2012 by Box Ha hc www .boxmat h.vn TUYN TP CC BI TP HA HC HAY V LI GII CA DIN N BOXMATH BoxMath, thng 1 2012 MC LC LI NG ...................................................................................................................................................................... 1 DANH SCH THNH VIN THAM GIA BIN SON ................................................................................. 2 PHN 1. CC BI TON V C .......................................................................................................................... 3 Bi 1 10 ................................................................................................................................................................... 3 Bi 11 20 ................................................................................................................................................................. 9 Bi 21 30 ............................................................................................................................................................... 13 Bi 31 40 ............................................................................................................................................................... 17 Bi 41 47 ............................................................................................................................................................... 22 PHN 2. CC BI TON HU C..................................................................................................................... 25 Bi 1 10 ................................................................................................................................................................. 25 Bi 11 20 ............................................................................................................................................................... 29 Bi 21 30 ............................................................................................................................................................... 33 Bi 31 40 ............................................................................................................................................................... 37 Bi 41 50 ............................................................................................................................................................... 41 Bi 51 54 ............................................................................................................................................................... 46 1ht t p:/ / boxmat h.vn LI NG C nhiu ngi nhn xt Ha hc l b mn ca tr nh, hc ha phi nh rt nhiu th iu c th ng, nhng cha phi l tt c. Ha hc giu trong lp o nng n ca nhng cng thc, nhng phn ng phc tp l mt v p tinh t ca s t duy logic. V p y c bit c hin din qua nhng bi ton ha, nhng bi ton cho bn rt nhiu con ng, nhng s ch c mt con ng p nht, ngn nht i n c chn l. Nhim v ca ngi hc ha l phi vn dng cc thao tc nh gi, phn on tm ra c con ng y, con ng khng ch a n kt qu m con a n mt nim vui, nim hng khi nh mt cht keo gn tri tim bn vi Ha hc. Tng c nhng giy pht v a trong sung sng tng c nhng khonh khc chi vi, b tc gia nhng sliu,nhngphngtrnhphnngChngtitinhnhlmtuyntpnyvimcchutinl nimmongmunngcm,lkhtkhaochiastnhyuHahccachngtiviccbnthngqua nhng bi tp ha mi l, nhng li gii hay. Mi bi tp khng ch n gin l tnh ton, ng sau l nhng tng. Mi li gii khng ch l p dng phng php m thc s l mt qu trnh phn tch v sng to. Ha hc ang c tin hnh thi theo phng php trc nghim, th nhng li gii u tin m chng ti aravnlunlunlmtligiiy,itnhngdkincabitonnktqumhonton khng ph thuc vo cc p n cho trc. Bi chng ti hiu v mong cc bn s hiu, ch c nh vy, chng ta mi c th i ht c v p ca ha hc, rn luyn c t duy suy lun logic cho bn thn, hc tp mt cch thc cht v sng to. Tuyntp chyul stng hpvchnlc cc biton ha hctrn dinn Boxmath.vn nm2011. Chnh v vy, ban bin tp xin c chn thnh cm n s ng h ca ban qun tr din n, s tch cc vit bi v gii bi ca cc thnh vin trong sut mt nm qua. S n ch ca cc bn vi tuyn tp l mt ng lc rt ln gip ban bin tp c th hon thin cng vic nhiu ln tng chng nh phi b d dang. Mcdcscgng,xemxtklng,nhngchcchnskhngtrnhkhinhngkhimkhuyt, mong cc bn thng cm v gi li nhn xt bnh lun ca cc bn v cho chng ti tuyn tp c hon thin hn.Mi kin xin gi v hoahoc.boxmath@gmail.com Thay mt ban bin tp ti xin chn thnh cm n. H Ni, ngy 18/1/2012 i din nhm bin son Ch bin F7T7 2ht t p:/ / boxmat h.vn DANH SCH THNH VIN THAM GIA BIN SON 1.Ph Tin Cng THPT chuyn i hc S phm H Ni 2.Thi Mnh Cng THPT chuyn Phan Bi Chu, Ngh An3.Trn Bo Dng THPT Ng Quyn, B Ra Vng Tu 4.Nguyn Th Thu Hi THPT Trn Ph, H Tnh5.Nguyn Quc Oanh THPT So Nam, Qung Nam6.Nguyn Hu Ph THPT Ty Sn, Lm ng7.Phan Qunh Nga THPT Hng Kh, H Tnh 3ht t p:/ / boxmat h.vn PHN 1. CC BI TON V C Bi 1 10 Bi 1. _________________________________________________________________________________Hn hp A gm mui sunfit, hidrosunfit v sunfat ca cng mt kim loi kim M. Cho 17,775 gam hn hp A vo dung dch 2( ) Ba OHd, to thnh 24,5275 gam hn hp kt ta. Lc kt ta, ra sch v cho kt ta tc dng vi dung dchHCld, thy cn 2,33 g cht rn. Kim loi kim M l A.LiB.KC.RbD.Na Li gii. Phng trnh phn ng 2 24 4Ba SO BaSO+ + 23 2 3OH HSO H O SO + +2 23 3Ba SO BaSO+ + Cht rn cn li l 4BaSO: 42, 330, 01( )233BaSOn mol = =3(24, 5275 2, 33)0,1023( )217BaSOn mol= =17, 7751580,1023 0, 01M = =+tr ung bi nh Ta c 80 158 2 96TBM M M + < < + = . Ch c M = 39 tha mn iu kin ny.Chn p n B. Bi tp tng t. Hn hp X gm 2 3 3, MCO MHCOvMCl(M l kim loi kim). Cho 32,65 gam X tc dng va vi dung dchHClthu c dung dch Y v c 17,6 gam 2COthot ra. Dung dch T tc dng vi dung dch 3AgNOd c 100,45 gam kt ta. Kim loi M l A.NaB. LiC. KD. Rb Bi 2. ______________________________________________________________________________ Cho mt hp cht ca st tc dng vi 2 4HSOc nng, to ra 2SO(sn phm kh duy nht). Nu t l 2 4HSOem dng v 2SOto ra l 2 4 2: 4:1HSO SOn n =th cng thc phn t ca X l: A.FeB.FeSC.FeOD.Fe3O4 Li gii. t ( )2SOn amol =Ta c:-Nu X l oxit ca st th ta c qu trnh kh l ( ) ( )6 42a2mol amolS e S+ ++ ( )2 44 3HSOn a a amol = = tao muoi ( ) ( ) 32 4 3( )2Fe SOFen amol n amol+ = =( ) ( )32 22 2(3 ) 2xamol amolFe x e Fe+ ++ 2 x =Ngoi ra khng cn nguyn t no nhng hay nhn e Vy cng thc X lFeO 4ht t p:/ / boxmat h.vn -Nu X l mui sunfua ca st th X c CTPT dng x yFe S (a mol) Ta c cc qu trnh oxi ha kh ( ) ( )6 422mo b b l molS e S+ ++ ( (2) ) ( )31ax mol ax mol ax molFe e Fe+ + +244( ) ( )2 ( )24ay mol ay molay ax moylxxS S ey++| | + |\ . Vy 2( )SOn ay bmol = + ,( ) 6 22 4432HSOS SO taomuoiaxn n n b mol+ = + = +Suy ra( )34( ) 12axb ay b + = +Theo nh lut bo ton electron ta li c( ) 2 (4 2 ) 2 b ax ay ax = + Gii h phng trnh (1) v (2) suy ra 10 3, : 10: 3b bx y x ya a= = =(khng tn ti hp cht no tha mn iu ny). Vy trng hp th hai khng tha mn. Kt lun: Ch c FeO tha mn bi. Chn p n C. Bnh lun: Vi trng hp X l oxit ca st, d nhn thy 2 3Fe Ophn ng khng to ra 2SO , ch xtFeOv 3 4Fe O .Tathy1molcahaichtnykhiphnngunhng1molelectron.Vynut ( )2SOn amol = thsmolcaFeO v 3 4Fe O ul2 ( ) amol .Nhngchc FeO mitoracmui cha 243 ( ) amol SO tha mn yu cu bi ra. Chn ngay FeO, p n B. Bi tp tng t. X l mt hp cht ca Fe. Cho X tc dng vi 2 4HSOc nng thy thot ra kh 2SOvi t l mol X v 2SOl 2:9. X l: A. 3 4Fe O B.FeS C. 2FeS D.FeOp s: B. FeS Bi 3. _____________________________________________________________________________Cho t t a gam st vo V ml dung dch 3HNO1M khuy u cho n khi tan ht thy thot ra 0,448 lt kh NO (ktc) ng thi thu c dung dchA. Dung dchA c kh nng lm my mu hon ton 10 ml dung dch 4KMnO0,3M trong mi trng axit. Gi tr caav Vl: A. a =1,4 gam; V = 80 ml B. a = 1,12 gam; V = 80 ml C. a = 0,56 gam; V = 56 mlD. a = 0,84 gam; V = 60 ml Li gii. Ta c cc phng trnh th hin qu trnh oxi ha kh: ( )( )5 20,06 0,023 1 N e N NO+ ++ ( )333 2x x xFe Fe e+ +( )222 3y y yFe Fe e+ +7 20,003 0,0155 (4) Mn e M+ ++ 2 31 (5)yyFe Fe e+ + +Theo phn ng kh 4KMnO , t (4) v (5) ta c0, 015 y =Theo phn ng kh 3HNO , t (1), (2) & (3), ta c3 2 0, 06 0, 01( ) x y x mol + = = 5ht t p:/ / boxmat h.vn Vy0, 025 1, 4Fen m g = =33 33 2 0, 02 0, 08( )HNONO NOn n n x y mol = + = + + =tao muoibikhu Suy ra( ) 80 V ml = Bi 4. _____________________________________________________________________________ Hn hp X gm c,x yAl Fe O . Tin hnh nhit nhm hon ton( ) m ghn hp X trong iu kin khng c khng kh thu c hn hp Y. Chia Y thnh hai phn. Phn 1. Cho tc dng vi NaOH d thu c 1,008 lt 2H(ktc) v cn li 5,04 gam cht rn khng tan. Phn 2 c khi lng 29,79 gam, cho tc dng vi dung dch 3HNOlong d thu c 8,064 lt NO (ktc, l sn phm kh duy nht). Gi tr ca m v cng thc ca oxit st l A. 39,72 gam &FeO B. 39,72 gam & 3 4Fe OC. 38,91 gam &FeOD. 36,48 gam & 3 4Fe OLi gii. Cch 1 2 32 3 3 (1)x yyAl Fe O yAl O xFe + + Phn ng nhit nhm hon ton m phn 1 to kh 2Hnn hn hp Y gm 2 3, , Al FeAl O d-Phn 1 ( )Aln amol = ; ( )Fen bmol = ; ( )2 3Al On cmol =Phng trnh phn ng 2 2 20,030,0452 2 2 2 3 Al NaOH H O NaAlO H + + +5, 04Fem m g = =ran. Suy ra ( ) 0, 09Fen mol =-Phn 2 ( )Aln kamol = ; ( )Fen kbmol = ; ( )2 3Al On kcmol =Theo nh lut bo ton electron ta c:( ) 3 3 .0,123. 0, 36.3 3NOk a b k n k + = = =29, 79 (27.0, 09 56.0, 27)0, 04( )3.102c mol += =29, 79.439, 723m g = =T (1) ta c0, 04.3 0, 09 4 3 x y x y = =Chn p n B Cch 2. -Phn 1 Sau khi tc dng vi NaOH to ra kh nn sau khi nhit nhm th Al cn d0, 03Aln = Cht rn cn li l Fe0, 09Fen mol =13AlFenn =-Phn 2 Gi s mol ca Al l a th s mol ca Fe l 3a.Bo ton electron ta c: 12 1, 08 0, 09 a a = =2 3 2 312, 24 0,12AlO Al Om n = = 6ht t p:/ / boxmat h.vn Suy ra phn 1 c 2 30, 04Al On =39, 72( ) m gam =Da vo s mol ca O v Fe suy ra CTPT ca oxit l 3 4Fe O . Bi 5. ______________________________________________________________________________ in phn in cc tr dung dch cha 0,2 mol 3AgNOvi cng dng in 2,68 A, trong thi gian t (gi) thu c dung dch X (hiu sut ca qu trnh in phn l 100%). Cho 16,8 gam btFevo X thy thot ra khNO (sn phm kh duy nht) v sau cc phn ng hon ton thu c 22,7 gam cht rn. Gi tr ca t l: A. 2,00B. 1,00C. 0,50D. 0,25 Li gii. 3 2 3 24 2 4 4 AgNO H O Ag HNO O + + +Gi x l s mol3AgNOd v y l s mol 3HNOto ra. Tng s mol ca 3AgNO d v 3HNOchnh bng s mol 3NO khng i. Do , ta c 30, 2NOx y n+ = =Agti a to ra cng ch 0,2 mol tc l khi lng cht rn sau phn ng nh hn 21,6 gam. Vy trong cht rn cnFed. 22 2 Fe Ag Fe Ag+ ++ +3 3 3 24 ( ) 2 Fe HNO FeNO NO H O + + + Suy ra s mol Fe phn ng l30.58yx + S mol Ag to ra l x, vy c 3108 (0.5 .56) 22, 7 16, 88yx x + = T tm c0,1 x y = =hay3600( ) 1( ). t s h = = Bi 6. _______________________________________________________________________________ Cho 8,64 gamAlvo dung dch X (X c to thnh bng 74,7 gam hn hp Y gm 2CuClv 3FeClvo nc). Kt thc phn ng thu c 17,76 gam cht rn gm hai kim loi. T l mol ca 3 2: FeCl CuCltrong hn hp Y l: A.2:1 B. 3:2C. 3:1D. 5:3 Li gii. -Nu d Al th chc chn hn hp kim loi phi cha c 3 kim loi Al, Cu v Fe. Do Al phi ht sau phn ng v hai kim loi cn li lCuvFe . -DoFenn chc chn c 3 phn ng sau xy ra theo th t: 3 3 233 e 3 exx xAl FCl AlCl FCl + +2 3232 3 2A 3yy yAl CuCl lCl Cu + +2 3322 3 3 3zx zAl FeCl AlCl Fe + + ( y 3 2x z > , do Al phn ng ht. Ta c 20, 32( ) 0, 323 3Alxn mol y z = + + =Khi lng hai mui ban u l 74,7 gam, suy ra.162, 5 .135 74, 7 x y + =Khi lng hai kim loi thu c l 17,76, suy ra 364 56. 17, 762x z + = 7ht t p:/ / boxmat h.vn Gii h ba phng trnh trn suy ra:0, 36; 0,12; 0,12 x y z = = =Vy: 3:1 x y = . Chn p n C. Bi 7. _______________________________________________________________________________Cho 240 ml dung dch 2( ) Ba OH1M vo 200 ml dung dch hn hp 3AlCla mol/lt v 2 4 3( ) Al SO2a mol/lt thu c 51,3 gam kt ta. Gi tr ca a l: A.0,12B. 0,15C. 0,16D. 0,2 Li gii. Trong cc dung dch c 0,24 mol 2; Ba +0,48 mol, OHa mol 3; Al +0,6a mol; Cl1,2a mol 24SO . Xt bng sau da theo cc gi tr ca a a 0,120,160,2 Nhn xt4 0, 48; a < 1, 2 0, 24 a + 0, 73469xy < . Vy ch c 2 3Fe Otho mn. Bi 38. _____________________________________________________________________________ Dung dchX cha 14,6 g HCl v 22,56g 3 2( ) CuNO . Thm( ) m gFe vo dung dchXsau khi phn ng xy ra hon ton thu c hn hp kim loi c khi lng 0,5m v ch to NO (sn phm kh duy nht). Gi tr cam l A. 1,92 g B. 20,48g C. 14,88 D. Khng tn ti m tha mn Li gii.Dung dch X cha 0,4 mol, H+ 0,12 mol 2Cu + v 0,24 mol 3. NO Do sau phn ng cn hn hp kim loi d nn khng cn d ng thiH+ v 3, NO ng thi ch cn mui 2, Fe +cn 2Cu + phn ng ht. Cc phn ng xy ra : 23 20,4 0,150,243 e 8 2 3 e 2 4 (1) F H NO F NO H O+ ++ + + + 21ht t p:/ / boxmat h.vn 2 20,12 0,12 0,12(2) Cu Fe Cu Fe+ ++ +Ta c 0, 4 0, 248 2< nn (1)30, 4. 0,15( )8Fen mol = =Gi m l khi lng Fe ban u Khi lng hn hp kim loi sau phn ng:64.0,12 56.(0,12 0,15) 0, 5 14, 88( ) m m m gam + + + = =Bi ton tng chng nh gii quyt xong nhng Khi lng st phn ng l ' 56(0,12 0,15) 15,1 2 14, 88 m = + = >(v l). Vy khng tn ti m tha mn. Chn p n D. Bi 39. _____________________________________________________________________________Cho11,15gamhnhpgm haikimloiAlvkimloikimMvotrong nc.Sau phnngthuch c dung dch B v V lt kh (ktc). Cho t t n 400 ml dung dch HCl 1M vo dung dch B. Trong qu trnh thu c lng kt ta ln nht l 15,6 gam, sau kt ta tan mt phn. Kim loi kim l: A.BaB. NaC. KD. Khng d kin Li gii.Nhn xt. Do sau phn ng thu ch c dung dch B v kh, vy nn Al phn ng ht v bazo c th ht hoc d. 12 2 2x xM HO MOH H + + ; 32 2 20,2 00 2,2,Al MOH MAlO H + + (d0, 2 x mol MOH ) 2OH H H O ++ ; 2 2 30,20,2 0,2( ) H H O MAlO Al OH M+ ++ + +Do kt thc phn ng th 2, kt ta tan mt phn nn0, 4 ( 0, 2) 0, 2HCln x x = > + =Ta c0, 2 0, 4, x < 33OH Aln n +>2 24Ba SOn n+ = > 2349, 5 m V+ = . 1 2168, 5 427, 5 V V =1 2: 2, 537 3 V V < (nhn) 2 1 2973, 5 156 427, 5 V V V =1 2: 3, 5 V V =(nhn) loi Kt lun. C hai gi tr 2,537 v 3,5 u ng. Chn p n C. Bnhlun.Nhnxtvkhilngkttatanhnthykhilngkttatcciti 1 2: 3 V V = v 2 1 (max)504 168 m V V+= = . Bi ton c th hi theo mt hng khc l tnh khi lng kt ta cc i. y, 22ht t p:/ / boxmat h.vn 1 2: 3 V V =khin ta ngh ti t l 3:OH Aln n + khi lng 3( ) Al OHt cc i .Tuy nhin y ch l mt s trng hp th v. Cc bn s mc sai lm khi p dng iu ny cho cc bi ton tng t. Bi 41 47 Bi 41. _____________________________________________________________________________ Phn ng no sau y sai. A. 2HI + Cu CuI2 + H2 B. 2HBr + 2FeCl3 2FeCl2 + Br2 + 2HCl C. H2O2 + KNO2 H2O + KNO3 D. 2SO2 + O2 o5 2V O , t 2SO3 Li gii.Phn ng B sai. 2HBr + 2FeCl3 2FeCl2 + Br2 + 2HCl. HBr khng tnh kh mnh kh c Fe3+. HI mi c phn ng ny. Thm ch ta cn c phn ng 2 2 3 36 3 4 3 FeCl Br FeCl FeBr + + Bi 42. _____________________________________________________________________________ Chn pht biu ng nht A. Trong kh than kh, hm lng CO nhiu hn kh than t B. Trong phng thi nghim, CO c iu ch bng cch thi CO2 qua C nung nng trn ngn la n cn C. CO c thu bng cch y khng kh (p ngc bnh) D. CO c dng lm nhin liu kh Li gii.A sai, v theo s liu c cho trong SGK, hm lng CO trong kh than t nhiu hn trong kh than kh.B sai, trong phng th nghim, ngi ta iu ch CO bng cch nhit phn HCOOH.C sai. Ta c CO = 28, ch hi nh hn khng kh, nn cch tt nht thu CO l y nc (CO khng tan trong nc). Chn p n D. Bi 43. _____________________________________________________________________________ (1) Ngi ta khng dng CO2 dp tt cc m chy c Al, Mg do Al, Mg c th chy trong CO2 to hp cht cacbua (2) Trong cng nghip, kh CO2 c thu hi trong qu trnh ln men ru t glucz (3) CO km bn nhit, d tc dng vi oxi to CO2 (4) Trong t nhin, Cacbon ch yu tn ti trong cc khong vt: canxit, magiezit, dolomit,... C bao nhiu pht biu ng trong cc pht biu trn? A. 1B. 2C. 3D. 4 Li gii.Pht biu sai l : (2) do Al, Mg chy trong CO2 to oxit, khng to hp cht cacbua; (4) sai v u tin, v CO rt bn nhit; (5) sai do C cn tn ti vi t l ln dng n cht (than), trong cc hp cht hu c. Cu 44. ____________________________________________________________________________Ha tanmgam hn hp X gm, Cu Alvo dung dchHCld thy c 2 gam cht rn khng tan. Nu trn thm 4 gamMgvo 0,5m gam X th c hn hp Y. Hm lng % theo khi lng caAltrong Y nh hn trong X l 33,33%. Khi cho Y tc dng vi dung dchNaOHth th tch 2Hthu c ktc nhiu hn 2 lt. Thnh phn phn trm theo khi lng caCu trong X l: 23ht t p:/ / boxmat h.vn A.40%B.16.67% C.18.64%D.30%Li gii. 2 gam cht rn khng tan chnh l Cu. Suy ra trong m gam X c 2 gam Cu, (m 2) gam Al Hm lng % theo khi lng ca Al trong Y l:240, 5( 2) 2 33, 330, 5 8 24 012 0, 5 4 100mm mm mm m m= = + = = + Li c 22( )HV lit =nn 0, 5( 2) 3 2. 5, 21427 2 22, 4mm> >Vy12 m =Suy ra 2% .100% 16, 67%Cumm= =Chn p n B. Bi 45. ____________________________________________________________________________ Nung10gamhnhpgm3 3 3, , MgCOCaCO BaCO thuckhA.DnkhAvodungdchncvi trong d thu c 5 gam kt ta v dung dch B. un nng B hon ton th thu c thm 3 gam kt ta na. Hi % khi lng 3MgCOtrong hn hp ban u nm khong no? A.52,5% n 86,69%C. 52,5% n 60,64% B.60,64% n 86,69%D. 86,69% n 90,45% Li gii. Cch 1. D dng tnh c: 20,11( )COn mol =Gi s mol ca 3 3 3, , MgCOCaCO BaCOln lt l, , xyzTa c h phng trnh: 0,11 0,11100 197 10 8484 100 197 10 100 197 10 84 0, 11x y z y z xy z xPx y z y z x y z x+ + = + = + = = + + = + = + Do0 z =nn:100 197100 1971yzyzy zPy z++= =++ t 100 197( 0) ( )1y tt t P f tz t+= > = =+ Kho st hmftrn(0; ) +cho ta: ( )10 84 84100 197 100 197 100 197 0, 0625 0,10327 52, 5% 86, 75%0,11 10x xf t P xx< < < < < < < < < < Chn p n A. Cch 2.Ta cng c xx 0,1 16797230,11 (1) 0,11 (1')84 100 197 10(2) 100 1 197 (2') 0 84 z xzx y z y z xx y z yy= + + = + = + + = += = Ta c( )84 84 588x%9 0, 23 1 1684 100 197 2168x 17484 100. 197.7 7Mgx xm f xx xx y zx= = = = + + ++ + Ta li c( ) ( ) ( ) ( ) 100 100. 0,11 10 84x 100 197z 197 197. 0,11 y z x y y z x + = < = + < + = .Suy ra0, 0625 0,1038 x < < . Kho st hm( ) f x trn( ) 0, 0625; 0,1038 ta c kt qu nh cch 1. 24ht t p:/ / boxmat h.vn Bi 46. _____________________________________________________________________________ Ha tan hon ton m gam hn hp gm 2FeSv 2Cu Strong dung dch 3HNOsau cc phn ng xy ra hon ton thu c dung dch X ch c hai cht tan khi lng cc cht tan l 72. Gi tr ca m l A. 20B. 40C. 60D. 80 Li gii. Hai cht tan l: 2 4 3 4( ) ( ); ( ) Fe SO a mol CuSO b mol 400 160 72 (1) a b + =Theo bo ton nguyn s Fe v Cu th:2 22 ;2FeS CuSba n n = = 42Sbn a = +M 243 4 3 (2)2SObn a b a a b = + + = +T(1)v(2) 0,1; 0, 2 a b = =Do : 2 .120 .160 402bm a = + = . Chn p n B. Bi 47. _____________________________________________________________________________Hin tng g xy ra khi cho 2HSli qua dung dch 3FeCl(mu vng)? A.Khng c hin tng g. B.Mu vng mt dn v c kt ta en xut hin. C.Mu vng mt dn v c kt ta mu vng xut hin. D.Mu vng mt dn v c kt ta mu nu xut hin. Li gii. Phn ng xy ra l: 2 3 2HS FeCl FeCl S HCl + + + +Nh vy, sau phn ng s c kt ta vng, chnh l lu hunh. Chn p n C. Bnh lun. y l mt cu hi l thuyt hay, c bit nhng p n nhiu thc s gy nhiu. Kt ta en c nhc ti cu B, ngi ra mun nhc ti FeS ( trnh nhm ln, bn ch cn nh phn ng ny to raHCl , mFeStan trongHClnn khng th c kt taFeS ). Kt ta nu cu D li mun m ch 3( ) , Fe OHnhiu bn c th nhm ln y l mt phn ng axit baz bnh thng, to ra 2 3, Fe Ssau 2 3Fe Sb thy phn to ra 3( ) . Fe OH (Nhng chc chn phn ng khng th xy ra theo hng ny, v 2HSl mt axit rt yu, khng th phn ng to raHCll mt axit mnh hn). 25ht t p:/ / boxmat h.vn PHN 2. CC BI TON HU C Bi 1 10 Bi 1. _____________________________________________________________________________ Oxi ha 9,2 gam ru etylic bngCuO un nng, c 13,2 gam hn hp (andehit, axit, nc v ru cha phn ng). Hn hp ny tc dng viNato ra 3,36 lt 2H(ktc). Phn trm khi lng ru b oxi ha l: A. 75%B. 25%C. 66,67%D. 33,33% Li gii. Cch 1 Ta c ( )( )( )2[ ]0, 20, 30,1513, 2 9, 20, 2516ruouOHHOn moln moln moln mol = === = Phng trnh phn ng2 5 3 23 3[ ][ ]C H OH O CH CHO H OCH CHO O CH COOH+ ++ Suy ra 3 2( ) [ ]0, 25( )OH trong CHCOOH va HO On n mol = =( )( )3 2( )0, 3 0, 25 0, 050, 2 0, 05 0,150,15% .100% 75%0, 2OH OHtrong CHCOOH va HOn n n moln molH = = = = == =ru dru phn ng Cch 2. Ta c cc phng trnh phn ng ( ) ( ) ( )( ) ( ) ( )2 5 3 22 5 3 2[ ]2[O] CH OOxmoyl xmol xmoly mol y mol molC H OH O CH CHO HOC H OH C H H O+ ++ + V d z (mol) 2 5C H OHTa c PTP ( )( )( )( )2 223 3 222 2 22 2 2x y x ymoly ymolH O Na NaOH H molCH COOH Na CH COONa H mol+ ++ ++ + Theo nh lut bo ton khi lng [ ] [ ]13, 2 9, 2 4( ) 0, 25( )O Om g n mol = = =Ta c h phng trnh0,150, 052 2 22 0, 25 0,10, 2 0, 05y x y zxx y yx y z z+ + + == + = = + + = = 26ht t p:/ / boxmat h.vn Vy 0, 05 0,1% 75%0, 2H+= = Bi 2. _____________________________________________________________________________Thy phn hon ton 60 gam hn hp hai ipeptit thu c 63,6 gam hn hp X gm cc amino axit (ch c 1 nhm amin v 1 nhm cacboxyl). Nu cho 110 hn hp X tc dng viHCl (d), c cn cn thn dung dch, th lng mui khan thu c l: A. 7,82B. 16,3C. 7,09D. 8,15 Li gii. 22.(63, 6 60)2 0, 418X HOn n mol= = =(do ipeptit thy phn ra 2 amino axit) 0,04 02 3,04NH R COOH HCl NH R COOH Cl + p dng LBTKL:36, 36 0, 04.36, 5 7, 82Cl NH R COOHm g= + =Chn p n A. Bi 3. _____________________________________________________________________________ HnhpXgm 2 1 2 1 2 1 2, ,n n n n n nC H CHO C H COOH C H CH OH (umchh,nnguyndng).Cho2,8 gamXphnngvavi 8,8gam bromtrongnc.Mtkhc choton blngXtrn phn ngvi lng d dung dch 3AgNOtrong 3NH , kt thc phn ng thu c 2,16 gamAg . Phn trm khi lng ca 2 1 n nC H CHO trong X l: A. 26,63%B. 22,22%C. 20,00%D. 16,42% Li gii. Cch 1. Gi s mol ca cc cht trong hn hp X ln lt l, , xyzTa c 2 10, 02( ) 0, 01( )n nAg CH CHOn mol x n mol= = =Xt phn ng vi dung dch brom, ta c mi cht trong X u phn ng vi brom lin kt i, ring 2 1 n nC H CHO cn c phn ng gcCHO Ta c 20, 055( ) 2 0, 055Brn mol x y z = + + =Vy suy ra tng s mol cc cht l0, 055 0, 01 0, 045 n x y z = + + = =Do khi lng phn t trung bnh ca cc cht l 2, 862, 20, 045TBM = ~Nn 2 1 2 114 28 62, 2 14 44 1, 3 2, 5 2n n n nCH CHO TB CH COOHM M M n n n n < < + < < + < < =Phn trm khi lng 2 1 n nC H CHO trong X l 0, 01.56% .100% 20%2, 8M = =Chn p n C. Cch 2. T phn ng vi Ag, suy ra0, 01andehitn =Xt phn ng vi dung dch brom, li suy ra &0, 035ancol axitn =Do axit ancolM M >Suy ra:( ) ( ) ( ) ( ) 0, 01. 14 28 0, 035. 14 30 2, 8 0, 01. 14 28 0, 035. 14 44 n n n n + + + < < + + + 27ht t p:/ / boxmat h.vn Dn n: 1, 5 2, 4 n < < . Li do n nguyn dng nn2 n = .Phn trm khi lng 2 1 n nC H CHO trong X l 0, 01.56% .100% 20%2, 8M = = . Chn p n C. Bi 4. ____________________________________________________________________________Choccchtlngkhngmu:dungdch 3, NaHCO dungdch 2, NaAlO dungdch 6 5, C H ONa6 5, C H OH6 5 2, C HNH3 2 5, CH COOC H2 5, C H OH3 7C H OH . Ch dng thmHClc th nhn ra s lng cht lng l? A.5B. 7C. 8D. 6 Li gii. Hin tng khi cho cc dung dch HCl vo cc cht lng khng mu l -Dung dch 6 5C H ONa : to ra kt ta 6 5C H OHkhng tan trongHCld. -Dung dch 2NaAlO : ban u xut hin kt ta 3( ) Al OH , sau kt ta tan trongHCld. -Dung dch 3NaHCO : xut hin bt kh 2( ) CO- 6 5C H OH : dung dch phn lm hai lp - 6 5 2C HNH : dung dch ban u phn lm hai lp ri dn dn ha tan vo nhau - 3 2 5CH COOC H : dung dch phn lp, un nng th to ra dung dch ng nht, c mi gim chua bay ln. -Cc cht 2 5C H OHv 3 7C H OHth to ra dung dch ng nht ngay lp tc. Vy nhn bit c 6 cht u tin, cn 2 5C H OHv 3 7C H OHkhng phn bit c bngHCl Bi 5. ______________________________________________________________________________ Oxi ha 9,2 gam ru etylic bng CuO un nng, c 13,2 gam andehit, axit, ru cha phn ng v nc. Hn hp nytc dngviNa dsinhra3,36lt 2H (ktc).Phn trmkhilngru boxi hal bao nhiu? Li gii. Phng trnh phn ng: 2 5 3 22 5 3 22[ ]2[ ]x x x xy y y yC H OH O CH CHO H OC H OH O CH COOH H O+ ++ + V d0, 2 x y mol 2 5C H OHCc cht phn ng vi Na l ru d, nc v axit to ra. Suy ra 22 (0, 2 ) 0, 2 0, 3 0,1Hn x y x y y x x = + + + = + = =p dng LBTKL nn s mol [O] l [ ]13, 2 9, 20, 25 2 0, 2516On x y= = + = , do 0, 05 y = . Vy phn trm khi lng ru phn ng l 0, 25.2 0,1.100% 75%0, 2H+= = . Bi 6. ______________________________________________________________________________ Hn hpAgm 4CH v 2H .ChoAiqua ngng btNi nung nngthuc hn hpB chgm3 hidrocacbon c t khi so vi 2Hl 21,5. T khi ca A so vi 2Hl:A.10,4B. 9,2C. 7,2D. 8,6 Li gii. Do hn hp ch gm cc hidrocacbon c 3 C, nn c th t CTPT chung ca B l 3 yC H . 28ht t p:/ / boxmat h.vn Ta c12.3 21, 5.2 7BM y y = + = =Phng trnh phn ng: 4 2 3 732CH H C H + Gi s c 1 mol 3 4C Hban u phn ng ht vi 1,5 mol 2Hto ra 1 mol B. Vy ban u c 2,5 mol kh, li c khi lng c bo ton nn1.21, 5.2 43( )A Bm m gam = = =Vy8, 6AM= Bi 7. ______________________________________________________________________________ Hp cht X cha C, H, O c khi lng mol phn t l 74 gam. X phn ng c viNaOH . S cht tha mn X l: A.3B. 4C. 5D. 6 Li gii. Do X c khi lng mol phn t l 74 gam nn X khng th l phenol. X li tc dng c viNaOHnn X l axit cacboxylic, este hoc l tp chc cha chc este. Suy ra X cha t nht hai oxi. NuXc2oxi:XcCTPTl 3 6 2C H O ,vccchtthamnl 3 2, CH CH COOH2 5, HCOOC H3 3CH COOCHNuXc3oxi:Xccngthcphntl 2 2 3C H O ,vccchtthamnl, OHC COOH v ( ) ( ) HC O O C O H (anhydrit fomic) X khng th c nhiu hn 3 oxi, v lc , s khng c C trong phn t. Vy c tt c 5 cht tha mn. Chn p n C. Bnh lun. Cht m cc bn d dng b qun khi lm bi ny chnh l anhydrit fomic, mt cht khng qu ph bin trong chng trnh ph thng. Bi 8. ______________________________________________________________________________ Hn hpMgmankenX vhaiamin no,n chc,mch hY,Z(Y ZM M < ).tchyhontonmt lng M cn dng 21 lt 2O , sinh ra 11,2 lt 2CO . Cng thc ca Y l A. 3 2 2 2CH CH CHNH C. 2 5 2C HNHB. 3 2 3CH CHNHCH D. 3 2CHNHLi gii. Ta c 20, 9375( )On mol =v 20, 5( )COn mol = . Bo ton nguyn t O, ta suy ra 20,875( )HOn mol =Gi CTPT trung bnh ca hai amin l 2 3 n nC H N+ Phn ng chy ca aken v amin2 2 2 232m mmCH O mCO mH O + +2 212 3 2,5a34 2 23 3kka akakkCO k C O O H HkN++| | + +

| |+|\ .+ |\ . T hai phng trnh ta c 2 21, 5. 0, 25HO CO amin aminn n n n = =Vy 0, 520, 25k < = . Suy ra amin Y ch c 1 C trong phn t, do Y l 3 2CHNH . Bi 9. ______________________________________________________________________________ Cho anilin tc dng vi cc cht sau: dung dch 2, Br2, H3, CHI dung dch, HCl dung dch, NaOH2HNO . S phn ng xy ra l: A.5B. 3C. 4D. 6 29ht t p:/ / boxmat h.vn Li gii. Anilin l mt amin c bit vi tnh baz yu ca n, v c mt vi phn ng c trng lin quan n gc 6 5C H (phn ng cng 2, H phn ng vi dung dch 2Br ) Anilin tuy vy vn l mt amin bc 1 nn vn th hin y tnh cht ca amin bc 1 nh: tc dng vi , HCl2, HNO3CH I . Vy anilin phn ng vi 5 cht. Chn p n A. Bi 10._____________________________________________________________________________ Cho hn hp 3,88 gam X gm 2 axit cacboxylic no, n chc, mch h. X tc dng vi NaOH to ra 5,2 gam mui. t chy hon ton 3,88 gam X cn bao nhiu lt 2O ? A. 2,24B. 3,36C. 1,12D. 5,6 Li gii. S dng phng php tng gim khi lng suy ra:( )5, 2 3, 880, 0623 1Xn mol= = Phn t khi trung bnh ca X l: 3,88 1940, 06 3tbM= =Do X l axit cacboxylic no, n chc, mch h nn c th t CTPT trung bnh ca X l 2 1 n nC H COOH+ Suy ra 43n =Phng trnh phn ng 4 11 2 2 23 35 7 72 3 3CHCOOH O CO H O + +Suy ra 2 25.0, 06 0,15( ) 3, 362O On mol V = = =Bnh lun. C th dng phng php bo ton nguyn t oxi tm th tch kh 2Ocn dng Bi 11 20 Bi 11. _____________________________________________________________________________ t chy hon ton mt hidrocacbon mch h X (X l cht kh iu kin thng, X nng hn khng kh) thu c 14,08 gam 2CO . Sc m gam hidrocacbon ny vo nc brom d n phn ng hon ton thy c 51,2 gam brom tham gia phn ng. Gi tr ca m l: A.2B. 4C. 4,053D. 4 hoc 4,053 Li gii. Ta c 2 2r0, 32( ); 0, 32( )CO Bn mol n mol = =Gi s c x mol hidrocacbon 2 2 2 n n kC H+ . Ta suy ra t bi:0, 32; 0, 32 xn xk = = . Suy ran k = , 2 nX C H =Li c X l cht kh v nng hn khng kh nn4 n s v 12 2 29 n + > . Suy ra3 n = hoc4 n = . Tuy nhin vi3 n = th X khng tn ti CTCT mch h. Vy X l 4 2( ) C H HC C C CH .D dng tm ra c khi lng l 4. Bi 12. ____________________________________________________________________________ t chy hon ton 29,6 gam hn hp X gm 3 2, , ( )x yCH COOHC HCOOH COOHthu c 0,8 mol nc v m gam 2CO . Cng 29,6 gam X tc dng vi 3NaHCOd thu c 0,5 mol 2CO . Gi tr ca m l: A. 11B.22C.44D.33 30ht t p:/ / boxmat h.vn Li gii. S molCOOHl 20, 5COOH COn n mol= =Theo nh lut bo ton nguyn t ta c: X C H Om m m m = + +229, 6 0, 8.2 0, 5.2.16 12 44C COm g m g = + + = = Bi 13. ____________________________________________________________________________ Cho hn hp X gm ancol metylic v 2 axt k tip nhau trong dy ng ng ca axt axetic tc dng ht vi Na gii phng 6,72 lt 2H . Nu un nng hn hp X c2 4HSOc xc tc th cc cht trong hn hp phn ng va vi nhau to thnh 25 gam este. t chy hon ton hn hp X ri dn sn phm qua bnh 2 4HSOc th khi lng ca bnh tng bao nhiu gam? A.25,2 B. 23,3C. 24,6D. 15,2 Li gii Ta c( )22 0, 6ancol axit Hn n mol+= =Phn ng va chng t( ) 0, 3ancol axitn n mol = =Suy ra 25 2500, 3 3TBM = = , do hn hp X gm : 3, CH OH3, CH COOH2 5C H COOHt s mol ca hai axit l ( )( )32 5::CH COOH amolC H COOH bmol T gi thit ta suy ra20, 374 0,188 25 0, 20, 3.2 0,1.2 0, 2.3 1, 4HOa b aa b bn mol+ = = + = = = + + = Khi lng ca bnh tng bng khi lng ca nc: 225, 2m HOm g A= =Bnh lun. T vic tm ra 2503TBM =c th suy ra CTPT trung bnh ca axit l 8 193 3CH COOH . Phn ng este to ra 20, 3( )HO ancoln n mol = = . Bo ton nguyn t H, suy ra 21 190, 3.6 1 .0, 3 0, 3.2 1, 4( )2 3HOn mol (| |= + + = |(\ . Bi 14. _____________________________________________________________________________a mol cht bo X c th cng hp ti a vi 4 a mol2Br . t chy hon ton a mol X thu c b mol 2H O v V lt kh 2CO .Biu thc lin h gia V vi a,b l: A.22.4( 7 ) V b a = + B.22.4(4 ) V a b = C.22.4( 3 ) V b a = + D.22.4( 6 ) V b a = +Li gii. Do a mol X c th cng hp ti a vi 4a mol 2Brnn X c CTPT dng 3 5 3 2 1 2.4( )n nC H OOC C H+ hay 2 14 6 m mCH O Phng trnh phn ng: 2 13 6 2 2 23 7 6( 7)2m mmCH O O mCO m H O + + Ta c 2 2.7 722, 4CO X HOVn n n a b = + = + 31ht t p:/ / boxmat h.vn Biu thc ny tng ng vi p n A. Bi 15. _____________________________________________________________________________ Cho hai hidrocacbon X v Y ng ng ca nhau, phn t khi ca X gp i Y. Cng thc tng qut ca hai hidrocacbon l: 2 2.n nAC H+ 2 2.n nBC H 2 6.n nCC H 2.n nDC HLi gii. Hai hidrocacbon c dng: 2*& ( ) ,nR CH n R N eTheo ra ta c22 ( )nR R CH = + , do 2 n nR C H = . Chn p n B. Bi 16. ____________________________________________________________________________t chy hon ton 2,76 gam hn hp X gm:3;x y x yC HCOOHC HCOOCHv 3CH OHthu c 2,688 lt 2CO (ktc)v1,8gam 2H O.Mtkhc,cho2,76gamhnhpXphnngvavi30mldungdch NaOH 1M, thu c 0,96 gam 3CH OH . Ly lngx yC HCOOHc trong X cho tc dng hon ton vi hn hp cha 0,04 mol 3CH OHv 0,06 mol2 5C H OH , xc tc 2 4HSOc, un nng. Gi s 2 ancol phn ng vi kh nng nh nhau th khi lung este ho to thnh l: A.0,88 gamB.0,944 gamC.1,62 gamD.8,6 gam Li gii. Gi s mol ca 3 3, ,x y x yC HCOOHC HCOOCH CH OHln lt l, , a b c p dng nh lut bo ton khi lng, suy ra 20,12.44 1, 8 2, 760,13532On+ = =Li bo ton nguyn t O suy ra2 2 0,135.2 0,12.2 0,1 0, 07 a b c + + = + + =T ta c h phng trnh ( )( )( )0, 012a 2 0, 070, 03 0, 020, 030, 01a molb cb c b mola bc mol= + + = + = = + == Suy ra x yC HCOOHl 2 3(0, 01 ) C H COOH mol . Do ,0, 944( )estem g = . Chn p n B Bi 17. _____________________________________________________________________________ Hn hp X gm vinyl axetat, metyl axetat , etyl fomat. t chy hon ton 3,08 gam X, thu c 2,16 gam 2H O. Phn trm s mol ca vinyl axetat trong X l: A.25% B.27,92%C.72,08%D.75% Li gii Hn hp gm 2 cht l ng phn 3 6 2C H O , cht cn li vilynaxetat c CTPT l 4 6 2C H OGi s mol vinyl axetat l x (mol) v tng s mol metyl axetat v etyl fomat l y (mol)S mol 2H O to ra l 23 3 0,12HOn x y = + =hay0, 04 x y + =(1) Ta cng c khi lng ca hn hp l:74 86 3, 08 x y + =(2) Gii h phng trnh (1) v (2) suy ra0, 03; 0, 01. x y = = Phn trm s mol ca vinyl axetat s l 25%. Bnh lun. im mu cht ca bi ton l pht hin metyl axetat v etyl fomat l ng phn, t a bi ton 3 n v 2 n. Mt bi tp tng t, nhng kh hn mt cht l: Bi tp tng t. ha tan va ht 15,2 gam hn hp X gm, , MgOCaOFecn 600 ml dung dch HCl1M thu c kh 2Hv dung dch Y. Khi lng 2MgCltrong dung dch Y l: A.11,4 gamB. 9,5 gamC. 7,6 gamD. 13,3 gam 32ht t p:/ / boxmat h.vn p s. B. 9,5 gam. Ch CaO v Fe c chung M = 56. Bi 18. ____________________________________________________________________________ Cho X l hn hp gm 2Hv hi ca hai anehit (no, n chc, mch h, phn t u c s nguyn t C nh hn 4), c t khi so vi heli l 4,7. un nng 2 mol X(xc tc Ni), c hn hp Yc t khi hi so vi heli l 9,4. Thu ly ton b cc ancol trong Y ri cho tc dng vi Na (d), cV lt 2H(ktc). Gi tr ln nht ca V l: A. 22,4B. 5,6C. 11,2D. 13,44 Li gii. Gi x l s mol hn hp lc sau Ta c p dng L Bo ton khi lng ta c: Khi lng ca hn hp khng i v bng ( ) 4.4, 7 37, 6 1 m x mol = = =S mol ancol to thnh bng s mol 2Hphn ng v bng s mol hn hp gim:1( )ancoln mol =Suy ra s mol 2Hto ra l210, 52H ancoln n = = . Vy V = 11,2 lt. Bi 19. ____________________________________________________________________________Hn hp X gm 2Ov 3Oc t khi so vi 2Hl 22. Hn hp kh Y gmmetylamin v etylamin c t khi so vi 2Hl 1076. t chy honton 1Vlt Y cn va 2Vlt X (bit sn phm chy gm 2CO , 2H O v 2N , cc cht kh khi o cng iu kin nhit , p sut). T l 1 2: V Vl: A. 3 : 5B. 5 : 3C. 2 : 1D. 1 : 2 Li gii. Gi s s mol ca hn hp kh l: ( )11 n mol = . t cng thc phn t trung bnh ca Y l 2 3 n nC H N+. Li c phn t khi trung bnh ca hn hp l: 107 107.26 3M = =nn 43n =Ta c phng trnh phn ng: 4 17 2 23 3[ ]13 21 4 176CH N CO H O O + +t s mol ca 2Ov 3Oln lt l a (mol) v b (mol). Ta c2 3 5, 5 a b + =Mt khc khi lng hn hp X l: 2 222.2. 32 48 16(2 3 ) 88 2( )Xm n a b a b n mol== + = + = =Suy ra 1 2 1 2: : 1: 2 V V n n = = Bi 20. _____________________________________________________________________________ Hn hp X gm 2 axit hu c no (mi axit cha khng qu 2 nhm) COOH c khi lng 16gtng ng vi 0,175 mol. t chy hon ton hn hp X ri cho sn phn qua nc vi trong d thu c 47,5 g kt ta. Mt khc, nu cho hn hp X tc dng va vi dung dch 2 3Na COthu c22, 6gmui. Cng thc cu to ca cc axit trong hn hp X l:2. , ( ) AHCOOH COOH3 2. , ( ) BCH COOH COOH2 5 2. , CC H COOHHOOC CH COOH 2 3. , DCH COOHHOOC CH OH CO Li gii. 33ht t p:/ / boxmat h.vn t CTPT trung bnh ca hai axit no l: 2 2( )n n k kC H COOH+ ( | | 1, 2 k e , do mi axit cha ko qu 2 nhm COOH ). Ta c 2190, 475( ) (1)7COn mol k n = + =Mui to ra l 2 2( )n n k kC H COONa+ , suy ra phn t khi ca mui l 22, 614 2 66 (2)0,175M n k = ++ =Gii h (1) v (2) ta c1 n = v 127k = . Do 127k = nn c mt axit A c 1 nhmCOOH v axit B cn li c 2 nhmCOOH . T l s mol c xc nh theo s ng cho: 27127571 22 5=Gi s axit A c x nguyn t C ( 1 x > ), axit B c y nguyn t C ( 2 y > ) Ta c 2 5 192 5 19(*)2 5 7Cx yn x y+= = + =+. Do2, y >nn y ch c th nhn cc gi tr 2, 3. Nu2 y =th xe. Vy3 y =v2 x = . Cc axit l 3CH COOHv 2HOOC CH COOH . Chn p n D. Bnh lun. D nhin, li gii trn l cn thit rn luyn v kh nng suy lun trong vic gii ton Ha hc. Nhng vi yu cu v thi gian trong cc k thi, ta c th chn cch nhanh hn sau.Ta c1691, 430,175M = ~nn loi A v B! Da theo p n C v D, suy ra c 1 axit l 2HOOC CH COOH v axit cn li l RCOOH. Li c( )247, 50, 475100COn mol = =0, 475 192, 70,175 7Cn = = ~ , nn phi c 1 axit c di 2,7 nguyn t C trong phn t. Chn p n D. Bi 21 30 Bi 21. _____________________________________________________________________________ Cho hp cht thm X: 6 8 2 3C HN O . Cho28, 08gX tc dng 200ml KOH2M. Sau phn ng thu c dung dch Y. C cn Y c m gam cht rn khan. Gi tr ca m l? A. 21,5gB. 30,5gC. 18,1gD. 38,8g Li gii. Phng trnh phn ng 6 5 3 3 6 5 2 3 20,40,180 C HNHNO KOH C HNH KNO H + + + Suy ra 30,18.101 0, 22.56 30, 5 .duran KOH KNOm m m g = + = + = Chn p n B Bi 22.____________________________________________________________________________ Cho hn hp XHCOOH , 3CH COOH(t l mol 1:1) v hn hp Y ch 3CH OH , 2 5C H OH(t l mol l 2:3). 16,96 gam hn hp X tc dng vi 8,08 g Y (c xc tc 2 4HSOc) thu c m gam este (Hiu sut cc phn ng este ha bng 80%). Gi tr ca m l? Li gii. S dng cng thc M trung bnh 1 2xM yMMx y+=+ 34ht t p:/ / boxmat h.vn t CT chung ca 2 axit l RCOOH v ca 2 ancol l R'OH. Khi da vo cng thc trn ta c: 46 60 46 6053( )2332 46.32 46240, 4( : 2: 3)32acidacolx yM Do x yx yx xx yM Do x yx yx x+ + = = = =+++= = = =++ T suy ra: 823, 4RR= ' =16, 960, 32538, 080, 240, 4acidancolnn= == = Da vo PTP ta suy ra c ancol phn ng ht, suy ra s mol ca este (Hiu sut 80%) l0, 2.0,8 0,16 =Suy ra khi lng este l:0,16( 12 32 ) 0,16.(8 44 23, 4) 12, 064 m R R' = + + + = + + = Bi 23. ____________________________________________________________________________ t chy m gam hn hp X gm 2 ancol n. Cho ton b sn phm chy qua bnh 1 ng 3 4H POc sau cho qua bnh 2 ng nc vi trong d. Sau th nghim thy bnh 1 tng 1,98 gam, bnh 2 c 8 gam kt ta. Mt khc, cho m gam hn hp X tc dng viCuO un nng c sn phm ha c Y. Cho Y tc dng vi lng d 3AgNOtrong 3NHthu c 2,16gamAg . Xc nh CTCT 2 ancol. Li gii. T cc gi thit v bnh 1 v bnh 2 suy ra: ( ) ( )2 200, 08 , 0,11CO Hn mol n mol = = 0, 03Xn =Ta c 0, 02Agn =Ta c:2AgXnn (loi) -TH2: to 3CH OHtrong phn ng viKOHKhi lng cht rn khan thu c l: 74 0, 7.0, 2.56 32 12, 88 0,12; 8, 88 m a a a m g = + = + = =p n 8,88 gam. Bnh lun. Bi ton t ra hai thao tc, trong vic xc nh c CTPT ca este l thao tc u tin v mu cht. tm c CTPT ca este ta phi gii mt bi ton v phn ng chy m hon ton c th ng ring ra lm thnh mt bi ton c lp. Bi tp trn thc ra khng kh, nhng li c t vo hai bi ton khc nhau, tnh ton li kh di dng, nn li tr thnh bi tp kh. Bi 26. ______________________________________________________________________________________Cho m gam tinh bt ln men thnh ancol (ru) etylic vi hiu sut81%. Ton b lng 2CO hp th hon ton vo dung dch nuc vi trong, thu c 550 g kt ta v dung dichY . un k dung dchYthu thm 100 gam kt ta. Khi lng m l bao nhiu ? Li gii Cc phng trnh phn ng: 2 2 3 25,5 5,5( ) Ca OH CO CaCO H O + +1 12 2 2 3 2( ) ( ) Ca OH CO HO CaHCO + + 213 3 2 21( ) CaHCO CaCO HO CO + +Tng s mol 2CO phn ng l 36ht t p:/ / boxmat h.vn ( )( ) ( )25, 5 1.2 7, 51257, 5: 2: 81% 75027COtinh bot tinh botn moln mol m gam = + = = = = Bi 27.______________________________________________________________________________________Cho 200 gam mt loi cht bo c ch s axit bng 7 tc dng va vi mt lng NaOH, thu c 207,55 gam hn hp mui khan. Khi lng NaOH tham gia phn ng l: A. 31 gamB. 32,36 gam C. 30 gam D. 31,45 gam Li gii 1 g cht bo cn trung ha bi 7 mgKOHSuy ra 200 g s cn 1400 mgKOH tc l cn 0,025 mol KOH 20,025 0,025R COOH NaOH R COONa H O + + x3 3 5 3 5 33( ) 3 3 ( )xRCOO C H NaOH RCOONa C H OH + +Ta c 14000, 025( )56.1000KOH NaOHn n mol = = =p dng LBT KL:200 0, 025.40 120 207, 55 18.0, 025 92 0, 025( ) x x x mol + + = + + =3.0, 25.40 0, 025.40 31NaOHm g = + = .Chn p n C. Bi 28. ______________________________________________________________________________________Khi thu phn kim 265,2 g cht bo to bi mt axitcacboxylic thu uc 288 gam mui kali. Cht bo ny c tn gi l: A. glixerol tristearat B. glixerol trioleat C. glixerol trilinoleatD. glixerol tripanmitat Li gii Dng phng php tng gim khi lng vi X l s mol ca este. (393 41) 288 265, 2 22, 8 0, 3 x x mol = = =17 33884 .axitM CHCOOH = Chn p n B. Bi 29. _____________________________________________________________________________________ Almt hn hpgm hai chtthuc dyng ng castiren ckhilng phnt hnkm nhau 14 vC. t chy hon ton m gam A bng 2Od. Cho sn phm chy hp th vo 300ml dung dch NaOH 2M.Khilng bnhng dung dchtng 22,44gamvthuc dung dchD.Cho 2BaCl dvo dung dch D thu c 35,46 gam kt ta. cc phn ng Xy ra hon ton. Tm CTPT ca 2 hidrocacbon trong A. Li gii Gi CTPT trung bnh ca A l 2 8 n nCH 3 22 2 235, 460,18( ) 0,18 (0, 3.0, 2 0,182) 0, 42( )1970, 42 0, 2244 18 22, 44 0, 22( ) 0, 0540, 428, 40, 05BaCO COCO HO HO An mol n moln n n mol nn= = = + =+ = = = == = Suy ra hai hidrocacbon cn tm l 8 8 9 10, C H C H Bi 30. _____________________________________________________________________________________ 37ht t p:/ / boxmat h.vn Hp cht X mch h c CTPT l 4 9 2C HNO . Cho 10,3 g X phn ng va vi dung dch NaOH sinh ra 1 cht kh Y v dung dch Z. Kh Y nng hn khng kh, lm qu tm m chuyn mu Xanh. dung dch Z c kh nng lm mt mu nc brom. C cn dung dch Z thu c m g mui khan. Gi tr m? Li gii Kh Y nng hn khng kh, lm qu tm m chuyn mu Xanh suy ra: 3 2CHNHDung dch Z c kh nng lm mt mu nc brom suy ra phi c lin kt iC C = . Suy ra X l 2 3 2CH CHCOONH CH = . Do 10, 394 .94 9, 4103muoi muoi muoi Xm n M n g = = = = Bi 31 40 Bi 31. ____________________________________________________________________________ Aminoaxit Ycng thc c dng ( )x ymNCH COOH . Ly mt lng axit aminoaxetic ( X ) v 3,104 gam Y. Bit X v Y c cng s mol. t chy hon ton lng X v Y trn, th tch 2Ocn dng t chy Y nhiu hn X l 1, 344l (ktc). CTCT thu gn ca Y l: 3 2. ACHNHCH COOH2 2 2. BHNCH CH COOH( )33. CNCH COOH ( )4 82. DNC H COOHLi gii. t: 1: : ( )x y zY C HO N kmol2 5 2: : ( ) X C H O N kmol1 2 2 2 21( )4 2 2 2x y zy z yC HO N x O xCO H O N + + + +2 5 2 2 2 2 25 2 5 1(2 ) 24 2 2 2C H O N O CO H O N + + + +5 2 1, 344(2 ). ( ). (1)4 2 22, 4 4 2y zk x k + + = + vi3,104 3,10412 16 14Ykx y z M= =+ + + Thay vo ta c: ( ) ( )3,104 9 4x 2z0, 06 15. 12x 16z 14 194 9 4x+ 2z12x 16z 14 4596x 179 628z 1956 0yy yyy + | | = + + + = + |+ + +\ . + = T y suy ra4 y , ch c p n C tha mn. Th li p n ny, ta chn C l ph hp. Bi 32. _____________________________________________________________________________Cho 0,02 mol mt este X phn ng va ht vi 200 ml dung dchNaOH0,2M, sn phm to ra ch 1 mui v mt ancol u c s mol bng s mol este, u c cu to mch thng. Mt khc khi x phng ho hon ton 2,58 gam este bng 20 ml dung dch KOH 1,5M va thu c 3,33 gam mui. Vy X l: A. etilenglicol oxalatB. etilenglicol adipat C. imetyl adipatD. ietyl oxalat Li gii. 38ht t p:/ / boxmat h.vn TN1. Do 12este NaOHn n = , nn este phi l este hai chc hoc l este ca phenol. Li c muoi ancol esten n n = = , nn este phi l este ca ancol hai chc v axit hai chc. Suy ra CTPT ca este l 2( ) R COO R' . TN2. Ta c 10, 0152este muoi KOHn n n mol = = =T tnh ra c 4 8R C H = v 2 4R C H ' = . Suy ra este l C. imetyl adipat. Bi 33. ____________________________________________________________________________Mt este A ( khng cha chc khc) c to nn t 1 axit hu c B v 1 ancol C. Ly m gam A cho tc dng vi KOH d thu c m1 gam mui. Ly m gam A cho tc dng vi 2( ) Ca OHd thu c 2mgam mui. Bit 2 1m m m < < . CT thu gn ca C l: A. 2 5C H OH B. 3CH OHC.3 7C H OH D. 4 8C H OHLi gii. 1m gam mgamRCOOR RCOOK ' Ta c 1m m >nn39 R nn F l ete, do ( )2 52 161, 61 2917RR C HR+= = +. Vy E hay ancol A chnh l 2 5C H OH . Loi p n B v C -Gi x, y, z ln lt l s mol ca 2 5, C H CHO2 5, C H OH2 1 2 5 m mCH COOC H (Lu , do axit to thnh este l ng ng ca axit acrylic nn3 m> ) Ta c58 46 (72 14 ) 38 x y z m + + + =-Oxi ha hn hp s to ra hn hp X gm x mol 2 5, C H COOH y mol 3, CH COOHz mol este S mol NaOH phn ng vi hn hp sn phm:0, 5.1, 5 0,15 0, 6( ) 0, 6NaOHn mol x y z = = + + = . -C cn D s to ra x mol 2 5, C H COONay mol 3, CH COONaz mol 2 1 m mCH COONa v 0,15 mol NaCl. Suy ra96 82 (66 14 ) 0,15.78, 5 64, 775 x y z m + + + + =Nh vy ta c h 3 phng trnh: 38 5858 46 (72 14 ) 38 (1) (1')140, 6 (2) 38 (2')96 82 (66 14 ) 0,15.5x 46 72zzx 36 6z 160 8, 5 62, 775 (3) (3') , 6myyx y zx y z mx y zx y z m + =+ + = =+ + + = + + = + + + + = T (2) v (3) suy ra 2 1736( ) 16 6 38( ) 36(0, 6 ) 16 6 38(0, 6 ) (*)15 110x y z x y z z z z + < + < + < + < < nn tm li hn hp X to ra ( ) 2 2 242.5CO CO Nn n+> =Loi thm p n A. Vy ta chn B. Bi ton thuc dng bi hn hp, phng php c s dng ch yu l tr s trung bnh. Nhng, hn hp c cho trong bi c cc cht khc loi nhau (amin v hidrocacbon), thm na tr s trung bnh c s dng kh c bit ( )2 2CO NXnn+| | |\ .. Vy nn, bi ton tr nn kh gii theo cch chnh thng (khng da vo p n trc nghim). Nhng c l mt cht mm ci, mt cht sung sng s l s tr cng xng ng cho nhng ai chu kh i tm mt chn l ton vn. Chng ti xin c gi cch gii y , p mtv ngn gn trn l mt chn l ton vn! Ti liu c ti xung min ph ti www.boxmath.vn Ti liu c th in ra phc v mc ch hc tp. Mi s sao chp, trch dn u phi c s cho php ca box Ha hc, thuc din n BoxMath.