De Tang Kem Hoa Hoc Megabook So 1

De Tang Kem Hoa De Tang Kem Hoa Hoc Megabook So 1De Tang Kem Hoa Hoc Megabook So 1De Tang Kem Hoa Hoc Megabook So 1De Tang Kem Hoa Hoc Megabook So 1De Tang Kem Hoa Hoc Megabook So 1Hoc Megabook So 1

www.facebook.com/groups/TaiLieuOnThiDaiHoc01 TNG KẩM S MễN HểA HC Thi gian: 90 phỳt w fa ce bo ok c om /g ro up s/ Ta iL ie uO nT hi D H oc 01 Cõu 1: un núng 0,1 mol este no, n chc mch h X vi 30 ml dung dch 20% (D = 1,2 g/ml) ca mt hiroxit kim loi kim A Sau kt thỳc phn ng x phũng hoỏ, cụ cn dung dch thỡ thu c cht rn Y v 4,6 gam ancol Z, bit rng Z b oxi hoỏ bi CuO thnh sn phm cú kh nng phn ng bc t chỏy cht rn Y thỡ thu c 9,54 gam mui cacbonat, 8,26 gam hn hp CO2 v hi nc Cụng thc cu to ca X l: A CH3COOCH3 B CH3COOC2H5 C HCOOCH3 D C2H5COOCH3 Cõu 2: Hũa tan ht 8,56 gam hn hp X gm Fe3O4 v CuO 400 ml dung dch HNO3 1M, kt thỳc cỏc phn ng thu c dung dch Y v 0,01 mol NO (sn phm kh nht) in phõn dung dch Y (in cc tr, khụng mng ngn, hiu sut 100%) vi cng dũng in khụng i 5A, gi 20 phỳt 25 giõy Khi lng catot tng lờn v tng th tớch khớ thoỏt (ktc) hai in cc kt thỳc in phõn ln lt l A 1,28 gam v 2,744 lớt B 2,40 gam v 1,848 lớt C 1,28 gam v 1,400 lớt D 2,40 gam v 1,400 lớt Cõu 3: un núng mt ru n chc X vi dung dch H2SO4 c trng iu kin nhit thớch hp sinh cht hu c Y , t ca Y so vi X l 1,4375 Cụng thc phõn t ca Y l A C3H8O B C2H6O C C4H10O D CH4O Cõu 4: t chỏy hon ton 0,05 mol hn hp M gm anehit X v este Y, cn dựng va 0,155 mol O2, thu c 0,13 mol CO2 v 2,34 gam H2O Mt khỏc, cho 0,1 mol M phn ng vi lng d dung dch AgNO3 NH3, kt thỳc cỏc phn ng thu c 21,6 gam Ag Cụng thc cu to thu gn ca X, Y ln lt l A CH3CHO v HCOOCH3 B CH3CHO v HCOOC2H5 C HCHO v CH3COOCH3 D CH3CHO v CH3COOCH3 Cõu 5: Dựng thờm mt thuc th hóy phõn bit cỏc cht rn mu trng Na2O, Al, MgO, Al2O3 ? A Qu tớm B Nc C Dung dch NaOH D Dung dch HCl Cõu 6: Cho 39,2 gam hn hp M gm Fe, FeO, Fe3O4, Fe2O3, CuO v Cu (trong ú oxi chim 18,367% v lng) tỏc dng va vi 850 ml dung dch HNO3 nng a mol/l, thu c 0,2 mol NO (sn phm kh nht ca N+5) Giỏ tr ca a l A 2,0 B 1,5 C 3,0 D 1,0 Cõu 7: Hp cht hu c X tỏc dng vi dung dch NaOH v dung dch Brom nhng khụng tỏc dng vi dung dch NaHCO3 Tờn gi ca X l A metyl axetat B axit acrylic C anilin D Phenol Cõu 8: Dn t t 5,6 lớt khớ CO2 (ktc) vo 400 ml dung dch cha ng thi cỏc cht NaOH 0,3M; KOH 0,2M ; Na2CO3 0,1875M v K2CO3 0,125M thu c dung dch X Thờm dung dch CaCl2 vo dung dch X ti d, s gam kt ta thu c l : A 7,5g B 27,5g C 25g D 12,5g Cõu 9: Hn hp M gm mt andehit v mt ankin (cú cựng s nguyờn t cacbon) t chỏy hon ton x mol hn hp M thu c 3x mol CO2 v 1,8x mol H2O Thnh phn % v s mol ca andehit hn hp M l : A 20% B 30% C 50 % D 40% w w Cõu 10: Mt mu nc cng cha cỏc ion: Mg2+, Ca2+, Cl , SO 24 Cht c dựng lm mm mu nc cng trờn l A NaHCO3 B BaCl2 C Na3PO4 D H2SO4 Cõu 11: Cho hn hp khớ X gm HCHO, C2H2 v H2 i qua ng s ng bt Ni nung núng Sau mt thi gian thu c hn hp Y (gm khớ v hi) t chỏy hon ton Y cn dựng va 0,07 mol O2, sinh 0,055 mol CO2 v 0,81 gam H2O Phn trm th tớch ca HCHO X l A 25,00% B 75,00% C 66,67%% D 33,33% Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Cõu 12: Cho cỏc chuyn húa sau: H ,t X + H2O X1 + X2 0 t X1 + 2[Ag(NH3)2]OH X3 + 3NH3 + 2Ag + H2O w w w fa ce bo ok c om /g ro up s/ Ta iL ie uO nT hi D H oc 01 t X2 + 2[Ag(NH3)2]OH X3 + 3NH3 + 2Ag + H2O X3 + HCl axit gluconic + NH4Cl Cht X l A xenluloz B mantoz C tinh bt D saccaroz Cõu 13: Thu phõn hon ton m gam mt pentapeptit mch h M thu c hn hp gm hai amino axit X1, X2 (u no, mch h, phõn t cha mt nhúm -NH2 v mt nhúm -COOH) t chỏy ton b lng X1, X2 trờn cn dựng va 0,1275 mol O2, ch thu c N2, H2O v 0,11 mol CO2 Giỏ tr ca m l A 3,17 B 3,89 C 4,31 D 3,59 Cõu 14: in phõn dung dch CuSO4 vi in cc tr v dũng in chiu cng 1A Kt thỳc in phõn catot bt u cú bt khớ thoỏt trung hũa dung dch sau in phõn cn dựng va 50 ml dung dch NaOH 0,2M Thi gian in phõn l : A 193s B 96,5s C 965s D 1930s Cõu 15: Dóy gm cỏc cht c sp xp theo chiu gim dn nhit sụi t trỏi sang phi l A C2H5COOH, C2H5CH2OH, CH3COCH3, C2H5CHO B C2H5COOH, C2H5CHO, C2H5CH2OH, CH3COCH3 C C2H5CHO, CH3COCH3, C2H5CH2OH, C2H5COOH D CH3COCH3, C2H5CHO, C2H5CH2OH, C2H5COOH Cõu 16: Hp th ht 0,15 mol CO2 vo dung dch cha 0,025 mol NaOH v 0,1 mol Ba(OH)2, kt thỳc cỏc phn ng thu c m gam kt ta Giỏ tr ca m l A 14,775 B 9,850 C 29,550 D 19,700 Cõu 17: Hn hp X gm kim loi kim v mt kim loi kim th Hũa tan hon ton 1,788 gam X vo nc thu c dung dch Y v 537,6 ml khớ H2 (ktc) Dung dch Z gm H2SO4 v HCl ú s mol ca HCl gp hai ln s mol ca H2SO4 Trung hũa dung dch Y bng dung dch Z to m gam hn hp mui Giỏ tr ca m l : A 3,792 B 4,656 C 2,790 D 4,46 Cõu 18: Hn hp X gm H2, C2H4 v C3H6 cú t so vi H2 l 9,25 Cho 22,4 lớt X (ktc) vo bỡnh kớn cú són mt ớt bt ung núng bỡnh mt thi gian,thu c hn hp khớ Y cú t so vi H2 bng 10 Tng s mol H2 ó phn ng l : A 0,015 mol B 0,075 mol C 0,07 mol D 0,05 mol Cõu 19: Hn hp khớ X gm O2 v O3 cú t so vi H2 l 22 Hn hp khớ Y gm metyl amin v etylamin cú t so vi H2 l 17,833 t chỏy hon ton V1 lớt khớ Y cn va V2 lớt X (bit sn phm chỏy gm CO2, H2O, N2, cỏc cht khớ o cựng iu kin nhit , ỏp sut) T l V1 : V2 l : A : B : C : D : Cõu 20: Hũa tan m gam kim loi M dung dch HCl d thu c 2,46 gam mui Mt khỏc cho m gam kim loi M tỏc dng vi Cl2 d thu c 3,17 gam mui Kim loi M l : A Cr B Cu C Fe D Al Cõu 21: Cho cỏc phỏt biu sau : (1) Glucozo cú kh nng tham gia phn ng bc (2) S chuyn húa tinh bt c th ngi cú sinh mantozo (3) Mantozo cú kh nng tham gia phn ng bc (4) Saccarozo c cu to t gc glucozo v fructozo Trong s phỏt biu trờn, s phỏt biu ỳng l A B C D Cõu 22: Hp cht hu c X, mch h cú cụng thc phõn t C5H13O2N X phn ng vi dung dch NaOH un núng, sinh khớ Y nh hn khụng khớ v lm xanh qu tớm m S cụng thc cu to tha iu kin trờn ca X l Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 w w w fa ce bo ok c om /g ro up s/ Ta iL ie uO nT hi D H oc 01 A B C D 10 Cõu 23: Hn hp X gm 3,92 gam Fe; 16 gam Fe2O3 v m gam Al Nung X nhit cao iu kin khụng cú khụng khớ, thu c hn hp cht rn Y Chia Y thnh phn bng Phn tỏc dng vi dung dch H2SO4 loóng d, thu c 4a mol khớ H2 Phn phn ng vi dung dch NaOH d, thu c a mol khớ H2 (bit cỏc phn ng u xy hon ton) Giỏ tr ca m l : A 7,02 B 4,05 C 3,51 D 5,40 Cõu 24: Cho 4,4 gam andehit n chc X phn ng hon ton vi lng d dung dch AgNO3/NH3 un núng, thu c 21,6 gam Ag Cụng thc X l : A CH3CHO B C2H3CHO C HCHO D C2H5CHO Cõu 25: Cho cỏc phng trỡnh phn ng : (a) Fe + 2HCl FeCl2 + H2 (b) Fe3O4 + 4H2SO4 Fe2(SO4)3 + FeSO4 + 4H2O (c) 2KMnO4 + 16HCl 2KCl + 2MnCl2 + 5Cl2 + 8H2O (d) FeS + H2SO4 FeSO4 + H2S (e) 2Al + 3H2SO4 Al2(SO4)3 + 3H2 (g) Cu + H2SO4 CuSO4 + SO2 + 2H2O Trong cỏc phn ng trờn, s phn ng m ion H+ úng vai trũ cht oxi húa l : A B C D Cõu 26: Mt hp cht X cha nguyờn t C, H, O cú t l lng mC : mH : mO = 48 : : Hp cht X cú cụng thc n gin nht trựng vi cụng thc phõn t S ng phõn cu to thuc loi ancol thm ng vi cụng thc phõn t ca X l A B C D Cõu 27: Hn hp X gm Fe, Fe2O3, Fe3O4 Cho khớ CO i qua m gam X nung núng, sau mt thi gian thu c hn hp cht rn Y v hn hp khớ Z Cho ton b Z vo dung dch Ca(OH)2 d, n phn ng hon ton thu c gam kt ta Mt khỏc hũa tan hon ton Y dung dch H2SO4 c núng d thu c 1,008 lớt khớ SO2 (ktc, sn phm kh nht) v dung dch cha 18 gam mui Giỏ tr ca m l : A 6,80 B 13,52 C 7,12 D 5,68 Cõu 28: Cho 200 ml dung dch Ba(OH)2 0,1M vo 300 ml dung dch NaHCO3 0,1M thu c dung dch X v kt ta Y Cho t t dung dch HCl 0,25M vo X n bt u cú khớ sinh thỡ ht V ml Bit cỏc phn ng u xy hon ton Giỏ tr ca V l : A 80 ml B 160 ml C 60ml D 40 ml Cõu 29: t chỏy hon ton 0,1 mol cht hu c X (cha C, H, O) cn dựng va 0,6 mol O2, sinh 0,4 mol CO2 S ng phõn cu to ca X l A B C D Cõu 30: Hũa tan hon ton m gam hn hp gm Ba v Al2O3 vo nc thu c dung dch X v 0,2 mol H2 Sc khớ CO2 ti d vo X, xut hin 11,7 gam kt ta Giỏ tr ca m l A 37,60 B 21,35 C 42,70 D 35,05 Cõu 31: Ho tan ht 2,32 gam hn hp X gm FeO, Fe2O3, Fe3O4, ú t l lng ca FeO v Fe2O3 l : 20 200 ml dung dch HCl 1M thu c dung dch Y Dung dch Y ho tan c ti a bao nhiờu gam st ? A 3,36 gam B 3,92 gam C 4,48 gam D 5,04 gam Cõu 32: t chỏy hon ton 0,05 mol hn hp M gm anehit X v este Y, cn dựng va 0,155 mol O2, thu c 0,13 mol CO2 v 2,34 gam H2O Mt khỏc, cho 0,1 mol M phn ng vi lng d dung dch AgNO3 NH3, kt thỳc cỏc phn ng thu c 21,6 gam Ag Cụng thc cu to thu gn ca X, Y ln lt l A CH3CHO v HCOOCH3 B CH3CHO v HCOOC2H5 C HCHO v CH3COOCH3 D CH3CHO v CH3COOCH3 Cõu 33: Hn hp M gm ancol no, n chc, mch h X v hirocacbon Y, v X cú s nguyờn t cacbon ln hn Y t chỏy hon ton mt lng M cn dựng va 0,07 mol O2, thu c 0,04 mol CO2 Cụng thc phõn t ca Y l Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Ta iL ie uO nT hi D H oc 01 A C3H8 B C2H6 C CH4 D C4H10 Cõu 34: Cho 6,44 gam mt ancol n chc phn ng vi CuO un núng, thu c 8,68 gam hn hp X gm anehit, nc v ancol d Cho ton b X tỏc dng vi lng d dung dch AgNO3 NH3, kt thỳc cỏc phn ng thu c m gam Ag Giỏ tr ca m l A 30,24 B 86,94 C 60,48 D 43,47 Cõu 35: Hp cht hu c X cú cụng thc C2H8N2O4 Khi cho 12,4g X tỏc dng vi 200ml dung dch NaOH 1,5M thu c 4,48 lớt( ktc) khớ X lm xanh qu tớm m Cụ cn dung dch sau phn ng thu c m gam cht rn khan Giỏ tr ca m l A 17,2 B 13,4 C 16,2 D 17,4 Cõu 36: Cho thớ nghim nh hỡnh v sau: Phn ng xy ng nghim nm ngang l: A Zn + 2HCl ZnCl2 + H2 B H2 + S H2S C H2S + Pb(NO3)2 PbS + 2HNO3 D 2HCl + Pb(NO3)2 PbCl2 + 2HNO3 up s/ Cõu 37: Cú 500 ml dung dch X cha cỏc ion: K+, HCO , Cl v Ba2+ Ly 100 ml dung dch X phn ng vi w w w fa ce bo ok c om /g ro dung dch NaOH d, kt thỳc cỏc phn ng thu c 19,7 gam kt ta Ly 100 ml dung dch X tỏc dng vi dung dch Ba(OH)2 d, sau cỏc phn ng kt thỳc thu c 29,55 gam kt ta Cho 200 ml dung dch X phn ng vi lng d dung dch AgNO3, kt thỳc phn ng thu c 28,7 gam kt ta Mt khỏc, nu un sụi n cn 50 ml dung dch X thỡ lng cht rn khan thu c l A 23,700 gam B 14,175 gam C 11,850 gam D 10,062 gam Cõu 38: Thc hin cỏc thớ nghim sau : (a) Nhit phõn AgNO3 (b) Nung FeS2 khụng khớ (c) Nhit phõn KNO3 (d) Cho dung dch CuSO4 vo dd NH3 d (e) Cho Fe vo dung dch CuSO4 (f) Cho Zn vo dung dch FeCl3 d (g) Nung Ag2S khụng khớ (h) Cho Ba vo dung dch CuSO4 d S thớ nghim thu c kim loi sau phn ng kt thỳc l : A B C D Cõu 39: Cho 9,55 gam hn hp gm Mg, Al v Zn tỏc dng va vi 870 ml dung dch HNO3 1M, thu c dung dch cha m gam mui v 0,06 mol hn hp khớ N2 v N2O T ca hn hp khớ so vi H2 l 20,667 Giỏ tr ca m l A 54,95 B 42,55 C 40,55 D 42,95 Cõu 40: Cho cỏc thớ nghim sau: Sc Cl2 vo dung dch Ca(OH)2 Sc CO2 vo dung dch cloruavụi Sc O3 vo dung dch KI Sc H2S vo dung dch FeCl2 Cho HI vo dung dch FeCl3 Cho dung dch H2SO4 c núng vo NaBr tinh th S trng hp xy phn ng oxi húa kh l: A B C D Cõu 41: Hũa tan hon ton Fe3O4 dung dch H2SO4 loóng d, thu c dung dch X Trong cỏc cht NaOH, Cu, Fe(NO3)2, KMnO4, BaCl2, Cl2, v Al, s cht cú kh nng tỏc dng vi dung dch X l : Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Cl2 (1:1) dungdichBr2 CuO,t NaOH KOH / ROH X Q Z Propen Y T HBr 5000 C uO nT hi D H oc 01 A B C D Cõu 42: Cho cao su buna tỏc dng vi Cl2 (trong CCl4 cú mt P) thỡ thu c polime no, ú clo chim 17,975% v lng Trung bỡnh c phõn t Cl2 thỡ phn ng c vi bao nhiờu mt xớch cao su buna? A B C 10 D Cõu 43: Cht hu c X cú cụng thc C6H10O4 ch cha loi nhúm chc un núng X vi dung dch NaOH d thu c mui ca1 axit cacboxylic Y v mt ancol Z Bit Y cú mch cacbon khụng phõn nhỏnh v khụng cú phn ng bc S cụng thc cu to ca X l : A B C D Cõu 44: Tng s proton ht nhõn nguyờn t X v Y l 25 Y thuc nhúm VIIA iu kin thớch hp n cht X tỏc dng vi Y Kt lun no sau õy ỳng? A X l kim loi, Y l phi kim B trng thỏi c bn X cú electron c thõn C Cụng thc oxit cao nht ca X l X2O D Cụng thc oxit cao nht ca Y l Y2O7 Cõu 45: Hp cht Q(cha C,H,O) c iu ch theo s : w w w fa ce bo ok c om /g ro up s/ Ta iL ie Nu ly ton b lng cht Q thu c t 0,14mol propen cho tỏc dng vi dung dch AgNO3/NH3 d thỡ lng kt ta thu c l bao nhiờu ? (Bit hiu sut cỏc phn ng l 100%) A 42,28 B 57,4 C 30,24 D 52,78 Cõu 46: in phõn cú mng ngn vi in cc tr 400ml dung dch hn hp CuSO4 aM v NaCl 1M vi cng dũng in 5A 3860s Dung dch to thnh b gim so vi ban u 10,4g Giỏ tr ca a l A 0,125M B 0,2M C 0,129M D 0,1M Cõu 47: Cho 200 ml dung dch Al2(SO4)3 0,5M tỏc dng vi 200 gam dung dch NaOH thu c 11,7 gam kt ta trng Nng dung dch NaOH ó dựng l : A 9% B 12% C 13% D Phng ỏn khỏc Cõu 48: t chỏy hon ton a mol hn hp X gm hai anehit, thu c a mol H2O Cụng thc ca hai anehit cú th l A HCHO v OHC-CH2-CHO B HCHO v CHC-CHO C OHC-CHO v CH3CHO D CH3CHO v CHC-CHO Cõu 49: Cho cỏc cht sau: CH3COOCH2CH2Cl, ClH3N-CH2COOH, C6H5Cl (thm), HCOOC6H5 (thm), C6H5COOCH3 (thm), HO-C6H4-CH2OH (thm), CH3CCl3, CH3COOC(Cl2)-CH3 Cú bao nhiờu cht tỏc dng vi NaOH c d, nhit v ỏp sut cao cho sn phm cú mui? A B C D Cõu 50: Hn hp X gm axit no n chc A v axit khụng no n chc cú liờn kt ụi B, C l ng ng k tip (MB < MC) u mch h X tỏc dng va vi 100ml dung dch NaOH 2M, thu c 17,04 gam hn hp mui Mt khỏc t chỏy hon ton X thu c tng lng CO2 v H2O l 26,72 gam % s mol ca B hn hp X l: A 20% B 30% C 22,78% D 34,18% Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 - LI GII Ta cú: nX = nAOH (p) = nZ = 0,1 mol MZ = 14m + 18 = 4,6 = 46 m = 0,1 H O mY + m O2 (p /-) = m Na 2CO m CO m H2O Ta iL ie Vy: uO Na CO C n H n COONa : 0,1 mol O2 ,t CO NaOH d -: 0,18 0,1 0,08 mol Y nT hi D Mt khỏc: nA = 30.1,2.20 = 9,54 MA = 23 A l Na nNaOH (ban u) = 7,2 0,18 mol 40 100.( M A 17) M A 60 oc X l este no, n chc, mch h : CnH2n+1COOCmH2m+1 ( n; m) H Cõu Bi gii : 01 GII CHI TIT V ễN TP, T LUYN ok c om /g ro up s/ (3n 1) Hay 0,1(14n+68) + 0,08.40 + 0,1.32 = 9,54 + 8,26 n = X : CH3COOCH3 ỏp ỏn A Nhn xột : Nu nNaOH phn ng = nEste Este n chc Nu RCOOR (este n chc), ú R l C6H5- hoc vũng benzen cú nhúm th nNaOH phn ng = 2neste v sn phm cho mui, ú cú phenolat: VD: RCOOC6H5 + 2NaOH RCOONa + C6H5ONa + H2O Nu nNaOH phn ng = .neste ( > v R khụng phi C6H5- hoc vũng benzen cú nhúm th) Este a chc Nu phn ng thu phõn este cho anehit (hoc xeton), ta coi nh ancol (ng phõn vi andehit) cú nhúm OH gn trc tip vo liờn kt C=C tn tai gii v t ú CTCT ca este Nu sau thy phõn thu c mui (hoc cụ cn thu c cht rn khan) m mmui = meste + mNaOH thỡ este phi cú cu to mch vũng (lacton): O bo C = O + NaOH HO-CH2CH2CH2COONa ce Nu gc hidrocacbon ca R, mt nguyờn t C gn vi nhiu gc este hoc cú cha nguyờn t halogen thỡ thy phõn cú th chuyờn húa thnh andehit hoc xeton hoc axit cacboxylic t VD: C2H5COOCHClCH3 + NaOH C2H5COONa + CH3CHO +NaCl w w w fa CH3-COO CH3-COO CH + NaOH CH3-COO Na + HCHO Bi toỏn v hn hp cỏc este thỡ nờn s dng phng phỏp trung bỡnh Cõu Bi gii: Ta cú: n HNO3 0, (mol) Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 CuO HNO3 Cu(NO3 )2 H O Phng trỡnh phn ng: 0, 02 0, 04 3Fe3O 28HNO3 9Fe(NO ) NO 14H O 0, 03 n CuO 0, 02 8, 56 0, 03.232 0, 02 (mol) ; n HNO3 80 cũn 0, 28 0, 09 0, 01 01 = 0, 0, 32 0, 08 (mol) oc Dung dch Y cha: Fe3 ;Cu ;H ; NO3 Fe3 1e Fe2 D kh) catot( cc õm, xy s anot ( cc dng, xy s oxi húa) 2HOH O2 H 4e hi H Nu Fe3+ ht uO Nu Cu2+ ht thỡ H+ bao gm lng ban u va lng sinh anot tham gia in phõn trc Fe2+ Ta iL ie nT Cu 2e Cu H 2e H2 up s/ Theo nh lut bo ton electron Tng s electron catot phúng bng tng s electron anot nhn vo: I.t 5.( 3600 20.60 25) ne 0, 25 (mol) F 96500 Ta cú n Fe3 2.n Cu2 0, 13 (mol) 0, 25 mol Fe3 ,Cu b in phõn ht ro Khi lng catot tng lờn l lng Cu bỏm vo: mcatot tng = mcu = 64.0,02=1,28(gam) 0, 25 0, 0625(mol) S mol H2 thoỏt catot l: n e(catot ) n Fe3 2.n Cu 0, 25 0, 13 n H2 0, 06 (mol) 2 Tng th tớch khớ thu c l: V (0, 065 0, 06).22, 2, 744 (lit) c om /g S mol O2 sinh anot l: n O2 n e(anot) fa ce bo ok ỏp ỏn A Cõu Bi lm : H2SO4 c l cht xỳc tỏc nh hng theo nhit nhit 1400C ancol n chc X xy phn ng to ete 2ROH ROR + H2O M M d Y Y ROR X M X M ROH w w w nhit 1800C ancol n chc X xy phn ng to anken M M d Y Y ROR X M X M ROH Theo bi t ca Y so vi X l 1,4375 , ú phn ng phn ng trờn l phn ng to ete M M d Y Y ROR 1, 4375 2R 16 1, 4375 ( R 17 ) R 15 ( CH ) X M X M ROH Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 ỏp ỏn D Cõu Bi gii: t s mol ca anehit X v este Y ln lt l x v y Bo ton nguyờn t O ta c : x 2y 0, 08 (mol)(1) Mt khỏc ta li cú : x y 0, 05 (2) oc 01 n CO2 n H2 O 0, 13 (mol) X, Y u l nhng hp cht no n chc mch h nT Ta iL ie Mt khỏc : CH3CHO Ag n Ag 0, 08 (mol) 0, (mol) uO hi D H Gii h (1)(2) ta c : x = 0,02 (mol) ; y = 0,03 (mol) t cụng thc ca anehit X v este Y ln lt l: CnH2nO v CmH2mO2 Bo ton nguyờn t C ta c: 0,02.n + 0,03.m = 0,13 2.n + 3.m = 13 Ch cú n = v m = l tha Andehit l: CH3CHO; este l C3H6O2 n CH3 CHO 0, 04 (mol) Khi cho 0,1 mol M phn ng vi AgNO3/NH3, ta cú n C3 H6 O2 0, 06 (mol) n Ag 0, (mol) C3 H6O2 cng gng Vy este cú cụng thc l HCOOC2H5 ỏp ỏn B ro Cỏc cht rn cũn li khụng tan nc l Al ; MgO ; Al2O3 Hũa tan ln lt cỏc cht rn khụng tan trờn vo dung dch NaOH va thu c + Cht no tan dung dch NaOH to khớ bay ú l Al om /g up s/ Cõu Bi gii : Hũa tan cỏc cht rn trờn vo nc d, ch cú Na2O tan to dung dch Na 2O H2O NaOH 2Al NaOH HOH NaAl(OH)4 H2 + Cht no tan dung dch to dung dch sut l Al2O3 Al2O3 2NaOH 3HOH 2NaAl(OH)4 bo ok c + Cht cũn li khụng tan dung dch NaOH l MgO Thuc th dung phõn bit cỏc cht rn l nc ỏp ỏn B Cõu Bi gii: fa ce Cỏch 1: Coi hn hp M gm Fe; Cu; O 18, 367.39, 7, Theo bi ta cú mO 7, (gam) n O 0, 45 (mol) 100 16 Gi s mol Fe; Cu ln lt l x mol ; y mol 56.x 64.y 39, 7, 32 ( 1) w w w Khi cho hn hp M tỏc dng va vi HNO3 : Bo ton electron ta c: 3.x 2.y 2.0, 45 0, 2.3 ( 2) Gii h (1);(2) ta c x 0, (mol); y 0, 15 (mol) n e ( ) 3.0, 0, 15 1, (mol) Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 n HNO3 = n NO (mui) + n NO = 1,5 + 0,2 = 1,7 (mol) a = 2(M) Cỏch 2: 18, 367.39, 7, 7, (gam) n O 0, 45 (mol) 100 16 Khi cho hn hp M tỏc dng vi HNO3 2.n O 4.n NO 1, (mol) a (M) oc n H 01 Theo bi ta cú mO D H Cõu Bi gii : Cht khụng tỏc dng vi dung dch NaHCO3 thỡ chng t X khụng cú chc COOH X tỏc dng c vi dung dch NaOH ; v dung dch Brom X l phenol C6 H 5OH NaOH C6 H 5ONa H 2O hi C6 H 5OH Br2 C6 H (OH) Br3 HBr nT ỏp ỏn D Theo bi ta tinh c : n NaOH 0, 12 (mol) 5, n KOH 0, 08 (mol) n OH 0, (mol) 0, 25 (mol) ; n 0, 075 (mol) n 0, 125 (mol) 22, Na CO3 CO23 n K CO 0, 05 (mol) Ta iL ie n CO2 uO Cõu Bi gii : up s/ Khi dn CO2 vo dung dch thỡ cú phng trỡnh phn ng nh sau OH CO 23 H 2O 0, 0, 0, ro CO n CO2 0, 0, 125 0, 225 (mol) CO 32 HOH HCO 0, 15 0, 15 om /g CO 0, 15 bo ok c Na : 0, 27 (mol) K : 0, 18 (mol) Dung dch X sau phn ng cú cha CO3 : 0, 225 0, 15 0, 075 (mol) HCO : 0, 15 (mol) Khi cho dung dch CaCl2 ti d vo X thỡ cú PTP : fa ce Ca CO32 CaCO3 n CaCO3 0, 075 (mol) mCaCO3 0, 075.100 7, (gam) w ỏp ỏn A w w Cõu Bi gii : Do hn hp M gm andehit v ankin ( cú cựng s nguyờn t cacbon) Gi s x = ( mol) Nh vy t chỏy mol hn hp M thu c mol CO2 v 1,8 mol H2O n CO2 S nguyờn t C = andehit v ankin u cú nguyờn t C nM Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Ankin cú cụng thc l C3 H ; andehit cú cụng thc l C3 H yOz S nguyờn t H = 2.n H2 O nM 2.1, 3, , ta cú s nguyờn t H ca ankin = > 3,6 , H % v s mol ca andehit hn hp M l oc mol hn hp M gm C3 H :x (mol) x y x 0, (mol) 2.x y 1, y 0, (mol) C3 H 2O :y(mol) 0, 100 20 % D nT hi ỏp ỏn A Mu nc cng trờn thuc loi nc cng vnh cu Dựng Na3PO4 lm mm mu nc cng trờn Khi ú s kt ta ht c cỏc cation uO Cõu 10 Bi gii: Mg2+; Ca2+ 01 ú y < 3,6 , m y l s chn ú y = Z l s nhúm chc (-CHO) Do ú z =1 CT ca andehit l C3H2O CH C CHO Ta iL ie 3Mg 2PO43 Mg (PO4 )2 3Ca 2PO43 Ca (PO4 ) ỏp ỏn C om /g ro up s/ Cõu 11 Nhn thy X u cú cụng thc chung dng CxH2Oy Hn hp khớ X qua ng s ng bt Ni c hn hp khớ Y õy ó xy phn ng hidro húa Do thnh phn nguyờn t X v Y nh nhau, nờn t hn hp Y cng chớnh l t hn hp X X O2 CO2 H2O Bo ton nguyờn t H n X n H2 O 0, 81 0, 045(mol) 18 ok n HCHO 0, 015 (mol) c Bo ton nguyờn t O ta c n O(HCHO) 2.n CO2 n H2 O n O2 0, 055.2 0, 045 0, 07.2 0, 015 (mol) bo Do t l th tớch cng l t l s mol : %VHCHO 0, 015 100 33, 33% 0, 045 w w w fa ce ỏp ỏn D Cõu 12 Bi gii : Do X thy phõn to X1; X2 u cú kh nng tham gia phn ng gng nờn X1 ; X2 l saccarozo C12 H22O11 C6 H12O6 (glucozo) C6 H12O6 (fructozo) Glucoz v fructoz l ng phõn ca Khi tham gia phn ng gng t C6 H12O6 2AgNO3 NH H 2O CH (OH)(CHOH) COONH 2Ag 2NH NO3 CH (OH)(CHOH)4 COONH HCl CH (OH)(CHOH) COOH NH 4Cl ỏp ỏn D Cõu 13 Bi gii: Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 10 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Gi cụng thc chung ca hn hp amino axit X1, X2 (u no, mch h, phõn t cha mt nhúm -NH2 v mt nhúm -COOH) l: Cn H2 n 1NO2 Cụng thc ca pentapeptit l C5n H10 n N5O6 Phng trỡnh chỏy: Cn H2 n 1NO2 (1, 5n 0, 75)O2 nCO2 (n 0, 5)H2O 0, N2 Cụng thc ca pentapeptit l C11H19O6 N5 Bo ton lng C, ta tớnh c n M 0, 01 (mol) m 3, 17 (gam) hi SO24 nT uO Quỏ trỡnh xy ti cỏc in cc nh sau: Cc Catot: Cu ;HOH Cc Anot: SO24 ;HOH Cu 2e Cu (1) 2HOH 4e O2 H Ta iL ie 2HOH 2e 2OH H2 (2) Kt thỳc in phõn catot bt u cú khớ thoỏt chng t mi bt u xy (2) Dung dch thu c sau in phõn tỏc dng vi NaOH Theo bi ta tớnh c : n NaOH 0, 2.0, 05 0, 01(mol) up s/ 11 Trong dung dch CuSO4 phõn li nh sau : CuSO4 Cu n CO2 D ỏp ỏn A Cõu 14 Bi gii : n O2 01 n 0, 11 n 2, 1, 5n 0, 75 0, 1275 oc Theo bi H n CO2 PTP : H OH H2O n e 0, 01 n e F 0, 01.96500 965 (s) I om /g p dng nh lut Faraday; t ro n H 0, 01(mol) ỏp ỏn C w fa ce bo ok c Cõu 15 Bi gii: Cht cú liờn kt H cú nhit sụi cao hn cht khụng cú liờn kt H Nhit sụi cũn ph thuc vo lng phõn t Dóy cỏc cht c sp xp theo chiu gim dn nhit sụi l: C2H5COOH, C2H5CH2OH, CH3COCH3, C2H5CHO ỏp ỏn A Cõu 16 Bi gii: n OH 0, 025 0, 1.2 0, 225 (mol);n Ba 0, 1(mol) Ta cú t l T w w n OH n CO2 0, 225 1, To 0, 15 CO 23 ; HCO n CO2 n OH n CO2 0, 075 (mol) n Ba 0, 1(mol) Kt ta thu c l BaCO3, n BaCO n CO2 0, 075 (mol) mBaCO3 0, 075.197 14, 775 (gam) ỏp ỏn A Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 11 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc 01 Cõu 17 Bi lm : Hai kim loi kim v kim th hũa tan nc to dung dch cha ion OH v gii phúng khớ H2 t chung kim loi l M n PTP : M nHOH M(OH) n H 0, 5376 0, 024 (mol) Theo bi ta cú : n H2 22, H Theo phng trỡnh phn ng n OH 2.n H2 2.0, 024 0, 048 (mol) Dung dch Z hi D HCl : 2x (mol) n H x (mol);n Cl 2x (mol);n SO2 x (mol) H 2SO4 :x (mol) nT Khi cho dung dch Y phn ng vi Z : H OH H2O PTP : uO x 0, 048 x 0, 012 (mol) mmuoi Ta iL ie Khi cụ cn mui thu c gm mKl mCl mSO2 1, 788 2.0, 012.35, 0, 012.96 3, 792 (gam) ỏp ỏn A up s/ Cõu 18 Bi gii : Trong phn ng hidro húa cỏc hp cht hu c , thỡ gim s mol hn hp chớnh l mol H2 phn ng hay cng chớnh l mol ó b phỏ v ro Theo bi cho ta cú : n X 1(mol);MX 18, (mol) mX 18, (gam) c Theo nh lut bo ton lng : mx = my = 18,5 (gam) 18, nY 0, 925 (mol) 20 S mol H2 ó phn ng l : n H2 n X n X 0, 925 0, 075 (mol) ỏp ỏn B bo Cõu 19 Bi gii : T M suy t l s mol ca amin l: nCH5N : nC2H7N = :1 T M suy t l s mol ca amin l: nO2 : nO3 = 1: fa ce Chn nX = w w w ok om /g M Y 20 suy nO2 = 1; n O3 = Nhng electron 2C 2CH5N 2x mol Nhn electron N2 +18e O2 4e O O3 6e 3O2 18x 4C 2C2H7N N2 +30e Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 12 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 15x p dng bo ton electron: 1.4+3.6=18x+15x x = VY VX H 2 n Y 3x 3 oc 01 x mol ỏp ỏn D 2, 46 123 M 123 71 52 0, 02 Vy kim loi l Cr ỏp ỏn A ro : ỳng vỡ phõn t glucozo cú nhúm CHO : ỳng vỡ c th ngi tinh bt c chuyn húa thnh mantozo bi cỏc enzim : ỳng vỡ phõn t Mantozo cú kh nng chuyn húa to nhúm CHO : Sai vỡ saccarozo c to bi - glucozo v - Fructozo om /g Cõu 21 Bi gii : (1) (2) (3) (4) up s/ M MCl2 Ta iL ie uO nT hi D Cõu 20 Bi gii : Nhn thy cng m gam kim loi M to mui clorua ; nhng lng mui thớ nghim khỏc iu ú chng t kim loi M th hin cỏc húa tr khỏc mi thớ nghim Loi ỏp ỏn B v D Vy M ch cú th l Cr hoc Fe Khi tỏc dng vi HCl to mui MCl2 ; Khi tỏc dng vi Cl2 to mui MCl3 Nh th chờnh lch lng mui ta tớnh c mol kim loi M 3, 17 2, 46 nM 0, 02 (mol) 35, c ỏp ỏn B l: ce bo ok Cõu 22 Bi gii: Khớ Y nh hn khụng khớ v lm xanh qu tớm m, chng t Y l NH3 Võy ng vi cụng thc C5H13O2N, X cú cụng thc CH3CH2CH2CH2COONH4 ;(CH3 )2 CHCH2COONH4 ;C2H5CH(CH3 )COONH4 ;(CH3 )3 CCOONH4 fa ỏp ỏn B w w w Cõu 23 Bi gii : S mol Fe = 0,07; Fe2O3 = 0,1 Do phn ng hon ton nờn: Fe2O3 + 2Al Al2O3 + 2Fe 0,1 0,2 0,2 Cht rn sau phn ng nhit nhụm bao gm : Al2O3 ; Fe v Al d phn : Khi hon tan vo NaOH d c Al2O3 v Al u tan , nhng Al tan to H2 Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 13 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc H D 3a 0, 07 0, 3a a 0, 045 mol a = 0,045 mol Theo nh lut bo ton nguyờn t Al: n Al 0, 045 0, 0, 26 (mol) mAl 0, 26.27 7, 02 (gam) hi 3a 01 NaOH Al H2 2a a (mol) phn : Khi hũa tan cht rn vo dung dch H2SO4 loóng d Theo bi mol H2 thu c l 4a (mol) H SO Al H2 2 a a H SO Fe H2 Ta iL ie uO nT n Fe = 3a = ỏp ỏn A Cõu 24.Bi gii : AgNO3 / NH3 HCHO Ag up s/ Do andehit l n chc tỏc dng vi AgNO3 TH1 : Andehit n chc l HCHO ro 21, 0, (mol) 108 0, n HCHO 0, 05 (mol) mHCHO 0, 05 30 1, (gam) Theo bi lng andehit l 4,4 gam Loi TH1 TH2 : Andehit n chc l RCHO ( R H) ok c om /g Ta cú n Ag AgNO3 / NH3 RCHO 2Ag 4, 44 CH 3CHO 0, ce ỏp ỏn A bo n Ag 0, (mol) n RCHO 0, 1(mol) M RCHO w w w fa Cõu 25 Bi gii : húa ỏp ỏn C Cht oxi húa l cht nhn electron , sau nhn electron s oxi húa gim Do vy ion H+ th hin tớnh oxi húa sinh H2 Vy cỏc phn ng (a) ; (e) sinh H2 l nhng phn ng m ion H+ úng vai trũ cht oxi Cõu 26 Bi Gii: X: CxHyOz cú x : y : z = Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 14 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 CTPT ca X l C8H10O CTCT loi ancol thm ng vi CTPT ca X l: HO CH 2CH 2OH CH 2OH CH 2OH CH3 CH 2OH 01 CH3 oc CH3 CH3 Cõu 27 Bi gii : D H ỏp ỏn B hi Ta cú s phn ng nh sau : uO nT CO Ca (OH)2 dZ CaCO Fe CO CO X Fe O 18 gam muối:Fe (SO ) H SO4 (dặcnóng d- ) Fe O Y SO2 ro up s/ Ta iL ie Theo bi ta tớnh c : 1, 008 18 n CaCO3 0, 04 (mol);n SO2 0, 045 (mol);n Fe2 (SO4 )3 0, 045 (mol) 100 22, 400 V bn cht õy l phn ng oxi húa kh , ta quan tõm cỏc quỏ trỡnh phn ng v i xỏc nh cht oxi húa v cht kh Quy i hn hp X v ch gm Fe v O Theo nh lut bo ton nguyờn t n Fe 2.n Fe2 (SO4 )3 2.0, 045 0, 09 (mol) n CO n CO2 n CaCO3 0, 04 (mol) c om /g Gi mol O l a ( mol) Ta cú cỏc quỏ trỡnh nh sau : Nhng e Fe0 Fe3 3e S6 2e S4 ok C2 C4 e Nhn e O 2e O2 bo Theo dnh lut bo ton electron ta cú : 3.0, 09 0, 04 a 2.0, 045 a 0, 13 (mol) ỏp ỏn C ce Khi lng ca hn hp X l m = 0,09.56 +0,13.16 = 7,12 (gam) fa Cõu 28 Bi gii : w w w Theo bi ta tớnh c : n Ba(OH)2 0, 02 (mol);n NaHCO3 0, 03 (mol) Phng trỡnh phn ng xy nh sau: OH HCO 0, 04 0, 03 CO32 H O 0, 03 n OH (du ) 0, 01 (mol) Ba CO 32 BaCO 0, 02 0, 03 0, 02 n CO2 (du ) 0, 01 (mol) Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 15 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 01 Na : 0, 03 (mol) Nh vy dung dch X bao gm CO 23 : 0, 01 (mol) OH : 0, 01 (mol) Khi nh t t dung dch HCl vo dung dch X cho ti bt u cú khớ thoỏt : H OH H 2O PTP: oc H CO23 HCO3 Theo phng trỡnh phn ng n H n OH n HCO 0, 02 (mol) H 0, 1, (lit) 160 ml 0, 125 ỏp ỏn B hi D VHCl 1, X cú dng Cn H2 n 2O Ta cú Ch s C ca X l: n S ng phõn ancol l: 2n ; s ng phõn ete: n CO2 nX 0, Cụng thc phõn t ca X l C4H10O 0, Ta iL ie n CO2 uO n O2 nT Cõu 29 Bi gii: (n 1)(n 2) ro up s/ Tng s ng phõn ng phõn ỏp ỏn B Cõu 30 Bi gii: Khi cho m gam hn hp gm Ba v Al2O3 vo nc thu c dung dch X iu ú chng t cht rn tan ht to thnh dung dch X Thnh phn ca dung dch X l Ba ; Al(OH)4 ; cú th cú OH d om /g ok c Sau ú CO2 ti d vo dung dch X thi xut hin 11,7 gam kt ta Do CO2 d nờn khụng th tn ti kt ta BaCO3 Do vy kt ta ch l Al(OH)3 11, 0, 15 n Al(OH)3 0, 15 (mol) Bo ton Al, ta tớnh c n Al2 O3 0, 075 (mol) 78 Khớ H2 Ba tan vo H2O sinh Do ú bo ton electron ta tớnh c n Ba n H2 =0,2 mol bo Vy giỏ tr ca m: m 0, 2.137 0, 075.102 35, 05(gam) ce ỏp ỏn D w w w fa Cõu 31 Li gii Qui hn hp X thnh FeO, Fe2O3 t s mol cỏc cht X l FeO : x(mol); Fe2O3 : y(mol) 72 x xy 160 y 20 Do n FeO n Fe3 O4 Qui X thnh Fe3O4 2, 32 0, 01(mol);n HCl 0, 2.1 0, 2(mol) 232 FeCl2 + 2FeCl3 + 4H2O PTP : Fe3O4 + 8HCl 0,01 0,08 0,02 mol n Fe3 O4 Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 16 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 01 nHCl d = 0,2 0,08 = 0,12 (mol) oc Bo ton mol e 2a 0, 14 a 0, 07(mol) mFe 56.0, 07 3, 92(gam) Kt hp ỏp ỏn chn ỏp ỏn ỳng PTP O2 nCO2 + nH2O CnH2nOm + 0,05 0,155 0,13 0,13 mol D up s/ nT Xột trng hp: uO hi T ỏp ỏn t cụng thc chung cho M l CnH2nOm Da vo phn ng chỏy tỡm m, n Tớnh nX, nY Ta iL ie Cõu 32 Bi gii : H ỏp ỏn B Cú c ok Phng trỡnh chỏy: Cn H n 2Ox Theo bi ta cú t l fa ce Theo bi M ch gm ancol no n chc mch h v hidrocacbon Y Da vo ỏp ỏn nhn thy Y l hidro cacbon no mch h t cụng thc chung ca hn hp M l Cn H2 n 2Ox bo Cõu 33 Bi gii: om /g ro 0,1 mol M + AgNO3/NH3 0,2 mol Ag Nu ch cú anehit tham gia phn ng thỡ nAg ti a to thnh = 4nX = 4.0,02 = 0,08 mol < 0,2 Chng t c X v Y u tham gia phn ng Loi C, D Chn ỏp ỏn B n CO2 n O2 3n x nCO2 (n 1)H 2O 2n n 2x 3n x w w w Do x nờn n < Do X li cú s nguyờn t C ln hn ca Y Do ú, hidrocacbon Y l CH4 ỏp ỏn C Cõu 34 Bi gii: Khi oxi húa ancol bng CuO m sn phm thu c cú andehit iu ú chng t bi cho l ancol n chc bc 1: Phng trỡnh oxi húa nh sau: t RCH2OH CuO RCHO Cu H2O o Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 17 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Ta cú th vit nh sau: RCH2OH O RCHO H2O n ancol (phn ng) p dng nh lut bo ton lng, ta tớnh c 8, 68 6, 44 = nO 0, 14 (mol) 16 Do ancol cũn d sau phn ng nờn : n ancol (ban u) > 0,14 (mol) M ancol 6, 44 46 0, 14 01 oc Vy nú ch cú th l ancol metylic: CH3OH Andehi thu c l HCHO Khi gng HCHO 4Ag H Vy n Ag 4.n HCHO 4.n CH3 OH (phn ng)= 4.0,14=0,56 mol = 0,1 mol, nkhớ = hi Theo bi ta cú : nX = = 0,2 mol, nNaOH = 0,2.1,5 = 0,3 mol nT D Giỏ tr ca m= 0,56.108=60,48 gam ỏp ỏn C Cõu 35 Bi gii: mcht rn khan PTP : 0,1 0,2 mol = m(COONa)2 + mNaOH d = 134.0,1 + 40.(0,3 0,2) = 17,4 g Zn 2HCl ZnCl2 H t H S H 2S om /g Cõu 37 Bi gii : Ba o ro ỏp ỏn B up s/ 0,1 02 ỏp ỏn D Cõu 36 Bi gii : Ta iL ie uO 0,1 mol X (C2H8N2O4) + NaOH 0,2 mol khớ ( lm xanh qu m) CTCT ca X: (COONH4)2 Phng trỡnh phn ng: (COONH4)2 + 2NaOH (COONa)2 + 2NH3 + 2H2O 100 ml dung dch X tỏc dng vi dung dch Ba(OH)2 d: HCO OH BaCO3 H 2O 0, 15 c 0, 15 ok 100 ml dung dch X + NaOH d: HCO OH CO 23 H O bo 0, 15 0, 15 ce Ba CO 32 BaCO 0, fa w w w 0, 200 ml dung dch X + AgNO3 d: Ag Cl AgCl 0, 0, Vy 50 ml dung dch X cha: 0, 075 molHCO ; 0, 05 molBa ; 0, 05 molCl ;K n K 2.n Ba Theo nh lut bo ton in tớch ta cú: n HCO n Cl n K n HCO n Cl 2.n Ba 0, 025 (mol) 3 Nung núng 50 ml dung dch X: Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 18 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 t 2HCO3 CO32 CO2 H 2O 0, 075 0, 0375 mcht rn = mBa mCO2 mCl 39.0, 025 137.0, 05 60.0, 0375 35, 5.0, 05 11, 85 (gam) mK 01 oc ỏp ỏn C (c) t : KNO3 KNO2 (d) O2 : CuSO4 NH3 (d-) Cu(NH3 )4 (SO4 ) (e) : Fe CuSO4 FeSO4 Cu (f) : Zn Fe3 Zn2 Fe2 (g) : Ag2S O2 2Ag SO2 (h) : D (b) to : AgNO3 Ag NO2 O2 : 4FeS2 11O2 2Fe2 O3 8SO2 (a) H Cõu 38 Bi gii : nT uO Ba(OH)2 CuSO4 BaSO4 Cu(OH)2 up s/ Cỏc phn ng to kim loi l : (a) ; (e) ; (g) ỏp ỏn A Gi s mol N2 v N2O ln lt l x v y mol x + y = 0,06 (1) ro Cõu 39 Bi gii : Ta iL ie Ba 2H 2O Ba(OH) H hi Theo bi ta cú M 41, 334 28.x 44.y 41, 334 0, 06 (2) Gii h (1) v (2) ta c x= 0,01 (mol) ; y = 0,05 (mol) Theo bi ta cú : n HNO3 0, 87 (mol) om /g c Gi s phn ng to c NH4NO3 Khi ú ta cú phng trỡnh sau : n HNO3 12.n N2 10.n N2 O 10.n NH4 NO3 n NH4 NO3 0, 87 12.0, 01 10.0, 05 0, 25 (mol) Khi lng mui thu c bo ok sau phn ng l : mmui = mkim loi + m NO mui ca kim loi + m NH4 NO3 = 9, 55 62.(10 0, 01 8.0, 05 8.0, 025 ) 80 0, 025 54, 95 (gam) ce ỏp ỏn A w w fa Cõu 40 Bi gii : Sc Cl2 vo dung dch Ca(OH)2 L phn ng t oxh kh 2Cl2 2Ca OH CaCl2 Ca(OCl)2 2H2O dung dich w Nu l vụi tụi hoc sa vụi (Ca(OH)2 c nh bt loóng) thỡ cho clorua vụi : voi sua Cl2 Ca OH CaOCl2 H2O Sc CO2 vo dung dch cloruavụi Khụng phi phn ng oxh kh 2CaOCl2 CO2 H2O CaCO3 CaCl2 2HClO Chỳ ý : cloruavoi l mui hn ca Cl v ClO Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 19 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Sc O3 vo dung dch KI 2KI O3 H2O I2 2KOH O2 L phn ng oxh kh Sc H2S vo dung dch FeCl2 Cho HI vo dung dch FeCl3 FeCl3 2HI FeCl2 I2 2HCl Khụng cú phn ng L phn ng oxh kh 01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 oc Cho dung dch H2SO4 c núng vo NaBr tinh th L phn ng oxh kh Chỳ ý : Phng phỏp ny khụng iu ch c HBr (tng t vi HI) t NaBr H 2SO4 dac NaHSO4 HBr 2HBr H 2SO4 dac SO2 Br2 2H O ỏp ỏn A D H uO nT hi Cõu 41 Bi gii : Dung dch X s gm mui Fe2+ v Fe3+.Vy cỏc cht tha l : NaOH ; Cu ; Fe(NO3)2 ; KMnO4 ; BaCl2 ; Cl2 ; Al C th nh sau : Fe3O4 4H2SO4 FeSO4 Fe2 (SO4 )3 4H2O up s/ Ta iL ie Fe Fe Dung dch X thu c gm Khi cho dung dch X vi cỏc cht , cú phn ng nh sau: SO H H OH H2O Cu 2Fe3 Cu Fe2 3Fe2 NO3 4H 3Fe3 NO 2H2O 5Fe2 MnO4 8H Fe3 Mn 4H2O SO24 Ba BaSO4 om /g ro Cl2 Fe2 (SO4 )3 FeCl3 Al 3Fe3 3Fe2 Al3 3FeSO4 c ce Chn B bo ok 2Al Fe2 2Al3 Fe w w w fa Cõu 42 Li gii PTP (C4H6)nCl2 (C4H6)n + Cl2 71 100 17, 975 n 54n 71 ỏp ỏn A Nhn xột: Chỳ ý mi mt xớch cao su buna cũn liờn kt pi (-CH2-CH=CH-CH2-)n Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 20 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Ngoi cng chỳ ý cao su buna-S to t phn ng ng trựng hp buta 1,3 ien v stiren cú CT l (-CH2-CH=CH-CH2-CH2-CH(C6H5)-)n; cao su buna-N to t phn ng ng trựng hp buta-1,3-ien vi acrolitrin (CH2=CHCN) cú CT l (CH2-CH=CH-CH2-CH2-CH(CN)-)n Y thuc nhúm VIIA, ZY < 25 up s/ Cõu 44 Bi Gii : om /g ro trng thỏi c bn X cú e c thõn ỏp ỏn B Cõu 45 Li gii S phn ng : Ta iL ie uO nT hi D H oc 01 Cõu 43 Li gii X + NaOH 1mui + 1ancol 6.2 10 kX X l este chc ca axit v ancol (ancol chc hoc axit chc) X khụng bc, ú X khụng l este axit fomic Chỳ ý Y mch thng Cỏc cụng thc cu to ca X l Axit chc : C2H5OOC-COOC2H5 ; CH3OOC-(CH2)2-COOCH3 Ancol chc : CH3COO-CH2-CH2-OOCCH3 Tng cú ng phõn ỏp ỏn A Nhn xột : Cht hu c X tỏc dng vi NaOH, KOH, Ba(OH)2, thu c mui v ancol thỡ X l este Ngoi tỏc dng vi NaOH, KOH, Ba(OH)2,cũn cú phenol, axit cacboxylic, aminoaxit, mui amoni, Chỳ ý este ca axit fomic tham gia phn ng bc Cl2 ( 500 C) NaOH CH CH CH ClCH CH CH HOCH CH CH Br2 KOH / ROH CuO,t HOCH CHBr CH Br HOCH C CH CH C CHO c o ok Theo s n CH C CHO 0, 14(mol) AgNO3 / NH CH C CHO CAg C COONH 2Ag ce bo 0, 14 0, 14 0, 28 m (12 108 12 44 18).0, 14 108.0, 28 57 , (gam) w w w fa ỏp ỏn B Nhn xột : Hp cht cú H liờn kt vi C ni ba s tỏc dng vi vi AgNO3/NH3 to kt ta, ú Ag s thay th K C ni ba to cht kt ta theo s sau AgNO3 / NH3 R C CH R C CAg Cõu 46 Li gii 3860.5 0, 2(mol) ; n Cl 0, 4(mol) 1.nCl 0, 96500 Cl cha in phõn ht, ú anot ch cú Cl in phõn Phn ng in phõn cỏc in cc ne Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 21 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 Cu 2e Cu 2Cl Cl2 2e 2H O 2e H 2OH 0, mdung dịch gi ả m Nu catot ch cú Cu2+ in phõn 71.0, 64.0, 13, 10, 01 0, H oc cú H2O in phõn Gi s mol Cu l x ; s mol H2 l y Bo ton mol e 2x y 0, (1) Khi lng dung dch gim 64x y 71.0, 10, 4(2) Ta iL ie uO nT hi D T hp (1) v (2) x = 0,05(mol) ; y = 0,05(mol) 0, 05 a 0, 125(M) 0, ỏp ỏn A Nhn xột : Nh th t in phõn: Catot(cc õm): xy quỏ trỡnh kh (cht oxi hoỏ) gm cation, H2O; th t in phõn theo dóy in hoỏ, cation in phõn ht mi n nc in phõn to H2 (trao i 2e) Riờng cation kim loi kim, kim th v Al khụng b in phõn H2O Anot (cc dng): xy quỏ trỡnh oxi hoỏ (cht kh) gm anion v th t in phõn I Br Cl H2O ; H2O in phõn O2 (trao i 4e); chỳ ý cỏc anion a nguyờn t khụng b up s/ in phõn nh SO24 , NO3 ,CO32 , Nh nh lut Faraday: It n e n e (F 96500) F Khi thi gian tng gp n ln thỡ n e n e tng gp n ln - om /g ro - Cõu 47 Bi lm : ce bo c ok n Al2 (SO4 )3 Theo bi ta cú : 21, 0, (mol) ; n Al(OH)3 0, 15 (mol) 78 Al(OH)3 : 0, 15 Al3 OH PTP 0, Al(OH)4 TH1 : Al3+ d , ú OH ht ch xy phng trỡnh Al3 OH Al(OH)3 w w w fa n OH 3.n Al(OH)3 3.0, 15 0, 45 (mol) 18 100 % 200 Al(OH)3 : 0, 15 TH2 :Phn ng to Al(OH)4 Bo ton nguyờn t Al n Al(OH) 0, 0, 15 0, 05 (mol) mNaOH 0, 45.40 18 (gam) C%dd NaOH Bo ton nhúm OH n NaOH n OH 0, 15.3 0, 05.4 0, 65 (mol) Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 22 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 C%dd NaOH 0, 65.40 100 13% 200 Cõu 48 Bi gii : 01 ỏp ỏn D t cụng thc chung ca andehit l : Cn H2 n k Ox a (n k ) a H oc O2 Cn H n k O x (n k ) H 2O t CH3COOCH2CH2Cl + 2NaCl CH3COONa + NaCl + C2H4(OH)2 0 t C6H5Cl + 2NaOH HCOONa + H2O hi Ta iL ie t ClH3N - COOH + 2Na H2N - CH2COONa + NaCl + H2O nT Cõu 49 : Bi gii : uO liờn kt Kt hp ỏp ỏn Cụng thc ca andehit l : HCHO v CHC-CHO ỏp ỏn B D n H2 O (n k )a a k n phõn t andehit thỡ s nguyờn t C bng s t HCOOC6H5 + NaOH HCOONa+C6H5COONa+H2O up s/ t CH3COONa + 3NaCl + 2H2O CH3CCl3 + 2NaOH Axit cacboxylic cú nhúm chc COOH Phn ng vi NaOH theo phng trỡnh nh sau om /g Cõu 50 Bi gii : : - ro t 2CH3COONa + 2NaCl + 2H2O CH3COOC(Cl) + 4NaOH ỏp ỏn D COOH + NaOH - COONa + H2O Ta cú n NaOH 0, (mol) n H2 O 0, (mol) c BTKL mX 0, 2.40 17, 04 0, 2.18 mX 12, 64 ok X don chuc n NaOH 0, n X 0, n Otrong X 0, 2.2 0, bo Khi t chỏy hon ton hn hp X thu c CO2 v H2O cú tng lng l 26,72 fa ce gam CO : a 44a 18b 26, 72 a 0, 46 12a 2b 12, 64 0, 4.16 (BTNT X) b 0, 36 H O : b Do Hn hp X gm axit no n chc A v axit khụng no n chc cú liờn kt ụi B, C Nờn ta tớnh c n B C n CO2 n H2 O 0, 46 0, 36 0, 1(mol) ; n A 0, (mol) w w w n CO2 0, 46 2, 0, Ta tớnh c ch s C trung bỡnh ca hn hp bng Do B , C l axit cacboxylic khụng no cha liờn kt ụi n chc nờn ch s C ti thiu n hh phi l Do vy s nguyờn t C ca A Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 23 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 w w w fa ce bo ok c om /g ro up s/ Ta iL ie uO Megabook chỳc cỏc em hc tt! nT hi D H oc 01 TH 1: A l CH3COOH ( 0,1 mol ) ú da vo tng mol CO2 ca hn hp ta tớnh c mol CO2 B , C to t ú tớnh c s nguyờn t C trung bỡnh ca B v C 0, 46 0, 1.2 2, loai ch s C ti thiu l ( to Ta t cụng thc ca B v C l : Cn H n 2O2 n 0, c liờn kt pi C vi C ) TH 2: A l HCOOH ( 0,1 mol )tng t nh trng hp ta cú : 0, 46 0, 1.1 B,C : Cn H n 2O2 n 3, 0, B v C li ng ng k tip , ỏp dng s ng chộo ta tớnh c s mol ca mi cht: CH CH COOH : 0, 04 CH CH CH COOH : 0, 06 % s mol ca B l 0, 04 100 20% 0, ỏp ỏn A Tuyt nh Luyn Húa Hc THPT Quc Gia 2015 24 www.facebook.com/groups/TaiLieuOnThiDaiHoc01 ... : n NaOH 0, 2.0, 05 0, 01( mol) up s/ 11 Trong dung dch CuSO4 phõn li nh sau : CuSO4 Cu n CO2 D ỏp ỏn A Cõu 14 Bi gii : n O2 01 n 0, 11 n 2, 1, 5n 0, 75 0, 12 75 oc Theo bi H n CO2 ... mBaCO3 0, 075 .19 7 14 , 775 (gam) ỏp ỏn A Tuyt nh Luyn Húa Hc THPT Quc Gia 2 015 11 www.facebook.com/groups/TaiLieuOnThiDaiHoc 01 www.facebook.com/groups/TaiLieuOnThiDaiHoc 01 oc 01 Cõu 17 Bi lm :... Gia 2 015 12 www.facebook.com/groups/TaiLieuOnThiDaiHoc 01 www.facebook.com/groups/TaiLieuOnThiDaiHoc 01 15x p dng bo ton electron: 1. 4+3.6 =18 x +15 x x = VY VX H 2 n Y 3x 3 oc 01 x mol