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Ngy 20/8/2012 T chn tit 1:N TP

I. Mc tiu- Lp phong trnh ha hc ca cc phn ng oxi ha kh bng phng phpthng bng electron

- Gii mt s bi tp c bn nh xc nh thnh phn hn hp hp, bi tp v chtkh- Vn dng cc phng php c th gii bi tp ha hc nh: p dng nh l;ut

bo ton khi lng, tnh tr s trung bnh, phng php ng cho,.. trong nhnmnh cung cp phng php hs vn dng lm bi tp v sau.II. Phng php: m thoiIII. Chun bGv: h thng bi tpHs: Vn dng tt cc kin thc vo gii bi tpIV. Thit k cc hot ng

1. n nh t chc:2. Bi mi:

Hot ng ca gio vin Hot ng ca hc sinhHot ng 1: Cn bng cc phn ngoxiho kh theo phng php thng

bng electron.a. FeSO4 + KMnO4 + H2SO4Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

b. Zn + HNO3 Zn(NO3)2 + NH4NO3

+ H2Oc. Al + HNO3 Al(NO3)3 + N2 + H2O

Ln bng lm theo cc bca. 10FeSO4 + 2KMnO4 + 8H2SO45Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

b. 4Zn + 10HNO3 4Zn(NO3)2 +NH4NO3 + 3H2Oc. 10Al +36 HNO3 10Al(NO3)3 + 3N2

+18 H2O

Hot ng 2:a. S FeS Fe2O3 Fe FeSO4Fe(OH)2 Fe2(SO4)3 Fe(OH)3,

b. SO2 SO3NaHSO4 Na2SO4BaSO4

c.MnO2KMnO4Cl2 NaCl NaNO3NaCl

d. FeCl3 Fe(OH)3 FeCl3 Fe(NO3)3e. Clorua viCaCl2CaCO3Hng dn Hs lm vo v.

2 Hs hon thnh chui phn ng ln bngCc Hs cn li lm vo v.

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Hot ng 3: gii bi tp ha hc theophng php thng thng nluyn phng trnh ha hcBi 1: Cho 12g hn hp gm Fe v Cutc dng vi dd HCl 0,5M thu c

2,24l kh ( kc)a. Xc nh % khi lng mi kim loitrong hn hp ?

b.Tnh th tch HCl tham gia phnng ?

Bi 2: Cho hn hp gm Mg v Al vodd H2SO4 thu c 2,24 lit kh ( kc).

Nu hn hp trn cho vo H2SO4 c

k thng th thu c 0,56 lit kh A(kc)a. Tnh % khi lng mi kim loitrong hn hp?

b. Dn kh A vo 28g dd NaOH 15%.Tnh C% cc cht trong dd sau phnng ?Gi Hs lm bi tp

Hs vn dng kin thc ha hc gii 2 bitpBi 1:a. Cu khng tc dng vi HClFe + 2HCl FeCl2 + H2

0,1mol 0,1 moln HCl = 0,1 mol => nFe = 0,1 mol=> mFe =5,6 g=> mCu = 6,4g

b.nHCl = 0,2 mol => VHCl = 0,2 / 0,5 = 0,4M

Bi 2. a) Phng trnh ha hc:2Al + 3H2SO4 Al2(SO4)3 + 3H2Mg + H2SO4 MgSO4 + H2Al khng tc dng vi H2SO4 c nhit thng .Mg + 2H2SO4 MgSO4 + SO2 + 2H2OGi x , y l s mol ca Al v Mg .Ta c h phng trnh :1,5x + y = 0,1 v y=0,15molb) tnh nNaOH lp t l nNaOH / nSO2mui to ra

Hot ng 4: gii bi tp ha hc

bng phng php ng cho- Phng php ny gip gii nhanhmt s dng bi tp nh: nng dungdch, khi lng trung bnh, k c %hn hp cht rn trong qung- Bi tp 1:Mt hn hp kh c tkhi so vi H2 l24. Tnh thnh phn% ca mi kh theo th tch?Gi Hs gii bi bi tpt V1 v V2 ln lt l th tch ca O2

v SO2 trong hn hp, theo bi- Gio vin cung cp thm phng

php ng cho

%V1 = %V2 = 50%- Bi tp 2: thu c dung dchHCl 25% cn ly m1 gam dung dch

(Hs thng gii theo cch dng khi lng

mol trung bnh ca hn hp) nh sau: tac: 1 2

1 2

32 6424 2 48

V VM

V V

+= = =

+

%V1 = %V2 = 50%

SO2

O2

M1 = 64

M2 = 31

M=48

16

16

Vn dng phng php ng cho gii bi tp

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HCl 45% pha vi m2 gam dung dchHCl 15%. T l m1/m2 l:A. 1:2 B. 1:3 C. 2:1 D. 3:1- Gv c th gi cch gii bi tpny theo phng php ng cho

m1

m2

45

15

25

10

20

p n A

3. Dn d: Hs rn luyn thm mt s bi tp sau, c gng dng nhng phngphp gii nhanh (nu thch hp)

Bi 1 : un nng hn hp gm 1,2g Mg v 2,4g S ( khng c khng kh ) . Snphm em ho tan vo 18,25g dd HCl 25%a. Tnh th tch kh bay ra kc ?

b. Dn kh trn vo 30g dd NaOH 20% . Tnh C% c trong dd sau phn ng ?Bi 2 : ho tan 11g hn hp gm NaBr v NaCl thnh dd . Cho dd trn tc dng

va vi 127,5g dd AgNO3 20% .a. Tnh khi lng kt ta to thnh ?

b. Tnh C% cc cht c trong dd thu c

Ngy 24/8/2012 T chn tit 2:LUYN TP S IN LI

I. Mc tiu- HS bit vit phng trnh in ly

- Bit cch tnh nng ion v nng phn t ca cc cht in ly .- Yu thch mn hc . Bit vn dng vo thc tinII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cu hi v bi tp vn dng tnh nng ion , nng phn t.Hs: Chun b kin thc v s in li.IV. Tin trnh

1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng

Hot ng ca gio vin Hot ng ca hc sinhHot ng 1:H thng kin thc v sin li? nh ngha qu trnh in ly v chtin ly. Cho v dB sung thm: mt s cht khi nng chycng phn li ra ion, nn trng thi nng

- S in ly: L qu trnh phn ly cc chttrong nc ra ion- Cht in ly : L cht khi tan trong nc

phn ly ra ionV d: NaCl Na+ + Cl-

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chy cc cht ny cng dn c inHot ng 2:Bi tp vn dng s in li1. Dd no sau y khng dn c in?a. dd KCl c. dd

NaHSO4

b. dd CH3COONa d. dd ruetylic2. Cht no khng in li ra ion khi hotan trong nc?a. CaCl2 c. glucozo

b. HNO3 d. KOH3. S in li l:a. s ho tan mt cht vo nc to thnhdd

b. s phn li mt cht thnh ion dng v

ion m khi cht tan trong nc haynng chyc. s phn li mt cht thnh ion dc tcdng ca dng ind. thc cht l qu trnh oxiho kh4. Cc dd axit, baz, mui dn c inl do dd ca chng c cc:a. ion tri du c. anion

b. caction d. cht

Chn phng n ng1. d

2. c

3. b

4. a

Hot ng 3: nguyn nhn tnh dn inca dd cht in li- Nc ng vai tr g trong qu trnhin li ca cc cht?- Ti sao cc dung dch: KOH, CuSO4,HClO4 dn c in?

Nc ng vai tr l dung mi phn cc

Vit phng trnh din li v gii thchKOH K+ + OH-

CuSO4 Cu2+ + SO42-

HClO4 H+ + ClO4-

Hot ng 4:Bi tp vn dng1. Vit phng trnh in li ca cc dd

sau: K3PO4, Al2(SO4)3, H2SO4 , Ca(OCl)2,MgCl2Gi 5 Hs ln bngHng dn Hs vit phng trnh in liv hon thin cc phng trnh 2. Vit cng thc phn t ca cht khiin ly ra ion sau :a. K+ v CrO42- ; b. Fe3+ v NO3-

Ln bng

Cc Hs cn li lm vo vHs khc nhn xt bi lm trn bng

2.a. K2CrO4

b. Fe(NO3)3c. MgMnO4d. Al2(SO4)3

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c. Mg2+ v MnO4-; c. Al3+v SO42-

Gi 4 Hs ln bngHng dn hs lm

4. Cng c dn d:

Cch vit phng trnh in li BTVN:1. Cc dung dch sau y c cng nng 0,10 mol/l, dung dch dn in km nht l

a. HCl b. HF c. HI d. HBrGii thch ?2. Dung dch dn in tt nht l :

a. NaI 0,002 M b. NaI 0,010 M c. NaI 0,100 M d. NaI 0,001 MGii thch ?3. Trong 800ml c 8gam NaOH

a. Tnh nng mol/l ca dd NaOH

b. Phi thm ? ml H2O vo 200 ml dd NaOH trn c dd NaOH 0.1M4. Tnh s ml H2O cn thm vo 2 lt dd NaOH 1 M thu c dung dch mi cnng 0.1 M .

Chun b bi phn loi cht in li

Ngy 30/8/2012 T chn tit: 3LUYN TP PHN LOI CHT IN LI

I. Mc tiu

Cng c kin thc v in li, vit phng trnh in li ca cht in li mnh,cht in li yu.Cng c kin thc v axit- baz theo thuyt A-r-ni-ut.

II. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cu hi v bi tp vn dngHs: Chun b kin thc v s in li, phn li cht in li, axit-bazIV. Tin trnh

1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp

3. Cc hot ng

Hot ng ca gio vin Hot ng ca hc sinhHot ng 1: Kin thc v in li- Th no l in li? Vit cng thc tnh in li

=o

n

n vi 0 1

Trnh by ln bngGi s trong 1 lt dd. Nng mol

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Yu cu Hs chng minh cng thc0

C

C =

trong C l nng mol cht phn li raion, C0 l nng mol cht ho tan

- Kim tra vic lm bi tp ca Hs nh

cht phn li ra ion l C C C.6,02 . 1023 phn t cht in li ra ion(n). Nng mol cht ho tan l C0 C C0 . 6,02 . 1023 phn t cht tan (n0)

23

23

0 0 0

C. 6,02 . 10

C . 6,02 . 10

n C

n C

= = =

Tr li v ln bng vit phng trnhin li

Hot ng 2:Phn loi cht in li- Th no l cht in li mnh? Cht in liyu? Cho v d- Vit phng trnh in li ca cc cht inli sau: HCl, H2SO4, Na3PO4, Mg(NO3)2,KOH, Ba(OH)2, Fe(OH)3, H2S, CH3COOH,CH3COOKGV Hng dn Hs cch vit ptrnh in li- Cht no l cht in li mnh? Cht in liyu?

1. Cht in li mnhNm c cht in li mnh u c =12. Cht in li yu

Nm c cht in li yu 0< < 1CH3COOH CH3COO

- + H+ (1)Nm cHng s cn bng ca (1) gi l hng s

phn li

[ ]

3

3C

CH COO H K

CH COOH

+ =

Kc ch ph thuc bn cht ca cht in li vnhit

Nm c cn bng in li tun theo nguynl chuyn dch cn bng L satlie

Trnh by nh ngha v phn loi chtin liCc Hs ln bngHCl H+ + Cl-

H2SO4 H+ + HSO4-

HSO4

-

H

+

+ SO42-

H2SO4 2H+ + SO42-

Na3PO4 3Na+ + PO43-

Mg(NO3)2 Mg2+ + 2NO3-

KOH K+ + OH-

Ba(OH)2 Ba2+ + 2OH-

H2S H+ + HS-

HS- H+ + S2-

H2S 2H+ + S2-

CH3COOH H+ + CH3COO-

CH3COOK K+ + CH3COO-

Hot ng 3:Mt s bi tp v in li-Trnh by phng php:

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o Gi x l nng mol cht in li bphn li ra ion

o t C l nng mol ca cht in liban u

o T cng thc 00

.x

x C

C

= =

o Cho bi tp vn dngBi tp 1: in li ca CH3COOH0,2M l 1,2%. Tnh nng cc ion v

phn t

Bi tp 2: in li ca HNO2 trong ddHNO2 0,01 M l 18%.a. Tnh hng s phn li ca HNO2.b. Thm H2O vo dd HNO2 trn th hng

s CB c thay i khng (coi nhit khngi)

Vn dng lm cc bi tpGii bi tp 1:Ptin ly

CH3COOH H+ + CH3COO-

u 0,2ly 0,2 0,2 0,2[ ] 0,2- 0,2 0,2 0,2

Ta c : in li 1,2% vy = 0,012Vy [CH3COOH] = 0,2- 0,2.0,012 =0,1976M[H+] = [CH3COO-] = 0,2.0,012=0,0024MGii bi tp 2:Gii bi tp 1:a. Ptin ly

HNO2 H+ + NO2-

u 0,01ly 0,01 0,01 0,01

[ ] 0,01- 0,01 0,01 0,01

Ta c : in li 18% vy = 0,18

Vy : Ka =

01,001,0

)01,0( 2

=

0082,0

10.24,35

= 3,95.10-4

b. khi thm nc vo th dd long inng H+ gim nn Ka gim

4. Dn d:- Nm c cch vit phng trnh in li ca cc cht- Nh c cc hiroxit lng tnh thng gp Zn(OH)2, Al(OH)3, Cr(OH)3, Pb(OH)2,Sn(OH)2 v vit phng trnh in li ca chng

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Ngy 6/9/2012 T chn tit: 4THUYT AXIT-BAZ CA ARRHENIUS V BRONSTED-LAURY, CHT

LNG TNH,

I. Mc tiuHS bit cc khi nim axit-baz theo bronstet, cc khi nim hng s phn li axit vbazHS vn dng cc hng s phn li axit v baz tnh [H+] trong cc dung dch axit v

baz 1 ncII. Phng php: m thoi gii bi tp.III. Chun b- HS chun b cc kin thc c lin quan n axit, baz, v cht lng tnh theo ccquan im khc nhau- GV: H thng cu hi v bi tp vn dng .

IV. Tin trnh1. n nh lp: Kim tra s s2. Kim tra bi c: Nu nh ngha axit-baz theo areniut.

Vit PTL ca HClO, H2CO33. Cc hot ng

Hot ng ca gio vin Hot ng ca hc sinhHot ng 1:Axit baz theo thuyt A-r-ni-ut

- Yu cu Hs nh ngha axit baz theothuyt A-r-ni-ut- Th no l hiroxit lng tnh?

Vit phng trnh in li ca cc hiroxitlng tnh sau: Zn(OH)2, Al(OH)3,Cr(OH)3, Pb(OH)2, Sn(OH)2, Ba(HSO4)2,

Na3PO4, K2HPHO4, [AgNH3]NO3- Hng dn Hs dng axit ca cc hiroxitlng tnh c ho tr II l H2MO2, cn hotr III l l HMO.H2O- Lu cc hiroxit lng tnh c lc axitv lc baz yuHot ng 2:Axit baz theo thuyt

Bronsted-laury

- Axit l cht phn li ra cation H+

- Baz l cht phn li ra cation OH-

- Hiroxit lng tnh l hiroxit va cth phn li nh axit, va c th phn linh baz

- Mui l cht phn li ra cation kim loihay NH4+ v anion gc axit

Vit ptrnh

-YC HS vit PTL ca HCl, HNO3. ?

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Axit l cht c kh nng cho proton (H+)Baz l cht c kh nng nhn proton (H+)Vy Axit Baz + H

+

HClO + H2O H3O+ + ClO-

Hs vit c PT trao i proton ca nhng

cht sau vi H2O : NH3, HCO3-.Rt ra nhnxt v kh nng cho nhn proton ca cc

phn t trong mi PTSuy ra nhn xt+H2O cth cho hoc nhn proton.N lcht lng tnh+Theo thuyt Bronstet th axit, baz c thl phn t hoc ion- u im ca thuyt Bronstet so vi thuyt

areniut l gii thch c tnh axit, bazca mt s dd m nhn vo khng thycha ion H+ hay OH-.Hot ng 3Chia nhm v yu cu tng nhm gii cc

bi ton bng cch vn dng LBTTBi 1: Trong mt dung dch c cha a molCa 2+ v b mol Mg2+, c mol Cl- v dmol

NO3-. Tnh khi lng mui to thnh nucho bit a=0.01, c=0.01, d=0.03

Bi 2: Trong 800ml c 8gam NaOHa. Tnh nng mol/l ca dd NaOHb. Phi thm ? ml H2O vo 200 ml dd

NaOH trn c dd NaOH 0.1M.

Hng dn HS gii theo phng phpng cho

Bi 3: Tnh s ml H2O cn thm vo 2 lt

chng thuc loi hp cht noDn dt HS suy ra nh nghaGV ly VD vi CH3COOH.?ly VD vHClO(H+ l cch vit n gin ca H3O+)

-C nhng cht va c kh nng choproton va c kh nng nhn proton.Chng l cht lng tnh. C nhng chkhng c kh nng cho proton khng ckh nng nhn proton .Chng l chtrung tnh.

-Phn tch c th v lyVD

Gii bi 1 :p dng LBTT ta c :2*a+2*b=1*c+1*d

b=0.01 (mol)Khi lng mui to thnh l :

2 23muoi Ca Mg Cl NO

m m m m m+ + = + + +

muoim =40*0.01+24*0.01+35.5*0.01+62*0.03 =2.855(gGii bi 2 :a. Tnh nng mol/l ca dd NaOH .

NaOHn = m/M= 8/40=0.2 (mol)MC =n/V= 0.2/0.8=0.25 (M)

b. Tnh V H2O cn thmCch 1: Ap dng cng thc C1V1=C2V2

0.25*0.2= V2* 0.1 V2 =0.5 (l) =500(ml)Th tch H2O cn thm l:

2H OV = 500-200=300 (ml)Cch 2 : p dng quy tc ng cho

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dd NaOH 1 M thu c dung dch mic nng 0.1 M .- GV : nhn xt chung, nh gi cch lmvic ca nhm

=2

NaOH

H O

V 0.1

V 0.15

2H OV = 300 (ml)- Hot ng theo nhm gii .V nhnxt gia cc nhm vi nhau .p n : V= 18 (l)

4. Cng c: Cc cng thc tnh pha long , nh lut bo ton in tch, quytc ng cho, cng thc tng quan gia C% v CM .

5. Bi tp v nh :1. Trn x (g) H2O vo y (g) dung dch HCl 30 % c dung dch HCl 12 % .

Tnh t l x: y2. Trong 1 dung dch cha 2 loi cation Fe3+ (0.1 mol) v Al3+(0.2 mol) cng 2

loi anion Cl-

(x mol) v SO42-

(y mol). Tnh x, y bit rng khi c cn dungdch v lm khan thu c 46.9 gam mui khan .3. Tnh s ml dd NaOH 2.5 %(D=1.03g/ml) iu ch 80 ml dd NaOH 35

% c (D=1.38g/ml)4. Lm bay hi 500 ml dung dch NaOH 20% (D=1.2g/ml) ch cn 300g

dd. Tnh nng % ca dung dch ny .6. Dn d : Lm bi tp SBT v BTVN.

Ngy 10/9/2012 T chn tit: 5pH CA DUNG DCH, S THY PHN CA MUI

I. Mc tiu- HS bit cch tnh pH ca 1 dung dch.- Bit cch tnh nng ion v nng phn t ca cc cht in ly trong hn hpnhiu dung dch.- Rn luyn k nng t duy, tnh ton gii cc bi ton nhanh.- Da vo cu to ca mui nhn bit c mi trng ca dung dch mui- Vit c phng trnh thu phn ca mui.- Gii mt s bi tp c bn.II. Phng php: vn p gii bi tp.

III. Chun b- Gv:H thng cu hi v bi tp vn dng tnh pH, nng ion, nng phn ttrong hn hp nhiu dung dch.

- Hs: chun b nhng kin thc c lin quan n pHIV. Tin trnh


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Hot ng ca gio vin Hot ng ca hc sinhHot ng 1:Bi ton tnh nng ion trong hn hp nhiu dungdch

Trnh by phng php- Tnh s mol ca cc cht in li- Vit phng trnh in li- Tnh [ ion]Yu cu hc sinh gii1. Tnh nng mol/l ca cc ion H+

trong dd HNO3 10 % (D=1.054g/ml)

2. Tnh nng ion trong cc dungdch bi tp 1,2 hot ng trc- GV: Nhn xt v b sung (nu c)- GV: Cng c li cch tnh nng ion trong hn hp nhiu dd

Lm nhanh vo v1/ - Nng axit HNO3

CM=10*10*1.054

63

=1.67 (M)

Nng ion H+ trong dd .H+ = 1.67 (M)

- HS : T gii v trnh by bi gii2 / Bi tp 1 :a/ Tnh nng cc ion c trong dd D:

- Dd D l dd HCl c CM=0.2 MNng ion H+ H+ = Cl

=0.2 (M)b/ Tnh nng ion trong dd A, ddB.

- Dd A l dd HCl c CM=0.142 MNng ion : H+ = Cl

=0.142 (M)- Dd B l dd HCl c CM=0.06M

Nng ion : H+ = Cl =0.6 (M)

3./ Bi tp 2 :- Tnh nng ion trong dd H2SO4 thu c

Nng mol/l ca H2SO4 : CM=3.53 (M)Nng ion : H+ =0.3.53*2=7.06 (M)

2

4SO

=0.3.53 (M)

Hot ng 2:Bi ton tnh pH cadd n ginTrnh by phng php gii- Tnh s mol ca cc cht in li- Vit phng trnh in li- Tnh [H+] da vo phng trnhin li hoc da vo tch s ion canc

pH = -lg([H+])Yu cu hc sinh vn dng gii cc

bi ton tnh pH n gin .( y lnhng dng bi ton n gin ch pdng cng thc )1. Tnh pH ca dd sau :a. 100 ml dd HCl 0.01 M .

Tng c nhn Hs gii.1 Tnh pH ca dd :a . Hn + = 0.01*0.1 =0.001 (mol)

H+ =0.001

0.1=0.01 (mol/l) pH=2

b. OH =0.01 (mol/l)

H+ = 10-12 (mol/l) pH=12

c. H+ =10-3 pH=3d. HCln =0.12*0.05 =6*10-3(mol)

NaOHn = 0.1* 0.05 =5*10-3(mol)PTPU :NaOH +HCl NaCl+H2O

- Khi phn ng xy ra hon ton th :HCln d = 6*10-3 - 5*10-3 = 10-3(mol)

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b. Dd KOH 0.01 M.c. Dd H2SO4 0.0005 Md. Trn ln 50 ml dd HCl 0.12 Mvi 50 ml dd NaOH 0.1 M thu cdd Y.

- GV : Nhn xt v nh gi phntrnh by ca hc sinh .Hot ng 3:Kin thc c bn v

phn ng thy phn ca mui.Th no l phn ng thu phnmui?

Nu cch xc nh mi trng cadung dch mui?Trnh by cch xc nh mi trngca dung dch mui.

- Vit phng trnh in li ca mui- Nu cation l cation ca cc bazyu s b thu phn:

n (n 1)

2M H O M(OH) H+ + ++ +

- Nu anion l anion ca cc axityu s b thu phn:

m (m 1)

2A H O HA OH + +

Mui trung ho to bi cation cabaz mnh v anion ca axit yu

Mi trng ca mui ny l g?Xc nh mi trng ca cc dungdch mui sau:

3 2 2 3 2 3CH COONa,Na S,Na CO ,K SO

Gi Hs ln bngHng dn Hs vit phng trnhthu phn ca cc anion ca chtin li yu

Mui trung ho to bi cation cayu v anion ca axit mnhMi trng ca mui ny l g?Xc nh mi trng ca cc dungdch mui sau:

H+ =10-3/(0.05+0.05) = 10-2(mol/l)

pH=2

Tm tt kin thc c

Nm vng phng php v phng trnh thuphn ca cc ion ca cht in li yu

Mui trung ho to bi cation ca baz mnh

v anion ca axit yu c mi trng kim3 3

3 2 3

CH COONa CH COO Na

CH COO H O CH COOH OH

+

+

+ + 2

2

2

2

2

2 3 3

2

3 2 3

2

2 3 3

2

3 2 3

Na S 2Na S

S H O HS OH

Na CO 2Na CO

CO H O HCO OH

K SO 2K SO

SO H O HSO OH

+

+

+

+

+ +

+

+ +

+

+ +

Mi trng ca cc dung dch trn l baz

Mui trung ho to bi cation ca yu v anionca axit mnh to mi trng axit

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3 4 3 3 4FeCl ,CuSO ,Al(NO ) ,ZnSO

Gi Hs ln bngHng dn Hs vit phng trnhthu phn ca cc cation ca chtin li yu

Mui trung ho to bi cation cabaz mnh v anion ca axit mnhkhi tan trong nc to dung dch cmi trng trung tnh do cc ionny khng b thu phnCho Hs mt s bi tp trcnghim

3

3

3 2

2

FeCl Fe 3Cl

Fe H O Fe(OH) H

+

+ + +

+

+ + 2 2

4 4CuSO Cu SO+ +

2

2Cu H O Cu(OH) H+ + ++ +

3

3 3 3

3 2

2

2 2

4 4

2

2

Al(NO ) Al 3NO

Al H O Al(OH) H

ZnSO Zn SO

Zn H O Zn(OH) H

+

+ + +

+

+ + +

+

+ +

+

+ +

Mi trng ca cc dung dch trn l axit

1. Dung dch no sau y c mi trng kim?A. NaClO B. NaClO3 C. KClO4 D.

NaCl2. Dung dch no sau y c mi trng axit?A. Na2PbO2 B. NaNO3 C. KClO4 D.

NH4Cl3. Dung dch no sau y c pH = 7A. CaCl2 B. NaAlO2 C. CuSO4 D. K2S

4. Tm nhn nh sai:A. dung dch NaHSO4 c pH7C. dung dch AgNO3 c pH=7D. dung dch K2 CO3 c pH>7

Hot ng 4 : cng c dn dCng c: Cng thc tnh pH, cch tnh nng phn t, ion trong hn hp nhiu dd.

Dn d : Lm bi tp SBT.

Bi tp v nh :Bi 1: A l dd KOH c pH= 13.

a. Tnh nng mol/l ca dd A .b. Nu pha long dd A 50 ln c dd B. Tnh pH ca dd B.c. Nu un 1 lt A bay hi bt mt lng nc thu c dd C c pH=13.062

Tnh nng mol ca KOH trong dd C . Tnh th tch dd C .

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Bi 2: Cho dd NaOH c pH= 12 (ddA) .Cn pha long dd A bao nhiu ln thuc dd B c pH=11.Bi 3: Cho dd A :HNO3 0.01 M .Tnh nng mol ca cc ion v pH ca dd A .

Ngy 17/9/2012 T chn tit: 6PHN NG TRAO I ION TRONG DUNG DCH CHT IN LI- LUYN TP

I. Mc tiu- Da vo iu kin xy ra phn ng trao i trong dung dch cc cht in li bit

c phn ng c xy ra hay khng xy ra .- Vit c phng trnh ion rt gn ca phn ng

II. Phng php: m thoi gii bi tp.III. Chun b

Gv:H thng cu hi v bi tp vn dng

Hs: Nm vng bng tnh tan ca mt s cht trong nc v iu kin xy ra phnng trao i ionIV. Tin trnh


Hot ng ca gio vin Hot ng ca hc sinhHot ng 1: iu kin phn ng

trao i ion xy ra c trong cc ddcht in liTL: p rao i ion trong dung dchcht in li xy ra khi c s kt hpgia cc ion trong dung dch tothnh t nht 1 trong 3 trng hp sau-Cht kt ta-Cht in li yu-Cht khHot ng 2:Phng php vit

phng trnh ion rt gnPhn ng trao i ion xy ra khi no?Trnh by phng php:- Vit phng trnh phn tPhn tch thnh ion ( lu : cht ktta, cht kh v cht in li yu vngi dng phn t )- Rt gn cc ion ging nhau hai v

Nu iu kin xy ra phn ng trao i ion

Ch nm vng phng php

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c phng trnh ion rt gnHot ng 2:yu cu Hs l bi tp 1

sgk

Kim tra vic lm bi tp nh ca

HsYu cu Hs nhn xtB sung v hon thin bi tp

4 Hs ln bng vit phng trnh phn t vion rt gn

3

3. 3 ( )a Fe OH Fe OH + +

b. Khng phn ng2

3 3 2

2

4 3 4

2

2 2 2

2

2

2

2 2

2

2 2

..

. ( ) 2 2

. 2

. ( ) 2 2

. ( ) 2 2

c HSO OH SO H Od HPO H H PO

e Cu OH OH CuO H O

g FeS H Fe H S

h Cu OH H Cu H O

i Sn OH H Sn H O

+

+ +

+ +

+ +

+ ++

+ +

+ +

+ +

+ +

Cc Hs khc xem li bi tp lm nh nhnxt bi lm trn bng

Hot ng 3: Trnh by phngphp iu ch CuS bng 3 phngphp khc nhau

Yu cu Hs vit phng trnh ion rtgn ca cc phn ng Bn cht ca cc phn ng l g?

4 2 2 4

3 2 2 3

2 2

( ) 2

2

CuSO K S CuS K SOCu NO Na S CuS NaNO

CuCl H S CuS HCl

+ ++ +

+ +

Phng trnh ion rt gn: 2 2Cu S CuS + + Bn cht l s kt hp ca ion Cu2+ v S2- to kt ta CuS

Hot ng 4:Yu cu Hs l bi tp 8sgk

Gii thiu v cc kt ta v mu scca chngKim tra vic lm bi tp nh caHs

Yu cu Hs nhn xtB sung v hon thin bi tpYu cu Hs l bi tp

Bi 1: Vit phn ng xy ra dngphn t v dng Ion khi cho d2

5 Hs ln bng vit phng trnh phn t vion rt gn

4 2 2 4.a CuSO K S CuS K SO+ +2 2Cu S CuS + +

3 2 2 3. ( ) 2b Cd NO Na S CdS NaNO+ +2 2Cd S CdS + +

3 2 2 3. ( ) 2c Mn NO Na S MnS NaNO+ +2 2Mn S MnS+ +

4 2 2 4.d ZnSO K S ZnS K SO+ +2 2Zn S ZnS+ +

3 2 2 3. ( ) 2e Fe NO Na S FeS NaNO+ +2 2Fe S FeS+ +

Cc Hs khc xem li bi tp lm nh nhnxt bi lm trn bng

Gii Bi 1 :NaHCO3 + HCl NaCl + H2O + CO2.

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NaHCO3 ln lt phn ng vi d2

HCl, d2 KOH, d2 Ba(OH)2 d, d2

H2SO4 thiu. Trong mi phn ng ion HNO3- ng vai tr l axt haybaz.

Bi 2: Vit ptp dng phn t v ionthugn ca cc p (nu c) xy ra

giaa.dd AgNO3 v dd HClb. dd Na2CO3 v dd Ca(NO3)2c.dd BaCl2 v dd K2SO4d. dd NaHCO3 v dd NaOH

e. FeS v dd HClf. dd NaHCO3 v dd HClg.dd BaCl2 v dd NaNO3h. dd FeSO4 v dd HCl

Bi 3 Cho dd AgNO3 d vo 500 mldd HCl a M thu c 143,5 gam ktta. Vit pt hh ca p xy ra dng

phn t v ion thu gn. Tnh aS. a = 1/0,5 = 2 M

HCO3- + H+ CO2 + H2O2 NaHCO3 + 2 KOH K2CO3 + Na2CO3+ 2 H2O

HCO3- + OH- CO32- + H2ONaHCO3 + Ba(OH)2 BaCO3 + NaOH +

H2OHCO3- + OH- + Ba2- BaCO3 + H2O

Hot ng 4 : cng c dn dDn d : Lm bi tp SBT.

Hot ng 5: Bi tp v nhBi 1: Vit phng trnh phn t v ion rt gn khi cho cc cht sau y tc dng vinhau tng i mt:

3 2 2 3 2 2 4( ) , , , , ,Ba NO Na CO MgCl NaOH K SO HCl

Bi 2: Vit phng trnh phn t ca cc phn ng c phng trnh ion rt gn sau:3

3. 3 ( )a Fe OH Fe OH + + 23 2 2.2b H CO CO H O

+ + +2

3 3 2.c HSO OH SO H O + + 2 24 4.d Ba SO BaSO

+ +

Ngy 1/10/2012 T chn tit: 7KHI QUT V NIT

I. Mc tiu- Cng c kin thc v tnh cht ca cc nguyn t nhm nit, ca nit v

amoniac, cch iu ch nit trong cng nghip v trong phng th nghim.

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- Vit c phng trnh phn ng lin quan ti tnh cht ca nit- Gii c cc bi tp tnh ton

II. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cu hi v bi tp

Hs: Cc bi tp phn khi qut cc nguyn t nhm nit, nit, amoniac trong SGK,SBTIV. Tin trnh


Hot ng ca gio vin Hot ng ca hc sinhHot ng 1:H thng l thuyt cbn

Nu h thng cu hi:- Vit cu hnh electron lp ngoicng ca cc nguyn t nhm nitv cho bit s oxi ho ca ccnguyn t ?- Nu tnh cht ho hc v cchiu ch nit?- Nu tnh cht ho hc caamoniac?

Tr li h thng cu hi ca gio vin- Nhm nit c cu hnh ns2np3, nit c s oxi

ho l: -3, +1, +2, +3, +4, +5. Cc nguyn t P,As, Sb, Bi c s oxi ho l -3, +3 +5.- Nit va c tnh oxi ho ( tc dng vi H2 vvi kim loi) va c tnh kh ( tc dng vioxi) trong tnh oxi ho c trng

Tnh baz yu: tc dng vi nc,vi axit, vi dd mui

- NH3 Kh nng to phc vi Cu(OH)2,Zn(OH)2, AgCl

Tnh kh: tc dng vi oxi, clo,oxit kim loi

Hot ng 2:Bi tp 5 trang 40SGK- Bng th nghim no c th nhn

bit nit ln clo, hiro clorua, hirosunfua? Vit phng trnh- Kim tra s lm bi ca Hs nh- Yu cu Hs nhn xt bi lm tren

bng v nhn mnh nhm khc sukin thc

Hs ln bngCc Hs cn li ch theo di bn lm v nhnxt- Dn hn hp kh vo ddch NaOH, dung dchthu c c tnh ty mu: Cl2 + 2NaOH

NaCl + NaClO + H2O- Nhn bit nit ln hiro clorua dn qua nc,

hiro clorua tan nhiu tong nc to thnh axitHCl lm qu tm chuyn - Nhn bit nit c ln H2S dn qua dung dchPb(NO3)2 to kt ta en: H2S + Pb(NO3)2PbS + 2HNO3

Hot ng 3: Bi tp 6 trang 40 Vn dng phng php ca gio vin p

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SGK- Hng dn Hs cc bc lmBc 1: tnh s mol tng chtBc 2: vit phng trnh phnng

Bc 3: tnh s mol cht cn dda vo phng trnh phn ngBc 4: tnh ton theo bi

dng lm bi tp

Hot ng 4:Bi tp 5 trang 47SGKHng dn Hs lm bi tp 5 SGK

Kh A l NH3, dung dch A l dung dch NH3NH3 + HCl NH4ClNH4Cl + NaOH NH3 + H2O + NaClNH3 + HNO3 NH4NO3NH4NO3 N2O + 2 H2O

Hot ng 5:Bi tp 2.10 SBTTrnh by phng phpPhn ng tng hp NH3 c cim g?Hiu sut phn ng:

tt

lt

VH

V=

Cng iu kn nhit v p sut t l v thtch bng t l v s mol nn ta c:

0t ,xt,p

2 2 3N 3H 2NH+

Theo bi (lit) 4 14Khi phn ng x 3x 2xSau phn ng 4 x 14 3x 2xTheo bi ta c: 18 2x = 16,4

x = 0,8 lt

Th tch NH3 thu c 1,6 ltTh tch NH3 theo l thuyt l 8 lt1,6

.100% 20%8

H= =

Hot ng 6:Bi tp 2.15 SBTTrnh by phng phpBc 1: tnh s mol ca NH3 vCuOBc 2: vit phng trnhBc 3: da vo phng trnh xc

nh NH3 v CuO cht no phnng htTnh ton theo yu cu ca bi- hng dn Hs lm bi tp

Vn dng phng php ca gip vin lmbi tp

Bi tp v nh:Cu 1: Hn hp N2 v H2 c t l th tch l 1:3 dn vo bnh phn ng c th tch 60lt, c xt thch hp, p sut hn hp kh ban u l 224,0 atm v nhit l 4270C

a.Tnh s mol N2 v H2 ban u

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b.Tnh p sut trong bnh v thnh phn phn trm cc kh sau phn ng bitrng hiu sut phn ng tng hp l 40%Cu 2: Cho dung dch NH3 ti d vo 20 ml dung dch Al2(SO4)3. ho tan kt tathu c cn ti thiu 10 ml dung dch NaOH 2M

a. Vit phng trnh dng phn t v ion rt gn

b. Tnh nng mol ca dung dch Al2(SO4)3 ban uCu 3: Dn 1,344 lt NH3 vo bnh cha 0,672 lt Cl2a. Tnh thnh phn phn trm cc kh sau phn ng

b. Tnh khi lng ca mui NH4Cl to thnh bit rng th tch cc kho dktc

Ngy 5/10/2012 T chn tit: 8NIT - AMONIAC - MUI AMONI

I. Mc tiu- H thng kin thc v tnh cht ca Nit, amoniac, mui amoni- Vit c phng trnh phn ng lin quan ti tnh cht ho hc ca N2, NH3,

v mui amoni- Gii c bi tp lin quan

II. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc v tnh cht ho hc v iu ch N2, NH3, mui amoni

IV. Tin trnh1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ngHot ng ca gio vin Hot ng ca hc sinh

Hot ng 1:GV: Chp ln bng, yu cu HSchp vo v.Bi 1:Trong mt bnh kn dung tch 10 lt

cha 21 gam nit. Tnh p sut cakh trong bnh, bit nhit ca kh

bng 250C.GV: Yu cu 1 HS ln bng gii, ccHS cn li lm nhp v theo di bi

bn lm.GV: Yu cu 1 HS nhn xt, GVnhn xt ghi im.

Bi 1:Trong mt bnh kn dung tch 10 lt cha 21gam nit. Tnh p sut ca kh trong bnh,

bit nhit ca kh bng 250C.Gii:

S mol kh N2: )(75,028

21 mol=

p sut ca kh N2:

p = )(83,110

)27325(082,0.75,0atm

V

nRT=

+=

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Hot ng 2:GV: Chp ln bng, yu cu HSchp vo v.Bi 2:

Nn mt hn hp kh gm 2 mol nit

v 7 mol hiro trong mt bnh phnng c sn cht xc tc thch hp vnhit ca bnh c gi khng i 4500C. Sau phn ng thu c 8,2mol hn hp kh.a/ Tnh phn trm s mol nit

phn ng .b/ Tnh th tch (kt) kh ammoniacc to thnh.GV: Yu cu HS tho lun.

GV: Hng dn HS cch lm biHS:Nghe ging v hiu

HS: T tnh phn trm s mol nit phn ng, th tch (kt) khammoniac c to thnh.

Hot ng 3:GV: Chp ln bng, yu cu HS

chp vo v.Bi 3:Cho lng d kh ammoniac i t tqua ng s cha 3,2 g CuO nungnng n khi phn ng xy ra honton, thu c cht rn A v mt hnhp kh. Cht rn A phn ng va vi 20 ml dung dch HCl 1 M

Bi 2:Nn mt hn hp kh gm 2 mol nit v 7mol hiro trong mt bnh phn ng c sncht xc tc thch hp v nhit ca bnh

c gi khng i 450

0

C. Sau phn ngthu c 8,2 mol hn hp kh.a/ Tnh phn trm s mol nit phn ng .

b/ Tnh th tch (kt) kh ammoniac c tothnh.

GiiN2 (k) + 3H2 (k) 2NH3(k)

S mol kh ban u: 2 70S mol kh phn ng: x 3x

2xS mol kh lc cn bng: 2-x 7 3x 2xTng s mol kh lc cn bng: 2 x + 7 3x +2x = 9 2xTheo ra: 9 2x = 8,2x = 0,4a/ Phn trm s mol nit phn ng

%202

%100.4,0=

b/ Th tch (kt) kh ammoniac c tothnh: 2.0,4. 22,4 = 17,9 (lt)HS:Nghe ging v hiu

HS: T tnh phn trm s mol nit phnng, th tch (kt) kh ammoniac c tothnh.

HS:Nghe ging v hiu

HS: Ln bng trnh byBi 3:Cho lng d kh ammoniac i t t qua ngs cha 3,2 g CuO nung nng n khi phnng xy ra hon ton, thu c cht rn A vmt hn hp kh. Cht rn A phn ng va vi 20 ml dung dch HCl 1 Ma/ Vit pthh ca cc phn ng.

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a/ Vit pthh ca cc phn ng.b/ Tnh th tch nit ( ktc) c tothnh sau phn ng.GV: Yu cu HS tho lun.GV: Hng dn HS cch vit pt.

GV:Yu cu HS ln bng trnh bycu b

GV: Gi HS nhn xt

b/ Tnh th tch nit ( ktc) c to thnhsau phn ng.

Giia/ pthh ca cc phn ng.2NH3 + 3CuO Ct N2 + 3Cu + 3H2O (1)

Cht rn A thu c sau phn ng gm Cu vCuO cn d . ch c CuO phn ng vi dungdch HCl.CuO + 2HCl CuCl2 + H2O

b/ S mol HCl phn ng vi CuO: nHCl =0,02( mol)Theo (2) s mol CuO d: nCuO = 1/2 s molHCl = 0,02: 2 = 0,01 (mol)S mol CuO tham gia phn ng (1) = s molCuO ban u s mol CuO d =

)(03,001,0802,3 mol=

Theo (1), s mol N2=3

1s mol CuO =

3

1

.0,03 = 0,01 (mol)Th tch kh nit to thnh : 0,01. 22,4 =0,224 (lt)

Hot ng 4: Cng c - dn d* Cng c:

Amoniac phn ng c vi tt c cc cht trong nhm no sau y.A. HCl, O2, Cl2, CuO, dd AlCl3 B. H2SO4, PbO, FeO, NaOHC. HCl, KOH, FeCl3, Cl2 D. KOH, HNO3, CuO, CuCl2

* Dn d:

Ngy 10/10/2012 T chn tit: 9

AXIT NITRIC - MUI NITRATI. Mc tiu- H thng kin thc v tnh cht ca axxit nitric, mui nitrat- Vit c phng trnh phn ng lin quan ti tnh cht ho hc ca HNO3,

v mui nitrat- Gii c bi tp lin quan


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Gv:H thng cc dng bi tpHs: Kin thc v tnh cht ho hc v iu ch HNO3, mui nitratIV. Tin trnh


3. Cc hot ngHot ng ca gio vin Hot ng ca hc sinh

Hot ng 1:GV: Chp ln bng, yu cu HSchp vo v.Bi 1:Khi cho oxit ca mt kim loi ha trn t dng vi dung dch HNO3 d thto thnh 34,0 g mui nitrat v 3,6 g

nc ( khng c sn phm khc ).Hi l oxit kim loi no v khilng ca oxit kim loi phn ngl bao nhiuGV: Hng dn HS cch vit pt, gi cch gii, yu cu HS lm

GV: Yu cu HS cho bit kt qu

GV: Yu cu HS vit pt v tnh khilng ca oxit kim loi phn ng

HS: Vit pt v tnh khi lng caoxit kim loi phn ng

Hot ng 2:GV: Chp ln bng, yu cu HSchp vo v.Bi 2:Chia hn hp hai kim loi Cu v Allm 2 phn bng nhau.

HS: Tho lun lm biBi 1:Khi cho oxit ca mt kim loi ha tr n tdng vi dung dch HNO3 d th to thnh34,0 g mui nitrat v 3,6 g nc ( khng csn phm khc ). Hi l oxit kim loi nov khi lng ca oxit kim loi phn ng

l bao nhiuGii:

PTHH.M2On + 2nHNO3 2M(NO3)n + nH2O (1)Theo phn ng (1), khi to thnh 1 mol ( tc(A + 62n) g ) mui nitrat th ng thi tothnh n/2 mol ( 9n gam ) nc(A + 62n) g mui nitrat 9n g nc34,0 g mui nitrat 3,6 g nc

Ta c: 6,39

34

62 nnA=

+

Gii pt: A = 23n.Ch c nghim n = 1, A = 23Vy kim loi M trong oxit l natri

Na2O + 2HNO3 2NaNO3 + H2O (2)Theo phn ng (2)C to ra 18 g nc th c 62 g Na2O phnngVy to ra 3,6g nc th c x g Na2O phn

ngx = (3,6.62) : 18 = 12,4 (g)

HS:Ln bng trnh byBi 2:Chia hn hp hai kim loi Cu v Al lm 2

phn bng nhau.+ Phn th nht: Cho tc dng vi dung dch

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+ Phn th nht: Cho tc dng vidung dch HNO3 c ngui thu c8,96 lt kh NO2 ( ktc)+ Phn th hai: Cho tc dng vihon ton vi dung dch HCl, thu

c 6,72 lt kh ( ktc)Xc nh thnh phn phn trm vkhi lng ca mi kim loi tronghn hp.GV: Yu cu 1 HS ln bng trnh

by. Cc HS cn li lm v theo dibi ca bn

GV: Gi HS nhn xt, ghi im


Cho 12,8 g Cu tc dng vi dungdch HNO3 c, sinh ra kh NO2.Tnh th tch NO2 ( ktc).GV: Yu cu 1 HS ln bng trnh

by. Cc HS cn li lm v theo dibi ca bnGV: Gi HS nhn xt, ghi im


Nhit phn hon ton 27,3 gam hnhp rn gm NaNO3 v Cu(NO3)2,thu c hn hp kh c th tch 6,72lt ( ktc).

HNO3 c ngui thu c 8,96 lt kh NO2( ktc)+ Phn th hai: Cho tc dng vi hon tonvi dung dch HCl, thu c 6,72 lt kh( ktc)

Xc nh thnh phn phn trm v khi lngca mi kim loi trong hn hpGii

Phn th nht, ch c Cu phn ng vi HNO3c.Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2H2O (1)Phn th 2, ch c Al phn ng vi2Al + 3HCl AlCl3 + 3H2 (2)Da vo (1) ta tnh c khi lng Cu c

trong hn hp l 12,8 g.Da vo (2) ta tnh c khi lng Al ctrong hn hp l 5,4 g.% khi lng ca Cu = 70, 33%% khi lng ca Al = 29,67%

HS:Ln bng trnh byBi 3:Cho 12,8 g Cu tc dng vi dung dch HNO3

c, sinh ra kh NO2. Tnh th tch NO2( ktc).Gii

Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2H2O0,2 0,4 (mol)

nCu = )(2,064

8,12mol=

)(96,84,22.4,02

lVNO ==

HS: Tho lun lm biHS: Ln bng trnh byBi 4:

Nhit phn hon ton 27,3 gam hn hp rngm NaNO3 v Cu(NO3)2, thu c hn hpkh c th tch 6,72 lt ( ktc).Tnh thnh phn % v khi lng ca mi

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Tnh thnh phn % v khi lngca mi mui trong hn hp X.GV: Hng dn HS cch vit pt, gi cch gii, yu cu HS lm

GV: Yu cu HS ln bng gii

GV: Nhn xt ghi im


Nung nng 27,3 g hn hp NaNO3

v Cu(NO3)2 ; hn hp kh thot rac dn vo 89,2 ml nc th cnd 1,12 l kh(ktc) khng b hp th.( Lng O2 ha tan khng ng k)a/ Tnh khi lng ca mi muitrong hn hp u.b/ Tnh nng % ca dd axt.

GV: Hng dn HS cch gii, yucu HS ln bng trnh by

mui trong hn hp X.

Gii:2NaNO3

0t 2NaNO2 + O2 (1)

x 0,5x ( mol)

2Cu(NO3)2

0t

2CuO + 4NO2 + O2 (2)y y 2y 0,5y( mol)Gi x v y l s mol ca NaNO3 v Cu(NO3)2trong hn hp X. Theo cc phn ng (1) v(2) v theo bi ra . Ta c.

85x + 188y = 27,30,5x + 2y + 0,5y = 0,3

x = y = 0,1

% %1,313,27

%100.1,0.853

==NaNOm

% %9,683.27

%100.1,0.18823 )(

==NOCum

HS:Ln bng trnh byBi 5:

Nung nng 27,3 g hn hp NaNO3 vCu(NO3)2 ; hn hp kh thot ra c dn vo89,2 ml nc th cn d 1,12 l kh(ktc)khng b hp th. ( Lng O2 ha tan khng

ng k)a/ Tnh khi lng ca mi mui trong hnhp u.b/ Tnh nng % ca dd axt

Gii2NaNO3

0t 2NaNO2 + O2 (1)

2 1 ( mol)2Cu(NO3)2

0t 2CuO + 4NO2 + O2 (2)

2 4 1 ( mol)4NO2 + O2 + 2H2O 4 HNO3 (3)4 1 4 ( mol)

a/ Theo pt (1), (2), (3) , nu cn d 1,12 l kh( hay 0,05 mol ) th l kh O 2, c th coilng kh ny do mui NaNO3 phn hy toraT (1) ta c: )(1,005,0.23 molnNaNO ==

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Nung mt lng mui Cu(NO3). Saumt thi gian dng li, ngui vem cn th thy khi lng gim i54g.

+ Khi lng Cu(NO3) bphn hy.+ S mol cc cht kh thot ra l

GV: Yu cu 1 HS ln bng trnhby. Cc HS cn li lm v theo dibi ca bn


)(5,885.1,03

gmNaNO ==

)(8,185,83,2723 )(

gm NOCu ==

)(1,0188:8,1823 )(

moln NOCu ==

T (2) ta c: )(2,04.2

1,02

molnNO ==

)(05,01.21,0

2molnO ==

( Cc kh ny hp th vo nc)T (3) ta c : )(2,023 molnn NOHNO ==

Khi lng HNO3 l: 0,2.63 = 12,6 (g)Khi lng ca dung dch = 0,2.46 + 0,05.32+ 89,2 = 100 (g)C% ( HNO3) = 12,6 %

HS:Ln bng trnh byBi 6:

Nung mt lng mui Cu(NO3). Sau mt thigian dng li, ngui v em cn th thykhi lng gim i 54g.

+ Khi lng Cu(NO3) b phn hy.+ S mol cc cht kh thot ra l

Gii2Cu(NO3)2 0

t 2CuO + 4NO2 + O2 + C 188g mui b phn hu th khi lnggim : 188 80 = 108 (g)Vy x = 94 g mui b phn hu th khi lnggim 54 gKhi lng mui b phn hu

)(9423 )(

gm NOCu =

+ )(5,0188:9423 )( moln NOCu ==

)(14.2

5,02 molnNO ==

)(25,0.2

5,02

molnO ==


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Ha tan 12,8 g kim loi ha tr II trong mt lng va dung dch HNO3 60%( d = 1,365g/ml), thu c 8,96 lt ( ktc) mt kh duy nht mu nu . Tn cakim loi v th tch dung dch HNO3 phn ng l

A. Cu; 61,5 ml B. Cu; 61,1 ml C. Cu; 61,2 ml D. Cu; 61,0 ml* Dn d:

Ngy 16/10/2012 T chn tit: 10PHOTPHO - AXIT PHOTPHORIC - MUI PHOTPHAT

I. Mc tiu- H thng kin thc v tnh cht ca P, H3PO4, mui photphat- Vit c phng trnh phn ng lin quan ti tnh cht ho hc ca P,

H3PO4, mui photphat- Gii c bi tp lin quan


Gv:H thng cc dng bi tpHs: Kin thc v tnh cht ho hc v iu ch P, H3PO4, mui photphatIV. Tin trnh


Hot ng ca gio vin Hot ng ca hc sinhHot ng 1:

GV: Chp ln bng, yu cu HSchp vo v.Bi 1:Cho 11,76 g H3PO4 vo dung dchcha 16,8 g KOH. Tnh khi lngca tng mui thu c sau khi chodung dch bay hi n khGV: Yu cu HS cch vit pt, gi cch gii, yu cu HS lmGV: Yu cu HS ln bng gii

GV: Nhn xt ghi im


HS: Chp , HS: Tho lun lm bi

HS: Ln bng trnh byBi 1:Cho 11,76 g H3PO4 vo dung dch cha 16,8g KOH. Tnh khi lng ca tng mui thuc sau khi cho dung dch bay hi n kh

Gii:H3PO4 + KOH KH2PO4 + H2O (1)H3PO4 + 2KOH K2HPO4 + 2H2O (2)H3PO4 + 3KOH K3PO4 + 3H2O (3)

S mol H3PO4 0,12 (mol)S mol KOH 0,3 (mol)Da vo t l s mol gia KOH v H3PO412,72 g K3PO4 v 10,44g K2HPO4

HS: Chp HS:Ln bng trnh byBi 2:

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Bng phng php ha hc, hyphn bit dung dch HNO3 v dungdch H3PO4GV: Yu cu HS ln bng trnh byGV: Gi HS nhn xt, ghi im

Hot ng 3:GV: Chp ln bng, yu cu HSchp vo v.Bi 3:Bng phng php ha hc phn bitcc mui: Na3PO4, NaCl, NaBr,

Na2S, NaNO3. Nu r hin tng

dng phn bit v vit phngtrnh ha hc ca cc phn ng

GV: Yu cu HS chia nhm tholun. Gi i din mt nhm lntrnh by


Hot ng 4:GV: Chp ln bng, yu cu HSchp vo vBi 4:Cho 62 g canxi photphat tc dngvi 49 g dung dch H2SO4 64%. Lm

Bng phng php ha hc, hy phn bitdung dch HNO3 v dung dch H3PO4

GiiCho mnh kim loi Cu vo dung dch catng axit

Cu + HNO3 ()

Cu(NO3)2 + 2NO2 +2H2OCu khng t dng vi H3PO4HS: Chp HS:Ln bng trnh byBi 3:Bng phng php ha hc phn bit ccmui: Na3PO4, NaCl, NaBr, Na2S, NaNO3.

Nu r hin tng dng phn bit v vitphng trnh ha hc ca cc phn ng

GiiDng dung dch AgNO3 phn bit ccmui: Na3PO4, NaCl, NaBr, Na2S, NaNO3.Ly mi mui mt t vo tng ng nghim,thm nc vo mi ng v lc cn thn ha tan ht mui. Nh dung dch AgNO3 votng ng nghim- dung dch no c kt ta mu trng khngtan trong axit mnh, th l dung dch NaCl

NaCl + AgNO3

AgCl + NaNO3- dung dch no c kt ta mu vng nhtkhng tan trong axit mnh, th l dungdch NaBr.

NaBr + AgNO3 AgBr + NaNO3- dung dch no c kt ta mu en, th l dung dch Na2S

Na2S + 2AgNO3 Ag2S + 2NaNO3- dung dch no c kt ta mu vng tantrong axit mnh, th l dung dch Na3PO4

Na3PO4 + 3AgNO3 Ag3PO4 + 3NaNO3

HS: Chp HS:Ln bng trnh byBi 4:Cho 62 g canxi photphat tc dng vi 49 gdung dch H2SO4 64%. Lm bay hi dungdch thu c n cn kh th c mt hn

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bay hi dung dch thu c n cnkh th c mt hn hp rn, bitrng cc phn ng u xy ra vihiu sut 100%

GV: Hng dn HS cch vit pt. Yucu HS gii


hp rn, bit rng cc phn ng u xy ravi hiu sut 100%

GiiCa3(PO4)2 + H2SO4 2CaHPO4 + CaSO4 (1)Ca3(PO4)2 + 2H2SO4 Ca(H2PO4)2 + 2CaSO4

(2)Ca3(PO4)2 + 3H2SO4 H3PO4 + 3CaSO4(3)

S mol Ca3(PO4)2 = )(2,0310

62mol=

S mol H2SO4 = )(32,098.100

64.49mol=

V t l s mol H2SO4 v Ca3(PO4)2 l 1,6Nn xy ra phn ng (1) v (2).Gi a v b l s mol Ca3(PO4)2 tham gia cc

phn ng (1) v (2)Ta c h pt:a + 2b =0,32a + b = 0,2

a = 0,08; b = 0,12)(76,21136.08,0.2

4gmCaHPO ==

)(08.28234.12,0242 )(

gm POHCa ==

)(52,45136).24,008,0(136).2(4

gbamCaSO =+=+=

Hot ng 5: Cng c - dn d* Cng c:Dung dch H3PO4 c cha cc ion ( khng k ion H+v OH- ca nc)A. H+, PO 34 B. H+, PO

3

4 , H2PO

4

B. H+, PO 34 , HPO

4 D. H+, PO3

4 , H2PO

4 , HPO

4

* Dn d:

Ngy 23/10/2012 T chn tit: 11PHN BN HA HC LUYN TP

I. Mc tiu- H thng kin thc v phn bn ha hc- Gii c bi tp lin quan n phn bn ha hc

II. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tp

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Hs: Kin thc v phn bn ha hc, cng thc tnhIV. Tin trnh


Hot ng ca gio vin Hot ng ca hc sinhHot ng 1:GV: Chp ln bng, yu cu HSchp vo v.Bi 1: Thnh phn khi lng ca

photpho trong Na3PO4.nH2O l10,69%. Tm nGV: gi cch gii, yu cu HS lm

GV: Yu cu HS ln bng giiGV: Nhn xt ghi im

Hot ng 2:GV: Chp ln bng, yu cu HSchp vo v.Bi 2:phn m ure thng ch cha46% N. khi lng (kg) ure cungcp 70kg N l

a. 152,2 b. 145,5 c. 50,9 d. 200GV: Yu cu HS ln bng trnh byGV: Gi HS nhn xt, ghi imHot ng 3:GV: Chp ln bng, yu cu HSchp vo v.Bi 3:Trn thc t, phn m

NH4Cl thng ch c 23%N.a)Tnh khi lng phn bn cungcp 60kg N?

b)Tnh hm lng % ca NH4Cltrong phn bn?GV: Yu cu HS chia nhm tholun. Gi i din mt nhm ln trnh

byGV: Gi HS nhn xt, ghi im

Hot ng 4:

HS: Chp , HS: Tho lun lm biHS: Ln bng trnh byBi 1:

mmui =31.100

29010,69

= (g)

290 1647

18n

= =

HS: Chp HS:Ln bng trnh byBi 2:Cng thc ure l: (NH2)2CO

ur

70.100152,2( )

46em g = =

HS: Chp HS:Ln bng trnh byBi 3:a, NH4Cl N253,5kg 53,5.0,23x kg 60 kg

x = 60.53,5/53,5.0,23 = 361gb, Gi s c 10 kg phn m. c 2,3 kg N2 nN2 = 2,3/28 nN = 2,3.2/28 mNH4Cl = nN.53,5 = 8,79 kgVy %NH4Cl = 8,79/10.100 = 87,9 %

HS: Chp

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GV: Chp ln bng, yu cu HSchp vo vBi 4: Mt th bt qung photphat ccha 35% Ca3(PO4)2. Tnh khi lng P2O5tng ng vi

10 tn btqung?GV: Hng dn HS Yu cu HS giiGV: Gi HS nhn xt, ghi imHot ng 5:GV: Chp ln bng, yu cu HSchp vo vBi 5: Phn ln Supephotphat kpthc t sn xut c thng ch c40% P2O5. Tnh hm lng % caCa(H2PO4)2trong phn bn ?

GV: Hng dn HS Yu cu HS giiGV: Gi HS nhn xt, ghi im

HS:Ln bng trnh byBi 4:10 tn bt qung cha 10.0,35 =3,5 tnCa3(PO4)2 nCa3(PO4)2 = nP2O5 = 3,5 / 310mol

mP2O5 = 3,5 / 310.142 = 1,603 tn.

HS: Chp HS:Ln bng trnh byBi 5:Xt 10 kg Supephotphat kp c 10.0,4 = 4gP2O5 nP2O5 = nCa(H2PO4)2 = 4/142 mol mCa(H2PO4)2 = 4/142.234 = 6,59 kg

Vy %Ca(H2PO4)2 = 6,59/10.100 = 65,9 %

Hot ng 6 : cng c dn dDn d : Lm bi tp SBT.

Hot ng 7: Bi tp v nhBi tp 1: Phn Kali sn xut c t qung sinvinit thng ch c 50% K2O. Tnhhm lng % ca KCl trong phn bn ?

/s: 79,2(%):Bi tp 2: Tnh khi lng NH3v dd HNO3 45% iu ch 100kg phn mNH4NO3 loi c 34%N?Bi tp 3: Tnh khi lng dd H2SO4 65% dng iu ch c 500kgSupephotphat kp? Bit rng trong thc tlng axit cn nhiu hn 5% so vi lthuyt./s: khi lng dd H2SO4thc t cn dng l: 677 (kg)Bi tp 4: Cho 13,44 m3 kh NH3(ktc) tc dng vi 49kg H3PO4. Tnh thnh phnkhi lng Amophot thu c?

Ngy 29/10/2012 T chn tit: 12LUYN TP CHNG 2

I. Mc tiu- H thng kin thc v nit, photpho v cc hp cht ca chng- Gii c bi tp lin quan n nit, photpho v cc hp cht ca chng


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Gv:H thng cc dng bi tpHs: Kin thc v nit, photpho v cc hp cht ca chngIV. Tin trnh


3. Cc hot ngHot ng ca gio vin Hot ng ca hc sinh

Hot ng 1: Hon thnh chuiphn ng sau:a.N2 NH3 NO NO2 HNO3NH4NO3 N2Ob.NO HNO3 Fe(NO3)3

Fe2O3 Fe(NO3)3Yu cu cc nhm tho lun v

trnh by ra bng phQuan st cc nhm tho lun

Nhn xt v b sung

Hot ng 2:Bi ton tnh chiu sut .

Gio vic nhc li cng thc tnhhiu sut .

Hc sinh : n tp cng thc .Gio vin chia nhm v phtphiu hc tp

Bi ton :1. Tnh hiu sut ca qu trnhtng hp amoniac t 6,72 lit kh

N2 thu c 3.36 lit kh NH3 ?

Cc nhm hon thnh vo bng pha. N2 + H2 , ,

oxt t p 2NH34NH3 + 5O2 ,850 900

o oPt C C 4NO + 6H2O2NO+ O2 2NO24NO2 + O2 +2H2O 4HNO3HNO3 + NH3 NH4NO3

NH4NO3ot N2O + 2H2O

b. Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O4Fe(NO3)3 2Fe2O3 + 12NO2 +3O2Fe2O3 + 6HNO3 2Fe(NO3)3+ 3H2O8HNO3long + 3Cu 3Cu(NO3)2 + 2NO+4H2O

1. Cng thc tnh hiu sut :+ Hiu sut ca cht sn phm :

*100%Thctien

HLythuyet

=

+ Hiu sut ca cht phn ng :*100%

LythuyetH

Thctien=

2. Gii ton :1. Tnh hiu sut ca qu trnh tng hp NH3 :Phng trnh : N2 + 3H2 2NH3

S mol kh N2 : 26.72

0.3( )22.4

Nn mol= =

Tnh theo phng trnh s mol kh NH3 :

32*0.3 0.6( )NHn mol= =

Th tch NH3 thu c theo phng trnh l :0.6*22.4 13.44( )V lit = =

Hiu sut ca phn ng l :H=3.36/13.44 *100 %=25 %.Vy hiu sut ca qu trnh tng hpNH3 l 25%

Hot ng 3:Bi tp vn dng1. Tnh th tch ca kh H2 v kh

1. Tnh th tch H2 v kh N2 :Phng trnh : N2 + 3H2 2NH3

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N2 cn dng iu ch c44,8lit NH3 vi hiu sut ca qutrnh l 20 %

2. Tnh th tch axit HNO3 t 22.4lit kh N2 bit rng ton b qutrnh iu ch c hiu sut l 85 %.

Hc sinh gii cc bi ton theonhm

Nhm so snh nhn xtGio vin nhn xt chung v choim tng ng vi kt qu hotng ca nhm

Th tch kh NH3 l :V=44.8/22.4 = 2 (mol)

Th tch kh N2 l :2* 22.4*100

112( )20*2

V l= =

Th tch kh H2 cn dng l : V=112*3=336 (l)2. Tnh th tch ca HNO3 :Chui phn ng : N2 NONO2 HNO3

N2 + O20

3000 C 2NO2NO+O2 2NO24NO2 + O2 + 2H2O 4HNO3

Theo phng trnh phn ng ta nhn thy s molkh N2 bng s mol HNO3 to thnh.

S mol N2 l :3 3

22.41( )

22.4NH HNOn n mol= = =

Th tch ca HNO3 tnh theo phng trnh :

V=1*22.4 =22.4 (l)Th tch ca HNO3 thc t thu c l :V=22.4*85/100=19.04 (lit)Vy th tch ca HNO3 thu c l 19.04 lit

Hot ng 4:Bi ton tnh phntrm khi lng bng cch lp h

v p dng nh lut bo tonelectron1. Cho 11.0 gam hn hp gm Alv Fe vo dung dch HNO3 longd thu c 6.72 lit kh NO dktc. Tnh phn trm khi lng cami kim loi trong hn hp .Gio vin c th b sung thmcch gii bng nh lut bo tonelectron.

gii bi ton bng lbt electroncnB1: Xt cc cht thay i s oxiha. Vit qu trnh oxi ha v qutrnh khB2 : t n s. Lp phng trnh

bo ton electron theo nh luttng s elctron nhng = tng s

Cch 1: Gii bng cch lp h phng trnh :Phng trnh phn ng :

Fe + 4HNO3

Fe(NO3)3 +NO + 2H2OXmol x mol

Al + 4HNO3 Al(NO3)3 +NO + 2H2Oy mol y molGi x, y l s mol ca Fe v AlTa c h phng trnh sau :

56 27 11.0

6.720.3

22.4

x y

x y

+ =

+ = =

0.1( )

0.2( )

x mol

y mol

=

=

Khi lng, % ca Fe v Al l :0.1* 56 5.6( )Fem g= = % 5.6/11*100% 50.1%Fe = =

27* 0.2 5.4( )Alm g= = % 49.9%Al =

Cch 2 : Dng nh lut bo ton electron :Ta c Fe Fe3++ 3e Al Al3+

+ 3eX mol 3x mol Y mol

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electron nhnLp h v gii hLu : u im ca phng phpngn gn, n gin vic vit

phng trnh trnh sai st

BTVN:1. Ha tan hon ton hn hp 30.0gam hn hp gm Cu v CuOtrong dung dch HNO3 1,0 M lyd, thy thot ra 6,72 lit kh NO ktc. Tnh khi lng ca Cu vCuO trong hn hp

2. Ha tan hon ton hn hp gm

Zn v ZnO bng dung dch HNO3long d, kt thc th nghimkhng c kh thot ra, dung dchthu c c cha 8 gam NH4 NO3v 113,4 g Zn(NO3)2 . Tnh phntrm khi lng Zn c trong hnhp

3y molN5+ + 3e N2+

0.9 mol 0.3 molGi x , y l s mol ca Fe v Al

S mol ca kh NO6.72

0.3( )22.4

NOn mol= =

AD nh lut bo ton eclectron c: 3x+3y=0.9Ta c h phng trnh :

56 27 11.0

3 3 0.9

x y

x y

+ =

+ =

0.1( )

0.2( )

x mol

y mol

=

=

Khi lng, % ca Fe v Al l :0.1* 56 5.6( )Fem g= = % 5.6/11*100% 50.1%Fe = =

27* 0.2 5.4( )Alm g= = % 49.9%Al =

Hot ng 5: cng c dn d

Dn d : Lm bi tp SBT.Hot ng 6: Bi tp v nhBi tp 1: Nhn bit cc dung dch mt nhn:a. NaNO3, NaCl, Na2SO4, Na2CO3.

b. Na2SO4, NaCl, NaBr, Na2S,NaNO3 .c. Na2SO4, NaNO3, Na2S, (NH4)2SO3 .d. KNO3, Zn(NO3)2, K2SO4, Al(NO3)3, KClBi tp 2: Chn cht thch hp in vo ch trng v lp phng trnh ha hc sau :

a. ? + HNO3 NH4NO3b. Na2CO3 + ? NaNO3 + ?

c. ? + NH3 (NH2)2CO + ?d. ? +H2SO4 Ca(H2PO4)2+ CaSO4

Ngy 8/11/2012 T chn tit: 13LUYN TP CACBON V CC HP CHT CA CACBON

I. Mc tiu

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- H thng kin thc v cacbon v cc hp cht ca cacbon- Gii c bi tp lin quan n cacbon v cc hp cht ca cacbon

II. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tp

Hs: Kin thc v cacbon v cc hp cht ca cacbonIV. Tin trnh1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng

Hot ng ca gio vin Hot ng ca hc sinhHot ng 1GV: Chp ln bng, yu cu HSchp vo v.

Bi 1:Nung 52,65 g CaCO3 10000C vcho ton b lng kh thot ra hpth ht vo 500 ml dung dch NaOH1,8 M. Khi lng mui to thnh l( Hiu sut ca phn ng nhit phnCaCO3 l 95% )GV: Yu cu HS tho lun lm bi.

GV: Cho HS xung phong ln bnggii

GV: Gi HS nhn xt ghi im

HS: Chp HS: Tho lun lm biHS: Ln bng trnh by, cc HS cn li ly

nhp lm biBi 1:

Nung 52,65 g CaCO3 10000C v cho tonb lng kh thot ra hp th ht vo 500 mldung dch NaOH 1,8 M. Khi lng mui tothnh l ( Hiu sut ca phn ng nhit phnCaCO3 l 95% )

Gii:CaCO3 Ct

0 CaO + CO2

)(5265,010065,52

32 molnn CaCOCO ===

V phn ng trn c h = 95% nn s mol CO2thc t thu c

)(5002,095.100

5265,02

molnCO ==

nNaOH = 0,5.1,8 = 0,9 (mol)T l s mol NaOH v CO2

1 < 25002,0

9,0

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Hot ng 2GV: Chp ln bng, yu cu HS

chp vo v.Bi 2: xc nh hm lng cacbon trongmt mu thp khng cha lu hunh,ngi ta phi t mu thp trong oxid v xc nh CO2 to thnh. Hyxc nh hm lng cacbon trongmu thp X, bit rng khi t 10g Xtrong oxi d ri dn ton b sn

phm qua nc vi trong d th thu

c 0,5 g kt taGV: Gi hng dn HS cch gii,yu cu 1 HS ln bng trnh by


Hot ng 3GV: Chp ln bng, yu cu HSchp vo v.Bi 3:C a gam hn hp bt X gm CuO,Al2O3 . Ngi ta thc hin cc thnghim sau:Th nghim 1: Cho X phn ng honton vi dung dch HCl, c cn dungdch thu c 4,02 g cht rn khan.

Th nghim 2: Cho X phn ng va vi bt cacbon nhit cao ththu c 0,112 lt kh (kt)Tnh a ?


2x + y = 0,9 y = 0,1004NaHCO3 8,438 g v Na2CO3 42,38 gHS: Chp HS: Ln bng trnh by, cc HS cn li lynhp lm bi

Bi 2: xc nh hm lng cacbon trong mtmu thp khng cha lu hunh, ngi ta

phi t mu thp trong oxi d v xc nhCO2 to thnh. Hy xc nh hm lngcacbon trong mu thp X, bit rng khi t10g X trong oxi d ri dn ton b sn phmqua nc vi trong d th thu c 0,5 g ktta

Gii

C + O2 CO20,005 0,005 (mol)CO2 + Ca(OH)2 CaCO3 + H2O0,005 0,005 (mol)

%6,0%100.10

06,0%

)(06,0005,0.12);(005,02

==

====

C

gmmolnn CCOC

HS:Ln bng trnh by

Bi 3:C a gam hn hp bt X gm CuO, Al2O3 .Ngi ta thc hin cc th nghim sau:Th nghim 1: Cho X phn ng hon ton vidung dch HCl, c cn dung dch thu c4,02 g cht rn khan. Th nghim 2: Cho X phn ng va vi

bt cacbon nhit cao th thu c 0,112lt kh (kt)Tnh a ?

GiiCuO + 2HCl CuCl2 + H2O0,01 0,01Al2O3 + 6HCl 2AlCl3 + 3H2O0,01 0,02

2CuO + C 2Cu + CO20,01 0,005 (mol)

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GV: Gi HS nhn xt, ghi im)(005,0

4,22

112,02

molnCO ==

)(02,05,133

67,2

)(67,235,102,4

)(35,1135.01,0

3

3

2

moln

gm

gm

AlCl

AlCl

CuCl

==

==

==

a = 80.0,01 + 102.0,01 = 1,82 (g)

Hot ng 4: Cng c - dn d* Cng c:Cho 224 ml kh CO2 (ktc) hp th ht trong 100 ml dung dch KOH 0,2M. Khilng ca mi cht trong dung dch to thnh l

A. KHCO3 0,3 g v K2CO3 1,28 g B. K 2CO3 1,28 gC. KHCO3 0,25 g v K2CO3 1,38 g D. K 2CO3 1,38 g

* Dn d: Chun b bi Silic v hp cht ca silicHot ng 5:Bi tp v nhBi tp 1: Hon thnh cc s phn ng sau;CO2

1 K2CO32 KOH 3 KHCO3

4 CO25 CO 6 CO2

Bi tp 2: kh hon ton 40,0 gam hn hp gm CuO v Fe2O3, ngi ta dng15,68 lt CO (kc). Xc nh % mi oxit trong hn hp ?Bi tp 3: ho tan hon ton 2,76 gam mui cacbonat ca kim loi kim R trong dungdch HCl, thu c 448 ml kh CO2 (kc). Xc nh cng thc ho hc ca muicacbonat trn ?

Ngy 28/11/2012 T chn tit: 14

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LUYN TP CHNG 3I. Mc tiu

- H thng kin thc chng 3 (cacbon, silic v cc hp cht ca chng)- Gii c bi tp lin quan n cacbon, silic v cc hp cht ca cacbon

II. Phng php: m thoi gii bi tp.

III. Chun bGv:H thng cc dng bi tpHs: Kin thc v cacbon, silic v cc hp cht ca cacbonIV. Tin trnh

1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ngHot ng ca gio vin Hot ng ca hc sinh

Gv cho hs tho lun theo nhm . Sau

trnh by .Hng dn lm cc bi tp sgk v sbt

Nhm 01i din nhm ln trnh byGV tng kt




Lm cc bi tp sgk v sbt

i din cc nhm ln trnh by

Nhm 11. Natrisilicat c th c iu ch bng cch nnng chy NaOH rn vi ct . Hy xc nh hmlng SiO2 trong ct , bit rng t 25 kg ct khosn xut c 48,8 kg Na2SiO3 .

Nhm 2

2. Khi nung 30 g SiO2 vi 30 g Mg trong iukin khng c kk , thu c cht rn A . B quas to x MgSiO3 trong qu trnh .

a. Hy vit cc PTPU ?b. Xc nh thnh phn nh tnh v nh

lng ca A ?

Nhm 33. Nu hin tng , cho bit cc cht c tothnh v vit ptpu xy ra khi cho t t n d c

TN sau ?- Ba vo dd (NH4)2CO3 - Na vo dd Ca(HCO3)- CO2 vo dd Ca(OH)2

Nhm 44. Ha tan hon ton 11,2 g CaO vo nc thuc dd A . Cho V lit CO2 ( ktc)li qua dd A thc 2,5 g kt ta . Tnh gi tr ca V ?

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Nhm 55 . Hp th ton b 2,24 lit CO2 (ktc) vo 2 litdd Ca(OH)2 0,03 M s thu c mt lng ktta l bao nhiu g ?

Nhm 66. Hp th h/ton V lit CO2 (ktc ) vo 100ml ddBa(OH)2 3M c 19,7 g kt ta . Tnh gi tr caV ?

Hot ng 4: Cng c - dn d* Cng c:1. Loi thy tinh kh nng chy cha 18.43 % K2O, 10.98 %CaO, 70.59 % SiO2. Xcnh cng thc ca thy tinh .

2. sn xut 100 kg loi thy tinh c cng thc Na2O.CaO.6 SiO2 cn phi dngn bao nhiu Kg natri Cacbonat vi hiu sut ca qu trnh l 90 %- Hng dn cc nhm tho lun3. Xi mng Pooclng c cng thc Ca3SiO5 . Ca2SiO4. Ca3(AlO3)2 . Tnh % khi lngca canxi oxit c trong xi mang .- Hc sinh tho lun theo bn tm cch gii .- Cc nhm nhn xt ln nhau .- Gio vin nhn xt chung .* Dn d: Chun b biM u v hp cht hu c

Hot ng 5:Bi tp v nhBi tp 1: 1. Ha tan hon ton 3.5 g hn hp 2 mui Na2CO3 v K2CO3 vo nc.Ri chia thnh 2 phn :Phn 1 : Cho tc dng vi dung dch HCl 3.65%cho n khi khng cn kh thot rath thu c 0.224 lit kh (kc)Phn 2 : Cho tc dng vi nc vi trong d, thu c 2 gam kt ta .a. Tnh khi lng dd HCl 3.65 % phn ng .

b. Tnh khi lng ca mi mui trong hn hpBi tp 2: . Ngi ta dn d kh CO i qua 16 gam bt st oxit . Sau dn sn phmkh i qua dung dch nc vi trong d thu c 30 gam kt ta . Hy xc nh cng

thc st oxit ?Bi tp 3: Trn u hn hp gm CuO v 1 oxit st vi 1 lng Cacbon d . Khi phnng kt thc thu c 2.8 lit kh CO2 (kc) v12 gam hn hp 2 kim loi . Xc nhcng thc st oxit bit rng nCuO : n oxit st = 2 : 1

Ngy 16/12/2012 T chn tit: 15

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DANH PHAP HP CHT HUC C,CU TRC PHN T HP CHT HU C

I. Mc tiu- Hc sinh nm c cch gi tn ca mt s hp cht hu c, cu to ca hp

cht hu c, vit c ng ng, ng phn

- Gii c bi tp lin quan n cch gi tn ca mt s hp cht hu c, cuto ca hp cht hu c, ng ng, ng phnII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc gi tn ca mt s hp cht hu c, cu to ca hp cht hu c, ngng, ng phnIV. Tin trnh


3. Cc hot ng

Hot ng ca gio vin v hc sinh Ni dungHot ng 1 : Vo bi . Nu s khc nhaugia cc hp cht hu c sau :* CCl4 , C6H5NO2 , CH3COOH* CH4 , C6H6 , C2H4 Rt ra cch phn loi?

Hot ng 2 : Hng dn HS nghin cu- Thnh phn phn t mt s cht hu c hc ? rt ra kt lun ? HS : nghincu tr li-Hydrocacbon l nhng hp cht c tothnh bi cc nguyn t ca hai nguyn tC v H-Dn xut ca hidrocacbon l nhng hpcht m trong phn t ngoi C, H ra cn cmt hay nhiu nguyn t ca cc nguyn tkhc nh O,N,S Halogen....

Hot ng 3 :Yu cu HS vit mt s phnng bit * C2H5OH + Na

I PHN LOI HP CHT HU C1 Phn loi :-Hirocacbon : l hp chhu c trong phn t ch c H v CHC no HC khng no HC thmCH4... C2H4 ... C6H6...

-Dn xut ca Hirocacbon : ngoi ngut C , H cn c nhng nguyn t khc .VD : Ancol , axit , dn xut halogen ,este ...2 Nhm chc :Nhm OH v COO gy ra cc phn ng ho hc c trn phn bit etanol ,axit axetiic vi imeete v vi cc loi hp cht khc - OHCOOH c gi l nhm chc

Nhm chc l nhm nguyn t gy ranhng phn ng c trng ca phn t hcht hu c .VD : C2H5 OH , CH3OH ... R - OH

II DANH PHP HP CHT HU C1 Tn thng thng :- t theo ngun

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* CH3COOH + NaOH * CH3 O CH3 + Na c p khng ?

HS vit phng trnh ho hc .Nhm nhng nguyn t no gy ra phnng ?

Kt lun v nhm chc .Hot ng 4 :Cho HS nghin cu SGK rt ra kt lun v danh php thngthng .HS nhn xt cho VD ?

Kin Gim Bc h

Hot ng 5 :- Ly mt s v d , gi tn ,

phn tch thnh phn tn gi?- c tn cc cht sau : CH3Cl , CH3CH2 Br , CH2=CH Cl .- Metyl clorua ,etyl bromua, vinyl clorua .HS rt ra kt lun v cch gi tn theo kiugc chc :

Hot ng 6:Cho HS nghin cu s m vtn ca mch cacbon ?Yu cu HS v hc thuc bng 4.1 , Tr. 109SGK.- Phn tch thnh phn mt s tn gi?

c tn : CHCCH = CH2HS nghin cu SGK v vn dng c tnmt s mch cacbon :Ap dng gi tn mt s hp cht hu c :

Hot ng 7: Gio vin yu cu hs nuthuyt cu to

Gio vin trnh by:

tm ra cht . - i khi phn ui trong tngi ch loi cht .VD : HCOOH : Axit fomic (fomica: kin )CH3COOH: Axit axetic (axetus: Gim)C10H20O: mentol (menthapiperita: Bc

h)2 Tn h thng theo danh php IUPAC :a) Tn gc chc :Tn phn gc + Tn phn nh chc .VD:CH3CH2Cl CH3CH2OCOCH3CH3CH2 O CH3(etylclorua) (etyl axetat ) (etyl metylete)

b) Tn thay th : Tn phn th + Tnmch cacbon chnh + Tn phn nh chcH3CCH3 H3CCH2Cl H2C=CH2

etan cloetan eten 1 2 3 4HCCH CH2=CH CH2 CH3( et + in ) but 1 en

OH 1 2 3 4 1 | 2 3 4CH3CH=CHCH3 CH3CH CH=CH2But2en but3en2ol

- gi tn hp cht hu c, cn bit tncc s m v mch cacbon chnh

III. Cu trc phn t hp cht hu c:

1. Thuyt cu to: 3 lun im

2. ng phn:C cc loi ng phn:+ ng phn cu to+ ng phn nhm chc+ ng phn hnh hc

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Cng CTPT

ng phn cu to ng phn hnh hc

Hot ng 8 :Cho hs vit CTCT ca C4H10O , t rt rakt lun v 3 loi ng phn ?

- HS Vit cc CTCT- Di s hng dn ca GV rt ra kt lun:Hot ng 4: Cng c - dn d* Cng c:1. Vit cc ng phn ca C6H14, C5H10* Dn d: Chun b bi luyn tp chng 4Hot ng 5:Bi tp v nhLm cc bi tp trong SGK v sch bi tp

Ngy 22/12/2012 T chn tit: 16LUYN TP CHNG 4

I. Mc tiu- Hc sinh vn dng kin thc hc ca chng 4 vo luyn gii cc bi tp

ha hc lin quanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc i cng v ha hc hu c

IV. Tin trnh1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng

Hot ng ca thy v tr Ni dungHot ng 1:GV: Chp ln bng, yu cu

Bi 1:Trong cc cht di y, cht no l ng ng

41

Khc nhauv cu to

Cng cu tokhc cu trckhng gian

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HS chp vo v.Bi 1:Trong cc cht di y, cht nol ng ng ca nhau? cht nol ng phn ca nhau?

1. CH3CH2CH32. CH3CH2CH2Cl3. CH3CH2CH2CH34. CH3CHClCH35. (CH3)2CHCH36. CH3CH2CH=CH27. CH3CH=CH28. CH2-CH2

CH2-CH29. CH3

C=CH2CH3

HS: Chp GV: Yu cu HS tho lun lm

bi.HS: Tho lun lm biGV: Cho HS xung phong ln

bng giiHS: Ln bng trnh by, cc HS

cn li ly nhp lm biGV: Gi HS nhn xt ghi imHot ng 2:GV: Chp ln bng, yu cuHS chp vo v.Bi 2:Khi t chy 1,5 g ca mi cht Ahoc B hoc D u thu c sn

phm gm 0,9 g nc v 2,2 g khCO2. Ba cht trn c phi l ng

phn ca nhau khng? Cho v d.HS: Chp GV: Gi hng dn HS cchgiiHS: ch nghe hiuGV: ly v d minh ha

ca nhau? cht no l ng phn ca nhau?10.CH3CH2CH311.CH3CH2CH2Cl12.CH3CH2CH2CH313.CH3CHClCH3

14.(CH3)2CHCH315.CH3CH2CH=CH216.CH3CH=CH217.CH2-CH2

CH2-CH218.CH3

C=CH2CH3

Gii:+ Cc cht ng ng:(1) v (3); (1) v (5); (6) v (7); (7) v (9)+ Cc cht ng phn:(2) v (4); (3) v (5); (6) v (7); (6), (8) v (9)

Bi 2:Khi t chy 1,5 g ca mi cht A hoc B hocD u thu c sn phm gm 0,9 g nc v 2,2g kh CO2. Ba cht trn c phi l ng phn canhau khng? Cho v d.

GiiV cc cht c cng s mol C ( cng khi lng

CO2), cng s mol H ( cng khi lng nc) vcng s mol oxi trong cng mt lng mi chtc ngha l 3 cht c cng thc n gin gingnhau. Nu 3 cht c cng phn t khi na thchng mi l ng phn ca nhau.V d: Ba cht l axit axetic C2H4O2, glucozC6H12O6 v anehitfomic khng phi l ng

phn ca nhau mc d u c cng thc n gin

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Hot ng 3:GV: Chp ln bng, yu cuHS chp vo v.

Bi 3:Hn hp kh A cha haihirocacbon k tip nhau trongmt dy ng ng. Ly 1,12 ltA (ktc) em t chy hon ton.Sn phm chy c dn qua

bnh (1) ng H2SO4 (c), sau qua bnh (2) ng dung dch

NaOH ( c d). Sau thnghim, khi lng bnh (1) tng

2,16 g v bnh (2) tng 7,48g.Hy xc nh CTPT v % v thtch ca tng cht trong hn hpA.HS: Chp

GV: Gi hng dn HS cch

gii

HS: ch nghe hiu

l CH2O; khi t 30 g mi cht u sinh ra 1 molCO2 v 1 mol nc.Bi 3:Hn hp kh A cha hai hirocacbon k tipnhau trong mt dy ng ng. Ly 1,12 lt A

(ktc) em t chy hon ton. Sn phm chyc dn qua bnh (1) ng H2SO4 (c), sau qua bnh (2) ng dung dch NaOH ( c d).Sau th nghim, khi lng bnh (1) tng 2,16 gv bnh (2) tng 7,48g. Hy xc nh CTPT v %v th tch ca tng cht trong hn hp A.

GiiHai hirocacbon k tip nhau trong dy ng

ng c CTPT l CxHy v Cx+1Hy + 2Gi a l s mol CxHyGi b l s mol Cx+1Hy + 2Ta c: a + b = 0,05 (1)

CxHy + OH2

yxCO)O

4

y(x 222 ++

a ax y/2a

OH2

2y1)CO(x1,5)O

4

y(xHC 2222y1x

++++++

++

b (x + 1)b

2

2+y b

S mol CO2: ax + b(x + 1) = 0,17 (2)

S mol H2O: 0,122

2)b(yay=

++

(3)

T (2) ta c (a + b)x + b =0,17b = 0,17 - 0,05x

b l s mol a mt trong hai ht nn 0 < b < 0,05Do 0 < 0,17 0,05x < 0,05

34,34,2 =

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% v th tch ca C4H6 trong hn hp l 40%

* Cng c:Cht no di y l ng phn ca CH3COOCH3?A. CH3CH2OCH3 B. CH3CH2COOH

C. CH3COCH3 D. CH3CH2CH2OH* Dn d: Chun b bi n hkIBi tp v nh: Hn hp M th lng,cha 2 hp cht hu c k tip nhau trong mtdy ng ng. Nu lm bay hi 2,58g M th th tch hi thu c ng bng th tchca 1,4 g kh N2 cng iu kin.t chy hon ton 6,45 g M th thu c 7,65 g H2O v 6,72 lt CO2(ktc). Xc nhCTPT v % khi lng ca tng cht trong hn hp M.

Ngy 28/12/2012 T chn tit: 17N TP HC K I

I. Mc tiu- Hc sinh vn dng kin thc hc ca hc k I vo luyn gii cc bi tp

ha hc lin quanII. Phng php: m thoi gii bi tp.III. Chun b

Gv:H thng cc dng bi tpHs: Kin thc chung ton hc k IIV. Tin trnh


Hot ng ca thy v tr Ni dungHot ng 1:GV: Chp ln bng, yu cu HS

chp vo v.Bi 1:Trn 200 ml dung dch HCl 0,1 Mvi 800ml dung dch HNO3 0,01M.Tnh pH ca dung dch thu cHS: Chp GV: Yu cu HS tho lun lm bi.

Bi 1:Trn 200 ml dung dch HCl 0,1 M vi 800ml

dung dch HNO3 0,01M. Tnh pH ca dungdch thu c

Gii:Nng cc cht sau khi pha trn

CHCl = M02,0800200

1,0.200=

+

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HS: Tho lun lm bi

GV: Cho HS xung phong ln bnggii

HS: Ln bng trnh by, cc HS cnli ly nhp lm bi


Hot ng 2:GV: Chp ln bng, yu cu HSchp vo v.

Bi 2:Ha tan hon ton 5,6 g Fe vo dungdch HNO3 c nng, thu c V ltkh (ktc). Tm VHS: Chp GV: Gi hng dn HS cch gii,yu cu 1 HS ln bng trnh byHS: Ln bng trnh by, cc HS cnli ly nhp lm bi

GV: Gi HS nhn xt, ghi imHot ng 3:GV: Chp ln bng, yu cu HSchp vo v.Bi 3:

Nhit phn 66,2 gam Pb(NO3)2 thuc 55,4 gam thu c 55,4 gamcht rn. Tnh hiu sut ca phn ng

phn hy.HS: Chp


HS:Ln bng trnh by

MCHNO 008,0800200

01,0.8003

=

+

=

Phng trnh in liHCl H+ + Cl-

0,02 0,02 (M)HNO3 H+ + NO3-

0,008 0,008 (M)Tng nng ion H+ = 0,028M

pH = -lg0,028 =1,55Bi 2:Ha tan hon ton 5,6 g Fe vo dung dchHNO3 c nng, thu c V lt kh (ktc).Tm V

Gii

Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O0,1 0,3(mol)

6,72(l)0,3.22,4V

0,3(mol)nn0,1(mol)56

5,6n

2

2

NO

FeNOFe

==

====

Bi 3:Nhit phn 66,2 gam Pb(NO3)2 thu c 55,4gam thu c 55,4 gam cht rn. Tnh hiusut ca phn ng phn hy.

GiiPb(NO3)2 PbO+ 2NO2 + 1/2O2x 2x x/2Gi x l s mol Pb(NO3)2 nhit phn

Khi lng kh thot ra = 2x.46 + 0,5x.32 =66,2 55,4 = 10,8 x = 0,1 (mol)Hiu sut ca phn ng l:

H = %50%100.2,0

1,0=

Bi 4:Ch dng mt ha cht duy nht phn bitcc l mt nhn ng cc dung dch sau:

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Hot ng 4:GV: Chp ln bng, yu cu HS

chp vo v.Bi 4:Ch dng mt ha cht duy nht

phn bit cc l mt nhn ng ccdung dch sau: NaCl, Na2SO4,

NH4Cl, (NH4)2SO4HS: Chp GV: Hng dn HS cch gii, yucu HS trnh byHS:Ln bng trnh by

GV: Gi HS nhn xt


Bi 5: Hon thnh s phn ngsau.A C03000 B C D NaOH NaNO3HS: Chp GV: yu cu 1HS trnh byHS:Ln bng trnh byGV: Gi HS nhn xt, ghi im.

NaCl, Na2SO4, NH4Cl, (NH4)2SO4Gii

- Trch mi l ra mt t lm mu th- Cho Ba(OH)2 ln lt vo cc mu th+ Mu th khng c hin tng: dung dch

NaCl+ Mu th c kt ta trng : dung dchNa2SO4

Na2SO4 + Ba(OH)2 BaSO4 + 2NaOH+ Mu th c kh mi khai : dung dch

NH4Cl2NH4Cl + Ba(OH)2 BaCl2 + 2NH3 +

2H2O+ Mu th c kt ta trng, c kh mi khai :dung dch (NH4)2SO4

(NH4)2SO4 + Ba(OH)2 BaSO4 +2NH3 + 2H2O

Bi 5: Hon thnh s phn ng sau.A C03000 B C D NaOH NaNO3

N2 + O2 C0

3000 2NO2NO + O2 2NO24NO2 + O2 + 2H2O 4HNO3HNO3 + NaOH NaNO3 + H2O

Vy A l N2,B l NO, C l NO2, D l HNO3

Hot ng 6: Cng c - dn d

* Cng c:- Cc bi tp tnh pH, lin quan ti HNO3, nhn bit, s phn ng- Cho 4,8 gam Cu kim loi vo dung dch HNO3 long d n khi phn ng xy rahon ton. Lng kh thot ra iu kin chun l

A. 2,24 lt B. 6,72 lt C. 1,12 lt D. 3,36 lt- nhn bit s c mt ca 3 ion Fe3+, NH+4 , NO

3 c trong dung dch ta c thdng cht no sau y

A. NaOH B. H2SO4 C. Qu tm D. CaO

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- Cho phn ng:2NH4Cl + Ca(OH)2 0t CaCl2 + A +2H2O. A l cht kh no diy

A. N2 B. NH3 C. H2 D. N2O* Dn d: n tp chun b kim tra hc k I

Ngy 5/1/2013 T chn tit: 18ANKAN V XICLOANKAN

I. Mc tiu- Hc sinh nm c kin thc v xicloankan- Hc sinh vn dng kin thc v ankan, xicloankan vo luyn gii cc bi tp hahc lin quanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tp

Hs: Kin thc chung v ankan, xicloankanIV. Tin trnh

1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng:

Mt s kin thc cn nm:Ankan xicloankan

CTTQ CnH2n+ 2 : n 1 Cm H2m : m 3

Cu trc

Mch h ch c lin kt n

C C .Mch cacbon to thnhng gp khc .

- Mch vng ch c lk n C C

- Tr xiclopropan(mch C phng) , Cc nguyn t C trong phnt xicloankan khng cng nmtrn mt mt phng .

Danh php S ch Tn - Tn mch ANv tr nhnh chnh

S ch Tn - Xiclo + Tn ANv tr nhnh mch chnh

Tnh cht vt l

C1 C4 : Th kh .t 0nc ,t0s, khi lng ring tngtheo phn t khi - nh hnnc , khng tan trong ncnc .

C3 - C4 : Th kh .t 0nc ,t0s , khi lng ring tngtheo phn t khi- nh hn nc , khng tan trongnc nc .

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Tnh cht hahc

- Phn ng th .- Phn ng tch .- Phn ng oxiha .

KL : iu kin thngankan tng i tr .

- Phn ng th .- Phn ng tch .- Phn ng oxiha .

Xiclopropan , xiclobutan cphn ng cng m vng vi H2 .

Xiclopropan c phn ng cngm vng vi Br2KL : Xiclopropan , xiclobutankm bn

iu ch ngdng

- T du m .- Lm nhin liu , nguyn

liu

- T du m .- Lm nhin liu , nguyn liu

Hot ng ca gio vin v hc sinh Ni dungHot ng 1 :HS in cng thc tng qut v nhn xt v

cu trc ankan , xicloankan .Hot ng 2 :HS in c im danh php v qui lut vtnh cht vt l ca ankan , xicloankan .Hot ng 3 : HS in tnh cht ha hc v ly VD minh ha.Hot ng 4 :HS nu cc ng dng quan trng ca ankan vxicloankan.

- Gio vin t h thng cu hi : Cng thc ca ankan v xicloankan ?

Quy tc gi tn ? Tnh cht ho hc ? ng dng ?

- Vn dng gii mt vi dng bi tpBi 1 : So snh ankan v monoxicloankanBi 2 / Nhn xt kt qu:

Propanvxiclopropan

Butan vxiclobuta

n

Pentan vxiclopentan

Hexan vxilohexan

C3H8C3H6 C4H10C4H8 C

5H12C5H10 C6H14C6H12

t0n/c ,0C

-42-33

-0,513

3649

6981

t0s,0C

-188-127

-158-90

-130-94

-957

Khilng

0,5850,689

0,6000,7303

0,6260,755

0,660,778

Hc sinh tho lun nhm :

- HS in cng thc tngqut v nhn xt v cu trcan kan , xicloankan- HS in c im danh

php v qui lut v tnhcht vt l ca ankan ,xicloankan- HS in tnh cht ha hcv ly VD minh ha .

- HS nu cc ng dngquan trng ca ankan vxicloankan1)Ging nhau : Thnh phnnh tnh ca ankan vmono xicloankan gm C vH

Khc nhau : Cng snguyn t C th mono

xicloankan c t s nguynt H hn. Cu trcmonoxicloankan c mchvng .Ankan c mch cav

bon to thnh ng gpkhc .2)

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ringg/cm3

Nhn xt :- Ging nhau : S nguyt C tng th t0s ,t0n/c d,tng .-Khc nhau : Cng s

nguyn t Cmonoxicloankan c t0n/c,v d ln hn .

Cng c:Bi 1: Cht A l mt ankan th kh. t chy hon ton 1,2 lt A cn dng va ht6 lt oxi cng iu kin.a/ Xc nh CTPT ca A.

b/ Cho cht A tc dng vi kh clo 250C v c nh sng. Hi c th thu c mydn xut monoclo ca A.Cho bit tn ca mi dn xut . Dn xut no thu c

nhiu hn.Bi 2: t chy hon ton 1,45 gam mt ankan phi dng va ht 3,64 lt O2( ktc)a/ Xc nh CTPT ca ankan

b/ Vit CTCT v gi tn tt c cc ng phn ng vi cng thc .Dn d: v nh chun b luyn tp chng 5Bi tp v nh: t chy hon ton 2,86 g hn hp gm hexan v octan ngi ta thuc 4,48 lt CO2 ( ktc). Xc nh % v khi lng ca tng cht trong hn hp.Ngy 8/1/2013 T chn tit: 19

LUYN TP CHNG 5I. Mc tiu- Hc sinh vn dng kin thc v hidrocacbon no vo luyn gii cc bi tp ha hclin quanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc chung v hidrocacbon noIV. Tin trnh


3. Cc hot ng:

Hot ng ca thy v tr Ni dungHot ng 1:GV: Chp ln bng, yu cuHS chp vo v.Bi 1: Gi tn cc CTCT sauCH3 CH2 CH CH2 CH3

Bi 1: Gi tn cc CTCT sauCH3 CH2 CH CH2 CH3

CH CH3

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CH CH3

CH3CH3

CH3 CH2 CH CH2 CH CH3 CH CH3 CH3

CH3

Hot ng 2:

GV: Chp ln bng, yu cuHS chp vo v.Bi 2:Vit CTCT thu gn caa/ 4-etyl-2,3,3-trimetylheptan

b/ 3,5-ietyl-2,2,3-trimetyloctan

GV: Chp ln bng, yu cuHS chp vo v.Bi 3: Gi tn cc CTCT sau

C2H5

H3C

CH3

GV: Yu cu HS tho lun lmbi.Hot ng 3:GV: Chp ln bng, yu cuHS chp vo v.Bi 4: Vit CTCT thu gn caa/ 1,1-imetylxiclopropan

b/ 1-etyl-1-metylxiclohexanGV: Yu cu HS tho lun lm

bi.

CH3CH3

CH3 CH2 CH CH2 CH CH3

CH CH3 CH3 CH3

Gii:+ 3-etyl -2-metylpentan.+ 4-etyl-2,2,5-trimetylhexan

Bi 2: Vit CTCT thu gn caa/ 4-etyl-2,3,3-trimetylheptan

b/ 3,5-ietyl-2,2,3-trimetyloctan

Giia/CH3 CH(CH3) C(CH3)2 CH(C2H5) CH2 CH2 CH3b/CH3C(CH3)2C(C2H5)(CH3)CH2 CH(C2H5)CH2CH2CH3

Bi 3:Gii:

4-etyl-1,2-imetylxiclohexan

Bi 4: Vit CTCT thu gn caGii

a/CH3

CH3 b/

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Hot ng 4:GV: Chp ln bng, yu cuHS chp vo v.Bi 5: Mt monoxicloankan c tkhi hi so vi nit bng 3.

a/ Xc nh CTPT ca A.b/Vit CTCT v tn tt c ccxicloankan ng vi CTPT tmcGV: Hng dn HS vit ccCTCT ca C6H12HS: Ch cch vit ng phn

GV: Yu cu HS gi tn cc ngphn

HS: Gi tn cc ng phn

Hot ng 5:

GV: Chp ln bng, yu cuHS chp vo v.Bi 5: Hn hp kh A cha mtankan v mt xicloankan. T khica A i vi H2 l 25,8. tchy hon ton 2,58gam A rihp th ht sn phm chy vodung dch Ba(OH)2 d, thu c35,46 gam kt ta. Xc nhCTPT ca ankan v xicloankan

GV: Gi hng dn HS cchgiiTm MAVit pthhGi x, yln lt l s mol caankan, xicloankanLp phng trnhGii phng trnh v bin lun

CH3CH2CH3

Bi 5:Gii

a/ CnH2n = 28.3 = 8414n = 84 n = 6CTPT: C6H12

b/ Cc CTCT

xiclohexan

CH3

metylpentan

CH3

CH3

1,1-dimetylxiclobutan

CH3


CH3


CH2-CH3

etylxiclobutan

CH3 H3C

CH3

CH3H3C

1,2,3-trimetylxiclopropan

CH3

CH3

1,1,2-trimetylxiclopropan

CH3

CH3

1-etyl-2-metylxiclopropan

1-etyl-1-metylxiclopropan

CH3

CH2CH3

CH2CH3

propylxiclopropan

CH2CH2CH3CHCH3

CH3

isopropylxiclopropan

Bi 5:Gii

Gi s trong 2,58g hn hp A c x mol CnH2n + 2(n1) v y mol CmH2m (m3) .MA = 25,8.2 = 51,6(g/mol)

x + y = )1(05,06,51

58,2=

CnH2n + 2 +2

13 +nO2 t nCO2 + (n+1)H2O

x nx(mol)

CmH2m +2

3mO2 t mCO2 + mH2O

y my (mol)CO2 + Ba(OH)2 BaCO3 + H2O

S mol CO2 = s mol BaCO3 = )(18,0197

46,35mol=

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tm n, m

HS: Lm bi theo cc bc GV

hng dn

nx + my = 0,18 (2)Khi lng hn hp A: (14n + 2)x + 14my =2,58 (3) 14(nx + my) + 2x = 2,58 2x = 2,58 14.0,18

x = 0,03; y = 0,02(2) ta c : 0,03n + 0,02m = 0,18 3n + 2m =18

Nghim thch hp m = 3; n = 4CTPT l C4H10; C3H6

Cng c - dn d* Cng c:

Nhc li cch gi tn ca xicloankan.Cch gii bi ton tm CTPT ca ankan vxicloankan* Dn d: Chun b bi Thc hnh 3 trang 124

Ngy 17/1/2013 T chn tit: 20ANKEN + KIM TRA 15 PHT

I. Mc tiu- Hc sinh vn dng kin thc v anken vo luyn gii cc bi tp ha hc lin quanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tp

Hs: Kin thc chung v ankenIV. Tin trnh1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng:

Hot ng ca thy v tr Ni dungHot ng 1:GV: Chp ln bng, yu cu HSchp vo v.

Bi 1: Gi tn cc CTCT sau

CH3 - C - CH2 - CH = CH2

CH3

CH3

CH3 - CH2 - C - CH2 -CH3

CH2

Bi 1: Gi tn cc CTCT sau

CH3 - C - CH2 - CH = CH2

CH3

CH3

CH3 - CH2 - C - CH2 -CH3

CH2

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HS: Chp GV: Yu cu HS tho lun lm bi.HS: Tho lun lm biGV: Cho HS xung phong ln bnggii

HS: Ln bng trnh by, cc HS cnli ly nhp lm biGV: Gi HS nhn xt ghi imHot ng 2:GV: Chp ln bng, yu cu HSchp vo v.Bi 2:Vit CTCT thu gn ca 2,4imetylhex-1-enHS: Chp

GV: Yu cu HS tho lun lm bi.HS: Tho lun lm biGV: Cho HS xung phong ln bnggiiHS: Ln bng trnh by, cc HS cnli ly nhp lm biGV: Gi HS nhn xt ghi imHot ng 3:GV: Chp ln bng, yu cu HS

chp vo v.Bi 3:Hn hp kh A cha mt ankan vmt anken. Khi lng hn hp A l9 gam v th tch l 8,96 lt. tchy hon ton A, thu c 13,44 ltCO2. Cc th tch c o ktc.Xc nh CTPT v % th tch tngcht trong A.HS: Chp

GV: Yu cu HS ln bng trnh byHS: Ln bng trnh by

Gii:

4,4 imetylpent 1- en2-etylbut-3-en

Bi 2:Vit CTCT thu gn ca 2,4imetylhex-1-en

Gii

CH2 = C - CH2 - CH - CH2 - CH3

CH3 CH3

Bi 3:Hn hp kh A cha mt ankan v mt anken.Khi lng hn hp A l 9 gam v th tch l8,96 lt. t chy hon ton A, thu c13,44 lt CO2. Cc th tch c o ktc.Xc nh CTPT v % th tch tng cht trongA.

GiiGi s hn hp A c x mol CnH2n + 2 v y molCmH2m.

)2(914)214(

)1(4,04,22

96,8

=++

==+

myxn

yx

CnH2n + 2 + 213 +n

O2 t

nCO2 + (n+1)H2Ox nx

(mol)

CmH2m +2

3mO2 t mCO2 + mH2O

y my (mol)

nx + my = 6,04,22

44,13= (3)

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Hot ng 4:GV: Chp ln bng, yu cu HSchp vo v.Bi 4:Dn 3,584 lt hn hp X gm 2anken A v B lin tip nhau trong

dy ng ng vo nc brom (d),thy khi lng bnh ng nc

brom tng 10,5 ga/ Tm CTPTca A, B ( bit th tchkh o 00C v 1,25 atm ) v tnh %th tch ca mi anken

b/ Tnh t khi c hn hp so vi H2HS: Chp GV: Gi hng dn HS cch gii

t cng thc 2 anken, cng thctrung bnhVit pthhTm gi tr xTm CTPT ca 2 ankenTnh % th tch ca mi ankenTnh t khi c hn hp so vi H2

HS: Lm bi theo cc bc GV hng dn

Hot ng 5: Kim tra 15 pht: bi:

T (1), (2), (3) ta c x = 0,3; y = 0,1Thay x, y vo (3) ta c: 3n + m = 6Chn m = 3, n =1CH4 chim 60% th tch A v C3H6 chim40%

Bi 4: Dn 3,584 lt hn hp X gm 2 ankenA v B lin tip nhau trong dy ng ngvo nc brom (d), thy khi lng bnhng nc brom tng 10,5 ga/ Tm CTPTca A, B ( bit th tch kh o 00C v 1,25 atm ) v tnh % th tch ca mianken

b/ Tnh t khi c hn hp so vi H2

Giia/ t ng thc ca 2 anken l CnH2n vCn+1H2n+2Cng thc chung ca 2 anken CxH2xvi n < x < n + 1CxH2x + Br2 CxH2xBr2 tng khi lng ca bnh ng dd chnh l

khi lng ca 2 anken.)(2,0

4,22

584,3.25,1moln ==

M= x145,522,0

5,10== 375,3 == nx

Hai anken l C3H6 v C4H8Gi a v b l s mol ca C3H6 v C4H8 tronghn hp. Ta c:a + b = 0,2 a = 0,0542a + 56b = 10,5 b = 0,15

%25%63

=HC

V

%75%74

=HC

V

b/ 25,262/ =HXd

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1. Hon thnh pthh theo s (ghi rk nu c) 4 C2H4(OH)2C4H10 1 C2H4

2

3

C2H5OH

5 6

C2H4Cl Polietilen2. C 3 ng phn anken . Thcnghim cho thy rng 7 g mi chtlm mt mu dd brm c cha 16 g

brom .a. Xc nh CTPT 3 ng phn ?

b. Vit CTCT 3 ng phn , gi tn .Bit 3 ng phn khi tc dng vihidro ( XT : Ni , to ) cho sn phm 2

Metylpentan .Hot ng 6: Cng c - dn d* Cng c:

Nhc li cch gi tn ca anken. Tnh cht ha hc ca anken. Cch gii bi tontm CTPT ca 2 anken ng ng lin tip nhau.* Dn d: Chun b bi ankadien

Ngy 23/1/2013 T chn tit: 21ANKAIEN

I. Mc tiu- Hc sinh vn dng kin thc v ankaien vo luyn gii cc bi tp ha hc linquanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc chung v ankaienIV. Tin trnh

1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng:

Hot ng ca thy v tr Ni dungHot ng 1:GV: Chp ln bng, yu cu HS

55CH2 = C - CH = CH2

CH3

2-metylbuta-1,3-dien

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chp vo v.Bi 1:Cht A l mt ankaien lin hp cmch cacbon phn nhnh. t chyhon ton 3,4 g A cn dng va ht

7,84 lt oxi (ktc). Xc nh CTPT ,CTCT, gi tnGV: Yu cu HS tho lun lm bi.

Hot ng 2:GV: Chp ln bng, yu cu HSchp vo v.Bi 2:Hn hp kh A cha mt ankan v mtankaien . t chy hon ton 6,72lt A phi dng va ht 28 lt O2 ( cc

th tch kh ly ktc). Dn sn phmchy qua bnh 1 ng H2SO4 c, sau qua bnh 2 ng dung dch NaOHd th khi lng bnh 1 tng p gam,

bnh 2 tng 35,2 gam.Xc dnh CTPT, tnh p.HS: Chp GV: Gi hng dn HS cch giit cng thc ankan, cng thcankaien

Vit pthhDa vo d kin ra tm CTPT, tnh

p

HS: Lm bi theo cc bc GV hng dn

Bi 1:Gii

CnH2n - 2 +2

13 nO2 t nCO2 + (n-1)H2O

13

7,0

n0,35 (mol)

(14n -2).13

7,0

n= 3,4 n = 5

CTPT: C5H8CTCT:

Bi 2:

GiiGi s hn hp A c x mol CnH2n + 2 v ymol CmH2m - 2.

)1(3,04,22

72,6==+yx

CnH2n + 2 +2

13 +nO2 t nCO2 + (n+1)H2O

x2

13 +n.x nx (n

+1)x

CmH2m - 2 +2

13 mO2 t mCO2 + (m-1)H2O

y2

13 m.y my (m-

1).y

S mol oxi:2

13 +n.x +

2

13 m.y = 1,25

(3n + 1)x + (3m -1)y =2,5 (2)S mol CO2: nx + my = 8,0

44

2,35= (3)

T (1), (2), (3) ta c x = 0,2; y = 0,1Thay x, y vo (3) ta c: 2n + m = 8Chn m = 4, n =2CTPT: C2H6 v C4H6

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Hot ng 3:GV: Chp ln bng, yu cu HSchp vo v.Bi 3:Cht A l mt ankaien lin hp cmch cacbon phn nhnh. t chy

hon ton 3,4 g A cn dng va ht7,84 lt oxi (ktc). Xc nh CTPT ,CTCT, gi tnHS: Chp GV: Gi hng dn HS cch gii


Bi 4:Vit cc ng phn ankaien c cngthc phn t C6H8 (k c ng phnhnh hc) v gi tn thay th.HS: Chp GV: Gi hng dn HS cch gii

S mol H2O = (n + 1)x + (m -1)y =0,9(mol)p = 0,9.18 = 16,2 (g)

Bi 3:

GiiCnH2n - 2 +

2

13 nO2 t nCO2 + (n-1)H2O

(14n -2).13

7,0

n= 3,4 n = 5

CTPT: C5H8 ; CTCT: CH2=CH(CH3)-CH=CH2


Nhc li tnh cht ha hc ca ankan v ankaien. Cch gii bi ton tm CTPT

ca ankan, ankaien .* Dn d: Chun b bi luyn tp

Ngy 31/1/2013 T chn tit: 22LUYN TP ANKIN

I. Mc tiu

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- Hc sinh vn dng kin thc v ankin vo luyn gii cc bi tp ha hc lin quanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc chung v ankin

IV. Tin trnh1. n nh lp: Kim tra s s2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng:

Hot ng ca thy v tr Ni dung

Hot ng 1:Bi 1:GV: Chp ln bng, yu cuHS chp vo v.HS: Chp GV: Yu cu HS tho lun lm

bi.HS: Tho lun lm bi


Hot ng 2:Bi 2:GV: Chp ln bng, yu cuHS chp vo v.HS: Chp GV: Yu cu HS tho lun lm

bi.HS: Tho lun lm biGV: Cho HS xung phong ln

bng gii

Bi 1: Trnh by phng php ha hc phn bit cccht sau: but -2 en, propin, butan. Vit cc phngtrnh ha hc minh ha.

Gii:- Dn tng kh qua dung dch bc nitrat trong amoniac:

bit c cht to kt ta l propin, do c phn ng:CH3 C = CH + AgNO3 + H2O CH3 C = CAg +

NH4NO3- Dn hai kh cn li vo dung dch brom: bit cht lmnht mu dung dch brom l but 2 en, do c phnng:CH3CH=CHCH3 + Br2 CH3CHBrCHBrCH3Kh cn li l butan.

Bi 2: Mt bnh kn ng hn hp kh H2 vi axetilenv mt t bt niken. Nung nng bnh mt thi gian sau a v nhit ban u. Nu cho mt na kh trong

bnh sau khi nung nng i qua dung dch AgNO3 trongNH3 th c 1,2 gam kt ta mu vng nht. Nu chona cn li qua bnh ng nc brom d thy khilng bnh tng 0,41 g. Tnh khi lng axetilen cha

phn ng, khi lng etilen to ra sau phn ng.Gii

C2H2 + H2 C2H4 (1)C2H2 + 2H2 C2H6 (2)C2H4 + H2 C2H6 (3)CH = CH + 2AgNO3 + 2H2O CAg = CAg +2NH4NO3 (4)C2H2 + 2Br2 C2H2Br4 (5)C2H4 + Br2 C2H4Br2 (6)

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Hot ng 3:Bi 3:GV: Chp ln bng, yu cuHS chp vo v.HS: Chp GV: Gi hng dn HS cchgii, yu cu HS ln bng trnh

byGV: Gi HS nhn xt ghi im

S mol C2Ag2 = 0,005 (mol)T (4) ta c s mol axetilen trong hn hp cn li l:2.0,005 =0,01 (mol)Theo (5), khi lng bnh ng brom tng 0,005.26 0,13 gam

Vy khi lng etilen phn ng (6) l: 0,41- 0,13 =0,28(g)Khi lng etilen to ra: 2.0,28 = 0,56 gamBi 3: t 3,4 gam mt hirocacbon A to ra 11 gamCO2. Mt khc, khi cho 3,4 gam tc dng vi lng ddung dch AgNO3 trong NH3 thy to ra a gam kt taa/ Xc nh CTPT ca A.

b/ Vit CTCT ca A v tnh khi lng kt ta tthnh, bit khi A tc dng vi hiro d, c xc tc Nto thnh isopentan.

Giia/ Gi CTPT ca A l CxHy.

CxHy + (x +y

4)O2 xCO2 +

y

2H2O

C

11m .12 3(g)

44= =

Hm 3, 4 3 0, 4(g)= =

x:y =3 0, 4

: 5 : 812 1

=

CTGN: C5H8 CTPT (C5H8)n

b/ V A tc dng c vi dung dch AgNO3 trongNH3, A c dng R - C = CHV A tc dng vi H2 to thnh isopentan nn A phic mch nhnh.CTCT: CHC CH(CH3) CH3CHCCH(CH3)CH3+AgNO3+H2O CAgCCH(CH3)CH3

+ NH4NO3S mol A = s mol kt ta = 3,4 : 68 = 0,05(mol)

Khi lng kt ta = 0,05 . 175 =8,75 (gam)Cng c - dn d* Cng c: CTPT chung ca ankin? Nu tnh cht ca ankin, gii thch?.

+ Hon thnh s phn ng sau:CaCO3 CaO CaC2 C2H2 vinylclorua PVC

+ Trnh by phng php ha hc nhn bit but 1-in, but-2-in, metan.+ Cht no khng tc dng vi dung dch AgNO3 trong amoniac?

A. but 1-inB. but 2-inC. Propin D. Etin

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* Dn d: Chun b bi t chn 23.BTVN: BT SGK v SBT.

Ngy 6/2/2013 T chn tit: 23

LUYN TP HIROCACBON KHNG NOI. Mc tiu- Hc sinh vn dng kin thc v chng 6 vo luyn gii cc bi tp ha hc linquanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc chung v chng 6IV. Tin trnh

1. n nh lp: Kim tra s s

2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng:

Hot ng ca thy v tr Ni dung

Hot ng 1:GV: Chp ln bng, yu cuHS chp vo v.Bi 1:t chy hon ton hn hp haihirocacbon mch h X, Y lin

tip trong dy ng ng thuc 11,2 lt CO2 (ktc) v 12,6gam nc. Tm CTPT ca X, YHS: Chp GV: Yu cu HS tho lun lm

bi.HS: Tho lun lm bi

Hot ng 2:GV: Chp ln bng, yu cuHS chp vo v.Bi 2:Cho 3,5 gam mt anken X tcdng hon ton vi dung dchKMnO4 long d, thu c 5,2gam sn phm hu c. TmCTPT ca X.

Bi 1:Gii:

2CO

11,2n 0,5(mol)

22,4

= = ;2H O

12,6n 0,7(mol)

18

= =

S mol nc > s mol CO2 X, Y thuc dy ngng ca ankan, t cng thc chung ca X v Y lCnH2n+2

CnH2n + 2 +3n 1

2

+O2 t nCO2 + (n+1)H2O

0,5 0,7Ta c : 0,5(n + 1) = 0,7n n = 2,5CTPT ca X, Y l: C2H6, C3H8

Bi 2:Gii3CnH2n + 2KMnO4 + 4H2O CnH2n(OH)2 + 2MnO2 +2KOH14n 14n + 343,5 5,2

Ta c: 3,5( 14n + 34 ) = 5,2.14n n = 5

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HS: Chp GV: Yu cu HS tho lun lm

bi.HS: Tho lun lm biHot ng 4:

GV: Chp ln bng, yu cuHS chp vo v.Bi 3:t chy hon ton hon haihirocacbon mch h M, N lintip trong dy ng ng ca cchirocacbon hc thu c22,4 lt CO2 (ktc) v 12,6 gamnc. Tm CTPT ca M, N.HS: Chp

GV: Gi hng dn HS cchgii, yu cu HS ln bng trnh

by


Hot ng 4:Bi 4:t chy hon ton a lt (ktc)

mt ankin X th kh thu cCO2 v H2O c tng khi lng12,6 gam. Nu cho sn phmchy qua dung dch nc vitrong d, thu c 22,5g kt ta.Tm CTPT ca X.HS: Chp GV: Yu cu HS tho lun lm

bi.


CTPT ca X l C5H10

Bi 3:Gii

2CO

22,4n 1(mol)

22,4= = ;

2H O

12,6n 0,7(mol)

18= =

S mol nc < s mol CO2 M, N thuc dy nng ca ankin hoc ankaien.

CnH2n - 2 +3n 1

2

O2 t nCO2 + (n -1)H2O

1 0,7Ta c : (n - 1 ) = 0,7n n = 3,3

CTPT ca M, N l: C3H4, C4H6

Bi 4:Gii

3CaCO

22,5n 0,225(mol)

100= =

2COm 44.0, 225 9, 9(gam)= =

2H O

12,6 9,9n 0,15(mol)

18

= =

CnH2n - 2 +2

13 nO2 t nCO2 + (n -1)H2O

0,225 0,15Ta c : 0,225(n - 1 ) = 0,15n n = 3CTPT ca X l: C3H4

Cng c - dn d* Cng c: Khi t chy hirocacbon thu c

- S mol H2O > s mol CO2 hirocacbon thuc dy ng ng ankan- S mol H2O = s mol CO2 hirocacbon thuc dy ng ng anken- S mol H2O < s mol CO2 hirocacbon thuc dy ng ng ankin,

ankaien ...

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* Dn d: Chun b bi kim tra vit..BTVN: BT SGK v SBT.

Ngy 26/2/2013 T chn tit: 24

BENZEN ANKYLBEZEN - STIRENI. Mc tiu- Hc sinh vn dng kin thc v hirocacbon thm vo luyn gii cc bi tp hahc lin quanII. Phng php: m thoi gii bi tp.III. Chun bGv:H thng cc dng bi tpHs: Kin thc chung v hirocacbon thmIV. Tin trnh

1. n nh lp: Kim tra s s

2. Kim tra bi c: Tin hnh trong lc luyn tp3. Cc hot ng:

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Hot ng ca thy v tr Ni dungHot ng 1:GV: Chp ln bng, yu cu HSchp vo v.Bi 1:A l mt ng ng ca benzen c tkhi hi so vi metan bng 5,75. Atham gia cc qu

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