10 Đề Thi Ôn Luyện THPT Quốc Gia 2016

7/25/2019 10 Thi n Luyn THPT Quc Gia 2016

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thi thTHPT Quc Gia nm 2016 Mn Ha Hc

Trn con ngthnh cngkhng c du chn ngi li bing!1

Cu 1: Phn bn nitro photka l hn hp ca:A.NH4H2PO4v KNO3 B.(NH4)2HPO4v NaNO3C. (NH4)2HPO4v KNO3 D. (NH4)3PO4v KNO3Cu 2: Cho hai hp cht hu cX, Y c cng cng thc phn tl C3H7NO2. Khi phn ng vi dungdch NaOH, X to ra H2NCH2COONa v cht hu cZ; cn Y to ra CH2=CHCOONa v kh T. Cc

cht Z v T ln lt lA.CH3OH v CH3NH2. B.CH3NH2v NH3.C.CH3OH v NH3. D.C2H5OH v N2.Cu 3: Thuphn hon ton tinh bt trong dung dch axit v clong, thu c cht hu cX. ChoX phn ng vi kh H2(xc tc Ni, to), thu c cht hu cY. Cc cht X, Y ln lt l:A. glucoz, fructoz. B.glucoz, sobitol.C.glucoz, saccaroz. D.glucoz, etanol.Cu 4: t chy hon ton mt hirocacbon mch hX cn va V lt kh O2(ktc), thu c 0,4mol CO2v 0,5 mol H2O. Gi trca V l

A.8,96. B.2,24. C.14,56. D.11,2.Cu 5: Anehit no mch hX c cng thc n gin nht C2H3O. Cng thc phn tca X lA.C4H6O2. B.C2H3O C.C8H12O4. D.C6H9O3.Cu 6: Cho m gam hn hp axit axetic, axit benzoic, axit adipic, axit oxalic tc dng va vidungdch NaOH thu c a gam mui. Nu cng cho m gam hn hp X ni trn tc dng vi Ca(OH)2va th thu c b gam mui. Biu thc lin hm, a, b l:A.m = 11b10a B.9m = 20 a11b C.8m = 19 a- 11b D.3m = 22b19aCu 7: Hirocacbon X khng lm mt mu dung dch brom nhit thng. Tn gi ca X lA. etilen B.isopren. C.hexan. D.stiren.

Cu 8: Hp cht hu cX tc dng c vi dung dch NaOH v dung dch brom nhng khng tcdng vi dung dch NaHCO3. Tn gi ca X lA.phenol B.metyl axetat. C.axit acrylic. D.anilin.Cu 9: Cho Fe tc dng vi dung dch H2SO4long to thnh kh X; nhit phn tinh thKNO3tothnh kh Y; cho tinh thKMnO4tc dng vi dung dch HCl c to thnh kh Z. Cc kh X, Y v Zln lt lA.Cl2, O2v H2S. B. H2, O2v Cl2. C.SO2, O2v Cl2. D.H2, NO2v Cl2.Cu 10: Cho m hnh th nghim iu chv thu kh nhhnh vsau:

THI THTHPT QUC GIA NM 2016

MN HA HC LN 1


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Phng trnh ha hc no sau y ph hp vi m hnh thu kh trn?A.CaC2+ H2O -> Ca(OH)2+ C2H2 B.FeS + HCl -> FeCl2+ H2SC.CH3COONa + NaOH -> CH4+ Na2CO3 D.NaHCO3+ HCl -> NaCl + CO2+ H2OCu 11: Cho X l metylamin. Ly 3,1 gam X tc dng ht vi dung dch HCl th khi lng mui thuc lA.11,46 gam B.12,82 gam C.14,38 gam D.6,75 gamCu 12: Ha tan mt kh X vo nc, thu c dung dch Y. Cho ttdung dch Y n dvo dung

dch ZnSO4, ban u thy c kt ta trng, sau kt ta tan dn. Kh X lA. HCl. B.NO2. C. SO2. D.NH3.Cu 13: Dy gm cc cht c sp xp theo chiu tng dn nhit si ttri sang phi l:A.CH3COOH, C2H5OH, HCOOH, CH3CHO. B.CH3COOH, HCOOH, C2H5OH, CH3CHO.C.CH3CHO, C2H5OH, HCOOH, CH3COOH. D.HCOOH, CH3COOH, C2H5OH, CH3CHO.Cu 14: Hon tan ht m gam gm Fe v mt oxit st (FexOy) trong 800ml dung dch HCl 1M (va)thu c dung dch X v 1,792 lt kh H2(ktc). Cho dung dch X tc dng vi dung dch AgNO3dthu c 132,08 gam kt ta. Gi trca m l:A.21,5472 gam B.25,6 gam C. 27,52 gam D. 23,04 gam

Cu 15: X phng ho mt hp cht c cng thc phn tC10H14O6trong dung dch NaOH (d), thuc glixerol v hn hp gm ba mui (khng c ng phn hnh hc). Cng thc ca ba mui l:A. CH3-COONa, HCOONa v CH3-CH=CH-COONa.B.CH2=CH-COONa, CH3-CH2-COONa v HCOONa.C.CH2=CH-COONa, HCOONa v CHC-COONa.D.HCOONa, CHC-COONa v CH3-CH2-COONa.Cu 16: Cc kh c thtn ti trong cng mt hn hp l:A.NH3v HCl B. H2S v Cl2 C.Cl2v O2 D.HI v O3.Cu 17: Sn phm ca phn ng nhit phn hon ton AgNO3l:

A.Ag2O, NO, O2 B.Ag2O, NO2, O2 C.Ag, NO2, O2 D.Ag, NO, O2Cu 18: Hai cht X v Y c cng cng thc phn tC2H4O2. Cht X phn ng c vi kim loi Nav tham gia phn ng trng bc. Cht Y phn ng c vi kim loi Na v ho tan c CaCO3.Cng thc ca X, Y ln lt l:A.HOCH2CHO, CH3COOH. B.CH3COOH, HOCH2CHO.C.HCOOCH3, HOCH2CHO. D.HCOOCH3, CH3COOH.Cu 19: Pht biu no sau y l ng?A.Saccarozlm mt mu nc brom.


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B.Amilopectin c cu trc mch phn nhnh.C. Glucozbkhbi dung dch AgNO3trong NH3.D.Xenlulozc cu trc mch phn nhnh.Cu 20: Cho dy cc oxit sau: SO2, NO2, NO, SO3, P2O5, CO, N2O5, N2O. Soxit trong cc dy tcdng c vi H2O iu kin thng l

A.8 B.5 C.7 D.6Cu 21: C 1 loi oleum X trong SO3chim 70% theo khi lng. Tnh khi lng nc cnthmvo 100 gam oleum trn thu c dung dch mi trong H2SO4chim 80% theo khi lngA.16,2. B.21,6. C.10,8. D.8,8.Cu 22:Trong phng th nghim, ngi ta thng iu chHNO3t:A.NH3v O2 B.NaNO3v HCl c. C.NaNO2v H2SO4 D.NaNO3v H2SO4cCu 23: Ha tan hon ton 11,2 gam mt kim loi X vo dung dch H2SO4long d. Sau phn ngthu c 4,48 lt kh H2(ktc). Tn gi ca X lA.Magie. B.Km. C.Canxi. D. St.

Cu 24: Cho cc hp cht hu c:(1) ankan; (2) ancol no, n chc, mch h; (3) xicloankan; (4) ete no, n chc, mch h; (5) anken;(6) ancol khng no (c mt lin kt i C=C), mch h; (7) ankin; (8) anehit no, n chc, mch h;(9) axit no, n chc, mch h; (10) axit khng no (c mt lin kt i C=C), n chc.Dy gm cccht khi t chy hon ton u cho smol CO2bng smol H2O l:A.(3),(4), (6),(7),(10). B.(3), (5), (6), (8), (9).C.(1),(3), (5),(6),(8). D.(2), (3), (5), (7), (9).Cu 25: Mt mu kh thi c sc vo dung dch CuSO4, thy xut hin kt ta mu en. Hintng ny do cht no c trong kh thi gy ra?

A.H2S B.CO2 C.NO2 D.SO2Cu 26: Dy gm cc cht u iu chtrc tip (bng mt phn ng) to ra anehit axeticl:A.C2H5OH, C2H4, C2H2. B.C2H5OH, C2H2, CH3COOC2H5.C.HCOOC2H3, C2H2, CH3COOH. D.CH3COOH, C2H2, C2H4.Cu 27: SO2lun thhin tnh khtrong phn ng viA.CaO. B.nc brom. C.H2S. D.dung dch NaOH.Cu 28: Cho phn ng trng thi cn bng: H2(k)+Cl2(k) 2HCl(k)( H


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A.FeCl2+ H2S->FeS + 2HCl B.O3+ 2KI + H2O->2KOH + I2+ O2C.3O2+ 2H2S-> 2H2O + 2SO2 D.Cl2+ 2NaOH->NaCl + NaClO + H2O.Cu 33: Cho 18 gam glucozphn ng hon ton vi AgNO3/ NH3, thu c m gam Ag. Gi trcam lA.10,8. B.21,6. C.32,4. D.43,2.

Cu 34: Hn hp kh X gm CH4, C2H2 cMX = 23,5. Trn V(lt)X vi V1(lt)hirocacbon Y c107,5 gam hn hp kh Z. Trn V1(lt)X vi V(lt)hirocacbon Y c 91,25 gam hn hp kh F. BitV1V = 11,2 (lit)(cc kh u o ktc). Cng thc ca Y l:A.C3H8 B.C2H6 C.C4H8 D.C3H6Cu 35: Cho mt stnh cht: c dng si (1); tan trong nc (2); tan trong nc Svayde (3); phnngvi axit nitric c (xc tc axit sunfuric c) (4); tham gia phn ng trng bc (5); bthuphn trongdung dch axit un nng (6). Cc tnh cht ca xenlulozl:A.(1), (2), (3) v (4). B.(3), (4), (5) v (6).C.(1), (3), (4) v (6). D.(2), (3), (4) v (5).

Cu 36: ung nng 18 gam CH3COOH vi 13,8 gam C2H5OH c mt H2SO4c lm xc tc. Sauphn ng thu c 12,32 gam este. Hiu sut ca phn ng este ha lA.92,35%. B.35,42%. C.70,00%. D. 46,67%.Cu 37: Hn hp A gm Fe, Cu, Al, Mg (c tlmol tng ng l 1 : 1 : 2 : 2). Ho tan 22,2 gamhnhp A cn va 950ml dung dch HNO32M sau cc phn ng hon ton thu c dung dch X vV lit (ktc) hn hp kh Y gm 4 kh N2, NO, N2O, NO2trong 2 kh N2v NO2c smol bngnhau. C cn rt cn thn dung dch X thu c 117,2 gam mui khan. Gi trV l:A. 8,86 B.6,72 C.7,84 D.5,04Cu 38: Ancol etylic (d = 0,8 gam/ml) c iu chttinh bt bng phng php ln men vi hiu

sut ton bqu trnh 80%. Hp thton blng CO2sinh ra khi ln men tinh bt vo 4 lt dungdch Ca(OH)21M th thu c 320 gam kt ta, lc bkt ta, un nng dung dch thu c thyxut hin thm kt ta. Thtch ancol etylic 460thu c lA.0,48 lt. B.0,75 lt. C.0,40 lt. D.0,60 lt.Cu 39: Cho 5,8 gam mui FeCO3tc dng vi dung dch HNO3va , thu c hn hp khchaCO2, NO v dung dch X. Cho dung dch HCl dvo dung dch X c dung dch Y, dung dch Yny ha tan c ti a m gam Cu, sinh ra sn phm khNO duy nht. Gi trca m l:

A.9,6 gam B.16 gam C.14,4 gam D. 11,2 gam

Cu 40: Trong tnhin ng v37Cl chim 24,23% snguyn tclo. Nguyn tkhi trung bnh caclo bng 35,5. Thnh phn phn trm vkhi lng ca 37Cl c trong HClO4l (vi 1H, 16O):A.9,82%. B.8,65%. C.8,92%. D.8,56%.Cu 41:Thuc thc dng phn bit Gly-Ala-Gly vi Gly-Ala lA.dung dch NaOH. B.dung dch HCl.C.dung dch NaCl. D.Cu(OH)2trong mi trng kim.


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Cu 42: Cho 32,25 gam mt mui c cng thc phn tl CH7O4NS tc dng ht vi 750 ml dungdch NaOH 1M un nng thy thot ra cht kh lm xanh qutm m v thu c dung dch X chcha cc cht v c. C cn dung dch X thu c bao nhiu gam cht rn khan?A.45,5 B.30,0 C.50,0 D.35,5Cu 43: Khi nung nng hn hp cc cht Fe(NO3)2, Fe(OH)3v FeCO3trong khng kh n khi

lng khng i, thu c mt cht rn l:A.FeO B.Fe2O3. C. Fe D.Fe3O4Cu 44: Cho m gam hn hp X gm mt peptit A v mt amino axit B ( MA> 4MB) c trn theo tlmol 1 : 1 tc dng vi mt lng dung dch NaOH va thu c dung dch Y cha (m + 12,24)gam hn hp mui natri ca glyxin v alanin. Dung dch Y phn ng ti a vi 360 ml dung dchHCl 2M thu c dung dch Z cha 63,72 gam hn hp mui. Cc phn ng xy ra hon ton. Ktlun no sau y ng?A.A c 5 lin kt peptit.B.B c thnh phn phn trm khi lng nitl 15,73%.

C.Tlsphn tglyxin v alanin trong phn tA l 3 : 2.D.A c thnh phn trm khi lng nitl 20,29%.Cu 45: Cho 33,2 gam hn hp A gm Fe3O4, Fe(NO3)3, Cu tc dng hon ton vi dung dch cha0,48mol H2SO4(long) th thu c kh NO duy nht v dung dch B chcha 2 mui sunfat. C cnB thu c m gam mui khan. m c gi trgn nht l:A.75,9 B.64,4 C.67,8 D.65, 6Cu 46: E l mt este 3 chc, mch h. un nng 7,9 gam E vi dung dch NaOH d, n khi phnng hon ton thu c ancol X v 8,6 gam hn hp mui Y. Tch nc tX c ththu cpropenal. Cho Y tc dng vi dung dch H2SO4thu c 3 axit hu no, mch h, n chc (trong 2

axit c khi lng phn tnhl ng phn ca nhau). Cng thc phn tca axit c khi lngphntln hn l:A.C5H10O2 B. C7H14O2 C. C6H12O2 D.C5H12O2Cu 47: Ha hi 8,64 gam hn hp gm mt axit no, n chc, mch hX v mt axit no, a chcY(c mch cacbon h, khng phn nhnh) thu c mt thtch hi bng thtch ca 2,8 gam N2(otrong cng iu kin). t chy 8,64 gam hn hp axit trn thu c 11,44 gam CO2. Phn trm khilng ca X trong hn hp ban u l :A.65,15% B.27,78% C.72,22% D.35,25%Cu 48: Pht biu no sau y l ng ?

A.Dung dch NaF ph

n

ng v

i dung dch AgNO3sinh ra AgF k

t t

aB.Iot c bn knh nguyn tln hn bromC. Flo c tnh oxi ha yu hn clo.D.Axit HBr c tnh axit yu hn axit HClCu 49: Cacbohirat nht thit phi cha nhm chc caA.ancol. B. xeton C. amin. D.anehit.Cu 50: Trung ho 5,4 gam X gm CH3COOH, CH2=CHCOOH, C6H5OH v C6H5COOH cn dngVml dung dch NaOH0,1M. C cn dung dch sau phn ng thu c 6,94 gam hn hp cht rnkhan. Gi trca V l.A.450 ml B.350 ml C. 900 ml D.700 ml


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P N V GII CHI TIT S1Cu 1: p n : C

Cu 2: p n : C

C3H7NO2+ NaOH H2NCH2COONa + Z

=> Z l CH3OH ( X l H2NCH2COOCH3)

Vi Y to ra CH2=CHCOONa

=> Y l CH2=CHCOONH4=> T l NH3

Cu 3: p n : B

Tinh bt -> X => X l C6H12O6

X + H2-> Y => Y l Sobitol

Cu 4: p n : C

Bo ton O : 2nO2= 2nCO2+ nH2O => nO2= 0,65 mol => V = 14,56 lit

Cu 5: p n : A

Cu 6: p n : D

Dng tng qut :

COOH + NaOH -> COONa + H2O

2COOH + Ca(OH)2-> (COO)2Ca + 2H2O

Bo ton khi lng : mmui Namaxit= (23 1)nCOOH

V mmui Camaxit= (20 1)nCOOH

=> nCOOH= (a m)/22 = (b m)/19

=> 3m = 22b 19a

Cu 7: p n : C Cu 8: p n : A

Cu 9: p n : B

Fe + H2SO4long -> H2; Nhit phn KNO3-> O2; KMnO4+ HClc-> Cl2


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Cu 10: p n : A

Kh phi i qua NaOH v H2O nn kh ny khng c phn ng vi NaOH v H2O hoc tan trong 2

dung dch trn. Mt khc th nghim trn khng cn un nng nn khng thl phn ng vi ti

xt

Cu 11: p n: DCH3NH2+ HCl -> CH3NH3Cl => mmui= 6,75g

Cu 12: p n : D

NH3+ H2O -> NH4++ OH-; Zn2++ 2OH--> Zn(OH)2 ; Zn(OH)2+ 4NH3-> Zn(NH3)4(OH)2

Cu 13: p n : C

Xt trc ht vkhi lng mol ( cng ln th nhit si cng tng )

Nu khi lng mol tng ng th xt khnng to lin kt hidro lin phn t

( axit > ancol > andehit )Cu 14: p n : D

nH2= 0,08 mol < nHCl= 0,8 mol => Fe phn ng ht vi HCl to Fe2+

Fe2++ Ag+-> Fe3++ Ag

Ag++ Cl--> AgCl

,mkt ta= mAg+ mAgCl=> nAg= 0,16 mol => nAg= nFe2+= 0,16 mol => nFeO= 0,08 mol

Nu l FeO -> nHCl= 2nFe+ 2nFeO = 0,32 mol < nHCl

=> oxit l Fe3O4vi smol l 0,08 mol => m = 23,04g

Cu 15: p n : B

Tng (pi + vng) = 4 => trong gc hydrocacbon sc 1 lin kt pi

=> Loi C ; D

Xt 2 p n A v B. V cc mui khng c ng phn hnh hc => Loi A

Cu 16: p n : C Cu 17: p n : C

Cu 18: p n : A

X + Na v X trng bc => X l HOCH2CHO

X + Na v CaCO3=> X l axit : CH3COOH

Cu 19: p n :B

Saccarozo khng lm mt mu nc brom

Glucozo boxi ha bi AgNO3/NH3

Xenlulozo c cu trc mch thng


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Cu 20: p n : B

Cc oxit tha mn : SO2; NO2; SO3; P2O5; N2O5

Cu 21: p n : C

Xt 100g Oleum c 70g SO3v 30g H2SO4

Khi ha tan vo nc thu c dung dch c H2SO4chim 80%Gi smol H2O tham gia l x = smol H2SO4(sinh ra)dd mi c (98x + 30) gam H2SO4.

(98x + 30) : (100+ 18x) = 0,8x = 0,6 molmH2O= 10,8g

Cu 22: p n : D

Cu 23: p n : D

Nu ha trca X l n. bo ton e :

2nH2= n.nX=> nX= 0,4/x (mol)

=> MX= 28n; Vi n = 2 th MX= 56(Fe) (TM)Cu 24: p n : B

Cc cht c 1 lin kt pi hoc 1 vng n trong phn tstha mn iu kin bi.

Cu 25: p n : A

H2S + CuSO4-> CuS en + H2SO4

Cu 26: p n : A Cu 27: p n : B

Cu 28: p n : A

Cn bng schuyn dch sang tri khi tng nhit

V phn ng thun ta nhit nn t0tng th chuyn dch theo chiu thun

Cu 29: p n : B

Cu 30: p n : B

Cc peptit : Gly-Gly ; Ala-Ala ; Gly-Ala ; Ala-Gly

Cu 31: p n : A

Phn ng va : nNaOH= nHCl= 0,1 mol

=> CM(HCl)= 1M

Cu 32: p n : A V FeS tan trong HCl

Cu 33: p n : B

Glucozo -> 2Ag

=> nAg= 2nGlucozo= 0,2 mol=> m = 21,6g

Cu 34: p n : C


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V1V = 11,2 => n1n = 0,5 mol

+) n.MX+ n1.MY= 107,5

=> 23,5n +MY.(n + 0,5) = 107,5 (1)

+)n1.MX+ n.MY= 91,25

=> (n + 0,5).23,5 + nMY= 91,25 (2)Ly (1) (2) : 0,5MY= 28 => MY= 56 (C4H8)

Cu 35: p n : C

Cu 36: p n : D

nCH3COOC2H5= 0,14 mol

,naxit= 0,3 mol = nancol= 0,3 mol=> H% = 0,14/0,3 = 46,67%

Cu 37: p n : D

Ta c : nFe: nCu: nAl: nMg =1 : 1 : 2 : 2=> nFe= nCu= 0,1 ; nAl= nMg= 0,2 mol

4 kh trong N2v NO2c smol bng nhau

=> c thqui vN2O v NO vi smol ln lt l x v y

,nNO3 mui KL= ne KL= 3nFe+ 2nCu+ 3nAl+ 2nMg= 1,5 mol

=> mmui= mKL+ mNO3 mui KL+ mNH4NO3=> nNH4NO3= 0,025 mol

Bo ton N : 2nN2O+ nNO+ 2nNH4NO3+ nNO3 mui KL= nHNO3

=> 2x + y = 0,35 mol

Bo ton e : 8nN2O+ 3nNO+ 8nNH4NO3= ne KL

=> 8x + 3y = 1,3

=> x = 0,125 ; y = 0,1 mol => V = 5,04 lit

Cu 38: p n : D

V khi un ln th dung dch vn cn kt ta xut hin

=> Ca(OH)2ht

,nCaCO3= 3,2 mol. Bo ton Ca : nCa(OH)2= nCaCO3+ nCa(HCO3)2

=> nCa(HCO3)2= 0,8 mol

Bo ton C : nCO2= nCaCO3+ 2nCa(HCO3)2= 4,8 mol => nC2H5OH= nCO2= 4,8 mol

=> mC2H5OH= 220,8g => VC2H5OH= 276 ml => Vru= 600 ml = 0,6 lit

Cu 39: p n : B

X gm c : 0,05 mol Fe(NO3)3


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=> Y c thm HCl

Khi cho Cu vo :

3Cu + 8H++ 2NO3--> 3Cu2++ 2NO + 4H2O

Cu + 2Fe3+-> Cu2++ 2Fe2+

=> nCu p= 1,5nNO3+ 0,5nFe3+= 0,25 mol=> m = 16g

Cu 40: p n : C

Cu 41: p n : DChtripeptit trln mi c phn ng mu biure

Cu 42: p n : A

Cht ban u + NaOH => mui v cv kh lm m qu tm

=> c thl mui ca amin

Cng thc ph hp : CH3NH3HSO4(X)

=> nX= 0,25 mol ; nNaOH= 0,75 mol => cht rn gm : 0,25 mol Na2SO4v 0,25 mol NaOH => mrn=

45,5gCu 43: p n : B

Cu 44: p n : A

Trong Y: NH2-CH2-COONa (a mol) v NH2-CH(CH3)-COONa (b mol) =>nHCl = 2a + 2b = 0,72

mmui = 110,5a + 124,5b + 58,5(a+b) = 63,72 =>a = 27/175 v b = 36/175 => nGly : nAla= 3 : 4 Nu A l

(Gly)3(Ala)3v B l Ala th A v B u ng nn loi. Nu A l (Gly)2(Ala)4v B l Gly => A ng

Cu 45: pn : C

Gi smol cc cht trong A ln lt l x ; y ; z

.mA= 232x + 242y + 64z = 33,2g

+) Nu dung dch sau phn ng cha Fe2(SO4)3v CuSO4

=> B c : (1,5x + 0,5y) mol Fe2(SO4)3v z mol CuSO4

=> nH2SO4= 4,5x + 1,5y + z = 0,48 mol

V sau phn ng chcha 2 mui => H+v NO3-u ht v NO3-> NO


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Bo ton e : 3nNO3= nFe3O4+ 2nCu=> 9y = x + 2z

=> x = 0,06 ; y = 0,04 ; z = 0,15 mol => mmui= 68g

Cu 46: p n : A

V Y gm 3 mui khc nhau => ancol X l triol

Mag khi tch nc thu c propenal CH2=CH-CHO => ancol l C3H5(OH)3Xt tng qut : 1 mol C3H5-> 3 mol Na

=> mmuimeste= (3.23 41)neste => neste= 0,025 mol => ME= 316g.

V trong Y c 2 mui ca 2 axit l ng phn ca nhau

=> 2R1+ R2+ 44.3 + 41 = 316

=> 2R1+ R2= 143

V R1< R2=> R1< 47,67 < R2

+) R1= 15 => R2= 113 ; +) R1= 29 => R2= 85 ; +) R1= 43 => R2= 57 (C4H9) c trong p nCu 47: p n : B

Cu 48: p n : B

AgF tan; Flo c tnh oxi ha mnh hn Clo; HBr c tnh axit mnh hn HCl

Cu 49: p n : A

Cu 50: p n : D

Ta thy c1 mol H i thnh 1 mol Na trong mui


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=> (23 1)nNaOH= mmui- mX=> nNaOH= 0,07 mol =>V = 0,7 lit = 700 ml


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Cu 1.trng thi cbn, nguyn tAl ( Z = 13) c selectron lp ngoi cng lA. 1 B. 2 C. 3 D. 4Cu hnh electron: 1s22s22p63s23p1 c 3electron lp ngoi cng CCu 2.Lin kt ha hc gia cc nguyn ttrong phn tHCl thuc loi lin ktA. cng ha trkhng cc B. ionC. cng ha trc cc D. hiroy l lin kt gia 2 nguyn tphi kim lin kt CHT phn cc CCu 3. Cho cc phng trnh phn ng:(a) Fe + Cl2 FeCl3. (b) NaOH + HCl NaCl + H2O.(c) CuO + CO Cu + CO2. (d) AgNO3+ NaCl AgCl + NaNO3.Trong cc phn ng trn, sphn ng oxi ha - khlA. 2. B. 1. C. 3. D. 4.Cc phn ng oxi ha khl (a) v (c) ACu 4. Cho phng trnh phn ng aAl +bHNO3 cAl(NO3)3+ dNO + eH2O. Tla : b l

A. 1 : 3 B. 2 : 3 C. 2 : 5 D. 1 : 4Cn bng: Al + 4HNO3 Al(NO3)3+ NO + 2H2O DCu 5.Cho cc cn bng ha hc sau:(a) H2(k)+ I2(k)2HI (k). (b) 2NO2(k) N2O4(k).

(c) 3H2(k)+ N2(k) 2NH3(k). (d) 2SO2(k)+ O2(k) 2SO3(k).

nhit khng i, khi thay i p sut chung ca mi hcn bng, cn bng ha hc no trnkhngbchuyn dch?A. (a). B. (c). C. (b). D. (d).Cn bng (a) c smol kh 2 bn bng nhau khngphthuc vo P ACu 6. Thc hin th nghim ha tan vi (CaCO3) bng dung dch HCl, bin php no sau y cthlm tng tc phn ng?A. Hnhit xung B. Gim nng ca dung dch HClC. Tng thtch ca dung dch HCl D. Nghin nh viKhi nghin nh vi slm tng din tch tip xc slm tng tc phn ng DCu 7. Gi trpH ca dung dch H2SO40,05M l:A. 1 B. 13 C. 7 D. 5[H+] = 2.0,05 = 0,1M pH = 1 A


MN HA HC LN 2


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Cu 8. Mt dung dch gm: 0,01 mol Na+; 0,02 mol Ca2+; 0,02 mol HCO3v a mol ion X (bqua sin li ca nc). Ion X v gi trca a lA. NO3v 0,03. B. Clv 0,01. C. CO32v 0,03. D. OHv 0,03.V trong dung dch c ion Ca2+ loi C (v c ion CO32- c thkt ta vi Ca2+)V trong dung dch c ion HCO3- loi D (v c ion OH- c thtc dng vi HCO3-)

V

y X c th

l NO3-

ho

c Cl-

. p d

ng nh lu

t b

o ton i

n tch ta c: 0,01.1 + 0,02.2 = 0,02.1 +a.1 a = 0,03 A

Cu 9. trung ha 20 ml dung dch HCl 0,1M cn 10 ml dung dch NaOH nngx mol/l. Gi trca x l:A. 0,1 B. 0,2 C. 0,3 D. 0,4Phng trinh ha hc: HCl + NaOH NaCl + H2O smol NaOH = smol HCl = 0,002 (mol) x = 0,2M BCu 10. Kim loi no sau y khng tc dng vi dung dch H2SO4long?A. Na B. Al C. Mg D. Cup n D v Cu ng sau H trong dy hot ng ha hcCu 11. Kh X c thlm c nc vi trong, dng cha chy v sn xut thuc gim au ddy, X l kh no sau y?A. CO2 B. N2 C. CO D. CH4p n ACu 12. Cho 5,6 gam Fe tc dng vi kh Cl2dth thu c m gam mui, gi trca m l:A. 16,25 B. 12,7 C. 9,15 gam D. 32,5Phng trnh phn ng: 2Fe + 3Cl2 2FeCl3 smol FeCl3= smol Fe = 0,1 m = 16,25 gam ACu 13.

Cho este no,

n ch

c, mch h

c cng th

c C

n

Hm

O2

. M

i quan h

gi

a n v

i m lA. m = 2n + 1. B. m = 2n. C. m = 2n - 2. D. m = 2n + 2.Este no, n chc, mch hc k = 1 m = 2n B Cu 14. t chy hon ton 2,8 gam hp cht hu c X th thu c 0,2 mol CO2v 3,6 gam hnc, CTPT ca X c thl:A. CH2 B. C2H4 C. C2H4O D. CH2OSmol CO2= 0,2 nC= 0,2 (mol) mC= 2,4 (gam)Smol H2O = 0,2 nH= 0,4 (mol) mH= 0,4 (gam) mC + mH = 2,8 gam = mX X khng c O loi C v DMt khc loi A v khng tha mn ha tr p n BCu 15. Hirocacbon X mch hc CTPT l C5H12, cho X tc dng vi kh Cl2(a/s) th chthu c 1sn phm monoclo duy nht. Tn gi ca X l:A. pentan B. 2-metylbutan C. 2,2-imetylpropan D. butanV X tc dng vi kh Cl2(a/s) th chthu c 1 sn phm monoclo duy nht X c CTCT ixng CCu 16. Hp thhon ton 1 lng anken X vo bnh ng nc brom th thy khi lng bnh tng5,6 gam v c 16 gam brom tham gia phn ng. SCTCT ca X l:A. 1 B. 2 C. 3 D. 4


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khi lng bnh tng = manken = 5,6 gamnBr2 = 0,1 mol = nanken Manken = 56 CTPT l C4H8Cc CTCT: CH3-CH=CH-CH3; CH2=CH-CH2-CH3; CH2=C(CH3)-CH3 CCu 17. Cho 6 gam ancol X (no, n chc, mach h, bc I) tc dng vi kim loi Na dth thu c1,12 lit kh H2(ktc). Tn gi ca X l:A. ancol etylic B. ancol metylic C. ancol propylic D. ancol isopropylicGi CT ca X l R-OH, phng trnh ha hc l: ROH + Na RONa + H2 nancol = 2nH2 = 0,1 mol Mancol = 60 R = 43 (C3H7) ancoll CH3-CH2-CH2-OH CCu 18. Phenol (C6H5OH)khngphnng vi cht no sauy?A. Na B. NaOH C. Br2(dung dch) D. NaHCO3Phenol phn ng c vi Na v c nhm OHPhenol phn ng c vi NaOH v c tnh axitPhenol phn ng c vi Br2to kt ta v c phn ng thvo vng p n D.Cu 19. Cho 6 gam HCHO tc dng vi lng ddung dch AgNO3/NH3, sau khi phn ng xy rahon ton thu c m gam Ag. Gi trca m l:A. 43,2 B. 86,4 C. 21,6 D. 32,4Sphn ng: 1HCHO 4Ag nAg = 0,8 mol m = 86,4 gam BCu 20. Cho anehit fomic phn ng vi H2(xc tc Ni, nun nng) th thu cA. CH3COOH. B. HCOOH. C. CH3CH2OH. D. CH3OH.Phng trnh phn ng: HCHO + H2 CH3-OH DCu 21. Dung dch axit acrylic (CH2=CH-COOH)khng phnngc vi cht no sauy?A. Na2CO3. B. Mg(NO3)2. C. Br2. D. NaOH.Axit acrylic phn ng c vi Na2CO3, NaOH v c nhm COOH; phn ng c vi Br2v clin kt i BCu 22: Trung ha 2,76 gam axit cacboxylic X n chc bng dung dch NaOH, thuc 3,772 gammui. Cng thc ca X lA. C3H7COOH. B. CH3-COOH C. H-COOH. D. C2H5COOHGi CT ca X l R-COOH, phng trnh ha hc l: R-COOH + NaOH R-COONa + H2O naxit = nmui 2,76/(R + 45) = 3,772/(R + 67) R = 15 (CH3) BCu 23. Este X c CTPT l C4H8O2. Khi thy phn X trong mi trng axit th thu c 2 cht hu cc scacbon bng nhau. Tn gi ca X l:A. metyl propionat B. propyl fomat C. isopropyl fomat D. etyl axetatR-COO-R + H2O R-COOH + R-OH axit v ancol u c 2 cacbonaxit l CH3-COOH v ancol l C2H5OH este l CH3-COO-C2H5 DCu 24. Thy phn hon ton 8,84 gam triolein bng dung dch NaOH dth thu c m gam muiGi trca m l:A. 9,12 B. 3,04 C. 9,18 D. 3,06Olein l (C17H33COO)3C3H5+ 3NaOH 3C17H33COONa + C3H5(OH)3

0,01 0,03


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m = 9,12 gam ACu 25. Cht tc dng vi H2to thnh sobitol lA. saccaroz. B. xenluloz. C. tinh bt. D. glucoz.p n D. C6H12O6+ H2 C6H14O6(sobitol)Cu 26. Cho cc pht biu sau:

(1)

Fructoz

v glucoz

u c kh

nng tham gia ph

n

ng trng bc;(2)Saccarozv tinh bt u khng bthy phn khi c axit H2SO4(long) lm xc tc;

(3)Tinh bt c to thnh trong cy xanh nhqu trnh quang hp;(4)Xenlulozv saccarozu thuc loi isaccarit;Pht biu ng l

A. (3) v (4). B. (1) v (3). C. (1) v (2). D. (2) v (4).(2) sai v csaccarozo v tinh bt u c phn ng thy phn(4) sai v saccarozo l issaccarit nhng xenlulozo l polisaccarit BCu 27. Sng phn cu to bc I ca amin c CTPT C3H9N l :

A. 1 B. 2 C. 3 D. 4Cc CTCT : CH3-CH2-CH2-NH2v CH3-CH(NH2)-CH3 BCu 28. C bao nhiu ipeptit (mch h) khi thu phn hon ton u thu c sn phm gmalanin v glyxin ?A. 1 B. 2 C. 6 D. 4V sn phm c alanin v glyxin nn ipeptit c 1 mt xch ala v 1 mt xch gly C 2 CT: Ala-gly v gly-ala BCu 29.Polime no sau y trong thnh phn cha nguyn tnit?A. Poli(vinyl clorua). B. Polibutaien. C. Nilon-6,6. D. Polietilen.

A. l (C2H3Cl)n; B l (C4H6)n; C l (-OC-(CH2)4-CO-HN-(CH2)6-NH-)n; D l (C2H4)n CCu 30. Cc polime thuc loi tnhn to lA. tvisco v tnilon-6,6. B. ttm v tvinilon.C. tnilon-6,6 v tcapron. D. tvisco v txenlulozaxetat.p n D (cu ny cn nh)Cu 31. Tin hnh cc th nghim sau:

(a)Cho l Fe vo dung dch gm CuSO4v H2SO4long;(b)t dy Fe trong bnh ng kh O2;(c)Cho l Cu vo dung dch gm Fe(NO3)3v HNO3;

(d)Cho l Zn vo dung dch HCl;Sth nghim c xy ra n mn in ha lA. 3 B. 2 C. 1 D. 4iu kin c n mn in ha l phi c 2 kim loi (hoc 1 kim loi v 1 phi kim) tip xc cinhau v tip xc vi 1 dung dch in li.(a) c v Fe + CuSO4 Cu + FeSO4, Cu sinh ra bm vo l st.(b) khng v chc 1 kim loi v khng c dung dch in li(c) khng v chc 1 kim loi(d) khng v chc 1 kim loi


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CCu 32. Cho cc pht biu sau:

(a) Trong cc phn ng ha hc, flo chthhin tnh oxi ha.(b) Axit flohiric l axit yu.(c) Dung dch NaF long c dng lm thuc chng su rng.

(d) Trong h

p cht, cc halogen (F, Cl, Br, I)

u c s

oxi ha: -1, +1, +3, +5 v +7.(e) Tnh khca cc ion halogenua tng dn theo tht: F, Cl, Br, I.

Trong cc pht biu trn, spht biu ng lA. 3. B. 5. C. 2. D. 4.Pht biu ng l a, b, c, e D(d) sai v F khng c soxi ha dngCu 33. Ha tan ht 0,2 mol FeO bng dung dch H2SO4c, nng (d), thu c kh SO2(sn phmkhduy nht). Hp thhon ton kh SO2sinh ra trn vo dung dch cha 0,07 mol KOH v 0,06mol NaOH, thu c dung dch cha m gam mui. Gi trca m l

A. 15,32. B. 12,18. C. 19,71. D. 22,34.p dung nh lut bo ton electron nSO2= 0,1 mol.Li c tng smol OH-= 0,13 mol phn ng to ra cmui HSO3-v mui SO32-.Phng trnh ha hc:

SO2+ OH- HSO3-a a a

SO2+ 2OH- SO32-+ H2Ob 2b b

Ta c hphng trnh: a + b = 0,1 v a + 2b = 0,13 a = 0,07 v b = 0,03 (mol)Khi lng mui = tng khi lng ca cc ion: K+, Na+, HSO3-v SO32- m = 0,07.39 + 0,06.23 + 0,07.81 + 0,03.80 = 12,18 gam BCu 34. Ha tan hon ton 24 gam hn hp X gm MO, M(OH)2v MCO3(M l kim loi c ha trkhng i) trong 100 gam dung dch H2SO439,2%, thu c 1,12 lt kh (ktc) v dung d ch Y chcha mt cht tan duy nht c nng 39,41%. Kim loi M lA. Zn. B. Ca. C. Mg. D. Cu.Cc phng trnh ha hc:

MO + H2SO4 MSO4+ H2OM(OH)2+ H2SO4 MSO4+ 2H2OMCO3+ H2SO4 MSO4+ H2O + CO2

V dung dch Y chcha 1 cht tan duy nht l MSO4 H2SO4phn ng va nMSO4 = nH2SO4 = 0,4 (mol) mMSO4 = 0,4(M + 96) gamLi c mddY = 24 + 100 mCO2= 121,8 gam C% = 0,4(M + 96)/121,8.100% = 39,41% M = 24 (Mg) CCu 35. Dy gm cc cht c xp theo chiu nhit si tng dn ttri sang phi l:A. CH3CHO, C2H5OH, C2H6, CH3COOH. B. C2H6, C2H5OH, CH3CHO, CH3COOH.

C. C2H6, CH3CHO, C2H5OH, CH3COOH. D. CH3COOH, C2H6, CH3CHO, C2H5OH.Nhit si gim dn theo chiu: axit > ancol > amin > cc cht khc (este, anehit, xeton, ete,


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hidrocacbon, v nhit si ca cc cht ny tlthun vi M) CCu 36. Trong cc cht: stiren, axit acrylic, axit axetic, vinylaxetilen, etanol v butan, scht c khnng tham gia phn ng cng hiro (xc tc Ni, un nng) lA. 4. B. 2. C. 5. D. 3.Gm c cc cht c lin kt i C=C, lin kt ba CC hoc lin kt i C=O (trong anehit vxeton) c stiren (C6H5-CH=CH2), axit acrylic (CH2=CH-COOH), vinylaxetilen (CHC-CH=CH2) DCu 37. Cho hnh vm tth nghimiu chkh Y tdung dch X:

Hnh vtrn minh ha phn ng iu chcht Y no sau y?A. NH3 B. HCl. C. C2H4 D. C2H5OH

Thnh vtrn suy ra kh Y phi khng tan trong nc chc C tha mn.Cu 38. Cht X c cng thc phn tC6H8O4. Cho 1 mol X phnng ht vi dung dch NaOH, thuc cht Y v 2 mol cht Z. un Z vi dung dch H2SO4c, thu c imetyl ete. Cht Y phnng vi dung dch H2SO4long (d), thu c cht T. Cho T phn ng vi HBr, thu c hai sn

phm l ng phn cu to ca nhau. Pht biu no sau y ng?A. Cht T khng c ng phn hnh hc.B. Cht X phn ng vi H2(Ni, to) theo tlmol 1 : 3.C. Cht Y c cng thc phn tC4H4O4Na2.D. Cht Z lm mt mu nc brom.un Z vi H2SO4c thu c imetyl ete Z l CH3OH, m 1 mol X + NaOH thu c 2 mol Z X l este 2 chc c dng R(COOCH3)2 CTCT ca X l C2H2(COOCH3)2 CTCT ca Y lC2H2(COONa)2v ca T l C2H2(COOH)2. M T + HBr thu c 2 sn phm T c cu to khngi xng CTCT ca T l CH2=C(COOH)2 CTCT ca X l CH2=C(COOCH3)2

ACu 39. Chdng Cu(OH)2 (mi trng v nhit tchn) c thphn bit c tt ccc dung

dch ring bit no sau y?A. glucoz, mantoz, glixerol, anehit axetic.B. lng trng trng, glucoz, fructoz, glixerol.C. saccaroz, glixerol, anehit axetic, ancol etylic.D. glucoz, lng trng trng, glixerol, ancol etylic.p n D.Cho Cu(OH)2 nhit thng vo 4 dung dch th c glucozo v glixerol (nhm I) to ra dung


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dch mu xanh cn lng trng trng v ancol etylic (nhm II) khng c hin tng g.Cho nhm (I) tc dng vi Cu(OH)2trong mi trng kim, un nng th c glucozo to kt tagch cn glixerol khng c hin tng gCho nhm (II) tc dng vi Cu(OH)2 trong mi trng kim, un nng th c lng trng trngto thnh mu tm cn glixerol khng c hin tng gCu 40. Tri cyc bo qun lu hn trong mi trng v trng. Trong thc t, ngi ta sdngnc ozon bo qun tri cy. ng dng trn da vo tnh cht no sau y?A. Ozon trvmt ho hc. B. Ozon l cht kh c mi c trng.C. Ozon l cht c tnh oxi ho mnh. D. Ozon khng tc dng c vi nc.Ozon c tnh oxi ha mnh nn c khnng st trng v ozon li khng to ra cht c hi do ozon dng khtrng cho cc loi thc phm nhhoa qu, tht, c, .... CCu 41. Hai cht hu cX v Y,thnh phn nguyn tu gm C, H, O, c cng snguyn tcacbon (MX < MY). Khi t chy hon ton mi cht trong oxi du thu c smol H2O bng smol CO2. Cho 0,1 mol hn hp gm X v Y phn ng hon ton vi lng ddung dch AgNO3

trong NH3, thu c 28,08 gam Ag. Phn trm khi lng ca X trong hn hp ban u lA. 39,66%. B. 60,34%. C. 21,84%. D. 78,16%.

V t chy X v Y u thu c smol CO2= smol H2O c k = 1.Cho 0,1 mol hn hp X v Y trng bc thu c 0,26 mol Ag c 1 cht l HCHOv 2 cht c cng sC cht cn li cng c 1 cacbon, m n phi cha oxi v c k = 1 chcthl axit HCOOH.Vy X l HCHO (a mol) v Y l HCOOH (b mol) a + b = 0,1 (1)Li c: HCHO 4Ag v HCOOH 2Ag tng smol Ag = 4a + 2b = 0,26 a = 0,03 v b = 0,07 (mol) %X = 21,84% CCu 42. Cho cc hp cht hu c:

(1) ankan; (2) ancol no, n chc, mch h;(3) xicloankan; (4) ete no, n chc, mch h;(5) anken; (6) ancol khng no (c mt lin kt i C=C), mch h;(7) ankin; (8) anehit no, n chc, mch h;(9) axit no, n chc, mch h; (10) axit khng no (c mt lin kt i C=C), n chc.

Dy gm cc cht khi t chy hon ton u cho smol CO2bng smol H2O l :A. (3), (5), (6), (8), (9) B. (3), (4), (6), (7), (10)C. (2), (3), (5), (7), (9) D. (1), (3), (5), (6), (8)

Cc cht khi t chy hon ton u cho smol CO2bng smol H2O Cc cht c k = 1 gm: xicloankan (CnH2n); anken (CnH2n); ancol khng no, mch h, c 1 lin kt i (CnH2nO)anehit no, n chc, h(C2H2nO); axit no, n chc, h(C2H2nO2) ACu 43. Cho cc nhn xt sau:

(1). C thto c ti a 2 ipeptit tphn ng trng ngng hn hp Alanin v Glyxin(2). Khc vi axt axetic, axt amino axetic c th tham gia phn ng vi axit HCl hoc phn ng

trng ngng(3). Ging vi axt axetic, aminoaxt c thtc dng vi bazto mui v nc


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(4). Axt axetic v axt -amino glutaric c th lm i mu qutm thnh (5). Thy phn khng hon ton peptit: Gly-Phe-Tyr-Gly-Lys-Gly-Phe-Tyr c ththu c 6 tripeptit

c cha Gly(6). Cho HNO3c vo ng nghim cha anbumin thy to dung dch mu tmSnhn xt ng l

A. 5 B. 3 C. 6 D. 4(1) sai v c thto ra 4 ipeptit: Ala-Ala ; Gly-gly ; Ala-Gly ; Gly-Ala(2) ng(3) ng

(4) ng (axit -amino glutaric c CTCT l HOOC-CH2-CH2-CH(NH2)-COOH)(5) sai v khi ct 3 mt xch cnh nhau sthu c 6 tripeptit u cha Gly nhng c 2 tripeptittrng nhau do chthu c c 5 loi.(6) sai v to ra kt ta mu vng. BCu 44. Cho tt c cc ng phn mch h (trong phn t ch cha cc loi nhm chc: axit, esteancol, anehit, xeton) c cng cng thc phn t C2H4O2 ln lt tc dng vi: Na, NaOH

NaHCO3. Sphn ng xy ra l

A. 4. B. 5. C. 3. D. 6.C 3 ng phn tha mn l CH3COOH, HCOOCH3v HO-CH2-CHO. Cc phn ng:

(1) CH3-COOH + Na CH3-COONa + H2(2) CH3-COOH + NaOH CH3-COONa + H2O(3) CH3-COOH + NaHCO3 CH3-COONa + H2O + CO2(4) HCOOCH3+ NaOH H-COONa + CH3OH

(5) HO-CH2-CHO + Na NaO-CH2-CHO + H2 BCu 45. Hn hp X gm FeCl2v NaCl c tlmol tng ng l 1: 2. Ha tan hon ton 2,44 gam Xvo nc, thu c dung dch Y. Cho Y phn ng hon ton vi dung dch AgNO3d, thu c mgam cht rn. Gi trca m lA. 2,87 B. 5,74 C. 6,82 D. 10,80Gi smol ca FeCl2 l a smol NaCl = 2a (mol) 127a + 58,5.2a = 2,44 a = 0,01 molCc phng trnh ha hc:

Ag++ Cl- AgCl

0,04 0,04Ag++ Fe2+ Ag + Fe3+

0,01 0,01 m = mAgCl + mAg = 6,82 gam CCu 46. Thuphn hon ton 0,1 mol peptit A mch h(to bi cc amino axit c mt nhm aminov mt nhm cacboxylic) bng lng dung dch NaOH gp i lng cn phn ng, c cn dungdch thu c hn hp cht rn tng so vi khi lng A l 78,2gam. Slin kt peptit trong A lA. 20 B. 10 C. 9 D. 19


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Nhn xt: GisA c x mt xch Khi A tc dng vi NaOH th lun xy ra theo phn ng sau(khng phthuc vo smt xch ca A):

1A + xNaOH mui + 1H2O0,1 0,1.x 0,1

mNaOH phn ng = 0,1x mol mNaOH ban u = 0,2x molBo ton khi lng ta c: mA + mNaOH b = mrn + mH2O mNaOH mH2O = mrn mA = 78,2 40.0,2x 0,1.18 = 78,2 x = 10 slin kt = 9 CCu 47. C 500 ml dung dch X cha Na+, NH4+, CO32-v SO42-. Ly 100 ml dung dch X tc dng vlng ddung dch HCl thu c 2,24 lt kh. Ly 100 ml dung dch X cho tc dng vi lng ddungdch BaCl2thu c 43 gam kt ta. Ly 200 ml dung dch X tc dng vi lng ddung dch NaOH thuc 8,96 lt kh NH3. Cc phn ng hon ton, cc thtch kh u o ktc. khi lng mui c trong300 ml X lA. 71,4 gam. B. 23,8 gam. C. 47,6 gam. D.119 gamgistrong 100ml dung dch X c Na+(a mol); NH4+(b mol); CO32-(c mol); SO42-(d mol)TN1: CO32-+ 2H+ CO2+ H2O

c c c = 0,1 molTN2: CO32-+ Ba2+ BaCO3

c cSO42-+ Ba2+ BaSO4d d 197c + 233d = 43 d = 0,1 mol

TN3: NH4++ OH- NH3+ H2O2b 2b 2b = 0,4 b = 0,2 mol

Bo ton in tch ta c: a + b = 2c + 2d a = 0,2 mol trong 300 ml dung dch c Na+(0,6 mol); NH4+(0,6 mol); CO32-(0,3 mol); SO42-(0,3 mol) mmui = tng khi lng cc ion = 23.0,6 + 18.0,6 + 60.0,3 + 96.0,3 = 71,4 gam ACu 48:Khi sc ttn dCO2vo dung dch hn hp gm a mol NaOH v b mol Ca(OH)2, ktquth nghim c biu din trn thsau:

3CaCOn

0 0,5 1,42CO

n

Tlb : a l:A. 5 : 1 B. 5 : 4 C. 5 : 2 D. 5 : 3

Cc phng trnh ha hc xy ra theo thtnhsau:(1) CO2+ 2OH- CO32-+ H2O(2) CO32-+ CO2+ H2O 2HCO3-

Lc u, khi nCO2tng dn th smol ion CO32-cng tng dn smol kt ta tng do phn ng:

0,5


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CO32-+ Ca2+ CaCO3 smol kt ta = smol CO32-v smol kt ta cc i = smol ion Ca2+ snol Ca2+= 0,5 mol b = 0,5Ta thy thc 1 on nm ngang, on ny, smol kt ta khng i lc ny vn xy raphn ng (1) nhng ion Ca2+ kt ta ht riSau khi OH-ht th bt u xy ra phn ng (2), lc kt ta tan dn cho n htKhi ht kt ta th smol CO2 dng l 1,4 molTheo (1) v (2) th tng smol CO2= 2 smol OH-+ smol CO32-= (a + 2b)/2 + (a + 2b)/2 = 1,4Vi b = 0,5 a = 0,4 a : b = 5 : 4 BCu 49. Cho X, Y l hai axit cacboxylic n chc mch h(MX< MY); T l este hai chc to bi X, Y vmt ancol no mch hZ. t chy hon ton 6,88 gam hn hp E gm X, Y, T bng mt lng va O2, thu c 5,6 lit CO2 (ktc) v 3,24 gam nc. Mt khc 6,88 gam E tc dng vi dung dchAgNO3/NH3dthu c 12,96 gam Ag. khi cho cng lng E trn tc dng vi 150 ml dung dchKOH 1M ri c cn th khi lng rn khan thu c lA. 10,54 gam B. 14,04 gam C. 12,78 gam D. 13,66 gamV E c phn ng trng bc X l HCOOH (v MX< MY), gi CT ca Y l R-COOH v CT ca Z lR(OH)2 CT ca T l HCOO-R-OOC-R.Gi smol ca X, Y, T trong hn hp E l a, b, c (mol)Smol Ag = 0,12 mol 2a + 2d = 0,12 a + c = 0,06 (1)Khi lng hn hp E = 46a + (R+45)b + (R + R+ 89)c = 6,88

46(a+c) + 45(b+c) + (b+c)R + (R-2)c = 6,88 (2)Khi t chy E, bo ton khi lng ta c: mE + mO2 = mCO2+ mH2O mO2 = 7,36 gam nO2= 0,23 molBo ton nguyn toxi ta c: nO(E) + 2nO2 = 2nCO2 + nH2O nO(E) = 0,22 mol

2a + 2b + 4c =0,22 a + b + 2c = 0,11 (3)T(1) v (3) b + c = 0,05 mol (4)Thay (1) v (4) vo (2) 005R = 1,87 - (R-2)c 0,05R < 1,87 R < 37,4Li c khi cho E tc dng vi KOH c cc phng trnh ha hc sau:

HCOOH + KOH HCOOK + H2Oa a a

RCOOH + KOH RCOOK + H2Ob b b

HCOO-R-OOCR + 2KOH HCOOK + RCOOK + R(OH)2c 2c c c

tng smol KOH phn ng l a + b + 2c = 0,11 mol smol KOH dl 0,15 0,11 = 0,04 mol cht rn gm: HCOOK (a+c mol); RCOOK (b+c mol); KOH (0,04 mol) m = 84(a+c) + (R + 83)(b+c) + 56.0,04 = 11,43 + 0,05Rvi R < 37,4 v R>0 11,43 < m < 13,35 CCu 50. Hn hp A gm 3 cht X, Y, Z l ba hidrocacbon mch hc cng CTGN (theo thttngdn vsnguyn tcacbon) trong C chim 92,31% vkhi lng. Khi t chy 0,01 mol Z thuc khng qu 2,75 gam CO2. Cho 3,12 gam hn hp A (c smol cc cht bng nhau) tc dng vlng ddung dch AgNO3/NH3thu c ti a m gam kt ta. Gi trca m gam l :A. 13,82 B. 11,68 C. 15,96 D. 7,98


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Gi CTGN ca 3 hidrocacbon l CxHy %C = 12x/(12x + y).100% = 92,93% x:y = 1:1 CTGN ca 3 hidrocacbon l CHLi c khi t chy 0,01 nol Z thu c smol CO2< 0,0625 scacbon ca Z < 6,25 cc CTPT c thc ca X, Y, Z l C2H2, C3H3, C4H4, C5H5, C6H6v shidro l chn X l C2H2, Y l C4H4, Z l C6H6.Tnh c smol ca 3 cht u l 0,02 molthu c lng kt ta ln nht th cX, Y, Z u phi tc dng c vi AgNO3/NH3v tcdng vi t l ln nht CTCT ca cc cht phi l: CHCH; CHC-CH=CH2; CHC-C2H4-CHCHKhi cc kt ta thu c l: AgCCAg (0,02 mol); AgCC-CH=CH2(0,02 mol) v AgCC-C2H4-CHCAg (0,02 mol) m = 240.0,02 + 159.0,02 + 292.0,02 = 13,82 gam A


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Cu 1:Mt mu nc cng c cha cc ion: Ca2+, Mg2+, HCO3

, Cl , SO 24

. Cht lm mm mu nc

cng trn lA. HCl. B. NaHCO3. C. Na3PO4. D. BaCl2.

Cu 2: Cho hn hp Mg v Cu vo dung dch HCl d. Kt thc phn ng, c cn dung dch thu c

cht rn gmA.Cu B.CuCl2; MgCl2 C.Cu ; MgCl2 D.Mg ; CuCl2Cu 3:Cho dung dch Ba(HCO3)2ln lt vo cc dung dch: CaCl2, Ca(NO3)2, NaOH, Na2CO3,

KHSO4, Na2SO4, Ca(OH)2, H2SO4, HCl. Strng hp c to ra kt ta l

A. 6. B. 7. C. 4. D. 5.

Cu 4: Lu hunh trong cht no sau y va c tnh oxy ha va c tnh khA.Na2S B.Na2SO3 C.FeS D.KHSO4Cu 5: Cht bo l trieste ca axit bo vi ancol no sau yA.ancol metylic B. etylenglycol C.Glyxerol D.Etanol

Cu 6: Ha tan m gam nhm vo dung dch NaOH va thu c dung dch X .Chia X thnh 2phn bng nhau .Phn 1 nh0,26 mol HCl vo thu c 1,8a gam cht rn .Phn 2 nhvo 0,5 molHCl thu c a gam cht rn .Ga trca m l

A.8,1 B.10,8 C.2,7 D.5,4Cu 7: Nhn xt no sau y khng ng vglucozoA.Phn tglucozo c 5 nhm OH B.Phn tglucozo c 1 nhm CHOC.Glucozo bthy phn trong mi trng axit D.t chy hon ton a mol Glucozo thu c 6amol CO2

Cu 8: Ha tan ht 4,6 gam Natri trong 100 ml dung d

ch HCl 0,5M thu

c H2v dung d

ch X. Ccn X c sgam cht rn l

A.10,2 gam B. 8,925 gam C. 8 gam D.11,7 gamCu 9: X phng ha hon ton 8,88 gam hn hp 2 este C3H6O2(c smol bng nhau) bng dungdch NaOH c bao nhiu gam muiA.9 gam B.4,08 gam C. 4,92 gam D.8,32 gam

Cu 10:Cho a gam P2O5 vo dung dch cha a gam KOH, thu c dung dch X. Cht tan c trongdung dch X l:


MN HA HC LN 3


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A. KH2PO4 v H3PO4 B. K2HPO4 v K3PO4 C. KH2PO4 v K2HPO4 D. K3PO4 v KOHCu 11: Hyrocacbon mch hno sau y phn ng vi Brom trong dung dch theo tlmoltng ng 1:2A.CnH2n+2 B.CnH2n-6 C.CnH2n D. CnH2n-2Cu 12: Hydrocacbon no sau y tc dng vi dung dch AgNO3trong NH3to thnh ta

A.Styren B.imetyl axetylen C.But-1-in D.But-1,3-dienCu 13: Cht no sau y khng tham gia phn ng thy phnA.Poli etylen B.Xenlulozo C.Mantozo D. TriaxylglyxerolCu 14: Thc hin cc phn ng nhit nhm hn hp gm m gam Al v 4,56 gam Cr2O3(trong iukin khng c O2), sau khi phn ng kt thc, thu c hn hp X. Cho ton bX vo mt lng ddung dch HCl (long, nng), sau khi cc phn ng xy ra hon ton, thu c 2,688 lt H2 (ktc)Cn nu cho ton bX vo mt lng ddung dch NaOH (c, nng), sau khi phn ng kt thc thsmol NaOH phn ng l:

A. 0,16 mol B. 0,06 mol C. 0,08 mol D. 0,10 mol

Cu 15: Nhttdung dch NaOH n dvo dung dch X. Sau khi cc phn ng xy ra hon tonchthu c dung dch trong sut. Cht tan trong dung dch X lA. CuSO4. B. AlCl3. C. Fe(NO3)3. D. Cu.

Cu 16:nhttdung dch H2SO4long vo dung dch K2CrO4th mu ca dung dch chuyn t:A. khng mu sang mu vng B. khng mu sang mu da camC. mu vng sang mu da cam D. mu da cam sang mu vng

Cu 17: Ha tan ht 0,56 gam Fe trong lng dH2SO4c nng thu c sn phm khduy nhtl bao nhiu lt SO2ktcA.0,56 lit B.0,448 lit C.0,224 lit D.0,336 litCu 18: Kim loi no sau y khi cho vo dung dch CuSO4bha tan ht v phn ng to thnh ktta gm 2 chtA.Na B.Fe C. Ba D.ZnCu 19: Hn hp X c cha 0,2 mol C2H2v 0,5 mol H2. Nung nng X sau mt thi gian (xc tc Ni)thu c hn hp Y c tkhi so vi H2l 6,2. Hn hp Y tc dng ti a vi a mol Br2trong dungdch. Gi tra l

A.0,2. B. 0,3. C. 0,1. D. 0,4.Cu 20: Ha tan hon ton 11,2 gam CaO v H2O thu c dung dch A. Sc kh CO2vo dung dchA, qua qu trnh kho st ngi ta lp thca phn ng nhsau:

x 15x Smol CO2

Khi lngkt ta


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Gi trca x l:A. 0,025 B. 0,020 C. 0,050 D. 0,040

Cu 21: Nguyn tc chung iu chkim loi l thc hin qu trnhA. khcc ion kim loi B. oxi ho cc ion kim loiC. khcc kim loi D. oxi ho cc kim loi

Cu 22:Cho dy cc cht sau: Al, NaHCO3, (NH4)2CO3, NH4Cl, Al2O3, Zn, K2CO3, K2SO4. C baonhiu cht trong dy va tc dng c vi dung dch HCl, va tc dng c vi dung dchNaOH?

A. 2 B. 3 C. 4 D. 5Cu 23: Cho 4,368 gam bt Fe tc dng vi m gam bt S. Sau phn ng c rn X. Ton bX tan httrong dung dch HNO3long d c sn phm khduy nht l 0,12 mol NO. Gi trm lA.0,672 gam B.0.72 gam C.1,6gam D.1,44 gamCu 24: Cho 2,24 lit ktc kh CO i ttqua mt ng snung nng cha m gam hn hp MgO,Fe2O3, CuO. Sau phn ng thu c (m - 0,8) gam cht rn v hn hp kh X. Tnh tkhi hi ca X

so vi H2A.14 B.18 C.12 D.24Cu 25:oxit lng tnh:

A. Cr2O3 B. CrO C. CrO3 D. CaOCu 26:Cho sphn ng sau: Al XYAlCl3. X, Y c thln lt l cp cht no sau y?

A. Al(OH)3, Al(NO3)3 B. Al(OH)3, Al2O3C. Al2(SO4)3, Al2O3 D. Al2(SO4)3, Al(OH)3

Cu 27: Mt l st c khi lng m gam nhng vo dung dch CuSO4. Kt thc phn ng thy khilng l st bng (m + 2,4) gam. Khi lng Cu do phn ng sinh ra bm ln l st l

A.12,8 gam B.9,6 gam C. 16 gam D. 19,2 gamCu 28: Ha tan hon ton m gam Na vo 100 ml dung dch Y gm H2SO40,5M v HCl 1M, thythot ra 6,72 lt kh (ktc). Hi c cn dung dch sau phn ng thu c a gamcht rn .a c gi trgn nht l:

A. 27,851 B. 29,852 C. 25,891 D. 28,952Na tc dng vi axit trc.Dung dch sau phn ng gm: Na+; Cl-; SO42-=> m = 28.952 => Chn DCu 29: un nng 66,4 gam hn hp 3 ancol n chc vi H2SO4c thu c 55,6 gam hn hp 6ete vi smol bng nhau. Smol ca mi ancol l:

A. 0,4mol B. 0,2mol C. 0,5 mol. D. 0,3 molCu 30: Dung dch X cha m gam hn hp CuSO4v NaCl. Thc hin in phn dung dch X chon khi nc bt u in phn c2 in cc th dng li, khi anot thot ra 4,48 lt kh (ktc)Dung dch sau in phn c thha tan ti a 2,7 gam Al. Gi trln nht ca m l

A. 47,4g B. 58,625g C. 55,4g D. 34,625g.Cu 31: Ha tan hon ton mt hn hp 2 kim loi Al v Fe trong mt lng va dung dch longHNO3long nng 20% thu c dung dch X (2 mui) v sn phm khduy nht l NO. Trong Xnng Fe(NO3)3l 9,516% v nng C % ca Al(NO3)3gn bng


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A.9,5 % B.4,6 % C. 8,4 % D. 7,32 %

Cu 32: Cho m gam NaOH vo 2 lt dung dch NaHCO3nng a mol/l, thu c 2 lt dungdch X. Ly 1 lt dung dch X tc dng vi dung dch BaCl2 (d) thu c 11,82 gam kt ta. Mtkhc, cho 1 lt dung dch X vo dung dch CaCl2 (d) ri unnng, sau khi kt thc cc phnng thu c 7,0 gam kt ta. Gi trca a, m tng ng l:

A. 0,08 v 4,8. B. 0,04 v 4,8. C. 0,14 v 2,4. D.

0,07 v 3,2Cu 33: Cho 1,2 gam mt axit cacboxylic n chc X tc dng va ht vi dung dch NaOH thu c1,64 gam mui. X l

A.H-COOH B.CH2 = CHCOOH C.C6H5-COOH D.CH3-COOHCu 34: Dy gm cc dung dch u ha tan c Cu(OH)2nhit phng l

A. glucoz, fructozv tinh bt B. glucoz, fructozv xenlulozC. glucoz, fructozv saccaroz D. glucoz, fomalin v tinh bt

Cu 35: Cho m gam Fe vo bnh ng dung dch H2SO4v HNO3thu c dung dch X v 1,12 ltkh NO. Thm tip H2SO4 dvo bnh c 0,448 lt NO v dung dch Y. Trong c2 trng hp u

c NO l sn phm khduy nht kin tiu chun. Dung dch Y ha tan va ht 2,08 gam Cu khngto sn phm khN+5. Cc phn ng u hon ton. Gi trm lA.4,2gam B.2,4gam C.3,92 gam D.4,06 gamCu 36: Phn ng gia 2 cht no sau y thu c sn phm trong Clo t mc oxy ha cao nhtca nA.MnO2+ HCl B.Cl2+ NaOH long nguiC.Cl2+ KOH c nng D.Cl2+ bt Ca(OH)2Cu 37: Trong cng ngip, sn xut NH3, phn ng xy ra to thnh mt cn bng ha hc. Cn bngha hc ny phi thc hin p sut cao, nhit thp nhng khng qu thp (khong 4500C). T

suy ra c im ca phn ng lA.Phn ng thun ta nhit, p sut tng B.Phn ng thun thu nhit , gim p sutC.Phn ng thun ta nhit gim p sut D.Phn ng thun thu nhit , p sut tngCu 38: Cho 66,2 gam hn hp X gm Fe3O4, Fe(NO3)2, Al tan hon ton trong dung dch cha 3,1mol KHSO4. Sau phn ng hon ton thu c dung dch Y chcha 466,6 gam mui sunphat trungha v 10,08 lit ktc kh Z gm 2 kh trong c mt kh ha nu ngoi khng kh. Bit tkhi ca Zso vi He l 23/18. Phn trm khi lng ca Al trong hn hp X gn nht vi gi trno sau y

A.15% B.20% C.25% D.30%Cu 39: Ba dung dch X,Y,Z, tha mn

- X tc dng vi Y th c ta xut hin- Ytc dng vi Z th c ta xut hin- X tc dng vi Z th c kh thot ra . X,Y,Z, ln lt lA.Al2(SO4)3, BaCl2, Na2SO4 B.FeCl2, Ba(OH)2, AgNO3C.NaHSO4, BaCl2, Na2CO3 D.NaHCO3, NaHSO4, BaCl2Cu 40: t chy hon ton m gam hn hp X gm 2 amin n chc trong oxy c 0,09 mol CO2,0,125 mol H2O v 0,015 mol N2. Cho m gam hn hp X tc dng va vi dung dch H2SO4longc sgam mui l


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CO

A.3,22 gam B.2,488 gam C.3,64 gam D.4,25 gamCu 41:. Cho 3,48 gam bt Mg tan ht trong dung dch hn hp gm HCl (d) v KNO3, thu cdung dch X cha m gam mui v 0,56 lt (ktc) hn hp kh Y gm N2v H2. Kh Y c tkhi so vH2bng 11,4. Gi trca m l

A.16,085. B. 14,485. C. 18,300. D. 18,035.

Don phn ng to mui ion NH4+

Smol N2=0,025(22,8-2):26=0,02; smol H2=0,005; dng bo ton e tnh mol NH4;nK=nNO3=0,01+0,04=0,05;Smol NH4+=(3,48:242-0,0210-0,052):8=0,01bo ton in tch tnh Cl-=0,145*2+0,05+0,01=0,35;(dng bo ton in tch ca mui)m mui = 3,48+0,05*39+0,35*35,5+0,01*18=18,035;(dng pp phn tch thnh phn);m=3,48:2495+(0,022+0,01)74,5+0,01*53,5=18,035(dng cng thc cc mui)Cu 42: Hn hp A gm 2 este n chc X, Y. Cho 0,05 mol A tc dng va vi dung dch NaOHthu c hn hp sn phm hu cB. t chy ht ton bB thu c 2,688 lt CO2; 3,18 gamNa2CO3. Khi lm bay hi B thu c m(g) cht rn. Gi trca m l

A.4,56 (g). B. 3,4(g). C. 5,84 (g) D. 5,62 (g).+ bo ton Na c 0,06 mol NaOHTa c nNaOH : nA = 1,2 , hn hp c 1 este ca phenol+ Bo ton C n 0,15 mol Ctb = 3.

Hn hp c HCOOCH3 v CxHyO2(cht ny tc dng vi NaOH theo tl1:2)

Ta c: a + b = 0,05 mol v a + 2b = 0,06 mol; nn a = 0,04 mol; b = 0,01 mol.Bo ton cacbon 0,04.2 + 0,01.x = 0,15 x = 7 chc C7H6O2 HCOOCH3 v HCOOC6H5. v tnh

c mCR =4,56gam

Cu 43: Trong Pho tan CuFeS2vi HNO3c, nng to ra cc mui v H2SO4th mt phn t

CuFeS2sA. nhng 11 electron B. nhng 17 electron C. nhn 17 electron D. nhn 11electronCu 44: Cho s:

2 2 23 1 .H O H O H O

metylbut en X Y Z

. Trong X,Y,Z u l sn phm chnh. Nhn xt no sau y ng :A.X l 2-metylbut-3-ol B.Y l 2-metylbut-1-enC.Z l 2-metylbut-2-ol D. Y l 2-metylbut-3-en

Cu 45: Cho dy cc cht : m-CH3COOC6H4CH3; m-HCOOC6H4OH; ClH3NCH2COONH4; p-C6H4(OH)2; p-HOC6H4CH2OH; CH3NH3NO3 .C bao nhiu cht ktrn tha mn iu kin: mt molcht phn ng ti a 2 mol NaOHA.2 B.4 C.5 D.3Cu 46: Ho tan hon ton 4,92 gam hn hp A gm (Al, Fe) trong 390ml dung dch HCl 2M thuc dung dch B. Thm 800 ml dung dch NaOH 1M vo dung dch B thu c kt ta C, lc kt taC, ra sch sy kh nung trong khng kh n khi lng khng i, thu c 7,5 gam cht rnPhn trm khi lng ca Al trong hn hp A c gi trgn nhtvi gi trno sau y?


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A. 65,8% B. 16,5% C. 85,6 %. D. 20,8%t nAl= x ; Fe = y => 27x +56y = 4.92Thm 0.8 mol NaOH vo dd NaOH d= 0,8 0,39.2 = 0,02Al3++ 3OH-=> Al(OH)3x 3x x

Al(OH)3+ OH-

=> Al(OH)4-

x-0.02Fe2++ 2OH-=> Fe(OH)2y y=> (x-0,02)/2. 102 +y/2.160 = 7,5=> x = 0,12 ; y = 0.03 => %Al = 65.85 => Chn ACu 47: Hp cht hu cX thun chc (chcha C,H,O). 5,8 gam X tc dng vi lng dAgNO3trong NH3c 43,2 gam Ag. Hydro ha hon ton 0,1 mol X c cht Y. Ton bY phn ng vaht 4,6 gam Natri. t chy hon ton X thu c

A.nCO2= nH2O B.nCO2=2nH2O C.nH2O=2nCO2 D.nH2O=3nCO2Cu 48: Khi tri sm chp ma ro, trong khng trung xy ra cc phn ng ha hc iu kin nhitcao c tia la in, to thnh cc sn phm c tc dng nhmt loi phn bn no di y, theonc ma ri xung, cung cp cht dinh dng cho cy trngA.m amoni B.Phn ln C. m nitrat D.Phn kaliCu 49: Cho hn hp 2 kim loi Al v Cu vo dung dch hn hp 2 mui AgNO3v Ni(NO3)2. Ktthc phn ng c rn X (tan mt phn trong dung dch HCl d) v thu c dung dch Y (phnng va vi dung dch NaOH c ta gm 2 hydroxit kim loi). Nhn xt no sau y khngng vth nghim trn

A.Rn X gm Ag ,Al , Cu B.Kim loi Cu cha tham gia phn ngC.Dung dch Ygm Al(NO3)3,Ni(NO3)2 D.Rn X gm Ag,Cu v NiCu 50: Cu 26:Ha tan hon ton m gam hn hp X gm Al v Al2O3 trong 200,0 ml dung dchNaOH 2M thu c dung dch Y v 3,36 lt kh H2(ktc). Thm 300,0 ml hoc 700,0 ml dung dchHCl 1M vo dung dch Y u thu c cng mt lng kt ta c khi lng m gam. Gi trgnnht ca m l

A.6,9. B.8,0. C.9,1. D.8,4.NaOH ban u = 0.4 molAl + NaOH => NaAlO2+ 3/2H2

0.1 0.1 0.15t Al2O3= aAl2O3+ 2NaOH => 2NaAlO2+ H2O

a 2a (d0.3-2a)=> m=2,7+102aLp hpt H+ = 0.3 v H+ = 0.7 => a = 0.05 => m = 7.8

P N & LI GII CHI TIT


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Cu 2: p n : CBut 1 in c CHC u mch nn c phn ng to kt ta vi AgNO3/NH3

Cu 4: p n : Ap-nitroanilin c dng : p- O2N C6H4NH2Do nhm NO2ht e mnh => cp e tdo ca N trong NH2bht vpha vng thm=> Tnh bazo gim r rt.Cn nhm metyl CH3- li l nhm y e => tnh bazo smnh hn.

Cu 5: p n : C

Cu 7: p n : D,bo ton e : 3nFe= 2nSO2=> nSO2= 0,015 mol=> V = 0,336 lit

Cu 8: p n : CKhi cho Ba vo th to Ba(OH)2sau phn ng vi CuSO4sto 2 kt ta l BaSO4v Cu(OH)2

Cu 9: p n : DCu 11: p n : Bphn ng khng c kh thot ra th phi to NH4NO3=> Kim loi phi c tnh khmnh=> MgCu 12: p n : CCu 13: p n : ACaC2+ 2HCl -> CaCl2+ C2H2Al4C3+ 12HCl -> 4AlCl3+ 3CH4=> Hn hp kh gm CH4v C2H2

Cu 17: p n : CGlucozo khng bthy phn

Cu 18: p n : B

,nNa= 0,2 mol ; nHCl= 0,05 molNa + HCl -> NaCl + H2Na + H2O -> NaOH + H2Sau phn ng c cn dung dch c cht rn : 0,05 mol NaCl v 0,15 mol NaOH=> m = 8,925g

Cu 21: p n : A


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,nNaOH= 0,018 mol3NaOH + triolein -> Glixerol + 3Natrioleat=> nGlixerol= 0,006 mol => mGlixerol= 0,552g

Cu 23: p n : A

Coi hn hp X gm Fe v S phn ng vi HNO3Fe 3e -> Fe3+S 6e -> S6+N5++ 3e -> N2+Bo ton e : 3nFe+ 6nS= 3nNO=> nS= 0,021 mol=> m = 0,672g

Cu 24: p n : BPhn ng tng qut : CO + Ooxit-> CO2=> mrn gim= mO p= 0,8g => nO p= nCO2= 0,05 mol=> X gm 0,05 mol CO v 0,05 mol CO2=> dX/H2= 18

Cu 27: p n : DFe + CuSO4-> FeSO4+ Cu,x -> x mol=> mtng= mCumFe p= 64x 56x = m + 2,4 m

=> x = 0,3 mol=> mCu bm= 0,3.64 = 19,2g

Cu 29: p n : BCH2=CHCOOCH3+ KOH -> CH2=CHCOOK + CH3OH=> Sau phn ng cht rn gm 0,2 mol CH2=CHCOOK=> m = 22g

Cu 31: p n : C


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Cu 35: p n : DTng smol kh NO sau cc phn ng l : 0,07 molGistrong Y c Fe3+v Fe2+=> bo ton e : 3nFe3++ 2nFe2+= 3nNOLi c : 2Fe3++ Cu -> Cu2++ 2Fe2+( Y ha tan Cu nhng khng c sn phm khca N+5)=> nFe3+= 2nCu= 0,065 mol => nFe2+= 0,0075 mol=> m = 56.( 0,065 + 0,0075 ) = 4,06g

Cu 36: p n : CCl t soxi ha +5 trong phn ng :

3Cl2+ 6KOH c nng -> KClO3+ 5KCl + 3H2O

Cu 37: p n : CV mc ch tng hiu sut => c yu ttc ng sao cho cn bng chuyeenrdichj theo chiu thun

+) p sut cao => phn ng thun lm gim p sut ca h+) Nhit thp nhng khng qu thp => phn ng thun ta nhit

Cu 38: p n : A10,08 lt (ktc) kh Z gm 2 kh trong c mt kh ha nu ngoikhng kh. Bit tkhi ca Z so vi He l 23 : 18

=> Z c 0,05 mol NO v 0,4 mol H2Bo ton khi lng : mX+ mKHSO4= m mui+ mZ+ mH2O

=> nH2O = 1,05 mol


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Bo ton H : nKHSO4= 2nH2+ 2nH2O+ 4nNH4+ => nNH4+= 0,05 mol.Bo ton N : 2nFe(NO3)2 = nNO+nNH4+ => nFe(NO3)2 = 0,05 molBo ton O : 4nFe3O4+ 6nFe(NO3)2 = nNO+ nH2O => nFe3O4= 0,2 mol%mAl= 16,3% gn nht vi gi tr15%

Cu 39: p n : C

X tc dng vi Y th c kt ta xut hin=> loi D v: NaHCO3+ NaHSO4khng c kt ta.X tc dng vi Z th c kh thot ra=> loi A v B v khng to kh.

Cu 40: p n : ABo ton khi lng :mX= m = mC+ mH+ mN= 0,09.12 + 0,125.2 + 0,015.2.14 = 1,75g, nH+= 2nH2SO4= nNH2= nN= 0,03 mol => nH2SO4 p= 0,015 mol=> mmui= m + mH2SO4 p= 3,22g

Cu 43: p n : B

Cu 47: p n : B, Khi 0,1 mol Y phn ng vi 0,2 mol Na => Y c tng snhm OH v COOH trong phn tl 2M X hidro ha to Y => X c 2 nhm CHO ( V X thun chc ). => nX= nAg= 0,1 mol => MX= 58g => X l (CHO)2 Khi t chy X th nCO2= 2nH2OCu 49: p n : Au tin Al phn ng vi AgNO3trc(1) Nu Al cn dth phn ng vi Ni(NO3)2; Cu khng phn ng(2) Nu Al ht v AgNO3dth Cu phn ng vi AgNO3, Ni(NO3)2ginguyn.X tan 1 phn trong HCl => Chng tgithuyt (1) ng=> Y gm Al3+; Ni2+( c th)M Y phn ng vi NaOH va to 2 hydroxit => Y phi c Ni2+=> Rn X gm Ag, Cu , Ni


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Cho bit nguyn tkhi ca cc nguyn t: H = 1; C = 12; N = 14; O =16; Na = 23; Al = 27; S = 32;Cr = 52; Fe = 56; Cu = 64; Zn = 65; Br = 80; Ag = 108; Sn = 119, Ba = 137; Pb = 207.

Cu 1: m in ca Al v Cl ln lt bng 1,6 v 3,0. Lin kt ha hc gia cc nguyn ttrongphn tAlCl3l

A. lin kt ion B. lin kt cng ha trc cc.C. lin kt kim loi D.lin kt cng ha trkhng cc

Cu 2: Xt phn ng ha hc: A (k) + 2B (k) AB2(k), H > 0 (phn ng thu nhit). Hiu sut qutrnh hnh thnh AB2stng khi

A. tng p sut chung ca h B. gim nhit phn ng.C. gim nng cht A D.tng thtch bnh phn ng

Cu 3: Dung dch X cha 0,1 mol 2Ca ; 0,3 mol 2Mg ; 0,4 mol Cl v a mol3

HCO . un dung dch X

n c cn thu c mui khan c khi lng l

A.49,4 gam B.28,6 gam C.37,4 gam D.23,2 gam

Cu 4: Xt phn ng: NaX (rn)+ H2SO4(c)0t

NaHSO4+ HX (kh). Cc hiro halogenua (HX)c thiu chtheo phn ng trn l

A. HCl, HBr v HI B. HBr v HI C. HF v HCl D.HF, HCl, HBr v HI

Cu 5: Cho 3,48 gam bt Mg tan ht trong dung dch hn hp gm HCl (d) v KNO3, thu cdung dch X cha m gam mui v 0,56 L (ktc) hn hp kh Y gm N2v H2. Kh Y c tkhi so vH2bng 11,4. Gi trca m l

A. 16,085 gam B. 14,485 gam C. 18,300 gam D.18,035 gam

Cu 6: Trong cng nghip, sn xut axit H3PO4c tinh khit v nng cao, ngi ta lm cch

no sau y?A. Cho photpho tc dng vi dung dch HNO3c, nng.

B. Cho dung dch H2SO4c, nng tc dng vi qung apatit.

C. t chy photpho trong oxi d, cho sn phm tc dng vi nc.

D.Cho dung dch axit H2SO4c, nng tc dng vi qung photphorit


MN HA HC LN 4 THPT CHUYN L QU N


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Cu 7: Hp thhon ton 3,36 lt kh CO2 (ktc) vo dung dch cha 0,15 mol NaOH v 0,1 moBa(OH)2, thu c m gam kt ta . Gi trca m l


Cu 8: Bn kim loi K, Al, Fe v Ag c n nh khng theo thtl X, Y, Z, v T. Bit rng X v Yc iu chbng phng php in phn nng chy; X y c kim loi T ra khi dung dch

mui; v Z tc dng c vi dung dch H2SO4c nng nhng khng tc dng c vi dung dchH2SO4c ngui. Cc kim loi X, Y, Z, v T theo thtl

A. Al, K, Fe, v Ag B. K, Fe, Al v Ag

C. K, Al, Fe v Ag D.Al, K, Ag v Fe.

Cu 9: C 4 ng nghim cng thtch, mi ng ng mt trong bn kh sau (khng theo tht): O2H2S, SO2, v HCl. Lt p tng ng nghim v nhng vo cc chu nc th kt quthu c nhcchnh vdi y:

Vy cc bnh a, b, c, v d ln lt cha cc khA. O2, H2S, HCl, v SO2 B. H2S, HCl, O2, v SO2

C.HCl, SO2, O2, v H2S D.SO2, HCl, O2, v H2S

Cu 10: Tin hnh th nghim nhhnh v. Kh A trong bnh c thl kh no di y?

A. H2S B. NH3 C. SO2 D.HCl

Cu 11: Kt lun no sau y vtnh cht ca hp kim l sai?A.Lin kt trong a stinh thhp kim vn l lin kt kim loi

B. Hp kim thng dnng chy hn so vi kim loi nguyn cht

C. Hp kim thng dn nhit v dn in tt hn kim loi nguyn cht

D.cng ca hp kim thng ln hn cng ca kim loi nguyn cht


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Cu 12: Pht biu no sau y l sai?A. Na2CO3l nguyn liu quan trng trong cng nghip sn xut thutinhB. nhit thng, tt ccc kim loi kim thu tc dng c vi ncC.Nhm bn trong mi trng khng kh v nc l do c mng oxit Al2O3bn bo vD.Theo chiu tng dn ca in tch ht nhn, nhit nng chy ca kim loi kim gim dn

Cu 13: Khi nhttdung dch AlCl3cho ti dvo dung dch NaOH v lc u thA. u tin khng xut hin kt ta, sau c kt ta trng keo

B. u tin xut hin kt ta trng keo, sau kt ta tan li.

C. u tin xut hin kt ta trng keo, sau kt ta khng tan li

D.khng thy kt ta trng keo xut hin.

Cu 14: Trng hp no sau y khng xy ra sn mn in ho?A. Sn mn vtu trong nc bin

B. Sgca gang trong khng kh m

C. Nhng thanh Zn trong dung dch H2SO4c nhvi git CuSO4.D.Nhng thanh Cu trong dung dch Fe2(SO4)3c nhvi git dung dch H2SO4

Cu 15: Sc ttkh CO2vo dung dch natri aluminat n dth

A. khng c phn ng xy ra

B. to kt ta Al(OH)3, phn dung dch cha Na2CO3

C. to kt ta Al(OH)3, phn dung dch cha NaHCO3D.to kt ta Al(OH)3, sau kt ta bha tan li.

Cu 16: Cho hn hp c a mol Zn tc dng vi dung dch cha b mol Cu(NO3)2v c mol AgNO3. Ktthc phn ng hon ton, thu c dung dch X v cht rn Y. Bit 0,5c < a < b + 0,5c. Kt lun nosau y ng?

A. X cha 1 mui v Y c 2 kim loi

B. X cha 3 mui v Y cha 2 kim loi

C. X cha 2 mui v Y cha 2 kim loi

D.X cha 2 mui v Y c 1 kim loi

Cu 17: Hp thV lt CO2(ktc) vo 200 mL dung dch NaOH 1M thu c dung dch X. Khi choBaCl2dvo dung dch X c kt ta v dung dch Y, un nng Y li thy c kt ta xut hinKhong gi trca V l

A. V 1,12 B.2,24 < V < 4,48


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C.1,12 < V < 2,24. D.4,48V 6,72

Cu 18: Ha tan hon ton m gam hn hp Al v Ba vi smol bng nhau vo nc c dung dchX. Cho t tdung dch HCl 0,5M vo dung dch X n khi lng kt ta t gi tr ln nht thydng ht 200 mL. Gi trca m bng


Cu 19: Thc hin phn ng nhit nhm hn hp gm Al v m gam hai oxit st trong kh tr, thuc hn hp rn X. Cho X vo dung dch NaOH d, thu c dung dch Y, cht khng tan Z v0,672 lt kh H2(ktc). Sc kh CO2dvo Y, thu c 8,58 gam kt ta. Cho Z tan ht vo dung dchH2SO4, thu c dung dch cha 20,76 gam mui sunfat v 3,472 lt kh SO2(ktc, l sn phm khduy nht ca H2SO4). Bit cc phn ng xy ra hon ton. Gi trca m l


Cu 20: Ha tan ht 1,69 gam oleum c cng thc H2SO4.3SO3vo 10g dung dch H2SO420% cdung dch X c nng a%. Gi trca a l

A. 33,875%. B. 11,292%. C. 22,054%. D.42,344%.

Cu 21: Nh t t n d dung dch NaOH long vo mi dung dch sau: FeCl3, CrCl3, AlCl3MgSO4. Sau khi cc phn ng xy ra hon ton, strng hp thu c kt ta l

A. 1 B. 2 C. 3 D.4

Cu 22: Cho hn hp gm 1 mol cht X v 1 mol cht Y tc dng ht vi dung dch H2SO4c, nng(d), to ra 2 mol kh SO2(sn phm khduy nht). Hai cht X v Y l:

A. Fe v Fe2O3 B. FeO v Fe3O4 C. Fe3O4v Fe2O3 D.Fe v FeO

Cu 23: C ba dung dch ring bit: H2SO41M; KNO31M; HNO31M c nh sngu nhin l (1)(2), (3).

Trn 5 ml dung dch (1) vi 5 ml dung dch (2), thm bt Cu d, thu c V1lt kh NO.

Trn 5 ml dung dch (1) vi 5 ml dung dch (3), thm bt Cu d, thu c V2 lt kh NO.

Trn 5 ml dung dch (2) vi 5 ml dung dch (3), thm bt Cu d, thu c 3V1 lt kh NO.

Bit cc phn ng xy ra hon ton, NO l sn phm khduy nht, cc thtch kh o cng iukin. So snh no sau y ng?

A. V2= V1 B. V2= 3V1 C. V2= 2V1 D.2V2= V1Cu 24: Trong cc cp cht sau: (1) AgNO3v NaCl; (2) NO2v NaOH; (3) FeS2v HCl; v (4) CaO vCO2. Scp cht xy ra phn ng oxi ha - khl

A. 1 B. 2. C. 3 D.4

Cu 25: Pht biu no sau y l sai?

A. CrO3l mt oxit axit


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B. Cr(OH)3 tan c trong dung dch NaOH

C.Cr phn ng vi axit H2SO4long to thnh Cr2+.

D.Trong mi trng kim, Br2 oxi ha 2CrO

thnh2

2 7Cr O

.

Cu 26: Cho dy chuyn ha sau: X 2 2CO H O

YNaOH

X. Cng thc ca Y l

A. CaO B. Ca(OH)2 C. CaCO3 D.Ca(HCO3)2

Cu 27: Tin hnh cc th nghim sau: (a) Cho dung dch NH3 n dvo dung dch Al2(SO4)3; (b)Sc kh SO2vo dung dch H2S; (c) Cho dung dch AgNO3vo dung dch H3PO4; (d) Cho dung dchAlCl3 vo dung dch Na2CO3; (e) Cho dung dch AgNO3 vo dung dch HF. Sau khi kt thc thnghim, strng hp thu c kt ta l

A. 3 B. 2 C. 5 D.4

Cu 28: Xt cc phn ng sau:

(a) F2+ H2O (hi)0

t

(d) dung dch AgNO3+ dung dch Fe(NO3)2

(b) Al + dung dch NaOH (e) Ca(NO3)20

t

(c) P2O5+ H2O (f) NaHCO30

t

Sphn ng sinh ra n cht lA. 2 B. 3 C. 4. D.5

Cu 29: Xt cc hp cht C10H14, C6H4(OH)2, C9H10BrCl v C6H6O(NO2)2. S cht c th cha vng

benzen trong phn tbngA. 1 B. 2 C. 3 D.4

Cu 30: Cho cng thc cu to sau: CH3CH(OH)CH=C(Cl)CHO. Soxi ha ca cc nguyn tcacbontnh tphi sang tri c gi trln lt l

A. +1;+1;-1; 0; -3 B. +1;-1;-1; 0; -3 C.+1;+1;0;-1; +3 D.+1;-1;0;-1; +3

Cu 31: t chy hon ton hn hp gm cc hidrocacbon thu c 2,24 lt (ktc) CO2v 2,7 gamH2O. Thtch oxi tham gia phn ng chy iu kin tiu chun l

A.5,6 lt. B. 2,8 lt C.4,48 lt. D.3,92 lt

Cu 32: S lng ng phn cha vng benzen (khng cha cc vng no khc) ng vi cng thcphn tC9H10l

A.5 B. 6 C. 7 D. 8

Cu 33: Xt bn ankan: metan, etan, propan, isobutan, v neopentan. Scht to c mt sn phmthmonoclo duy nht l

A. 2 B. 2 C. 3 D.4


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Cu 34: Xt su nhn nh sau: (1) Phn ng monobrom ha propan (bng Br2, un nng) to snphm chnh l n-propyl bromua; (2) Phn ng ca isobutilen vi hidro clorua to sn phm chnh lt-butyl clorua; (3) Phn ng dehidrat ha 2-metylpentan-3-ol to sn phm chnh l 4-metylpent-2-en(4) Phn ng ca buta-1,3-ien vi brom c thto c3,4-ibrombut-1-en v 1,4-ibrombut-2-en; (5)iclo ha benzen bng Cl2 (xc tc bt Fe, un nng) u tin to sn phm l o-iclobenzen v p-

iclobenzen; v (6) Monoclo ha toluen bng Cl2(chiu sng) u tin to sn phm l o-clotoluen vp-clotoluen. Snhn nh ng trong ssu nhn nh ny bngA. 2 B. 3 C. 4 D.5

Cu 35: Dn V lt (ktc) hn hp X gm axetilen v hiro i qua ng sng bt niken nung nngthu c kh Y. Dn Y vo lng dAgNO3 trong dung dch NH3 thu c 12 gam kt ta. Kh ra khi dung dch phn ng va vi 16 gam brom v cn li kh Z. t chy hon ton kh Zc 2,24 lt kh CO2(ktc) v 4,5 gam H2O. Gi trca V bng

A. 11,2. B. 13,44 C. 5,60 D.8,96

Cu 36: Dy gm cc cht u iu chtrc tip (bng mt phn ng) to ra anehit axetic lA. C2H5OH, C2H2, CH3COOC2H5 B. HCOOC2H3, C2H2, CH3COOH

C. C2H5OH, C2H4, C2H2 D.CH3COOH, C2H2, C2H4

Cu 37: Hn hp X gm ancol metylic, etylen glycol v glixerol. Cho 43,2 gam X phn ng hon tonvi Na d, thu c 15,68 lt kh H2(ktc) v hn hp mui X. t chy hon ton 43,2 gam X, rthi sn phm chy qua bnh cha CuSO4khan d, th khi kt thc th nghim khi lng bnh nytng

A. 9 gam B. 18 gam C. 36 gam D.54 gam

Cu 38: Axit cacboxylic n chc mach h phn nhanh (A) co phn trm khi lng oxi la 37,2%Pht biu no di y l sai?

A. A lam mt mau dung dich brom B. A languyn liu tng hp polime

C. A cong phn hinh hoc D.A cohai lin kt trong phn t

Cu 39: Hn hp X gm CH3COOC2H5, C2H5COOCH3v C2H5OH. t chy hon ton m gam hnhp X thu c 0,6 mol CO2v 0,7 mol H2O. Thnh phn phn trm vkhi lng ca C2H5OH bng

A. 34,33% B. 51,11% C. 50,00% D.20,72%

Cu 40: Axit cacboxylic X hai chc (c phn trm khi lng ca oxi nhhn 70%), Y v Z l haancol ng ng k tip (MY < MZ). t chy hon ton 0,2 mol hn hp gm X, Y, Z cn va 8,96 lt kh O2(ktc), thu c 7,84 lt kh CO2(ktc) v 8,1 gam H2O. Phn trm khi lng ca Ytrong hn hp trn l

A. 14,95%. B. 12,60%. C. 29,91%. D.29,6%.

Cu 41: t chy hon ton 1 mol cht bo, thu c lng CO2v H2O hn km nhau 8 mol. Mtkhc a mol cht bo trn tc dng ti a vi 600 ml dung dch Br21M. Gi trca a l


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A. 0,15 B. 0,10 C. 0,30 D.0,20

Cu 42: Cho cc pht biu sau: (1) Glucozc khnng tham gia phn ng trng bc; (2) Schuynha tinh bt trong c thngi c sinh ra mantoz; (3) Mantoz c khnng tham gia phn ngtrng bc; (4) Saccarozc cu to thai gc -glucozv -fructoz; (5) Hiro ha hon tonglucozto ra axit gluconic; (6) iu kin thng, glucozv saccarozu l nhng cht rn, d

tan trong nc; (7) Xenluloztrinitrat l nguyn liu sn xut tnhn to v chto thuc sngkhng khi; (8) Amilopectin trong tinh bt chc cc lin kt -1,4-glicozit; (9) Sacarozbha entrong H2SO4 c; v (10) Trong cng nghip dc phm, saccarozc dng pha ch thucTrong cc pht biu trn, spht biu ng l

A. 7 B. 8 C. 9 D.10

Cu 43: Cho 11,8 gam amin n chc X tc dng va vi dung dch HCl, sau khi phn ng xy rahon ton thu c dung dch Y. Lm bay hi dung dch Y c 19,1 gam mui khan. Scng thccu to ng vi cng thc phn tca X l

A.2 B. 3 C. 4 D.5

Cu 44: Cho 58,8 gam mt tetrapeptit Ala-Gly-Val-Phe tc dng vi dung dch cha 0,7 NaOH thuc m gam cht rn. Gi trca m l


Cu 45: Pht biu no sau y l ng?

A. Trng ngng buta-1,3-ien vi acrilonitrin c xc tc Na c cao su buna-N.B. Trng hp stiren thu c poli(phenol-fomanehit).C. Poli(etylen terephtalat) l polime trng ngngD.Tvisco l ttng hp

Cu 46: Dy no di y gm cc cht c khnng ha tan Cu(OH)2 to hp cht mu tan trongtrong nc?

A. etilen glycol, axit axetic, v gly.ala.gly

B. ancol etylic, fructoz, v gly.ala.lys.val

C.glixerol, glucoz, v gly.ala

D.ancol etylic, axit fomic, v lys.valCu 47: Xt cc pht biu sau: (1) metan, metanol, metanal v metanamin u l nhng cht kh iu kin thng. (2) metanol, metanal, metanoic, alanin v sacaroz u tan tt trong nc. (3)xyclopropan, propen, etanal, metanoic, v mantozu c khnng lm nht mu nc brom; (4)axetilen, anehit axetic, axit fomic, v fructozu to kt ta mu trng bc khi tc dng vi dungdch AgNO3 trong NH3. Spht biu ng bng

A. 1 B. 2 C. 3 D.4


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Cu 48: Xt cc tc nhn phn ng gm Na, dung dch NaOH v dung dch Na2CO3. Trong sbncht l (1) ancol etylic, (2) phenol (C6H5OH), (3) axit axetic, v (4) glyxin, c bao nhiu ch t c thphn ng c vi cba tc nhn?

A. 1 B. 2 C. 3 D.4

Cu 49: ng vi cng thc phn tC3H6O2, c x ng phn lm qu tm ha ; y ng phn tc

dng c vi dung dch NaOH, nhng khng tc dng vi Na; z ng phn va tc dng c vdung dch NaOH v va tc dng c vi dung dch AgNO3/NH3v t ng phn va tc dngc vi Na, va tc dng c vi dung dch AgNO3/NH3. Nhn nh no di y l sai?

A. x = 1 B. y = 2 C. z = 0 D.t = 2

Cu 50: Xt cc pht biu: (1) SO2v NO l nhng nguyn nhn chnh gy hin tng ma axit; (2)CFC v NO l nhng nguyn nhn chnh ph hy tng ozon ca tri t; v (3) Ngoi CO2, freonmetan v initoxit cng tham gia ng kvo hiu ng nh knh. Spht biu ng l

A. 0 B.1 C. 2 D.3

------------------------------------H t------------------------------------


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BNG P N

PHN LI GII CHI TITCu 1 : Chn p n CTa c hiu m in l : 0, 4 I 3 1, 6 1, 4 1, 7 Vy lin kt trong AlCl3l CHT c cc.

Cu 2: Chn p nATheo nguyn l chuyn dch cn bng khi p sut tng cn bng sdch vbn (lm gim p sut)hay bn c t phn tkh.

Cu 3: Chn p n C

BTDT0,1.2 0, 3.2 0, 4 a a 0, 4

0t BTKL2

3 3X HCO CO m 0,1.40 0,3.24 0, 4.35,5 0, 2.60 37, 4(gam)

Ch : Khi bi ni nung mui khan ti khi lng khng i th0 0

t t2 2

3 3HCO CO O

Cu 4: Chn p n C

Ch : Khng thiu chc HI hay HBr v :

0

0

c,t

2 4 4

c,t

2 4 2 2 2

NaBr H SO NaHSO HBr

2HBr H SO SO Br 2H O

0

0

c,t

2 4 4

c,t

2 4 2 2 2

NaI H SO NaHSO HI

8HI H SO H S 4I 4H O

Cu 5: Chn p n D

Ta c :4

2 BTE

Mg NH2

N : 0,02 0,29 0,02.10 0,005.2n 0,145 0,025 n 0,01

H : 0,005 8

2

4

BTNT.Nito

BTDT

Mg : 0,145

NH : 0,01dd X m 18,035

K : 0,02.2 0,01 0,05

Cl : 0,35

Ch : V Y c H2nn trong dung dch X chc chn khng cn N+5

Cu 6: Chn p n C

01.B 02.A 03.C 04.C 05.D 06.C 07.D 08.A 09.D 10.D

11.C 12.B 13.A 14.D 15.C 16.C 17.B 18.A 19.B 20.A

21.B 22.D 23.C 24.B 25.D 26.D 27.A 28.C 29.C 30. A

31.D 32.C 33.C 34.B 35.A 36.C 37.C 38.C 39.A 40.C

41.B 42.A 43.C 44.A 45.C 46.A 47.B 48.C 49.C 50.D


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Theo SGK lp 11.

Cu 7: Chn p n D

Ta c :2

2

2CO 3BTDT Ba :0,1

OH

n 0,15(mol) CO :0,15m 0,1.197 19,7(gam)

n 0, 35(mol) OH : 0,05

Cu 8: Chn p n A+ X v Y c iu chbng phng php in phn nng chy Loi B.

+ X y c kim loi T ra khi dung dch mui Loi C.

+ Z tc dng c vi dung dch H2SO4c nng nhng khng tc dng c vi dung dchH2SO4c ngui Z l Fe hoc Al.

Cu 9: Chn p n D

+ Theo m hnh (c) kh khng tan trong nc l O2.

+ Theo m hnh (b) kh tan rt nhiu trong nc l HCl.

+ Theo m hnh (d) kh tan t trong nc l H2S

Cu 10: Chn p n D

Th nghim trn chng t:

+ A phi tan nhiu trong nc Loi A v H2St tan trong nc.

+ Dung dch A c tnh axit mnh (dung dch mu ) nn NH3loi ngay.Vi SO2cng tan nhiu

trong nc tuy nhin tnh ty mu ca SO2rt mnh nn qutm smt mu khi gp SO2.

Cu 11: Chn p n C

Cu 12: Chn p n B

Theo SGK th Be v Mg khng tc dng c vi nc nhit thng.

Cu 13: Chn pn A

Cu 14: Chn p n DCh : c n mn in ha th phi tha mn 3 iu kin

iu kin 1 : C 2 cc (2 kim loi khc nhau hoc 1 kim loi 1 phi kim)iu kin 2: 2 cc ny phi tip xc (trc tip hoc gin tip)iu kin 3: Cng c nhng vo dung dch cht in ly

Dthy cc trng hp A, B, C u tha mn c3 iu kin trn.

Cu 15: Chn p n C

Cu 16: Chn p n C

Cu 17: Chn p n B


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Ddng suy ra X phi cha2

3 BTDT

3

CO : a (mol)2a b 0, 2

HCO : b (mol)

2

BTNT

CO0,1 n a b 0, 2 2, 24 V 4, 48

Cu 18: Chn p nA

Khi kt ta ln nht th Cl-schy ht vo BaCl2.

Do ta c ngay :2

BTNT.Clo

BaCl

0,5.0,2n 0, 05(mol)

2

BTNT.Ba BTKL m 0,05(137 27) 8, 2(gam)

Cu 19: Chn p n B

X + NaOH c kh H2nn Al c d BTE DuAl0,03.2

n 0, 02(mol)3

3 2 3

BTNT.Al

Al(OH) Al O

0,11 0,02n 0,11(mol) n 0, 045(mol)

2

Z chl Fe : 22 4

BTE BTKL

SO FeSOn 0,155 n 0,155 m 20, 76 0,155.96 5,88(gam)

BTKL

Fe Om m m 5,88 0,045.3.16 8,04(gam)

Cu 20: Chn p n A

Ta c : 2 4 3

2 4

H SO .3SO

H SO

n 0, 005(mol) 0,005.4.98 2a% .100% 33,875%

m 2(gam) 1,69 10

Cu 21: Chn p n B

Strng hp thu c kt ta l : FeCl3, MgSO4

Cu 22: Chn p n D

Dthy en 2.2 4(mol) chc D hp l.

Cu 23: Chn p n C

Nhn th

y th

tch v n

ng

cc dung d

ch

u b

ng nhau v

kiu g th 3

NO

vn c d

. Do vn chlin quan ti H+. Da vo phn ng :

3 24H NO 3e NO 2H O

(1) l dung dch KNO3.

(2) l dung dch HNO3.

(3) l dung dch H2SO4.

Cu 24: Chn p n B

Scp cht xy ra phn ng oxi ha - khl: (2) v (3)


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(2). 3 2 22 NaNO H O22NO NaOH NaNO

(3).2 2Fe S2FeS 2HCl Cl + S + H

Cu 25: Chn p n D

Cu 26: Chn p n D

Cu 27: Chn p n A

strng hp thu c kt ta l: (a), (b), (d).

(a). 2H O 33 3NH OH Al 3OH Al OH

(b).2 2 2

SO H S 3S 2H O

(d).2 3 3 2 3 2

3Na CO 2AlCl 3H O 2Al(OH) 3CO 6NaCl

Cu 28: Chn p n C

(a). C0

t

2 2 22F 2H O 4HF O

(b). C2 2 2

3Al NaOH H O NaAlO H

2

(c). Khng3 4

2H PO2 5 2P O 3H O

(d). C 2 3Fe Ag Fe Ag

(e). C 0

t

2 2 2Ca(NO ) O

3 2Ca NO

(f). Khng0

t

2 3 2 2Na CO CO H O

32NaHCO

Ch : Khi nhit phn hai mui ca hai kim loi kim th l Ca(NO3)2v Ba(NO3)2th :

i vi Ca(NO3)2nhit t450-500oC:Ca(NO3)2 Ca(NO2)2+ O2nhit cao hn:Ca(NO3)2 CaO + 2NO2+ 1/2O2

i vi Ba(NO3)2:nhit 594-620oC:Ba(NO3)2Ba(NO2)2+ O2nhit 621-670oC:Ba(NO3)2 BaO + 2NO2+ 1/2O2

Cu 29: Chn p n CMun c vng benzen cc cht phi tng ng c t nht 4 lin kt trong phn t. Bao gm :

C10H14, C6H4(OH)2, C9H10BrCl

Cu 30: Chn p n A

Ch : Khi tnh soxi ha ca C trong cc HCHC th ngi ta tnh theo cc nguyn tkhc dnh voC m khng quan tm ti C khc.


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Cu 31: Chn p nD

Ta c : 22

2

CO BTNT.O

O

H O

n 0,1(mol)n 0,175 V 3, 92(l)

n 0,15(mol)

Cu 32: Chn p n C

CH=CH2

CH3

CH=CH2 CH=CH2

CH3

CH3

CH=CH-CH3 C=CH2 CH2-CH=CH2

CiS - Trans

CH3

Cu 33: Chn p n C

Scht to c mt sn phm thmonoclo duy nht l : metan, etan, neopentan.Cu 34: Chn p n B

(1). Sai, sn phm chnh u tin thvo C bc cao l CH3CHBrCH3.

(2). ng, 2 3 3 3 3 3CH CH(CH ) CH HCl CH CHCl(CH ) CH .

(3). Sai, sn phm chnh l 2-metylpent-2-en.

(4). ng. Theo SGK lp 11.

(5). Sai, u tin to sn phm l o-clobenzen vp- clobenzen.

(6). ng. Theo SGK lp 11.

Cu 35: Chn p n A


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Ta c :2

BTNT.CCAg CAg

2 2 BTNT.HBr 2BTNT.C

2 6

BTNT.H

2

CH CH : 0, 05n 0, 05(mol) CH CH : 0, 2

Y CH CH : 0,1 Xn 0,1 H : 0, 3

C H 0, 05Z

H 0,1

V 0,5.22,4 11, 2(l)

Cu 36: Chn p n C0

t

2 5 3 2C H OH CuO CH CHO Cu H O

2 2PdCl ;CuCl

2 2 2 32CH CH O 2CH CHO

2 0Hg / 80 C

2 3CH CH H O CH CHO

Cu 37: Chn p n CTa c :

2

Trong X Trong X

H OH On 0,7 n n 0, 7.2 1, 4(mol)

thy sC trong X bng sO trong X nn ta c :

2

BTKL BTNT.HTrongX

H H Om 43, 2 1, 4.12 1, 4.16 4 n 2(mol)

2H Om m 2.18 36(gam)

Cu 38: Chn p n C

Ta c : 3 532%O 0, 372 R 41 C H C HR 45

OO

C sai v A c thl 2 2CH CH CH C HOO

Cu 39: Chn p n ACh : CH3COOC2H5, C2H5COOCH3l ng phn ca nhau.

Ta c :BTNT.C

4 8 2

BTNT.H2 5

C H O : a(mol) 4a 2b 0, 6 a 0,1

C H OH : b(mol) b 0,18a 6b 1, 4

2 5

0,1.46%C H OH 34,33%

0,1(46 88)

Cu 40: Chn p n C

Ta c :2

2

2

O

BTKL

CO hh

H O

n 0, 4

n 0,35 m 8,1 0,35.44 0, 4.32 10, 7(gam)

n 0, 45


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1BTNT.O Tronghh

O

2 2

R OH : 0,15(mol)n 0, 35.2 0, 45 0, 4.2 0, 35

R (C H) :0,05(mol)OO

Tsmol CO2suy ra hai ancol phi l CH3OH v C2H5OH v axit l HOOC CH2COOH

2

BTNT.C3 3

3 2

H C CH C H : 0, 05

0,1.32CH OH : 0,1 %CH OH 29,91%10,7

CH CH OH:0,05

OO OO

Cu 41: Chn p n B

Dsuy ra trong cht bo c tng cng 9 lin kt hay 6 lin kt trong mch cacbon.

0, 6a 0,1(mol)

6

Cu 42: Chn p n A

C 3 pht biu sai l (4), (5), (8)

(4). Saccarozc cu to thai gc -glucozv -fructoz

(5). Hiro ha hon ton glucozto ra sobitol.

(8). Amilopectin trong tinh bt c cc lin kt -1,4-glicozit v -1,6-glicozit

Cu 43: Chn p n C

Ta c :

3 7 2

BTKL

X X 3 2 5

3 3

C H NH (2dp)19,1 11,8

n 0, 2 M 59 CH NHC H36,5CH N

Cu 44: Chn p n ACc bn cn phi nhmt sloi aminoaxit quan trng sau :Gly :

2 2NH CH COOH c M = 75

Ala : 3 2CH CH NH COOH c M = 89

Val : 3 3 2CH CH(CH ) CH NH COOH c M = 117

Lys : 2 2 24H N CH CH(NH ) COOH c M = 146

Glu : 2 22HOOC CH CH(NH ) COOH c M = 147

Tyr :6 4 2 2

HO C H CH CH(NH ) COOH c M = 181

phe : 6 5 2 2C H CH CH NH COOH c M = 165

Ta c :58,8

n 0,15(mol)89 75 117 165 3.18

Ala Gly Val Phe

NaOH c d

BTKL58,8 3.0,15.18 0,7.40 m 0,15.4.18 m 84,1(gam)


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Cu 45: Chn p n C

A. Sai trng hp chkhng phi trng ngng.

B. Sai thu c PS

C. ng ng trng ng-ng6 4 2 2HOOC C H COOH HO CH OH lapsan Poli(etylen terephtalat)

D. Sai visco l tbn tng hp

Cu 46: Chn p n A

Cu 47: Chn p n B

(1). Sai metanol l cht lng k thng.

(2). ng.

(3). ng. Cc cht c lin kt khng bn, vng 3 cnh v nhm CHO c khnng tc dng vi ncbrom.

(4). Sai axetilen to kt ta vng.

Cu 48: Chn p n C

Cc cht c thphn ng c vi cba tc nhn l :

(2) phenol (C6H5OH), (3) axit axetic, v (4) glyxin

Cu 49: Chn p n C

Sng phn ca C3H6O2l :

3 2

CH CH C HOO

, 2 5HCOOC H

, 3 3CH COOCH

, 2 2HO CH CH CHO

, 3CH CH(OH) CHO

x sgm :

3 2CH CH C HOO

y sgm :2 5

HCOOC H v3 3

CH COOCH

z sgm :2 5

HCOOC H

t sgm :2 2

HO CH CH CHO v 3CH CH(OH) CHO

Cu 50: Chn p n DCh : NO ngoi khng kh gp Oxi bin ngay thnh NO2l tc nhn chnh gy ra ma axit


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Cho KLNT: C = 12; N = 14; P = 31; O = 16; S = 32; H = 1; Li = 7; Na = 23; K = 39; Mg = 24; Ca = 40; Sr =

87,5; Ba = 137; Al = 27; Zn = 65; Cu = 64; Pb = 207; Ag = 108; Fe = 56; Cu = 64; Mn = 55; Cl = 35,5; Br =

80.

Cu 1:t chy hon ton 17,4 gam hn hp Mg v Al trong kh oxi (d) thu c 30,2 gam hn hp

oxit. Thtch kh oxi (o ktc) tham gia phn ng lA. 11,20 lt. B. 17,92 lt. C. 4,48 lt. D. 8,96 lt.

Bo ton khi lng

Cu 2:Cng thc n gin nht ca mt hirocacbon l CnH2n+1. Hirocacbon thuc dy ngng ca

A. anken. B. ankan. C. ankaien. D. ankin.

Da vo bn cht SH=2*sC + 2

Cu 3:Ngnh sn xut no sau y khng thuc vcng nghip silicat?

A. Xi mng. B. Thy tinh thng. C. Thy tinh hu c. D. gm.Cu 4: no sau y ng?

A. Chc nhng phn ng thun nghch mi c trng thi cn bng ha hc.B. Bt cphn ng no cng phi t n trng thi cn bng ha hc.C. Khi phn ng thun nghch t trng thi cn bng th phn ng dng li.D. trng thi cn bng, thtch cc cht hai vphng trnh ha hc phi bng nhau.

Cu 5:Este X khng no, mch h, c tkhi hi so vi metan bng 6,25 v khi tham gia phn ng xphng ho to ra mt anehit v mt mui ca axit hu c. C bao nhiu cng thc cu to ph hpvi X?

A. 2. B. 4. C. 3. D. 5.M=100; cng thc dng RCOOCH=CHR => R+R=30;

Cc cht: HCOOCH=CHC2H5; HCOOCH=C(CH3)2; CH3COOCH=CHCH3; C2H5COOCH=CH2

Cu 6:Dung dch X gm HCl v H2SO4c pH = 2. trung ha hon ton 0,59 gam hn hp haamin no, mch h, n chc bc mt (c snguyn tcacbon nhhn hoc bng 4) phi dng 1 ltdung dch X. Hai amin c thl

A. CH3NH2 v C4H9NH2.B. C3H7NH2v C4H9NH2.


MN HA HC LN 5 THPT CHUYN H LONG


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C. C2H5NH2v C4H9NH2.D. CH3NH2 v C4H9NH2hoc C2H5NH2v C4H9NH2.

[H+]=0,01=n amin; M amin = 59; R = 43; R l C3H7; loi p n B.

Cu 7:Cho 0,15 mol ancol X phn ng vi kim loi natri (d) th thu c 3,36 lt kh hiro (o ktc). Snhm chc ancol trong X l

A. 1. B. 2. C. 3. D. 4.Cng thc: nH2= (snhm OH/2)* n ancol = > nH2= n ancol; nn ancol c 2 nhm OH;Cu 8:Cho cc cht sau:

C2H3

(1)

CH3

(2) (3)

C2H5C2H5

CH3

(4)

C2H3

C2H5

(5) C bao nhiu cht l ng ng ca Benzen?

A. 2. B. 5. C. 4. D. 3.Cht 2, 3, 4, cht 1 v 5 c gc vinyl nn loi;

Cu 9:Khi cho 5,6 gam Fe tc dng vi 250 ml dung dch AgNO31M th sau khi phn ng kt thcthu c bao nhiu gam cht rn?

A. 27,0 gam. B. 20,7 gam. C. 37,0 gam. D. 21,6 gam.

nAgNO3/nFe=2,5 nn to 2 mui Fe2+v Fe3+; bo ton Ag;

Cu 10:Mt mu nc cng c cha cc ion: Ca2+, Mg2+, HCO3

, Cl , SO 24

. Cht c khnng lm

mm mu nc cng trn l

A. HCl. B. NaHCO3. C. Na3PO4. D. BaCl2.Cu 11:Cho cc gluxit: mantoz, glucoz, fructoz, saccaroz, tinh bt v xenluloz. C bao nhiucht va lm nht mu nc brom va c phn ng trng bc?

A. 4. B. 1. C. 3. D. 2.

Mantozv fructoz

Cu 12:Cho dy cc cht sau: Al, NaHCO3, (NH4)2CO3, NH4Cl, Al2O3, Zn, K2CO3, K2SO4. C baonhiu cht trong dy va tc dng c vi dung dch HCl, va tc dng c vi dung dchNaOH?

A. 2. B. 3. C. 4. D. 5.Al, NaHCO3, (NH4)2CO3, Al2O3, Zn.

Cu 13:Cht no sau y khngphn ng vi dung dch NaOH?A. CO2. B. CuO. C. Cl2. D. Al.

Cu 14:Cng thc cu to thu gn ca glixerol lA. CH2(OH) CH2CH2(OH). B. CH2(OH) CH2(OH).C. CH2(OH) CH(OH) CH2(OH). D. CH2(OH) CH2(OH) CH2(OH).


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Cu 15:Cho cc cp oxi ha khsau: Sn4+/Sn2+; Cu2+/Cu; Fe3+/Fe2+. Cho bit tnh oxi ha tng dntheo tht: Sn4+, Cu2+, Fe3+; tnh khgim dn theo tht: Sn2+, Cu, Fe2+. Don cc phn ng sauy c xy ra khng?

1. Cu + FeCl3 2. SnCl2 + FeCl3A. 1 ( c), 2 ( c). B. 1 ( khng), 2 ( c).

C. 1 ( c), 2 ( khng). D. 1 ( khng), 2( khng).Dng quy tc anpha;

Cu 16:Anken thch hp iu ch3-etylpentan-3-olbng phn ng hirat ha lA. 3-etylpent-2-en. B. 3, 3-imetylpent-2-en.C. 3-etylpent-3-en. D. 3-etylpent-1-en.

Vit CTCT ancol v dng pp tch nc tm anken;

Cu 17:Dn kh than t qua m gam hn hp X gm cc cht Fe2O3, CuO, Fe3O4 (c smol bngnhau) un nng thu c 36 gam hn hp cht rn Y. Cho Y phn ng ht vi dung dch HNO3long dthu c 11,2 lt kh NO (sn phm khduy nht o ktc). Gi trca m l

A. 48,2. B. 54,2. C. 47,2. D. 46,4.

Coi smol mi oxit l a mol; => nFe=5a, nCu=a, nO d=b ; 344a + 16b = 36; bo ton e cho hnhp 36 gam;

3*Fe + 2*nCu = 3*nNO + 2* nO d=> 15a + 2a = 2b + 3*0,3; a=0,1; b=0,1; m = 472*0,1=47,2 gam

Cu 18:khhon ton 17,6 gam hn hp gm Fe, FeO, Fe2O3, Fe3O4cn dng va 2,24 lt H2(o ktc). Khi lng Fe thu c l

A. 15 gam. B. 17 gam. C. 16 gam. D. 18 gam.

m Fe = 17,6 mO = 17,6 0,1*16=16 gam

Cu 19:C bao nhiu hp cht hu cmch hc cng thc phn tC3HxO va phn ng vi H2(xc tc Ni, to), va phn ng vi dung dch AgNO3 trong NH3 un nng?

A. 2. B. 4. C. 5. D. 3.Cht phi c lin kt 3 u mch hoc c nhm CHO;TH1: x= 2; C3H2O; CHC-CHO

TH2: x=4; C3H4O; (k=2) CH2=CH-CHO; CH

C-OCH3; CH

C-CH2OHTH3: x=6; C3H6O; C2H5CHO;Cu 20:Thc hin cc th nghim sau:

(1) Sc kh SO2vo dung dch KMnO4.(2) Sc kh Cl2vo dung dch H2S.(3) Sc hn hp kh thu c khi nhit phn Cu(NO3)2vo nc.(4) Cho Na2CO3vo dung dch AlCl3.(5) Cho HCl vo dung dch Fe(NO3)2.(6) Cho Fe2O3vo dung dch HI.

Sth nghim c phn ng oxi ha khxy ra lA. 5. B. 6. C. 3. D. 4.


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Cu 21:Sc kh clo dvo dung dch cha mui NaBr v KBr thu c mui NaCl v KCl, ng ththy khi lng mui gim 4,45 gam. Thtch kh clo tham gia phn ng vi 2 mui trn (o ktc) l

A. 4,48 lt. B. 3,36 lt. C. 2,24 lt. D. 1,12 lt.

Phng php tng gim khi lng: nCl2=1/2n mui =4,45/(80-35,5)=0,05 mol;

Cu 22:Khi x phng ha hon ton tristearin ta thu c sn phm lA. C17H35COOH v glixerol. B. C15H31COONa v etanol.

C. C17H35COONa v glixerol. D. C15H31COOH v glixerol.

Cu 23:Hnh v sau do mt hc sinh vm tli th nghim n mn in hahc khi cm hai l Cu v Zn (c ni vinhau bng mt dy dn) vo dung dchH2SO4long:Trong hnh v bn chi tit no cha

ng?A.Chiu dch chuyn ca ion Zn2+.B. Bmt hai thanh Cu v Zn.C. Chiu chuyn dch ca cc electrontrong dy dn.D. K hiu cc in ccKim loi c tnh khmnh hn lun l cc m v n bn mn;Cu 24:Hstrng hp ca tnilon-6,6 (M = 2500 vC) v tcapron (M = 15000 vC) ln lt l

A. 11 v 123. B. 11 v 133. C. 22 v 123. D. 22 v 133.

Cu 25:Cht rn X mu thm tan trong nc thnh dung dch mu vng. Mt scht nhS, P, CC2H5OH bc chy khi tip xc vi X. Cht X l

A. CrO3. B. P. C. Cu. D. Fe2O3.Cu 26:Cho cc bc tin hnh th nghim trng bc bng andehit fomic:

(1). Nhtip 3-5 git dung dch HCHO vo ng nghim.

(2). Nhtttng git dung dch NH32M cho n khi kt ta sinh ra bha tan ht.

(3). un nng nhhn hp 60 -70 oC trong vi pht.

(4). Cho 1 ml AgNO31% vo ng nghim sch.

Thttin hnh ng lA. (4), (2), (1), (3). B. (4), (2), (3), (1). C. (1), (4), (2), (3). D. (1), (2), (3), (4).

Cu 27:Cho 44 gam NaOH vo dung dch cha 39,2 gam H3PO4. Sau khi phn ng xy ra honton, em lm kh. Mui no c to thnh v khi lng mui khan thu c l bao nhiu?

A. 14,2 gam Na2HPO4v 49,2 gam Na3PO4. B. 15 gam Na3HPO4.C. 19,2 gam NaH2PO4v 14,2 gam Na2HPO4. D. 50 gam Na3PO4.

Dng pp tlv bo ton nguyn ttm smol tng mui;

Zn C

e

Dung dich H2SO4

Zn2

H+


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Cu 28:Thy

10 Đề Thi Ôn Luyện THPT Quốc Gia 2016

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Transcript of 10 Đề Thi Ôn Luyện THPT Quốc Gia 2016