cÁc chuyÊn ĐỀ quan trỌng trong hÓa hỌc - nguyỄn ĐÌnh ĐỘ

8/10/2019 CC CHUYN QUAN TRNG TRONG HA HC - NGUYN NH

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NGN NH

c n u y e n aQUAN TRNHP I

l i :

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CH B THNG GP TRONG K THI BI Hm m

jmp"' NH XUT BNl HC QUC GIA H NI

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WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU

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W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


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NGUYN NH at ^t 2/

TRONG/ r

o PHNG PHP TRC NGH

o CH THNG GP TRO

M

NG K THI AI HC

NH XUT BN I HC QUC GIA H NI



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NHft XUflT BRN m \ HC u c Gift Hfo NI

16 Hng Chlui -- Hai B Trng H Ni

in thoai: Bin tp-Ch' bn: (04) 39714896:

Hnh chnh: (04) 39714899: Tng bin tp: (04> 39714897

Fax: (04> 39714899

Chiu trch nh im xu t bn

Gim c:

Tng bin tp:

Bin tp:

PHNG QUC BO

PHM TH TRM

TRN VN HNG

i tc lin kt xut bn:

NH SCH HNG N

SCH LIN KT

CC CHUYN QUAN TRNG TRONG H A HC

M s: 1L-47 H20 11 I

In 2000 cun, kh 16 X 24cm, ti Cng ti TNHH In Bao BI Phong Tn

S xut bn: 216 - 2011/CX B/46 -23 HQ GHN, ngy 3/3/2011 rQuyt nh xut bn s: 47 LK-TN /Q -NXB H QG HN .

In xong v np lu chiu qu H nm-2011- '



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' M I c c u

Cc em hc sinh thn mn!

Mn Ha hc i vi hu ht cc hc sinh, l mt mn hc nng n

vi nhng l thuyt di dng, kh nh. c bit trong cc k thi tuynsinh, cc em lun xem mn Ha hc l mt mn kh khn vi cc

bi tp t n h to n r m r , phc tp v nhng cu l thuyt khng hn gin nh trong thu n ty Sch Gio Khoa hc

gii ta nhng bn khon ny, tc gi bin son cun Cc

c h u y n q u a n tr n g tr o n g H a h c nhm trc ht gip cc emh thng ha kin thc hc, sau trang b k nng gip cc emtng bc tin ti lm ch cc dng bi tp l thuyt ln bi tp tnh

ton thng gp tron g cc thi tuyn sinh Ha hcCun sch c bin son cng phu theo tng chuyn . Mi

chuyn nghin cu k tng ch v thuyt hoc bi tp s gptrong cc thi. C tt c 23 chuyn , tri rng t lp 10 ti lp 12,s gip cc em nm vng tng tn trng tm v l thuyt ln bi tp

b m n H a hc

Quan im ca tc gi khi vit cun sch ny l mun cc em nhnra Ha hc l mt mn hc khng h kh nh tng suy ngh. c

cun sch ny, cc em s thy cch tt nht hc Ha hc khngphi, l .hc th c, m l hc th eo kiu nm bn ch t vn : V saocht X ch c cc phn ng Y; 2...?. V sao cht A tc dng vi chtB to ra cht c...?. Tt c u c cu tr li, l: c im cu tocht quyt nh tnh cht ha hc cht , v cc cht khi phn ngu to ra sn phm theo mt qui lut nht nh.

Chc cc em thnh cng trn con ng hc vn. Nht l phi thayi c t duy hc Ha hc. y cng l bc i u tin ca vic t

hc ni cc emTrn trng n nhn nhng kin ng gp ca cc bn c gn, xa.

Xin Chn thnh cm n

TC GI

N g u y n n h



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Q CHUYN 1

C C C H S Q U f t f t T R N G c f C H T B O

cc triglixerit m thng co khng ln axt bo t do

Cht bo tht ra khng phi hon ton lln m t lng nh cc ax it bo t do. C ht b|

gi l cht bo trung tnh.Mt cht bo c c trng bi cc ch cc s' sau:

- Ch s axit: l s mg KOH dng trung ha ht lng axt bo tdo c trong 1 gam ch t bo. I

- Ch s' x p hng ha: l s mg KOH dng x phng ha h on ton1 gam c ht bo (bao gm KOH dng trung ha axit bo t do v KOHdng x phng ha cc triglixerit)

- Ch s' iot: l s gam .I2 dng cng hp vo 100 gam ch t bo

L i i : Nguyn t c trong nhm CH3- nm. ui ca mt axit bo cgi l cacbon omega. Khong cch t cacbon omega n ni i gnnht nu c 3 cacbon th gi omega- 3 va axit bo c cu trc oi ny

gi l axit omega- 3, cn nu c 6 cacbon th gi axit omega- 6. Nhvy, con s sau ch omega nhm ch v tr ca ni i trong cu trc.Cc axit omega thng gp nh: axt omega- 3 L axit anpha- inoeic C17H29COOH; axt omega- 6 l axit linoleic C17H21COOH; axt omega- 9l axit oleic c17H33COOH. DHA (docosahexaenoic acid) cng l mt axitomega' 3. DHA c cng thc CoiiCOOHl l mt ajx.it bo c 6 ni i

c = c trong phn t. Cc nghin cu. ch ra rng DHA-cn thit cho sph t tr in hon th in chc nng nhn caca h thn kinh. Nhng th nghim trn cao trong cc t chc thn kihh nh vnlm gim lng rigycerit mu, gim lonsuy ng mch vnh, gi m chng nhi mu c tim ...

V d. 1: trung ha 15 gam mt cht bcdch cha a gam NaOH. G i tr a l

A. 0,200 B. 0,280 c . 0,075

G i iC trung ha lng axt bo t do c trong

KOH. Vy trung ha lng axt bo t docn 15.7 = 105 mg

K O H , t c 1Q = 1,875.1c r^ m o l K O H .56000

mt, s pht trin hon hochut cho thy DHA c nng

g mc mt, no b... DHA cnnhp tim, gim t l b bnh

c ch s axit l 7 cn dung

D. 0,15

lgam cht bo trn cn 7 mgc trng lgam cht bo trn



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Vy nNa0H= nKOH = 1,875.10"3m


7/296

G i inh lu t bo to n khi lng cho:

&A+ BIkOH = phng + llglixerol + linc

Vy Ilx phng ~ + ^KOH) (rHghxerol TOnc)

= (300 + 1.56 ) - (9 2 ----- 56000 +18 30-7 ) = 3 2 5 , 8 1 gam

3 56000Lu + Ta c phn ng ca axt bo vi KOH xy ra theo phng trnh:

RCOOH+ KOH RCOOK + H sO

fl>KOH laxi t f t nc

+ Nu bi trn thay KOH bng NaO H (cc s liu gi nguyn) th khing x ph n g th u c thay i do khi lng Na O H ph i khc khilng KOH, trong khi khi lng glixerol v nc vn khng di. Do khi lng x ph ng thu c l:

yiT-x ph n g = ( m A WINclOh) (m-glii:erol Hl/iT.)

( 1 - M i )= (300 + 1.40 ) - 92 ----- S600 + I8. 300 '7

V 3 56000;

V d 6: Cht bo A c ch s" axit v ch s^ x phng ha ln lt l s,8v 210. Tnh khi lng x phng thu c khi x phng ha honton 300 gam cht bo ny bng dung dch KOH va .

Gi i

nh lut bo ton khi lng cho:l lA + Ik OH = Hx phng frlglixerol "t" llnc

V y m x phng = ( iA ixerol t" lnc)

300.210

\

= 309,81 gam

= 300 +V 1000

300(210 - 9,8) , 0 300.9,892----_ . ..... - + 18 _

3.56000 56000= 329,165 gam

Lu : N u bi trn thay KOH bng NaO H (cc s liu gi nguyn) th ikhi lng x phng thu c thay i do khi lng NaOH phi khc khi lng KOH, trong khi khi lng glixero v nc vn khng i.

Do khi lng x phng th u c l:

^.rphng ~ (niA + m NaOH.) - (m g xerol 117-nc)

56000

= 311,165 gam

300(210 -9 ,8 ) J 300.9.8

3.56000 + 5600

7



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V d 7: Mt loi cht bo c ch s' it bng 7,62. Gi s mu chtbo ch gm triolein v tris te a rin . T hnh phn phn tr m khilng tristearin trong cht bo trn l bao nhiu?

G i i rng ch c triolein tc dng vi I '2 (theo t l mol tng ng 1:3)

Trong 100 gamcht bo trn c n trinem = n, = = 0 , 0 1 moi

3 2 3.254mlrioiem = 0,01.884 = 8,84 g

Vy %triolein = 8,84% nn 9Hristearin = 91,16%

C H I T R C N G H I M

1. Khi lng NaOH cn thit trung ha ht lng axit bo t dotrong 10 gam cht bo c ch s axit l 5,6 l bao nhiu?

A. 0,04 gam B. 0,056 gam c . 0,2 gam D .lgam

2. x phng ha hon ton 300 gam ch t bo A c ch s' x phngha l 210 cn dng va bao nhiu ml dung ch NaOH 2M?

A. 562,5 B. 2250 c . 1125 D . 625

3. Mt mu cht bo c ch sx phng v ch s' axit ln lt l 198,8v 8,4. Lng x phng thu c khi cho 300 gam cht bo ny tcdng ht vi NaOH l

A. SlO .lgam B. 311,32g-am C'342,6gam D. 248,76gam

4. Cht bo A c ch s axit v ch s' x phng ha ln lt l 9,8 v

210. Khi lng x phng thu c khi x phng ha hpn ton 600gam cht bo ny bng dung dch NaOH va l bao nhiu?

A. 625,90 B. 680,56 c . 641,89 D. 622,33

5. Mt loi m cha 40% olein, 20% panm itin v 40% stea rin (theo khilng). X phng ha hon ton m gam m trn c 138 gam glxerol.Gi tr m l

A. 1281 B. 1304 c . 1326 D. 1283

6. x phng hon ton 5,04 gam ch't bo A cn 90ml dung dch

KOH 0,2M. Ch s x phng ha ca cht bo A l:A. 100 B. 180 c . 200 D. 108

7. x phng ha hon ton l,68gam cht bo A cn 30ml dung dchKOH 0,2M. Sau phn ng thu dc 0,1766 gam glixerol. Ch s x

phng ha v ch s axit ca cht bo A l n lt l

A. 200 v 8 B. 200 v 192 c . 200 v 15 D . 108 v 12

8



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8. Mt loi cht bo c ch s it bng 7,62. Gi s mu cht bo trn ch gmtrilinolein v tristearin . Thnh phn phn trm khi lng triste arin l

A. 8,84% B. 95,58% c . 26 52% D. 73,48%

9. X phng ha hon ton 8,68gam cht bo X cn dng 62ml dung dchKOH 0,5M. Sau phn ng thu c mui kali ca cc axit bo v 0,92 gamglixerol. Ch s x phng ha v ch s axit ca cht bo X ln t l

A. 200 v 193,45c . 198,8 v 5,6 D. 19 v 8,4

T .

1 0 . Th tch dung dch NaOH 0,1M ti thiu cn dng trung haht lng axit bo t do c trong 100 gam cht bo c ch s axit l 8,4

l bao nhiu?

A. 0,15 lt B. 0,1 lt c . 0,3 lt

B. 20 v 6,45ID. 196 v 8,4

D. 0,015 lt

11. 300 gam cht bo A chi gm triolein; tripanmitin v tristearin tcdng va vi dung dch cha 11,43 gm lo.Ch s' iot ca A l

A. 4,42 B. 95,61 c . 8,84 D. 2,21

12. A l cht bo ch gm trilinolein; tristearin v tripanmitin. Bit A

c ch s iot l 15,24. Thnh phn phn trm khi lng trilinolein

trong A l

A. 4,42% B. 95,58% c . 8/j8% D. 17,56%

13. X phng ha hon ton 30 gam cht bo A c ch s axit l 8,4cn va 500 ml dung dch KOH 2M. Khi lng x phng thu c

sau phn ng l _ A. 325,90 B. 380,06 - 3 c i; 3 4

14. A l cht bo ch gm trilinolein V

axit l 8,4. Vy ch sx phng ha cz A l

A. 225,90 B. 180,56 c . 211,89

15. x phng ha h on ton 300- gam cht bo A c ch sx phng

ha l 210 cn dng va bao nhieu ml dung dch hn hp gm

NaO H 2M v KOH 1M? j

A. 562,5 B. 225 c . -37p D. 625

16. x phng ha hon ton 200 gam cht bo A cn dng va

250 ml dung dch hn hp gm NaOH _2M v KOH IM? Ch s xphng ha ca A l

A. 180 B. 225 c_ 210 D. 200

1,89 - D. 327,54

axit stearic. Bit A c ch s'

. 191,58

9



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N ' G DS* G I I

3 1. Ta c mNa0H = 40 0,0

2. Ta c nNaOH = nKOH =

56000300.210

56000

4 gam

= 1,125 moi.

Vy V = 0,5625 lt

3 Ta co TT x phngfWlj\+ niN aO ) ~ (wiglixerol !i;


11/296

Ch Saxit = 56000-

2 A Q 9

(0.062.0,5 - i -)92

8,68= 6,45

10. Ta c V = 8? - 100 = 0,15 lt56000.0,1

11. V ch c trio lein tc dng c vi I2 n n ch s iot:

11,43.100A = 8 8 4 -2 5 4 . 3 0 0

= 4,42 (chn A)

_ _ _ _ 15 24 1 2 . Ta c %trilinolein = 8 7 8 - ^ ^ = 8,78

6 . 2 5 4

13. nh lut bo ton khi lng cho:

I I + m KOH phng I^glixerol I^nc

V y m x phng (^A ** 3HKOh) (ift-giixero] + m-nc)

300.8,4= (300 + 1.56 ) - (9 2 ------ 56000 + 18.30 4 )

56000

= 325,9 gam

14. Xt 1 gam cht bo A, ta c

8,4n K O H / a xit - n c l7H lsC OOH

56000= 1, 5 . 1 0 m o l

1 - 2 8 4 . 1, 5 . 1 0 '

~ 1,0904.10~?(C H COO CHs - g7g

n K O H/ir:o slo = n cr orto = 3 , 1 , 0 9 0 4 - . 1 0 m o l

mKOH/iricste =56000.3 .1 ,0 904.10~3 = 183,1872 mg

Ch s x phng ha = mKOH/trieste + rriKOH/axit

= 183,1872 + 8,4 = 191,5872

300.210 15. Ta c V = = 0,375 lt

(chn B)

(chn A)

(chn C)

(chn A)

(chn D)

(chn C)56000(2 + 1)

1 6 . T a c n N a 0 H = 0 , 5 m o l ; i k o h = 0 , 2 5 m o l .

Suy ra x phng ha hon ton 200 gam cht bo A cn va

0 , 7 5 m o l K O H

- r o , - - * I , ' V ' 0 , 7 5 . 5 6 0 0 0 _ . 1 a . .Ch s x phng ha = ---------------= 210 (chn C)

11



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o CHUYN 2---------- - ----- ------------------------

B I N LC 1 N T M C N G T H C P H N T T C N G

T H C P N G I N N H T

Khi bit cng thc n gin nht, tm cng thc phn t cnbit h s" t l n.

bit h st n}cn phi bit d kin v Mcht hu c- Trong trnghp khng c d kin v Mcht hu c, ta phi bin lun tm gi tr n.

C th bin lun theo 2 cch sau y

A. CCH DNG IU KIN TN TI

Hp cht hu c CxHyOzNtClu mun tn ti phi tha ng thi 2 iu kin:

(y + u + t) phi chn

y + u - t < 2x + 2 *Da vo iu kin trn s bin lun tm c n

V d 1: A l cht hu c cha C; H; Cl, trong d %c v %H (v khilng) ln t l 24,4% v 3,39%. A c tng s ng phn cu to l

A. 3 B.4 C.5 D. 6

Gii

Ta c %C1 = 100 - 24,4 - 3,39 = 72,2124 4 3 39 72 21 '

t cng thc A l CxHyClz ta c x: y: z = : - : 1y 12 1 35.5= 2,033 : 3,39 : 2,034 = 1 : 1,667 : 1 = 3 : 5 : 3

Vy A c cng thc (C3H5Cl3)n

Ta ph i c 5n + 3n < 2 .3n + 2 n < 1.

Vy n = 1, tc A c cng thc phn t l C35CI3

A c cc ng phn cu to sau:

C H 2 C 1C H C 1C H 2 C 1 . C H 3 C H 2 C C 13 C H 3 C H C I C H C I 2

CH2C1CH2CHC12 CH2C1CC12CH3

A c 5 ng ph n cu to (chn C)V d 2: A l cht hu c cha C; H; o, trong %c v %H (theo khi

lng) ln lt l 64,86% v 13,51%. Tng- s ng phn cu to ca A l

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

*Bi phng trnh ny c ti khng ig i vi hp cvl c licrt it ion. V i d CH^NH j+c t

12



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13/296

Gi iTa c % 0 = 100- 64,86 - 13,51 = 21,63

t cng thc A l CxHyOz ta c x: y: z = 64,86 13,5] 21,63

= 5,405 : 13,51 :Vy A c cng thc (C4HioO)n- Ta phi c

Do n = 1, tc A c cng thc phn t l C4 10OA c th c cc ng phn cu to sau y:

12 1 161,351 = 4 : 10 : 1lOn < 2.4n + 2 o n < 1

CH3CH2CH2CH2OHCHsCHzCHXCH^OHCH3OCH2CH2CH3CH3CH2OCH2CH3

Vy A c 7 ng phn cu to

CH3CH(pH3)CH2OH(CH3)3CHOHCH30C(CH3)CH3

(chn C)V d 3: t chy hon ton 11,3 gam cht hu c A c hn hp X ch

gm CO2, hi nc v kh HC1. Dn ton b X qua bnh ng dung

dch AgN3 d thy khi lng bnh tng 10,9 gam v xut hin 28,7gam kt ta. Kh thot ra khi bnh dn tip qua bnh nc vi trongd thy xut hin 30 gam kt ta. Tng s ng phn cu to ca A lA. 4 B. 5 C.7

Gii

Ta c me = 12nCOi = 12.0,3 = 3,6 gam

mCi= 35,5nHci 35,5 28,7 = 35,5.0,2 = [7,1 gam143,5

n i H = n iH /H C ] + m K /H 2o = l - n H C i +

2.nH;0 , 1.28,7 -.,-2.(10,9 365:Q,2)

D. 6

' 143,5m o= 11,3 - 3,6 - 7,1 - 0,6 = 0

Vy A ch cha c, H, Cl.t cng thc A l CxHyClz ta c

36 0,6 7,112

x: y: z =

18= 0 ,6 gam

= 0,3: 0,6: 0,2 = 3: 6: 21 35,5

Do A c cng thc (CHsC^n.Ta ph i c 6n + 2n < 2.3n + 2 n A : C2H30(1)

n = 2 ^ A : C 4H60 7

l C4I6O2

3H502)n. Tm cng thc phn t ca A.

Gi i

Cng thc A c th vit C3nH5n02n hay C2nH4n (COOH)n

n = l=> A :C 3H(iO:,(l)

Ta phi c 2n + n < 2n + 2

Vy A c cng thc phn t

V d 2: A l axit cacboxylic (C

Ta phi c 4n + n < 2.2n +

Vy A c cng thc phn t

.11 < 2

2 n < 2 n = 2 => A : C6H]()0 4

l C6H 10O4

V d 3: A l axit cacboxylic (GsH403 )n- Vy cng thc phn t ca A l

A. CO B. CHi 12 c. CO D. CgHaOgI Gii

Cng thc A c th v it C3nH4n0 3 n hay C3nH5n(COOH)3? y '2 2

n = 1=> A :.C_i H4Os(1)n ~2 =>A :C 6Hs0 6

Vy A c cng thc ph n t i CfiHgOs (chn D)

N h n xt: Thc ra c 3 vi d trn u c th gii nhanh hn nh dngcng thc sau: \

Ta phi c + < 2 + o n < 2 o2 2 2

S 7Tmax ca CxHyOz -2 x + y - 2

Th t vy xt v d 3 tre, rng s nhm chc COOH - snguyn t oxi, th

+ Khng th chon A ; axit cacboxyic ph i c s chn oxi

+ Khng th chn B v s 7Cr

phi c t nht 6 n) \

24 -16 + 2=5(v l v axt C12H1(p 12

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15/296

^ J. t _ L /~r ' 36 24+2 v+ K h n g , t h c h n c VI s 7t(rU(X = ------------- = 7 ( l v a x i t Cj sH 240 18

ph i c t nht 9 n )

=> A l C6H80 6 (chn D)

V d 4: A l hp cht hu c c nhiu trong qu nho, phn t A ch chang thi cc nhm chc -OH v -COOH. A c cng thc n gin nht

l C2H

3Os. Vy cng thc phn t ca A lA. C2-H30 3 B. C8H 12o 2 c . C12H 180 i8 D. C4Hg0 6

G i iTheo , cng thc A c th vit

C2nH30 3 n hay C2n - mHrn(OH)3ri _ 2m(COOH)m

Ta ph i c m + (3n - 2m) + m < 2(2n - m) + 2 2 m - 2 < n (1)

Mt khc A phi csnhm -OH < s c gc 3 n - 2m < 2 n - m n < m (2 )

(1) v (2 ) cho 2m ~ 2 < n < 1 o m < 2 .'n= l=> A :C2H30 3(I)

M n < m nn n < 2 on = 2 => A :C 4H6Oft

Vy cng thc ph n t ca A l C4He06 (chn D)

V d 5: A l amino axit c cng thc (CsHgC^lSQn. Vy cng thc phn tca A l

A. C5H9O4N B. C10H 18O8N2 c . C15H27O12N3 D. C2oH36016N4

Gi i

Cng thc A c th vit C5nH9n0 4nNn hay C3nH5n(COOH)2n(NH2)n

^ Ta ph i c n + 2n + n < 2 .3n + 2


16/296

+ A axit cacboxylic nh chc

Khi A phi c 4 nguyn t ox trong phn t

Vy cng thc phn t A phi l C4H4O4

V d 7: A l ankylbenzen, cng thc (C3H4)n. Tm cng thc phn t ca A.

G i iCng thc A cho l C3i>H4h

A ankyib enzen nncng thc A phi c dng CmH&H - 6

Suy ra m o n = 3. Vy A c cng thc phn t l C9H 124n = 2 m -6 O

C U H I T R C N G H I M

1 . t chy hon ton 4,4 gam cht hu c A cn va 8,4 lt 0 2 (ktc).Hp th ht sn phm chy (ch gm CO2v H20) vo nc vi trong dc 25 gam kt ta. Tng s ng phn cu to ca A l

A. 8 B. 7 c . 9 D. 14 2. A l cht hu c cha C; H; C, trong %cv %H(v khi lng)

ln lt l 14,28% v 1,19%. A c tng s" ng phn cu to

A. 3 B. 4 c . 5 . 2

3. t chy hon ton 5,1 gam cht hu c A cn va lt 10,08 ltO2 (ktc). Hp th-ht sn phm chy (ch gm CO2 v H 20) vo ncvi trong c 20 gam kt ta. Lc b kt ta ri un: nng nc lc lic 5 gam kt ta na. Bit A khng tc dng c vi Na. Vy A cth c bao nhiu cng thc cu to?

A. 8 B. 9 c . 15 D. 16 4. t chy hon ton 14,75 gam cht hu c A c hn hp X ch gm

CO2, hi nc v kh HC1. Dn ton b X qua bnh ng dung dchAgN03 d thy khi lng bnh tng 12,75 gam v xut hin 43,05 gamkt ta. Kh thot ra khi bnh dn tip qua bnh nc vi trong d thyxut hin 30 gam kt ta. Tng s ng phn cu to ca A l

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

5. A cht hu c cha C; H; Br, trong %c v /cH (theo khi lng)

ln lt l 2 2 ,2 2 % v 3,7%. Vy tng s' ng ph n cu to ca A lA. 10 B. 11 C.9 D. 8

6 . A l anehit (CsHgOXj. Cng thc phn t ca A l

A. C3H5O3 B. C6H 10O2 c . C12H20O4 D. C18H3oC>6

7. A l axit cacboxylic (C2H302)n- Vy cng thc phn t ca A l

A, C2H3O2 B. C4I 6O4 c . CsH^Og D. C12H 18O12

16



B

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17/296

8 . t chy ho n ton '5,2 gam axit cacboxylic A cn va V lt Oo(ktc), sau ph n ng thu c H20 v js,6 gam C02. Mt khc trungha hon ton cng ng axit trn cn va 50 ml dung dch NaOH2M. Gi tr V l !

A. 1,12 B. 2,24 C.3,36 D. 1,68

9. t chy hon ton nit lng axit cacboxylic A cn va 5,6 lt

2 (ktc), sau phn ng thu c 3,6 gkm H 2O v 11 gam CO2. A ccng thc ph n t l

A. C5I8O2 B. CioHieOg c . CsgO D. C4I6O4

1 0 . t chy hon ton mt lng anehit A cn va 2,688 lt 0 2(ktc), sau phn ng th u c H2O v 4,4 gam C2. Mt khc cnglng anehit trn nu tc dng ht vpi dung- ch AgNCVNHs c8,64 gam bac. Cng thc phn t ca A

A. C3H4O2 B. CeHi0O2 c . C5H8O2 D. C10H 1 IX. A l amin bc mt, cng thc (CH4NL. A c cng thc phn t l

A .C H 4 N B . C 2H s N 2 C . C 3H 12N3 D . c 4h 16n 4I

1 2 . A l amino axit, cng thc n gin nht l C3H7NO. A c cngthc phn t l

A.C3H7ON b . c 6h 14o 2n 2 c . c 12h 28o 4n 4

13. A l amino axit, cng thc n gi

thc phn t l

A .C 3 H 7 O 2 N : B . c h 14o 4n 2 G . C i

14. A l amino axit c cng thc (C3H

ca A l

A. c h 6o 2n b . .c 9h 18o 6n 3 c . c 6h 12o 4n 3

D. CisH^ONs

n. nht l C3H7NO2. A c cng

H 2s0 s % ; ; ; .

502N)n. V cng thc phn t

15. A l hp cht hu c c nhiu trongnhnh, phn t ch cha ng thi cc E hm chc -O H v -COOH. BitA c cng thc n gin nh t l C4H6O5. ]Vy cng thc phn t ca A

. A. C46O5 B. C8 12O10 c* Ci

D. C12H 24O8N4

qu to. A c mch khng phn

ifeHisOis D. C16H 24O20 16. A l hp cht hu c c nhiu trong qu xanh cha chn, phn t

ch cha ng thi cc nhm chc -OH v -COOH. A c cng thc ngin nht l C2H4O3. Vy cng thc phan t ca A l

A. C2H4O3 B. C4I8OS c . C J 1I 12 O 9 D. C816O12

17



B

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18/296

HNG DN GII l . n h lu t bo ton khi ling cho' raA+ m0> = mco^ + mH o

8,4 mHQ = m A+m 0; - m c;0 = 4,4 + 32- - 44.0,25 = 5,4 gam

Rt ra mc = 3 gam; mH = 0 js gam; mo = 0,8 gamT tm c cng thc cua A l (CsHi2>0 )n

Bin lun tm c n = 1. Vy A c cng thc phn t l C5H 12ODo cng thc ny c S7U = 0 nn A ch c th l ancol n chc no

hoc ete n chc noSancol CnH2ft + 20 = 2n- 2 |n < 5) = 2 5 " 2= 8

(n - l)(n|S ete CnH2n +2O .=

Vy s ng ph n cu to cua A l 14 (chn )

2. T cc d kin cho tm c cng thc phn t A l C2H2CI4, ng

vi cc ng phn cu to: CHC2CHCI2CH 2CICCI3Vy s" ng ph n cu to cua A l 2 (chn D)

3. T cc d kin cho, tm c cng thc phn t A l C6H 140

V A khng tc dng vi Na nn. A ch c th l 15 ete (chn C)

4. T cc d kin cho, lm c: mc = 3,6 gam ; mci - 10,65 gam;EiH = 0,5 gam; mo = 0. Sau .tm c cng thc phn t C 3H 5CI3,ng vi 5 ng ph n cu tao (chn B)

5. T cc d kin cho, tm c cng thc phn t A l C4H 8Br2, ng

vi 9 ng phn cu to sau yCH3CH2CH2CHBr2

CH 3CH2CHBrCH2Br

CH 2BrCH2CH2CH2Br

CHsCH(CH3)CHBr2

CH 2BrCH(CHs)CH2Br

6 . Dng phng php tch

CH3CH2CBr2CH3 ;CH 3C H B r C H B r H ^ g w ^ f 5C3CHBrCH2CH2Br

CH3CBr(CH3)CH2Br

(chn C)

ring nhm chc hoc phng php tnh s'7max u tm c cng thc phn t A l C6H 10O2 (chn B)

7. Dng phng php tchu tm c cng thc phn t A l C 4H 6CXj

8 . t cng thc A l CxHyOz, ta c cc phn ng:

CxHyOz + (x + ) xC02 + ^ H 20 .y 4 2 2

ring nhm chc hoc phng php tnh s(chn B)

a mol

4 2

ax mol mol

1a



B

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N

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O

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19/296

CxHyOz + ~ NaOH - CXH ,OzNa7 + - H 20y 2 2 2

a molaz

mol

a(12x + y + 16z) = 5,2

Suy r a ax = = 0,1544

= 0,05.2 = 0,1I 2

ax = 0,15

ay = 0,2

az = 0 ,2

Do X : y : z = 3 : 4 : 4, tc A c cng thc (CsHiOJnhay CnH2n(COOH)2nTa phi c 2n + 2n < 2n + 2 n < 1.

Vy 11=1, tc A c cng thc C3I4O4Phn ng chy ca A: C3H4O4 + 2 O2 -> 3 CO2 + 2H20

0,05 moi 0,1 molVy V = 2,24 lt (chn B)

9. Tng t nh cu 8, ta c h:

a(x + - - ) = = 0,254 2 22,4

ax = = 0,2544

^ = M = 0,22 18

Do X : y : z = 5 : 8 : 4, tc A c cng thc ( Cs H s C^Xhay C3nH6n(COOH)2n

Ta ph i c 6n + 2n < 2.3n + 2 o n < 1.Vy n = 1, tcA cr CQng th c C5g04 (chn C)

10. t cng thc A l CxHyOz, ta c cc phn ng:

CxHyOz + (x + ^ - * ) ~ x C 0 2 + l 20 .4 2

a mol ax mol mol

CxHyOz > 2zAg

a mol

Suy ra

2az mola(x _ ) = ^ = o ,i2

4 2 22,44 4

ax= = 0,144

2az = = 0,08108

fax = 0,1

o


20/296

Ta phi c 2 n + n < 2n + 2 n < 2

Do X : y : z = 0,1 : 0,16 : 0,04 = 2,5 : 4 : 1 = 5 : 8 : 2

Vy A c cng thc (CHgC^n tc C3nH6n(CHO>2

Ta phi c 6n + 2 n < 2.3n + 2 o n < '1.

Vy n = 1, tc A c cng tbc C58O2 (chn C)

11. Cng thc A c th v it CnH4nNn hay CnH2n(NH2)n'n = l=> A:C H4N(1)

n = 2 => A : C2HSN2

Vy A c cng thc phn t l C28N2 (chn B)

1 2 . Cng thc A c th vit C3nH7nNuOn hay C2,5iiH4,5n(NH2)n(COOH)o,5nTa phi c 4,5n + n + 0,5n < 2.2,5n + 2 n < 2Nhng amino axit phi c s nguyn t oxi > 2.

Vy A phi l C6H 14N2O2 (chn B)

13. Cng thc A c th vit C3nH7nNn0 2 n hay C2aH4n(NH2)i(COOH)n

Ta phi c 4n + n + n < 2.2n + 2 ci> 1 < 1 .Vy A c cng thc phn t l C3H7NO2 (chn A)

14. Cng thc A c th vit C3nH6nNn0 .2n hay C2riH 3n(NH2)n(COOH)11

Ta phi c 3n + n + n < 2.2n + 2 o n < 2.

n = 1=> A : C3H6N 0 2(1)

n = 2 A : C6H12N20 4

Vy A c cng thc phn t l C6H 12N2O4 (chn C)

15. Theo , A c cng thc C4DH6n0 5 n.

V A khng ph n nh nh nn ch! c 2 kh nng: ;:

+A c nhm COOH

Cng thc A khi c dng C4n - iHn +i(COOH)(OH)5ri _ 2

Ta phi c: s' nhm - 0 H < s c d gc 5n - 2 < 4n- 1 n A :

n < 2 n = 2=> A : CxH120 10

20



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21/296

Tuy nhin cng thc C8H i2O10 trong trng hp ny phi c 2 nhm-COOH v 6 nhm -OH . V A c mch khng phn nh nh nn A phi c

2812+ 2s l in kt 71 = ------- ----------= 3, trong CI 2 lin kt rc nm 2 nhm

2

-COOH v 1 lin kt K nm gc hidro^icbon. Gc ny c 6C, cha 1lin kt 71 v c 6 nhm OH s khng bn (loi)

Vy A c cng thc phn t l C4I6O5 !

N h n x t; ph hp vi t duy trc nghim khch quan, bi ny cLl - 1----- 1 ^ ----- s lin k.t7Ckt hp vi iu

cn s nhm OH chnh l s"th gii nhanh nh dng cng thc tnhkin A ch cha 1 hoc 2 nhm COOH,nguyn t oxi cn lai. Tht vy, nu A chi cha 1 nhm COOH, ta thytt c trng hp u loi v khi cc nhm -OH u phi gn trn tngc ( thm ch hp cht Ci2H 80 i5 v C1H24O20 th cc nhm -OH phign cng 1C) ca gc hirocacbon cha no. Nu A cha 2 nhm -COOHth c 3 phng n B; C; D a l CgHigOio; C12H 1SO15v C16H24O20u xy ra h in tng ny (loi) I

16. Theo , A c cng thc C2nH4n0 3n. V s H gp 2 ln s c nn Ach cha ti a 1 lin kt 7C, tc A ch c th cha 1 nhm -COOH.

Vy cng thc A c dng c2n- iHn +1(CCOH)(OH)3a _ 2Ta p hi c: s' nhm OH < s c gc

Do A c cng thc C2H4O3 (chn A)

o 3n - 2 < 2n - 1

cx> n < l.Vy n = 1

Q CHUYN 3T O f i N V fT M I N O iX IT

A. NGUYN TC

iu trc-ht cn nm l amiio axit A, cng thc (NH2)xR(COOH)y , tc

dng c vi HG1 theo t t mol tng ng 1 . X, to mui c M = M a + 36,5x.Amino axit A cng tc dng c vi NaOH teo t t mol tng ig 1 : y, tomui c M = Ma + 22y. I

Ton v amino axit (NH2)KR(COOH)y tng gp v cng thng gyI *cho hc sinh nhieu lng tng l dng cho mt amino axit (hoc hn hp

amino axit) vo cc ag dung dch HCI, sau thm tip dung dchNaO H va . hoc d vo dung dch sau pnn ng (hoc tin hnh ngclai). Dang ny tht ra rt n gin, v HC1 hc NaOH c trong cc banu s chuyn h t th n h NaCl, cn amino axit s chuyn h t th n h mui

21



B

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22/296

(NH2)xR(COONa)y {n u dng N aOH g ia i o n c u i) hoc chuyn ht

thnh (NH3Cl)xR(COOH)y {nu drig HC l g iai o n cui). iu ny c

ngha nu ban u cho a mol amino axit (NHj^RCCOOH)^vo cc cha b

mol HC1 th s moi NaOH phn ng va vi cc cht trong dung dch sauphn ng ph i l (ay + b) mol. Vi trng hp ngc li, cho a mol aminoaxit (NH2)xR(COOH)yvo cc cha b mol NaOH th smol HC1 phn ng

va vi cc cht trong dung dch sau phn ig s l (ax + b) mol.B. CC V D

V d 1 : Cho m gam gyxin vo dung dch cha 0,3 mo HC1. Dung dchsau phn ng tcdng va vi 0,5 mol NaOH. Tm m

Gi i

Ta c 0,3 + rigiyxin = 0,5. Vy Hgiyxin = 0 ,2 moi tc m = 75.0,2 = 15 gam

V d 2: Cho m gam axit glutamic vo dung dch cha 0,3 mol HC1. Dung

dch sau phn ng tc dng ^a vi 0,5 moi NaOH. 11111 m! G i i

T a co 0 ,3 + 2naxil glutamic = 0,5.

Vy riaxit glutamic = 0,1 mol tc m = 147.0,1 = 14,7 g

V d 3: Cho m gm lysin vo dung dch cha 0,3 mol HC1. Dung ch

sau phn ng tc dng va cj vi 0,5 moi NaOH. Tm m

G i i

Ta c 0,3 + niysia = 0,5. Vy riiysin = 0,2 mol, tc miysin = 146.0,2 = 29,2 g

L u : Lysin l mt amino ax

N l am ino axit khng th

hp c m phi ly trc

it rt cn th it cho s tng, trng c th,

thay th v c th ngi khng th t tngtip t thc n. Nht Bn nc tng

thm ly sin vo gao, bt m . xc tin s tng hp protein trong cth. Lysin c cng thc l N2(CH2)4CH(NI3)COOH

V 4: Cho m gam lsin vol dung dch cha 0,3 mo NaOH. Dung ch

sa phn ng tc dng va vi 0,5 mol HC1. Tm m

I Gi iTa c 0,3 + 2niysi = 0,5.

Vy niysia= 0,1 moi, tc miysL= 146.0,1 = 14,6 gam

V d 5: Cho mt lng axit glutamic vo dung dch cha 0,2 mol HC1.

Dung dch sau phn ng tc dng va vi 0,3 moi NaOH, sau ccn dung dch c bao nhiu gam rn khan? '

22



B

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O

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-

L

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23/296

Gi iRn khan phi gm: 0 ,2 irol NaCl

v '- z0 2 = 0,05 mo H 2NC3H 5(COONa)2

Vy mrnkhaii = 58,5. 0 ,2 + 191. 0,05 = 21,25 gam

V d 6 : Cho mt lng axit glutamic vo dung dch cha 0,2 mol NaOH.

Dung dch sau phn ng tc dng va vi 0,3 mol HC1, sau ccn dung dch sau phn ng c bao nhiu gam rn khan?

Gi iRn khan phi gm:

0,2 mol NaCl v


24/296

Gi iGi a, b ln lt l s mo alain CH3CH(NH2>COOH v axit glutamic

H2NC3H5(COOH)2 dng, ta c h:

Vy m = 112,2 (chn A)V d 10: Cho 1 mol amino axit X phn ng vi dung dch HC1 (d), thuc mi gam mui Y. Cung 1 mol amino axit X phn ng vi dung dch

NaOH (d), th u c m2 gam mui z. Cho bit m 2 - mi = 7,5. Cngthc phn t ca X lA. C4H i0O2N2. B. c 5h 9o 4n . c . c 4h 8o 4n 2. D. C5H 21O2N

Nu chn A (c 2 nhm -N H 2; 1 nhm -COOH) th

mi = Mx + 73 v m 2= Mx + 2 2 .Nh vy m 2mi < 0 (loi).Tng t chn c th m 2 - mi = (Mx + 44) - (Mx + 73) < 0 (loi)v chn D th m 2 - mi = (Mx + 2 2 ) - (Mx + 36,5) < 0 (loi)Vy X C5H90 4N (chn B)

V d 11: Hn hp X gm 1 amino axit no, mch h v 1 amin no, mchh. X c kh nng tc dng ti a vi 2 mol HC1 hoc 2 mol NaOH.'t chy hon ton X c 6 moi C2, X mol H20 v y mol N2. Cc gitr X, y tng ng l

V X c kh nng tc dng ti a vi 2 mol HC1 hoc 2 mol NaOH nnamino axit cho phi c 1 nhm -NH 2 v 2 nhm -COOH, cng thcCnH2n _ 1NO4; cn amin cho c 1 nhm -NH2, cng thc CmI2m+3N.

cc phn ng chy ca chng:

CnH2n_ 1NO4 ? > nC2+ (n - 0,5)H20 + 0,5N21 mol n mol (n - 0,5) moi 0,5 moi

C mH 2m, sN mC Og + ( m + 1,5) H20 + 0 ,5N 21 mol m mo (m + 1,5) mo 1 0,5 mol

V nco^ = (n + m) = 6 mol nn nH>o = (n - 0,5) +(m + 1,5)

(i h c k h i A / 2009)

Gii

A. 8 v 1,0 B. 8 v 1,5 c . 7 v 1,0 D. ,7 y 1,5( i h c kh i A / 2010)

Gi i

= n + m + l = 7 mo vn N , =C0,5 + 0,5) = 1 mol (chn D)

24



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25/296

CU HOI TRC NGHIE3V[|

1. Cho m gam alan in vo dung dch cia 0,375 mol NaOH. Dung dchsau phn ng tc dng va vi 0,575 mol HC1. Gi tr m l

A. 15,0 B. 17,8 c . 8 ,9 D. 14.7D. 14,7dch hn hp gm NaOH IM v

A. 15,0 B. 17,8 c . 8, 2 . Cho m gam alanin vo 400 ml dung

KOH 0,5M. Dung dch sau phn ng tc dng va vi 600 m dungdch HCI 1,5M. Gi tr m l

A. 26,7 B. 85,65 c. 75,95 3. Cho m gam axit glutamic vo dung

dch sau phn ng tc dng va vi

A. 14,70 B. 7,35 c . 9 i

D. 87,15

dch cha 0,4 mol NaOH. Dung0,5 mol HCl. Ch ra gi tr m

55 D. 29,40

4. Cho m. gam a xit g lutamic vo 400 mi dung dch hn hp gm NaOHIM v KOH 0,5M. Dung dch sau phndung dch HC1 1,5M. Ch ra gi tr m

A. 44,10 B. 22,05 c . 9,^5 i

ng tc dng- va vi 600 m

D. 29,40 5. Cho m gam ax it glutamic vo 400 ml dung dich hn hp gm HC

IM v HB r 0,5M. Dung dch sau phn) ng tc dng va vi 600 mldung dch NaOH 1,5M. Ch ra gi tr V

A . 44,10 B . 22,05 c . 9 , 6 . Cho 21,9 gam a - amino axit A vo

Dung dch sau phn ng tc dng vacng thc ph n t l:

; c 5H9N 0 4 : . . . C . 5H u N^ 2 7. CEo 11,025 gam amino axit vo

Dung dich sau phn ng tc dung va vi 0,23 mol NaOH. Vy A ccng thc ph n t l:

A. C2H5NO2 B. C5H9NO4 c . cUnNOa D. C6H12N204 8 . 0,1 mol amino axit X, cng thc R(NH 2)n(COOH)m tc dng va

vi 0 ,2 mol HC1, sn phm to thnh phn ng va ht vi dung dchcha 0,3 mol NOH. Gi tr ca n v m ln lt l

A- 2 ; 1 B. 2 ; 3 c . 1 ] 2 D. 1 ; 3 9. 0,15 mol amino axit X, cng thc R(NH2)n(COOH)m tc dng va

vi 0,3 mol NaOH, sn phm to thanh phn ng va ht vi dungdch cha 0 ,6 mol HC1. Gi tr ca n v m ln lt l

55 D. 29,40

dung dch cha 0,12 mol NaOH. vi 0,42 mol HC1. Vy A c

i'- C6H 14N2O2 ; -dung dch cha 0,155 mo HC1

A. 2; 1 B, 2; 2 c. lj D. ly 3

25



B

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26/296

1 0 . Cho 0,15 mol H 2NC3Hs(COpH)2 vo 175 ml dung ch NaOH 2 M,thu c dung dch X. Cho dung dch HC1 d vo dung dch X. Sau khicc phn ng xy ra hon ton, s" mol HC1 phn ng l

A . 0,50 B. 0,65 ,c. 0,55 D. 0,70

1 1 . Cho 14,7 gam axit glutamic vo dung dch cha a mol HC1. Dungdch sau phn ng tc dng jva vi 300 ml dung dch NaOH 2M.Gi tr a l

A . 0,30 B . 0,35 c . 0,45 D . 0,40 1 2 . Cho 14,7 gam axit glutamic vo dng di ch cha a mol NaOH. Dung

dch sau phn ng tc dng va vi 300 ml dung dch HC1 2 M. Gitr a l

13. Cho 0,15 mol H2NCsH5(OOH)2 vo 170 ml dung dch hn hpgm NaOH 2M v KOH IM, thu c dung dch X. Cho ung dch HCd vo dung dch X. Sau khiI cc ph n ng xy ra hon ton, s molKC1 phn ng l

A. 0,36 B. 0,66 c . 0,81 D. 0,70 14. Cho mt lng axit glutamic vo 400ml dung dch NaOH IM. Dung

dch sau ph n ig tc dng vi 250m l dung dch HC1 2M, saurn khan?

c . 42,5 D. 41,75

15. Cho mt lng axit glutamic vo 400ml dung dch HC1 IM. Dungdch sau phn ng tc dng va vi 250ml dung dch NaOH 2M,sau c cn. c bao nhiu gam rn khan?

c cn c bao nhiu gam

A. 23,4 B. 59,1

A. 45,75 B. 32,95

16. Cho mt lng alanin vdch sau phn ng tc dng

c. 42,5 D. 41,75.0 dung dch cha 0,25 mol NaOH. Dungva vi 0,45 mol HC1, sau c cn

dung dch sau phn ng c bao nhiu gam rn khan?

A. 39,725 B. 32,425 c. 36,825 D. 51,75017. Cho mt lng axit glutamic vo 2 0 0 ml dung dch hn hp gm

NaOH IM v KOH 1,5M. Dung dich sau phn ng tc dng va vi300ml dung dch HC1 2M, sau c cn c bao nhiu gam rn khan?

A. 52,40 B . 59,15 c. 62,55 D. 58,75 18. Cho mt lng amino axit A vo dung dch cha 0,15 mol HC1.

Dung dch sau phn ng tc dng va vi 0,25 moi NaOH, sau ccn c 18,325 gam rn khan. Vy A c cng thc phn t l:

A. C 4 H 7 N O 4 B. C 4 H 10 N 2O 2 c . C5H nN 0 2 D. C 5 H 9 N O 4

26



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27/296

19. Cho m gam hn hp X gm glyxin v alanin (t l mol 1:1) vo ccng 200 ml dung dch HC1 IM. Dung dch sau phn ng tc dng va vi 150 ml dung dch NaOH 2M. Gi tr m lA. 8,20 B . 9,15 c . 6,55 . 8,75

1. m = 89(0,575 - 0,375) = 17,8 gam

2. T a c n NaOH = 0 ,4 m o l;.nKoH = 0,2 mol; n Hci = 0,9 molm = 89(0,9 0,4 0,2) = 26,7 gam

Q 3 Ta co 0,4 + Haxit glutamic = 0j5.V y iW t glutamic = 0 , 1 m o l, t c m = 14,7 g a m

4. Ta c riNaOH = 0,4 moi; 1K0H = 0,2 mol; nHci = 0 ,9 niolV y 0,4 + 0,2 + riaxit glutamic = 0,9

w n axit glutamic = 0 ,3 m o i, LC 1X1 4 4 , 1 ^ 3.rl

5. Ta c nHci = 0 ,4 moi; riHBr = 0,2 mol; riNaOH= 0 ,9 mol

Vy 0,4 + 0,2 + llaxt glutamic = 0,9

n axit glutamic = 0,1 5 m ol , tc m = 22 ,0 5 g am (ch n B)

6 . Theo cc phng n tr li th A c 1 hoc 2nhm -N H 2+Nu A c 1 nhm -N 2Ta c 0,12 + IIA = 0,42 n A= 0,3 moi.Vy 0,3Ma = 21,9 MA = 7 3 (loi, v amino ax it c M nh nht l 75,

ng vi glyxin)+Nu A c 2 nhm -N 2Ta c 0,12 + 2iA = 0,42 nA = 0,15 mol. Vy 0,15MA = 21,9

o Ma = 146, tc A c cng thc C6H 14N2O2 (chn D) 7. Theo cc phng n tr li th A c 1 hoc 2nhm -COOH. + N u A c 1 nhm-COOH

Ta c 0,155 + Ha = 0,23


28/296

11. Ta c naxit glutamic = 0,1 mol; nNa0H = 0,6mol.Vy 2.0,1 + a = 0,6 Ci> a = 0,4 mol (chn D)

o . T s c n axit glutamic = TO0I5 IHC m o i .

Vy 0,1 + a = 0,6 a = 0,5 mol (chn A)

13. Ta c riNaOH = 0,34 mol: ik o h = 0,17 molVy nHC = 0,34 + 0,17 + 0,15 - 0,66 mol (chn B)

14. V nNaOH = 0A moi; nHcr= 0,5 mol nn naxit glutamic = 0,1 mol Rn khan gm 0,4 mol NaCl v 0 ,1 mol NHsC1C3H5(COOH)2Vy mrn = 58,5.0,4 + 0,1.183,5 = 41,75 gam (chn D)

15. V iiNaOH = 0,5 moi; riHci = 0,4 mol n n r w glutamic = 0,05 mol

Rn khan gm 0,4 moi NaCl v 0,05 mol NH 2C3H5(COONa)2Vy mrn = 58,5.0,4 + 0,05.191 = 32,95 gam (chn B)

16. Ta c laianin = 0,2 molRn khan gm 0,25 moi NaCl v 0,2 mol CH3CH(NH3C)COOH

Vy mrn = 58,5.0,25+ 0,2.125,5 = 39,725 gam (chn A) 17. Ta c nNa0H = 0,2 mol; riKOH = 0,3 moi; IIHCI ==0,6 mol

n n . Igxit glutamic = 0 ,1 m o

Rn khan gm 0,2 mol NaCl;

0,3 mol KC1v 0 ,1 mol NHsClCsHsCOOH^Vy irn = 58,5.0,2 + 74,5.0,3 + 0,1.183,5 = 52,4 gam (chn A)

18. Theo cc phng n tr li th A c 1 hoc 2 nhm -COO H

+Nu A c nhm -COOH

Ta c 0,15 + nA = 0,25 nA = 0,1 mol.Rn khan gm 0,15 niol NaCl v 0,1 moi mui (NH2)xRCOONa.

A/T _ * * . _ 1 8 , 3 2 5 - 5 8 , 5 . 0 , 1 5Vy Ma M ( n h , ) n r c o o h - Q 22 73,5 (loi)

+Nu A c 2 nhm -COOH

Ta c 0,15 + 2nA = 0,25 o nA= 0,05 mol.

Rn khan gm 0,15 moi NaCl v 0,05 mol mui (NH2)xR(COONa)2.

1 7 A T\yr \ / i 1 8 , 3 2 5 5 8 , 5 . 0 , 1 5 . u m ^

Vy Ma - M(NH R(COOH)j ~5 ~ ^ 5- 9

(chn )

19. D dng t h c igyxin = Iiaianin = 0,05 mol.

Vy m = 75.0,05 +89.0,05 = 8,2 (chn A)

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29/296

o CHUYN 4Q C I I T f f C a

IA. NGUYN TC

vn dng c qui tc a , nh t llhit phi nm c th t dy

in ha (cn gi l dy th in cc chuan). Dy in h a l m t dy cccp oxi ha kh (mi cp oxi ha kh c biu din di dng mt phns', trong t S' l cht oxi ha, mu s l cht kh), th t cc cptrong dy cng l th t tng dn tnh oxi ha

Qui tc a dng d on xem mt ph n ng oxi ha kh c xy rac khng? sn phm to th nh l g? I

Gm cc bc: I

+ Xp cc cp oxi ha- kh lin quan jn phn ng cn kho st theochiu tn g dn tn h oxi ha (tc chiu trong tr t t dy in ha)

+ C gia 2 cp s xut hin -mt ch a . y chnh l chiu biu dinphn ng oxi ha- kh xy ra c

Chng han cn kho st xem trong dung dch, phn ng sau c xy rakhng?

Sn + Fe 3+

Quan st trong dy in ha, ta thyquan n phn ng trn c xp theo th t (tc chiu tng dn tnh oxiha) l:

th t 2 cp oxi ha- kh lin

Sn Fe2+Sn Fe

Chn ch a vo gia 2 cp trn:

Sn4a

FeSn?+ Fe"

Ta c chiu i ca ch a t phi ssig tr i cng l chiu biu dinphn ng oxi ha- kh xy ra c:

Sn2+ + 2 Fe3+ ----- Sii4+ + 2Fe2+

L u : P h n ng c a cng ln cng d xy ra, v l p hn ng u tintrong s cc ph n ng c th xy ra tro ng mt th nghim. V d trongth nghim cho bt magie vo dungCuCl2, ta c th t phn ng l:

Mg + CuCl2 -> MgCl2+ Cu (1 )

Mg + FeC2 -------> MgC2+ Fe (2 )

dch cha ng thi FeCl2 v

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30/296

S di (1) xy ra trc (2 ) vi theo th t 3 cp oxi ha kh trong dyin ha .............................-....

Mg2 CuMg Fe Cu

th a phn ng (1 ) ln hi a phn ng (2 )

B. CC V D

Vi d 1; Hon thnh cc phng trnh phn ng sau (trong dung dch):a/ Cu + FeC3 ------>

b/ Fe + F eC ls____

cAgNOa + Fe(N03)2

sTa c theo qui tc a :Cuf

Ci

Vy phn ng sau xy ra

G i i

F e 3+

Fe 2 +

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

Hay: Cu + 2FeCl3 ------ CuCl2+ 2FeCl3

b/ Ta c theo qui tc a :Fe Fe

Fa

Fe2+

Vy phn ng sau xy rai c: Fe + 2 Fe"

Hay: Fe + 2FeCl3 ------> FeCI3

c/ Ta c theo qui tc a :FeFe

aAg+

Vy phn ng sau xy ra

Hay: AgN3 + Fe(N3)3

Ag + Fe: c: Fe2+ + Ag+

----- > F (N 03)2 + Ag

V d 2: Cho bit th t cc cp oxi ha kh trong dy in ha:

Al3* Fe2+ |Fe3+

AI Fe' Fe:+

Hon th n h cc phng trrh phn ng- sau .(trong ung dch):

a/ AI + Fe: () b/ Al(d) + Pe'

Rt ra kt lun v 2 ph n ng trn.Gi i

l-Ja/ Ta c theo qui tc a :

Fea

A Fe2+

Vy phn ng sau xy ra c: I + 3 Fe3+

V Fe u d nn AI s ph n ng h t

Al',+ + 3 Fe2

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31/296

b/ Trc ht cng xy ra phn ng do qui tc a nh trn:

AI Fe2T

AI + 3 Fe3+ ------ AI3* +3 Fe2+

Nhng khc vi trng hp a, y Fe2" phn ng tip vi AI cn dtheo qui tc a sau:

Al3+ Fe2+aAI Fe

2AI + 3Fe2+ ------> 2 A l3+ + 3 F e

Do trong trng hp ny, phn ng din ra l:

Al(d) + Fe3+ -------* Al3+ + Fe

Kt lun: thu c st k im loi t ph n ng ca AI vi Fe3+, cndng AI d

V d 3: Cho bi t th t cc cp oxi ha kh trong dy in ha:

L Ag+Fe Fe2+ Ag

Hon thnh cc phng trnh phn ng sau (trong ung dch):

s / Fe+ Ag+ -------- > b / F e + A g +(d) -------- >

Rt ra kt lun v 2 phn ng trn.

Gi i

Fe2+ Ag+

a/ Ta c th eo qui tc a : r otFe Ag

Vy phn ng su xy ria c: Fe + 2 Ag+ ------> Fe2+ + 2Ag

V Fe d nn Ag+ s phn ng ht

b/ Trc h t cng xy r phn ng do qu tc a nh trn:

Fe2+ . Ag+a

Fe Ag

Fe + 2 Ag+ ------> Fe2+ + 2Ag

Nhng khc vi trng hp a, y Fe2+ sinh ra s phn ng tip viAg+cn d theo qui tc a sau:

Fe3+ Ag+'i a

. Fe Ag

Fe2+ + Ag+ ------ Fe3+ + Ag

31



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32/296

Do trong trng hp ny7phn ng din ra l:

Fe + 3Ag"(d) ------ Fc 3+ + Ag

Kt lun: Hu ht Fe phn ng vi cc mui to Fe2"1 , tuy nhin trongtrng hp AgN03 d th p hn ng s to Fe?+

V d 4: C 2 phn ng sau:

2 FeB*2 + Br2 ----- > 2FeBr3

2NaBr + Cl2 ----- > 2NaCl + Br2

Pht biu ng lA. Br2 c tnh oxi ha mnh hn C2

B. Cl c tnh kh mnh hn Br"

c . Fc 2+ c tnh kh yu hn C1"

D. Fe3+ c tn h oxi ha yu hn Cl2i h c k h i B / 2008)

Gi i 2 phn ng trn din ra c, th t cc cp oxi ha- kh trong

u Br2 Cl2day in hoa phai l: _ - ~r~

F e 2Br 2c r

Ch rng t s" l cc cht oxi ha, vi tnh oxi ha tng dn t trisang phi mu s l cc cht kh, vi tnh kh gim dn t tri sang phi

pht biu ng l D

V d 5: C mt mu bc b ln tp cht l ng v st. ngh 2 phngphp ha hc khc nhau tinh ch mu bc trn

G i i

Cch 1: Ngm mu bc tr n vo lng d ung dch Fe3+ , ton b Fe vCu s tan ht, cn li l Ag

Fe + 2 Fc3* ------ 3 Fe3+

Cu + 2 Fc 3+ ------ Cu2* + 2Fe2+

Cch 2 : Ngm mu bc trn vo lng d dung dchAgNO, ton b Fev Cu s tan ht, cn li l Ag

Fe + 3AgN03() ------> Fe(N

03)3+ 3AgCu + 2AgN03 ----- > Cu(N0 3>2+ 2Ag

V d 6 : Cho 8,1 gam bt nhm vo cc ng 300 ml dung dchFeC3. Saukhi phn ng xong, khi lng ung dch tng hay gim bao nhiu gam?

'Gii

Ta c nA) = 0,3 mol; n r,cC1 =- 0,6 moi

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33/296

AI T SFeC3 ACI3 + 3FeC

0,2 mol 0,6 moi 0,6 mol

Sau phn ng ny, I cn d 071 mol s tc dng tip vi 0,6 moi FeCl2:

2A + 3FeC2 ------ 2A1C1? + 3Fe I

0,1 moi 0,15 moi

N h vy nhm tan ra 8,1 gam v to 0,15.56 = 8,4 gam s t nn khilng un g dch gim (8,4 - 8,1) = 0,3 gani

V d 7: Cho 19,3 gam hn hp bt gm Zn|v Cu (t l moi tng ng 1 : 2 .)

vo dung dch cha 0.2 mol Fe2(S0 4)3. su khi cc hn ng xy ra honton thu c m gam rn. Gi tr m l

A. 12,00 B. 16,53 c . 6 ,4

Trc ht xy ra phn ng: .

D. 12,8

(i hc kh i A / 2010)

Gi iT a c: 65 a + 64.2a = 19,3 a = 0,1 moi

Trc ht xy ra phn ng:

Zn + Fe2(S04)3 ------> ZnS04 + 2FeS4

0,1 mol 0,1 mol

Sau : Cu + Fe2(S0 4)s ------> CS04 + 2FeS04

0 ,1 mo 0 ,1 moi

(chn C)Vy rn cn li l 0,1 mol Cu, tcirn

= 6,4 gamV d. 8 : Nhng mt l kim loi M vo dng dch Fe(NC>3)3. Sau mt thi

gian ly M ra cn thy khi lng khng i. M l

A . F e B . C u c.M D. PbGii I

th t cc cp oxi ha- kh trong (y in ha sau -dy:

M g 2+ F e 2* Pb 2+ Ci j2* F e 3*

Mg Fe Pb CU Fe2+ *'**

T thy rng c 4 kim loi cho u tan ra khi nhog vo dung

dch Fe3+, nhng sau ch c Mg c kha nng phn ng tip vi mui

Fe2+ va to th n h to ra st kim loi Dm trn b m t Mg, nh

khi lng khng gim i nh 3 l kim loi cn li.

Vy M l Mg (chn C)

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34/296

!CTJ; H TRC NGHIM

1. Cho 5,6 gam bt st phnngl

IA. Sau phn ng thu c 21,6 gam bc

B. Sau phn ng thu c 32,-11 gam bc

n ht vi dung dch AgN03. Pht biu

c. Sau phn ngthu c toiD. Sau phn ng thu c ti a2 1 ,6

gam bca 32,4 gam bc

2. Cho hn hp bt st, ng o cc ng dung dch HNO 3. Khuy ud phn ng xy ra hon toan c dung dch X v rn Y. Ngm Ytrong dung dch HC1d khng thy c si bt kh. Dung dch X cha:

A. Fe(N0 3>3v C(N03)2 I B. Cu(N03)2v Fe(N03)2

c . Fe(NOs)2v HNOs D. Cu(N03)2; Pe(N Os)2v HNOs

3. Nhng mt l kim Loi M, ha tr II, khi lng 9,6 gam vo dungdch cha ng thi 0,1 moi PeS0 4 v 0,12 mol Fe2{S0 4 >3. Sau mt thi

gian ly M ra cn thy khi lng khng i. Ha tan ht l kim loi Msau khi ly ra vo dung dch HC1 d c 6,272 lt H 2 (ktc). Kim loi Mv s mol FeS 04c trong dung dch sau kh i ly M ra cn, ln lt l

B. Mg v 0,3 mo

D. Zn v 0,4 mol

A. Mg v 0,24 moi

c. Mg v 0,25 mol 4. Cho 3 phn ng sau:

2FeC2+ Cl2----- > 2FeCls !

H2S + I2 ------ 2 HI + s I2FeCl3+ 2 KI ----- 2FeCli+ 2+ 2KC1

T cho bit c bao nhiu plin ng di y xy ra c:

a/ 2HC1 + sb/ 2Fe2+ I2

c/ 2HC1 + I2

d / 2FeCl3 + H 2S

e/ Cl2 + H2S

VCl2+ 2 KI ----

A .2

-> H2S + Cl2

-> 2F el3

2HI + Cl2

----- 2FeCl2

s + 2 HCII2+ 2KC1

B. 5

+ s + HC

C.-4 D .3

5. Cn dng t n h t bao nhiu m dng dch FeC32M oxi ha tonb lng ion Sn2+ trong 300 ml dung dch S11CI2 IM thnh ion Sn4+?

A. 150 B. 300 c . 450 D. 600

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35/296

6 . Kim loi M tan. trong dung dch HC1 to dung dch mui M2+. tkim loi M trong kh clo ri ho tan sn phm vo nc c dungdch mui M4+. Dung dch mui M2+ tc dng dc vi dung dch FeC3to dung dch cha M4+. Vy M l

A. Sn B. Fe c . Pb D. Cr

7. Cho 5,6 gam bt st vo cc ng1dung dch AgN3- Khuy u

phn ng xy ra hon ton c 16,8 gam r n X. Phn trm khilng Ag trong X l?

A. 100% B. 90% c . 60% D. 80%

8 . Cho 5,6 gam bt st vo cc ng 200 m dung dch AgN03. Khuyu phn tLg xy ra hon ton c 29,7 gam rn X. Nng moldung dch AgNC>3 l?

A, 1,250m B. l',125M c . 2,250M D. 1 ,375M' 9. Cho 8 gam rn X gm Cu v Fe20 3 vo dung dch HC1 d. Sau khi

phn ng xong thy en 1 gam rn. Vy %khi lng Cu trong X

A. 25,0% B. 12,5% c . 62,5% D. 37,5%

1 0 . Cho 50 gam rn X gm Cu v Fe0 4vo dung dch HC1 d. Sau khiphn ng xongthy cn 5,6 gam rn. Vy % khi lng Cu trong X l

A. 11,2% B. 19,2% c . 30,4% D. 37,5%

1 1 . Cho rn X gm MgO; Fe 23; ZnO; AI2O3 v Cu vo dung dch HC1d. Sau khi phn ng xong c dung dch Y v rn z. Dung- dch Ycha bao nhiu c ht tan?

A. 3 B. 4 c. 5 D. 6

1 2 . Cho 10 gam bt st vo dung dch HNO 3. Sau khi phn ng xongthy thot ra 2,24 lt NO (ktc) v cn li m gam rn, Bit ch xy ramt qu t rn h kh N +5. Gi tr m l

A. 4,4 B. 1,6 c . 2,8 D. 1,4

13, Cho 23 gam bt st vo dung dch HNO3. Sau khi phn ng xong thytho t r a 6,72 lt (ktc) hn hp NO; NO2c t khi so vi H 2 l 19 v cnli m gam rn. Bit ch xy ra hai qu trn h kh N+5. Gi tr m l

A. 4,4 B . 0,6 c . 2,8 D. 6,2

14. Khi lng st cc di tan c trong 400 ml dung dch HNO 3 2Ml bao nhiu?. B it N +s ch b kh th nh N +2

A. 4,2 B. 5,6 c . 16,8 D. 11,2o 15. Ha tan h t 9 gam bt st trong HNO3 c 3,36 lt NO (ktc) v

dung dch cha m gam mui.?. Bit 1ST5ch b kh th nh N+2Gi tr m l

A. 38,89 B. 35,60 c . 36,9 D. 36,3

35



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36/296

16. Ha tan ht 7 gam bt st trong KNO 3 c 3,36 lt (ktc) hn hpNO; NO2 c t khi so vi X2 19 v dung dch cha m gam mui?Bit ch xy ra 2 qu trn h kh N+5Gi tr rn l

A. 30,25 B. 25,60 c . 26,9 D. 36,3 17. Cho 4 gam rn X gmMg? Fe vo 20Gb dung ch CuCl2. Sau khi

phn ng xong c 6,2gam rn Y. Thm dung dch NaOH d vo phnnc lc, ri thu kt ta nung trong khng kh n khi lng khng ic 4 gam hn hp rn z. Vy nng mo dung dch CuCl2 cho l

A. 0.375M B. 0.5M c . IM D. 2 M

18. Chia rn X gm AI, ZnvCu lm 2 phn bn g nhau:

+ Cho ph n 1 vo dung dch NaOK d, sau ph n ng th y cn12,4 gam rn.

-V- Cho phn 2 vo 500rnl dung dch AgN03IM, sau ph n ng cdung dch Y v m gm rn 2. Thm dung dch NaOH d vo dung dchY, lc ly kt ta nung n khi lng khng i c 8gam rn T

Ch ra gi tr m

A. 62 B. 60 c . 84,4 D. 6

19. Cho m gan?, rn X gm Cu v Fe (t l khi lng tng ng 7: 3)vo 350 ml dung dch HNO3 2M, Sau khi cc phn ng xy ra xong thuc kh NO l sn phm kh duy nht v cn li 0,8m gam rn chatan. Gi tr m l

A. 73,5 B. 32,6 c 43,2 . 50,4

20. Cho m gam rn X gm Cu v Fe (t l khi lng'tngng -3: 1 )vo 400 ml ung dch HNO 3 2 M. Sau khi ph n ng xy. ra xon g .dc5,6 'lt (ktc) hn hp NO; NO2 v cn li 0,8m gam rn cha tail! Cho

bi t phn ng ch xy ra 2 qu trnh kh N+5, vy gi tr m l:

A. 77 B. 50,4 c . 61,6 D. 82,4

21. Cho 10,8 gam bt nhm vo 300 ml dung dch FeC 3 2M. Sau khiphn ng xong, khi lng dung dch tng hay gim bao nhiu gam?

A. Gim 11,6 gam B. Tng 2fl gam

c . Gim 6,0 gam D. Tng b ,l gam

22. Cho rn X gm Al, Zn v Cu vo dung dch AgNCV Sau khi phnng xong c hn hp rn Y v dung dch z. Cho dung dch NaOH dvo dng dch z c k t t a T. Vy y gm:

A. AI, Zn, Cu, Ag B. Zn, Cu, Ag

c . Cu, Ag D. Zn, Ag

36



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37/296

23. Cho hn hp l, Fe v Cu vo cc ng ung dch HNOs iong. Saukhi cc phn ng xy ra xong c rn X v dung ch y. Rn Xkhng thy si bt kh khi cho vo duiig dch HC1 d. Thm dung dch

NaO H d vo dung dch Y c kt ta z. Nung 2 trong khng kh nkhi lng khng i c rn T gm

A. CuO; Fe20 3 B. A2O3 v Fe23

c. AI2O3; Fe203 v CuO . Cjhc Fe2 0 3

24. Cho dung dch AgN0 3 d vo V m ung dch FeCl2 2 M. Sau phnng thu c 39,5 gam rn. Chi ra gi ;r V

A. 30 . B. 45 c . 5 D. 100

25. C bao nhiu tc nhn di y khc ion Fes+ trong dung dch?

1. Fe 2. Ag+ 3. Cij

4. r 5.Cr 6 . Si2+

A. 3 B. 4 c .

26. Cho 3 ph n ng sau: - 2 FeC3 + 2 K ------ 2FeCl2+ 2+ 2 CC1H2S + 2 ---- > 2HI + s

2_beCl2 + C2 ---------- ^ 2 FeC3

Phn ng no di y xy ra c?

A. I2 + 2HC1 ----- > 2 HI +CI2

B. s + 2HC1 ----- > H2S + Cl2

c . 2 FeCl#V H i S 2 F e C l 2^ S 2HC1

D. 2

D. 2FeI2 + I2 2FeI,

27. Cho hn hp bt Mg, Zn vo ung ch hn hp AgNC>3 vCu(N0 3 )2. Sau phn ng thu c dung dieh Y (cha 2 cation) l

A. Mg2+, Ag+. B. Ag+, Zn2+. c . Zn3+, Cu2+. D. Mg2\-Zn2+.

28. Cho hn hp bt Mg, Zn, AI voCu(N0 3 )2. Sau phn ng thu c dung

A. Mg , Ag+, Cu2+, Zn

c. Mg2+, Zn , AI3+, Cu2+

dung dch hn hp AgN03 vdch Y (cha 4 cation) l

B .A g , Cu2+,Z n 2\.A I3+

D . Z n 2YCu2\-A g+, Mg2*

29. Cho hn hp bt Fe, AI vo dung dch hn hp gN03 v Cu(N0 3 >2.Sau phn ng thu c dung dch Y (cha 2 cation) l

A. Als+, Ag+. B. Ad\ Fe3+ .2+ c. AI3+ Cu2+ D. Fe2+, Cu2+

37



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38/296

LJ o. V>no I1UI1 iiup Juu ivi-g, lu vau (auiig U-L\ LYJb' rx5i-'"~'z tu VU\A>'Sau phn ng thu c rn Y (cha 2 kim loi) l

A. Mg, Ag. B. Ag, Cu. C. Ag, Zn. D. Zn, Cu.

31. Cho hn hp bt Mg, Zn, AI vo dung ch hn hp AgN0 3 vCu(NC>3)2. Sau ph n ig tliu c r n Y (cha 3kim loi) l

A. Mg, Ag, Cu B. Ag, Cu, Zn c . Ag-, Zn, AI D. Zn, Cu, Mg

32. Cho 2,8 gam bt st vo 60 ml dung dch AgNC>3 2 M. Sau khi ccphn ng xy ra xong th

A. 10,80 B. 16,20 khi lng bc thu c l bao nhiu gam?c. 12,96 D. 12,4

33. Cho 5,6 gam bt st vo 160 ml dung dch gN03 2M. Sau khi ccphn ng xy ra xong th khi lng bc thu c l bao nhiu gam?

A. 32,40 B. 16,20

34. Cho 5,6 gam bt st \phn ng xy ra xong thu

A. 20,00 B. 19,44

c . 12796 D.21,60o 90 ml dung dch AgN03 2M. Sau khi ccc rn X c khi lng l

c . 32,40 D. 21,60

HNG DN GII

1. Trc ht xy ra ph n ng:

Fe + 2AgN03 ------> FeN03)2 + 2Ag

0 ,1 mol I 0 ,2 mol

Lng Ag to ra trn lvn cn d, s xy ra tip phn ng:

AgNC>3 + Fe(N03>3

0 ,2 mol tc 21 ,6 gam. Tuy nhin nu AgN3

Fe(N0 3>2 + AgLng bc sinh ph n togny ty thuc vo Ag N03cn d nhiu hay

t, nhng ti a ch l 0,1 mol tc 10,8 gam. Vy sau khi cc phn ngxy ra xong, thu c ti a 32,4 gam bc (chn D)

2 Ngm. Y trong dung dch HC1 d khng thy c si bt kh cho thyY ch l Cu, vy Fe phan ng ht.

V Cu cn d nn dung dch X khng th c HNO3

V Cu cn d nn Fe(N03>3 trong X chuyn h t th n h Fe(N03>2theo

phn ng: ICu + 2Fe(N03)3 ------> 2Fe(N03)2 + Cu(N03)2

Vy X gm Fe(N 03)2v CiKN03)2(chn B)

3. Ch rn g Fe3+ d b h thnh Fe2+ so vi Fe2+ b kh th nh Fe.Do trc ht xy ra phan ng:

38



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39/296

M + Fe2(S04)3 ------> MSO4 + 2FeS04. 0,12 mol 0 ,12 mol 0,24 mol

Nhung nu ch xy ra phn ng ny th l M lc ly ra cn phi ckhi lng gim. Theo khi lng l M khng i chng t phi xy ratip phn ng:

M + F eS 04 ----- MSO4 + Fea mol a mol a mol

9 6L M lc ly ra cn gm c: ( -0 ,1 2 -a ) mol kim loai M v aM

mol Fe bm vo

V M v Fe u tc dng vi HC1 gii phng H 2 theo t l mol 1: 1 nnta c:

( - 0 , I 2 -a ) + a = ^2 2 1 = 0,28 c=>M = 24(Mg)M 22,4 s

M t kh c khi lng Mg khng i lc ly ra cn nn ta phi c:

tan ra = bm vo 24(a + 0,12) = 56a a = 0,09Vy s mol FeS 04tron g dung dch sau khi ly l Mg r a = 0,24 + 0,1 - a

= 0,25 mol (chn C)

4. T 3 phn ng cho, ta c th t cc cp oxi ha kh trong dyin ha l:

s I2 Fe3+ Cl,s 2' 2 1 ~ Fe2+ 2c r

p dng qui tc (Xth c 3 phn ng sau y xy ra c (chn D):

2FeCl3+ H2S -----> 2FeCl2 + s + 2HCCl2 + H2S -------- s + HCCl2+ 2 KI ------> l + 2KC1

5. Ta c ph n ng:

Sn2+ + 2Fe3+ -> Sn4+ +2Fe2+

0,3 moi 0,6 moi

=> v = 0,3 lt (chn B) .2

6 Ta c th t cc cp oxi ha kh trong dy in ha l:Sn2* 2HT Sn4+ Fe3+

Sn H2 Sn2* Fe2"

Nh vy Sn tan c trong dung dch HC1 to Sn2\*

Sn + 2H+ ------ > Sn2+ + H2

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40/296

Sau Sn2" b Fe3+ oxi ha thnh Su4-:

SnJf -f 2 Fc ^ > Sri4 + 2 Fc"+

Vy M l Sn (chn Aj

7. rng nu 0,1 mol Fe phn ng ht th rn thu c phai nngt nht 2 1 , 6 gam. Rn v ch nng 16,8 gam chng t Fe ch phnng mt phn theo phng trnh:

Fe + 2 AgN0 3 ------> Fe(N0 3)2 + 2 Aga mol 2 a mol

=> 56(0,1 - a) + 108.2a + 16,8 a = 0,07

1OH.2.0,07.100 ' / I EM=>%Aff = ---------------- = 90 (chn B)16,8

8 . rng nu 0,1 mol Fe phn ng ht th rn thu c phi nngt nht 2 1 , 6 gam. Rn y nng n 29,7 gam chng t Fe phnng ht theo phng trnh:

Fe + 2AgN03 ------> Fe(N03)2 + 2 Ag

0 , 1 moi 0 ,2 mol 0 ,2 mo

Sau o AgN03 cn d nn xy ra tip phn ng:

AgNOs + F(N0 3)3 > Fe(N 03)2 + Ag

a mol a mol

=> 0,2.108 + 108a = 2S,7 o a = 0,075

0,2 + 0,075

9. Gi a l s mol Fe2C>3 c trong X, ta c phn ng:

Fe20 3 + 6HC1 ------ 2FeC]s + 3H20

a mol ' 2 a iol

Cu + 2FeCl3 ----- CuCl2 + 2FeCl2

a mol 2 a mol

=> 64a + 160a +1 = 8 c=> a = 0,03125

_ fl/ri (64.0,03125 + 1)100 __ _ / u=> %Cu = --------------------- = 37,5 (chn D)8

10 . Gi a s' mol Fe30 4 c trong X, ta c phn ng:

Fe30 4 + 8HC1 ------> FeCla + 2FeCl3 + 4H20

a mol 2 a mol

Cu + 2FeCls ------> CuCls + 2FeCl2

a mol 2 a mol

^ , i i T U , u / j ,CAgNO = -------- ----- = 1,375M (chn D)

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41/296

64a + 232a + 5,6 = 50 a = 0,15%Cu= ( 6 4 , 0 1 5 ^ 1 0 0 ^

8(chn C)

1 1 . V r n z l Cud nn FeCls b kh i th nh FeCU.

Vy dung dch Y cha: MgCl2; FeCl2; CuCl2;A1C13;ZnCi2v H.C d,tc gm 6 cht tan (chon D)

1 2 . Cc phn ng xy ra;

Fe + 4 HNO3 ------ Fe{N03)3 + NO * 2H20

0,1 mo 0,1 mol 0,1 jmol

Fe + 2F e(N 03)3----- ->3Fe(NOs

0,05 moi 0,1 mol i

=> 56.0,15 + m = 10 o m = 1,6 gam (chn B)

C c h k h c : V Fe cn d nn sn phm thu dc l Fe2+. Theo nh lut

bo to n electron th :CHO

Fe ------ Fe2+ + 2e

0,15 mo 0,3 mi

N+5

NHN

+ 3e ------> N+2

0,3 moi 0,1 moi

=> m = 10 - 56.0,15 = 1,6 gam !

13. Bng phcmg php ng cho tm dc nNO= nNO> - 0,15 mol

V Fe cn d nn sn phm thu .c l Fe2+. Theo nh lut bo tonelectron th: ;

Fe ----

0,3 mol

CHO

Fe2+ + 2e

0 ,6 mol

N'

N"

=> m = 23 - 56.0,3 = 6,2 gam

14. rng kh i s t tan cc i trong H

3Fe + 8HNO3 3Fe(N0 3)2+ 2N0,3 mol 0,8 mol

=> mF(max) = 56.0,3 = 16,8 gam

15. Cc phn ng xy ra:

Fe + 4HNO3 ------> Fe(N3 )3 + N

0,15 mol

o + 2 H 20

0,15 mol 0,15 xno

NHN

*5 + 3e ----- -> N+2

. 0,45 mol 0,15 moi

+ le -------> N+4

0,15 moi 0,15 moi

(chn D)

N03 th sn phm phi l Fe2+:

o + 4H2O

(chn C)

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42/296

Fe

( -0,15)56

+ 2Fe(JMOs)3 -

2 ( -0 ,15)56

3Ke(N03)2

3 ( ^ - 0 , 1 5 )56

mmui = 180. 3 ( ~ - 0,15) + 242[0,15 - = 36,9 gam56

C ch kh c Dng cng thc

NHANH TRC NGHIMNXBHQGH H Ni 2010)

3,36= 9 + 62.3.

15

56(chn C)

gii nhanh (xem CC CNG THC GII

1HA HC, tc gi Nguyn nh -

36,9 gam

16. Dng cng thc gi inr ian h (xem t i liu nu, cng tc gi) th

^Mu'i ^Kim Jo;ii + 62.(3.nN+ nNOi + 8 .nN>0+ 10.nN)

Vy m mui = 7 + 62(3.0,075

(khng to kh. no thi s mo kh bng khng)+ 0,075) = 25,6 gam (chn B)

17. Mg phi phn ng trc Fe. Nu ch c Mg phn ng th z ch lMgO, tri vi gi thit z l hn. hp. Vy Fe c phn ng. Tuy nhinnu c Mg v Fe u phn ng ht th ta phi c mz > 4 gam (tri githit). Do , Mg phn ng ht, Fe ch phn ng mt phnGi a, b l s mol Mg v Fie phn ng, ta c h:

(64(a + b) + [4 - (24a + 56b)] = 6,2. \ os

40a + 160= 4I 2

b = 0.025

0,075(chn A)=>ncuci, = a + b = 0,075. V CCuCl = - ^ ~ = 0,375M

18. Theo th nghim 1 , mi phn c 12,4 gam C

Thep th nghim 2, AI v Zn phn ng ht, sau c c Cu phn ngth ung dch Y cho NaOH .vo mi c kt ta. Tuy nhin vi IIT = 8 gamcho thy Cu ch mi phn ng 6,4 gam. Do z gm 0,5 mol Ag tc 54 gamAg v 6 gam Cu d, tc ,mz = 60 gam (chn B)

19. Theo , X gm 0,7m gm Cu v 0,3m gsm FeFe phi phn ng trc Cu. V sau phn ng c 0,8m gam rn

chng t Fe mi phn ng 0,2m gam. Vy Fe vn cn d sau phn ng,Cu cha phn ng, v mui thu c ch l Fe(NQ3>2. Vy ta c phn ng:

3Fe + 8 HNO3 SFe(N03)2+ 2 NO + 4H200,2625 mol 0,7 moi

=> 5 ^ - = 0,2625 m = 73,5 gam56

(chn A)

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43/296

L.I is. I neo ae , A gom u, /om gam (Ju v 0,2 5m gsm Fe

Fe phi phn ng trc Cu. V sau phn ng c 0,8m gam rn chngt Fe mi phn ng 0 ,2m gam. Vy Fe vn cn d sau phn ng, Cu cha

phn ng, v mui th u c ch l - m Fe(N 03)2.56

in h lu t b o to a n N cho n N/HNOj = flN/Fe(N0 ,)-> + n N/No + n N/NOj

^ n HNO, = 2 nFc(N0 ,);> + n N0 + n N0 ,

_ _ O 2.0,2m 5,6 0,8 = ~ + -- 2, nnr n T gm CuO v Fe 23 (chn A)

24. V ml dng dch FeC2 2M c 0 ,0 0 1 V mol Fe2+ v 0 ,0 0 2 V mol Cl'

V Ag+d nn Fe2+ v c r ph i phn ng ht:

Ag+ + c r ----- > AgCl0,002Vmol 0,002Vmol

Ag+ + Fe2+ ------> Ag + Fe3+

O.OOlVmol 0,001Vmol

Vy 143,5.0,002V + 108.0,001V = 39,5 o V = 100 (chn D)

25. Cc ph n ng xy ra:

Fe + 2Fes+ ----- >3Fe2+

Cu + 2Fes+ ------ 2Fe2+ + Cu2"

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44/296

2 " + 2 2Fe2t + l2

Sn2' + 2Fe+ ----- > 2Fe2+ + Sn4+

Vy c 4 tc nh n kh c Fe3-* thnh e'z* (chn E)

Q 26. T 3 phn nE cho, ta c th t cc cp oxi ha kh trong dyin ha :

s F c ^ CU

S- 21' Fc 2t 2Cp dng qui tc a th ch c phn ng sa y xy ra c (chn C)

2FeC3+ I2S i> 2FeCI2+ 's + 2HC1

27 Vi dng ny, dung dch sau phn ng lun cha cation ca cc kimloi c tnh kh mnh nht trong s cc kim oi c lin quan trong bi

=> 2 ca tion cn tm l Mg2*, Zn2+ (chn D)

28. Theo nguyn tc trn, 4 cation cn tm l Mg2+, Zn'2+, Al3+, Cu2+

(chn C)

29. Theo nguyn tc trn, 2 cation cn tm A3+, Fe2* (chn B)

30. Vi dng ny, cht rn sau phn ng lun cha cc kim loi ctnh kh yu nht trong s' cc kim lo i c lin quan trong b i

=> 2 kim loi cn tm l Ag, Cu (chn B)

31 Theo nguyn tc tr n, 3 kim loi cn tm l Ag, Cu v Zn (chn B)

D 32. Ta c npe = 0,05 mol; nA.,NO = 0,12 moi

Cc phn ng xy ra;

Fe + 2Ag+ > Fo 2+ + 2Ag .0,05 mol 0,1 mo 0,05 moi 0,1 mo .

Sau phn ng ny, Ag+cn 0,02 moi s tc dng vi 0,05 mol Fe2+:

Fe2+ + Ag+ > Fc+ + Ag

0 ,0 2 mol 0,02 moi

Vy mAg = 108.0,12 = 12,96 gam (ehn C)

N h n xt: Ton v Fe tc dng vi dung dch Ag* th:

n

+ < 2: Rn sau phn ng gm Fe d v Ag, vi IA - n VnFc '*

n .+ - > 3: Rn sau phn ng ch l Ag, vi nA =3nK.

n,c

nA. -r 2 < - < 3: Rn sau ph n ng ch l Ag,- vi n. . =n

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45/296

33. Ta c nFe = 0,1 moi; nAyNOi =0,32 ioi

V 3 n n nA -3 n,.c = 0,3 mol.

Vy iAg = 32,4 gam chn A)G 34. Ta c nFe = 0 , 1 mo; nAt;NO =0,18 m

!n . '

V < 2 nn Fe cn d sa phn ng.n t.-.Rn. X gm 0,18 mol Ag v (0,1 - 0,09) = 0,01 mol Fe dVy mx = 108.0,18 + 56.0,01 = 20 gain (chn A)

I

o CHUYN B 5 "

%T O N V T H I S TRONG B K H c sA. NGUYN TC .

Cng thc t rung b nh l cng thc i din cho mt hn hp. y cch thng ng gii cc bi ton hn jhp cc c h t hu CO' c ng h o c khn g cng dy ng ng. :

Hn hp cc cht hu c cha c, H (hoc t , H, 0 hay c, H, o, N) s c:

Tr s" c trng bnh = - c: ; Tr 's H trung b nh = ________________________nhh ' __________ rchh

(nu cng iu kin nhit , p sut, c th thay s mol bng th tch)

- M n h p c e c h t h u c c n g d y n g n g s c c n g

th c t r u n g b n h t r n g v i c n g t h c c h u n g c a d y - ig n g .-:... Mon h p c e -a n c o C M ), CHeO; CsH sO s ::&:eng Jc -

t ru n g h h h ia Cn-2n *20,Hn hp cc cht hu c khc dy ilg ng s c cng thc trung

bnh khc vi cng thc chung ca tng cht trong hn hp. Tuy vy, nup h n t c a chng c cng s nguyn t ca m t nguyn t th cng thctrung bnh s gi nguyn s nguyn t ca nguyn t ny. V d hn hpcc cht hu c C2H4Q2; C3H4O2; C4H8O2; C5I12O2 s c cng thc trung

bnh l CxHy0 2

, v t t nh in l ta phi c i < x


46/296

B. CC V D

V d 1 : t chy hon ton 0,25 moi hn hp 2 este cng dy ngngc H2O v 0 ,6 mol c*'Hm khi lng hn hp este t.

Gi

Ta c s c trung bnh = 0,60,25

= 2,4

c mt este c 2C, este ny ch c th l HCOOCH3

chng l 2 este n chc no, cng thc trung bnh l CnH 2n02

mhBa hp = 0,25(14n + 32) = 0,25(14.2,4 n+ 32) = 16,4 gam

V d 2: t chy hon ton 0ng ng c m gam nc

Ta c s c trung bnh = ,|0 ,12I5

,125 moi hn hp 3 hirocacbon cng dyv 8 ,8 gam c o 2- Tm m,

G i i

= 1,6

c mt hirocacbon c 1C, hirocacbon ny eh c th l CH4chng l 3 an kan, cng c trung bnh l CDH2n +2

Hu rt H"n,H-O **ankai> 1 CO->

L u : Hn hp ankan hoc ete

=0,125 0,2 = 0,325. Vy m = 18.0,325 = 5,85 gam

no, hoc ancol no lun chy eho

HjO - nhh + nco.

V d 3: i; chy hon ton |o,35 mol hn hp 3 axit cacboxylic ngng cn va V lt 0 2 (ktc). Hp th ht sn phm chy vo nc

ta. Tm V

G i i

vi trong d c 63 gam kt

Ta c s c trung bnh = - 1,80,35

c mt axit cacboxylic c c, axit ny ch c th l HCOOHchng l 3 axit cacboxylic n chc no, cng thc trun g bnh l CnH2n02

C H 2 n0 2 + ( ^ r ^ ) 0 2 J > 1 C O 2 + 1H2O

0,35mol 0,3 5(

2

3 n - 2)mol

/ 3.1 ,8-2 .V = 22,4.0,35. ( ) = 13,328 lt

V d 4; X l hn hp gm ankn Y v ankin z e cng s H trong phnt. t chy hon ton 0,25 |mol X c H20 v 30,8 gam CO2. Tnhkhi lng X t.

46



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47/296

GiiV c cng s H trong phn t nn ankin phi hn ankan 2C

Ta c s' c trun g bn h = = 2,80 , 2 5

=> ankan l C2He; ankin l C4H6

=> chng c cng thc trung bn h l c 2sH6

=> mx = 0,25(12.2,8 + 6 ) = 9,9 gamV d : X l hn hp gm C3H4; C3H6; C4H 10 v hirocacbon CxHy. t

chy hon ton 2 lt X c 5 lt C0 2v 3 lt hi nc (cc th tch ocng iu kin). Cng thc ph n t ca CxHy l: .

A. CH4 B. C2H2 c. C2 H4 D. c 4h 6

Gii

Ta c s c tru ng b nh = = 2,5 X. 2,5)2 3

s H trung bnh = = 3 => y < 3 (v cc cht cn li u c s H > 3)

Vy CxHy l C2H 2 (chn B)

V d 6 : X l hn hp gm C2H6; C36 v C4H6j c t khi so vi H2 24. t chy hon ton 0,1 mol X cn ti thiu bao nhiu mol O2?

Gii t cng thc tru ng b nh 3 hirocacbon l CXH6

Ta c 12x + 6 = 24.2= 48 X= 3 ,5 .Vy chng c cng thc trun g bnh l C35I6

C3,5H6 + 5 O2 ----- > 3,5C02 + 3HaO

0,1 moi 0,5 mol

Do cn ti thiu 0,5 mol O2

V d 7: X l hn hp 2 este n chc (to bi cng mt anco vi haiaxit lin tip trong y ng ng). t chy hon ton 35 gam X

c 1,9 mol CO2 v 1,3 mol H

2O. nh cng thc phn t

2 este v %khi lng mi este trong X.

Gii

t cng thc 2 este trn l RCOOR v RCH 2COOR ha y

Suy ra cng thc 2 e ste c th vit CxHyO v c x +iHy +2O2.

Vy ch ng c cng thc t rung b nh l CUHV0 2

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48/296

;i{' 2 u + V + 32) - 35 ia 0.3

Gi a l smoi 2 ese, ta c h: Ha ' B. CH4 c . C3Hg D. C2X4( i li c k h i B / 2 01 0)

Gi iX chv phi cho nco ~ n H V , m theo th hn hp M chy cng cho

nco - n H() n n Y phi chy cho nCOi = nH(), vy l o i A v B

Ta c s" c tru ng bnh = 7= 2 . Nu chn. Y l C3H6 th khi X phi

KCHO v hn hp khi ny mun c sc trun g bn h l 2 th ta phi c

rx = P-Y, tri vi gi thit l nx < iy, l o i .Vy Y l C2H 4 ' : (chn )

Vi d 9: t chy hon ton m gam hn hp icol n chc cng dyng ng c 3,808 lt C02Cktc) v 5,4 gam HO. Gi tr m lA. 5,42 B. 4,72 c. 7,42 . D. 5,72

(i h c kh i A / 20X0)

G i iV nC(X = 0,17 < nHio =0,3 nn y l 3 ancol n chc no, cng thc

trung bnh l Cn 2n +lOHGi a l s' moi hn hp, ta c phn ng:

CnI i2ll + jOH + 2 ------> nCOs + (n + DH2O

a an a(n + 1)

Vy m = a(14n +18) = 18a(n + 1) - 4h a =1 8.0 ,3 - 4.0,17 = 4,72 gam(chn B)

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49/296

C U H I T R r N G H I K M

1 . t chy hon ton mt lng hn ip X gm 2 ancol (u no, achc, mch h, c cng s nhm OH) cn va V lt 0 2, thu c1 1 , 2 lt C2 v 1 2 , 6 gam H20 (cc th tch o kc). Gi tr ca V l

A. 14,56 B. 15,68 c . I 1 J20 . D. 4,48( h c kh' B / 2010)

2 . t chy hon ton 1 lt hn hp . kh gm C2H2 v hirocacbon Xsinh ra 2 lt kh GO2v 2 lt hi H 2O. Cong thc phn t ca X l

A. CH4. B. C2H6. c . C3Hs. D. C2H4.

(i hc kh i B / 2008)

3. Cho m gam hn hp X gm hai ancol no, n chc, k tip nhautrong dy ng ng tc dng vi CuO (d) nung nng, thu c mthn hp rn z v mt hn hp hi Y (c t khi hi so vi H 2 l13,75). Cho ton b Y phn ng vi mt lng d Ag20 (hoc AgNC>3)

trong dung dch NHS un nng, sinh ra 64,8gam Ag. Gi tr ca m l

A. 7,8 B' 8,8. c. 7,4. D. 9,2.(i hc khi B / 2008)

4. Hn hp X gm hai ancol no, an chc, mch h, k tip nhau trongdy ng ng. Oxi ho hon ton 0,2 mol hn hp X c khi li^g mgam bng CuO nhit thch hp, thti.c hn hp sn phm huc Y. Cho Y tc dng vi mt lng d Iung dch AgN3 trong NH3,thu c 54 gam Ag. Gi tr ca m l I

. 15,3 B. 8,5 c . 8,1 X). 13,5

I (i hc khi B / 2009)

5. Hir ho hon ton m gam hn hp X gm hai anehit no, cmchc, mch h, k tip nhau trong dy Ing ng th u c (m + 1)gam hn hp hai ancol. Mt khc, khi t chy hon ton cng m gamX th cn va 17,92 lt kh O2 ( ktc), Gi tr ca m l

A. 10,5 B. 17,8 c . 8,8 D. 24,8

(i h c kh i B / 2009) 6 . t chy hon ton 5 lt hn hp kh gm 2 anken lin tip cn 18 ltc>2 (o cng iu kin).Cng lng- hn hp kh ny nu hirat ha

. hon ton c hn hp aneol z trong cacbon t hn l:

A. 60% B. 53,48% c . 72%

% khi lng ca ancol c s

D. 33,33%

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50/296

7. t chy hon ton 0,25 moi hn' hp X gm mt anken v mtankin (cng s' H trong phan t) c H 2O v 35,2 gam CO2. T khihi ca X so vi H 2 l

A. 18,8 B. 22,2 c.'21,6 D. 24,4 8 . t chy hon ton 1 lt hn hp kh gm C2H2 v hirocacbon A

c 5 lt hn hp kh v ni c t kho'i so vi H 2 l 15,5 (cc th tch,o cng iu kin). Vy cng thc phn t ca A l:

A. CsH6 B. C2H6 I c . C3I8 D. CH, 9. Hn hp kh X gm etylmin v hai hirocacbon ng ng lin tip.

t chy hon ton V m hn hp X bng mt lng oxi ya , thu cVi ml hn hp Y gm kh hi nc. Nu cho Y i qua dung dch axitsunfurie c (d) th cn li v 2 ml kh (cc th tch kh v hi o cngiu kin). Bit V = Vj - 2V2. Vy cng thc phn t ca hai hirocacbon

A. C3H2 v C3H4 ; B. CsH6v C4HS

c . CH4v C 2H6 D. C2H4 v C 3H6

1 0 . Hn hp kh X gm irnetylamn v h i hirocacbon hg ng lintip. t chy hon ton xo mhn hp X bng mt lng oxi va ,thu c 550 ml hn hp Y gm kh v hi nc. Nu cho Y i qua dungdch axit sunfuric.c (d) th cn li 250 ml kh (cc th tch kh v hio cng iu kin). Cng thc phn t ca hai hirocacbon l

A C2I6v c$$ I B. C36v C4I8

c . CH4 v C2H6 D. C2H4 v C3H6

(i hc kh i A / 2010)

. X l hn hp gm amn A (ng ng-ca metylamin) v amin B(ng ng ca anilin). [t chy hon ton 22 >5gam X b ng mt lngO2 va c H2O; 1,3 riol C O2 v 0,15 mol N2. .

Cho bit cng thc phn t ln lt ca A v B?

A. C3H9N v C7H9N B. C3H9N v CsHn N

c . C4H11N v C7H9N D. C5H13N v C7H9N 12. t chy hon ton 2 amin n chc no, mch h l ng ng

lin tip c N2, C02 v H20 trong nCQ2 :n Hi0 = 1:2. Phn trm

khi lng m t .ainm trong hn hp lA. 59,21 B. 43,27 c . 55,30 D. 50,00

13. X l hn hp gm metan, eta n v propan , c t khi hi so v1 s < w o ' I A _t_ _7m. _ .L._____ J. 1 A1 V ImA A IkA.Inl 17,8. t chy hon tcchy vo nc vi trong dhay gim bao nhiu gam?

n 0,1 mol X ri hp th ton b sn phm th sau hp th, khi lng dung dch tng

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51/296

A. Tng 24 gam. B Gim 7,32 gam

c . Gim 16,68 gam D. Tng 30,12 gam

. t chy hon ton ,gam X gm 2 mui natri ca 2 axit ankanoiclin tip nhau trong dy ng ng thu c Na2C03j H20, v 0,085 molC02. Cng thc ca 2mui l:

A. C2H5COONa v C3H 7COONa. B. HCOONa v CH3COONa.

c . C4H9COONa v C5H u COONa. D. C3H7COONa v C4H9COONa.

15. t chy 10 lt hn hp gm 5 hircacbon th kh l C4H 10; C3H8;C3H4; C4H6 v CxHy c 22 lt CO2 v 14 t hi nc (cc th tch o cng iu kin). Vy CxHy l

A. C4H2 B. CH4 c . C2H2 D. C3H4 16. X l hn hp 2 ancol n chc cng dy ng ng (t l khi lng

1 : 1 ). t chy hon ton X c 21,45 gam CO2 v 13,95 gam.H 20.Vy X gm 2 ancol c cng thc p hn t l

A. CH40 v C2H60 B. CH4O v C3H80

c . C2HeO v C3HsO D. C2HeO v C4H10O 17. t chy hon, ton 7 lt hn hp kh X gm 2

cÁc chuyÊn ĐỀ quan trỌng trong hÓa hỌc - nguyỄn ĐÌnh ĐỘ

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