Gio n dy thm 12 GV: Hong Trng

Bui 1 Ngy son: 07/09/2014Este

A. Mc ch yu cu.

1. V kin thc

Hc sinh nm vng cc khi nim trong chng, tnh cht ha hc ca este v lipit, phng php iu ch.

2. V k nng.

Hc sinh vit c ng phn, danh php ca este.

Phng php gii cc dng bi tp v este v lipit.B. Kin thc cn nm vng.

I.Este.

1.Khi nim- Danh php

a.Khi nim v este:

+Khi thay nhm OH nhm cacboxyl ca axit cacboxylic bng nhm OR th c este.

+Este l sn phm ca phn ng gia axitcacboxylic vi ancol.

CTPT ca este n chc: RCOOR

Trong : R l gc hidrocacbon hoc H

R l gc hidrocacbon

Ch : axit v ancol tng ng vi ctpt trn l: RCOOH v ROH.

CTPT ca este no n chc mch h:CnH2nO2 (v y cng l ctpt ca axit no

n chc mch h).

b.Tn gi=tn gc hidrocacbon R+tn anion gc axit(ui at).

Vd: CH3COOC2H5 etyl axetat HCOOCH3 metyl fomat

C6H5COOCH3 metyl benzoat CH3COOCH2C6H5 benzyl axetat

2.Tnh cht ha hc ca este.

Este l mt loi hp cht hu c bao gm c phn gc v chc nn s th hin tnh cht ha hc hai phn :

a. Phn ng nhm chc.

+ Phn ng thy phn

-Mi trng axit:

RCOOR + H2O RCOOH + ROH

-Trong mi trng kim:

RCOOR + NaOH RCOONa + ROH

Mt s ch trong ptpu thy phn este:

+ Mt s este khi thy phn sn phm thu c khng phi l mui v ancol m c th thu c sn phm khc:

Thy phn este ca phenol cho ta sn phm l hai mui v nc

RCOOC6H5 + 2 NaOH RCOONa + C6H5ONa + H2O

-Khi thy phn mt s este cho ta sn phm l ancol khng bn v chuyn ngay thnh cht khc

Vd: RCOOCH=CH2 + NaOH RCOONa + CH2=CH-OH

CH2=CH-OH CH3CHO

V t ta xc nh c cu to ca este

b. Phn ng kh

RCOOR RCH2OH + ROH

c. Phn ng gc hidrocacbon khng no

-Phn ng cng vo gc hidrocacbon khng no

-Phn ng trng hp

Vd: nCH2=CH-C-O-CH3 (- CH-CH2-)n O COOCH3

II. LIPIT

1. Khi nim: Lipit l trieste ca glyxerol vi cc axit bo

Vd: C3H5(C17H35COO)3 tristearin

Mt s loi axit bo thng gp:

C17H35COOH Axit stearic

C17H33COOH Axit oleic

C17H31COOH Axit linoleic

C15H31COOH Axit pamitic

2. Tnh cht ha hc ca lipit

V lipit l mt loi este nn n c y tnh cht ca mt este

a. phn ng thy phn( Phn ng x phng ha)

ptpu tq C3H5(RCOO)3 + 3NaOH C3H5(OH)3 + 3RCOONa

Ch : T l s mol 1:3:1:3

Cc bi ton v sn xut x phng thng s dng ptpu ny nn i lc ta cn nh t l ny gii ton nhanh hn.

b. Phn ng cng hidro

lipit khng no(lng) + hidro lipit no(rn)

B. Bi tp

Dng 1: GII TON ESTE DA VO PHN NG T CHY.

Este no n chc mch h:

CnH2nO2(s mol CO2 = s mol H2O.

Este khng no c 1 ni i, n chc mch h:

CnH2n-2O2(s mol CO2 > s mol H2O v neste = nCO2 nH2O.

Este no 2 chc mch h:

CnH2n-2O2(s mol CO2 > s mol H2O v neste = nCO2 nH2O.

Bi tp minh ha:Bi 1: t chy hon ton 1,48g este A thu c 2,64gCO2 v 1,08g H2O. Tm CTPT ca A ?.

Hng dn gii:

Ta c: nCO2 = 0,06 mol; nH2O = 0,06 mol.(A l este no n chc mch h.

PTP. CnH2nO2 + O2 ( n CO2 + nH2O.

(mol) 0,06 0,06.

( (14n + 32) = 1,48. ( n = 3 ( CTPT A l: C3H6O2.

Dng 2: GII TON ESTE DA VO PHN NG X PHNG HA.

1.X phng ha este n chc:

- Tng qut: RCOOR/ + NaOH RCOONa + R/OH.

Cht hu c A khi tc dng vi NaOH, trong sn phm c ancol ( A phi cha chc este.

Lu :

-Este + NaOH 1 mui + 1 anehit este ny khi phn ng vi dd NaOH to ra ancol c OH lin kt trn C mang ni i bc 1, khng bn ng phn ha to ra anehit.

RCOOCH = CH2 + NaOH RCOONa + CH2 = CH- OH.CH3CHO.

Este + NaOH 1 mui + 1 xetoneste ny khi phn ng vi dd NaOH to ra ancol c OH lin kt trn C mang ni i bc 2, khng bn ng phn ha to ra xeton.

-Esste + NaOH 2 mui + H2O (Este ny c gc ancol l phenol hoc ng ng ca phenol

RCOOC6H5 + 2NaOH RCOONa + C6H5ONa +H2O.

2. gii nhanh bi ton este nn ch :

* Este c s C 3 hoc este M < 100 (Este n chc.

* Trong phn ng x phng ha: Este + NaOH mui + ancol.

+ nh lut bo ton khi lng: meste+ mNaOH = mmui + mancol.

+ C cn dd sau phn ng c cht rn khan, ch n khi lng NaOH cn d hay khng?3.Bi tp minh ha:

Bi 1: t chy hon ton 1 mol este X thu c 3 mol kh CO2. Mt khc khi x phng ha 0,1 mol este trn thu c 8,2g mui cha Natri.Tm CTCT ca X?

Hng dn gii:

t 1 mol este (3 mol CO2(X c 3C trong phn t (X l este n chc.

Gi cng thc tng qut ca este l: RCOOR/.

PTP. RCOOR/ + NaOH ( RCOONa + R/OH

(mol) 0,1 0,1.

Ta c: Mmui = = = 82(MR + 67= 82(MR = 15(R l CH3(R/ phi l CH3(v X c 3 C). Vy CTCT ca X l: CH3COOCH3.

Bi 2:Thy phn 4,4g est n chc A bng 200ml dd NaOH 0,25M (va ) th thu c 3,4g mui hu c B. Tm CTCT thu gn ca A?

Hng dn gii:

Ta c nNaOH = 0,2. 0,25= 0,05mol.

PTP. RCOOR/ + NaOH (RCOONa + R/OH.

(mol) 0,05 ( 0,05 ( 0,05.

(Mmui = =68( MR + 67= 68 (MR=1(R l H.

( Meste= =88( MR+ 44+ MR/ = 88( MR/ = 43. ( R/ l C3H7.

Vy CTCT thu gn ca A l: HCOOC3H7.

Bi 3: Cho 0,1 mol este A vo 50g dd NaOH 10% un nng n khi este phn ng hon ton(cc cht bay hi khng ng k).Dung dch thu c c khi lng 58,6g.C cn dd thu c 10,4g cht rn khan. Tm CTCT ca A?

Hng dn gii:

Ta c mdd sau ng = meste + mddNaOH (meste=58,6 50 = 8,6g.

(Meste = 86.< 100( A l este n chc.(RCOOR/)

M nNaOH= = 0,125 mol.

PTP. RCOOR/ + NaOH ( RCOONa + R/OH.

Ban u: 0,1 0,125 0

P/ 0,1 0,1 0,1 0,1

Sau p/ 0 0,025. 0,1 0,1

(mNaOH d = 0,025.40 = 1g.

M mcht rn khan = mNaOH d + mmui. ( mmui = 10,4 1 = 9,4g.

(Mmui = =94(MR = 27( R l C2H3.

Mt khc MA= 86. ( MR/ = 86-44-27=15. ( R/ l CH3.

Vy CTCT ca A l: CH2=CHCOOCH3.

Cu 4: X phng ha 8,8 g etyl axetat bng 200 ml dd NaOH 0,2M. Sau khi phn ng xy ra hon ton , c cn dd thu c cht rn khan c khi lng l:

A. 8,56 g. B. 3,28 g. C. 10,4 g. D. 8,2 g.

Dng 3: Hiu sut phn ng este ha- Hng s cn bng.

Xt phn ng: RCOOH + R/OH RCOOR/ + H2O.

Trc P: (mol) a b

P: ( mol) x x x x

Sau P: ( mol) a x b x x x

1. Tnh hiu sut ca phn ng:

Nu a b ( H tnh theo ancol v H = . 100% ( x = ( b = .

Nu a < b (H tnh theo axit v H = .100% ( x = ( a = .

2. Hng s cn bng:

Kc =

3. Bi tp minh ha :

Cu 1: Cho 3g CH3COOH phn ng vi 2,5g C2H5OH (xt H2SO4 c, to) th thu c 3,3g este. Hiu sut ca phn ng este ha l:

A.70,2%. B. 77,27%. C.75%. D. 80%.

Hng dn gii:

PTP: CH3COOH + C2H5OH CH3COOC2H5 + H2O.

(mol) <

EMBED Equation.DSMT4 Tnh theo axit.

(mol) 0,05 0,05

Meste = 0,05. 88 = 4,4g.( h = = 75%.

Cu 2: un nng 6 g CH3COOH vi 6 g C2H5OH( c H2SO4 xt) hiu sut phn ng este ha bng 50%. Khi lng este to thnh l:

A. 6 g. B. 4,4 g. C. 8,8 g. D. 5,2 g.

Cu 3: Khi thc hin phn ng este ha 1 mol CH3COOH v 1 mol C2H5OH, lng este ln nht thu c l 2/3 mol. t hiu sut cc i l 90% ( tnh theo axit) khi tin hnh este ha 1 mol CH3COOH cn s mol C2H5OH l ( bit cc phn ng este ha thc hin cng nhit )

A. 0,342. B. 2,925. C. 2,412. D. 0,456.

SBT CB.Cu 1:ng vi cng thc phn t C4H8O2 c bao nhiu este ng phn?

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

Cu 2: Hp cht X n chc c cng thc n gin nht l CH2O. X tc dng vi dd NaOH nhng khng tc dng vi Natri. Cng thc cu to ca X l:

A.CH3CH2COOH.

B.CH3COOCH3.C.HCOOCH

D.OHCCH2OH.

Cu 3: Hp cht X c cng thc cu to : CH3OOCCH2CH3. Tn gi ca X l:

A.Etyl axetat.

B.Metyl propionat.C.Metyl axetat.D.Propylaxetat.

Cu 4: Thu phn este E c cng thc phn t C4H8O2 ( c mt H2SO4 long) thu c 2 sn phm hu c X v Y.T X c th iu ch trc tip ra Y bng 1 phn ng duy nht. Tn gi ca E l:

A.Metyl propionat.B.propyl fomat.C.ancol etylic.

D.Etyl axetat.

Cu 5:Thu phn hon ton hn hp gm 2 este n chc X, Y l ng phn cu to ca nhau cn 100 ml dd NaOH 1M, thu c 7,85 g hn hp 2 mui ca 2 axit l ng ng k tip v 4,95g 2 ancol bc 1. Cng thc cu to v phn trm khi lng ca 2 este l:

A.HCOOCH2CH2CH3, 75%; CH3COOC2H5, 25%.

B.HCOOC2H5, 45%; CH3COOCH3, 55%.

C.HCOOC2H5, 55%; CH3COOCH3, 45%.

D.HCOOCH2CH2CH3, 25%; CH3COOC2H5, 75%.

Cu 6: Este X c cng thc n gin nht l C2H4O. un si 4,4 g X vi 200g dd NaOH 3% n khi phn ng xy ra hon ton. T dd sau phn ng thu c 8,1g cht rn khan. Cng thc cu to ca X l:

A.CH3CH2COOCH3

B.CH3COOCH2CH3.

C.HCOOCH2CH2CH3

D.HCOOCH(CH3)2.

CU 7: Pht biu no sau y khng ng?

A.Cht bo l trieste ca glixol vi cc axit mncacboxylic c mch C di, khng phn nhnh.

B.Cht bo cha ch yu cc gc no ca axit thng l cht rn nhit phng.

C.Cht bo cha ch yu cc gc khng no ca axit thng l cht lng nhit phng v c gi l du.

D.Phn ng thu phn cht bo trong mi trng kim l phn ng thun nghch.

Cu 8: Cht bo c c im chung no sau y?

A.Khng tan trong nc, nng hn nc, c trong thnh phn chnh ca du, m ng ,t/vt.

B.Khng tan trong nc, nh hn nc, c trong thnh phn chnh ca du, m ng ,t/vt.

C. L cht lng,khng tan trong nc, nh hn nc, c trong thnh phn chnh ca du, m ng ,t/vt.

D.L cht rn,khng tan trong nc, nh hn nc, c trong thnh phn chnh ca du, m ng ,t/vt.

Cu 9: Khi thu phn cht bo X trong dd NaOH, thu c glixol v hn hp 2 mui C17H35COONa, C15H31COONa c khi lng hn km nhau 1,817 ln. Trong phn t X c :

A. 3 gc C17H35COO.B.2 gc C17H35COO.C.2 gc C15H31COO.D.3 gc C15H31COO.

Cu 10:X phng v cht git ra c c im chung l:

A.Cha mui natri c kh nng lm gim sc cng b mt ca cc cht bn.

B.Cc mui c ly t phn ng x phng hocht bo.

C.Sn phm ca cng ngh ho du.

D. C ngun gc t ng vt hoc thc vt.

Cu 11: Trong thnh phn ca x phng v cht git ra thng c mt s este. Vai tr ca cc este ny l:

A.Lm tng kh nng git ra.

B.To hng thm mt , d chu.

C.To mu sc hp dn.

D.Lm gim gi thnh ca x phng v cht git ra.

Cu 12:Cho cc pht biu sau:

a) Cht bo thuc loi hp cht este.

b).Cc este khng tan trong nc do chng nh hn nc

c) Cc este khng tan trong nc v ni trn nc do chng khng to c lin kt hidro vi nc v nh hn nc.

d) Khi un cht bo lng trong ni hp ri sc dng kh hidro vo( c xc tc niken) th chng chuyn thnh cht bo rn.

e) Cht bo lng l cc triglixit cha gc axit khng no trong phn t.

Nhng pht biu ng l:

A. a,d,e.

B.a,b,d.C.a,c,d,e.D.a,b,c,d,e.

Cu 13: Khi cho 1 t m ln(sau khi rn, gi s l tristearin) vo bt ng dd NaOH, sau un nng v khuy u hn hp mt thi gian. Nhng hin tng no quan st c sau y l ng?

A. Ming m ni; sau tan dn.

B. Ming m ni; khng thay i g trong qu trnh un nng v khuy.

C. Ming m chm xung; sau tan dn.

D. Ming m chm xung; khng tan.

Cu 14:SGK CB Cht X c cng thc phn t C4H8O2. Khi tc dng vi dd NaOH sinh ra cht Y c cng thc C2H3O2Na. Cng thc cu to ca X l:

A. HCOOC3H7.

B.C2H5COOCH3.C.CH3COOC2H5.D.HCOOC3H5.

Cu 15. Thu phn este X c cng thc phn t C4H8O2 trong dd NaOH thu c hn hp 2 cht h c Y v Z trong Z c t khi hi so vi H2 bng 23. Tn ca X l:

A.Etyl axetat.

B.Metyl axetat.C..Metyl propionatD.Propylfomat.

Cu 16: Pht biu no sau y khng ng?

A.Cht bo khng tan trong nc.B. Du n v m bi trn c cng thnh phn nguyn t.

C.Cht bo khng tan trong nc, nh hn nc nhng tan nhiu trong dung mi hu c.

D.Cht bo l este ca glixerol v cc axit cacboxylic mch C di, khng phn nhnh.

Cu 17: Thu phn hon ton 8,8 gam este n chc, mch h X vi 100ml dd KOH 1M (va ) thu c 4,6g mt ancol Y. Tn gi ca X l:

A.etyl fomat.

B.etyl propionat.C. etyl axetat.

D.propyl axetat.

Cu 18: t chy hon ton 3,7 gam mt este n chc X thu c 3,36 lt kh CO2(ktc) v 2,7 g nc. Cng thc phn t ca X l:

A.C2H4O2.

B.C3H6O2.

C.C4H8O2.

D.C5H8O3.

Cu 19: 10,4 gam hn hp gm axit axetic v etyl axetat tc dng va vi 150 gam dd NaOH 4%. Phn trm khi lng ca etyl axetat trong hn hp l:

A.22%.

B.42,3%.

C.57,7%.

D.88%.

SBT NC. Cu 20: Trong phn t este X no, n chc, mch h, oxi chim 36,36% khi lng. S cng thc cu to tho mn cng thc phn t ca X l:

A.2.

B.3.

C.4.

D.5.

Cu 21: Thu phn 8,8 g este X c cng thc phn t C4H8O2 bng dd NaOH va thu c 4,6 g ancol Y v:

A. 4,1 g mui.

B.4,2 g mui.

C.8,2 g mui.

D. 3,4g mui.

Cu 22: un si hn hp X gm 12 g axit axetic v 11,5 g ancol etylic vi axit H2SO4 lm xc tc n khi kt thc phn ng thu c 11,44 g este. Hiu sut phn ng este ho l:

A.50%.

B.65%.

C.66,67%.

D.52%.

Cu 23: Thu phn 4,3 g este X n chc, mch h (c xc tc axit) n khi phn ng hon tan thu c hn hp 2 cht hu c Y v Z. Cho Y, Z phn ng vi dd AgNO3/ NH3 d thu c 21,6 g bc. Cng thc cu to ca Xl:

A. CH3COOCH=CH2.

B.HCOOCH=CHCH3.

C.HCOOCH2CH=CH2.

D.HCOOC(CH3)=CH2..

Cu 24: un a gam hn hp 2 cht X v Y l ng phn cu to ca nhau vi 200 ml dd NaOH 1M(va ) n khi phn ng hon ton thu c 15 g hn 2 hp mui ca 2 axit no, n chc l ng ng k tip nhau v 1 ancol. Gi tr ca a v cng thc cu to ca X v Y ln lt l:

A.12; CH3COOH v HCOOCH3.

B.14,8; HCOOC2H5 V CH3COOCH3.

C.14,8; CH3COOCH3 V CH3CH2COOH.

D.9; CH3COOH v HCOOCH3.

CU 25:un hn hp glixerol v axit stearic, axit oleic ( c axit H2SO4 lm xc tc) c th thu c my loi trieste ng phn cu to ca nhau?

A.3.

B.4.

C.5.

D.6.

Cu 26: Cho cc cht lng sau: axit axetic, glixerol, triolein. phn bit cc cht lng trn, c th ch cn dng:

A.Nc v qu tm.

B.Nc v dd NaOH.

C.dd NaOH.

D.nc brm.

Cu 27: Cht git ra tng hp c u im:

A.D kim.

B.R tin hn x phng.

C.C rh dng git ra trong nc cng.

D. C kh nng ho tan tt trong nc.

Cu 28:T cc ancol C3H8O v cc axit C4H8O2 c th tao ra bao nhiu este l ng phncu to ca nhau:

A.3.

B.4.

C.5.

D.6.

Cu 29: C bao nhiu cht ng phn cu to ca nhau c cng thc phn t C4H8O2 u tc dng vi NaOH?

A.8.

B.5.

C.4.

D.6.

BT CHUN KT.Cu 30: Phn ng thu phn este trong mi trng kim khi un nng c gi l g?

A. X phng ho.

B.Hidrat ho.C.Crackinh.

D.S ln men.

Cu 31: Este c to thnh t axit no, n chc v ancol no n chc c cng thc cu to nh p n no sau y?

A.CnH2n-1COOCmH2m+1.

B.CnH2n-1COOCmH2m-1.

C.CnH2n+1COOCmH2m-1.D.CnH2n+1COOCmH2m+1.Cu 32: Mt este c cng thc phn t l C3H6O2, c phn ng trng gng vi dd AgNO3 trong NH3. Cng thc cu to ca este l cng thc no?

A.HCOOC2H5.B.CH3COOCH3.C.HCOOC3H7.D.C2H5COOCH3.

CU 33: phn ng este ho gia ancol etylic v axit axetic to thnh sn phm c tn gi l g?

A.Metyl axetat.

B.Axyl axetat.C.Etyl axetat.D.Axetyl etylat.

Cu 34: Khi thu phn este vinyl axetat trong mi trng axit thu c nhng cht no?

A. Axit axetic v ancol etylic.

B.Axit axetic v andehit axetic.

C.Axit axetic v ancol vinylic.D.Axetat v ancol vinylic.

Cu 35: Dy cht no sau y c sp xp theo chiu nhit si cc cht tng dn?

A.CH3COOH, CH3COOC2H5,CH3CH2CH2OH, ..

B.CH3COOH, CH3CH2CH2OH, CH3COOC2H5C.CH3CH2CH2OH,CH3COOH, CH3COOC2H5.

D.CH3COOC2H5,CH3CH2CH2OH,CH3COOH .

CU 36:Mt este c cng thc phn t l C4H8O2, khi thu phn trong mi trng axit thu c ancol etylic. Cng thc cu to ca C4H8O2 l:

A.C3H7COOH.

B.CH3COOC2H5.C.HCOOC3H7.D.C2H5COOCH3.

Cu 37: Mt este c cng thc phn t l C3H6O2 c phn ng trng gng vi dd AgNO3 trong NH3, cng thc cu to ca este l:

A.HCOOC2H5.

B.HCOOC3H7.C.CH3COOCH3.D.C2H5COOCH3.

Cu 38: Tnh khi lng este metyl metacrylat thu c khi un nng 215g axit metacrylic vi 100 g ancol metylic. Gi thit phn ng este ho t hiu sut 60%.

A.125g.

B.150g.

C.175g.

D.200g.

Cu 39: Metyl propionat c cng thc no sau y?

A.HCOOC3H7.B.C2H5COOCH3.C.C3H7COOH.D.C2H5COOH.

Cu 40: Thu phn este C4H6O2 trong mi trng axit ta thu c mt hn hp cc cht u phn ng trng gng. vy cng thc cu to ca este l:

A.CH3COOCH=CH2.

B.HCOOCH2CH =CH2.

C.HCOOCH=CHCH3.D.CH2=CHCOOCH3.

Cu 41: Dun 12 g CH3COOH vi mt lng d C2H5OH (c H2SO4 xc tc).n khi phn ng dng li thu c 11 g este Hiu sut ca phn ng este ha l:

A. 70%.

B. 75%.

C. 62,5%.

D.50%.

Cu 42: un mt lng d axit axetic vi 13,8 g ancol etylic ( c H2SO4 xc tc). n khi phn ng dng li thu c 11 g este. Hiu sut phn ng este ha l:

A. 75%.

B. 62,5%.

C. 60%.

D.41,67%.

Cu 43: Cho 6,6 g CH3COOH phn ng vi hn hp gm 4,04 g CH3OH v C2H5OH t l 2 : 3 v s mol, ( H2SO4 c xc tc) th thu c a(g) hn hp este. Hiu sut chung l 60%. Gi tr ca a l:

A. 4,944 .

B. 5,103.

C.4,4.

D.8,8.

Bui 2 Ngy son: 13/09/2014CACBOHIDRATA. Mc ch yu cu.

1. V kin thc.

Nm c cc khi nim trong chng, cu to ca mi loi cacbohidrat.

Nm c tnh cht ha hc ca tng loi cacbohidrat

2. V k nng.

Vit cc phng trnh phn ng, chui phn ng.

Lm cc bi tp v nhn bit, bi tp nh tnh v nh lng.B. Kin thc cn nm vng. I. KIN THC B TRa) Tnh cht ho hc ca anehit.

- Phn ng cng hiro.

Phn ng oxiho khng hon ton

VD: R-CHO + 2AgNO3 + H2O + 3NH3 R-COONH4 + 2Ag(b) Tnh cht ho hc ca ru: Tc dng vi Cu(OH)2 to dung dch mu xanh lam trong sut

II. KIN THC C BN:a) Khi nim: Cacbohirat l nhng hp cht hu c tp chc v thng c cngthc chung l Cn(H2O)m.

V d: Tinh bt (C6H10O5)n hay (C6(H2O)5(n hay C6n(H2O)5n, glucoz C6H12O6 hay C6(H2O)6b) Phn loi: Gm 3 loi ch yu sau

+) Monosaccarit: L nhm cacbohirat n gin, khng th thu phn c. Th d: glucoz, fructoz

+) isaccarit l nhm cacbohirat m khi thu phn mi phn t sinh ra hai phn t monosaccarit. Th d: saccaroz, mantoz

+) Polisaccarit l nhm cacbohirat phc tp, khi thu phn n cng mi phn t sinh ra nhiu phn t monosaccarit. Th d: tinh bt, xenluloz.

c) Cu trc: C nhiu nhm hiroxyl (-OH) v c nhm cacbonyl ( >C=O) trong phn t

d) Cc cht c th

- Glucoz: CTPT: C6H12O 6 5 4 3 2 1 CTCT dng mch h: CH2OH CHOH-CHOH-CHOH-CHOH-CHO

- Fructoz: CTPT: C6H12O6 6 5 4 3 2 1 CTCT dng mch h: CH2OH CHOH-CHOH-CHOH-CO-CH2OH

Trong mi trng baz ta lun c:

OH- Glucoz Fructoz

- Saccaroz: CTPT C11H22O11- Cu trc phn t: Saccaroz l mt isaccarit c cu to t mt gc glucoz v mt gc fructoz lin kt vi nhau qua nguyn t oxi.

+ c im: Khng c nhm chc CHO

C nhiu nhm ancol (OH)

- Tnh cht ho hc:

Khng tham gia phn ng trng bc

Tham gia phn ng vi Cu(OH)2 cho dd ng saccarat mu xanh lam

2C12H22O11 + Cu(OH)2 -> (C123H22O11)2Cu + 2H2O

Phn ng thu phn: H+, t0 (hoc enzim) C12H22O11 + H2O C6H12O6 + C6H12O6 Saccaroz glucoz fructoz

-Tinh bt: CTPT (C6H10O5)nCu trc phn t: Gm nhiu mt xch (-glucoz lin kt vi nhau to thnh 2 dng: amiloz v amilopectin.

Amiloz gm cc gc (-glucoz lin kt vi nhau to mch khng nhnh, di, xon li, c KLPT ln (khong 200.000 u).

Amilopectin gm cc gc (-glucoz lin kt vi nhau to thnh mch nhnh. Amilopectin c KLPT rt ln, khong 1000000 -> 2000000 u. Chnh v vy amilopectin khng tan trong nc cng nh cc dung mi thng thng khc.

S to thnh tinh bt trong cy xanh: Nh qu trnh quang hp

C6H12O6 (C6H10O5)n - Tnh cht ho hc:

Phn ng thu phn. (C6H10O5)n + nH2O nC6H12O6

Phn ng mu vi iot, h tinh bt khi tip xc vi iot s cho mu xanh lc.

Nguyn nhn: Do h tinh bt c cu to dng xon c l rng nn hp th iot cho mu xanh lc.

- Xenluloz: CTPT (C6H10O5)n- Tnh cht vt l:

Khng tan trong nc v nhiu dung mi hu c khc

Tan trong nc Svayde (dd Cu(OH)2/NH3)

- Cu trc phn t: Xenluloz l mt polisaccarit, phn t gm nhiu gc (- glucoz lin kt vi nhau thnh mch ko di, khng phn nhnh, c phn t khi rt ln, vo khong 2000000. Nhiu mch xenluloz gp li vi nhau thnh si xenluloz.

- Cu to 1 gc glucoz trong xenluloz: (C6H7O2(OH)3(- Tnh cht ho hc.

Phn ng thu phn trong dung dch axit v c c, nng thu c glucoz

(C6H10O5)n + nH2O nC6H12O6

Phn ng vi axit nitric

(C6H7O2(OH)3(n + 3nHNO3(c)(C6H7O2(ONO2)3(n +3nH2O

* BI TP NH LNG V CACBOHIRAT BI TP T LUN

Cu 1. Ln men m gam glucoz c cha 20% tp cht, thu c 500ml ancol etylic 400. Bit khi lng ring ca ancol etylic nguyn cht l 0,8g/ml, hiu sut qu trnh ln men l 60%. Tm m.

Cu 2: Cho 112,5 kg glucoz cha 20% tp cht ln men thnh ancol etylic. Tnh khi lng ancol etylic thu c. (H=50%)

Cu 3. T khi lng kt ta => lng CO2 => lng glucoz =?. V H = 80% => khi lng m.

Cu 3: Ngi ta chia 200 gam dung dch hn hp glucoz v fructoz thnh hai phn bng nhau. Phn mt tc dng vi dung dch AgNO3 trong NH3 d thu c 86,4 gam Ag kt ta. Phn hai phn ng va ht vi 35,2 gam Br2 trong dung dch. Tnh nng phn trm mi cht trong dung dch A.

Hng dn gii:

Ch glucoz tc dng vi dung dch nc brom:

CH2OH[CHOH]4CHO + Br2 + H2O ( CH2OH[CHOH]4COOH + 2HBr

( n(glucoz) = n(Br2) =

C hai cht u tham gia phn ng trng gng:

C6H12O6 + 2Ag(NH3)2OH (

( CH2OH[CHOH]3COONH4 + 2Ag + 3NH3 + H2O

( n(glucoz) + n(fructoz) =

( n(fructoz) = 0,4 mol - 0,22 mol = 0,18 mol

( C%(glucoz)

v C%(fructoz)

Cu 5. Cho ln men 1 m3 nc r ng glucoz thu c 60 lt cn 96o. Tnh khi lng glucoz c trong thng nc r ng glucoz trn, bit khi lng ring ca ancol etylic bng 0,789 g/cm3 20oC v hiu sut qu trnh ln men t 80%.

Hng dn gii:

m(C2H5OH) =

Phng trnh phn ng:

C6H12O6 ( 2C2H5OH + 2CO2( m(glucoz) =

Bu i 3 Ng y son: 16/09/2014 AMINA. Mc ch yu cu.

1. V kin thc.

Nm c khi nim, phn loi v danh php ca amin.

Nm c tnh cht ha hc ca amin.

2. V k nng.- Vit ng phn amin, so snh tnh bazo gia cc amin.

- Lm cc bi tp nh tnh v nh lng v amin.B. KIN THC C BN1. Khi nim, Phn loi :

H - N - R AMIN BC 1

H - N H H

H H - N - R AMIN BC 2

R1

R2 - N - R AMIN BC 3

R1

(R, R1, R2 c th ging hoc khc nhau, c th no, khng no hoc thm.)

- Khi nim : SGK

- Xc nh bc ca amin.

2. Danh php :

Tn gc - chc : Tn gc hirocacbon + amin.

Tn thay th :

+ amin bc 1 : Tn hirocacbon tng ng + amin.

+ amin bc 2 : N- tn gc R1 + tn hirocacbon mch chnh + amin

Gi tn theo quy tc.

CH3 NH2 Metylamin (Metanamin)

CH3 NH CH3 imetylamin ( N-Metylmetanamin )

CH3CH2-NH-CH3 Etylmetylamin ( N-Metyletanamin )

3 . Tnh cht vt l :-Amin c kh nng to lin kt hiro vi nc nn d tan trong nc, nht l cc amin u dy.

- Khi M tng, tan gim.

- Amin to lin kt hiro lin phn t nhng km bn hn ancol nn amin c nhit si thp hn ancol tng ng.

4. Tnh cht ho hc :

Tnh baz :

- Dung dch metylamin, propylamin lm qu tm chuyn sang mu xanh, dung dch anilin khng lm i mu qu tm

- Tc dng vi axit :

R NH2 + HCl R NH3Cl

- So snh lc baz ca cc amin: metylamin> amoniac > anilin

Phn ng th nhn thm anilin :nh hng qua li gia nhm NH2 v nhn thm.

KL : - Anilin c tnh baz rt yu, khng lm i mu qu tm.

- Anilin c phn ng th nhn thm .

Kt ta trng ( Dng nhn bit anilin)

C. BI TP NH LNG V AMIN Dng 1: Xc nh cng thc phn t ca amin.1. Kin thc cn nh:

a. Bi tp t chy hp cht hu c A.

A CO2 + H2O + N2

A cha C, H, N c th c cha O hoc khng.

mc + mH + mn = mA A khng cha oxi.

mc + mH + mn < mA A cha oxi.

Mo = mA (mc+ mH + mn )

Gi CTTQ ca A: CxHyOzNv mc mH mo mN

x : y : z : v = : : : = a : b : c : d.

12 1 16 14

(a, b, c, d ti gin)

CTG: CaHbOcNd CTTN: (CaHbOcNd)n

(CaHbOcNd)n = M n Lp CTPT A.

b. Bi ton lp CTPT amin da vo tnh cht ho hc ca amin.

2. Bi ton v d:

Cu 1: t chy hon ton 1 amin n chc X, ngi ta thu c 10,125g H2O, 8,4 lt kh CO2 v 1,4 lt N2 (cc th tch o ktc). S ng phn ca amin trn l:

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

Cu 2. t chy hon ton hn hp 2 amin no n chc l ng ng lin tip thu c 2,24 lt CO2 (ktc) v 3,6g H2O. Cng thc phn t ca 2 amin l:

A. CH3NH2 v C2H5NH2B. C2H5NH2 v C3H7NH2

C. C3H7NH2 v C4H9NH3D. C4H9NH2 v C5H11NH

II. Dng 2. Tnh theo phng trnh, s dng cc kin thc lin quan, tnh cht ho hc ca amin.

2. Bi ton v d:

Cu 1: Tnh th tch nc brm 3% (d = 1,3g/ml) cn dng iu ch 4,4gam Tribromanilin. Tnh khi lng anilin c trong dung dch A, bit khi A tc dng vi nc brom thu c 6,6 g kt ta trng.

Cu 4. Khi t chy hon ton mt amin n chc X thu c 8,4 lt C02, 1,4lt N2 (ktc) v 10,125 g H2O. Cng thc phn t ca X l :

A. C3H7N B. C2H7N C. C3H9N D. C4H11N

Cu 5. Cho 20,6 gam hn hp amin n chc tc dng va vi HCl, sau phn ng thu c 37,6 g hn hp mui khan. Khi lng ca axit HCl tham gia phn ng l:

A. 15 g B. 17 g C. 14 g D. 13 g

Cu 6. Cho 9,3 g anilin tc dng vi dd HCl d. Lng mui khan thu c l:

(hiu sut l 70%)

A. 9,065 g B. 8,506 g C. 9,605 g D. 9,506 g

Cu 7. Cho 3,1 g metylamin tc dng vi 7,3 gam axit HCl (hiu sut l 80 %).Khi lng mui thu c l :

A. 5,4 g B. 4,5 g C. 6,5 g D. 5,6 g

Cu 8. Cho 7,6 g hn hp gm 2 amin no n chc l ng ng k tip tc dng vi dd HCl d thu c 0,1 mol hn hp mui khan. Cng thc phn t ca gm 2 amin l

A. CH3NH2 v C2H5NH2 B. C2H5NH2 v C3H7NH2

C. C3H7NH2 v C4H9NH2 D. C 4H9NH2 v C5H11NH2Bu i 4 Ng y son: 20/09/2014 AMINOAXITA.Mc tiu bi hc

1.Kin thc :

- Cng c v khc su kin thc v amino axit,tnh cht ca amino axit

2.K nng:

- Rn luyn k nng lm bi tp3. Thi :- Rn luyn thc hc tp tt, t gic tm hiu bi r l thuyt suy ra tnh cht, say m vi b mn.

B. KIN THC C BN

1. Cu to cc nhm c trng

Aminoaxit c 1 nhm NH2; 1 nhm COOH:

H2N-R-COOH

2. Tnh cht ho hc:

a. Tnh cht ca nhm NH2 - Tnh baz: HOOC- R -NH2 + H+ HOOC- R -NH3+b. Tnh cht ca nhm COOH

- Tnh axit: H2N-R -COOH + NaOH H2N-R -COONa + H2O

- Phn ng este ho

c. Aminoaxit c phn ng chung ca 2 nhm

- COOH v -NH2

- To mui ni

H2NCH2COOH H3N+-CH2-COO-- Phn ng trng ngng ca cc v amino axit to poliamit.

d. Phn ng chy: to sn phm CO2 + H2O + N2

III. BI TP VN DNG

Cu 1: Vit cng thc cu to v gi tn cc cht c cng thc phn t C2H7NO2, bit mi cht d dng phn ng vi dung dch HCl v vi dung dch NaOH

Hng dn gii:

Nhng cht hu c c cng thc phn t C2H7NO2 d dng phn ng vi dung dch HCl v dung dch NaOH l :

CH3COONH4 : Amoni axetat

HCOONH3CH3: Metyl amonifomiat

CH3COONH4 + HCl (( CH3COOH + NH4Cl

CH3COONH4 + NaOH (( CH3COONa + NH3( + H2O

HCOONH3CH3 + NaOH (( HCOONa + CH3NH2 + H2O

HCOONH3CH3 + HCl (( HCOOH + CH3NH3Cl

Cu 2. un 100ml dung dch mt amino axit 0,2M tc dng va vi 80 ml dung dch NaOH 0,25M. Sau phn ng ngi ta chng kh dung dch th thu c 2,5 g mui khan. Mt khc ly 100 g dung dch amino axit ni trn c nng 20,6% phn ng va vi 400 ml dung dch HCl 0,5M. Xc nh cng thc phn t ca amino axit.

Hng dn gii:

Phng trnh ho hc :

(H2N)n(R((COOH)m + mNaOH (( (H2N)n(R(COONa)m + mH2O

1 mol m mol

0,2(0,1 = 0,02 0,25(0,08 = 0,02

( m = 1.

b) T phng trnh trn ta cng suy ra M ca (H2N)n(RCOONa :

0,02 mol mui c khi lng 2,5 g

1 mol mui c khi lng 125g

M ca (H2N)nRCOOH = 125 ( 23 + 1 = 103 (g)

(H2N)nRCOOH + nHCl (( ( ClNH3)nRCOOH

1 mol n mol

0,2 0,2

( n = 1. Vy cng thc tng qut ca amino axit : H2N(CxHy(COOH

= 103 (g) ( = 103 ( 61 = 42 (g) ( 12x + y = 42

Lp bng :

x1234

y30 (loi)18 (loi)6 (hp l) (2) > (4)

C. (5) > (4) > (3) > (5) > (1) > (2)

D. (5) > (4) > (2) > (6) > (1) > (3)

Cu 4: A + HCl ( RNH3Cl. Trong A (CxHyNt) c %N = 31,11%. CTCT ca A l

A. CH3 - CH2 - CH2 - NH2 B. CH3 - NH - CH3C. C2H5NH2

D. C2H5NH2 v CH3 - NH - CH3Cu 5: Cho 5,9 g hn hp X gm 3 amin: propylamin, etylmetylamin, trimetylamin X tc dng va vi V ml dung dch HCl 1M. Gi tr ca V l

A. 100ml B. 150 ml

C. 200 ml

D. Kt qu khc

Cu 6: t chy hon ton a mol hn hp X gm 2 amin no n chc thu c 5,6 lt CO2 (ktc) v 7,2 g H2O. Gi tr ca a l

A. 0,05 mol B. 0,1 mol

C. 0,15 mol

D. 0,2 mol

Cu 7: t chy hon ton hn hp X gm 2 amin no n chc k tip nhau trong dy ng ng, thu c 22 g CO2 v 14,4 g H2O. CTPT ca hai amin l

A. CH3NH2 v C2H7N B. C2H7N v C3H9NC. C3H9N v C4H11N D. C4H11N v C5H13N

Cu 8: 9,3 g mt amin n chc cho tc dng vi dung dch FeCl3 d thu c 10,7 g kt ta. CTCT ca amin l

A. C2H5NH2

B. C3H7NH2

C. C4H9NH2

D. CH3NH2Cu 9: Cho 500 gam benzen phn ng vi HNO3 (c) c mt H2SO4 c, sn phm thu c em kh thnh anilin. Nu hiu sut chung ca qu trnh l 78% th khi lng anilin thu c l

A. 564 gam. B. 465 gam. C. 456 gam.

D. 546 gam.

Cu 10: Mt hn hp X gm 2 amin no A, B c cng s nguyn t C. Phn t B c nhiu hn A mt nguyn t N. Ly 13,44 lt hn hp X ( 273oC, 1atm) em t chy hon ton thu c 26,4 gam CO2 v 4,48 lit N2 (ktc). Bit rng c hai u l amin bc 1. CTCT ca A v B v s mol ca chng l:

A. 0,2 mol CH3NH2 v 0,1 mol NH2CH2NH2.

B. 0,2 mol CH3CH2NH2 v 0,1 mol NH2CH2CH2NH2.

C. 0,1 mol CH3CH2NH2 v 0,2 mol NH2CH2CH2NH2.D. 0,2 mol CH3CH2NH2 v 0,1 mol NH2CH2NHCH3AMINOAXITCu 1: Ch dng qu tm nhn bit c c dd no sau y?

A. Glyxin; ax glutamic v ClH3N-CH2COOH B. Glyxin; ax glutamic v H2N-CH2COONa

C. Glyxin; H2N-CH2COOCH3 v H2N-CH2COONa D. Glyxin; H2N-CH2COONa v alanin

Cu 2: Ch dng qu tm nhn bit c cc cht trong dy cc cht sau y ?

A. axit glutamic, alanin, glyxin B. axxit glutamic, alanin, valin

C. axit glutamic, alanin, lysinD. alanin, lisin, glysin

Cu 3 Pht biu no di y khng ng khi ni v amino axit ?

A. trng thi rn, amino axit tn ti dc dng lng cc B. Hp cht H2N-COOH l amino axit n gin nht

C. Amino axit ngoi dng phn t (H2N-R-COOH) cn dng lng cc (+H3N-R-COO-)

D. Amino axit l hp cht hu c tp chc phn t cha ng thi nhm amino va nhm cacboxyl

Cu 4: Trong s cc dd cht sau, cht no lm qu tm chuyn mu hng ?

A.(H2N)2CH2-COONa B. H2N-CH2-COOH C. ClH3-CH2COOHD. N2H-C2H3-(COONa)2

Cu 5: Trong s cc dd cht sau, cht no lm qu tm chuyn mu xanh ?

A.(H2N)2C2H3-COONa B. H2N-CH2-COOH C. ClH3-CH2COOHD. ClH2N-C2H3-(COOH)2

Cu 6: Cho 13,35 gam alanin vo 100 ml dd HCl a M thu c dd X, tc dng ht cc cht trong X cn 400 ml dd NaOH 1M. tnh gi tr ca a.

A. 1,5 MB. 2 M C. 2,5 MD. 3 M

Cu 7: Cho 15 gam glixin vo 200 ml dd H2SO4 x M thu c dd X, tc dng ht cc cht trong dd X cn 500 ml dd NaOH 1M. Tnh x ?

A. 0,5 MB. 0,75 M C. 0,1 MD. 1,5 M

Cu 8: Cho 8,9 gam glixin vo 200 ml dd NaOH x M thu c dd X, tc dng ht cc cht trong dd X cn 500 ml dd HCl 1M. Tnh gi tr ca x.

A. 2MB. 3M C. 1,5MD. 2,5M

Cu 9: Cho s chuyn ho sau: .X, Y ln lt l.

A. C6H5NH3Cl, C6H5ONa. B. C6H5Br, C6H5CH2NH3Cl. C. C6H5ONa, C6H5CH2NH3Cl. D. C6H5ONa, C6H5NH3Cl

Cu 10. X l mt - amioaxit no ch cha 1 nhm -NH2 v 1 nhm -COOH. Cho 23,4 gam X tc dng vi HCl d thu c 30,7 gam mui. Cng thc cu to thu gn ca X l cng thc no?

A. CH3-CH(NH2)-COOH B. H2N-CH2-COOH

C. H2N-CH2CH2-COOH D.CH2=C(CH3)CH(NH2)COOH

Cu 11.Cht A c % khi lng cc nguyn t C, H, O, N ln lt l 32%, 6,67% 42,66%, 18,67%. T khi hi ca A so vi khng kh nh hn 3. A va tc dng NaOH va tc dng dd HCl, A c cng thc cu to nh th no?

A. CH3-CH(NH2)-COOHB. H2N-(CH2)2-COOH

C. H2N-CH2-COOHD. H2N-(CH2)3-COOH

Cu 12 Cht A c thnh phn % cc nguyn t C, H, N ln lt l 40,45%, 7,86%, 15,73% cn li l oxi. Khi lng mol phn t ca A ne nhn Fe cn d dung dch X c Fe2+ v Fe3+ - Cc phn ng xy ra l: Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O 0,1 0,4 0,1 Fe (d) + 2Fe3+ 3Fe2+ 0,02 0,04 Cu + 2Fe3+ (d) Cu2+ + 2Fe2+ 0,03 0,06 mCu = 0,03.64 = 1,92 gam p n A V d 11: Ho tan hon ton 12,42 gam Al bng dung dch HNO3 long (d), thu c dung dch X v 1,344 lt ( ktc) hn hp kh Y gm hai kh l N2O v N2. T khi ca hn hp kh Y so vi kh H2 l 18. C cn dung dch X, thu c m gam cht rn khan. Gi tr ca m l: A. 38,34 gamB. 34,08 gamC. 106,38 gamD. 97,98 gam Hng dn: nAl = 0,46 mol ne cho = 1,38 mol ; nY = 0,06 mol ; Y = 36 - D dng tnh c nN2O = nN2 = 0,03 mol ne nhn = 0,03.(8 + 10) = 0,54 mol < ne cho dung dch X cn cha mui NH4NO3 nNH4+ = NO3 = mol - Vy mX = mAl(NO) + mNHNO = 0,46.213 + 0,105.80 = 106,38 gam p n C (Hoc c th tnh mX = mKl + mNO to mui + mNH = 12,42 + (0,03.8 + 0,03.10 + 0,105.8 + 0,105).62 + 0,105.18 = 106,38 gam) II BI TP V KIM LOI TC DNG VI OXIT KIM LOI (PHN NG NHIT NHM) 1) Mt s ch khi gii bi tp: - Phn ng nhit nhm: Al + oxit kim loi oxit nhm + kim loi(Hn hp X)(Hn hp Y) - Thng gp:+ 2Al + Fe2O3 Al2O3 + 2Fe+ 2yAl + 3FexOy y Al2O3 + 3xFe+ (6x 4y)Al + 3xFe2O3 6FexOy + (3x 2y)Al2O3- Nu phn ng xy ra hon ton, ty theo tnh cht ca hn hp Y to thnh bin lun. V d:+ Hn hp Y cha 2 kim loi Al d ; oxit kim loi ht+ Hn hp Y tc dng vi dung dch baz kim (NaOH,) gii phng H2 c Al d+ Hn hp Y tc dng vi dung dch axit c kh bay ra th c kh nng hn hp Y cha (Al2O3 + Fe) hoc (Al2O3 + Fe + Al d) hoc (Al2O3 + Fe + oxit kim loi d) - Nu phn ng xy ra khng hon ton, hn hp Y gm Al2O3, Fe, Al d v Fe2O3 d - Thng s dng:+ nh lut bo ton khi lng: mhhX = mhhY+ nh lut bo ton nguyn t (mol nguyn t): nAl (X) = nAl (Y) ; nFe (X) = nFe (Y) ; nO (X) = nO (Y) 2) Mt s v d minh ha: V d 1: Nung nng m gam hn hp Al v Fe2O3 (trong iu kin khng c khng kh) n khi phn ng xy ra hon ton , thu c hn hp rn Y. Chia Y thnh hai phn bng nhau: Phn 1: tc dng vi dung dch H2SO4 long (d) sinh ra 3,08 lt kh H2 ( ktc) Phn 2: tc dng vi dung dch NaOH (d) sinh ra 0,84 lt kh H2 ( ktc) Gi tr ca m l: A. 22,75 gamB. 21,40 gamC. 29,40 gamD. 29,43 gam Hng dn: nH2(1) = 0,1375 mol ; nH2(2) = 0,0375 mol - Hn hp rn Y tc dng vi NaOH gii phng H2 Al d v v phn ng xy ra hon ton nn thnh phn hn hp rn Y gm: Al2O3, Fe v Al d - Gi nFe = x mol ; nAl d = y mol c trong 1/2 hn hp Y - T ta c h phng trnh: - Theo lbt nguyn t i vi O v Fe: nAl2O3 = nFe2O3 = = 0,05 mol - Theo lbt khi lng: m = (0,05.102 + 0,1.56 + 0,025.27).2 = 22,75 gam p n A V d 2: Nung nng m gam hn hp gm Al v Fe3O4 trong iu kin khng c khng kh. Sau khi phn ng xy ra hon ton, thu c hn hp rn X. Cho X tc dng vi dung dch NaOH (d) thu c dung dch Y, cht rn Z v 3,36 lt kh H2 ( ktc). Sc kh CO2 (d) vo dung dch Y, thu c 39 gam kt ta. Gi tr ca m l: A. 45,6 gamB. 57,0 gamC. 48,3 gamD. 36,7 gam Hng dn: nH2 = 0,15 mol ; nAl(OH)3 = 0,5 mol - T suy ra thnh phn hn hp rn X gm: Fe, Al2O3 (x mol) v Al d (y mol) - Cc phn ng xy ra l: 2Al + 2NaOH + 6H2O 2Na[Al(OH)4] + 3H2 Al2O3 + 2NaOH + 3H2O 2Na[Al(OH)4] CO2 + Na[Al(OH)4] Al(OH)3 + NaHCO3 - nH2 = 0,15 mol y = 0,1 mol - Theo lbt nguyn t i vi Al: 2x + y = 0,5 x = 0,2 mol - Theo lbt nguyn t i vi O: nO(FeO) = nO(AlO) nFe3O4 = mol - Theo lbt nguyn t i vi Fe: nFe = 3nF3O4 = 3.0,15 = 0,45 mol - Theo lbt khi lng: m = 0,45.56 + 0,2.102 + 0,1.27 = 48,3 gam p n C V d 3: Thc hin phn ng nhit nhm hn hp X gm Al v mt oxit st FexOy (trong iu kin khng c khng kh) thu c 92,35 gam cht rn Y. Ha tan Y trong dung dch NaOH (d) thy c 8,4 lt kh H2 ( ktc) thot ra v cn li phn khng tan Z. Ha tan 1/2 lng Z bng dung dch H2SO4 c, nng (d) thy c 13,44 lt kh SO2 ( ktc) thot ra. Bit cc phn ng xy ra hon ton. Khi lng Al2O3 trong Y v cng thc oxit st ln lt l: A. 40,8 gam v Fe3O4B. 45,9 gam v Fe2O3 C. 40,8 gam v Fe2O3D. 45,9 gam v Fe3O4 Hng dn: nH2 = 0,375 mol ; nSO2(c Z) = 2.0,6 = 1,2 mol - T suy ra thnh phn cht rn Y gm: Fe, Al2O3, Al d v phn khng tan Z l Fe - nH2 = 0,375 mol nAl d = 0,25 mol - nSO2 = 1,2 mol nFe = mol - mAl2O3 = 92,35 0,8.56 0,25.27 = 40,8 gam (1) nAl2O3 = 0,4 mol - Theo lbt nguyn t i vi O nO(FeO) = 0,4.3 = 1,2 mol - Ta c: cng thc oxit st l Fe2O3 (2) - T (1) ; (2) p n C V d 4: Trn 5,4 gam bt Al vi 17,4 gam bt Fe3O4 ri tin hnh phn ng nhit nhm (trong iu kin khng c khng kh). Gi s ch xy ra phn ng kh Fe3O4 thnh Fe. Ha tan hon ton cht rn sau phn ng bng dung dch H2SO4 long (d) thu c 5,376 lt kh H2 ( ktc). Hiu sut phn ng nhit nhm v s mol H2SO4 phn ng l: A. 75 % v 0,54 molB. 80 % v 0,52 mol C. 75 % v 0,52 molD. 80 % v 0,54 mol Hng dn: nAl = 0,2 mol ; nFe3O4 = 0,075 mol ; nH2 = 0,24 mol - Phn ng xy ra khng hon ton:8Al + 3Fe3O44Al2O3 + 9Fex 0,5x(mol) - Hn hp cht rn gm: - Ta c phng tnh: .2 + (0,2 x).3 = 0,24.2 x = 0,16 mol Hphn ng =% (1) - nH+phn ng = 2.nFe + 3.nAl + 6.nAl2O3 + 8.nFe3O4 = 0,36 + 0,12 + 0,48 + 0,12 = 1,08 mol nH2SO4phn ng = mol (2) - T (1) ; (2) p n D Bui 9 Ngy son: 15/11Dy in ho ca kim loi

A. Mc ch yu cu.

1. V kin thc.

Nm c th t trong dy in ha ca kim loi,quy tc anpha

2. V k nng.

S dng thnh tho quy tc anpha, vit cc phng trnh phn ng di dng phn tv ion.

Gii quyt cc bi tp v kim loi tc dng vi dung dch mui.

B. Kin thc cn nm vng.I BI TP V KIM LOI TC DNG VI DUNG DCH MUI 1) Kim loi tc dng vi dung dch mui: - iu kin kim loi M y c kim loi X ra khi dung dch mui ca n: xM (r) + nXx+ (dd) xMn+ (dd) + nX (r) + M ng trc X trong dy th in cc chun+ C M v X u khng tc dng c vi nc iu kin thng+ Mui tham gia phn ng v mui to thnh phi l mui tan - Khi lng cht rn tng: m = mX to ra mM tan - Khi lng cht rn gim: m = mM tan mX to ra - Khi lng cht rn tng = khi lng dung dch gim - Ngoi l:+ Nu M l kim loi kim, kim th (Ca, Sr, Ba) th M s kh H+ ca H2O thnh H2 v to thnh dung dch baz kim. Sau l phn ng trao i gia mui v baz kim+ trng thi nng chy vn c phn ng: 3Na + AlCl3 (khan) 3NaCl + Al+ Vi nhiu anion c tnh oxi ha mnh nh NO3-, MnO4-,th kim loi M s kh cc anion trong mi trng axit (hoc baz) - Hn hp cc kim loi phn ng vi hn hp dung dch mui theo th t u tin: kim loi kh mnh nht tc dng vi cation oxi ha mnh nht to ra kim loi kh yu nht v cation oxi ha yu nht - Th t tng dn gi tr th kh chun (Eo) ca mt s cp oxi ha kh: Mg2+/Mg < Al3+/Al < Zn2+/Zn < Cr3+/Cr < Fe2+/Fe < Ni2+/Ni < Sn2+/Sn < Pb2+/Pb < 2H+/H2 < Cu2+/Cu < Fe3+/Fe2+ < Ag+/Ag < Hg2+/Hg < Au3+/Au 2) Mt s ch khi gii bi tp: - Phn ng ca kim loi vi dung dch mui l phn ng oxi ha kh nn thng s dng phng php bo ton mol electron gii cc bi tp phc tp, kh bin lun nh hn hp nhiu kim loi tc dng vi dung dch cha hn hp nhiu mui. Cc bi tp n gin hn nh mt kim loi tc dng vi dung dch mt mui, hai kim loi tc dng vi dung dch mt mui,c th tnh ton theo th t cc phng trnh phn ng xy ra - S dng phng php tng gim khi lng tnh khi lng thanh kim loi sau phn ng, - T s mol ban u ca cc cht tham gia phn ng bin lun cc trng hp xy ra - Nu cha bit s mol cc cht phn ng th da vo thnh phn dung dch sau phn ng v cht rn thu c bin lun cc trng hp xy ra - Kim loi kh anion ca mui trong mi trng axit (baz) th nn vit phng trnh dng ion thu gn - Kim loi (Mg Cu) y c Fe3+ v Fe2+. V d: Fe + 2Fe3+ 3Fe2+ ; Cu + 2Fe3+ Cu2+ + 2Fe2+ - Fe + 2Ag+ Fe2+ + 2Ag. Nu Fe ht, Ag+ cn d th: Fe2+ + Ag+ Fe3+ + Ag 3) Mt s v d minh ha: V d 1: Nhng mt thanh kim loi M ha tr II nng m gam vo dung dch Fe(NO3)2 th khi lng thanh kim loi gim 6 % so vi ban u. Nu nhng thanh kim loi trn vo dung dch AgNO3 th khi lng thanh kim loi tng 25 % so vi ban u. Bit gim s mol ca Fe(NO3)2 gp i gim s mol ca AgNO3 v kim loi kt ta bm ht ln thanh kim loi M. Kim loi M l: A. PbB. NiC. CdD. Zn Hng dn: Gi nFe2+p = 2x mol nAg+p = x mol M + Fe2+ M2+ + Fe 2x 2x 2x m = 2x.(M 56) %mKl gim = (1) M + 2Ag+ M2+ + 2Ag 0,5x x x m = 0,5x.(216 M) %mKl tng = (2) - T (1) ; (2) M = 65 Zn p n D V d 2: Cho m gam hn hp bt cc kim loi Ni v Cu vo dung dch AgNO3 d. Khuy k cho n khi phn ng kt thc thu c 54 gam kim loi. Mt khc cng cho m gam hn hp bt cc kim loi trn vo dung dch CuSO4 d, khuy k cho n khi phn ng kt thc, thu c kim loi c khi lng bng (m + 0,5) gam. Gi tr ca m l: A. 15,5 gamB. 16 gamC. 12,5 gamD. 18,5 gam Hng dn: Gi nNi = x mol ; nCu = y mol c trong m gam hn hp Ni + 2Ag+ Ni2+ + 2Ag (1) Cu + 2Ag+ Cu2+ + 2Ag (2) Ni + Cu2+ Ni2+ + Cu (3) - T (3) (64 59).x = 0,5 x = 0,1 mol (*) - T (1) nAg(1) = 0,2 mol mAg(1) = 21,6 gam mAg(2) = 54 21,6 = 32,4 gam nAg(2) = 0,3 mol y = 0,15 mol (**) - T (*) ; (**) m = 0,1.59 + 0,15.64 = 15,5 gam p n A V d 3: Ha tan hn hp bt kim loi gm 8,4 gam Fe v 6,4 gam Cu vo 350 ml dung dch AgNO3 2M. Sau khi phn ng xy ra hon ton thu c m gam cht rn. Gi tr ca m l: A. 70,2 gamB. 54 gamC. 75,6 gamD. 64,8 gam Hng dn:nFe = 0,15 mol ; nCu = 0,1 ; nAg+ = 0,7 mol Fe + 2Ag+ Fe2+ + 2Ag (1) 0,15 0,3 0,15 0,3 Cu + 2Ag+ Cu2+ + 2Ag 0,1 0,20,2 Fe2+ + Ag+ Fe3+ + Ag (3) 0,15 0,150,15 T (1) ; (2) m = (0,3 + 0,2 + 0,15).108 =70,2 gam p n AV d 4: Cho 2,24 gam bt st vo 200 ml dung dch cha hn hp gm AgNO3 0,1M v Cu(NO3)2 0,5M. Sau khi cc phn ng xy ra hon ton, thu c dung dch X v m gam cht rn Y. Gi tr ca m l: A. 2,80 gam B. 4,08 gamC. 2,16 gamD. 0,64 gam Hng dn: nFe = 0,04 mol ; nAg+ = 0,02 mol ; nCu2+ = 0,1 mol Th t cc phn ng xy ra l: (Fe2+/Fe < Cu2+/Cu < Fe3+ < Fe2+ < Ag+ < Ag) Fe + 2Ag+ Fe2+ + 2Ag (1) 0,01 0,02 0,02 Fe + Cu2+ Fe2+ + Cu (2) 0,03 0,03 T (1) ; (2) mY = 0,02.108 + 0,03.64 = 4,08 gam p n B V d 5: Cho hn hp gm 1,2 mol Mg v x mol Zn vo dung dch cha 2 mol Cu2+ v 1 mol Ag+ n khi cc phn ng xy ra hon ton, thu c mt dung dch cha ba ion kim loi. Trong cc gi tr sau y, gi tr no ca x tho mn trng hp trn: A. 1,8B. 1,5C. 1,2D. 2,0 Hng dn: - Dung dch cha 3 ion kim loi Mg2+, Zn2+, Cu2+ - ne cho = (2,4 + 2x) mol v ne nhn = 1 + 2.2 = 5 mol - Yu cu bi ton tha mn khi ne cho < ne nhn hay (2,4 + 2x) < 5 x < 1,3 x =1,2 p n C V d 6: Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO3)2 0,2M v H2SO4 0,25M. Sau khi cc phn ng xy ra hon ton, thu c 0,6m gam hn hp bt kim loi v V lt kh NO (sn phm kh duy nht, ktc). Gi tr ca m v V ln lt l: A. 17,8 v 4,48B. 17,8 v 2,24C. 10,8 v 4,48D. 10,8 v 2,24 Hng dn: nCu2+ = 0,16 mol ; nNO3 = 0,32 mol ; nH+ = 0,4 mol - Cc phn ng xy ra l: Fe + 4H+ + NO3 Fe3+ + NO + 2H2O (1) 0,1 0,4 0,10,10,1 VNO = 0,1.22,4 = 2,24 lt (*) Fe + 2Fe3+ 3Fe2+ (2) 0,05 0,1 Fe + Cu2+ Fe2+ + Cu (3) 0,16 0,16 - T (1) ; (2) ; (3) nFep = 0,1 + 0,05 + 0,16 = 0,31 mol - Hn hp bt kim loi gm Fe d v Cu (m 0,31.56) + 0,16.64 = 0,6m m = 17,8 gam (**) - T (*) ; (**) p n B

Bui 11 Ngy son: 13/12

BI TP V XC NH TN KIM LOIA. Mc ch yu cu.

Nm c cc phng php xc nh tn kim loi.

Luyn tp cc phng php gii bi tp ha hc.

B. Kin thc cn nm vng1) C th tnh c khi lng mol nguyn t kim loi M theo cc cch sau: - T khi lng (m) v s mol (n) ca kim loi M = - T Mhp cht Mkim loi - T cng thc Faraday M = (n l s electron trao i mi in cc) - T a < m < b v < n < tm M tha mn trong khong xc nh - Lp hm s M = f(n) trong n l ha tr ca kim loi M (n = 1, 2, 3), nu trong bi ton tm oxit kim loi MxOy th n = kim loi M - Vi hai kim loi k tip nhau trong mt chu k hoc phn nhm tm tn 2 kim loi 2) Mt s ch khi gii bi tp: - Bit s dng mt s nh lut bo ton nh bo ton khi lng, bo ton nguyn t, bo ton mol electron, Bit vit cc phng trnh ion thu gn, phng php ion electron - Khi bi khng cho kim loi M c ha tr khng i th khi kim loi M tc dng vi cc cht khc nhau c th th hin cc s oxi ha khc nhau t kim loi M c cc ha tr khc nhau - Khi hn hp u c chia lm hai phn khng bng nhau th phn ny gp k ln phn kia tng ng vi s mol cc cht phn ny cng gp k ln s mol cc cht phn kia 3) Mt s v d minh ha: V d 1: Cho 3,024 gam mt kim loi M tan ht trong dung dch HNO3 long, thu c 940,8 ml kh NxOy (sn phm kh duy nht, ktc) c t khi i vi H2 bng 22. Kh NxOy v kim loi M l: A. NO v MgB. NO2 v AlC. N2O v AlD. N2O v Fe Hng dn: M(NxOy) = 44 nN2O = 0,042 molM Mn+ + ne2NO3- + 8e + 10H+ N2O + 5H2O Theo lbt mol electron: ne cho = ne nhn 3,024 No duy nht n = 3 v M = 27 Al p n C V d 2: Hn hp X gm Mg v kim loi M. Ha tan hon ton 8 gam hn hp X cn va 200 gam dung dch HCl 7,3 %. Mt khc cho 8 gam hn hp X tc dng hon ton vi kh Cl2 cn dng 5,6 lt Cl2 ( ktc) to ra hai mui clorua. Kim loi M v phn trm v khi lng ca n trong hn hp X l: A. Al v 75 %B. Fe v 25 %C. Al v 30 %D. Fe v 70 % Hng dn: nHCl = 0,4 mol ; nCl2 = 0,25 mol ; nMg = x mol ; nM = y mol 24x + My = 8 (1) - X tc dng vi dung dch HCl (M th hin ha tr n) 2x + ny = 0,4 (2) - X tc dng vi Cl2 (M th hin ha tr m) 2x + my = 0,5 (3) - T (2) ; (3) y(m n) = 0,1 m > n No duy nht m = 3 v n = 2 x = y = 0,1 mol - T (1) M = 56 Fe v % M = 70 % p n D V d 3: Hn hp X gm hai mui cacbonat ca 2 kim loi kim th hai chu k lin tip. Cho 7,65 gam X vo dung dch HCl d. Kt thc phn ng, c cn dung dch th thu c 8,75 gam mui khan. Hai kim loi l: A. Mg v CaB. Ca v SrC.Be v MgD. Sr v Ba Hng dn: - t cng thc chung ca hai mui l CO3. Phng trnh phn ng: CO3 + 2HCl Cl2 + CO2 + H2O - T phng trnh thy: 1 mol CO3 phn ng th khi lng mui tng: 71 60 = 11 gam - Theo bi khi lng mui tng: 8,75 7,65 = 1,1 gam c 0,1 mol CO3 tham gia phn ng + 60 = 76,5 = 16,5 2 kim loi l Be v Mg p n C V d 4: Ha tan hon ton 6 gam hn hp X gm Fe v mt kim loi M (ha tr II) vo dung dch HCl d, thu c 3,36 lt kh H2 ( ktc). Nu ch ha tan 1,0 gam M th dng khng n 0,09 mol HCl trong dung dch. Kim loi M l: A. MgB. ZnC. CaD. Ni Hng dn: nH2 = 0,15 mol - nX = nH2 = 0,15 mol X = 40 - ha tan 1 gam M dng khng n 0,09 mol HCl 22,2 < M < 40 < 56 M l Mg p n A V d 5: ha tan hon ton 6,834 gam mt oxit ca kim loi M cn dng ti thiu 201 ml dung dch HCl 2M. Kim loi M l: A. MgB. CuC. AlD. Fe Hng dn: Gi cng thc oxit l MxOy ; nHCl = nH+ = 0,402 mol - Ta c nO2 (trong oxit) = mol nMxOy = mol (Mx + 16y) = Mx = 18y M = No duy nht v M = 27 Al p n C

Bui 12 Ngy son: 25/12/2014Bi tp tng hp

A. Mc ch yu cu.

Tng hp cc bi ton v kim loi v cc phng php gii bi tp.

Rn luyn k nng vit cc phng trnh v cn bng cc phng trnh phn ng.

B. Kin thc cn nm vng.I BI TON V KIM LOI TC DNG VI NC, KIM LOI TC DNG VI DUNG DCH BAZ KIM 1) Mt s ch khi gii bi tp: - Ch c kim loi kim, Ca, Sr, Ba mi tan trong nc nhit thng - Cc kim loi m hiroxit ca chng c tnh lng tnh nh Al, Zn, Be, Sn, Pbtc dng c vi dung dch kim (c) - Nu bi cho nhiu kim loi tc dng vi nc to dung dch kim, ri sau ly dung dch kim tc dng vi dung dch hn hp axit th:+ Gii bng cch vit phng trnh ion thu gn+ nOH = 2nH2 - Nu bi cho hn hp kim loi kim hoc kim th v kim loi M ha tr n vo nc th c th c hai kh nng:+ M l kim loi tan trc tip (nh kim loi kim, Ca, Sr, Ba)+ M l kim loi c hiroxit lng tnh (nh Al, Zn)M+ (4 n)OH + (n 2)H2O MO2n 4 + H2 (da vo s mol kim loi kim hoc kim th s mol OH ri bin lun xem kim loi M c tan ht khng hay ch tan mt phn) 2) Mt s v d minh ha: V d 1: Hn hp X gm Na, K, Ba ha tan ht trong nc d to dung dch Y v 5,6 lt kh ( ktc). Tnh V ml dung dch H2SO4 2M ti thiu trung ha Y A. 125 mlB. 100 mlC. 200 mlD. 150 ml Hng dn: nH2 = 0,25 mol Ta c nOH = 2nH2 m nOH = nH+ nH2SO4 = = nH2 = 0,25 mol V = 0,125 lt hay 125 ml p n A V d 2: Thc hin hai th nghim sau: Th nghim 1: Cho m gam hn hp Ba v Al vo nc d, thu c 0,896 lt kh ( ktc) Th nghim 2: Cng cho m gam hn hp trn cho vo dung dch NaOH d thu c 2,24 lt kh ( ktc) Cc phn ng xy ra hon ton. Gi tr ca m l: A. 2,85 gamB. 2,99 gamC. 2,72 gamD. 2,80 gam Hng dn: nH2 th nghim 1 = 0,04 < nH2 th nghim 2 = 0,1 mol th nghim 1 Ba ht, Al d cn th nghim 2 th c Ba v Al u ht - Gi nBa = x mol v nAl = y mol trong m gam hn hp - Th nghim 1: Ba + 2H2O Ba2+ + 2OH + H2x2x xAl + OH + H2O AlO2 + H22x 3x nH2 = 4x = 0,04 x = 0,01 mol - Th nghim 2: tng t th nghim 1 ta c: x + = 0,1 y = 0,06 mol m = 0,01.137 + 0,06.27 = 2,99 gam p n B V d 3: Ha tan hon ton 7,3 gam hn hp X gm kim loi Na v kim loi M (ha tr n khng i) trong nc thu c dung dch Y v 5,6 lt kh hiro ( ktc). trung ha dung dch Y cn dng 100 ml dung dch HCl 1M. Phn trm v khi lng ca kim loi M trong hn hp X l: A. 68,4 %B. 36,9 %C. 63,1 %D. 31,6 % Hng dn: nH2 = 0,25 mol ; nHCl = 0,1 mol - Gi nNa = x mol v nM = y mol 23x + My = 7,3 (1) - Nu M tc dng trc tip vi nc nH2 = nOH = 0,5 > nHCl = 0,1 loi - Nu M l kim loi c hiroxit lng tnh (n = 2 hoc 3): M + (4 n)OH + (n 2)H2O MO2n 4 + H2y(4 n)yny/2 - Do OH d nn kim loi M tan ht v nOH d = x (4 n)y mol x (4 n)y = 0,1 (2) v x + ny = 0,5 (3) y = 0,1 mol - Thay ln lt n = 2 hoc 3 vo (1) ; (2) ; (3) ch c n = 3 ; x = 0,2 ; M = 27 l tha mn %M = 36,9 % p n B II MT BI TON KINH IN 1) Ni dung tng qut: M hn hp rn (M, MxOy) M+n + sn phm khm gamm1 gam(n l s oxi ha cao nht ca M) (M l kim loi Fe hoc Cu v dung dch HNO3 (H2SO4 c nng) ly va hoc d) - Gi: nM = x mol ; ne (2) nhn = y mol ne nhng = x.n mol - Theo lbt khi lng t (1) nO = mol - ne nhn = ne (oxi) + ne (2) = .2 + y = + y mol - Theo lbt mol electron: ne nhng = ne nhn x.n = + y - Nhn c hai v vi M ta c: (M.x).n = + M.y m.n = m.= m = (*) - Thay M = 56 (Fe) ; n = 3 vo (*) ta c: m = 0,7.m1 + 5,6.y (1) - Thay M = 64 (Cu) ; n = 2 vo (*) ta c: m = 0,8.m1 + 6,4.y (2) (Khi bit 2 trong 3 i lng m, m1, y ta s tnh c i lng cn li) 2) V d minh ha: V d 1: Cho 11,36 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung dch HNO3 long (d), thu c 1,344 lt kh NO (sn phm kh duy nht, ktc) v dung dch X. C cn dung dch X thu c m gam mui khan. Gi tr m l: A. 38,72 gamB. 35,50 gamC. 49,09 gamD. 34,36 gam Hng dn: nNO = 0,06 mol y = 0,06.3 = 0,18 mol Theo cng thc (1) ta c: nFe = mol nFe(NO3)3 = 0,16 mol mmui khan = 0,16.242 = 38,72 gam p n A V d 2: kh hon ton 3,04 gam hn hp X gm FeO, Fe3O4, Fe2O3 cn 0,05 mol H2. Mt khc, ha tan hon ton 3,04 gam hn hp X trong dung dch H2SO4 c thu c V ml kh SO2 (sn phm kh duy nht ktc). Gi tr ca V l: A. 112 mlB. 224 mlC. 336 mlD. 448 ml Hng dn: Thc cht phn ng kh cc oxit l: H2 + O(oxit) H2O. V vy nO(oxit) = nH2 = 0,05 mol mFe = 3,04 0,05.16 = 2,24 gam Theo cng thc (1) ta c: ne nhn (S+6 S+4) = y = mol nSO2 = 0,01 mol V = 0,01.22,4 = 0,224 lt hay 224 ml p n B V d 3: Nung m gam bt Cu trong oxi thu c 37,6 gam hn hp rn X gm Cu, CuO v Cu2O. Ha tan hon ton X trong dung dch H2SO4 c, nng (d) thy thot ra 3,36 lt kh ( ktc). Gi tr ca m l: A. 25,6 gam B. 32 gamC. 19,2 gamD. 22,4 gam Hng dn: nSO2 = 0,15 mol y = 0,15.2 = 0,3 mol Theo cng thc (2) ta c: m = 0,8.37,6 + 6,4.0,3 = 32 gam p n B

+ 3Br2 + 3HBr

EMBED Unknown

+

EMBED Equation.DSMT4

PAGE 23

_1475176299.unknown

_1475176301.unknown

_1475176804.unknown

_1475176300.unknown

_1475176298.unknown