2011 - Lời giải chi tiết & phân tích đề thi Hóa - Khối A - năm 2011

7/31/2019 2011 - Li gii chi tit & phn tch thi Ha - Khi A - nm 2011

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TT LUYN THI & BI DNG KIN THC NGY MI

B GIO DC & O TO-----------------------

CHNH THC( thi c 06 trang)

HNG DN GII PHN TCH THI TUYN SINH I HC NM 2011

Mn: HA HC Khi AThi gian lm bi: 90 pht, khng k thi gian pht

M 318

H v tn th sinh: .

S bo danh:

Cho bit nguyn t khi o ca cc nguyn t:H = 1; C = 12; N = 14; O = 16; Na = 23; Mg = 24; Al = 27; S = 32; Cl = 35,5; K = 39; Ca = 40; Cr = 52; Mn =55; Fe = 56; Cu = 64; Zn = 65; Br = 80; Ag = 108

I. PHN CHUNG CHO TT C TH SINH (40 cu, t cu 1 n cu 40)Cu 1. Hn hp X gm axit axetic, acid fomic v acid oxalic. Khi cho m gam X tc dng vi NaHCO 3 (d) th

thu c 15,68 lt CO2 (ktc). Mt khc, t chy hon ton m gam cht X cn 8,96 lt kh O 2 (ktc), thuc 35,2 gam CO2 v y mol H2O. Gi tr ca y l:

A. 0,8 B. 0,3 C. 0,2 D. 0,6Hng dn:y l bi ton cho t chy v bi cho d liu lin quan ti oxi th dng ton ny ta thng s

dng nh lut bo ton nguyn t oxi v c s nhn xt v bn cht trong cc phng trnh phn ng.* i vi cc em hc sinh cha c s phn tch:

CH3COOH

HCOOH

(COOH)2

CO2

OH2

CO2

CO2

NaHCO3 CO

2

CO2

CO2

NaHCO3

NaHCO3

O2

O2

O2

OH2

OH2

2+

2

+

+

2

2

Gi s vi a, b v c ln lt l s mol ca axit axetic, acid fomic v acid oxalic

Theo phn ng hnh thnh ra kh CO2 (phn ng vi NaHCO3) ta c phng trnh: a + b + 2c =15,68

0,722,4

mol=

Theo phn ng t chy ta c2CO

n = 2a + b + 2c =35,2

0,844

mol= v2

2H On a b c= + +

Theo nh lut bo ton nguyn t oxi ta c:2 2 2/ / / /hh XO O O O CO O H O

n n n n+ = +

a + b + 2c + 0,4 = 2a + b + 2c +2

2

a b c+ +=>

2

2

a b c+ += 0,4 a ( a = 0,8 0,7 = 0,1)

Vy2

2H On a b c= + + = 0,6 mol

* i vi cc em hc sinh c s phn tch:2 /ch yCO

n = 0,8 mol;2 3/CO NaHCO

n = 0,7 mol;2 /

0, 4chy molOn =

Khi cho axit tc dng vi NaHCO3 th bn cht ca phn ng l H linh ng trong nhm -COOH tc dng vi

ion 3HCO theo phn ng: ( )3 2 2COOH HCO COO CO H O

+ + +

Qua phn ng trn ta c nhn xt:2 3/ /

2 2S nguyn t H linh ngO axit CO NaHCOn n n= =

nO/axit = 0,7.2 = 1,4Axit + O2 CO2 + H2O0,7 0,4 0,8 y

Bo ton nguyn t O : 0,14 + 0,4.2 = 0,8.2 + y y = 0,6Nhn xt: Qua bi ton ny, cc em hc sinh cn lu v vn khi cho acid phn ng vi mui cha ion

3HCO th ta lun c:

2/2 2S nguyn t H linh ngO axit COn n n= = ( )3 2 2COOH HCO COO CO H O

+ + +


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Cu 2. Hn hp X gm C2H2 v H2 c cng s mol. Ly mt lng hn hp X cho qua cht xc tc nung nng,thu c hn hp Y gm C2H4, C2H6, C2H2 v H2. Sc Y vo dung dch brom (d) th khi lng bnh bromtng 10,8 gam v thot ra 4,48 lt hn hp kh (ktc) c t khi so vi H2 l 8. Th tch O2 (ktc) cn tchy hon ton hn hp Y l

A. 33,6 lt. B. 22,4 lt. C. 26,88 lt. D. 44,8 lt.Hng dn: y l dng ton c bn ca hidrocacbon, trong bi ton ny chng ta thng s dng nh lut bo

ton nguyn t, qua gip chng ta => iu quan trng nht l chng ta phi hiu rng t chy hn hp Ycng tng ng t chy hn hp X

Ta c s phn ng:0

2 2

22 2 ,

2 4 2 2 2 2 6

2 4210,8 0,2

162 6

,xt Brm

gam molM

hh X hh Y + ,t

m n

C H

HC HC H C H H C H

C HH

C H

+

= =

=

1 442 4 43 1 4 2 4 3

Theo nh lut bo ton khi lng ta c mx = 10,8 + 16.0,2 = 14 gamTheo bi ta li c 26a + 2a = 14 => a = 0,5 mol (vi a l s mol ca C 2H2 v H2)

Theo phn ng t chy ta c: 2 22 2 2 2 2

2 2 2

52

2,5 0,5 3 1,5 33,622 2

mol =>V lit

O O

C H O CO H O

n a a aO H H O

+ + = + = = =+

* Do s mol ca C2H2 v H2 bng nhau v bng 0,5 mol => c th coi hn hp trn l C2H4: 0,5 mol

2 4 2 2 23 2 2C H O CO H O+ +

Vy =>2 2

0,5.3 1,5 33,6mol =>V litO On = = =

Cu 3. un nng m gam hn hp Cu v Fe c t l khi lng tng ng 7 : 3 vi mt lng dung dch HNO3.Khi cc phn ng kt thc, thu c 0,75m gam cht rn, dung dch X v 5,6 lt hn hp kh (ktc) gm NOv NO2 (khng c sn phm kh khc ca N+5). Bit lng HNO3 phn ng l 44,1 gam. Gi tr ca m l

A. 44,8. B. 40,5. C. 33,6. D. 50,4.

Hng dn: y l mt cu kh c bn trong chng kim loi, trong cu ny chng ta phi hiu c+ bi cho r lng HNO3 => ta thng p dng nh lut bo ton nguyn t Nv p dng nh lut boton electron+ 0,75m gam cht rn thu c sau phn ng l g? N l 1 phn cha kha gii quyt bi ton haTheo bi ta c 0,7m gam Cu v 0,3m gam Fe => m 0,75m = 0,25m cht phn ng=> C 0,05m gam Fe cn d v 0,25m gam st tham gia phn ng v chuyn t Fe => Fe(NO3)2

22Fe e Fe + HNO3 1e+ NO2 + 3 (mui)NO

HNO3 3e+ NO + 3 (mui)NO

Theo nh lut bo ton nguyn t N ta c: 3 23 30,7 0,45(mui) (mui)mol molHNO NO NONO NOn n n n n = + + = =

Ta c cng thc3

0,45 0,225(mui)

Trao i mol =2.n n mol=>m =e

Fe FeNOn n = = => = 50,4 gam

Hoc c th s dng theo hng:

( )

23

2

32

0,252. 0,45 50,4

56(mui)

3 2Fe NO

=2.n mol gamNO Fe

Fe NO

mn m +

+ +

= =

Nhn xt: y l bi tp kh hay, khi bi ch cho hn hp kh NO v NO2 m khng cho bit t l v s molkh lm mt s hc sinh lng tng bi cc em hc sinh thng c khuynh hng tm ra s mol tng kh

Nhiu hc sinh c cho rng khi Fe tc dng vi HNO3 th quan nim Fe v Fe3+ th s ra p n C. 33,6

Nu hc sinh vn coi l v Fe2+ nhng khi bo ton nguyn t N m ly s mol hn hp kh nhn 2 ri thay vos c p n A.44,8Cu 4. t chy hon ton x mol axit cacboxylic E, thu c y mol CO 2 v z mol H2O (vi z = y x). Cho x mol

E tc dng vi NaHCO3 (d) thu c y mol CO2. Tn ca E l

-2-


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A. axit oxalic. B. axit fomic. C. axit aipic. D. axit acrylic.Hng dn:Cu ny i hi chng ta phi ghi nh tn ng vi nhng cng thc l gaxit acrylic: CH2=CH-COOH axit oxalic : (COOH)2axit aipic: C4H8(COOH)2 axit fomic: HCOOHTheo bi da vo d kin: Cho x mol E tc dng vi NaHCO 3 (d) thu c y mol CO2 v t chy hon

ton x mol axit cacboxylic E, thu c y mol CO2=> s nguyn t C trong acid E bng s nhm chc COOH => loi p n A v CMt khc t: z = y x => p n l B

Mt hng lm khc: Goi cng thc: CxHyOz x CO2 +2

yH2O

a ax2y

a 2y

a = ax a y = 2x 2 => Axit khng no hay 2 chc

2COn = nCaxit hai chc HOOC COOH

Mt hng gii khc mt nhiu thi gian hn

Gi cng thc tng qut ca axit l CnH2n+2-2a-k(COOH)kVi n l s nguyn t cacbon trong gc axit, a l s lin kt pi v vng, k l s ln axit.CnH2n+2-2a-k(COOH)k 2O+ (n+k)CO2 + (n + 1 a)H2O

x mol (n+k)x = y mol (n + 1 a)x = z molCnH2n+2-2a-k(COOH)k 3NaHCO+ kCO2

x mol kx molTheo bi ta c: (n+k)x = y v kx = y => n = 0Vy Axit cn xc nh khng cha gc hidrocacbon => Loi c p n A v C

Mt khc ta c:(1 ).

z y x

y kxa x z

=

= =

2k a = +

+ k =1, a =1 axit cho l khng no n chc. Loi+ k=2 , a =0 axits cho l axit no hai chc v khng cha gc hidrocacbon. Ch c axit oxalic l tha mn iukin bi ton. Chn ACu 5. S ng phn amino axit c cng thc phn t C3H7O2N lA. 3. B. 4. C. 1. D. 2.Hng dn: y l 1 bi tng i n gin, i hi mt cht v mng kin thc khi nim v amino acid (chang thi 2 nhm COOH v NH2)

Bi ny chng ta cn tun th ng bc vit ng phn c hc trong chuyn : ng phn danhphp c hc th s d dng xc nh c.

C2H4(NH2)(COOH)C C COOH

Phn mi tn l phn ch v tr in nhm NH2 trong phn tH2N CH2- CH2 - COOH v H2N CH(CH3) COOH.

Cu 6. Dy gm cc cht u c th lm mt tnh cng tm thi ca nc l:A. HCl, NaOH, Na2CO3.

B. NaOH, Na3PO4, Na2CO3.

C. KCl, Ca(OH)2, Na2CO3.

D. HCl, Ca(OH)2, Na2CO3.Hng dn:Khi nim v nc cng: Nc cng l loi nc t nhin cha cation canxi (Ca2+) v magie (Mg2+).

- Nc cha nhiu Mg2+ c v ng


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Cu 10. Trong c th nghim sau :(1) Cho SiO2 tc dng vi axit HF. (2) Cho kh SO2 tc dng vi kh H2S.(3) Cho kh NH3 tc dng vi CuO un nng. (4) Cho CaOCl2 tc dng vi dung dch HCl c.(5) Cho Si n cht tc dng vi dung dch NaOH. (6) Cho kh O3 tc dng vi Ag.(7) Cho dung dch NH4Cl tc dng vi dung dch NaNO2 un nng.

S th nghim to ra n cht l:

A. 4 B. 7 C. 6 D. 5Hng dn: y l dng bi tp i hi hc sinh phi ghi nh l thuyt c bit l v tnh cht ha hc, phngtrnh phn ng iu ch ca cc cht t nm hc lp 10 ti 12.

2 4 2

2 2 3 2

2 2 2

3 2 2

3 2 2

4 2 2 2

2 2 2 2c

iF (P.ng n mn thy tinh)

SO

SiO HF S H OSi NaOH H O Na SiO H

H S S H OO Ag Ag O O

NH CuO Cu N H ONH Cl NaNO NaCl N H O

CaOCl HCl Cl CaCl H O

+ ++ + +

+ ++ +

+ + ++ + +

+ + +

Cu 11. Cho 13,8 gam cht hu c X c cng thc phn t C7H8 tc dng vi mt lng d dung dch AgNO3trong NH3, thu c 45,9 gam kt ta. X c bao nhiu ng phn cu to tha mn tnh cht trn?

A. 5. B. 6 C. 4. D. 2.Hng dn: Theo bi X tc dng vi AgNO3/NH3 => kt ta => X phi l hp cht cha lin kt ba ( )

S lin kt pi + vng = 4

7 80,15 0,15 306molC Hn n mol M = = => = => trong kt ta c 2 nguyn t Ag => X cha 2 lin kt ( )

u mch

C C C C C CHCH C C CHC C

C

CH C C CHCCH

C C

C C CHCCH

C

C

Cu 12. Nung m gam hn hp X gm FeS v FeS2 trong mt bnh kn cha khng kh (gm 20% th tch O 2 v80% th tch N2) n khi cc phn ng xy ra hon ton, thu c mt cht rn duy nht v hn hp kh Y cthnh phn th tch: 84,8% N2, 14% SO2, cn li l O2. Phn trm khi lng ca FeS trong hn hp X l

A. 59,46%. B. 26,83%. C. 19,64%. D. 42,31%.Hng dn:

2 2 3 2 2 2 2 3 2

7 112 2 2 4

2 2FeS O Fe O SO FeS O Fe O SO+ + + +

Gi s ban u c 1 mol khng kh: =>2 2

0,2 0,8mol; n molO Nn = =

=> S mol ca hn hp Y l nY =100.0,8

0,943484,8

mol= =>2

0,1321 molSO

n = v2

0,0113 molO

n =

Vy s mol O2 tham gia phn ng l: 2 0, 2 0, 0113 0,1887 molOn = =

Gi x l s mol FeS, y l s mol FeS2 ta gii h pt:1,75x + 2,75y = 0,1887 v x + 2y = 0,1321 => x = 0,0189 mol v y = 0,0566 mol%FeS = (0,0189.88)/(0,0189.88 + 0,0566.120) = 19,67% (Sai s 0,03% c th chp nhn c)

-----------Nhng khng mt tnh tng qut v gii s sai s khi tnh ton th ta nn chn theo hng:

Chn 1 mol hn hp sn phm 2Nn = 0,848; 2SOn = 0,14 v 2On = 0,012 mol

T 20% th tch O2 v 80% th tch N2 2On =0,848.20

80 = 0,212 tham gia 0,212 0,012= 0,2

vi2SO

n = 0,14 2FeS + 3,5O2 Fe2O3 + 2 SO2 v 2FeS2 + 5,5O2 Fe2O3 + 4 SO2

x 1,75x x y 2,75y 2y


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Lp h: 1,75x + 2,75y = 0,2 v x + 2y = 0,14 x = 0,02 v y = 0,06 %FeS =12006088020

88020

.,.,

.,=19,64%

Cu 13. Hp cht hu c X cha vng benzen c cng thc phn t trng vi cng thc n gin nht. TrongX, t l khi lng cc nguyn t l mC : mH : mO = 21 : 2 : 8. Bit khi X phn ng hon ton vi Na th thuc s mol kh hir bng s mol ca X phn ng. X c bao nhiu ng phn (cha vng benzen) thamn cc tnh cht trn?

A. 3. B. 9. C. 7. D. 10.Hng dn: Theo bi ta c t l nC : nH nO = 1,75 : 2 : 0,5 = 7 : 8 : 2 => CTPT: C7H8O2X tc dng c vi Na => H2 => c cha nguyn t H linh ng, mt khc s mol kh hir bng s mol ca X phn ng => cha 2 nguyn t H linh ng

CH2

OH

CH3

OH

CH3

OH

Phn mi tn ch l phn gn v tr nhm OH cn li

Cu 14. Cho dy cc cht: NaOH, Sn(OH)2, Pb(OH)2, Al(OH)3, Cr(OH)3. S cht trong dy c tnh cht lngtnh l

A. 1. B. 2. C. 4. D. 3.Hng dn:Cht lng tnh l nhng cht va c kh nng phn li ra ion H+, OH-

2( )Sn OH [ ]2

22

2

2( )

2

OH SnPb OH

H SnO

+

+

+

+

3( )Al OH [ ]3

3

2 2

3( )

OH Al Cr OH

H AlO H O

+

+

+

+ +Nhn xt: y l bi kim tra l thuyt n gin, tuy nhin Sn(OH)2, Pb(OH)2 c tnh lng tnh th SGK ban c

bn khng ni r, nn nhiu em c th sai cu hi ny.Tuy nhin nu thay Cr(OH)3 bng Cr(OH)2 th c l nhiu em hc sinh cng d b sai.Cu 15. in phn dung dch gm 7,45 gam KCl v 28,2 gam Cu(NO3)2 (in cc tr, mng ngn xp) n kh

khi lng dung dch gim i 10,75 gam th ngng in phn (gi thit lng nc bay hi khng ng k).Tt c cc cht tan trong dung dch sau in phn l

A. KNO3 v Cu(NO3)2.B. KNO3 v KOH.

C. KNO3, KCl v KOH.D. KNO3, HNO3 v Cu(NO3)2.

Hng dn:

-6-

CH2OH

OH

CH2OH

OH

CH3

OHHO

CH3

OH

OH

CH3

OHHO

OH

CH3

OH

CH2OH

CH3

OH

HO

CH3

OHHO


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Cc em cn nh rng: "Khi lng dung dch gim chnh l khi lng ca cht tch ra khi dung dch - l cht kt ta v cht bay hi" Nh vy 10,75 gam l khi lng ca Cu to thnh v khi lng ca khthot ra ( c th l Cl2, O2 , H2)

Theo bi ta c:

( ) ( )

3 2

2

3 2 2

2( ) 2

0,1 ; ; , ,

0,15 2 2 2

KCl

Cu NO

n mol Cl NO H O K Cu H O

n mol Cl e Cl Cu e Cu

+ +

+

+

= = +

+ Gi s Clo b in phn ht => khi lng dung dch gim sau l:mGim = 0,05.71 + 0,05.64 = 6,75 gam < 10,75 gam => c thm qu trnh nc b in phn

2 22 4 4H O e H O+ +

+ Gi s Cu b in phn ht => khi lng dung dch gim l:mgim = 0,15.64 + 0,05.71 + 0,05.32 = 14,75 gam > 10,75 gam => Cu2+ cha b in phn ht

Vy dung dch sau in phn s gm c: KNO3, HNO3 v Cu(NO3)2.Gii theo hng s dng phng trnh phn ng

nKCl = 0,1 ; 3 2( ) 0,15Cu NOn mol =

2KCl + Cu(NO3)2 Cu + 2KNO3 + Cl20,1 ---------0,05-------0,05----------------0,05

KCl ht , Cu(NO3)2 cn = 0,15 0,05 = 0,1Cu(NO3)2 + H2O Cu + 2HNO3 + 1/2O2

x---------------------x---------------------1/2x

mdung dch gim = mCu kt ta + ( )2 2Cl Om +

(0,05 + x)64 + 0,05.71 + 1/2x.32 = 10,75 x = 0,05 Cu(NO3)2 vn cn d dung dch sau p cha KNO3; HNO3 v Cu(NO3)2.Nhn xt:+ Nu bi tnh chnh xc nng cht tan c trong dung dch sau mt thi gian th cc em phi xt lng Cuto thnh sau phn ng phn ng vi lng HNO3 to ra bao nhiu nh.+ V trong qu trnh in phn Cu khng th tc dng c vi HNO 3. Nhng khi ngt dng in th c phn ng nh.Cu 16. Khi ni v peptit v protein, pht biu no sau y l sai?

A. Lin kt ca nhm CO vi nhm NH gia hai n v -amino axit c gi l lin kt peptit.B. Tt c cc protein u tan trong nc to thnh dung dch keo.C. Protein c phn ng mu biure vi Cu(OH)2.

D. Thy phn hon ton protein n gin thu c cc -amino axit.Hng dn:

+ Lin kt peptit: lin kt ng ho tr (-CO-NH-) gia nhm - caboxyl ca mt axit amin (amino axit) nyvi nhm - amin ca mt axit amin khc, loi tr mt phn t nc.

NH2

CH

R1

C

O

OH

N CH

H

H R1

C

OH

O

NH2

CH

R1

C

O

N CH

H

R1

C

OH

O

OH2

Sn phm ca phn ng ny l 1 ipeptit, nu c 3, 4, 5... hoc nhiu axit kt hp vi nhau, to thnh cc peptitc tn tng ng l tripeptit, tetrapeptit, vv.Nu ta vt t "" trong mnh A. Lin kt ca nhm CO vi nhm NH gia hai n v -amino axit c gi

l lin kt peptit. Ri hi c bao nhiu pht biu ng th c th nhiu em hc khng chc chn s sai.+Phn ng mu biure: phc cht mu tm c trng- Amino axit v ipeptit khng cho phn ng ny. Cc tripeptit tr ln tc dng vi Cu(OH) 2 to phc cht mutm


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+ Protein c phn thnh 2 loi:- Protein n gin: c to thnh ch t cc -amino axit- Protein phc tp: c to thnh t cc protein n gin kt hp vi cc phn t khng phi protein (phi

protein) nh axit nucleic, lipit, cacbohiratV c th tn ti hai dng: dng hnh si v dng hnh cu

Dng protein hnh si nh : keratin ca tc, mng, sng hon ton khng tan trong nc

Dng protein hnh cu nh : anbumin ca lng trng trng, hemoglobin ca mu tan c trong nc todung dch keoKhi t chy protein s cho ta mi kht

Cu 17. Thc hin cc th nghim sau:(1) t dy st trong kh clo.(2) t nng hn hp bt Fe v S (trong iu kin khng c oxi).(3) Cho FeO vo dung dch HNO3 (long, d).(4) Cho Fe vo dung dch Fe2(SO4)3.(5) Cho Fe vo dung dch H2SO4 (long, d).

C bao nhiu th nghim to ra mui st (II) ?A. 3 B. 2 B. 4 D. 1Hng dn:Fe + Cl2 => FeCl3Fe + S => FeSFeO + HNO3(long, d) => Fe(NO3)3 + NO + H2OFe + Fe2(SO4)3 => FeSO4Fe + H2SO4 (long d) => FeSO4 + H2=> 3 phng trnh phn ng to ra mui st (II)Cu 18. Sn phm hu c ca phn ng no sau y khng dng ch to t tng hp?

A. Trng hp vinyl xianua.B. Trng ngng axit -aminocaproic.C. Trng hp metyl metacrylat.D. Trng ngng hexametyleniamin vi axit aipic.

Hng dn:+ T nitron (hay olon) c tng hp t vinyl xianua (thng gi l acrilonitrin) CH2=CH-CN

CH2=CH-CN CH

2CH

CN

** nn

+ Nilon -6 hay t capron

H2N-CH

2-CH

2-CH

2-CH

2-CH

2-COOH NH CH

2C *5

O

*n

OH

2+

+ Thy tinh hu c ( plexiglas) hay poli(metyl metacrylat) dng lm cht do

CH2

C COO CH3

CH3

CH2* C

COO CH3

CH3

*n

+ Nilon 6,6

NH2

CH2

NH26 HOCO CH2 COOH4

C CH2

C4

O

NH CH2

NH6*

O

*n

+

-8-


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1


Cu 19. Phn chua c dng trong ngnh cng nghip thuc da, cng nghip giy, cht cm mu trong ngnhnhum vi, cht lm trong nc. Cng thc ha hc ca phn chua l

A. Li2SO4.Al2(SO4)3.24H2O.C. K2SO4.Al2(SO4)3.24H2O.

B. (NH4)2SO4.Al2(SO4)3.24H2O.D. Na2SO4.Al2(SO4)3.24H2O

Hng dn:Theo sch gio khoa: Li2SO4.Al2(SO4)3.24H2O; (NH4)2SO4.Al2(SO4)3.24H2O; (NH4)2SO4.Al2(SO4)3.24H2O c

gi chung l phn nhmNhn xt: y l mt cu hi c trong SGK, nu hi c bao nhiu cht thuc loi phn chua th hay hn.+ Cc em hc sinh no hay mc qun Jean th nn ngm qun mt m qua nc phn chua , khi qun s t b

phai mu.+ Nu chiu ci hoc chiu tre v ma ma, m cao hay b mc. Chng ta c th git bng nc phn chualong, khi chiu s khng b mc na+ Phn chua cn cha c bnh hi nch na.+ Mt s cng dng khc ca phn chua: Cha cao huyt p; Cha vim tai gia mn tnh; Cha st rt; Chaho ra mu; .Cu 20. Trung ha 3,88 gam hn hp X gm hai axit cacboxylic no, n chc, mch h bng dung dch NaOH,

c cn ton b dung dch sau phn ng thu c 5,2 gam mui khan. Nu t chy hon ton 3,88 gam X thth tch oxi (ktc) cn dng l

A. 1,12 lt. B. 3,36 lt. C. 4,48 lt. D. 2,24 lt.Hng dn:Gi cng thc chung ca hn hp acid no, n chc, mch h l: CnH2nO2Ta c c 1 mol CnH2nO2=> mui CnH2n-1O2Nath khi lng tng thm 22 gam

x mol ------------------------------- tng thm 5,2 3,88 = 1,32 gam=> S mol ca acid CnH2nO2 l: 0,06 mol v 14n + 32 = 64,67 => n = 2,33

CnH2nO2 +3 2

2

n O2 => nCO2 + nH2O

Vy th tch kh O2 cn dng t chy l: 0,06.3 2

2

n = 0,15 mol =>

23,36 molOV =

Cu 21. Ha tan 13,68 gam mui MSO4 vo nc c dd X. in phn X (vi in cc tr, cng dngin khng i) trong thi gian t giy, c y gam kim loi M duy nht catot v 0,035 mol kh anot. Cnnu thi gian in phn l 2t giy th tng s mol kh thu c c hai in cc l 0,1245 mol. Gi tr ca yl

A. 3,920. B. 1,680. C. 4,480. D. 4,788.Hng dn:

2 2

4 4MSO M SO

+

+(+)Anot2

4 2; HSO O

(-)Catot2

2; HM O+

2 22 4 4H O e O H+ + 2

2 2

2

2 2 2

M e M

H O e H OH

+

+

+ +

in phn trong thi gian t giy thu c 0,035 mol kh anot l kh O2 vy 2t giy ta s thu c 0,035.2= 0,07 mol kh, nhng thc t ta thu c 0,1245 mol kh, s chnh lch s mol l do in phn nc to khH2

2H

n = 0,1245 0,07 = 0,0545

H2O H2 + 1/2O20,0545----0,02725

2O

n to ra do mui in phn = 0,07 0,02725 = 0,04275


10/24

MSO4 + H2O M + H2SO4 + 1/2O20,0855-----------------------------------0,04275

Mmui = 13,68/0,0855 = 160 M = 64 mCu tnh theo t giy l mCu = 2.0.035.64 = 4,480 gamNhn xt:

lm c bi tp ny cc em phi hiu bn cht ca bi ton in phn

y l bi tp tng i kh nu cc em khng bit bn cht v in phn, hoc ch hc theo sgk cng khlm c bi tp ny.Cu 22. Xenluloz trinitrat c iu ch t phn ng gia axit nitric vi xenluloz (hiu sut phn ng 60%

tnh theo xenluloz). Nu dng 2 tn xenluloz th khi lng xenluloz trinitrat iu ch c lA. 3,67 tn. B. 2,97 tn. C. 1,10 tn. D. 2,20 tn.Hng dn:

( ) ( )60%6 7 2 3 3 6 7 2 2 3 2( ) 3 ( ) 3H

n nC H O OH nHNO C H O ONO nH O=+ +

Theo bi ta c mxenluloz trinitrat =2 60

. .297 2,2162 100

= tn

Cu 23.Hp th hon ton 0,672 lt kh CO2 (ktc) vo 1 lt dung dch gm NaOH 0,025M v Ca(OH)20,0125M, thu c x gam kt ta. Gi tr ca x lA. 0,75. B. 1,25. C. 1,00. D. 2,00.Hng dn:

{

{ {

2

2

2 23

2 3

2

2 3 2

2

2 2

3

0,031,67

0,05 2

0,030,02 0,0125

2 0,05

mol n

n mol

mol; n mol

Ca

xCO xOH

COOH

y yy

CO Ca

CO OH HCOn

Tn CO OH CO H O

x yn y

x y

CO

+

+

+ = = = = + +

+ = = = = + =

+

123

123

23 3

3

0,0125 1,25n gamCaCO CaCOCa

CaCO

n m+

= = =

Nhn xt :y l bi tp c ra i ra li nm 2008, 2009 v 2011

Nhng hai nm 2008 v 2009 l ly hiu OH- - tr i CO2 ri nhn vi 100 ra CaCO3 haynhn vi 197 ra BaCO3

Nm nay em no khng ch chc s chn p n A. 2 gam, v c ngh ging nm trc.Phi chng sau 3 nm n s lp li tun hon theo th hnh sin chng?

y l dng bi tp quen thuc m ch cn bm my tnh cng cho kt qu.Cu 24. Qung st manhetit c thnh phn chnh lA. Fe2O3. B. FeCO3. C. Fe3O4. D. FeS2.Hng dn:FeS2: Qung pirit (Him c trong t nhin) Fe3O4: Qung manhetitFe2O3: Qung hematit ; Fe2O3.nH2O qung hematit nu FeCO3 : Qung xideritCu 25. Khi lng ring ca canxi kim loi l 1,55 g/cm3. Gi thit rng, trong tinh th canxi cc nguyn t l

nhng hnh cu chim 74% th tch tinh th, phn cn li l khe rng. Bn knh nguyn t canxi tnh theo lthuyt l

A. 0,155 nm. B. 0,185 nm. C. 0,168 nm. D. 0,196 nm.Hng dn:y l bi thuc v mng ha hc 10, i hi cc em phi ghi nh nhiu v mt cng thc ton hc

mD

V= 3

43

V R= N = 6,02.1023 nguyn t

-10-


11/24

1


+ Th tch 1 mol tinh th Ca : V = 40/1,55 = 25,81 cm3

+ Th tch 1 mol nguyn t Ca : V = 25,81.74% = 19,1 cm3

+ Th tch 1 nguyn t Ca : V = 19,1/(6,02.1023) = 3,17.10-23

p dng cng thc : V = 4.R3/3 R =3( 3V/4.) = 1,96.10-8 cm = 0,196 nmCu 26. Este X c to thnh t etylen glicol v hai axit cacboxylic n chc. Trong phn t este, s nguyn

t cacbon nhiu hn s nguyn t oxi l 1. Khi cho m gam X tc dng vi dung dch NaOH (d) th

lng NaOH phn ng l 10 gam. Gi tr ca m lA. 14,5. B. 17,5. C. 15,5. D. 16,5.Hng dn:Este c dng

CH2

CH2

RCOO

R'COO

S nguyn t O = 4 s nguyn t C = 5 => Vy R = 1 v R = 15CH

2

CH2

HCOO

CH3-COO

nNaOH = 10/40 = 0,25 neste = 1/2.0,25 = 0,125 m = 0,125.132 = 16,5 gamCu 27. Cho cn bng ho hc: H2 (k) + I2 (k) 2HI (k); H > 0. Cn bng khngb chuyn dch khi

A. tng nng H2. B. gim nng HI.C. gim p sut chung ca h. D. tng nhit ca h.

Hng dn:y l mng kin thc ha hc lp 10 cui k II.Phn ny chng ta cn phi ghi nh nguyn l chuyn dch cn bng:

Mi s thay i ca cc yu t xc nh trng thi ca mt h cn bng lm cho cn bng chuyn dch v

pha chng li nhng thay i .=> Yu t khng lm cho cn bng dch chuyn l p sut (do tng h s phn t kh trc v sau phn ng lnh nhau)Cu 28. Chia hn hp X gm K, Al v Fe thnh hai phn bng nhau.- Cho phn 1 vo dung dch KOH (d) thu c 0,784 lt kh H2 (ktc).

- Cho phn 2 vo mt lng d H2O, thu c 0,448 lt kh H2 (ktc) v m gam hn hp kim loiY. Ho tan

hon ton Y vo dung dch HCl (d) thu c 0,56 lt kh H2 (ktc).

Khi lng (tnh theo gam) ca K, Al, Fe trong mi phn hn hp X ln lt l:A. 0,39; 0,54; 1,40. B. 0,39; 0,54; 0,56. C. 0,78; 0,54; 1,12. D. 0,78; 1,08; 0,56.

Hng dn:+ Phn 1:

2Hn = 0,035 + Phn 2: hn hp kim loi Y l Al d v Fe

K 1

2H2 K

1

2H2

a-------2

a(I) a-------

2

a

Al 3

2H2 Al

3

2H2

y--------3

2 y a--------3

2 a

Ta c2

a+

3

2a = 0,02 mol a = 0,01


12/24

Th a = 0,01 vo (I) 1/2a + 3/2y = 0,035 y = 0,02nAl trong hn hp Y = y a = 0,02 0,01 = 0,01

2Hn thu c khi Y p vi HCl = 0,025 mol

Al 3

2H2

0,01------0,015 2H

n do Fe to ra = 0,025 0,015 = 0,01 = n Fe

Khi lng mi kim loi trong mi phn: mAl = 0,02.27 = 0,54 ; mK= 0,01.39 = 0,39 ; mFe = 0,01.56 = 0,56Cu 29. Cho 0,87 gam hn hp gm Fe, Cu v Al vo bnh ng 300 ml dung dch H2SO4 0,1M. Sau khi cc

phn ng xy ra hon ton, thu c 0,32 gam cht rn v c 448 ml kh (ktc) thot ra. Thm tip vobnh 0,425 gam NaNO3, khi cc phn ng kt thc th th tch kh NO (ktc, sn phm kh duy nht) to

thnh v khi lng mui trong dung dch lA. 0,224 lt v 3,750 gam. B. 0,112 lt v 3,750 gam.C. 0,112 lt v 3,865 gam. D. 0,224 lt v 3,865 gam.

Hng dn:

2 4

H SOn = 0,03

H

n + = 0,06;2

Hn = 0,02 < 2.

H

n + => Fe v Al tan ht v 0,32 gam kim loi l Cu

nCu = 0,005Fe + 2H+ Fe2+ + H2

x 2x x x

Al + 3H+ Al3+ +3

2H2

y 3y y 1,5yTa c : x + 1,5y = 0,02 (1)

56x + 27y = 0,87 0,32 = 0,55 (2)(1) V (2) x = 0,005 v y = 0,01

Dung dch sau p c :

2Fen + = 0,005 v Hn +cn li = 0,06 2x 3y = 0,06 2.0,005 3.0,01 = 0,02 ;

3NaNOn = 0,005

3Fe2+ + 4H+ + NO3- 3Fe3+ + NO + 2H2O

0,0051

150

0,005

3

0,005

3

Hn +cn = 0,02

1

150=

1

75;

3NOn = 0,005

0,005

3=

1

3003Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O

0,0051

75 1

300 1

300Sau phn ng H+ v NO3- va ht

nNO =0,005

3+

1

300= 0,005 VNO = 0,005.22,4 = 0,112 lt

mmui = mcc kim loi ban u + 24SO Na

m m ++ = 0,87 + 0,03.96 + 0,005.23 = 3,865gam.

Cu 30. t chy hon ton x gam hn hp gm hai axit cacboxylic hai chc, mch h v u c mt lin kti C=C trong phn t, thu c V lt kh CO2 (ktc) v y mol H2O. Biu thc lin h gia cc gi tr x, y vV l

A. V = 28 ( 30 )55

x y . B. V = 28 ( 62 )95

x y C. V = 28 ( 30 )55

x y+ . D. V = 28 ( 62 )95

x y+ .

Hng dn :Cng thc chung ca cc axit trn l : CnH2n-4O4

-12-


13/24

1


CnH2n-4O4 nCO2 + (n-2)H2O

T phng trnh ta thy : naxit = ( 2 22

CO H On n

) naxit = 22,42

Vy

Khi lng axit = x gam = mC/axit + mH/axit + mO/axit

x = 12 22,4

V

+ 2y + 22,42

Vy

.4.16 V = 28/55(x + 30y)

Cu 31. Cho dy cc cht v ion: Fe, Cl2, SO2, NO2, C, Al, Mg2+, Na+, Fe2+, Fe3+. S cht v ion va c tnh oxiho, va c tnh kh lA. 4. B. 5. C. 6. D. 8.

Hng dn:Cht va c c tnh oxi ha va c c tnh kh => cht v ion phi c s oxi ha trung gian+ Fe c th c c cc s oxi ha: +3; +2; 0+ Cl c th c c cc s oxi ha: +7, + 5, + 3, +1, 0, - 1

+ S c th c c cc s oxi ha: +6, +4, 0, -2+ N c th c c cc s oxi ha: + 5, + 4, + 3, +2, +1, 0, -3+ C c th c c cc s oxi ha: +4, +2, 0, -4+ Al c th c c cc s oxi ha: +3, 0+ Mg c th c c cc s oxi ha: +2, 0+ Na c th c c cc s oxi ha: +1, 0=> Vy c : Cl2, SO2, NO2, C, Fe2+

Cu 32. Pht biu no sau y l sai?A. Bn knh nguyn t ca clo ln hn bn knh nguyn t ca flo.B. Tnh axit ca HF mnh hn tnh axit ca HCl.

C. m in ca brom ln hn m in ca iot.

D. Tnh kh ca ion Br- ln hn tnh kh ca ion Cl- .Hng dn:+ Trong mt nhm t F => I th bn knh nguyn t tng dn => A ng+ i t HF => HI th tnh acid tng dn (do bn knh nguyn t tng dn t F => I) => B sai+ Trong mt nhm t F => I th m in gim dn (Flo l nguyn t c m in ln nht)

+ Theo trn th t F => I th tnh oxi ha ca F > Cl > Br > I => Tnh kh ca I Br Cl F > > >Nhn xt: Nhiu em hc sinh thng nhm ln axit HF c tnh axit mnh hn HCl v m in ca F ln hn m in ca Cl (Mt nhm ln cht ngi)

Khi so snh tnh axit v c khng cha oxi ta khng dng m in m phi so snh di lin kt hahc gia nguyn t trung tm v H (xem thm chuyn so snh tnh axit-bazo ca axit v c; hu c)Cu 33. t chy hon ton 3,42 gam hn hp gm axit acrylic, vinyl axetat, metyl acrylat v axit oleic, ri hp

th ton b sn phm chy vo dung dch Ca(OH)2 (d). Sau phn ng thu c 18 gam kt ta v dung dch

X. Khi lng X so vi khi lng dung dch Ca(OH)2ban u thay i nh th no?

A. Tng 2,70 gam. B. Gim 7,74 gam. C. Tng 7,92 gam. D. Gim 7,38 gam.Hng dn:

+ axit acrylic: CH2=CH-COOH+ vinyl axetat: CH3COOCH=CH2+ metyl acrylat: CH2=CH-COOCH3+ axit oleic: CH3(CH2)7CH = CH(CH2)7COOHChng u c cng thc chung dng: CnH2n-2O2CnH2n-2O2 => nCO2 + (n-1)H2O


14/24

( )

2

2

2

2 2

6 10,18 .18 2,73,42

60,18 14 30 . 60,18 0,18.44 7,92

7,38

gammol

m gam

gam

H O

CO

CO

CO H O

mn n n n

m m m m

= == + = = = =

= + =

Vy khi lng dung dch gim i 7,38 gam

Hoc c th theo phng php bo ton khi lng v bo ton nguyn t: T pt chy ta thy

2 2 2 2 2n nC H O CO H On n n

=

2 3CO CaCOn n= = 0,18

gi2H O

n = a 2 2 2n nC H On = 0,18 a.

gi s mol O2 l y+ Bo ton nguyn t O : (0,18 a)2 + 2y = 0,18.2 + a -3a + 2y = 0 (1)+ Bo ton khi lng : 3,42 + 32y = 0,18.44 + 18a -18a + 32y = 4,5 (2)

(1) v (2) a = 0,15 ( )2 2CO H Om + = 0,18.44 + 0,15.18 = 10,62

Khi lng dung dch gim = 18 10,62 = 7,38gam.Ch : Cn phn bit c khi lng bnh tng so vi ban u v khi lng dung dch trong bnh tng haygim so vi ban u.

Nu cn bng phng trnh trn sai s dn n cc kt qu khc ( Nu hc sinh no cnbng h s ca oxi l 1,5n s dn n gim 7,74 gam ( khng tin th m xem)Cu 34. t chy hon ton hn hp X gm C2H2, C3H4 v C4H4 (s mol mi cht bng nhau) thu c 0,09

mol CO2. Nu ly cng mt lng hn hp X nh trn tc dng vi mt lng d dung dch AgNO 3trong NH3, th khi lng kt ta thu c ln hn 4 gam. Cng thc cu to ca C 3H4v C4H4 trong X ln

lt l:

A. CHC-CH3, CH2=C=C=CH2. B. CH2=C=CH2, CH2=C=C=CH2.C. CHC-CH3, CH2=CH-CCH. D. CH2=C=CH2, CH2=CH-CCH.

Hng dn:

{ {

{ {

{ {

{

{

{

2 2 2 22 2 2

2

3 4 2 3 4

3

4 4 2 4 4

4

2

3 2 3 4 0, 09 0, 01

4

m=240.0,01=2,4 gam Loi p n B+ Gii s ch c C3H4 to ra kt ta => khi lng kt ta l: = 2,4 + 1,47 = 3,87 gam < 4 => loi p n A+ Gii s ch c C4H4 to ra kt ta => khi lng kt ta l: = 2,4 + 1,59 = 3,99 gam < 4 => loi p n DVy c C3H4 v C4H4 u to ra kt ta => Cng thc cu to: CHC-CH3, CH2=CH-CCH

Cu 35. Cho dy cc cht: phenylamoni clorua, benzyl clorua, isopropyl clorua, m-crezol, ancol benzylic, natriphenolat, anlyl clorua. S cht trong dy tc dng c vi dung dch NaOH long, un nng lA. 4. B. 3. C. 5. D. 6.

Hng dn:C6H5NH3Cl + NaOH => C6H5NH2 + NaCl + H2O

6 5 2 6 5 2C H CH Cl NaOH C H CH OH NaCl + +3 2 3 2

( ) ( )CH CH Cl NaOH CH CH OH NaCl + +

3 6 4 3 6 4 2CH C H OH NaOH CH C H ONa H O + +

-14-


15/24

1


6 5 2C H CH OH NaOH +

6 5C H ONa NaOH+

2 2 2 2CH CH CH Cl NaOH CH CH CH OH NaCl = + = +

Vy c 5 phng trnh phn ng

Cu 36. Khi so snh NH3vi 4NH+ , pht biu khng ng l:

A. Trong NH3v 4NH+ , nit u c s oxi ha -3.

B.NH3 c tnh baz, 4NH+ c tnh axit.

C. Phn t NH3v ion 4NH+ u cha lin kt cng ha tr.

D. Trong NH3v 4NH+ nit u c cng ha tr 3.

Hng dn: Phn t NH3 c cng ha tr l 3 v ion 4NH+ u cha lin kt cng ha tr l 4.

Cu 37. Cho 7,68 gam Cu vo 200 ml dung dch gm HNO 30,6M v H2SO40,5M. Sau khi cc phn ng xy ra

hon ton (sn phm kh duy nht l NO), c cn cn thn ton b dung dch sau phn ng th khi lng

mui khan thu c lA. 19,76 gam. B. 20,16 gam. C. 19,20 gam. D. 22,56 gam.Hng dn:y l dng ton kh quen thuc, chng ta thng s dng phng trnh ion rt gn xt v phn ng

{ { {3 2 4

3

24

2

3 2

0,320,120,12

0,12

2 0,323 8 2 3 2 40,12

0,1

mol

n mol

n mol

n mol

Cu

HNO H SOH

NO

SO

n

n nCu H NO Cu NO H O

+

+ +

= = + =

+ + + += =

Dung dch sau p c :0,12 mol Cu2+ ; 0,1 mol 24SO

; v (0,12 0,08) = 0,04 mol NO3-

mmui = 0,12.64 + 0,1.96 + 0,04.62 = 19,76 gam.Cu 38. Cho axit salixylic (axit o-hiroxibenzoic) phn ng vi anhirit axetic, thu c axitaxetylsalixylic (o-CH3COO-C6H4-COOH) dng lm thuc cm (aspirin). phn ng hon ton vi 43,2 gamaxit axetylsalixylic cn va V lt dung dch KOH 1M. Gi tr ca V l

A. 0,48. B. 0,72. C. 0,24. D. 0,96.Hng dn:

OH

COOH

(CH3CO)

2O

OCOCH3

COOH

CH3COOH

++

OCOCH3

COOH

KOH

OK

COOK

CH3COOK OH

2+ ++3 2

Theo phng trnh phn ng ta c nKOH = 3naspirin =43,2

3. 0,24.3 0,72

180

mol= =

y l mt cu kh hay, i hi hc sinh phi lm c l thuyt. Qua phn bi ny chng ta cnphi ghi nh v tnh acid yu ca nhm OH khi n c gn trc tip vo vng benzene (hp cht caphenol)


16/24

Nu coi axit axetylsalixylic (o-CH3COO-C6H4-COOH) ch tc dng vi NaOH nh l axit hoc nhl este th s dn n p n D. 0,24

Nu coi c axit phn ng v este phn ng th s c p n B. 0,48p n C. 0,96 c a vo mt cch ngu nhin hay l nhn i p n B. 0,48

Axit axetylsalixylic (o-CH3COO-C6H4-COOH) dng lm thuc cm (aspirin) l cha chnh xclm. M thc cht n l thuc gim au (Thng thng nhng ngi au u, hay au tim th hay phi dng

aspirin )Ch : Loi ny c hi cho d dy.Aspirin, hay acetylsalicylic acid (ASA), (acetosal) l mt dn xut ca acid salicylic, thuc nhm thucchng vim non-steroid; c tc dng gim au, h st, chng vim; n cn c tc dng chng kt tp tiucu, khi dng liu thp ko di c th phng nga au tim v hnh thnh cc nghn trong mch mu.

Cu 39. Tin hnh cc th nghim sau:(1) Cho dung dch NaOH vo dung dch Ca(HCO3)2.

(2) Cho dung dch HCl ti d vo dung dch NaAlO2 (hoc Na[Al(OH)4]).

(3) Sc kh H2S vo dung dch FeCl2.

(4) Sc kh NH3 ti d vo dung dch AlCl3.(5) Sc kh CO2 ti d vo dung dch NaAlO2 (hoc Na[Al(OH)4]).

(6) Sc kh etilen vo dung dch KMnO4.

Sau khi cc phn ng kt thc, c bao nhiu th nghim thu c kt ta?A. 3. B. 5. C. 6. D. 4.

Hng dn:y l dng cu hi tng i d nhng hc sinh li thng d nhm ln v cc em hc sinh li xem l thuyt.

1:1

3 2 3 3 2 2

3 3 22:1

3 2 2 3 3 2

( )

2 ( ) 2

NaOH Ca HCO NaHCO CaCO H OOH HCO Ca CaCO H O

NaOH Ca HCO Na CO CaCO H O +

+ + + + + ++ + +

2 2 3 3

2 2

3 3 2

( )4 2

3 ( ) 3d-

d-

HCl NaAlO H O NaCl Al OHH AlO Al H O

HCl Al OH AlCl H O+ +

+ + + + ++ +

2 2H S FeCl+ (do FeS c kh n ng tan trong HCl)FeS HCl+

3 2 3 3 43 3 ( ) 3NH H O AlCl Al OH NH Cl+ + +

2 2 2 3 32 ( )CO H O NaAlO NaHCO Al OH + + +

2 2 4 2 2 4 2 2( )CH CH KMnO H O C H OH MnO KOH = + + + +

Vy => c 4 th nghim to ra kt ta.Trong phn ny thy xin lu thm:+ CuS l kt ta khng tan trong acid+ Cc mui ca Cu, Zn c kh nng b ha tan trong dung dch NH3+ Dung dch HCl nu dng ng theo t l th c kh nng to ra kt ta khi cho n tc dng vi 22 2, ZnOAlO

Cu 40. t chy hon ton anehit X, thu c th tch kh CO2bng th tch hi nc (trong cng iu kin

nhit , p sut). Khi cho 0,01 mol X tc dng vi mt lng d dung dch AgNO 3 trong NH3 th thu c

0,04 mol Ag. X lA. anehit no, mch h, hai chc. B. anehit fomic.C. anehit axetic. D. anehit khng no, mch h, hai chc.

Hng dn:T d kin:+ Khi cho 0,01 mol X tc dng vi mt lng d dung dch AgNO 3 trong NH3 th thu c 0,04 mol Ag =>

loi c p n C

-16-


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+ Th tch kh CO2bng th tch hi nc => 2 2CO H On n= andehit no, n chc

- Cng thc tng qut ca andehit no, mch h, hai chc: CnH2n-2O2 => 2 2H O COn n 2 2H O COn n p n l B: anehit fomic (HCHO => CO2 + H2O)y l mt cu tng i d, n ch i hi kin thc c bn :

HCHO => 4Ag RCHO => 2Ag R(CHO)2 => 4Ag

II. PHN RING [10 Cu]Th sinh ch c lm mt trong hai phn (phn A hoc phn B)

A. Theo chng trnh chun (10 cu, t cu 41 n cu 50)Cu 41. Cho 2,7 gam hn hp bt X gm Fe v Zn tc dng vi dung dch CuSO 4. Sau mt thi gian, thu c

dung dch Y v 2,84 gam cht rn Z. Cho ton b Z vo dung dch H2SO4 (long, d), sau khi cc phn ngkt thc th khi lng cht rn gim 0,28 gam v dung dch thu c ch cha mt mui duy nht. Phntrm khi lng ca Fe trong X l:

A. 58,52% B. 51,85% C. 48,15% D. 41,48%Hng dn:Rn Z p vi dung dch H2SO4 long v cht rn gim 0,28 gam v thu c mt mui duy nht Z c Fe dv Cu to raVy trong Z c 0,28 gam Fe d v 2,84 0,28 = 2,56 gam Cu mhn hp X p vi Cu2+ = 2,7 0,28 = 2,42 gamGi x v y ln lt l s mol ca Fe v Zn => ta c:

Zn + CuSO4 => Cu + ZnSO4Fe + CuSO4 => Cu + FeSO4

56x + 65y = 2,42 (1) v 64x + 64y = 2,56 (2)(1) v (2) x = 0,02 mFe(p vi Cu2+) = 0,02.56 = 1,12 mFe ban u = 1,12 + 0,28 = 1,4

%mFe = 1,4/2,7 = 51,85%.Cu 42. Cu hnh electron ca ion Cu2+ v Cr3+ ln lt l :

A. [Ar]3d9 v [Ar]3d3 . B. [Ar]3d74s2 v [Ar]3d14s2.C. [Ar]3d9 v [Ar]3d14s2. D. [Ar]3d74s2 v [Ar]3d3.

Hng dn:Cu c s th t Z = 29 c 29e : 1s22s22p63s23p63d104s1 Cu2+ c 27e [Ar]3d9

Cr c s th t Z = 24 c 24e : 1s22s22p63s23p63d54s1 Cr3+ c 21e [Ar]3d3

( Ar c 18e)Mt electron th mt lp ngoi cng

Cu 43. Ancol etylic c iu ch t tinh bt bng phng php ln men vi hiu sut ton b qu trnh l 90%,Hp th ton b lng CO2, sinh ra khi ln men m gam tinh bt vo nc vi trong, thu c 330 gam kt tav dung dch X. Bit khi lng X gim i so vi khi lng nc vi trong ban u l 132 gam. Gi tr cam l:

A. 405 B. 324 C. 486 D.297Hng dn:

2COm = m - mdung dch gim = 330 132 = 198g 2COn = 198/44 = 4,5 mol

C6H10O5 2C2H5OH + 2CO2m = 4,5/2 . 162 . 100/90 = 405 gam.

Nhn xt: Bi tp ny nu chn thm vo ru, tp cht tr, hiu sut mi qu trnh ln mem th hay hnCu 44. Dung dch no sau y lm qu tm i thnh mu xanh?A. Dung dch alanin B. Dung dch glyxin C. Dung dch lysin D. Dung dch valin


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Hng dn:Trong phn t lysin c 2 nhm NH2 v 1 nhm COOH

NH2 OH

NH2

O

c mi trng baz lm qu tm ha xanh.Cn cc dung dch alanin, valin, glyxin c 1 nhm NH2 v 1 nhm COOH

CH3

NH2

OH

O

NH

2 OH

O

mi trng trung tnh khng lm i mu quCu 45. Cho hn hp X gm Fe2O3, ZnO v Cu tc dng vi dung dch HCl (d) thu c dung dch Y v phn

khng tan Z. Cho Y tc dng vi dung dch NaOH (long, d) thu c kt ta:A. Fe(OH)3 v Zn(OH)2 B. Fe(OH)2, Cu(OH)2 v Zn(OH)2C. Fe(OH)3 D. Fe(OH)2 v Cu(OH)2

Hng dn:Phn khng tan Z l Cu (d) dung dch Y cha cc ion Fe 2+; Cu2+ v Zn2+ do lng NaOH d kt taZn(OH)2 to ra b tan ht, cn li 2 kt ta Fe(OH)2 v Cu(OH)2.Nhn xt:

Nu coi Cu khng phn ng, Zn(OH)2 khng tan th s c p n A.Nu coi Cu c phn ng vi Fe3+ , Zn(OH)2 khng tan th s c p n BNu coi Cu khng phn ng vi Fe3+ v Zn(OH)2 tan trong NaOH d th s c p n C. bi khng ni r hn th nghim trn lm trong iu kin nh th no? c khng kh hay khng c khng

kh, quan st sau th nghim hay l mt lc sau mi quan st.Nn xt mt kha cnh khc c th c p n l Fe(OH)2, Fe(OH)3 v Cu(OH)2 ??????????

Cu 46. X, Y ,Z l cc hp cht mch h, bn c cng cng thc phn t C3H6O . X tc dng c vi Na vkhng c phn ng trng bc. Y khng tc dng vi Na nhng c phn ng trng bc, Z khng tc dng cvi Na v khng c phn ng trng bc. Cc cht X, Y, Z ln lt l: A. CH2=CH-CH2-OH, CH3-CH2-CHO, CH3-CO-CH3.B. CH2=CH-CH2-OH, CH3-CO-CH3, CH3-CH2-CHO.C. CH3-CH2-CHO, CH3-CO-CH3, CH2=CH-CH2-OH.

D. CH3-CO-CH3, CH3-CH2-CHO, CH2=CH-CH2-OH.

Hng dn:C3H6O c th l ru khng no, andehit va xeton no+ X tc dng c vi Na v khng c phn ng trng bc X l ru+ Y khng tc dng vi Na nhng c phn ng trng bc Y l andehit+ Z khng tc dng c vi Na v khng c phn ng trng bc Z l xeton.Cu 47.Nhm nhng cht kh (hoc hi) no di y u gy hiu ng nh knh khi nng ca chng trong

kh quyn vt qu tiu chun cho php?A. N2 v CO B. CO2 v O2 C. CH4 v H2O D.CO2 v CH4

Hng dn:CO2 v CH4 u gy hiu ng nh knh.

Cu 48. Khi in phn dung dch NaCl (cc m bng st, cc dng bng than ch, c mng ngn xp) th:A. cc dng xy ra qu trinh oxi ha ion Na+ v cc m xy ra qu trnh kh ion Cl -.B. cc m xy ra qu trnh kh H2O v cc dng xy ra qu trnh oxi ha Cl -.C. cc m xy ra qu trnh oxi ha H2O v cc dng x ra qu trnh kh ion Cl -.

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D. cc m xy ra qu trnh kh ion Na+ v cc dng xy ra qu trnh oxi ha ion Cl -.Hng dn:

Vi dng ny ta ch cn vit cc qu trnh xy ra in cc v hiu cc khi nim: " Cht kh, cht oxiha, cht b kh, cht b oxi ha, s kh, s oxi ha, qu trnh kh, qu trnh oxi ha"

Trong bnh in phn, ion Na+ tin v cc m, do ion Na+ c tnh oxi ha rt yu nn khng b kh mnc s b kh, cn cc dng do Cl- c tnh kh mnh hn nc nn b oxi ha.

Nhn xt:Nu ta nh iu kin v in phn dung dch th ta thy ion Na+ khng b in phn trong dung dch viin cc tr, than ch ( tr in cc thy ngn) th loi c p n A v C.

V bi khng ni r in phn n giai on no, nc b in phn hai in cc cha? Nu nc b in phn c hai ic cc th cc dng cn xy ra qu trnh oxi ha c H 2O na.Cu 49. Ha hi 15,52 gam hn hp gm mt axit no n chc X v mt axit no a chc Y (s mol X ln hn s

mol Y), thu c mt th tch hi bng th tch ca 5,6 gam N2 (o cng trong iu kin nhit , p sut).Nu t chy ton b hn hp hai axit trn th thu c 10,752 lt CO2 (ktc) . CTCT ca X, Y ln lt l:A. CH3-CH2-COOH v HOOC-COOH B. CH3-COOH v HOOC-CH2-CH2-COOHC. H-COOH v HOOC-COOH D. CH3-COOH v HOOC-CH2-COOH

Hng dn:

2Nn = nX = 5,6/28 = 0,2 v 2COn = 0,48

Ctrung bnh = 0,48/0,2 = 2,4 (loi C)Dng quy tc ng cho da vo s C v C trung bnh v d kin s mol X ln hn s mol Y X l CH3COOHDa vo d kin s mol mi cht theo quy tc ng cho v khi lng hn hp l 15,52 gam Y l HOOC-CH2-COOHCu 50. Cho buta-1,3 - ien phn ng cng vi Br2 theo t l mol 1:1. S dn xut ibrom (ng phn cu to v

ng phn hnh hc) thu c l:A. 3 B. 1 C. 2 D. 4

Hng dn:Buta-1,3-dien phn ng cng vi Br2 cho hai sn phm cng ( sn phm cng 1,2 v sn phm cng 1,4 ) ringsn phm cng 1,4 c thm ng phn cis trans.

Nhn xt: Cc em cn ch i vi hp cht cha lin kt i th khi bi xt v ng phn, ta cn phi ch xem xt v ng phn hnh hc

ng phn hnh hc: mt loi ng phn khng gian m trong phn t c phn cng nhc (ni i, vngno) lm cn tr s quay t do ca phn t. Vd. Cab = Ccd vi a b v c d (a, b, c v d l cc nguyn t hocnhm nguyn t) c hai ng phn:

Cc PHH khc nhau v tnh cht vt l. Chng c th chuyn t dng ny sang dng kia di tc dng canhit, nh sng, tc nhn ho hc hoc t pht.

B. Theo chng trnh nng cao (10 cu, t cu 51 n cu 60)Cu 51. Thy phn hon ton 60 gam hn hp hai ipetit thu c 63,6 gam hn hp X gm cc amino axit (cc

amino axit ch c mt nhm amino v mt nhm cacboxyl trong phn t). Nu cho1

10hn hp X tc dng

vi dung dch HCl (d), c cn cn thn dung dch, th lng mui khan thu c l :A. 7,09 gam. B. 16,30 gam C. 8,15 gam D. 7,82 gam.

Hng dn:


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2H O

m = 63,6 60 = 3,6 2H O

n = 3,6/18 = 0,2

NH2-R-CO-NH-R-COOH + H2O NH2-R-COOH + NH2-R-COOH0,2----------0,2------------------0,2

naminoaxit = 0,41/10 hn hp X c 0,4/10 = 0,04 mol aminoaxit v 63,6/10 = 6,36 gam

nHCl = naminoaxit = 0,04

mmui = maminoaxit + mHCl = 6,36 + 0,04.36,5 = 7,82 gam.Cu 52. Ha tan hn hp bt gm m gam Cu v 4,64 gam Fe3O4 vo dung dch H2SO4 (long, rt d) sau khi cc

phn ng kt thc ch thu c dung dch X. Dung dch X lm mt mu va 100 ml dung dch KMnO40,1M. Gi tr ca m l:A. 1,24 B. 3,2 C. 0,64 D.0,96

Hng dn:

3 4Fe On = 0,02 ;

4KMnOn = 0,01

Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O0,02--------------0,02-----0,04

Cu + 2Fe3+ Cu2+ + 2Fe2+x----------------------------2x

2Fe

n + = 0,02 + 2x

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O0,05-------0,01

0,02 + 2x = 0,05 x = 0,015 ; Vy mCu = 0,015.64 = 0,96 gam.

Nhn xt: Nu dng mt lng d chnh xc H2SO4 sao cho khi Fe2+ tc dng ht vi 4MnO

m lng H+ ht, sau trong mi trng trung tnh Mn+7 b kh v Mn+4 (MnO2) th chc chn bi tp s hayhn v nhiu em hc sinh s sai

Cu 53. Pht biu no sau y v anehit v xeton l sai?A.Hiro xianua cng vo nhm cacbonyl to thnh sn phm khng bn.B. Axeton khng phn ng c vi nc brom.C. Axetanehit phn ng c vi nc brom.D. Anehit fomic tc dng vi H2O to thnh sn phm khng bn.

Hng dn:+ Lin kt i C=O fomandehit c phn ng cng nc nhng sn phm to ra c 2 nhm OH cng nh vo 1nguyn t C nn khng bn.+ Hidro xianua cng vo nhm cacbonyl to thnh sn phm bn gi l xianohidrin.

Cu 54. Cho s phn ng:CH CH HCN+ X; X tr ng h p polime Y; X + CH2=CH-CH=CH2 ng tr ng hp polime Z

Y v Z ln lt dng ch to vt liu polime no sau y?A.T capron v cao su buna. B. T nilon-6,6 v cao su cloropren.C. T olon v cao su buna-N. D. T nitron v cao su buna-S.

Hng dn:X : CH2=CH-CN trng hp to poliacrilonitrin dng ch to t nitron hay cn gi l olon.ng trng hp CH2=CH-CN + CH2=CH-CH=CH2 ta thu c caosu buna - NCu 55. Dung dch X gm CH3COOH 1M (Ka = 1,75.10-5) v HCl 0,001M . Gi tr pH ca dung dch X l:

A. 2,43 B. 2,33 C. 1,77 D. 2,55Hng dn:HCl H+ + Cl-

10-3 -----10-3

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CH3COOH CH3COO- + H+

B 1P x-----------------x-----------xSp 1-x ----------------x-----------x x(x + 10-3)/(1 x) = 1,75.10-5 x = 3,705.10-3

[H+] = 3,705.10-3 + 10-3 = 4,705.10-3 pH = -lg[H+] = 2,33

Nhn xt: Vi bi tp ny hc sinh no gii gn ng bng cng thc s sai s tng i ln.Cu 56. Khng kh trong phng th nghim b nhim bi kh clo. kh c, c th xt vo khng kh dd nosau y?

A.Dung dch NH3 B. Dung dch NaCl C.Dung dch NaOH D. Dung dch H2SO4 longHng dn:NH3 d kt hp vi Cl2 to sn phm khng c : 8NH3 + 3Cl2 6NH4Cl + N2.Vi NaOH th ta c phng trnh

2NaOH + Cl2 NaCl + NaClO + H2ONhn xt: Theo kin c nhn ca ti th c hai u hp th c, u kh c c, kh nng ca NaOH l lnhn, tuy l trong khng kh, nhng kh Clo nng nn s bay l st mt ca phng th nghim. Trong thc t trnh nhim mi trng v nhim c kh clo trong qu trnh lm th nghim chng ta cng thng dng bngtm dung dch NaOH long.

Nu dng dung dch NH3 th phi dng lng d mi kh c ht clo, m dng d th ha ra kh NH3 d lithot ra lm nh hng mi trng chng? thay nhim c kh clo bng nhim c kh NH 3.Theo ti: Hi cht no kh c tt nht th hay hn. Tuy nhin ng t "Xt" cng lin tng cho chng ta lha cht a vo s d bay hi.Cu 57. Hin tng xy ra khi nh vi git dung dch H2SO4 vo dung dch Na2CrO4 l:

A. Dung dch chuyn t mu vng sau khng muB. Dung dch chuyn t mu da cam sang mu vng.

C. Dung dch chuyn t mu vng sang mu da cam.D. Dung dch chuyn t khng mu sang mu da cam

Hng dn:2CrO42- + 2H+ Cr2O72- + H2O

(mu vng) (mu da cam)Cr2O72- + 2OH- 2CrO42- + H2O

(mu da cam) (mu vng)Cu 58. Cho cc phn ng sau:

Fe + 2Fe(NO3)3 3Fe(NO3)2

AgNO3 + Fe(NO3)2 Fe(NO3)3 + AgDy sp xp theo th t tng dn tnh oxi ha cc ion kim loi l:

A. Ag+, Fe2+, Fe3+ B. Fe2+, Fe3+, Ag+ C. Fe2+, Ag+, Fe3+ D. Ag+, Fe3+, Fe2+

Hng dn:Fe3+ oxi ha Fe thnh Fe2+ Fe3+ c tnh oxi ha mnh hn Fe2+

Ag+ oxi ha c Fe2+ thnh Fe3+ Ag+ c tnh oxi ha mnh hn Fe3+

Vy : Ag+ > Fe3+ > Fe2+.Cu 59. Cho dy chuyn ha sau

Benzen 2 40,

t

C H

xt

+ X 2 ,Br as

Tlmol 1:1+ Y 2 50/KOH C H OH

t Z (trong X, Y, Z l sn phm chnh)

Tn gi ca Y, Z ln lt lA. benzylbromua v toluen B. 1-brom-1-phenyletan v stirenC. 2-brom-1pheny1benzen v stiren D. 1-brom-2-phenyletan v stiren.

Hng dn:


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C6H6 2 40,

t

C H

xt

+ C6H5CH2CH3

C6H5CH2CH3 2,Br as

Tlmol 1:1+ C6H5CHBr-CH3

C6H5CHBr-CH3 2 50/KOH C H OH

t C6H5CH=CH2.

Cu 60. t chy hon ton 0,11 gam mt este X ( to nn t mt axit cacboxylic n chc v mt ancol nchc) thu c 0,22 gam CO2 v 0,09 gam H2O. S este ng phn ca X l:

A. 5 B. 4 C. 6 D.2Hng dn:

2COn = 0,005 ;

2H On = 0,005 =>

2COn =

2H On este no, n vi cng thc tng qut l CnH2nO2

CnH2nO2 nCO20,005/n -----0,005

M = 0,11n/0,005 = 22n 14n + 32 = 22n n = 4 este C4H8O2 c 4 ng phn este.

OCH

2

CH2

CH

O

CH3

OCH

CH3

CH

O CH3

O

CH2

CH3

CCH

3

O

O

CH3

C

CH2

O

CH3

Nhn xt: Nu hi c bao nhiu ng phn c kh nng tham gia phn ng trng gng hay c bao nhiu ngphn cho ancol bc 1 hoc ancol bc hai th s hay hn. Hc sinh khng s dng c cng thc tnh ng phn

LI KT

Trn y l li gii chi tit thi i hc mn Ha, khi A nm 2011.Tuy li gii hi di, nhng chngti c trnh by cch t duy trong qu trnh gii, vn phong d hiu. Rt mong nhn c kin ng gp cacc em hc sinh v ng nghip bi gii hay hn v n nhiu bn c.

Trong qu trnh nh my khng th trnh khi sai st, mong bn c gp cho chng ti hoc tho lunti :

DIN N HA HC VIT NAMhttp://hoahoc.org

Din n ca cng ng yu Ha Hc Vit Nam

Cc em hc sinh mun c nhng li gii ngn gn hn, nhanh hn cng nh phn tch nhng sai lmtrong qu trnh gii xin vui lng hc ti

TT Luyn Thi & Bi Dng Kin Thc Ngy Mi

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http://hoahoc.org/http://hoahoc.org/


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Trung tm ca chng ti nhn ging dy1. Lp luyn thi i hc2. Lp 11 ln 123. Lp 10 ln 114. Luyn thi vo 10 cc trng THPT chuyn Ha

S theo maps.google.com


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2011 - Lời giải chi tiết & phân tích đề thi Hóa - Khối A - năm 2011