GIẢI BÀI TẬP HÓA HỌC 11 CƠ BẢN - NGUYỄN ĐÌNH ĐỘ

8/10/2019 GII BI TP HA HC 11 C BN - NGUYN NH

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GII BI TP HA HC 11 (Ti bn)

NGUYN NH

Chu trch nhim xut bnGim c - Tng bin tp

NGUYN TH THANH HNG

Bin tp : THANH NGC - CAO B NH

Sa bn in : CAO TH BCH THY

Trnh by : Cng ty KHANG VIT

Ba : Cng ty KHANG VIT

NH XT BN TNG Hp THNH PH H CH MINH62JSguyn Th Minh Khai, Q.l, TP.HfcM

I : 38225340 - 38296764 - 38247225 - Fax: 84.8.38222726Email: [email protected]

Sch online: www.nxbhcm.com.vn - Ebook: www.sachweb.vnNH SCH TNG HP 1

62 Nguyn Th Minh Khai, Q. 1, TP.HCM - T: 38 256 804NH SCH TNG HP 2

' - 86- 88Nguyn Tt Thnh, Q.4, TP.HCM - T: 39 433 868 tc lin kt v t ng p h t hnh' .............. ... ^

CNG TY TNHH MTVDCH V VN HA KHANG VIT

r ia 'ch 71 inh Tin Hong - p a Kao - Q 1- TP HCM ''Nin thoai 08 39115694-39105797-39111969-39111968Fax 08 3911 0880 _ .Email khangvietbookstore @yahoo com.vn

Website: www.nhasachkhangviet.vn - J^ .............................................. .........

..

...

S lng 2.000 cun, kh 16x24cm.Ti: Cty TNHH MTV IN n m a i t h n h ca ch: 71, Kha Vn Cn, p. Hip Bnh Chnh, Q. Th c, TP. H Ch MinhS KKHXB: 141 7-2014/CXB/10-138/THTPHCM ngy 16/07/2014.Quyt nh xut bn SV 975/Q-THTPHCM-2014 do NXB Tng HpThnh Ph H Ch Minh cp ngy 05/08/2014.M s ISBN: 978-604-58-2585-3

ln xong v np lu chiu Qu IV nm 2014.

WWW.FACEBOOK.COM/DAYKEM.QU

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W.DAYKEMQUYNHON.UCOZ.COM

ng gp PDF bi GV. Nguyn Thanh T


3/96

Chng .s IN LI

BI 1: S IN LI

1 . Cc dung dch axit nh HC1, baz nh NaOH v mui nh NaCl dnin c, cn cc dung dch nh ancol etylic, sacqaroz, -glixrol khngdn in l do nguyn nhn g?

GiiCc dung dch HC1, NaOH, NaCl dn in c V trong dung dch cs hin din ca .cc ion. Cc dung dch ancol etylic, saccaroz, glixerolkhng dn in c v trong dung dch khng c s hin din ca ccion.

2. S in li, cht in li l g?Nhng loi cht no l cht in li? Th no l cht in li mnh,

cht in li yu? L v d v vit phng'trrih in li ca chng.Gii

Qu trnh phn li cc cht trong nc ra ion gi l s in li.Nhng cht ta n trong nc phh li ra ion gi l nhng cht in liAxit, baz, mui l nhng cht in li.Cht in li mnh l cc cht khi tan trong nc, cc phn t ha tanu phn li ra ion. V d:NH4Cl NH4+ + c rCht in li u cht khi tan trong nc, cc phn t ha tan chphn li mt phn ra ion, phn cn li vn tn ti di dng phn t-trng Hng dch. V : CH3COOH GH3COO" + H+

3. Vit phng trnh n li ca nhng cht sau:a) Cc cht in li mnh: Ba(N3)2 0 ,10 M; HNO3 0,020M; KOH

0,010M. Tnh nng mol tng ion trong cc dung dch trn.b) Cc cht in li yu: HC10, HNO2.

Gii

a) Ba(N03)2 > Ba2+ + 2NO3

0,1M 0,1M 0,2M

... HNO3 H+ + 1

co

0 ,02M,02M 0,02M

KOH , ^ - V. K+ + OH"0,0 IM 0 ,0 1M . 0,01M

b) HC10 H+ + CIO'

HN02 H+ + n 2

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4/96

4. Chn cu tr li ng trong cc cu sau y:

Dung ch cht in i dn in c l doA. S chuyn dch ca cc electron.B. S chuyn dch ca cc cation.c. S chuyn dch ca cc phn t ha tan.. S chuyn dch ca c cation v anion.

GiiDung ch cht in li dn in c l do s chuyn ch c hng cacc cation v anion di tc dng ca in trng.

5. ('ht no sau v khng n in c?A K( rn, khan. B. CaC2nng chy.

( . NaOH nng chy. D. HBr ha tan trong nc.GiiKC1 khan khng n in.

BI 2: AXIT, BAZ V MUI

1. Pht biu cc nh ngha axit, axit mt nc v nhiu nc, baz,hiroxit lng tnh, mui trung ha, mui axit.Ly cc th d minhha v vit phng trnh in li ca chng.

- Gii- Axit l ch t khi tan trong nc phn li ra cation H+- Axit mt nc l axit khi tan trong nc ch phn li mt nc ra ion H+

V d: HNOa- H+ + NOa"- Axit nhiu nc l axit khi tan trong nc phn li nhiu nc ra ion H+

V d: H2SC>4 -> H++ HS(VHSO4 -H++ SO42

- Baz l cht khi tn trong nc phn li ra anion OH"V d: KOH K+ + OH

- Hiroxit lng tnh l hiroxit khi tan trong nc va c th phn li

nh axit, va c th phn li nh baz.V d: Zn(OH)2 ^ Zn2+ + 20H'

Zn(OH)2 ^ 2H+ + Zn022-- Mui m anion gc axit khng cn hiro c kh nng phn li ra H+

gi l mui trung ha. V d: NaCl ; NaHP3- Mui m anion gc axit vn cn hiro c kh nng phn li ra H+gi

l mui axit. V d: NaHC03 ; NaH2P0 42. Vit phng trnh in li ca cc cht sau:

a) Cc axit yu: H2S, H2CO3.

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5/96

b) Baz mnh: LiOHc) Cc mui: K2C3, NaClO, NaHSd) Hidroxit lng tnh: Sn(OH)2

Giia i M l v S : H,s H+ + HS'

HS H+ + s 2\

* H0CO3 h , c o 3 H+ + HCO3HCO3 H+ + CO3

b)* LiOH LiOH > Li+ + OHc)* K2CO3 k2c o 3 2K+ + CO32

* NaClO NaClO Nat + cicr* NaHS NaHS -> Na+ + HS"

d)* Sn(OH)2 Sn(OH)2 Sn2+ + 2QH

Sn(OH>2 Sn022- + 2H+

3. Theo thuyt A - r - ni - ut, kt lun no sau y l ng?A. Mt hp cht trong thnh phn phn t c hiro l axit.

B. Mt hp cht trong thnh phn phn t c nhm OH l baz.c. Mt hp cht c kh nng phn li ra cation H+trong nc l axit.D. Mt baz khng nht thit phi c nhm OH trong thnh phn

phn t.

Mt hp cht c kh nng phn li ra cation H+ trong nc l axit.Chn c *

4. i vi dung dch axit yu CH3COOH 0,10M, nu b qua s in li canc th nh gi no v nng mol ion sau y l ng?A.[H+] = 0 ,10 M B.[H+] < [CH3COO-]G.[H+] > [CH3COOI . [H+3 < 0,10M

GiiC h n D. CH3COOH ^ CH3COO" + H+

0 ,1 M a.0,lMV a < 1 nn [H+] < 0,'1GM

5. i vi dung dch axit mnh HNO3 0,10M, nu b qua s in li canc th nh gi n v nng mol ion sau y l ng?A. [H+] = 0,10M B, [H+] < [NOS-]C.[H+]> [N 0 31 D. [H; ]< 0 ,10 M

GiiChnA. HNO3 -> H++ NCV

0,1M 0,1M[H+] = 0,10M

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6/96

BI 3: S IN LI CA Nc. pH.CHT CHI TH AXIT - BAZ

1 . Tch s ion ca nc l g v bng bao nhiu 25c?Gii

Tch nng ion H+v nng ion OH~ trong nc gi l tch s" ionca nc. 25c, tch s" ny c gi tr 1014

2. Pht biu cc nh ngha mi trng axit, trung tnh v kim theonng H+v pH.

Gii- Mi trng axit l mi trng trong [H+] > [OH~] hay [H+]

> 107M hay pH < 7- Mi trng baz l mi trng trong [H+] < [OH-] hay [H+]

< 10 'M hay pH > 7Mi trng trung tnh lmi trngtrong [H+] = [OH~] hay [H+]= 10 7 M hay pH = 7

3. Cht ch th axit - baz l g? Hy cho bi t mu ca qu vphenolphtalein trong dung dch cc khong pH khc nhau.

GiiCht ch th axit - baz l cht c mu bin i ph thuc vo gi trpH ca dung dch

T Xa------------------------------------------------------------v --------V-------------------------A >,* n . 1 f--- 1--- 1------------1------ 1---*------------------- >

W 0 6 7 8 7 14K H

* Phenolphtalein f ------- ^ 4

4. Mt dung dch c [OH~] = 1,5.10"5M. Mi trng ea dung dch ny l:A. Axit B. Trung tnh c. Kim D. khng xc nh c

GiiChn c .

Ta c: [H+] =-----

=-=6,6.10-10M < 10'7M =>mi trng kim1,5.10

5. Tnh nng H+, OH" v pH ca dung dch HC1 0,10M v dung dchNaOH 0,010M.

Gii* Dung dch HC1 0,1M: HC1 ->H* + c r

0,1M 0,1M

=> [H+] = 0 ,1M => [OH] = = 1 0 13M => pH = lg 0,1 = 10 ,1

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7/96


8/96

3. Ly mt s th d chng minh rng: bn cht caphn ng trong dung

dch cc cht in li l phn ng gia cc ion.

GiiBn cht ca phn ng trong dung dch cc cht in li l phn ng

gia cc ion. Tht vy:* NaCl + AgN03 -> AgCl + NaNOa

Bn cht Ag++ Cl' -> AgCl

* MgCl2 + 2NaOH - Mg(OH)2 + 2NaClBn cht: Mg2+ + 20H~ -> Mg(OH)2'l

4. Phng trnh ion rt gn ca phn ng cho bit:A. Nhng ion no tn ti trong dung dch.

B. Nng nhng on no trong dung dch ln nht.c. Bn cht ca phn ng trong dung dch cc cht in li.

D. Khng tn ti phn t trong dung dch cc cht in li.

Gii

Chn c

Phng trnh ion rt gn ca phn ng cho bit bn cht ca phn ng

trong dung dch cc cht in li.

5. Vit cc phng trnh phn t v ion rt gn ca. cc phnng (nu c)

xy ra trong dung dch gia cc cp cht sau:a) Fe2(S04)3+ NaOH b) NH4C1 + AgNOa c) NaF + HC1

d) MgCl2+ KNO3 g)HC10 + KOH e) FeS (r) + HC1

G

a) Fe2(S04)3+ 6NaOH > 2Fe(OH)34 + 3Na2S042Fe3* + 60H ^ 2Fe(OH)34

b) NH4CI + AgN03^ AgCU + NH4NO3

Ag* + c r A gCl

c) NaF + H C lX) MgCls + KNOaX

e) FeS (r) + 2HC1 -> FeCl2+ H2s t

FeS + 2H+ Fe2* + H2s t

g) HC10 + KOH >KCIO + H20

H+ + OH' H20

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9/96

6 . Phn ng n di y xy ra trong dung dch to c kt ta

Fe(OH)3{hnh 1 .6 )?

A. FeS04 + KMn04+ H2SO4 B. Fe2(S04)3+ KI

c. Fe(N03)3+ Fe D. Fe(N03)3+ KOH

Gii

Ch n D.

Fe(N03)3+ 3KOH -> Fe(OH)3 + 3KN037. Lv th d v vit cc phng trnh ha hc di dng phn t v ion

rt gn cho cc phn ng sau:a) To thnh cht kt ta.

b) To thnh cht in li yu.c) To thnh cht khi

Gii- To thnh cht kt ta: NH4C1 + AgN03 NH4NO3+ AgCl

Ag++ Cl" ->gCiTo thnh cht in li yu: NaOH + HC1 - NaCl + H20

H+ + OH > H20

- To thnh cht kh: K2CO3+ 2HC1 -> 2KC1 + C02T + H202H++ C032'-> CO2 1 + H20

BI 5. LUYN TP AXIT, BAZ V M. PHN NGTRAO I ION TRONG DUNG CH CC CHAT IN LI

1 . Vit phng trnh in li ca cc cht sau:K2S, Na2HP04, NH2P04,Pb(OH)2 , HBrO, HF, HCIO4.

Gii* K2S -> 2K+ +s2

* Na2HP04-2Na++ HPO42"* NaH2P 04 Na+ + H2P04~

* Pb(OH)2 vPb 2+ + 20H "v Pb(OH)2 ^ P b 0 22' + 2H+

* HBrO -> H+ + BrCr* HF -> H++ F"

* HCIO4 H + + CIO4"

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10/96

2. Mt dung dch c [H+] = 0,010M. Tnh [OH-] v pH ca dung dch. Mi

trng ca dung dch ny l axit, trung tnh hay kim? Hy cho bit

mu ca qu tm trong dung dch ny.Gii

[H+] = 0 ,0 1M th [OH-] = ----= 10"12M v pH = - lg0,01 = 20,01 p 6

Do[H+]=0,01M > 10_7M nn y l mi trng axit, qu tm s ha .

3. Mt dung dch c pH = 9,0. Tnh nng mol ca cc ion H+v OH" trongdung dch. Hy cho bit mu ca phenolphtalein trong dng dch ny.

Gii

V pH = 9,0 nn [H+] = 10 9M v[OH"] = 10_5M10Do [H+] = 1(T9M < 10_7M nn y l mi trng baz. pH = 9 nn

phenolphtalein ha hng.

4. Vit cc phng trnh phn t v ion rt gn ca cc phn ng (nu c)xy ra trong dung dch gia cc cp cht sau:

a) Na2C03 + Ca(N03)2 b) FeSC>4+ NaOH (lng)c) NaHC03+ HC1 d) NaHC03+ NaOHe) K2CO3 + NaC l g)Pb(OH)2 (r) + HNO3

h) Pb(OH)2 (r) + NaOH i) C11SO4+ Na2SGii

a) Na2C03+ Ca(N03)2->CaC03 + 2NaNOsCa + CO32" ->CaC03

b) FeSO,j + 2NaOH (long) -> Fe(OH>2 + NaSO

Fe2* + 20H" -> Fe(OH)2 lc) NaHCOa + HC1 NaCl + C02+ H20

H* + HCOa' > C02t + H20d) NaHCOa + NaOH -> Na2C03+ H20

HCOa- + OH -> C032- + H20e) K2C0 3 + NaCl X

g) Pb(OH)2 (r) + 2HNO3 > Pb(N03)2 + 2H20Pb(OH)2 + 2H+ > Pb2* + 2HsO

h) Pb(OH)2 (r) + 2NaOH-> Na2Pb02 + 2H20

Pb(OH)2 + 20H > Pb022' + 2H20i) CuSO + Na2S -> CuSi + Na2S04

Cu2+ + s 2 ^ CuSi

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11/96

5. Phn ng trao i ion trong dung dch cc cht in li ch xy ra khi

A. Cc cht phn ng phi l nhng cht d tan.

B. Cc cht phn ng phi l nhng cht in li mnh.

0 . Mt s' ion trong dung dch kt hp c vi nhau lm gim nng

ion cua chng.

D. Phn ng khng phi l thun nghch.

Gii

Chn c

Phn ng trao i ion trong dung dch cc cht in li ch xy ra khi

mt s ion trong dung dch kt hp c vi nhau lm gim nng

ion ca chng.

6. Kt ta CdS (hnh 1.7a) c to thnh trong dung dch bng cc cp

cht no di y?

A.C dCl2 + NaO H B. Cd(N0 3)2 + H2S

C.Cd(N03)2+ HC1 D. CdCl2+ Na2S04

Gii

Chn B. Cd(N03)2+ H2S -> C d S i + 2HNO3

7. Vit phng trnh ha hc (di dng phn t v ion rt gn) ca phn

ng trao i ion trong dung dch to thnh tng kt ta sau: Cr(0 H)3 ;

A1(0H)3 ; Ni(OH)2 (hnh 1.7 b, c, d).

Gii

Cr(OH)a i + 3NaCl

Cr(OH)a i

A1(0H)3 + 3KC1

A1(0H)3 4-

Ni(OH)2i +K2SO4

-> Ni(OH)2

a) CrCla + 3NaOH

Cr3++ 30H~

b) AICI3 + 3K0H

Al3++ 30H" >

c) N1SO4 + 2K0H

Ni2+ + 20H" ->

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12/96

('.hing II.

NIT - PHOTPHO

BI 7. NIT

1 . Trnh by cu to ca phn t N2. V sao iu kin thng, nit lmt cht tr? iu kin no nit tr nn hot ng hn?

GiiPhn t nit gm 2 nguyn t, gia chng hnh thnh mt lin kt ba.Lin kt ba trong phn t nit r t bn nn nit tr iu kin thng,

nhit cao (trn 3000C), nit hot ng hn v c th phn ngvi nhiu cht khc.

2 . Ni t khng duy tr s h hp,nit c phi l kh c khng?

GiiNit khng duy tr s chy v s h hp, tuy nhn nit khng phi lkh c

3. a) Cp cng thc ca liti nitrua v nhm nitrua lA. LiNs v A13N B. L3N v A1N

c. L12N3v AI2N3 D. L13N2v AI3N2b) Vit phng tr n h ha hc ca phn ng to th nh liti nitrua vnhm nitrua khi cho liti v nhm tc dng trc tip vi nit. Trong ccphn ng ny ni t l cht oxi ha hay cht kh'?

Giia) Liti nitrua v nhm nitrua c cng thc phn t ln lt l L 13N v

A1N0 3

b) 6Li + N2 -2L3N0 -3

2AI + N2-> 2A1NTrong phn t ny, nt ng vai tr cht oxi ha v s oxi ha gim

sau phn ng.

4. Nguyn t' nit c s oxi ha l bao nhiu trong cc hp cht sau: NO,

N 02, NH3, NH4CI, N20, N2O3, N2O5, Mg3N2?

Gii+2 +4 -3 -3 +1 +3 +5 -3

N O , N 0 2 , N H3 , N H 4C1, N2 0 , N2 O3 , N2 O5 , Mg3N2

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13/96


14/96

MBs + H20 NH4+ + OH"

NHs + HCl > NH4CINH4CI + NaOH -> NH3t + NaCl + H20NHg + HNO3 > NH4NO3

NH4NO3 > N2O + 2H2O

3. Hin nay, sn sut amoniac, ngi t iu ch nit v hiro bngcch chuyn ha c xc tc mt hn hp gm khng kh, hi nc vkh metan (thnh phn chnh ca kh thin nhin). Phn ng gia khmetan v hi nc to ra hiro v cacbon ioxit. loi kh oxi v thukh nit, ngi ta t kh metan trong mt thit b kn cha khng kh.

Hy vit cc phng trnh ha hc ca phn ng iu ch hiro, loikh oxi v tng hp kh amoniac.

Gii

- iu ch H2 :C H 4+ 2H20 C02+ 4H2

- Loi 0 2 : CH4 + 2O2 - > C02+ 2H20

- Tng hp NH3 : N2+ 3H2 ... V . 2NH3 -;

4. Trnh bv phng php ha hc j phn b it cc dung dch: NH3,Na2S04, NH4CI, (NH4)2S04. Vit phng trnh h hc ca cc phn

ng dng.Gii.

- a giy qu tm vo 4 mu th, mu lm qu tm ha xanh i dungdch NH3; hoa i NH4C v (NH4)2S0 4; khng i mu l Na2S04

- Cho dung dch BaC2 v hai mu lm qu tm ha , mu to kt

ta trng l (NH4)2S04, khng hin tng l NH4CI(NH4)2S04 + BaCl2-> BaS04i + 2NH4CI

5. Mun ch cn bng ca phn ng tng hp amoniac chuyn dch sangphi, cn phi ng thi: .A. Tng p sut v tng nhit .B. Gim p sut v gim nhit .

c. Tng p sut v gim nhit .

D. Gim p sut v tng nhit .

7;0Vi . G ii

C h n c .......N2 + 3H2 >.... 2NH3 AH < 0 ..................

cn bng chuyn dch sang phi, theo nguyn l Lsatli, ta cnng thi gim nhit v tng p sut.



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15/96

6. Trong phn rig nhit phn cc mui NH4NO2 v NH4NO3, s" oxi ha

ca nit bin i nh th no? Nguyn t nit trong ion no ca muing vai tr cht kh v nguyn t nit trong ion no ca mui ngvai tr cht oxi ha?: N2 + 2H2O3 +3

=> N ng vai tr cht kh ; N ng vai tr cht oxi ha

N H 4NO 3 > N2 0 +2H 203 +5 ' 'N ng vai tr cht kh ; N ng va tr cht oxi ha

7. Cho dung dch NaOH d vo 150,0 ml dung dch (NH4)2S04 1,00M, unnng nh.a) Vit phng trnh ha hc dng phn t v dng ion rt gn.

b) Tinh th tch kh (ktc) thu c.Gii

n(NH4)2S04= 0,15 .1 = 0,15 moi => nNH+ = 0,3 moi

(NH4)2S04+ 2NaOII > Na2S4+ 2NH3t + 2H20 ; ,Nir.,4 + o i r -> n h 3T + II200,3 mol 0,3 mol ...

Vy VNH3= 0,3 >22,4 = 6,72^

8 . Phi dng bao nhiu lt kh nit v bao nhiu lt kh hiro iuch

17,0 gam NH3? Bit rng hiu sut chuyn h thn h amoniac l25,0%. Cc th tch kh c o ktc.

A. 44,8 lt N2v 134,4 lt H2 . : B. 22,4 lt N2v 134,4 lt H2c. 22,4 lt N2v 67,2 lt H2 D. 44,8 lt N2v 67,2 lt H2

Gii .

Chn A

N2 + 3H2 2NH3 " --,7 ^

22,4^ -> 3. 22,4 -> 2.17 g ^ r

11,2 f Vn2 c = 11,2 . = 44,81

VK2cin = 3 3 f i . = 134>4

15



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16/96


17/96

4 . a) Trong phng trnh ha hc ca phn ng nhit phn st (III)

nitrat, tng cc h s bng bao nhiu?

A. 5 B.7 c 9 ' ; D 2

b) Trong phng tr nh ha hc ca phn fng nbi t phn th ngn (II)

nitrat, tng cc h s bn g bao nhiu?

A. 5 B.7 C.9 .:: D. 21

Gii , j ; ; 1

a) Chn D. 4Fe(N03)3, ; * >2Fo20;+ 12NQ2 + 3Q2 ,,

b) Chn A. IlglNO:;!:! -> llg + 2NO; + ()2 ^

5. Vit phng trnh ha hc cacc phn ng thc hin dychujen ha

sau y:

- (ff>Cu(N03)2 - (3- - >Cu(OH)2 (4) - > Cu(N03)2

C >CC2 V 0 8 ;

Gii

NO (1) > h n o 3

^ C u O

4N02 + O2 + 2H2O

CuO + 2HNO3

Cu(N03)2+ 2NaOH

Cu(OH)2 + 2HNO3

-> 4HNO3

>Cu(N03)2 + H2O

- Cu(0 H)2 i + 2NaNC>3

-> Cu(N03)2+ 2H20.r v:

Cu(N03)2

CuO +

Cu +

H2

Cl2

CuO + 2NO0T + - OT

-> Cu + H20

-> CuCl26 . Khi ha tan 30,Og hn hp ng v ng (II) oxit trong 1,50 lt dung

dch axit nitric 1,0QM (long) thy thot ra 6,72 lt nit monooxit

(ktc). Xc nh hm irig phn trm ca ng (II) oxt tirong hn

hp, nng moi ca ng (II) ntrat v axit nitric trong dung dch sau

phn ng, bit rng th tch dung dch khng thay i.

% . ^ ^ O Gii3Cu + 8HNO3 -> 3Cu(N03)2 + 2NOT + 4H20

. 8a ,..

2a : " : 0'a Cu(N03)2+ H20b 2b

17



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18/96

, a + b;MCu(N 03)2 " T T

1,5 f8 a

=0,31 M;

+ 2b= 0,18 M^HNC^ ! 5

7. iu ch 5,000 tn axit nitric nng 60,0% cn dng bao nhiu

tn amoniac? Bit rng s hao ht amoniac trong qu trnh sn xut l3,8c/c

Gii

m, = 5. = 3 tnl H NC >3 1 0 0

Ta c s :

NH3 -> N O -> N 02 -> H N03

1 mol -> 1 mol

17 g -------> 63 g?


19/96


20/96

Gii

Photpho c dng sn xut bom, n chy (do photpho trng d chy

trong khng kh), sn xut dim qut (do d b oxi ha bi KCIO3), sn

xut axit photphoric (do d b oxi ha to P 2O5 l anhiric photphoric)

5 . t chy hon ton 6 ,2g photpho trong oxi d. Cho sn phm to

thnh tc dng va vi dung dch NaOH 32% to ra mui Na2HP04.

a) Vit phng trnh ha hc ca cc phn ng xy ra.

b) Tnh khi lng dung dch NaOH dng.

c) Tnh nng phn trm ca mui trong dung dch thu c sau

phn ng.

Giia) 4P + 502 2P2O5

0,2mol 0 ,1 mol

P2O5 + 4NaOH 2Na2HP4 + H 20

0,1 mol 0,4 mol 0,2 mol, , _ 0,4.40.100 _b) mddNaOH = ;-----r---- = 50 gam

oZ

, ^ 0,2.142.100c ) C N a 2 H P0 4 ~ 6 2 + 5 0 =

BI 11. AXIT PHOTPHORIC V Ml PHOTPHAT

1 . Vit phng trnh ha hc dng phn t v dng ion rt gn ca phn

ng gia H3PO4vi lng d ca:

a) BaO b) Ca(OH)2 c) K2CO3

Gii

a) 3BaO + 2H3PO4 > Ba3(P04)2 + 3H2O

b) 3Ca(OH)2+ 2H3PO4 Ca3(P 04)2+ 6H20

c) 3K 2CO 3 + 2H3P 4 > 2K3PQ4 + 3C O2 + 3H 2O

2. Nu nhng im ging nhau v khc nhau v tnh cht ha hc gia axt

nitric v axit photphoric. pn ra nhng phn ng ha Hc minh ha

-V-, v:s,:,r i z .*J-. Xifryh 'OfrV; ii _* Ging: - Trong dung dch nc u in i r ion H*

- u c th tc dng vi kim loi, oxit baz, baz, ml.

20



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21/96

V d: Na20 -r 2HNO3 -> 2NaN03+ H20

3Na20 + 2H3PO4 2Na3P 04 + 3H20

* Khc: HNO3 c tnh 0X1 ha ; H3PO4 khng c tnh oxi ha

V d: c + 4HNO3 C02+ 4N02+ 2H20

c + H3PO4 X

3. Phng trnh in li tng cng ca H 3PO4 trong dung dch :H3PO4 3H+ + PO43"

Khi thm HC1 vo dung dch,

A. Cn bng trn chuyn dch theo chiu thun.

B. Cn bng trn chuyn dch theo chiu nghch,

c. Cn bng trn khng b chuyn dch.D. Nng PO43 tng ln.

Gii

C h n B Khi thm HC1 vo, tc thm H+ vo th theo nguyn lLsatli, cn bng phi chuyn dch theo chiu nghch.

4. Lp cc phng trnh ha hc sau y:

a) H3PO4 + K2HP4 ->

lmol 1 mol -

b) H3PO4 + Ca(OH)2

1 mol 1 mol

c) H3PO4 + Ca(OH)2

2moi 1 mold) HsP0 4 + Ca(OH)2 ->

2mol 3 mol

Gii

) H3PO4 + K2HPO4 2KH2P 0 4

lmol 1 mol

b) H3PO4 + Ca(OH)2 -> CaHP04+ 2H20

1 moi 1 mol

c) 2H3PO4 + Ca(OH)2 -> Ca(H2P 4)2+ 2H22 moi 1 mol

d) 2H3PO4 + 3Ca(OH)2-> Ca3(P0 4)2+ 6H20

2moi 3 mol

21



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22/96

5. thu c mui photphat trung ha, cn ly bo nhiu ml dung dch

NaOH 1,00M cho tc dng vi 50,0 ml dung dch H3PO40,50M?Gii

Ta c nH3po4 = 0,05 . 0,5 = 0,025 mol

3NaOH + H3PO4 -> Na3P 0 4+ 3H20

0,075 mol 0,025 mol

=> VddNaOH = M 75 = 0, 07 5 1

BI 12. PHN BN HA HC

1 . Cho cc mu phn m sau y: amoni sun fat, amoni clorua, natri

nitrat. Hy dng cc thuc th thch hp phn bit chng.Vit

phng tr n h ha hc ca cc ph n ng dng.

Gii

- Ph t hin amoni sunfat bng dung dch BaC2

- Ph t hin amoni clorua bng dung dch AgN03. Cn li l natri n itra t

(NH4)2S04+ BaCla -> BaS04 i + 2NH4CINH4CI + A g N O s A g C l + NH4NO3

2. T khng kh, than, nc v cc cht xc tc cn thit, hy lp s

iu ch phn m .NH4NO3

Gii

- Chng ct phn on khng kh lng c N2v O2

c + H20 , 1050c > CO + H2

N2+ 3H2 2NH3

4NH3 + 502 xt>t >4NO + 6H2O

2NO + 0 2 -> 2N02

4N02+ 0 2+ 2H20 ^ 4HNO3

NH3 + HNOs > NH4NO3

3. Mt loi qung photphat c cha 35% Ca3(P0 4)2. Hy tnh hm lng

phn trm P2O5c trong qung trn.

22



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23/96

Gii

10 0 gam qung trn c 35 gam Ca3(P0 4)2, tc e gamo XU

P2O5. Vy hm lng P2O5 trong qung l 16%

4. sn xut mt lng phn bn amophot dung ht 6.103 mol

H,P04a) Tnh th tch amoniac (ktc) cn dng, bit rng loi amophot ny

c t l v s mol nNH4H2po4 : n(NH4)2Hpo4 = 1 : 1

b) Tnh khi lng amophot thu c.

Gii

a) NH3 + H3PO4 > NH4H2PO4

a a a mol

2NH3 + H3PO4 -> (NH4)2 HPO4

2a a a mol

Theo ta C2a = 6000 VNH3 = 22,4.3a = 201600 litttlamophot = mNH4H2po4 + m(NH4)2Hpo4

= 115a + 132a = 247a = 247. 3000 = 741000 g tc 741'kg

BI 13. LUYN TP TNH CHT CA NIT, PHOTPHO

V CC HP CHT CA CHNG

1 . Hy cho bit s oxi ha ca N v p trong cc phn t v ion sau yNHs, NH4\ NCV, NOa', NH4HC0 3> P2O3, PBr6, P043', KH2P04>

Zn3(P04)2-

Gii-3 -3 +3 +5 -3 +3 +5 +5

N H 3, N H / , N 0 2-, NO3-, NH 4H C03> p 203> p Br5, p O4 ,

KH2 P O4, Zn3( P 0 4)2

2. Trong cc cng thc sau y, chn cng thc ha hc ng ca magie

photphua:

A. Mg3(P 04)2 B. Mg(P03)2

C. Mg3P2 D. Mg2P207

Gii

Magi photphua: Mg3? 2. C hn c

23



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24/96

V ' | '

3. a) Lp cc phng trnh ha hc sau y:

NH3+ Cl2 (d) ---->N2+ ... (1)NH3(d) + Cl2 ->NH4C1 + ... (2 )

.w,:n NH3+ H 3CO0H---- > ... (3)

jori;C N ^^ro4a - ^ - - H3P04 + ... (4)

Zn(N03)2 i =. * > ... (5)b) Lp cc phng trnh ha hc dng phn t v dng ion r t gn

ca phn ng gia cc cht sau ytrong ung ch:

K3PO4 v Ba(N03)2 (1)Na3P0 4y CaCl2 (2 )

Ca(H2P0 4)2 v Ca(OH)2vi t l1 : 1 (3)

(NH4)3P04 + Ba(OH)2 (4)

Gii

a) 2NH3+ 3C12 (d) -> N2+ 6HC1

8NHa(d>:+ 3C12 -N2+ 6NH4CI

NH3 + CH3COOH > CH3COONH4

:n ^ l i l p o l _ J lL ^ H3PO4 + 3NH3

ZnfNOaJi -> ZnO + 2KO2+ 022

b) 2K3PO4 + 3Ba(N03)2 -> Ba3(P04)2 i +6KNO3

2Na3P 0 4 +3CaCl2 -> Ca3(P04)2 + 6NaCl

Ca(H2P 0 4)2 + Ca(OH)2 -> 2CaH P04 + 2H20

2(NH4)3P 0 4 + 3Ba(OH)2 -> Ba3(P04)2 + 6NH3+ 6H20

4 . T hiro, clo, nit v cc ha cht cn thit, hy vit cc phng trnh

ha hc (c ghi r iu kin phn ng) iu ch phn m amoni

clorua.

Gii

H2+ Cl2 2HC1

n 2+ 3H2 ' - :t-0,p 2NH3

NH3 + HCI > NH4CI

24



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25/96

5. Vit phng trnh ha hc thc hin cc dy chuyn ha sau y:

a) (1) .... (2)N2 -(4)

VNO

>NH3

(5)

b) Photpho

h) N-; + 3H2

+Ca,t

->NO

(3)

(6)

NH4NO3

%2

(1)>B

(7)

+HC1

HNO,

-c +02,t

n h 3 + HNO3

NH4NO3+ KOH-----

N2 + 0 2 _ 2 c

2NO + 0 2

4N02+ 0 2+ 2H20

Cu + 4HNO3

NH3+ HNOs ----

b) 3Ca + 2P i-

Ca3P2 + 6HC1 2PH3 + 4 O 2 >

(2) (3)

G i i

2NH3

NH4NO3

KNO3+ NH3 + H2O

2NO

2N 02

> 4HNO3

> P2O5

()

(2)

(3)

(4)

(5)

(6)

(7)

(8)

Cu(N03)2+ 2N02+ 2H20

NH4NO3

Ca3P2

3CaCl2+ 2PH 3P2O5.+3H2O

6 . Hy a ra nhng phn ng hc c s tham gia ea n cht

photpho, trong s oxi ha ca photpho:

a) Tng b) Gim.

Gii

Tng: 4P + 502 2P2O5

* Gim: 2P + 3Ca------- > a^P27. Khi cho 3,00 g hn hp Cu v AI tc dng vi dung dch HNO 3 c d,

un nng, sinh ra 4,48 lt kh uy nht NO2 (ktc). Xc nh phn trm

khi lng ca mi kim loi trong hn hp.

25



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26/96

Gii

Cu + 4HNO3 - Cu(N03)2+ 2N02+ 2H20a 2a

AI + 6HNO3 -> A1(N03)3+ 3N02+ 3H20

b 3b

64a + 27b = 3

2a + 3b = | ^ | = 0 2 ~ a = l 5

=* *A 1 . ^ g . 23,4115.3

%Cu = 76,6

8 . Cho 6,00 g P2O5 vo 25,0 ml dung dch H3P04 6,00% (D = 1,03 g/ml).

Tnh nng phn trn H3PO4 trong dung dch to thnh.

Gii

mdd = 25. 1,03 = 25,75 g =* m H3P04 = 1,545 g100

P2O5+ 3H20 ^ 2H3PO4

142g > 2.98g

6g -------- 8,28g

(1,545 + 8,28)100c%dung dch mi = 30,95%6+ 25,75

9. Cn bn bao nhiu kg phn m amoni nitrat cha 97,5% NH 4NO3 cho

10,0 hecta khoai ty, bit rng 1,00 hecta khoai ty cn 60,0 kg nit

Gii

10 hecta cn 600 kg nit

C 80g NH4NO3 cung cp 28g nit

1714,28kg


27/96


28/96

I

5. t mt mu than (cha tp cht khng chy) c khi lng 0,600

kg trong oxi d thu c l,06m3 (ktc) kh cacbonic. Tnh thnh phnphn trm khi lng ca cacbon trong mu than trn .

Gii

c + o2 .- > c o 2

1 2 g ----> 22,4^

567,85gr/rC = ---------= 94,6%600

BI 16: Hp c h t C a c a c b o n

1 . Lm th no loi hi nc v kh C2c ln trong kh co? vit ccphng trnh ha hc.

GiiDn hn hp qua bnh ng NaOH c, d, kh tho t ra l c o

CO2+ 2NaOH -> Na2C03+ H202 . C ba cht kh gm CO, HC1, v S0 2 ng trong ba binh ring bit.

Trnh by phng php ha hc phn bit tng kh. Vit cc phng

trnh ha hc.Gii

Cho cc mu th tc dng vi nc vi trong d., mu lm c nc vil S02

SO2 + Ca(OH)2C&SO3 'l' + H2OHai mu cn li cho tc ng vi dung dch AgN3, mu to kt ta

trn g HCI. Cn li l c o

HC1 + AgNOs > AgCl + HNO3

3. iu no sau y khan g ng cho phn ng ca kh c o vi kh O2?

A. Ph n ng thu nhit.

B. Phn ng ta nhit.c. Phn ng km theo s gim th tch.D. Phn ng khng xy ra iu kin thng.

Gii

Chn A 2CO + O2 - > 2CO2Phn ng ny xy ra khi t nng, km theo s gim th tch v tanhit mnh

28



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29/96

4 . a) Khi un nng dung dch canxi hirocacbonat th c kt ta xut hin.Tng cc h s t lng trong phng trnh ha hc ca phn ng l:A. 4 B. 5 C. 6 D. 7

b) Khi cho d kh C02 vo dung dch cha kt ta canxi cacbont thkt ta s tan. Tng cc h s t lng trng phng trnh ha hccua phan ng l:

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

Giia) Chn A Ca(HC03)2 > CaC03 + C02+ H20

=>-Tng h s l 4b) Chn A C2+ CaCOs + H20 -> Ca(HC03)2

=> Tng h s" l 45. Cho 224,0 mi kh C2 (ktc) hp th ht trong 100,0 ml dung dch kali

hiroxit 0,200M. tnh khi lng ca nhng cht c trong dung dch to thnhGii

nC02= 0,01 mol; ]1 koh = 0,02mol

C02 + 2KOH ------- K2CO3 + H200 ,0 1 mol 0,02mol 0,01 mol^ mK2C03 = 138. = 1>38g

6. Nung 52,65 g CaC03 1000c v cho ton b lng kh thot ra hpth ht vo 500,0 ml dung dch NaOH 1,800M. Hi thu c nhngmui no? Khi lng l bao nhiu? Bit rng hiu sut ca phn ngnhit phn CaCO'3 l 95%.

GiinCaC03 = M = 0,5265 mol

3 100

CaC03 > Ca 0 + C020,5265 mol 0,5265 mol

95^C2thu c 0,5265. = 0,5 iol

Ta c nNaOH = 0,5,1,8 = 0,9 molCO2 + 2N aO H -------> Na2C03+ H2Oa 2a va

C02+ NaOH ------ NaHCOs

b b b a + b = 0,5 a = 0,4

=> - c=> 1[2a + b = 0,9 [b = 0 ,1

=> mNa2C3 = 106 . 0,4 - 42,4 g ; mNaHC03 = 84 . 0,1 = 8,4 g

29



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30/96

BI 17. SILIC V HP CHT CA SILIC

1 . Nu nhng tnh cht ha hc ging nhau v khc nhau gia silic vcacbon. Vit cc phng trnh ha hc minh ha.Gii

* Ging - u c cc s" oxi ha 4 ; 0; + 2 ; + 4- Va c tnh kh, va c tnh oxi ha* Tnh kh

V d c + 0 2 -"> CO:

Si + 02 > S i02* Tnh oxi ha

V d 4A1 + 3C ... --> A14C3

2Mg + Si > Mg2Si

* Khc - Cacbon ch yu th hin tnh kh- Silic tc dng d dng vi dung dch kim- S" oxi ha + 2 t c trng i vi silic

2 . S oxi ha ca silic th hin hp cht no sau y?A. SiO B. Si0 2c. SiH4D.Mg2Si

GiiChn B. Trong SiC>2, silic th hin s oxi ha cao nht + 4

3. Khi cho nc tc dng vi oxit axit th axit s khng c to thnh,nu oxit axit lA. Cacbon ioxit B. Lu hunh ioxitc. Silic ioxit D. init pentaoxit

GiiC h n c S1O2 + H2O X

4. T Si02v cc cht cn thit khc, hy vit pho'ng trnh ha hc cacc phn ng iu ch axit silixic.

Gii

Si0 2 + 2NaOH ------- Na2Si0 3 + H20Na2Si0 3+ CO2+ H2O > H2S1O3 'i + Na2C3

5. Phng trnh ion rt gn: 2H+ + Si032- -H2Si0 3 ng vi phn nggia cc cht no sau y?A. Axit cacbonic v canxi silicat B.Axit cacbonic v natri silicatc. Axit clohiric v canxi silicat D.Axit clohiic v na tri silicat

GiiCh n D Na2Si0 3+ 2HC1-> H2S1O3 i + 2NaCl

30



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31/96


32/96


33/96

h) Si02 + HC1 X

i) Si + 2NaOH + H2O -> Na2Si03+ 2H2

3. C cc cht sau: C02, Na2CC>3, c, NaOH, Na2SiQ3, H2S1O3. Hy lp

thnh mt dy chuyn ha gia cc cht v vit cc phng trnh ha

hc.

Gii

c C02 -> Na2C03 NaOH Na2Si0 3 -> H2S1O3

c + 0 2 > C02

C02 + 2NaOH ------ > Na2C03 + H20

Na2C03 + Ba(OH)2 ------ > BaCOs I + 2NaOH

Si0 2 + 2NaOH ------ > Na2Si0 3+ H20

NC2SO3 + CO2 + H2O > Na2CC>3 + H2S1O3 'i'

4. Cho 5,94 g hn hp K2CO3 v Na2C03 tc dng vi dung dch H2SO4 d

thu c 7,74g hn hp mui khan K2SO4 v Na2S4. Thnh phn ca

hn hp u l:

.3,18g Na2C03v 2,76g K2CO3 B. 3,81g Na2C03v 2,67g K2CO3

c. 3,02g Na2C03v 2,25g K2CO3 D. 4,27g Na2C03v 3,82g K2CO3

Hy chn p s" ng.

Gii

Chn A

Gi a, b l s'mol K2CO3v Na2C3

K2CO3 + H2SO4 > K2SO4 + C0 2 + H20

a a

N&2C03 + H2SO4 > N3.2SO4 + CO2 + H2O

b b

138a + 106b = 5,94 a = 0,02

174a + 142b = 7 ,7 4 |b = 0,03

^ mK2C03 = 138 . 0,02 = 2,76 g

m N a2C 0 3 = 5. t chy 6,80 g hn hp X gm hiro v cacbon monooxit cn 8,96

lt oxi (o ktc). Xc nh thnh phn phn trm theo th tch v

theo khi lng ca hn hp X.

33



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34/96

Gii

Gi a, b l s' mol H2v c o trong X

2H2+ O2 2H2Oa- a

2

2CO + 0 2 2C0 2

b I22a + 28b = 6,8

+b = M = 0,42 2 22,4

c:> a = 0, 6b = 0,2

0 , 6.100%H2- = 75% ;0,8%CO = 25%

6 . Mt loi thy tinh c thnh phn ha hc c biu din bng cng

thc K20.Pb0.6Si02. Tnh khi lng K2CO3, PbC03v S1O2 cn dng

sn xut c 6,77 tn thy tinh trn. Coi hiu sut ca qu trnh

l 100%.

Gii

Trong 677 gam K20 .Pb0 .6Si0 2c 94 g K2O ; 223 g PbO v 360 g S12

^ 6,77 tn K20.Pb0.6Si02c 0,94 t K20 ; 2,23 tPbO v 3,6 t Si02

Trong 138 g K2CO3 c 94 g K2O

1,38 tn K2CO3 0,94 tn K2O

- Trong 267 g PbC3 c 223 g PbO

2,67 tn PbC3 Phi dng 1,38 tn K2CO3 ; 2,67 tn PbC3v 3,6 tn S1O2

34



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35/96

Chng IV.

BI 20. M U V HA HC HU c

1. So snh hp cht v c v hp cht hu C v: thnh phn nguyn t,c im lin kt ha hc trong phn t.

GiiKhc vi hp cht v c, thnh phn hp cht hu c nht thit phic cacbon, thng gp l hiro, oxi, nit, sau l halogen, lu hunh ...

Khc vi hp cht v c, lin kt ha hc ch yu trong phn t hpcht hu c l lin k t cng ha tr -

2 . Nu mc ch v phng php tin hnh phn tch nh tnh v nhlng nguyn t.

GPHN TCH NH LNG- Mc ch

Xc nh hm lng cc nguyn ttrong phn t cht hu c

- Phng phpNung a gam cht hu c (C, H, o, N)vi CuO d. Hp th hi H20 v C2ln lt bng bnh ng H2SO4 c

d v KOH c d. tng khilng mi bnh chnh l khi lngH20 v CO2 tng ng. Kh N2 thotra c xc nh th tch ( ktc). T

mH20>mco2vVN2 rt ra mc> mH,

rriN v mo ri suy ra hm lng cac, H, N v 0

3. Oxi ha hon ton 0,600 gam hn hp hu c A thu c 0,672 lt GO2

(ktc) v 0,720 gam H2O. Tnh thnh phn phn trm khi lng ca

cc nguyn t" trong phn t cht A

Gii

m = 12 =: 0,36 g => %c.= 0, 3HP-- = 60(%)22,4 5 0,6

35

PHN TCH NH TNH- Mc ch

Xc nh nguyn t no ctrong thnh phn phn tcht hu c

- Phng php xc nh nh tnh c vH, nung cht hu c viCuO chuyn c thnh

CO2, H thnh H20, ripht hin CO2 bng ncvi trong v H20 bngCuS04 khan. Cn vi N thchuyn thnh NH3 rinhn ra bng giy qu m



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36/96

mH = M ?. . 2 = 0,08 g=>%H = Q,Q81- - := 13,33(%)18 0,6

> %mH= = 10,44 (%)l o u, fc>/

mc = . 12 = 0,6 g => %mc = 610-= 89,56 (%)100 0,67

p - caroten ch cha c , H.*Lu p caroten c cng thc phn t C40H56, c nhiu trong c rt,c chua, gc -

BI 21. CNG THC PHN T Hp CHAT h u c

1. Tnh khi lng mol phn t ca cc cht sau:a) Cht A c t khi hi so vi khng kh bng 2,07.

b) Th tch hi ca 3,30 gam cht X bng th tch ca 1,76 gam kh oxi(o cng iu kin v nhit , p sut).

G i ia) Ma = 2,07 . 29 = 60

b ) T a c n A= n o 2 ^ = ^

M x = ^ = 6 0

1,762 . Limonen l mt cht c mi thm du c tch t tinh du chanh. Ktqu phn tch nguyn t" cho thy limonen c cu to t hai nguynt' c v H, trong c chim 88,235% v khi lng. T khi hi calimonen so Vi khng kh gn bng 4,690. Lp cng thc phn t calimonen.

%H = 100 - 88,235 = 11,765t cng thc cn tm l CxHy ta c:

36



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37/96

12x y 4,69.29 = 136x= 10 . y = 16

t cng thc cn tm l CxHyOz ta c (lu M = ..-- =60)

88,235 11,765 100Vy limonen c CTPT l C10H 16

3. t chy hon ton 0,30 gam cht A (phn t ch cha c, H, O) thuc 0,44 gam kh cacbonie v 0,18 gam nc. Th tch hi ca 0,30gam cht A bng th tch ca 0,16 gam kh oxi ( cng iu kin v

nhit , p sut). Xc nh cng thc phn t ca cht A.Gii

mc = 0,44 .^1 = 0,12g44

_ 0,18 _mu = 2 = 0,02 g

mo = 0,3 - (0,12 + 0,02) = 0,16 g0,3.320,16

1 2 x y = 16z = 60 _ 200

0,12 0,02 0,16 0,3o x = 2 ;y = 4 ; z = 2. Vy CTPT ca A l C2H4O2

4. T tinh du hi, ngi ta tch c anetol - mt cht thm c dngsn sut ko cao su. Anetol c khi lng moi phn t bng 148,0g/mol. Phn tch nguyn t cho thy, anetol c %c = 81,08%; %H =8,10% cn li l oxi. Lp cng thc dofn gin nht v cng thc phn tca anetol.

%0= 100- (81,08+8,1) = 10,82%

t cng thc cn tm l CxHyOz ta cX: y : z = = 6,75 : 8,1 : 0,675 = 10 : 12 : 1

12 1 16

=> Cng thc nguyn ca anetol l (CioHi20 )n=> Cng thc n gin ca anetol l C10H 12OV M = 148 nn anetol cng c CTPT l C10H 12O

5. Hp cht X c phn trm khi lng cacbon, hiro v oxi ln lt bng54,54%, 9,10%, 36,36%. Khi lng mol phn t ca X bng 88 ,0g/mol.Cng thc phn t no sau y ng vi hp cht X?

A. C4H10O B. C4H8O2 c. C5H12O D.C4H10O2Gii

Chn B rng MC4H10o =: 74 * 88 v MC4H10o2 = 90 * 88nn X ch c th

l C4H80 2hoc C4H 10O2

37



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38/96

Nhng C4H 10O2c %c= 68,18 * 54,54 nn khng ph hpVy X l C4H80 2 :

6 . Hp cht z c cng thc n gin nht l CH3O v c t khi hi so vihiro bng 31,0. cng thc phn t no sau y ng vi hp cht z ?A. CH3O B .C 2H60 2 C.C 2H60 D.C3H9O3

.... Gii .Chn B.Tli(M) ta c 31 n = 62 o n = 2. Vy z l C2H60 2

Bi 22 : CU TRC PHN T HP c h t h u c

1. Pht biu ni dung c bn ca thuyt cu to ha hc.GiiC 3 lun im chnh- Trong phn t hp cht hu c, cc nguyn t lin k t vi nhau theo

ng ha tr v theo mt tr t t nh t nh. Trt t lin kt gil cu to ha hc. S thay i trt t lin kt , tc l thay icu to ha hc, s to ra hp cht khc

- Trong phn t hp cht hu c, cacbon e ha tr 4. Nguyn tcacbon khng nhng c th lin kt vi nguyn t ca cc nguyn t"

khc m cn lin kt vi nhau to thnh mch cacbon (mch vng,mch khng vng, mch nhnh, mch khng nhnh)

- Tnh cht ca cc cht ph thuc vo thnh phn phn t (bn cht,s lng cc nguyn t) v cu to ha hc (tr t t lin kt ec -nguyn t d)

2 . So snh ngha ca cng thc phn t v cng thc cu to. Cho thd minh ha.

G* Ging: u cho bit th nh phn nguyn t v s lng nguyn t

ca mi nguyn t trong phn t* Khc: Cng thc phn t khng biu din th t v cch thc linkt gia ec nguyn t, trong khi cng thc cu to ni ln r iuny.

3. Th no l lin kt n, lin kt i, lin kt ba?G i i

- Lin kt n () l lin kt do mt cp electron chung to nn.- Lin kt i (ln; l) l lin kt do,hai cp electron chung to nn.- Lin kt ba (2t ; l) l lin kt do ba cp electron chung to nn.

38



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39/96

4. Cht no sau y trong phn t ch c lin kt n?A.CH4 B.C2H4 C.C 6H6 D.CH3COOH

GiiChn A: CH,J ^

Phn t CH4 ch cha ton lin kt n H c H

H5. Nhng cht no sau y l ng ng ca nhau, ng phn ca nhau?

a) CH3 CH = CH CH3 e) CH 3 CH = CH GH2 CH3

b) CH2 = CH CH 2 CH3 g) CH2 = CH CH CH3

c )CH3 CH2 CH2 CH? CH3 g h 3

d) CH 2 = CH CH.3 h) CH3 CH 2 CH 2 CH 2 CH 2 CH3

i) CH3 - CH2 - CH - CH3

CH3 G i i-ng phn lC H3 CH = C H C H 3 CH3 CH2 CH2 CH2 CH3

| c H2 = CH CH2 CH3 c h 3 CH CH2 CHg

|C H ; CH = CH GH2 CH 3 CH3

cH = CH CH CH3

c h 3

- ng ng l

c h 3 - CH = CH - c h 2 - CHg r c h 2 = CH - c h 3

[CH3- CH = CH - CH3 ; cHg = CH - CH2- CHg

CHg - CH2 - CH2 - CH2 - CH3

CH3 CH2 CH2 CH2 CH2 CHg

6 . Vit cng thc cu to c th c ca cc cng thc phn t nh sau:C2H60, c 3h 6o, c 4k 10.

Gii* C2H60 : CH 3 CHS r H ; CH3 o CH3* C3H60 : CH2= CH -C H 2- OH ; C 2= CH - 0 - CH3

CHs - C - c h 3 c h 2- c h 2 I__ iO CH2 0

CH3 CH2 CHO ; H2C CH CH3 CH2 CH2

\ / \ /o CHI

=*=C4H10: c h 3 c h 2 c h 2 c h 3 c h 3 C H c h 3 o h

c h 3

39



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40/96

7. Nhng cng thc cu to no di y biu th cung mt cht?

L , 1 1Cl - b - H (II) H - c - c - H (III)pH 2 - Vit phng trnh ha hc ca cc phn ng xy ra.b) Tnh thnh phn phn trm khi lng ca mi cht trong hn hp

Gii

a) C2H5OH + Na -> C2H6ONa + - H 22

a

CH3- CH2- CH2OH + Na CH3- CH2- CHgONa + - H 22

46a + 60b = 5,3 ra = 0,05

=> i a b 1 , 1 2 -__ S2 2 ! = I b = 0>05

%C2H5OH= 4 6 0 L0! 10 Q^43,4 %C3H7OH = 56,6

BI 23. PHN NG HU c1 . Th no l phn ng th, phn ng cng, phn ng tch? Cho th d

minh ha.Gii

- Ph n ng th l phn ng trong nguyn t hoc nhm nguyn ttrong phn t cht hu c b thay th bi mt nguyn t hoc nhmnguyn t khc.

V d: CH4+ Cl2 > CH3CI + HC140



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41/96


42/96

(4) 3C2H2 - 4 CyFt;I

(5) C 6H6 + B r 2 C 6H5Br + HB rt

(3), (5) l cc phn ng th

(1), (2 ), (4 ) l cc phn ng cng

4. Kt lun no sau y l ng?

A, Phn ng ca cc cht hu c thng xy ra nhanh.

B. Phn ng ca cc cht hu c thng xy ra rt chm v theo nhiu

hng khc nhau.

c. Phn ng ca cc cht hu c thng xy ra rt chm v ch theomt hng xc nh.

D. Phn ng ca cc cht hu c thng xy ra nhnh v khng theo

mt hng nht nh.

GiiChn B.

Phn ng cc cht hu c thng xy ra chm v theo nhiu hngkhc nhau.

BI 24. LUYN TP HOP CHAT HU c CNG THC PHN T V CNG THC CU TO

1 . Cht no sau y l hirocacbon? L dn xut ca hirocacbon?

a) CH20 b) C2H5Br c) CH20 2

d) C6H5Br e)C6H6 g) CH3COOH

Gii

C6H6l hiro caebon

CH2O; C2H5Br; CH20 2; C6H5Br; CH3COOH l dn xut ea hirocacbon

2 . T genol (trong tinh du hng nhu) iu ch c metylgenol (M =

178g/mol) l cht dn d cn trng. Kt qu phn tch nguyn t ca

metylgenol cho thy: %c = 74,16%; %H = 7,86%,cn li l oxi.

Lp cng thc n gin nht, cng thc phn t ca ometylgenol

Gii

%0 = 100 - 74,16 - 7,86 = 17,98%

42



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43/96


44/96

7 . Cc phn ng sau y thuc loi phn ng no (phn ng th, phn

ng cng, phn ng tch) ?

a) C 2H 6 + C l 2 as -> C 2 H 5 C I + H C 1

b) c 4h 8 + h 20 dd axit > c 4h 10o

c) c 2h 5c i ddNaOH/C2H5OH > c i+ HC1

d) 2C2H5OH t , c 2h 5o c 2h 5 + h 20

Gii

a ) c 2h 6 + C l 2 as > C 2 H 5 C I + HC1 l phn ng th

b ) c 4h 8 + h 20dd

C 4 H 10 O l p h n n g c n g

c) c 2h 5c i dd NaOH/C2H5OH > Hh H C1 l p h n n g t c h

d ) 2C 2H 50 Hxt,

C 2 H 5 O C 2 H 5 + H 2 O l p h n n g t c h

8 . Vit phng trnh ha hc ca cc phn ng sau v cho bit cc phn

ng thuc loi phn ng no (phn ng th, phn ng cng, phn

ng tch) ?

a) Etilen tc dng vi hiro c Ni lm xc tc v un nng.

b) un nng axetilen 600c vi bt than lm xc tc thu c

benzen.c) Dung dch ancol etylic lu ngoi khng kh chuyn thnh dung

dch axit axetic (gim n)

Gii

a) C24 + H2 C2H6 l phn ng cngt

b) 3 C2H2 > C66 l phn ng cng

c) C2H5OH + 0 2 mengm > CH3COOH + H20 l phn ng th

44



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45/96

Chng V.IrllROCACBON NO

BI 25. ANKAN1 . Th no l hirocacbon no, ankan, xicloankan?

GiiHirocacbon no l hirocacbon ch cha lin kt n trong phn t

Ankan l hirocacbon mch h ch cha lin kt n trong phn t

Xicloankan l hirocacbon mch vng (1 vng) ch cha lin kt n

trong phn t

2 . Vit cng thc phn t ca cc hirocacbon tng ng vi cc gc ankyl

sau: CH3, C3H7, C6H13

Gii

H

* CH4 : H - b - H * C3H8: CH3 - CH2 - CH3

H

* C6H14 : CH3- CH2- CH2- CH2- CH2- CH3

CHa - CH - CH2- GH2 - CH3

c h 3

CH3 - CH2 - CH - CH2 - CH3

c h 3

c h 3c h 3 - c - c h 2 - c h 3

c h 3

CH3 - c h - CH - CH31 ICH3 cr6

3. Vit phng trnh ha hc ca cc phn ng sau :

a) Propan tc dng vi clo (theo t l 1:1) khi chiu sng.

b) Tch mt phn t hiro t phn t propan.

c) t chy hexan. Gii

a) CH3 - CH2 - CH3 + Cl2 ~ p 2 ." l2CI + HC1^ U1I3 UJ \J3

C145



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46/96

bi e,Hs -xi-> C,Ii6 + H2t ;IQ ___

c) C6Hi4 + 0 2 ^ 6CO2 + 7H2024 . Cc hirocacbon no c dung lm nhin liu l do nguyn nhn no

sau y?

A. Hirocacbon no c phn ng th.

B. Hirocacbon no c nhiu trong t nhin,

c. Hirocacbon no l cht nh hn nc.

D. Hirocacbon no chy ta nhiu nhit v c nhiu trong t nhin.

Gii

Chn D. Hirocacbon no dng lm nhin liu l do n chy ta ranhiu nhit v c nhiu trong t nhin

5. Hy gii thch:

a) Ti sao khng c cc bnh cha xng, du (gm cc ankan) gn

la, trong khi ngi ta c th nu chy nha ng (trong thnh

phn cng c cc ankan) lm ng giao thng.

b) Khng dng nc dp cc m ehy xng, du m phi dng ct

hoc bnh cha kh cacbonic.

Gii

a) Ankan c phn t lng ln th kh bay hi v kh chy hn ankan

Gphn t lng nh, do nha ng c th nung chy, cn xng

du rt d bt la v chy.

b) Xng, du khng tan trong nc, nh hn nc nn khng th dng

nc dp tt m chy xng du

6. Cng thc cu to CH 3 CH CH2 CH2 CH3 ng vi tn gi no

sau y? CH3

A. Neo pentan B. 2metylpentn

c . Isobutan D. 1 , 1imetylbutanGii

Chn B

2 - metyl pentan CH3 - CH - CH2- CH2- CH3

CHs7. Khi t chy hon ton 3,60 g ankan X thu c 5,60 lt kh CO2(ktc).

Cng thc phn t ca X l trng hp no sau y?

A.C3I8 b ;c 5h 10 C.C5H12 d .c 4h 10

46



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47/96

Gii

ChnC: CnH2n +2+

a mola(14n + 2) = 3,6

^ i 0 2 > n C 0 2+ (n + DHO

an5,6 OK => CTPT (X) C5H12

BI 26. XIGLOANKAN

1. Nhn nh no sau y l ng?

A. Xicloankan ch c kh nng tham gia phn ng cng m vng.

B. Xicloankan ch c kh nng tham gia phn ng th.

c. Mi xicloankan u c kh nng tham gia phn ng th v phnng cng.

D. Mt sxicloankan c kh nng tham gia phn ng cng m vng.

Chn D.

Mt s xicloankan ( xiclopropan, xiclobutan) Gn cho phn ng cng

m vng. V d xicloankan C36 v C4H8 u cho phn ng cng H2

(xc tc Ni), v C3H6cn lm mt mu nc brom

2 . Sc kh c xiclopropan vo dung dch brom. S quan st thy hin

tng no sau y?

A. Mu dung dcH khiig i.

B. Mu dung dch m ln

c. Mu dung dch b nht dn.D. Mu dung dch khng mu chuyn thnh nu .

Chn c. Mu dung dch b nht dn: C3H6+ Br2-> C3H6Br23. Vit phng trnh ha hc ca phn ng xy ra khi:

) Sc kh xiclopropan vo dung dch brom.b) Dn hn hp xiclopropan, xiclopentan v hiro i vo trong ng c

bt niken, nung nng.

c) un nng xiclohexan vi brom theo t l liol 1 : 1.

Gii

G

47



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48/96

Gii

a) \ 7 + Br 2 ----- - BrCH2- CH2 - CH2Br

b) ' \ 7 7 + H 2 __ > CHa - CH2- CH3

CI o + Br 2 > 0 ' Br +HBr

4. Trnh by phng php ha hc phn bit hai kh khng mu propan

v xiclopropan ng trong cc bnh ring bit.

Gii

Mu lm mt mu nc brom l / \

5. Xicloankan n vngX c t khi so vi nit bng 2,0. Lpcng thcphn t ca X. Vit phng trnh ha hc ( dng cng thccu to)

minh ha tnh cht ha hc ca X, bit rng X tc dng vi H 2 (xc tc

Ni) ch to ra mt sn phm.

Gii

Theo ta c 14n = 28.2 o n = 4. Vy X l C43

CTCT (X) l Q . Tht vy

n + h 2 _ NJ > c h 3 - c h 2- c h 2 - c h 3t

BI 27. LUYN TP ANKAN V XIGLOANKAN

1. Vit cng thc cu to ca cc ankan sau: pentan, 2-metylbutan,

isobutan. Cc cht trn cn c tn gi no khc khng?

Gii

CH3 - CH2 - CH2 - CH2 - CH3 : Pentan

CH3 - CH - CH2- CH3 : 2 - metyl butan (iso pentan)

CH3

CH3 - CH - CH3 : iso butan (2 - metyl propan)

CH3

2. Ank an Y mch k hng nhnh c cng thc n gin nh t C2H5

a) Tm cng thc phn t, vit cng thc cu to v gi tn cht Y.

b) Vit phng trnh ha hc phn ng ca Y khi chiu sng, ch r

sn phm chnh ca phn ng.

48



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49/96

Gii

a) A c cng thc (C2H5)n tc C2nH5n :

Ta phi c s H < 2 ln s c + 2

5n < 2 . 2n + 2 n < 2Nhng n = 1 cho cng thc C2H5khng tn ti

Vy n = 2, tc A c CTPT l C4H 10

b) A c th c CTCT l CH3- CH2- CH2GH3hoc CHs - CH - CHg

CH3

^ 9H3- CH2- CHC1 - CH3(chnh)CH3CH2CH 2CH3 + CH _ CH2 CH2 CH2 ()

C CHa - cpci - CH3(chnh)

CH3 _ +HC1

C H 3 H C H 2C 1(p h )CH, I

CH33. t chy hon ton 3,36 lt hn hp kh A gm metan v etan thu c

4,48 lt kh cacbonic. Cc th tch c o ktc, Tnh thnh phn

phn trm v th tch''ca mi kh trong hn hp A.

, , Gii , 7;; :.

Gi a, b l th tch CH4v C2I6trong A

CH4 + 2 0 2 -> C0 2 + 2H20a a . :

CgHg + O2 > 2CO2 + SHgO2 .......

b 2b : ^

fa + b = 3,36 a = 2,24 1

^ ja + 2b = 4,48 b = 1,12 - ' 3 r ^ -

=> %CH4 = ' - 4'10-0- = 66,6%; %C2H6 = 33,4%3,36 ~

4. Khi 1,00 gam metan chy ta ra 55,6 kJ. cn t bao nhiu lt kh

metan (ktc) nhit sink ra un 1,00 lt nc (D = 1,00 g/cm3) t

25c ln 100c?bit rng mun nng 1,00 gam nc ln r c cn tiu

tn 4,18 J v gi s nhit sinh ra ch dng .lm tng nhit ca

nc. ... _ . '. r

49



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50/96

Gii

1 gam nc mun nng ln lc cn 4,18 J

=> 1 gam nc mun nng ln 75c cn 4,18 . 75 = 313,5 J=> 1 000 gam nc mun nng ln 75c cn 313,5 . 1 000 = 313 500J

= 313,5 kJ1 gam CH4chy ta ra 55,6 kJ

5,638 g CH4


51/96

Chng VI.

HIROCACBON KHNG NO

BI 29, ANKEN

1. So snh anken vi ankan v c im cu to v tnh cht ha hc.Cho v d minh ha

Khc vi ankan l phn t ch cha lin kt ) phn t anken c cha1 lin kt t km bn, gy, do khng ging vi ankn l ch phnng th' l phn ng c trng, aiiken cho phn ng cng l phn ngc trng

V d: C2H4 + H 2 NL-> C2H6t ,

C2H4 + Br2 -----> C2H4Br2C2H4 + HBr -> C2H5Br

Ngoi ra anken cn eho phn ng trng hp, phn ng lm mt mudung dch thuc tm

V d: 3C2H4 + 2KIvI4 + 4H2O *43C2H4(OH)2 + 2Mn02 + 2KOH

nCH2= c h 2 *u0,p > (- CII2- CH2-) . ,2. ng vi cng the phn t C5H10 c bao nhiu anken ng phn cu

to? . ^

A. 4 B. 5 c. 3 ' D. 7Gii

C h n B

GH2 = CH - CH2 - CH2 - CH3; CHS - CH = CH - CH2 - CH3

CH2 = C - GH2 - CH3 ; CH2 = CH - CH - 0H2;

CH3 ; ', f CH3

CH3- c = C H - CH3 - - ;

c h 3 ......

3 . Vit phng trnh ha hc ca phn ng xy ra khi:a) Propilen tc dng vi hiro, un nng (xc tc Ni).b) But-2-en tc ng vi hio clorua.

c) Metylpropen tc dng vi nc c xc tc axit.

d) Trng hp but-1-e n

51



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52/96

Gii

a) C3H6+ H2 N0 > C3H8t

b) CH3 - CH = CH - CH3+ HCI GH3- GH2- CHC1 - CH3

OH

CH2= C - CH3+ H20

CH3

nCHa - CH2 - CH = CH2

CH3 - C - C H 3

CHaCH3- CH - CH2OH

CHa

rr CH2-CH2 -C H 3 -

4. Trnh by phng php ha hc :a) Phn bit metan v etilen.h) Tch ly kh metan t hn hp vi etilerLc) Phn bit hai bnh khng dn nhn ng hexan v hex-l-en.

Vit phng trnh ha hc ca cc phn ng; dng.

Giia) Dn mi mu th qua nc brom, mu lm mt mu nc brom l

C2H4, cn li l CH4b) Dn hn hp qua bnh nc brom d, etilen s b gi li, kh thot

ra l CH4c) Cho nc brom vo hai mu th, mu lm mt mu nc brom l

hex-l-en, cn li l hexan.

5. Cht no sau y lm mt mu dung dch brom?A. Butan B .But- l-enc. cacbon ioxit D. Metylpropan

G

Chn BCH2 = CH CH2 GH3 + Br2 > CH2Br CHBr CH2 CH3

6 . Dn t t 3,36 lt hn hp gm etilen v propilen (ktc) y dung dchbrom thy dung dch b nht mu v khng cn kh thot ra. Khilng dung dch sau phn ng tng 4,90 gam.a) Vit cc phng trnh ha hc v gii thch cc hin tng th

nghim nu trn.b) Tnh th nh phn phn trm v th tch ca mi kh trong hn hp

ban u. ....... _

52



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53/96

Giia) Gi X, y l s' mol C2H4 v C3H6

Khng c kh thot ra khi bnh, chng t hai kh phn ng htvi brom v khi lng bnh tng chnh l khi lng hai kh.

C2H4 + Br2> C2H4B r2X X X

C3H6+ Br2> C3H6Br2y y y

b) Ta c h x + y = = 0, 15 * 3" 0 122 ,4 %C2H4 = L-1 =66 ,6 ; %C3H6= 33,40,15

BI 30. ANKAIEN

1 . Th no l ankaien, ankaien lin hp? Vit cng thc cu to v gitn cc ankaien lin hp c cng thc phn t C4I6, C5H8.

GiiAnkaien l hirocacbon mch h c cha hai ni iAnkaien lin hp l hirocacbon mch h c 2 ni i cch nhau 1 ni

n. V d

* C4H6 : CH2= CH - CH = CH2

* C5

H8

: CH2

= CH CH = CH CH3

; GH2

= c CH = CH2

CH3

2 . Vit phng trnh ha hc ( dng cng thc cu to) ca cc phn ng

xy ra khia) Isopren tc dng vi hiro (xc tc Ni)

b) Isopren tc dng vi brom trong (trong CCI4). Cc cht c ly theo

t l s' moi 1 :1 , to ra sn phm theo kiu cng 1, 4.c) Trng hp isopren theo kiu 1,4.

Gi

a) CH2 = c - CH = CH2 + H2 ---> CH3 - c = CH - CH3c h 3 c h 3

b) CH2= - CH = CH2+ Br2 >CHBr - = CH - CH2Br

CH3 h3

53



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54/96

c) nCHa = c - CH = CH2 # > f - CH2- c = GH - CH2}I_ t,p I I Jn

c h 3 c h 3

3. Oxi ha hon ton 0,680 gam ankaien X thu c 1,120 lt C2 (ktc).a) Tm cng thc phn t ca X

b) Tm cng thc cu to c th c ca X

Gii

a) CnH2n - 2+ (- n ~ - 0 2-> nC02+ (n - 1 ) H20

CH2- c := CH - CH2- CH3

CH3- ch = c = ch - ch 3

a anra(14n - 2) = 0, 68 r ^

1 1 = 51 1 2 _ CH3 - CH2 - CH2 - CH3

5. Hp cht no sau y cng hp H2to thnh isopentan?

A. GH2= CH - CH = GH rr CH3 B- CH2= GH - c - CH2

T ^ I3c. c h 2 = CH - c h 2 - CH = c h 3 d . _ Gii

Chn B.

CH2= c - CH = CH2 + 2H2 CHa - CH - CH2 - C3CHa CH3

54



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55/96

BI 31. LUYN TP ANKEN V ANKAIEN

1 . Vit cc phng trnh ha hc minh h:

a) tch metan t hn hp vi mt lng nh etlen, ngi ta dn

hri hp kh i qua ung dch brom d.

b) Sc kh propiln vo dung dch KMn04 , thy mu ca dung chnht dn, c kt ta nu en xut hin.

Gii

a.) C2I4 + B r2 C2H 4Br2

b) 3C3H6 + 2KMn04+ 4H20 -> 3C3H6(OH)2+ 2Mn02+ 2KOH

2. Trnh by phng php ha hc phn bit ba bnh ng ba kh ring

bit l metan, etilen v cacbonie. Vit phng tr n h ha hc minh ha.

Gii

n 3 mu th qua nc vi trong , mu lm c nc vi trbng l

C02:

C02+ Ca(OH)2 -> CaCO i +H20

Dn 2 mu th qua nc brom, mu lm mt mu nc brom l C2H4,

cn li l CH4

C2I4 + B r2 > C2H4Br2

3. Vit phng trnh ha hc ca cc phn ng thc hin s chuyn

ha sau:

CH4 > C2H2 C2H4 > C2H6 > C2H5CI

Gii2CHt 1500c > C2H2+ 3H2

C2H2+ H2 _ E i- C2H4

C2H4+ H2 - i i ^ C2Hr,

C2H6 +C12 C2H5C1+ HC14. Vit phng trn h ha hc ca cc phn ng iu ch: 1,2 - icloetan;

1 ,1 - icloetan t etan v cc cht v c cn thit.

G

* C2I6 >C2H4 + H2

c h 2= c h 2+ C12 h 2 - c h 2 *.Cl C1

55



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56/96

* CH2 - C H 2 ddNaOH/C2H5OH ^ CH CH + 2HC1I I

Cl Cl

CH s CH + 2HC1 > C H 3 H C ICl

5 Cho 4,48 lt hn hp kh gm metan v etilen i qua dung dch brom

d, thy dung dch nht mu v cn 1,12 lt kh thot ra. Cc th tch

kh o iu kin tiu chun. Thnh phn phn trm th tch ca kh

metan trong hn hp l

A. 2 > m B. 50,0% c. 60,0% D. 37,5%

Gii

Chn A_ TT 1,12.100%CH4= -..= 25 (%)

4,48

6 . Vit phng trnh ha hc ca cc phn ng iu ch buta -1,3 - ien

v cao su buna t but - 1 -en.

Giixt - c h 3- c h 2- c h 2- c h 3

xt -> CH2= CH - CH = CH2+ 2H2

CH2= c h - CH2- CH3 +H2

c h 3- c h 2- c h 2- c h 3 t~

nCH2 = CH - CH = CH2 xt;tg- > (- CH2- CH = C H -C H 2-)n

7. 't chy hon ton 5,40 g ankaien lin hp X thu c 8,96 lt kh

CO2 (ktc). Cng thc no sau y l cng thc cu to ca X?

A. CH2= CH - CH = CH2 B. CH2 = C H -CH = CH -GH 3

c. CH2= c - CH2 - CH3 D. CH2= c = CH - CH3

CH3Gii

ChnA: CnH2n2 +

a mola(14n - 2) = 5,4

8,9622,4

3n - 1

an =

0 2 -> 1CO2+ (n 1)H20

an moi

a = 0 ,1CTPT (X) l C4H6

56



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57/96


58/96

Giia) n mi mu th qua dung dch AgN03/ NHs, mu to kt t a vng

l C22, cn li l C2H4 .... ....b) Dn mi mu th qua dung dch AgN03/ NH3, mu to kt ta vng

l C2H2. Hai mu cn. li dn qua nc forom, mu lm mt munc brom l C2H4, cn li l CH4

4. Cho cc cht sau; metan, etilen, but-2-in v axetilen. Kt lun no sauy ng?A. C 4 cht u c kh nng lm mt mu dng dch brom.

B. C hai cht to kt ta vi ung dch bc nitrat trong amoniac.c. C ba cht c kh nng lm mt mu ung dch brom.D. Khng c cht no lm nht mu dung ch kali pemanganat.

GiiChn c.

C 3 hirocacbon khng no lm mt mu nc brom

5. Dn 3,36 lt hn hp A gm propin v etilen i vo mt lng dungdch AgNC>3 trong NH3 thy cn 0,840 lt kh thot ra v c m gam kt

ta. Cc th tch kh o ktc.a) Tnh phn trm th tch etilen trong A.

b) Tnh m. Gii

a) %C2H4= = 25 (%)3,36

IX 3,36 0,84 110- .b) nc3H4 = --- 22 ,4 '= 0,1125 moi

=> n = 0,1125 moi =>' m =0,1125. 147 = 16,5375 g

6 . Trong s cc ankin c cng thc phn t CHg c my cht tc dng

c vi dung dch AgNOa trong NH3?

A. 1 cht B. 2 cht c. 3 cht D. 4 chtHy chn p n ung.

Gii

Chn B

C hai cht l: CH s c - CH2 - CH2 - CH3 v CH B c - CH - CH3

CH3

58



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59/96

BI 33. LUYN TP ANKIN

1 . Dn hn hp kh gm metan, etilen, axetilen i vo mt lng d ung

dch bc nitrat trong dung dch amoniac. Kh cn li c, dn vo dungch brom (d). Nu v gii thch cc hin tng xy ra trong thinghim.

GiiC s xut hin kt ta vng v ung dch brom b nht mu.

C2H2 + 2AgN0 3 + 2NT3 >C2Ag2 ^ + 2NH4NO3

2 . Vit phng trnh ha hc ca cc phn ng thc hii s chuynha sau:

CH4 >C2H2 C4H4 ^ >G4H6 (4) > polbutaien

Gii

2CH4 ----1500 c ) C22 + 3H^

2C 2H 2 > C H C H = C H 2

CH 3 c - CH = CH2+ H2 Pd/P c 3 > GH2 =CH - CH = GH2t

nCH2 = CH - CH = CH2xt't0p > (- CH2- CH = CH - CH2 - ) n

3. Vit phng trnh ha hc ca ccphn ng t axetilen v cc cht v

c cn thit iu ch cc cht sau:

a) 1,2 - icloetan b) 1 4 - icloetan c) 1,2 - ibrometen

d) Buta - 1,3 - ien e) 1,1,2 - tribrometan

Giia) C2H 2 Ha pd /^ co9 >C2H ,

t;

c h 2= c h 2 + C12 ------- > c h 2- c h 2I* \

Cl C1

b) CH = CH + 2 HC1----> CHs - c h - C1L

c) CH = CH + Br2 ------------> CH = CHI X

Br Brd) 2C2H 2 > C4H4

C 4H4 + Ha .... f f ' S g g a > C4H6



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60/96

e) CH 3 CH + r2

CH = CH + HBrT IBr Br

-> GH = GH

Br Br

> CH2 - CH1 1Br Br

Br

4. Khi thc hin phn ng nhit phn metan iu ch axetilen thu c

hn hp X gm axetilen, hiro v metan cha phn ng ht. T khi

ca X so vi H2bng 4,44. Tnh hiu sut ca phn ng.

Gii

Gi s nhit phn 1 mol CH4

2CH4 ------- > C2H2+ 3H23a2

Hn hp sau phn ng gm

CH4 :1 - a

C2H2 : 0, 5a

H2 :1,5a

16(1 - a) + 26.0,5a + 2 .1 ,5a= 4,44.2 = 8,88

1 - a + 0,5a + 1 ,5a

16 = 8,88 (1 + a) a = 0,8

Hiu sut phn ng t 80 %5. Dn 6,72 lt hn hp kh X gm propan, etilen, axetilen qua dung dch

brom d, thy cn 1,68 lt kh khng b hp th. Nu dn 6,72 lt kh X

trn qua dung dch bc nitrat trong amoniac thy c 24,24 gam kt ta.

Cc th tch kh o iu kin tiu chun.

a) Vit cc phng trnh ha hc gii thch qu trnh th nghim

trn.

b) Tnh th nh phn phn trm th tch v theo khi lng ca mi kh

trong hn hp. Gii

a) Gi a, b, c ln lt l s' mol C3H8, C2H4 v C2H2 trong X, ch c

C24 v C2H2 tc dng vi nc brom.

C2H4 + B r2 - C2H4Br2

b

C2H2+ 2Br2 C2H2Br4

60



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61/96


62/96

Chng VII.

BI 35. BENZEN V NG ANG.MT S HIROCACBON THM KHC.

1 . ng vi cng thc phn t CsHio c bao nhiu ng phn hircacbonthm?A. 2. B. 3. . 4. D. 5.

Hy chn p n ng.

Gi i

Chn c Hs CH3 CHa

P CH3 I& C H 3 CH,

2. Toluen v benzen cng phn ng c vi cht no Su y: (1) dungdch brom trong CCI4; (2) dung dich kal pemanganat; (3) hiro xc tc

vi Ni. un nng; (4) Br2 c bt Fe, n nng? Vit phng trnh hohc ca cc phn ng xy ra.

V G . Berizen v tolueri cng tc dng c vi

* H2 (Ni, t)

101 + 3H2 0

CH. CHsO + 3H2

* B r2 (Fe , t) Br

] + B r 2 - Fe^ - > ( ^ 1 + H B r

CHs Hs

A + Br2 (0 f Br + HBr

(hoc ng phn par)



B

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63/96


64/96


65/96

Gii* Ging- Etylbenzen v stren u c th cho phn ng th vo vng benzen- u c th cng hp vi H2cho cng mt sn phm l etylxielohexan

V d: -C 2H6 + 3H2 p - * < ^ C 2H5

CH = CH2 + 4H2 N; > O ** Khc- Stiren lm mt mu nc brom v dung dch KMnC>4 ngay nhit

thng, trong khi etylbenzen khng lm mt mu nc brom v

chi lm mt mu dung dch KMn 0 4khi nung nng.

Stiren cn cho c phn ng trng hp.

9. Dng cng thc ho hc vit phng trnh ho hc ca stiren vi.

a) H20 (xc tc H2SO4)

b) HBr.

c) H 2 (theo t l mol 1 : 1 , xc tc Ni)Gii

< ^ - C H = CH2 + H20 a g g i l > -CH - CH3

M 3 H = CH2 + HBr ------- > -CH - CH3

-CH = CH2 + H2 - r c f - * -CH2 -C H 3

10. Trnh by phng php ho hc phn bit 3 ch t lng sau: toluen,benzen, Stiren. Vit phng trnh ho hc ca cc phn ng dng.

Gii- Cht lm mt mu nc brom l stiren

- Cht lm mt mu dung dch KM11O4 khi un nng l toluen. Cnli l benzen

11. Khi tch hiro ca 66,25 kg etylbenzen thu c 50,00kg stiren. Tin

hnh phn ng trng hp ton b lng stiren ny thu c hn hp A

va lm mt mu ca 60,00 ml dung dch brom 0,15M

a) Tnh hiu sut ca phn ng tch hiro ca etylbenzen.b) Tnh khi lng stiren trng hp.c) Polistiren c phn t khi trung bnh bng 3,12.105. Tnh h s

trng hp trung bnh ca polime.

65



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66/96

a) -C 2H5 xt,t- -> -CH = CH2 + Ha106g - - > 104 g66,25 kg 65 kg

=> H = 5- - = 80 (%)65

b) 5,2g A lm mt mu dung dch cha 0,009 mol Br2

52kg A tc 52 OOOg A > 90 mol Br 2

M n.sUren d - n Bi'2 ^-stiren dif 9 0 . 0 4 9 3 6 0

Vy mstiren trng hp = 52 000 - 9360 = 42 640 g

c) H s" trng hp trung bnh ca PS = = 3 000

12. Trnh by cch n gin thu c naphtalen tinh khit t hn hp

naphtalen e ln tp cht khng tan trong nc v khng bay hi

un nng naphtalen thng hoa. Thu ly hi naphtalen ri lm lnh

ta c naphtalen sch

13. T etilen v benzen, tng hp c stiren theci s :

a) Vit phng trnh ho hc thc hn cc bin i trn.

b) Tnh khi lng stiren thu c t 1,00 t n benzen nu hiu sut

ca qu trnh l 78%.

Gii

C6Ha - S & - > C #H5C2H,H

6 su2 5 >c6h 5 - c h = c h 2

G

)

b) C 1 mol benzen c 1 mol stiren

Hay 78 g benzen c 104g stiren

==> 1 tn --------------> tan stren



B

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67/96

BI 36. LUYN TP HIROCACBON THM

1. Vit cng thc cu to v gi tn cc hirocacbon thm c cng thc

phn t C8H 10, C8H8. Trong s cc ng phn , ng phn no phn

ng c vi: dung dch brom, dung dch bromua? Vit phng trnh

ho hc ca cc phn ng xy ra.

Gii* CgHio

2H5 CHs H3 CH3

" , . " . CH3

Etylbenzen 1,2imetyl benzen 1,3 imetyl benzen 1,4 imetyl benzen

*C 8H8 C H Br CH2Br

-C H = CHa + HBr - -CHBr - CH3

2. Trnh by phng php ho hc phn bit cc cht lng sau: benzen,

stiren, toluen v hex: -- 1 - in.

Gii

- Cht to k t ta vi dung dch AgNC>3 / NH3 l hex - 1 - in

- Cht lm mt mu nc brom l stiren

Ch t lm m t mu dung dch KM1O4 khi un nng l toluen

Cn li l ,benzen

3. Vit phng trnh ho hc ca cc phn ng iu ch etilen, axetilen t

metan; iu ch clobenzen v nitrobenzen t benzen v cc cht v c

khc.

Gii

* C2H2: 2CH4 ....> C2H2 + 3H2

* C2H4: C2H2 + H Pd S 2a _ C2H4 C6H5C1: C6H6+ Cl2 Fe;t...> 6H5C1 + HC1

* C6H5N 02: C6H6- HNOa H2so4d.'t....-.>C6H5N 02+ H20

67



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68/96

4. Cho 23,0 kg toluen tc dng vi hn hp axitHN03 c, d (xc tc

H2SO4 c). Gi s ton b toluen chuyn thnh2,4,6-trinitrotoluen

(TNT). Hy tnh:a) Tnh khi lng TNT thu c.

b) khi lng axit HNO3 phn ng

GiiCH3

Hs o o 2r X \ n o 2(Q + 3HNO3 H2SO4M lT +3 H 20

N 02

92g ------- 3. 63g ----------------- > 227 g

23 kg >47,25 kg ----- 56,75 kg

5. Ankylbenzen X c phn trm khi lng cacbon bng 91,31%.

a) Tm cng thc phn t ca X.

b) Vit cng thc cu to v gi tn cht X.

Gii

X c cng the CnH2n - 6m , 12 n .l00 o-, _Ta c -= 91,31 n = 7

14n - 6

Vy X l C78, ng vi cng thc cu to ^ C H 3

6. Hirocacbon X th lng c t l phn trm khi lng H xp x 7,7%.

X tc dng c vi dung dch brom. Cng thc no sau y l cng

thc phn t ca X?

A. C2H2; B.C4H4; c. C6H6; D. C8H8.

Gii

Chn D

V X th lng nn X l C6I6hoc CgHs theo X c th l CsHg, ngviCTCT: -CH = CH2

(Lu : thc ra X cng c th l C6H6,

V C H = C H 2 C H 2 = C H )

68



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69/96

BI 37. NGUN HIROCACBON THIN NHIN

1. Hy cho bit thnh phn ca du m. Ti sao du m li khng c

nhit si nht nh? C th biu th thnh phn ca du m bng

mt cng thc phn t nht nh c khng? Ti sao?

GiiDu m l hn hp ca rt nhiu hirocacbon khc nhau (ankan;

xicloankan v hirocacbon thm). Ngoi ra cn c mt lng nh cc

hp cht hu c cha oxi, nit, lu hunh v mt lng rt nh cc

cht v c dng ha tan.

Du m khng c nhit si nht nh v n l hn hp nhiu cht

vi thnh phn thay i ty theo khu vc. Do l hn hp nhiu cht

nn du m khng th c cng thc phn t nht nh.

2 . Kh thin nhin, kh du m, kh l cc l g? Nu thnh phn chnhca mi loi kh ny v ng dng ca chng.

Gii

- Kh thin nhin l cc m kh tch t trong cc lp t xp

nhng su khc nhau v c bao bc bi cc lp t khng

thm nc v kh,

- Kh m du (hay kh ng hnh) l kh c trong cc m du, phn

ln tch t li thnh lp kh pha trn lp du

- Khi nung than m 1000c trong iu kin khng c khng kh sthu c phn kh gi l kh l cc.Thnh phn ch yu ca kh thin nhin l metan, cn li l etan,

propan, butan v mt s kh v c nh N2, CO2; H2S, H2 ...

Thnh phn ca kh m du gn ging nh kh thin nhin nhng hm

lng metan thp hn, cc thnh phn ankan cn li cao hn.

Thnh phn ca kh l cc ph thuc vo nguyn liu ban u, nhng

hm lng H2 l cao nht (khong 59%), ri n CH4 (klong 25%),

cn li l CO, N2, C(>2 . . .3. Trnh by tm tt quy trnh chng ct du m, cc phn on v ng

dng ca chng. C my loi than chnh? Thnh phn v cch ch bin

chng.

69



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GIẢI BÀI TẬP HÓA HỌC 11 CƠ BẢN - NGUYỄN ĐÌNH ĐỘ