September 2011 Trang 1 AOTRANGTB.COM KHI A-2007 M 182 Gii Ta c: mbnh tng = mX = 6,7 gam nX = 0,2 mol 2Brnphn ng = 0,35nX < 2Brn < 2nX X chc chn c 1 ankin, hirocacbon cn li c th l anken hoc ankan.TH1: X gm 1 anken (a mol) + 1 ankin (b mol) n 2n 2 n 2n 2n' 2n'-2 2 n' 2n'-2 4C H +BrC H BraaC H+ 2Br C H Brb2bTa c: a + b = 0,2 a = 0,05a + 2b = 0.35 b = 0,15 m= 6,7 gam14na + (14n'-2)b = 6,7 X0,7n + 2,1n' = 7 n + 3n' = 10 Chn n = 4 v n = 2 (X): C2H2 + C4H8TH2: loi Gii 33HNO2 2 4 3HNO2 42FeS Fe (SO ) 0,12 0,06 Cu S 2CuSOa2a Theo nh lut bo ton nguyn t: 2-4SSOn=n= 0,24 + aTheo nh lut bo ton in tch: 3+ 2+ 2-4Fe Cu SO3n+ 2n = 2n= 3.0,12 + 2.2a = 2.(0,24 + a)a = 0, 06 molGII BI TP TRONG CC THI I HC T NM 2007 N NM 2011 Download ti liu hc tp, xem bi ging ti : http://aotrangtb.com September 2011 Trang 2 AOTRANGTB.COM Gii Cng thc amin n chc: CxHyN 2 x y 2N C H N Nn = 0,0625n= 2n = 0,125 2COn= 0,0625 s C = 2COXnn= 2 Lu : A. t chy hp cht hu c CxHy hoc CxHyOz (khng cha N): 1. Ankan hay hp cht c lin kt n: CnH2n + 2 + O2nCO2 + (n + 1)H2O 22 2 n 2n+22C H OH OOO C C Hn< nn- n= n 2.Anken hay hp cht c 1 lin kt (1 vng): CnH2n + O2nCO2 + nH2O 2 2CO H On= n 3.Ankin ankaien hay hp cht c 2 lin kt : CnH2n 2 + O2nCO2 + (n 1)H2O 2 22 2 n 2n-2CO H OCO H O C Hn> nn- n= n 4.Hp cht c 3 lin kt : CnH2n 4 + O2nCO2 + (n 2)H2O 2 22 2 n 2n-4CO H OCO H OC Hn> nn - n = 2n 5.Hp cht c 4 lin kt (aren): CnH2n 6 + O2 nCO2 + (n 3)H2O 2 22 2 n 2n-6CO H OCO H O C Hn> nn- n= 3n B. t chy hp cht hu c CxHyNt hoc CxHyOzNt: 1. Amin n chc no: CnH2n + 3N + O2nCO2 + (n + 32)H2O + N2 2 22 2 n 2n+3CO H OH O CO C H Nn< nn - n= 1,5n Download ti liu hc tp, xem bi ging ti : http://aotrangtb.com September 2011 Trang 3 AOTRANGTB.COM 2.Amin n chc khng no c 1 lin kt : CnH2n + 1N + O2nCO2 + (n + 12)H2O + N2 2 22 2 n 2n+1CO H OH O CO C H Nn< nn- n= 0,5n 3.Amin n chc khng no c 2 lin kt : CnH2n 1N + O2nCO2 + (n - 12)H2O + N2 2 22 2 n 2n-1CO H OCO H O C H Nn> nn- n = 0,5n C.t chy hn hp hai hirocacbon bt k: -Nu 2 2CO H On>n Hai hirocacbon c th l: 2 ankan1 ankan + 1 anken1 ankan (x mol) + 1 ankin (y mol) (x > y)

-Nu 2 2CO H On n = Hai hirocacbon c th l: 2 anken1 ankan (x mol) + 1 ankin-ankadien (y mol) (x = y)

-Nu 2 2CO H On 2+2+nCanBa n = 2+2+nCanBa-Thng thng ch cn nh TH3 th c th p dng tng qut cho hai trng hp trn.-Ton SO2 cng thc tnh tng t. Gii Ta c: 22233COBa(OH)BaCOCOn= 0,12 moln= 2,5a moln= n= 0,08 mol3 2BaCO COn< n Xy ra hai trng hp Theo cng thc, ta c: 3 2 2BaCO Ba(OH) COn= 2n- n2.2,5a - 0,12 = 0,08 =a = 0, 04 molGii IN PHN in phn dung dch: - Catot cc (-): cc cho e cation tin v nhn e - C mt cc cation kim loi Mn+ v H+ (do nc hoc axit in li). - Th t nhn electron: K+ < Ca2+ < Na+ < Mg2+ < Al3+ < H2O < Mn2+ < Zn2+ < Cr3+ < Fe2+ < Ni2+ < Sn2+ < Pb2+ S2- > I- > Br- > Cl- > OH- (baz) > H2O > Gc axit c oxi (SO42-,NO3-, CO32-,F-) Quy lut:Cht c tnh kh mnh s b oxi ho trc. - CcanionSO42-,NO3-,CO32-,SO32-,PO43-,F-,thctkhnginphn,thayvo nc s in phn. - Cc dng khng phi l in cc tr (than ch, Pt) th s b in phn trc tin ti ccdng(hintngdngcctan).Khiccdngbtandn,ccmcbipdn bi kim loi do cc dng tan ra. y l hin tng xy ra khi m, trng kim loi. - Thc t anion gc axit c oxi khng b in phn, thay vo nc s b in phn (boxi ho) +2 21H O 2H + O +2e2Cng thc Faraday: A Itm =.n FHoc dng cng thc tnh s mol kh thot ra mi in cc: Itn = 96500nTrong : A: nguyn t khi. n: s e tham gia phn ng in cc.I: cng dng in. t: thi gian in phn. F: s Faraday ph thuc vo t. Nu t(s) F = 96500.Nu t(h) F = 26,8. Cc kiu mc bnh in phn: 1.Mc ni tip:-Cng dng in qua mi bnh l bng nhau.-S thu v nhng e cc in cc cng tn phi nh nhau cc cht sinh ra cc in cc cng tn phi t l mol vi nhau.V d: September 2011 Trang 10 AOTRANGTB.COM Bnh 1: catot Cu2+ + 2e Cux 2x Bnh 2: catot Ag+ + 1e Agy y y = 2x 2.Mc song song: Cng dng in I qua n bnh bng In. in phn nng chy C 3 loi hp cht in phn nng chy: 1.Oxit kim loi: MxOy dpnc xM + y2O2 2.Mui clorua: MCln

dpnc M+ n2Cl2 3.Hiroxit kim loi (nhm IA): 2MOHdpnc 2M + O2 + H2O Phng php ny dng iu ch nhng kim loi t Al tr v trc t cc oxit thay v dngCO, H2. Gii 2+ -2 Catot (-)Anot (+)Cu+ 2eCu 2ClCl + 2e0,010,0050,0050,01 2 22NaOH + ClNaClO + NaCl + H O0,010,005nNaOH cn li = 0,01 mol nNaOH ban u = 0,01 + 0,01 = 0,02 mol [NaOH] = 0,1M Gii 2 3NH -R-COOH + HCl ClNH -R-COOH Chnh lch khi lng: September 2011 Trang 11 AOTRANGTB.COM Xm 13,95 - 10,3n= = = 0,1molM 36,5 MX = 103 Chn C Lu : Cng thc tnh s mol da vo s chnh lch khi lng mn = Mng trong trng hp h s cn bng trong phng trnh bng nhau.V d 1: 2+ 2+Fe + CuFe+ Cu Cu FeFe Cum- mn= n= 64 - 56V d 2: 2+ 3+2Al + 3Cu2Al+ 3Cu Cu AlAlm- mn= 2.3.64 - 2.27 phi nhn h s cn bng ca cht cn tnh s mol. Gii 2 4 H SO (l)4Fe FeSO0,1 mol0,1 mol4 2 4KMnO+ H SO4 2 4 3 4FeSO Fe (SO )+ MnSO p dng LBT e ta c: 2+ 3+ 7+ 2+FeFe+ 1e Mn+ 5e Mn0,10,1 0, 02 0,1 4KMnOn= 0,02 mol V = 0,04 lt = 40 ml Gii un dung dch X thu thm kt ta trong X c Ca(HCO3)2 September 2011 Trang 12 AOTRANGTB.COM 3 2 3 2 2Ca(HCO )CaCO+ CO+ H O1 mol1 molot3CaCOn= 5,5 molBo ton C: 2 3 2 3CO Ca(HCO ) CaCOn= 2n+ n7,5 mol =(6 10 5C H O )enzim2 5 2n 2nC H OH + 2nCO3,75 mol 7,5 mol mtinh bt = 3, 75.162 = 750 gam0, 81Gii Ta c: S C2COYn2a= = = 2n aS nhm chc COOH = NaOHYn 2a =2n a = Chn D Gii (X): C2H7NO2 tc dng vi dung dch NaOH thu c 2 kh (X) gm: CH3COONH4 (mui amoni) v HCOONH3CH3 (mui to bi HCOOH v CH3NH2) Ta c: 33 2NHCH NHn= 0,05n= 0,15Phng trnh: NaOH3 4 3 3NaOH3 3 3 2CH COONHCH COONa + NH0, 05 0,05HCOONH CHHCOONa + CH NH 0,150,15 mmui = 14,3 gam September 2011 Trang 13 AOTRANGTB.COM Gii Gi s nng hai cht l aM HCl l cht in li mnh nn in li hon ton + -HCl H+ Claa pH = x = lg[H+] = lg(a) (1) 100 phn t CH3COOH th c 1 phn t in li in li o = 0,01 - +3 3CH COOH CH COO+ Ha0,01a pH = y = lg[H+] = lg(0,01a) lg(a) = y + 2 (2) T (1) v (2) x = y + 2 y = x 2 Lu : in li cho bit phn trm cht tan phn li thnh ion v c biu din bng t s nng mol ca phn t cht tan phn li thnh ion (C) v nng ban u ca cht in li (C0). MA M+ + A- + -0 0 0C [M ] [A ] = = = C C Ck: 0 1 -= 0: cht khng in li.-=1: cht in li hon toninli phthucvobnchtcachttan,nhitvnngdungdch.Dungdchcng long (C0 cng nh) th cng ln.-i vi dung dch axit yu (o < 1) HA H++A- (Ka)Ban u:C00 0 in li:oC0 oC0 oC0 Cn bng: (1-o)C0 oC0 oC0 2 + -0a C [H ][A ]K= = [HA] 1 - o M2) September 2011 Trang 31 AOTRANGTB.COM 4.Dung dch H+ tc dng vi hn hp dung dch OH- v [Al(OH)4]-: Tng t TH3 nhng cng thm lng OH-S mol kt ta cn li: TON KM: Dung dch Zn2+ tc dng vi dung dch kim: t: Kho st: Zn2++2OH-Zn(OH)2 a 2a a Zn(OH)2+2OH-[Zn(OH)4]2- (b 2a) S mol kt ta: p dng cho bi ton trn Ta c: Phng trnh phn ng: u tin H+ trc cn li = 2V 0,2 Gi tr ln nht ca V ng vi trng hp kt ta to ra sau b ho tan mt phn Ta c: 3+ -Al OHn= 4n- n= 4.0,2 - (2V - 0,2) = 0,1+ V = 0, 45 ltGii Oxit tc dng vi CO hoc H2:(i vi oxit ca kim loi sau nhm) September 2011 Trang 36 AOTRANGTB.COM Vkh thin nhin = 4CHV = 80%3448 mGii Bo ton e: 3++2+0,5 molAl Al+3eAg+ 1e Ag0,10,3 0,5 0,5 0,5Fe Fe+2e0,1 0,1 0, 2 d = 0,05 tip tc tc dng vi Fe2+ Phng trnh: nAg = 0,55 mAg = 59,4 gam Gii Phn 2: tc dng vi dd NaOH thu c kh H2 Al d v Fe2O3 ht nAl d = 0,025 molPhn 1: Phng trnh: September 2011 Trang 38 AOTRANGTB.COM Gii 3CH OHnban u = 0,0375 mol 3 3AgNO /NHH%3 2HCHO 4Ag0, 03 mol0,12 molCH OH + [O] HCHO + H O0,03 mol0,03 mol3CH OHn phn ng = 0,03 mol H = 80% Gii ru =

Th tch ru nguyn cht = 5.0,46 = 2,3 lt Khi lng ru nguyn cht = 2,3.0,8 = 1,84 kg nru = 0,04 kmol S phn ng +H ln men6 10 5 n 6 12 6 2 5 2(C H O )nC H O2nC H OH + 2nCO0,02 kmol0,04 kmol Khi lng tinh bt thc t = 0,02.162 = 3,24 kg Khi lng tinh bt l thuyt = 3,24:0,72 = 4,5 kg Gii September 2011 Trang 40 AOTRANGTB.COM Lu : Kim loi + HNO3 Mui + +H2O - Kim loi t Mg Al c th to ra tt c cc sn phm khtrn. - Kim loi sau hiro hu nh to ra NO2, NO. - HNO3 c, ngui khng tc dng vi Al, Fe, Cr, Au, Pt. Gii Ta c: X cha1N nn Xc th l estecaamino axit hoc mui ca amin hoc mui amoni, dcht no X cng tc dng vi NaOH theo t l mol 1:1 Nhn thy s mol NaOH d = 0,05 mol 11,7 gam cht rn gm NaOH d v mui mmui = mrn mNaOH d = 9,7 gam Ta c: nX = nmui Mmui = 97 (RCOONa) R = 30(H2NCH2-) Chn D. Gii Ta c: phn ng = 0,025 mol loi D. Th tch kh cn li l CH4 Th tch hirocacbon phn ng vi Br2 = 0,56 lt nhcb = 0,025Nhn thy s mol hirocacbon = s mol Br2 phn ng cht cn li l anken loi B S phn ng t chy: September 2011 Trang 46 AOTRANGTB.COM Gii Ta c: 2COhhV2C = =2 V 1 = S C ca X l 2 loi C, D S H ca X l 6 trong hai p n cn li Chn A. Cch khc: p dng lu cu 9 khi A-2007 Trng hp t chy 2 hirocacbon, s mol H2O bng s mol CO2 -TH1: Hn hp 2 anken loi, v c C2H2 -TH2: Hn hp gm 1 ankin v 1 ankan c s mol bng nhau X: CnH2n+2

Ta c: S phn ng t chy Gii S phn ng Chn C. Gii t: Bo ton e: September 2011 Trang 47 AOTRANGTB.COM 2+2+2+ Cu +2e Cu a2aa Zn Zn+ 2e(a + b)(a + b)2(a + b)Fe +2e Fe b 2bb : Khi lng cht rn thu c nh hn khi lng Zn ban u l 0,5 gam, ngha l khi lng Znphn ng ln hn khi lng cht rn sinh ra l 0,5 gam hay chnh lch 1 lng m = 0,5 Zn CuFem- (m + m ) = 0,565(a + b) - (64a + 56b) = 0,5 a + 9b = 0,5 (1)Mt khc: mmui = 13,6 gam 2ZnClm= 13,6 136(a + b) = 13,6 a + b = 0,1 (2) T (1), (2) suy ra a = b = 0,05 mol mX = 13,1 gam KHI A-2009 M 175 Gii t: 4ZnSOn= x molTa c: C hai trng hp u cho lng kt ta bng nhau, nn TH2 kt ta to ra sau ho tan 1 phntrong kim d Phng trnh phn ng Lng kt ta hai phn ng bng nhau nn: 2x 0,14 = 0,11 x = 0,125 mol September 2011 Trang 51 AOTRANGTB.COM Gii t x l s mol Cu(NO3)2 b nhit phnPhng trnh phn ng: Bo ton khi lng, ta c mX = 6,58 4,96 = 1,62 gam 32.0,5x + 46.2x = 1,62 x = 0,015 molHp th X vo nc [HNO3] = 0,1 pH = 1 Gii Ta c: 22CO CH O HV Vn= n= 22,4 22,4a an= n= 18 9Hn hp hai ancol no, n chc nn 2 2ancol H O CO [O] trong ancola Vn= n- n= - = n18 22,4Bo ton khi lng, ta c ancol C H OV a a Vm= m = m+ m+ m= 12.+ 1.+ 16.- 22,4 9 18 22,45V= a - = 28| | |\ .Vma - 5, 6Gii September 2011 Trang 55 AOTRANGTB.COM Gii Ta c: 22 22COCO H O n 2nH On= 0,4 mol n= nete: C H On= 0,4 mol Trong hn hp ban u c 1ancol khng no c 1t loi B, C S phn ng t chy n 2n2C H OnCO0,4 0,4 molnete0,4 m=.(14n + 16) = 7,2 n = 4n Tng s C trong hai ancol bng 4 Chn D. Gii Ta c: +3FeCuHNOn= 0,02n= 0,03n= 0,4n 0, 08=Bo ton e: ne nhng = 0,12 mol < ne nhn = 0,24 mol Fe v Cu tan ht Vit li phng trnh nhn e September 2011 Trang 62 AOTRANGTB.COM Ta c: Phng trnh ion: OH- d 0,02 mol [OH-] = 0,1M (th tch dung dch lc sau l 200 ml) pOH = 1 pH = 13 Gii Gii Cho kh X vo Ca(OH)2 thu c kt ta trong X c CO2 Do khi lng mol trung bnh ca X bng 32 nn Xgm CO, CO2 v O2 d (do in cc lmbng than ch nn tc dng vi O2) n = 0,02 mol trong 22,4 lt X c Trong 67,2 m3 X c t: Ta c: Phng trnh phn ng September 2011 Trang 67 AOTRANGTB.COM Nhn thy khng c natri halogen no c khi lng mol > 201 Trng hp ny loi Vy trng hp cn li l NaF v NaCl mNaF = 6,03 mNaCl = 2,52 gam % = 41,8% Gii Sau phn ng Cu cn d 2,4gam, vy trong dung dch tn ti ionFe2+v Cu2+ (do Cu d nntip tc kh Fe3+ thnh Fe2+) mX phn ng = 61,2 mCu d = 58,8 gam t: Bo ton e, ta c: T (1) v (2), suy ra a = 0,375 v b = 0,15S chuyn ho thnh mui mmui = 151,5 gam September 2011 Trang 68 AOTRANGTB.COM Gii X phn ng vi Cu(OH)2 trong mi trng kim loi B, C Ta c: 22 22COH O COH O n = 0,0195 n= nn = 0,0195 X c 1 t loi A Chn D. Gii Ta c:nX = nNaOH = a mol X cha 1 chc COOH (hoc OH gn trc tip vo vng benzen) 2X Hn= n= a mol X c 2 H linh ng Chn C. Gii Ta c:2++3CuHNOn= 0,16n= 0,4n= 0,32SauphnngthuchnhpbtkimloitcCuvFed,vytrongdungdchchcionkim loi Fe2+ Bo ton e, ta c +3 224H +NO +3e NO+2H O0,40,3 0,1V =Cu +2eCu0,16 0,320,16 + 2, 24 lt ne nhn = 0,62 mol mFe phn ng = 17,36 gam mFe d = m 17,36 0,6m gam hn hp gm Cu v Fe d (m 17,36) + 0,16.64 = 0,6m m = 17,8 gam September 2011 Trang 70 2 2C HX0,15 %n= = 0,250, 6c = 0,25 (3)a + b + c(phn trm s mol khng thay i) T (1), (2) v (3), suy ra a = 0,2, b = 1 v c = 0,1 Ch:Khicho2thnghimkhngngnhtvnvthphngphpgiillptlmol Gii Gi s: -Hn hp ch gm axit Y: RCOOH -Hn hp ch gm Z: R(COOH)2 0,2 < nX < 0,4 Tnh s C (do Y v Z c cng s C) S C = 2: CH3COOH (Y) v HOOC-COOH (Z) loi A, B t:(1) Ta c: T (1) v (2) suy ra a = 0,2 v b = 0,1 %mZ = 42,86%Gii September 2011 Trang 71 AOTRANGTB.COM XNaOHn= 0,25 moln= 0,3 molnNaOH d = 0,05 mol (do n chc nn t l mol 1:1) Ta c:Mancol > 32 gc ancol trong X phi l C2H5- X: NH2-CH2-COO-C2H5 2 2 X+NaOHNH CH COONa0,250,25 molTrong dung dch Y c 0,25 mol mui v 0,05 mol NaOH d mrn = 26,25 gam Gii t (X): n 2nC H OPhng trnh Chnh lch khi lng: Phng trnh t chy X m = 0,5.(14n + 16) = 17, 8 gamGii Ta c: 2++CuAgn= 0,02n= 0,02Nhng thanh Fe vo dung dch theo th t Ag+ b kh trc September 2011 Trang 77 AOTRANGTB.COM 2 2 4223 22 + O H SOd2x y 22250 ml500 ml (X)550 ml (Y)H OCO(CH ) NHCOH OC HNN V= 550 - 250 = 300 ml t: 3 2x y(CH ) NHC HV= a ml a + b = 100 a = 100 - b (1)V= b ml 3 2 2 2 22 2x y(CH ) NH 2CO +3,5H O+0,5Na 2a3,5a0,5aC HxCO +0,5yH Ob xb 0,5yb2 2CO NV +V= 2502a + xb + 0,5a = 250 2,5a + xb = 250 (2) Th (1) vo (2) ta c: loi A, D Ta c s H trung bnh ca hn hp X = S H ca amin bng 7 nn ta loi C, v C2H6 v C3H8 lun c s H trung bnh ln hn 6 Chn B. Gii Ta c: Mt khc, s lin kt t < 3, theo quy tc v s H ti a ta c September 2011 Trang 80 AOTRANGTB.COM Gii Ta c: 33BaCOCaCOn= 0,06n= 0,07TH1:Ch c Na2CO3 tc dng vi BaCl2 cho kt ta (V th tch gim mt na) TH2: Khi un nng c hai mui u tc dng vi CaCl2 theo phng trnh y = 0,04 mol Bo ton C, ta c: Lu : Mui CaCl2, BaCl2 khng cho kt ta vi nhit thng. Khi un nng b nhit phn cho 2-3CO cho kt ta vi hai mui trn ot 23 3 2 22 23 3HCO CO + CO +H OCa +COCaCO + +Gii September 2011 Trang 84 AOTRANGTB.COM n++ 24 2 2Fe Fe +ne x nx mol 4H +SO +2e SO +H O5x2,5x nx = 2,5x = y Chn D. Gii X phn ng ti a vi 2 mol HCl hoc 2 mol NaOH (X): NH2-R-(COOH)2 v R-NH2 S 2COXn6C = =3n 2 =Do s mol mi cht trong X bng nhau nn s C trung bnh cng chnh l s C ca mi cht (X): NH2-CH-(COOH)2 v C3H7-NH2 Bo ton H v N, suy ra: x = 7 v y = 1 Gii Ta c h phng trnh: m = 10,9 gam Gii S phn ng: September 2011 Trang 110 AOTRANGTB.COM mmui = Lu : Ngoi ion kim loi ban u cn c ion Na+ Gii T khi lng ring ta c: 1,55 gam Ca chim th tch l 1 cm3 mol Ca chim th tch l 1 cm3 1 mol Ca chim th tch lcm3 (th tch ny gm nguyn t Ca v khe rng) Th tch thc m cc nguyn t Ca chim ly trong 1 mol = Vy th tch 1 nguyn t Ca = Ta c: Gii Nhn xt: Cc cht u c cng thc tng qut l CnH2n-2O2Ta c: Phng trnh t chy < m = 18 gam Vy dung dch X gim 18 10,62 = 7,38 gam so vi ban u September 2011 Trang 88 AOTRANGTB.COM Gi s X l amin no n 2n+3 2 2 2C H N nCO +(n + 1,5)H O+0,5NV nV (n + 1,5)V 0,5V2 2 2CO H O NV+ V + V= nV + (n + 1,5)V + 0,5V = 8V n = 3 Chn A. Gii Tch nc 2 ancol thu c 2 anken Hai ancol c cu to i xng hoc ancol bc 1 loi A. Nhn xt p n Y l ancol no, n chc 2 5 2n 2n+1 2C H OH3H O (1) 1 mol3 molC H OH(n + 1)H O (2) 1 mol(n + 1) molTa c: Chn C. Gii Ta c: Phng trnh phn ng nCuO d = 0,2 0,03 = 0,17 mol mX = mCu + mCuO d = 15,52 gam%mCu = 12,37% September 2011 Trang 90 AOTRANGTB.COM 16 32 2 2 218 36 2 2 218 32 2 2 2a.panmiticC H O16CO +16H Oa.stearic C H O18CO +18H Oa.linoleicC H O18CO +16H ONhn xt a.pamitic v a.stearic trong phn t c 1t khi t chy cho 2 2CO H On= n , nn s chnh lch s mol CO2 v H2O l ca a.linoleic 2 2CO H O a.linoleica.linoleic n- n= 0,03 = 2n n= 0, 015 molCch khc: Gii phng trnh 3 n Gii Ta c: [O] COoxitn = n= 0,8molM + H2SO4 0,9 mol SO2Bo ton e, ta c: n c hai gi tr 2 v 3 Gii Ta c: September 2011 Trang 94 AOTRANGTB.COM 42 2 2 41CuSO + H O Cu +O+ H SO2a a 0,5aa molm dd gim = 2 2 4CuO H SOm+m= 8 gam 64a + 32.0,5a = 8 n= 0,1 mol a = 0,1 molGi d = b mol Vy trong dung dch Y c b mol CuSO4 d v 0,1 mol H2SO4 u tc dng vi Fe d = 0,3 (0,1 + b) = 0,2 b12,4 gam kim loi gm c Cu v Fe d ban u = a + b = 0,25 mol [CuSO4] = 1,25M Gii 8Al +3Fe3O44Al2O3+9Fe Ban u:0,40,15 Phn ng: x 0,375x 1,125x Kt thc: (0,4 x)(0,15 0,375x) 1,125x Hn hp sau phn ng gm c Al2O3, Fe3O4 d, Al d (0,4 x) mol, Fe 1,125x mol 2 422 4 H SOl2HH SO l2Fe H 1,125x1,125x n= 1,125x + 1,5.(0,4 - x) = 0,48 3 Al H2(0,4 - x)1,5.(0,4 - x) x = 0, 32 mol0, 32H == 80%0, 4Gii X l anehit no, n chc X c 1t September 2011 Trang 100 AOTRANGTB.COM Ta c: Theo th t trong dy in ho Cu s b oxi ho trc Gii p dng LBTKL, ta c HCl HClm= 17,64 - 8,88 = 8,76 n= 0,24 2 x 3 xR(NH ) +xHCl R(NH Cl)0,24 0,24xamin8,88 M= = 37x0,24x ( vi x l s nhm chc NH2) Chn x = 2 M = 74 (H2NCH2CH2CH2NH2) Gii Ta c:X no, n chc (do X n chc c 1O) Phng trnh ete ho September 2011 Trang 101 AOTRANGTB.COM KHI A-2011 M 273 Gii Ta c: 3 2KClCu(NO )n= 0,1n= 0,15Phng trnh in phn: 3 2 2 32KCl+Cu(NO )Cu+Cl +2KNO 0,10,05 0,05 0,053 2Cu(NO )n d = 0,1 molmdd gim = 2Cu Clm+ m= 0,05.64 + 0,05.71 = 6,75 gam < 10,75 gam Cu(NO3)2 tip tc b in phn mdd gim = 6,75 + 64a + 0,5a.32 = 10,75 a = 0,05 mol < d = 0,1mol Vy trong dung dch cn Cu(NO3)2, KNO3 v HNO3. Gii Axit hai chc, mch h c 1 lin kt C=C CTTQ: CnH2n-4O4 Ta c:Bo ton nguyn t September 2011 Trang 102 AOTRANGTB.COM ( )axit C H OV 5V ym== m+ m+ m=.12 + 2.y + -.4.16 =22, 4 224 2 | | |\ .55Vx - 30y2828V =x + 30y55Gii Ta c: Khi cho Fe v Cu vo dung dch theo th t trong dy in ho Fe s phn ng trc Nhn xt khi lng cht rn thu c sau phn ng l 0,75m ln hn 0,7m khi lng ca Cu,vy chng t Cu khng tham gia phn ng. mFe phn ng = 0,25m gam Do Fe d nn mui thu c cha ion Fe2+S phn ng Bo ton nguyn t N 3 2 x[N] [N] [N]axit Fe(NO ) NOn= n+ n0, 7x.2 + 0,25 = x = 0, 225 mol nFe phn ng = 0,225 mol mFe phn ng = 12,6 gam = 0,25m m = 50,4 gam Gii 22 62 6 2 2 2 4 dd Br Ni,t2 2 2 2X 0,2 mol2YC HC H C Hx mol C H M = 16Hx mol C H HH Ta c: tng == 10,8 gam September 2011 Trang 103 AOTRANGTB.COM 2 6 2C H Hm + m= 16.0,2 = 3,2 gam2 4 2 2 2 6 2C H C H C H H= m+ m+ m+ m= 14 gamX Ym= m2 2 2X C H Hm= m+ m= 26x + 2x = 14 x = 0,5 molt chy trng thi Y cng nh t chy trng thi X 22 2 2 2 2O2 2 25C H+O2CO + H O2 0,5 1,25 mol n= 1,5 mol 1H +OH O20,5 0,25 mol 2OV= 33, 6 ltGii Ta c: 2-2+COOHCan= 0,03n= 0,05n= 0,0125-2OHCOn1 c 2 phn ng xy ra: p dng cng thc: September 2011 Trang 123 AOTRANGTB.COM 2 22 2RCHO+HRCH OH 11 mol1RCH OH H210,5 molNa V = 11,2 lt Gii Ta c: R 2 2R14.1%N = 15,73=0,1573 M= 89 X: NH -CH -COO-M 3CHS phn ng: m = 2,67 gam Gii Lu : Tnh oxi ha tng dn theo th t: C ngha l Zn s kh Fe3+ xung Fe2+ ri xung FePhng trnh phn ng: mdd gim = mZn mFe = 9,6 65(0,12 + x) 56x = 9,6 x = 0,2 mol nZn = 0,32 mol m = 20,8 gam Gii September 2011 Trang 124 AOTRANGTB.COM C4H10 C4H8+H2 ~ Br2 C4H10 C4H6+ 2H2~ 2Br2 C4H10Ta c: XM= 23,2Gi s 4 10C Hnban u = 1, p dng LBTKL, ta c: 4 10C H Y YY 58.n= n .M1.58 = n .23,2 4 10C H XYm= mn= 2, 5Vy 2,5 mol Y cn 1,5 mol Br2 0,6 mol Y cn 0,36 mol Gii p dng LBT e: Vy 20 ml c 0,015 mol Fe2+ 150 ml c 0,1125 mol Fe2+ Gii Ta c: S H = Y c 1 nhm COOH v Z a chc, do khng c axit hu c no c 1H nn c hai axit u c2H Y: HCOOH (x mol) v HOOC-COOH (y mol) September 2011 Trang 125 AOTRANGTB.COM Gi s a = 1, ta c h phng trnh: 2 2HCOOHHCOOH CO (COOH)m= 18,4 x + y = 1x = 0,4 %m =x + 2y = n = 1,6 m= 54 y = 0,6 25, 41%Gii 2 2 2CO ( )+H O ( ) CO( ) + H( ) k k k kBan u: 0,20,3 Phn ng:x xx x Kt thc:(0,2 x)(0,3 x) x x 22 2C2[CO ][H ] x K=1[CO][H O] (0,2 - x)(0,3 - x) x = 0,12 mol = =2COH O 20, 08n= 0,08 [CO] = 100,18n =0,18 [H O] = 10 = =0, 008M0, 018MGii Gi x l s mol ca Cu tham gia phn ng: + 2Cu+2AgCu +2Ag x 2xx+Trong X c (0,08 2x) mol Ag+ v x mol Cu2+Cho Zn vo X, p dng LBT e ta c: +2+2+ Ag+ 1eAg(0, 08 - 2x) (0,08 - 2x) Zn Zn +2e 0,040,08 Cu+2eCux 2x nZn d = 0,09 0,04 = 0,05 mol mZn d = 3,25 gam Ta c khi lng Ag kt thc 2 phn ng = 0,08.108 = 8,64 gam (do Zn d nn AgNO3 ht)mcht rn c 2 phn ng = 10,53 + 7,76 = 18,29 = 0,08 Download ti liu hc tp, xem bi ging ti : http://aotrangtb.com September 2011 Trang 126 AOTRANGTB.COM mCu + mAg + mZn d = 18,29 mCu = 6,4 gam Nhn xt: Cu tham gia phn ng (1) sau c to thnh phn ng (2) vi khi lng bngnhau, c th xem Cu khng thay i khi lng trong phn ng. ----------The End---------- Muc luc: Nam Trang A-2007.......................................... 1B-2007 .......................................... 17 A-2008.......................................... 27 B-2008 .......................................... 37 A-2009.......................................... 47 B-2009 .......................................... 60 A-2010.......................................... 74 B-2010 .......................................... 89 A-2011.......................................... 101 B-2011 .......................................... 114 Trn y l bi gii cc thi i hc cc nm, hi vng n s gip ch cho cc bn trong ma thi i hc sp n, trong bi gii t nhiu cng c sai st, cc bn hy gp v gi phn hiv: LangQuang@yahoo.com Khoa Y H Y Dc Tp.HCM Chuc cac ban thanh cong trong k thi sap en !Download ti liu hc tp, xem bi ging ti : http://aotrangtb.com