C H C H C C H O H va C H C H C H O H D C2H'' Cr 0,01.2 ^ 0,06 (mol) ; " ' Qua trinh nhucmg electron: Fe - 0,16 ') ndiH' Cau : Cac tinh chat cua X: ' * ^' + X la hop chat thorn (c6 vong benzen phan tir) +3 le ^ Fe ^ X (mol) §iuib ^:, , :u ^'tSfcn£>;iniA' Theo djnh luat bao toan electron, ta c6: x = 0,1+0,06 =^ x = 0,16(mol) ;tl Dap an dung la C Cr 0,02 -> 0,1 (mol) X ,^ Di CO ket tua (CaCOj i ) thi a/2 < b = i ' a / b < «' + CTPT: C,H,oO ( n , + , = ) VddFeSO4=Vr 1,5 = 0'106(l) Dap an dung la C ChuJ,: Cac PTHH cac phan ung xay ra: 10FeSO4 + 2KMn04 + 8H2SO4 + CgH,oO^Y-*PS(polistiren) * ^'''^^ ''^ 5Fe2 (SO4 )^ + K2SO4 + 2MnS04 + 8H2O 6FeS04 + K2Cr207 + 7H2SO4 -> 3Fe2 (SO4), + K2SO4 + Cr2 (SO4 \ IH^O => Y la C^H^ - CH = C H (stiren) ; , jjitj,; Vay X phai c6 ca'u tao: 1) C^Hj - CH(OH) - CH3; 2) CgH^ - C H - CH2(OH)»j => Co ca'u tao thoa man Dap an diing la C Chu y; Cac so phan ling xay ra: C^H, - CH (OH) - CH3 + 3e +2 + Ca(OH)2 -> Ca(HC03)2 Ban dau: + 5e - * M n Cau 7: So d6: C5HSO2 + H2O CH3 - CO - CH3 + este E dimetyl xeton => E phai CO cau tao: CH3COO - CCCHj) = CH2 'K Dap an diing la B Chii v.- PTHH: > C,H5 - CH - C H C^H, - C H - C H ( H ) - ^ C H , - CH = C H - PS ) ,,/) ^ ' CH3COOC(CH3) = CH2 + H O — — C H C O O H + CH3 - CO - CH3 C^u 8: Khi xay cac qua trinh: PS Cr202- + H2O 2CrO^- + H + Cau 5: Theo bai ra: Ui^^ =0,03 (mol) 2H++20HM(Na,K) + H - ^ M O H + ^ H T Phan ling: 0,048 Con: 0,012 • : Phan ling: ^ Con: AICI3 M [ A ( H ) J 0,012 0,004 Dung dich chuyen sang mau vang A l ( H ) i + 3MC1 0,016 ^ ' Ci20^- + H - -> 2CrO^- + H2O 0,06 ^ 0,03 3MOH + -^2H20 r| , tJ) 0,012 > ^ n i A ( O H ) i = ' 0 = 0,312(g) , Dap an diing la A i , K2Cr207 + K H ^ 2K2Cr04 + 2H2O (da cam) (vang) ' '' Cau 9: Khi H i (du) chi khir dugc cac ion kim loai trung binh va yeu oxit d nhiet cao: H2 + MgO > kh6ng xay H2 + AI2O3 — ^ khong xay I hi U H inign B S C - Trung - Nam mOn H6a hgc - Cu Thanh ToSn Cty TMiili M I V DVVH Khang Vi^t C&u 38: Thirc hien phan ling nhiet nh6m h6n hop g6m m gam A l va 4,56 gam CuO} (trong di^u kiSn khdng c6 O,), sau phan Ung kfe't thiic, thu dugc h6n hop X Cho loan b6 X vao m6t luong du dung djch HCl (loang, nong), sau Cau 45: So dOng phan ca'u tao ciia C H phan utig dupe vdi dung djch brom la A B Cau 46: Phat bi^u nao sau day la sai? C D * (b) A Trong y hoc, ZnO dupe dung lam thu6c giam dau day thin kinh, ehua benh eczema, benh ngiJa B Nhom !a kim loai dan di^n t6't hon vang C Chi (Pb) C O ling dung de che tao thiet bj ngan can tia phong xa t -K- cac phan dug xay hoan loan, thu duoc 2,016 lit Hj (dktc) Con n6'u cho toan bo X vao mot lugng du dung djch NaOH (dac, nong), sau cac phan iing thiic thi s6' mol NaOH da phan vtng la A 0,16mol B 0,06 mol C 0,14 mol D 0,08 mol Cau 39: Cho cac to sau: to xenlulozo axetat, to capron, to nitron, to visco, to nilon6,6 Co bao nhi6u to thu6c loai to poliamit? ,, A B C D Cau 47: Chat hiru co X mach hd cd dang H^N-R-COOR' (R, R' la cac gdc Cau 40: Nhiet phan 4,385 gam h6n hop X gom K C I O va KMn04, thu dupe O, va ling hoan toan vdi dung djch NaOH, toan bp lupng ancol sinh cho tac dung m gam chat rSn g6m K2Mn04, MnOj va KCl Toan bo lupng Oi tac dung het voj het vdi CuO (dun nong) dupe andehit Y (ancol chi bj oxi hoa andehit) cacbon ncng do, thu dupe 0,896 lit h6n hop Y (dktc) c6 ti khd'i so v6i B 74,92% C 72,06% hidrocacbon), phSn tram khdi lupng nito X la 15,73% Cho m gam X phan Cho toan bp Y tac dung vdi mot lupng du dung djch AgNOj NH3, thu la 16 Thanh ph&n % theo kh6'i lupng ciia KMn04 X la A 62,76% D Thiec c6 th^ dung de phu len hi mat ciia sSt de ehdng gi D 27,94% II PHAN R I E N G [10 cau] dupe 12,96 gam Ag ket tiia Gia trj ciia m la A 3,56 B 5,34 C 2,67 Cau 48: Trong qua ga'c chin rat giau ham lupng A vitamin A B (3 — caroten C ete ciia vitamin A D este ciia vitamin D 4,45 ; ' Thi sink chiduac Idm mot hai phan (phan A hoqc B) A Theo chuong trinh Chuan (10 cau, tir cau 41 den cau 50) Cau 41: Cho butan qua xiic tac (o nhiet dp cao) thu dupe h6n hpp X g6m C4H10, C4H8, C4H6 va H , Ti kh6'i cua X so vdi butan la 0,4 N6'u cho 0,6 mol X vao dung dich brom (du) thi so mol brom toi da phan ling la A 0,60 mol B 0,36 mol C 0,48 mol D 0,24 mol u 49: Nhiet phan mdt lufpng AgN03 dupe cha't ran X va hdn hpp Y DSn toan Cau 42: Cho m gam bot Zn vao 500 ml dung dich Fe2(S04)3 0,24M Sau cac phan ling xay hoan toan, kh6'i lirpng dung djch tang thdm 9,6 gam so vdfi khoi lupng dung dich ban d^u Gia tn ciia m la A 75% A 20,80 B 48,75 Cau 43: Cho cac phat bi^u sau: C 32,50 bd Y vao mdt lupng du H^O, thu dupe dung djch Z Cho toan bp X vao Z, X chi tan mdt phdn va thoat NO (san ph^m khix nha't) Bie't cac phan vtng xay hoan toan Phdn tram khdi luong ciia X da phan ling la lol X (xue tac Ni), dupe hdn hpp Y cd ti khdi hoi so vdi heli la 9,4 Thu la'y toan bd cac ancol Y rdi cho tac dung vdi Na (du), dupe V lit Hj (dktc) Gia trj Idn nha't ciia V la A 22,4 (du) c6 xiic tac Ni dun nong, thu dugfc ancol bac nio' B.2 cau 44: Cho day cac chat sau: A l , NaHCOj, C B D 13,44 H2O Mat khae, n6'u cho a mol X tac dung vdi lupng du dung djch NaHCOs, thi D.4 (NH4),C03, N H C I , Al,03, Zn, K.CO.v C.4 C 5,6 Y nhd hon ciia Z) Dd't chay hoan toan a mol X, sau phan ling thu dupe a mol tac dung dupe vdi dung djch NaOH? A B 11,2 B Theo chuong trinh Nang cao (10 cau, tur cau 51 den cau 60) Cau 51: Hdn hpp X gdm hai axit cacboxylic no, mach hd Y va Z (phan tir khdi ciia K2SO4 Co bao nhieu cha't day vira tac dung dupe vori dung djch HCl, v^f^ ' D 60% tir ddu cd sd nguydn tir C nhd hon 4), cd ti khdi so vdi heli la 4,7 Dun ndng D 29,25 (d) Dung dich axit axetic tac dung dupe voi Cu(0H)2 (e) Dung djch phenol nude lam quy tim hoa (g) Trong cong nghiep, axeton dupe san xua't tir cumen S6' phat bieu diing la A C.70% p a u 50: X la hdn hpp gom H^ va hoi ciia hai andehit (no, don chiic, mach hd, phan (a) Andehit vira c6 tinh oxi hoa vira c6 tinh khir (b) Phenol tham gia phan utig the brom kho hon benzen (c) Andehit tac dung vdi B.25% , D I thu dupe 1,6a mol CO2 Thanh phSn % theo khdi lupng ciia Y X la " A 74,59% B 46,67% Cau 52: Thuc hien cac thi nghidm sau: (a) Nhiet phan AgNO, (b) (1 Nung FeSi khdng C 40,00% D 25,41% ^ ; 539 cairn Cty TNHH MTV DWH Khang Viftt nang On luy$n thi DH mjgn Bjc - Trung - Nam mOn H6a hgc - Cu Thanh Tpan (c) Nhiet phan KNO, (d) Cho dung dich CUSO4 vao dung dich NH,(dif) • u (e) Cho Fe vao dung dich CUSO4 ^iiid (g) Cho Zn vao dung dich FeCl, (dir) , , (h) Nung Ag.S ichong i Y, ^ ' ' Cac chat huu co X,, X,, Y,, Y,, Y, la cac san ph&m chinh Hai cha't X,, Y, Ian lugt la A axit axetic va ancol propylic B axit 2-bidroxipropanoic va axit propanoic C axit 3-hidroxipropanoic va ancol propylic D axit axetic va axit propanoic HI/ONG DAN r^^, GIAI Caul Theobai ra: n^^o = , I / I = 0,12(mol) CH3COOCH = CH2 ' 4CO2 + 3H2O Vinyl axetat (x) A 0,012M va 0,024M B 0,018M va 0,008M C 0,08M va 0,18M D 0,008M va 0,018M Cau 56: Thuy phan h6n hop gom 0,02 mol saccarozo va 0,01 mol mantozo m*'' thcri gian thu duoc dung djch X (hieu suA't phan iJng thijy phan m6i cha't ddu I'' 75%) Khi cho toan bo X tac dung vdi mot lugng du dung djch AgNO., NH, thi luong Ag thu duoc la A 0,12 mol B 0,090 mol Cau 57: Phat bieu khong diing la: I D E Protein la nhung polipeplit cao phan tu c6 phin tir khdi tir vai chuc nghin den vai trieu ^ HCOOC2H, z 3x » 3CO2 +3H2O CH^COOCH, y -> 3y ^"^ 3CO2 + 3H2O 3z |3x + 3y + 3z = 0,l2 la CO he: • [86x + 74y + 74z = 3,08 =>x + y H z = 0,04 va 12x+ 74(x + y + z) = 3,08 12x +74.0,04 = 3,08 => 12x = 0,12 => x = 0,01 541 I thi H mi§n Bjc - T r u n g - N a m m n H a h g c - C u T h a n h Toan x.100% %n^iny,axetal Vay " 0,01.100% x + y +z 0,04 Cty T N H H M i V DVVH Khang Vi$t (5: PTHH: = 25% 1) SO2 + H O ;=! H2SO3 Chu y: Cac chat tren c6 dac di^m chung la c6 cung s6' nguydn t u H phan tir 3) SO3 + H O Cdu 2: Xet cac bieu thiic: 4) C r + H O afirnufR (a) : Sai v i X c6 t h ^ la xicloankan (b) : D u n g (1) (c) : Dung (2) T h i du M = 60 C O C H C O O H ( C H O ) ; (f) : Sai, VI Cdu 3: Theo bai ra: nco2 = 2,24/22,4 = , l ( m o l ) n K C = 0,1.0,2 = , ( m o l ) ; n K O H = 0,1.x = , l x ( m o l ) Ke't tiia la BaCOj => n B a c o = 11,82/197 = 0,06(mol) - V i nco2 = ^ CO phan > " B a c o j - n K C =^ ling tao K2CO3 ^6 phan utig tao K H C O 2K0H + CO2 -> K2CO3 + H2O K O H + C O -> KHCO3 a ^ a - * a b * - b - ^ b K C O + B a C l -> BaCOj [ + K a ( a + 0,02) Ta CO h6: -> =0,1 2a+ b vay: X= 5) P2O5 + H O ^ H P O (axit photphoric) 6) N2O5 + H O H N O (axit nitric) I'k qi:"'! 2) A i + H S O (1) ^ AI2 (SO4 )3 + 3H2 D a p an dung l a B Cdu 7: X e t c a c phat bi^u (a) : D i i n g (1) (b) : D u n g (2) (c) : D i i n g (3) (d) : Sai, v i : {C,H,,0,l + n H - ^ n C H , , Tinh b6t glucoza C,2H220,,+H20-S± ^C6H,20,+C6H,20, Saccaroza Glucoza fructoza (e): Dung (4) ( a + 0,02) a + 0,02 = 0,06 a+ b (axit cromic) Cdu 6: H* dong vai tro la cha't oxi hoa phan ihig tao san ph^m l a H , : 'in "BaC03 H2Cr04 1) Sn + H C l ( l ) ^ S n C i + H Dap an d u n g la B V i n^jcoj < H2SO4 hay vdri baza) hOu co thucmg cham n^+^ = - '-lOj't Q l i i : N O , N , C O l a c a c cha't oxit trung tinh (khdng tac dung vdi H^O, vdri axit C H ( C H C H C H H ; [CH^]^ C H O H ; C H C H C H j ling ,0 D a p an diing la B (d) : Sai, v i d6ng phan phai c6 cung CTPT (e) : Sai, v! phan 2HNO3 + N O 2) N O + H O Dap an diing la D (axit sunfuro) fructoza >a = 0,04; b = 0,06 (g): Sai, V) saccaroza khdng tac dung vdi H , Dap an diing la A 2.0,04 + 0,06 = — > glucoza-* C H H ( C H H ) ^ C H O ^ A g = 1,4 0,1 0.1 Dap an dung la D Cdu 4: Cac k i m loai ki6m th6 (Be, M g , Ca, ) tac dung vdi nu6c nhOng dieu kien khac nhau: Trong tinh the phan tiir (thf du tinh thd' H^O, tinh thd' I , tinh the' CO,, ), cac phan tir lien ke't vdi bang cac lien ke't ye'u nen tinh the phan tiJ de ndng chay, de bay hai => Phat bie'u C sai Dap an diing la C + Be: Kh6ng tac dung + M g : Chi tac dung k h i or dang b6t vdi hofi nir6c + Ca, Sr, Ba- Tac dung v6i nu6c a diiu kien thucmg s6 nguyen tir H phan tir => Phat bi^u C sai T h e o b ^ i r a : M x = 17.2 = 34=i>12n + = 34 =^ 12n = 30=^n = 2,5(C2,5H4) Cac chat X (C2H4,CH4,C3H4,C4H4) cd dSc die'm chung la cd ciing Dat cdng thiic chung la C„H4 Dap an dung la C 542 543 C^m nang On luyQn thi DH mi^n B^c - Trung - Nam mOn H6a hpc - Cu Thanh Toan C2,,H4 + , - > , C 0,05 + ^0,125(mol) ^ Cty TMHH M!V DWH Khang Vigi 2H2O 200 i O,l(mol) "•^^^ =^ m = 0,125.44+ 0,1.18 = 7,3(g) Cu + O , 0,025 ^ 0.nf.i- Dap an dung la D Cdu 10: Cac thi nghiem: (a) : Ca Cu, A g deu phan iJng v6i O,: , 5 + 28x = 7,55 + 28x =^x = 0,25 Vay khoi liromg N a O H da tham gia phan iJng: mNaOH ^ Cdu Cu + 2FeCl, - ^ C U C I T + 2FeCl2 Dap an diing la A Chu y: Trong day dien hoa c6 thu tir: Cu^"^/Cu-.Fe"^"^/Fe^+;Ag+/ A g nen tlicu quy tac anpha: Cu2+ + 2Fe'+ f CgH^ONa + (x) natri phenolat HCl ^ CgHjOHi+NaCl (Y) fZ) axit clohidric , ^ phenol Dap an diing la C nNaOH = 12/40 = 0,3(mol) I T a t h ^ n x : n N a O H = , : , = 1:2 11: PTHH: CH ,COOCf,H, + 2NaOH ^ CH^COONa + Cf,H,ONa + H O Mat khac X la este don chiic nen X la este ciia phenol: R C O O C ^ H j khong c6 ancol 1) C H C O O C H C H = C H + NaOH ^ C H C 0 N a + C H = C H - C H j O H anlyl axetat 13: P T H H : C^HJ^: Theo bai ra: Fe-^+ + A g (khong xay phan ling ngugc lai) Phenyl axetat = (0.025 + 3x).40 = (0,025 + 3.0,25).40 = 31gam Dap an dung la C (d) : Chi CO Cu tac dung (bi oxi hoa) A g khong tac dung R C O O C g H , + N a O H ^ RCOONa + C^H^ONa + H O ancol anlylic 0,15 2) C H C O O C H + N a O H ^ CH^COONa + C H O H metyl axetat =0,55 Theo bai ra: m , g = 207,55 - 200 = 7,55(g) C u ( N O ) , + 2NO2 + H O ^ 0,15 ^ 0,15(mol) T a c o : , ' ( R + + 116) = 29,7 ^ R = 15(CH3) ' Ca'u tao ciia X dang este ciia phenol: ancol metylic 3) H C O O C H , + N a O H -+ 3x(mol) m,g = , ( - l ) + x.(3.23-41) (c) : Cii Cu va A g deu khong phan irng Cdu 0,025(inol) Khd'i luomg mu6'i tang so \di khd'i luong cha't beo: Ag + H N O , ^ A g N O , + NO2 + H2O + Fe'+ ^ 0,025 X (b) : Ca Cu va A g dau phan urng vdi H N O , dac: Ag+ "NaOH , 2Ag + , - > A g + Cu + 2Fe'+ ^ 56 " ( R C O O ) ^ C H + 3NaOH ^ R C O O N a + C3H5 ( O H ) O,'' CuO + O2 Cu + H N O ^ 1000 R C O O H + N a O H - > RCOONa + H j O Kh6'i lucmg binh tang bang khoi luonig C , H sinh phan utig chay Dodo: m - m c o , + ' " H ( ) " HCOONa + C H O H 1) C H j C O O C ^ H , etyl fomat ancol etylic 4) ( C | , H | C O O ) C H , + N a O H ^ C H , ( O H ) + C , H , C O O N a 2) - H C O O C H C H 3) m - H C O O C H C H tripanmitin glixerol 4) p - H C O O C H C H Dap an dung la B Cdu Dap an diing la A 12: Chat beo gom c6: - Triglixerit: (RCOo)^C3Hy = 0,12 ^"Ai2(so4)3=0'4.0,12- 0,048mol * Tim x: "NaOH > ^•"AI(OH)^ ^ ^ ' ( ^ ^ ) ^' AlCl3+3NaOH->Al(OH)3i+3NaCl ''^^ * 0,4x l,2x -» 0,4x AI2 (SO4 )j + 6NaOH ^ 2AI (OH)^ j +3Na2S04 0,048 ^ , 8 ^ 0,096 Al(OH)3 +NaOH-^Na[Al(OH) a a(mol) 0,4x+0,096-a = 0,108 0,4x-a = 0,012 Ta c6: l,2x+0,288 4-a = 0,612 l,2x+a = 0,324 =>l,6x = 0,336 =4>x = 0,21 Vay x:y = 0,21:0,12 = 7:4 Dap an dung la D Cdu 16: So d6 qua trinh san xua't Mandehit —»Fe304 —> 3Fe —> Gang 800 95 Khtfi lirong Fe c6 gang: mp^ = ^— = 760 (ta'n) 100 ""H-)- =hut767,68 VI luong sat1 0bj0 hao 1% ntn(ta'n) khd'i luong sat thuc td'can la: +CO 760.100 »3Fe 232 ta'n -> 168 ta'n m, x = r = 1325,16 (tan) 100.168.80.(100-1) Cdu 17: Amin va ancol cung bac la: C6HsNHCH3 (bac II) va C H C H ( H ) C H (bac U) Dap an dimg la B Cdu 18: - Cac kim loai ki^m 66u c6 cung kie'u mang tinh the' lap phuong tam kh6'i: Na, K, - Cac kim loai kidm th6 c6 kie'u mang tinh the' khac nhau, chi c6 kim loai Ba CO kie'u mang tinh the' lap phirong tam kh6'i vay cac kim loai la Na, K, Ba Dap an dung la C Cdu 19: Tac6:-„=2:2ai£=ll*i=3,6 Phan tif andehit va ankin diu c6 nguyen tij cacbon =>C00ankin la C3H4 andehit MatPhan khac:tir HH = s6' nguyen tir H be thua 3,6 andehit la C3H2O (CH = C - CHO) ^u;, " M C3H4 a ^ +"2 , C + H 3JS C3H2O -> 2a(mol) b =^a = 0,8x;b = 0,2x Ta c6: ^ + '^-'^ 2a + b = l,8x b.100% 0,2x.l00% v a y %n =—-—- = = 20% a +D X Dap an diing la A -""Z •? 3CO2+H2O , b(mol) i l Cty TNHH MTV DVVH Khang Vipt Cdu 20: Vay p H ( ) < p H ( l ) < p H ( ) Ph^n I : nco2 =0,25mol; I I H ^ O = 0-35(mol) Dap an diing la A "H20 - "ROH - nc02 nco2 d-r,a iu 22: PTHH: Vi ^ > " C O =^ancol no, hd, dom chirc: ROH -1 -2 +7 SC^H, - CH = C H + 10KMnO4 = , - , = 0,1 (mol) +4 C e H C O O K + K C O + 10MnO2 + KOH + 4H2O 0,25 -1 =>Haiancol la: C H H ( X ) va C H H ( Y ) 3x -2 c - c +3 +4 - C + C + lOe +4 +7 Goi X, y \in luot la s6' mol X , Y m6i phSn Ta c6: lOx M n + 3e -> M n vay t6ng h6 s6': + + + + + + = 2x + 3y ^ ^ = 0-05; y = 0,05 ,>^g Dap an diing la B x+ y PhSn I I : n , , , = n^^ =0,42/28 = 0,015(mol) Goi a, b lan luot la s6' mol X, Y tao ete (oartB 4v mrnA : V •^ >1 >» C2H5OH + C3H7OH ^ C2H5 - O - C3H7 + H2O 2C2H50H->C2H5-0-C2H5+H20 2C3H7OH £ A C3H7 - O - C3H7 + H2O ^ ' "ancol = 2nete (l) « Ma m^,g = mc2H^_ + m c H ^ _ + m _ o _ =^ 1,25 = a.29 + b.43+ 0,015.16 29a + 43b = 1,01 Tiir(l,2) 0,05 Dap an diing la A Cdu 21 * Dung djch (1): H2NCH2COOH c6 pH ^ * Dung djch (2): C H j C O O H ^ H+ + C H C O O " C O pH < 7: (m6i trucmg axit) * Dung djch (3): CH3CH2NH2 + H2O ^ C H C H N H I + O H ' C O pH > (m6i trucmg baza) „ q n»io Phan tiJr Y c6 lien ke't n => Phan tir X c6 lien ke't 71 (C = C,C = O ) \-A oil- , Dap an dung la D u v: Cac PTHH xay ra: CH, = C H - C H + - 2 xt.t" ->CH2=CH-COOH >CH3-CH2-CH20H a = 0,02; b = 0,01 , m6t niJra ciia phan tiJf Y ) => X c6 nguydn tu cacbon phan tir (X) Hidu sua't tao ete ciJa Y : H,v^ = " ' " ' ' ^ ^ ^ ^ = 20% Tir (3) va CTPT Y j => Y , , Y , d^u c6 nguydn tir cacbon phan tu (bang (2) 02 100% Hieu sua't tao ete cua X: H,^^ = ^ — = 40% 0,05 548 Tir 1,2 => Y | , Y2 C O cung s6' nguyfin tir cacbon phan tir vay X la CH2 = CH - CHO (andehit acrylic) g =4>a + b = 2.0,015 = 0,03 Cdu 23: Tir 1,2 => X la andehit CH2 = C H - C H O + 2H2 xt,t ^CHj =CK-C]0OH+CH3CH2CH2OH , « (Y.) (Y^) CH2 = CH - OOOCH2CH2CH3 + H2O N du 24: * Xac dinh ti le s6' mol O2 (x mol), O3 (y mol) ^ i ± ^ 2 = 44 x+ y ^_ , =J>32x + 48y = 44x + 44y 4y = 12x =»y = 3x Gia sir V , lit X c6 Imol O, thi c6 mol O3, (trong V2 c6 mol khQ => c6 llmolO Cty TNHH MTV DWH Khang Vigt d m nanq On luy^n thi BH mi^n B^c - Trung - Nam mOn H6a hpc - CCi Thanh Toan Taco: 4(x + y)/3 = 0,4 * Xac dinh ti 16 s6'mol CH^NHj (amol) va C2H5NH2 (bmol) Ta c6: =:>x + y - , ^ i i ± ^ ^ 17,833.2 = 35,666 a+b la + 45b = 35,666a + 35,666b 9,334b = 4,666a =^2b x.100% =a 2b ^ b f.^pt*- ^ v ',' " ^ 7,5b 16,5b = 11 =j> b = 2/3 " I " Trong V, lit Y c6 3b mol (= mol) Vay V, : V2 = H Y : Hx = :4 = 1:2 =>nK, ^ =^m 100—" ^ =0,12(mol) = n ' NOJ Cdu 25: Co can bang: 2SO2 + O2 ;=i 2SO3, A H < Cac bien phap lam can bang chuyen dich theo chi^u thuan: (2) : Tang ap sua't chung ciia ht phan ling (3) : Ha nhidt d6 (5): Giam ndng d6 SO3 '''' Dap an dung la D '!•' Chu v: Phan ling thuan c6: fi \ + S6 mol Ichi giam (3 -> 2) => tang ap suSft + AH < (toa nhiet) => giam nhi6t dd + La'y bdt'san ph^m + Thdm cha't tham gia d6u lam can bang chuy^'n djch theo chieu thuan Cdu 26: Theo bai ra: H X = 15,68/22,4 = 0,7(mol) nNo= 8,96/22,4 = 0,4(mol) \ NO3 ' ria'i? i'l *'" _ n i x - m ^ ^ _ =14,16-7,44 = 6,72(gam) Dap an dung la B Cdu 28: Theo bai ra: n^oH =0,144mol; n^Q^^^ =0,012mol =>n = 0,144+ 0,012.2 = 0,168mol;n 2+=0,012mol if SlftCI Theo djnh luat bao loan dien tich, ta c6: 0,1.1 + z.3 = 1.1 + 0,02.2 =>t-3z = 0,06 Vi n ,2+ < n ,2Ba so (1) nBaS04l= =^ mB,so4i = 0,012.233 = 2,796(g) < 3,732(g) => Trong ice't tua c6 Al ( O H ) ^ : '"A.(OH)3 =3,732-2,796 = 0,936(g) = > n A i ( O H ; = ' / = 0,012(mol) X -> x(mol) y-> 2y(mol) H+ + H " ^ H 0,1 =^2x + 3y = 0,7 Al^+ + H - ^ A l ( O H ) i (1) CO + CuO->Cu + C ^2 + C u O - ^ C u X x + 2y xmol 3Cu(NO3 \ N + 4H2O 4(x + y)/3 + H2O -^(x + 2y)mol ^0,1 (mol) Taco: x + y + x + 2y = 0,7 2x + 2y J IK* 2H20 + C ^ C + H HjO + C ^ C O + Hj 3Cu + 8HNO3 ' =0,12.62 = 7,44(g) ^mki,.M= j, Dap an diing la D 0,2.100% = ~~QJ~ " Dap an dung la A Cdu 27: Khdi luong nguy6n t6' nito 14,16 gam X: 14,16.11,864 , ^ 2C2H,NH2 +150 ^ 4CO2 + 7H2O + N2 Taco: 9b + 7,5b = l l I Tir(l,2) =^x = , ; y - , l Vay %Vco/x = %nco/x = C H N H + - > C + H + N2 b (2) • J ! ' j; , ID , ^ , ^ , (mol) Al^+ + H - ^ [ A ( H ) ^ - , 0,008^(0,168 - 0,1-0,036) =i>z = , + 0,008 = , (2) 551 dm nang On luyQn thi DH mign BJc - Trunp - Nam mOn H6a hQC - CO Thanh ToAn Tir(l,2) Cty TNHH MTV DVVH Khang Vi^t = 0,06 + 3.0,02-3 0,120 Chu v: Vay z, t ISn luot la 0,020 va 0,120 Dap an dung la C Chu v: M6t each gSn dung, coi kh6ng nuoc phkn li (b6 qua sir diSn li cua nuorc) Cdu 29: • i\ 0,035 (mol) =J>mY „,i fit =0,035.30 = l,05(g) Dat Z la RCHO (y mol) RCHO + AgNOj + 3NH3 + H2O RCOONH4 + 2Ag j +2NH4NO3 - R - C ( = 0)-|0-H+ir|-0-R'?:^ R - C ( = ) - - R ' + H Axit cacboxylic ancol este nude jtK 31: Cac chat tac dung duoc vdri NaOH (dac, nong): 1) Si02 + 2NaOH ^ Na2Si03 + H2O -fx' 5) NaHC03 + NaOH -^Na2C03 + H2O 6) AI2O3 +2NaOH + 3H20^2Na[Al(OH)^ Na2S + FeCl2 -> FeS i +2NaCl 2FeCl3 + Fe(NO3 )3 + NO + 2H2O li y: Dung dich NajS phan ling vdri FeCU c6 the sinh ca FeS va Fe(OH)2 84 = 56(C2H3CHO) 'i^'*"' ' + +s2 ^FeSi V a y Z la CH2 = C H - C H O (andehit acrylic) S^^+H^O^HS+OH" Dap an diing m D Fe^++20H- ^ F e ( O H ) , i Cdu 30: Xet cac phuong an: 'du33: Trio'ein (glixerol trioleat) (C,7H33COO)3C3H5 phan ling duoc vy = 0,015 C H - / t'^ ' -) O: 4) Zn(OH)2+2NaOH^Na2[Zn(OH)^] 4HNO3 ^ 3FeCl2 2y Taco: 4x + 2y = 0,17 Suy ra: Mz = Trong phan ling este hoa: Dap an diing la A Cdu 32: Cac chat phan ling vdi dung dich FeCU: Cl2+2FeCl2 ^ F e C l =»mz = 1,89-1,05 = 0,84(g) y - 3) Cr03 ^ 2NaOH^Na2Cr04 + HjO X • X Isoamyl axetat (CH3COOCH2CH2CH (CH3 ) ^ ) c6 mui thorn ciia chuO'i chin ; 2) C r ( O H ) + N a O H - N a [ C r ( O H ) J HCHO + 4AgN03 + 6NH3 + 2H2O -> 4Ag + 4NH4NO3 + (NH4 )^ CO3 x(mol) ^0-CH3 +H2O H2 (Ni, t°) VI g(5c axit kh6ng no NaOH (dd, t°) vi este n6n bi xa phong hoa H2O (H"*",!"! VI este nen b| thuy phan Kh6ng phan ling duoc vdri Cu(0H)2 Dap an diing la A luil + Tripanmitin: (C,5H3,COO)3 C H , + Tristearin: (C,7H35COO)3 C3H5 + Trilinolein: (CpH3,COO)3 C H , rinb silirij %()/•) Ga'in n a n g o i l l u y f i r i t h i D H m i e n B l c - tiuiiy N a m 111611 H a hpc - Cu Thanh ToAn Cty TNHH MTV DWH Khang Vi$t Cdu 34: PTHH xay ra: Trong X: Cu(4xmol);Ag(xmol) N2O T +2H2O 1) N H N O , •* 2) NaCl (r) + H2SO4 (d) — ^ HCl T +NaHS04 ^ 3) C l + H ? : l H C l + H C Taco: 4x.64 + x 108 = 1,82 =^ 364x = 1,82 3Cu •; ^ :.^lv/id HCl + N C H C O - > N a C l + CO2 T + H O oii < Phan ling: Con: 5) Na2S03 + H S ^ — • N a S +SO2 T + H O • Dap an diing la A * '* Chu v: (d) CO2 + Ca(OH)2(du) ^ CaCOj i +H2O (h) PbS + HCl axit HCl, JQJ, Khong xay (vi PbS, cung nhir Ag2S,CuS,HgS, khong tan H2SO4 loang) Cdu 35: %^^C\ 100 - 24,23 = 75,77% t6ng s6' nguyfen tiJr - 37.24,23 + 35.75,77 ^ 0,005 Phan ihig: 0,005 4H+ 3Ag+ + NO + H O 0,14/3 • :^ 2NO _^ + O2 ^ 0,005/3 2NO2 Bandau: 0,015 0,1 Phanirng: 0,015 0,0075 ^ 0,015 Con: 4NO ^ 0,0925 0,015 ' + O + H O ->4HN03 0,015 0,015 (mol) HNO3 H+ 0,015 0,015(mol) +NO3 H+ =0,015/0,15 = 0,1M =^pH = z = l Dap an diing la A "CI/HC104 =, 896,51.100% ^ (1 + 35,485+ 16.4) 100 - »'" nAg= Vi 2.nx day da cha'p nhan nguydn tu kh6'i c6 tri s6' bang s6' khd'i; nguydn tiSr khd'i cua hidro bang 1; cua oxi bang 16 , „ C6ng thuc tinh dung: %m-i-i « ^ 37c,/HC104 • « - Cong thuc tinh sai: %m-.7 "c,/Hcio4 Cdu 36: %-^^C1.37.100% —1 = ^ i + Mci+16.4.100 %"Cl.Ma = i + Mci+4.16 Theobaira: nH2S04 =0.015(mol);nHNO3 =0,06mol _ = 0,06 (mol) NO3 S54 , 8,64/108 = 0,08(mol) 1,5R + R ' = 48 Ta = > R = 14(CH2);R' = 27(CH2 = C H - ) =>x=0,01; =>no2 = X Cdu 38: :u>,L •istm tv Theobaira: n^^o^ = 4,56/152 = 0,03mol + Cr203 - > A I O + 2Cr 0,06 ^ 0,03 ^ C r C l + H2 0,06 MS' a —*• + 6HC1 ^ 0,02 2AICI3 + 3H2 ^ (0,09-0,06) * X + dd NaOH (dac, nong, du): Al + NaOH + H O ^ Na|Al(OH)^ + N a O H + H O ^ N a A1(0H)^ vay HNaOH S1.0-^ : • b/2(mol) b Ta CO 1,5a+ b/2 = 0,025 he: a = 0,01; b = 0,02 • nib :b:, 122,5a + 158b = 4,385 vay n._,=^^^;|P = 72,06^ Dap an dung la C Cdu 41: Mx =0,4.58 = 23,2 —>C4Hi5 + H a —> a —> a(mol) C4H10 0,03 ^ 0,06 + y / = 0,01 +0,03/2 =0,025(mol) 2KMn04 ^ K2Mn04 +Mn02 + O 0,02-* 0,02 AI2O3 y = 0,03 l,5a(mol) 0,06 2A1 :R\d Xac dinh phan cua X: 0,03 ^ 0,06 + 2HC1 = 16.2 = 32 KCIO3 ^ K C l + - = 2,016/22,4 = 0,09(mol) VI nj^2 > ^-^CriOj ^ Trong X c6 Al dir + y = 0,04 44x + 28y x+y Dap an diing la C Cr he: 0,0 Vay hai andehit la OHC - C H - CHO va C H = CH - CHO 2A1 CO y(mol) b —*C4H6 + H ^ b -^2b(mol) = , + 0,06 = 0,08 (mol) Dap an dung la D Cdu 39: Cac loai tor poliamit (chiJa lidn kd't amit - C O - N H - ) 1) To capron: - [ H N - ( C H j \ C O J - C4H,o ^ C H , o (du) c ^ HP, c (mol) C4H8+Br2->C4H8Br2 a —> a 2) Tor nilon 6,6r|HN - ( C H j )^ - NH - O C - ( C H )^ - COJ- C4H6+2Br2^C4H6Br4 Dap an dung la D b Cdu 40: * Xac djnh so mol O, sinh ra: ny = 0,04(mol) C + O2 C O X ZnS04 + Fe j X x(mol) 5) m = (0,12 + 0,2).65 = 20,80(g) Z n + 2HC1 ZnCl2 + Dap an dung la B ~du 45: C„H2„ phan u"ng duoc vdd dung djch Br, k h i : 43: Xet cac phat bia'u: (a) : D u n g ( l ) R-cboH ^ A I C I + 3H2O Zn + N a O H + H - > N a Z n ( O H ) ^ + H Dap an diing la A Cdu A I O + 6HC1 AI2O3 + N a O H + 3H2O ^ N a A I ( O H ) ^ Ta c6: 65x - 56x = 9,6 - 7,8 = 1,8 9x = 1,8 x = 0,2 « , => thoa man) Vay ill y ( N H )^ C O + N a O H ^ 2NH3 + Na2C03 + 2H2O => Phai CO them P T H H sau: —> a 3) ( N H ) C + H C ^ N H C + C + H T h e o P T H H : m ^ ^ =0,12.65 = , ( g ) < , ( g ) X NaHC03+ HCl->NaCl + C + H O ill H :i.i NaHC03 + N a O H - ^ N a C + H2O Nd'u chi CO phan ling thi "^^(tg) ~ "'zn X ^ • CH3OH 1w n :crk mi t>•>(-''( ancol andehit fomic Theo djnh nghTa bac ciia ancol thi C H , O H la ancol bac "kh6ng" , ZnS04 (dd) + 2FeS04 (dd) Ni,r u v: H C H O + H2 D i p an diing la B I'fomV/ Chu y: Phuong phap giai bai la g6p ^n Kh6ng giai he phuong trinh duoc vi s6' in nhi^u hon s6' phuong t n n h Cdu 42: i' + Anken + V o n g canh xicloankan Cac C r C T CjHio tac dung duoc v6i dung djch Br,: 1) C H = C H - C H - C H - C H RCHO-^RCH^OH (b) : Sai, v i : 2) C H - C H = C H - C H - C H C6H6 + Br2 (nguyen cha't) 3) CH2 iC^HsBr + HBr C g H j O H 4- 3Br2 (dd) ->C6H2Br3 ( O H ) [ + H B r 4) C H = C ( C H ) - C H - C H (c) : Dung (2) C n H n + - k - a ( C H O ) ^ + (k + a)H2 - ,X 5) C H - C H = C ( C H ) C„H2„+2_, ( C H H ) ^ -CH2-CH3 6) andehit =CH-CH(CH3)2 ancol bac I (d) : Dung (3) 7) 2CH3COOH + C u ( H ) ^ (CHjCOO)^ Cu + 2H2O ^ H 8) H C (e) : Sai, vi phenol la axit rat yS'u (g): Dung (4): \ CH, Dap an dung la C C,H,CH(CH3)^ cumen Dap an dung la D 558 ;;?aH2S04 'CH3COCH3 + C,H,OH axeton phenol iCdu 4& Phat bi^u C sai Dap an dung la C 559 Ca'm rang 6n luy^n thi DH mi^n BSc - Trung - Mam m6n Hoa hgc - Cii I loSn Cty Cdu 47: 0,75.108.100% T h e o b a i r a : n , , = 12,96/108 = l , ( m o l ) Ta c6: % m , , ^ = ^ o l ,uM,„ = 15,73 R + R' = - - 4 = 29(C2H5) Dap an diing la A T a c o : m ^ = 2.(4,7.4) = 37,6(g) ^ V l khoi krcnig duoc bao toan, nen: m y = m ^ = 37,6(g) X la H o N - C H - C O O C H , mv H j N - C H - COOCH3 + N a O H ^ H N - C H C O O N a + C H O H :(mol) —> X X X V HCHO + 2Ag20 > X 4x I} j ; ^ n,.„|j^i,i,ph.i„,i„g) = n,|,|j|.„pi,j,n„,j,, = — ^ ' ' ( a i i c i i l dcm c h r r t l;io r a ) — R C H O + H2 JjSA ã ~>j:ôằI.;>!' ' (mol) ^ i~i * * (mol) ,'' ' RCH.OH R C H O H + Na ^ R C H , O N a + 0,5H2 T •3 0,5(mol) Imol ? nS 4x = 0,12 =^ X = 0,03 ( m o l ) 37,6 = l ( m o l ) 9,4.4 Suy so mol khf giam - = (mol) ,>j.|/l, ^"^ » A g i + C + H ^ ,^j,, , ,„ - / Cdu 50: , : ) H d 4-!A:: C H O H + CuO — — > H C H O + Cu + H O — Vay %m^(^^) = ' Suyra:RI»-CH,-;R'laCH,- =J> TNHH IV, TV DVVH Khang Viet Vay V = 0,5.22,4 = 11,2 (lit) oS Dap an diing la B Vay m = 89.0,03 = 2,67 (g) Chu y: So mol h6n hop khf giam - so mol H , phan ling = so m o l andchit don chuc, Dap an diing la C no, phan iTng = so mol ancol don churc tao Cdu 48: Trong qua gfc chin rat giau ham lucmg P-caroten (ti6n vitamin A ) Cdu 51: Dap an dung la B * Chu v: I X + NaHC03 CO2 + - Qua ga'c chu-a P-caroten cao ga'p 68 I5n ca chua (SGK hoa hoc 12 - Trang 190) V l n ^ = a < n(-Q^ = 1,6a < 2n^ = 2a - Trong ru6t non va tuyd'n giap trang xay qua trinh chuy^n hoa: => Trong X gom axit no, ho, don chuc va axit no, hc>, da chiic » enzimcarotenaza » Dat cong thirc chung cua hai axit la C„H2„_,_2-x ( C O O H ) ^ ( (X = va x = ) caroten > VttaminA Cdu 49: Gia sir nhiet phan m o l A g N O j : AgNOj—^Ag Imol + > Imol a(mol) H C O O H N a H C - > HCOONa + C O + H O y 560 _^ y HOOC - C O O H + 2NaHC03 -> NaOOC - COONa + C O + H O Dung djch Z: chiia m o l H N O 3 A g + H N O ^ A g N -I- N O + H O baft/ Z la (n = 0, X = 2): H O O C - C O O H ^l(mol) f^ ti, • ,«»? i (n + l ) a (mol) V a y a x i t Y l a ( n = 0, x = 1): H C O O H va 0,5 m o l O, 2HNO3 0,75 V - I m o l , Theo bai ra: (n + l).a = a => n + = =^ n = 0,5mol Y : mol N O , 2NO2 + ^ + H2O l->0,25 N02T+^02T Imol => X : m o l A g ; C„H2„^2-x(COOH)^-.(n + l ) H z Ta CO he: -> y + z =a y + z = 1,6a z = 0,6a; y = 0,4a 2z dm nang On luyPn thi BH mi^n BJc - Trung - Nam mOn H6a hpc - Cii Thanh To4n r» 46.0,43.100% Vay % m i ^ n n n ^ v ^ / v = y HCOOH(Y)/x Dap an dung la D Cdu Cdu = 25,41% - =, , ; ,,, Theobai ra: nAgN03 = 0,4.0,2 = 0,08(mol) , a + 90.0,6.a n^,, = , / = 0,09(mol) ' A !; Ni f : i, 52: P T H H xay k h i thuc hidn thi nghiem: (1): A g N ^ A g i + N O 4FeS2 54: + 1IO2 — ^ F e ^-(^^•^-f^).^ ; ^rr ^l-:' Vi 2nz„ > n A g N ( , > 0,08) =»Zn d u sau k h i cac phan ling xay hoan toan Cu + A g N ^ C u ( N \ A g i =4> X (ran): A g , Cu (chua phan ling) +8SO2 X :^ffj| Y (dung djch): A g N O , (chua phan ling), C U ( N ) K N O — ^ K N + ^ Zn + A g N O , ^ Z n (NO3 \ A g j CuS04+6NH3+2H20-^|Cu(NH3)J(OH)2+(NH4)2S04 (2) : Fe + C U S O ^ FeS04 + Cu j ,.' - Zn + 2FeCl3 (du) ^ 2FeCl2 + Z n C l j Zn + Cu (NO3 )2 ^ C U i + Z n (NO3 )2 => Z (ran): A g , Cu, Z n (du) Vay thuc chat ciia t h i nghiem trdn la: (3) : A g S + ^ A g i + S Zn + A g N O , ^ Zn (NO3 )2 + A g i Ba + C U S O + H O ^ C u ( O H ) + BaS04 + H 0,04 ^ 0,08 - » my.}: 0,08(mol) Dap an dung la B Do do: m x + m = m + " l A g + " ^ Z n ^^^^ Cdu 53: Xet cac phat bie'u: (a) : D u n g ( l ) 7,76 + , = m + 0,08.108 + (0,09 - , ) Glucozo (R - C H O ) + Brj + H O -> A x i t gluconic + 2HBr Dap an diing la A "AgNOj ' Glucozo (c) : Sai, v i : Trong mdi tru6ng k i ^ m ye'u (dung djch NH3) thi c6 sir chuyfi'n hoa fructozo , Cach khac: Frutozo +Br2 + H O -> Kh6ng • Ay (b) : Sai, v i : Frutozo < glucozo Do do, dung djch fructozo va glucozo d^u c6 kha nang tham gia phan ung trang bac -XI ?7 • AgN03, Cu(N03)3 canh Zn(N03) Zn (NO3 \ A g i 2AgN03 + Zn (g):Dung(3) 0,08 Dap an diing la C Theo djnh luat bao toan khdi luong, ta c6: Ml: ^ 0,04(mol) mAgN03 + " I Z n + " I C u = nix + Trong dung dich glucozo tdn tai: dang mach hcf (0,003%) ¥, + Zn (e) : Sai, v i : Trong dung dich, fructozo tdn tai chii yd'u dang P , vong canh hoac (36%) fl So ton-, tat: V i glucozo, fructozo d^u c6 tinh chat cua ancol da chiic Dang vong p - * =0,08mol;nz„ = , ( m o l ) V i 2n2„ = 0,18 > n ^ g N o , = 0,08 =^ Z n (du) AgN03 (d) : Dung (2) Dang vong a ^ C H H ( C H H ) ^ C H O ^ m = 6,40 ^ (64%) Con k h i kd't tinh, glucozo tao dang tinh t h ^ litig v6i dang cSu true vong (dang vong canh a va p) + m2n(N03)2 = m = m x + m z + m^^(^Q^)^ - iriAgNOj - "Zn =^ m = 7,76 + , + 0,04.189 - 0,08.170 - 5,85 = > m = 6,4(gam) •0,0 Trong dung dich, fructozo tdn tai chii yS'u or dang p, vong canh hoac canh) trang thai tinh the, fructozo dang P , vong canh AO 563 Chu y: Can 55: Theo bai ra: = 5,6/28 = 0,2(mol) H H ^ O =5,4/18 = 0,3(mol) '^^^ ' {' f ; d \ n =^ Ceo = 0,2/10 = 0,02M; C^^^^^^^^^ = 0,3/10 = 0,03M ^ Band: Pliiin iJng: Can bang: Ta c6: C0(k) + H ( k ) ; r i C ( k ) + H , ( k ) 0,02 x ^ 0,02-x 0,03 x -» 0,03-x +) C,,H220,, + H O - , ^ , „ v ' » - sf'i- ''^ C,2H220|, + H O ^H- 2C,H,20, glucozo fructozo ^"-^^"2" ' glucozo Luong mantozo chua bi thuy phan se tham gia phan urng trang bac Saccarozo khong tham gia phan urng trang bac •'I Co lien ket peptit ( - C O - HN - ) m3 Jiff § 1 Jdt => Phat bieu A khong diing Diip an diing la A uj^: ; =^ x^ = 0,0006 - 0,03x - 0,02x + x^ => 0,0006 = 0,05x =>x = 0,012 C2H,NH2 + HNO2 ^ C2H5OH + N2 + H2O Vay [CO] = 0,02-0,012 = 0,008M etyl amin H2O] = 0,03-0,012 = 0,018M etanol CH 3NH2 + H2O ^ CH^NH^ + O H " Dap iin diing la D raw 56: PTHH: metyl amin moi triromg bazo Can 58: xt,r Dung djch chudn la KMn04 : nKM„04 = ^'03.0,1 = 0,003mol -2C,H,20, 10FeSO4 + 2KMn04 + 8H2SO4 ^ 5Fe2 ( S O + K2SO4 + 2MnS04 + 8H2O 0,06 (mol) (0,02+0,01) _ 0,06.75 Vi H = 75% ntn n ^6^1206 ~ NH^ = 0,045(mol) [ 0,015 ^ , 0 ( m o l ) => Trong 20ml dung dich Y c6 0,015 mol FeS04 C,H,207+2Agi 0,045 => Trong 150ml dung dich Y c6 0,015.150/20 = 0,1125 mol FeSO 0,090(mol) Luong mantozo chira bi thuy phan: , = 0,01 - 0,01 75/ 100 = 0,0025 (mol) Trong dung djch, vi g6c glucozo thir hai c6 nhom OH hemiaxetal ntn c6 kha nang mo vong tao nhom chuc andehit CHO, do no c6 kha nang tham gia philn iJng trang bac n,.a„,o.„ ^"lFe.S()4 =0,1125.152 = 17,1(g) w „ 17,1.100% • = 68,4% vay %mFeso4/x = 25 Diip iin diing la D Can 59: ^*i»s Phiin lirng hoii hoc xiiy ra: C,2H220,, (mantozo) + A g - ^ ^ C , H 2 , + A g i Zn + Cu^+ ^ Cu i + Zn 2+ 0,0025 Suy ra: + Kh6'i luong circ Zn giam + Kh6'i luong cue Cu tang -> vay n A g = 0,090+ 0,005 = 0,095(mol) Dap an diing la C ; Dipeptit glyxylalanin (mach ho): H,N-CH2 -CO-HN-CH(CH,)-COOH x.x =^1 = ( , - x ) ( , - x ) 54 fructozo ^CduJZ- CO, • [ " ] = CO H20 CI2H220.1+H20- xt.r glucozo mantozo x(M) X C6H,206+C„H,20(, Saccarozo win,&Hl ' xl,l 0,005 (mol) 565 + N6ng d6 ion Cu"* giam + N6ng ion Zn"* tang Dap an dung la D Cdu 60 PTHH: : motainl mowMa CH3CHO + HCN-^CH3CH(OH)(CN) CH3CH(0H)(CN) + 2H-f> Q ,i ,v H,< , I C H C H ( Q H ) C Q O H + NH3 (X,) (x^) => X, la axit - - hidroxipropanoic (axit lactic) C2H,Br + M g - ^ C H , M g B r , '' (^') CjH^MgBr + CO, ^C.H^COOMgBr (Y.) (Y2) C2HsCOOMgBr + H C l C j H s C O O H + ClMgBr ' (Y2) (Y3) => Y, la axit propanoic Dap an diing la B :hu v: Thirc te phan ling thiiy phan hcfp chat xianohidrin m6i trtircfng axit CH3CH(OH)(CN) + 2H20 + H + — ^ C H C H ( H ) C 0 H + N H J Chi cin xac djnh diing X, la chon duoc dap an diing ^* Tinh chat ciia hap chat ca magie: tac dung nhanh v6i nhirng hap chat c6 hidro linh d6ng (nhu niroic, ancol, ) va tac dung vcfi cacbonic (COi), ^ RMgX + H2O ^ R H + HOMgX RMgX + R ' O H ^ R H + R'OMgX RMgX + CO2 ^ RCOOMgX Chu6i (1, 2) la cac phuang phap tang mach cacbon dieu che axit cacboxylic 56 MUC LUC Phan I Cac phuang phap giai nhanh bai tap hoa hoc Phan II Cac bo de thi thu Dai hoc va Hudng dfin giai chi tiet Bo de thi so De thi thu \k\ nam 2013 - Tnxcmg THPT Chuyfin DHSP Ha N6i 85 Bo de thi so De thi thir Idn nam 2013 - Truomg THPT Chuyen KHTN 105 Bo de thi so De thi thir ISn nam 2013 - Truotig THPT Phan Oinh Phung 123 Bp de thi so D^ thi thir Idn nam 2013 - Trircmg THPT Chuyen BSc Ninh 150 Bp de thi so De thi thir l^n nam 2013 - Trucmg THPT DL Hoang Xudn Han 176 Bp de thi so D^ thi thuf \in nam 2013 - Truomg THPT Chuyen Le H6ng Phong 205 Bp de thi so De thi thir \in nam 2013 - Trucmg THPT Chuyen Le H6ng Phong 218 Bp de thi so thi thir Mn nam 2013 - Truomg THPT Chuyen Le H6ng Phong 230 Bp de thi so Dl thi thir dai hoc Mn Trircfng THPT Chuyen Dai hoc Vinh 242 Bp de thi so 10 De thi thir dai hoc \in trircmg THPT Chuyen Dai hoc Vinh 272 Bp de thi so 11 De thi thir dai hoc \in tiircmg THPT Chuyen Dai hoc Vinh 300 Bp de thi so 12 Y)i thi thir dai hoc l^n truomg THPT Chuyen Dai hoc Vinh 328 Bp de thi so 13 De thi thu dai hoc Ian trucmg THPT Phan Dinh Phung 358 Bp de thi so 14 De thi thir dai hoc truomg THPT Phan Chau Trinh 386 Bp de thi so 15 De thi thu Vin trucmg THPT DL Hoang XuSn Han 407 Bp de thi so 16 De thi thu dai hoc Idn truomg THPT Le Quy Don 428 Bp de thi so 17 De thi thu dai hoc \in truomg THPT Le Quy Don 453 Bp de thi so 18 De thi thir dai hoc \in nam 2013- TT LTDH Luang The'Vinh 477 Bp de thi so 19 De thi thiJ dai hoc \in nam 2013- TT LTDH Luong The' Vinh 506 Bp de thi so 20 De thi thir dai hoc Idn nam 2013- TT LTDH Luong The' Vinh 532 ... ,fr + Fe203 Ban ddu: Phan ling: 0 ,25 2x 0,15 Con: (0 ,2 5 -2 x) 2Fe + AI2O3 X 2x (0,15-x) 2x X 2A1 + 2NaOH + 6H2O (0 ,2 5 -2 x) -> 2Na Al(OH) I + 3H2 T 1,5(0 ,2 5 -2 x) ' AI2O3 + 2NaOH + 3H2O X 2Na Al(OH)^... + l , ) x - n x = 0,375 =^x = 0 ,25 Q , , 1) - H O - C H - C H O H 2) m - HO - C6H4 - CH2OH 3) P - H O - C H -CH2OH 4) - H - C H - - C H 5) m - H O - C H - - C H 6) p - HO - C6H4 - O - CH3 ( ,... 1 ,25 5 gam muO'i C6ng thiic ca'u tao ciia X la: A H2N - CH2 - CH2 - CH2 - COOH B H2N - CH2 - COOH C H2N - CH2 - CH2 - COOH D CH2 - CH(NH2) - COOH Cau 20 : Cho cac dung dich khOng mau: HCOOH, H^N-CHrCOOH