Phân dạng và phương pháp giải hóa học 11 Phần hữu cơ Dành cho học sinh lớp 11 ôn tập và nâng cao kĩ năng làm bài

Viet phufdng trinh hda hoc cua cdc phan iJng xay ra trong cac triTdng hdp sau va cho biet cac phan u'ng dd thuQC loai phdn uTng n^o phan vJng the, phan uTng cpng, phan u'ng tach... Tinh

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D O X U A N H U N G Gi^o vign chuy^n luy^n thi DH-CD H6a

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PHAN DANG VA PHLfCJNG PHAP GIAI HOA HQC 11 HQU C d

D6 X U A N H U N G

Chju trach nhi^m xu^t ban

N G U Y E N T H I T H A N H Hl/OfNG Bientap : H A I A U

Tong phdt hanh

C 6 N G TY T N H H M T V

D|CH vg VAN H6A KHANG VI|T

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T?i: Cong ty T N H H MTV in ^n MAI T H j N H DLfC '

Dja chl: 71, Kha V^n CSn, P.HiOp Binh Ch^nh, Q.Thii Dufc, Tp.HCM

So DKKHXB: 1482-12/CXB/12-181/THTPHCM ngSy 06/12/2012

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Thanh p h ^ H 6 Chi Minh cS'p ng^y 28/12/2012

In xong nOp lUU chi^u quy' I nanri 2013

(HuMtta 4,

p D A I CI/CJNG V E H 6 A H p C HtfU CCJ

A TOM TAT Lf THUY^T

I M d D A U V E H b A HOC HirUC(]fVAH0P CHAT H0UCCf

1 H^p chaft hffu ccf la hdp chat cua cacbon (trilf CO, CO2, muoi cacbonat xianua,

cacbua, )

2 PhSn loaii h^p chSt hffu ccf ^ 9!^: : J^'l • ,

Hop cha't hQu CO

Hidrocacbon ^ ^ ' Din xuS't cua hidrocacbon

Hidrocacbon Hidrocacbon Hidrocacbon

3 Danh phap h^p chS't hffu cd

a) Ten thong thiidng (nguon goc flm ra chat) ^ ' - ' ' - ^ M s g « * a v * f «

b) Ten h^ thong (lUPAC) ' ' ' : i , ' 1 ^ ' / ^ ( r y i

Ten goc chiJc = Ten phan goc + T6n phan dinh chiJc Ten thay the = Ten phan thay the + Ten mach C chinh + Ten phan dinh

II PHAN T I C H NGUYEN TO

* Xdc dinh C : Dot chdy chat hSu cd CO2 CaCOjl

* Xac dinh H : Dot chdy chat hihi cd H2O ^CuS04khan^ CUSO4.5H2O

'••'^ ^ - (xanh)

* Xic dinh nitd : C^HyCN, (NH4)2S04 + jV- v!* ? n f •

* « ^ • • * (NH,)2S04 + 2NaOH >Na2S04 + 2NH3t + 2H2O

* Xac dinh halogen : CxHyO,N, > CO2 + H2O + HCl 1:4 ; ?

HCl + AgNOj > AgCli + HNO3 vi ,

2 PhSn tich djnh Irf^ng (A la chat hffu cd c6 HIA)

Phan djing \ik phuong ph^p g'Ai H6a hpc 11 HJu co - B8 Xuan Hung

=> mo = niA - (mc + mH + IHN)

hay % 0 = 1 0 0 % - (%C + %H + %N) ^ ; ,

I I I C O N G T H L T C P H A N T l T H O P C H A T H 0 U COf '^'iMj^J

1 C6ng thtfc defn g i a n nhS't ! Y i t M i * t ,vN ' (

C o n g thu-c ddn g i a n nhat ( C T D G N ) : (CpH,OrN,)„ ( n = 1,2, 3, ) ,|

* C a c h thiet lap : T i m t i 1$ (C,HyO,) so, ^^.,^

* C o n g thiJc cau tao k h a i t r i e n ;| >fiX

* C o n g thiJc ca'u tao thu g o n

2 D d n g d a n g

Nhffng h d p chat c6 t h a n h p h a n p h a n tuT hdn k ^ m nhau mpt hay n h i e u nhom CH2

nhiftig c 6 t i n h chat hoa hpc tiidng tif nhau "V''';^;! ' ' ,4^

Dac diem cua phan iJng h6a hpc trong hoa hoc hffu cd : - 1

- Phan iJng thifdng xay ra cham \'t } -\{^ An

- Phan iJng thudng sinh ra hon hdp san pham ; '.^^

B PHAN LOAI V A PH JCJNG PHAP G I A I C A C D A N G BAI TAP

- (Vm fihan ling^hau c ^ r i r i O ' - v H ^ ^ - T H ; } ,HI4 -,H;:HHD-;.ii:):-,-i-'' •

- <Jjhaii loai fditut itiig hitii eet ' t

- (Vi& eMiq Ihiie eaii tim, ctmtg^ phdn^ ddtig^ dqttQ^ ede ehdt

B a i 1 V i e t phifdng trinh hoa hoc ciia cac phan iJng xay ra theo sd do sau f.^

Phan djing va phUdng ph^p giai H6a hpc 11 HOu cO - D5 XuSn HiJng

Vay chat cd nhieu dong phan nhat Ik CiH^N roi den C4H,„0, C 4 H y C l

Bai 3 Cho ba hdp chat : A (cau tao tit C, H); B (cau tao tuf C, H, N); C (cau tao tit

C, H, O) Chaft nao thupc loai hidrocacbon, cha't nao thuQC loai dan xuat cua

hidrocacbon ' ' ,

* A la hidrocacbon \\ chi cSu tao gom hai nguyen to C H 1;,) • j i l

* B va C la dan xuat cua hidrocacbon vi ngoai hai nguyen to C, H cdc hcJp chat B,

C con chiJa cdc nguyen to khac (N va O)

a) Trong cac hdp chaft sau hc(p chat n^o la hffu c d , hcJp cha't nao la v 6 c d CH4,

CHCI3 C2H7N, HCN, CHjCOONa, C,2H220|,, f C2H3Cl}-„, AI4C3

b) Chat nao sau day la hidrocacbon ? la dan xuat cua hidrocacbon : CH2O, C2H5Br,

CH2O2, CfiHsBr, CfiHfi, CH3COOH

Gidi

a) Hdpchathi?ucd:CH4,CHCl3,C2H7N.CH3COONa,C,2H220,i,{C2H3Cl}„ *

Hdp chat v 6 c d : H C N , AI4C3 '

b) Hidrocacbon: CfiHfi f r t ' )

Dan xua't cua hidrocacbon : CH2O, C2H5Br, CH2O2, CfiHjBr, CH3COOH

Bai 5 Cho 6 hdp cha't hi?u cd ddn chiJc, mach hcl cd cung cong thiJc phan tuT 1^

C4HXO2 Viet cdc cong thiJc ca'u tao thu gpn cua cdc chat dd • , • ,

CH3-CH2-CH2-COOH ; CH3-COO-CH2-CH3 ; CH3-CH-COOH

ir CH3

HCOO-CH2-CH2-CH3 ; CH3-CH2-COOCH3 ; HCOO-CH-CH3

CH

Bai 6 Cac phan uTng sau day thuOc loai phan iJng nho ? (phan iJng the, phan iJng

cong, phan iJng tach) ? a) C2H6 + CI2 ^ C2H5CI + HCl b) C4HX + H2O C4H,oO c) CzHsCl ) C2H4 + HCl d) 2 C 2 H 5 0 H ^ ^ C 2 H 5 0 C 2 H 5 + H20 e) CH2=CH2 + H2O CH3-CH2-OH f) CfiHft + Brj CfiHsBr + HBr

g) C4HH, + 5F2 > 4 C + lOHF h) CH3OH + HCl CH3CI + H2O

a) Phan iJng the b) Phan iJng c p n g c) Phan tfng tach ^' d) Phan iJng tach e) Phan iJng c p n g f) Phan iJng the g) Phan ifng tach h) Phan u-ng the '

Bai 7 Dtfa vao tinh chat hoa hoc cua CH2=CH2 v a CH^CH (da hoc d Idp 9) Hay

Viet phiTdng trinh hoa hoc khi cho CH3-CH=CH-CH3 va C H 3 - C ^ - C H 3 lac dung vdi Br2, H2 va cho biet nhufng nhdm nguyen tuT nao trong phan tuf cua hai h d p cha't

tren da gay nen cac phan u'ng dd , , : ( ;

* CH3-CH=CH-CH3 + Br2 > CH3-CH-CH-CH3 -mwaot/rh sn6:; ?iv (

^;,H"^ ' Br Br 'hi: , ' ' CH3-CH=CH-CH3 + H2 CH3-CH2-CH2-CH3 -

=> Nhdm nguyen tuT gay nen phan uTng 1^ : - C = C -

Bai 8 Viet phufdng trinh hda hoc cua cdc phan iJng xay ra trong cac triTdng hdp sau

va cho biet cac phan u'ng dd thuQC loai phdn uTng n^o (phan vJng the, phan uTng cpng, phan u'ng tach)

Phan dgng vji phuong phap giSi H6a hgc 11 HDu co - D8 XuSn Hung

a) Etilen tac dung vdi hidro c6 N i lam xuc tdc va dun n6ng

b) Dun n6ng axetilen d 600"C vdi bgt than lam xuc tac thu diTdc benzen '^f '

c) Dung dich ancol etylic d ^ lau ngoSi khong khi chuy^n thanh dung dich axit

axetic (giam an) • t i > i f i J ;^W;> ; / :i•0-.-,^K;;» ,•

' a) CH2=CH2 + H2 CH3-CH3 (phan iJng cOng) ,li::).:.:H':;).- H':) i : X ) ' » '

^ ^ (CfiHfi) [ Q ] (phan iJng cpng)

b) 3 C H = C H

600° C •

c) C2H5OH + 02(kk) ^^"^ > CH3COOH + H2O ^ phan ifng oxi h6a

,, , r); ^ <.': , , , khong hoan toan

Bai 9 Cho cac chat sau, nhffng chat nao la dong d i n g cua nhau ? NhCng cha't nao la

ddng phan cua nhau ? ,

f£;Cic chat la dong d i n g : ( 1 ) v a (6); ( 4 ) v a (9) rt' rirt 4 t\i& '•^'cf f , ' I I }H

* C d c chaft dong phan cua nhau : ( 2 ) v a (8); ( 4 ) (7), ( 3 ) ( 5 ) ^ < h

B a i 10 '

a) Hay viet cong thufc electron va cong thiJc cau tao c^c phan tuT sau : > ~ , r ) i }f'3 , *

C H 3 C I , CH4O, C H 2 O , CH5N

b) H a y vie't cong thufc cau tao khai trien v a cong thiJc cau tao thu gon nhat cula cac

hdp chat sau : C3H6, C H 3 C H O , CH3COOC2H5, C H 3 C N , biet r i n g trong phan tiJf

cua chiing d i u c6 lien ket bpi : ' ' r f !"'v •"^'H^;-^'' ijik I'-'iy '•^•"•.i-r., , •

^han dang phuong phiip giai H6a hgc 11 HOu co - D8 Xufln Ht/ng

•)) V i e l c o n g thu'c ca'u t a o c 6 the c d c u a c a c d o n g d a n g c u a a n c o l e t y l i c c d c o n g

thu'c p h a n tuT CsHxO v a C4H|„0

hay CHg - C H O ^>|ri; i;^ q i j, 1

i.iaii! K u H f i i d q s n o b i i i uVifi } « b j j r f M f i iftrfj

a) C H 3 - C H = C H - C H 3 b) C H 2 = C H - C H 2 - C H 3 c) C H 3 - C H 2 - C H 2 - C H 2 - C H 3 d) C H 2 = C H - C H 3 e) C H 3 - C H = C H - C H 2 - C H 3 g) CHg = C H - C H - C H 3 -)

a) H a y vie't c o n g thu'c p h o i canh cua C H 3 O H (metanol) va ciia C H C I 3 ( c i o r o f o m )

b) H a y v i e t c o n g thiJc p h o i c a n h cua C2H6 (etan) va C2H5OH (etanol)

- f£gfi etniff thiie ittfii qlun iihal ,,),,., „,,,,,v,y,) > 4 , j j.^j;,^;^

, , ^ BAI TAP MAU VA BAI TAP N A N G CAO

Bai 1 Oxi h6a ho^n toan 0,6g hdp chat hi?u cd A thu duTdc 0,672 lit CO2 (dktc) v j 0,72g H2O Tinh thanh phan % khoi liTdng cua cac nguyen to trong phan tuf cha't A

Phan d?ng va phudng ph^p giSi H6a hpc 11 HQu co - D8 XuSn Hung

Thanh phan % khoi lifdng cic nguyen 16' trong A : d, i jỵ

Bai 2 Tinh khoi lifdng m o l phSn tuf cua cdc chat sau : jj,;., ,,,,,

a) Chat A c5 t i khoi hdi so v d i khong khi bkng 2,07

b) The tich hdi cua 3,3g chat X b^ng the tich cua l,76g k h i oxi (do cf ciang dieu k i f n

ve nhiet do, ap suát)

Gidi '•^''''^ '^''''^ *

a) dA/kk = ^ = 2,07 ^ M A = 29.2.07 = 60 (dvC) , ^g^,, ^.^

b) Ta c6: no2 = ^ ^ = 0,055 mol => nA = = 0,055 m o l = > M A = = 60 (g/mol)

32 0,055

Bai 3 D o t chay hoan toan 0,282g hdp chat hiJu cd A, san pham sinh ra cho qua binh

dtrng CaCl2 khan va binh dUng NaOH, thay binh diTng CaCla tSng them 0,194g,

binh dufng N a O H tSng them 0,8g M a t khac dot chay liTdng chat A tren thu difdc

22,4ml khi nitd (dktc) Xac dinh thanh phan % khoi liTdng cac nguyen to trong Ạ

Gidi

" Cho san pham chay qua binh dtfng CaCb : mi = 0,194g =>mH20 = 0 ' 1 9 4 g

= Qua binh dirng N a O H : m, = 0,8g=> m c o , = 0.8g

Khói lifdng cac nguyen to :

!ai 4 O x i h6a hoan toan 4,92mg mot hdp chat A chiJa C, H , N O roi cho san

pham Ian liTdt qua binh chiJa H2SO4 dam dac, binh chiJa K O H thi thay khoi

liTdng binh chiJa H2SO4 dSc tang them 1,8 Img, binh chiJa K O H tSng them

10,56mg d thi nghie m khac, khi nung 6,15mg hdp chat A do v d i CuO thi thu

difdc 0,55ml (dktc) k h i nitd Hay xdc djnh h^m liTdng phan trSm cua C, H , O va

N d hdp chat Ạ

Gidi

* Khoi lifdng binh chiJaH2S04 tang them 1,8Img chinh la mH20=>mH20 =1.81mj

* Khoi liTdng binh chiJa K O H tang them 10,56mg chinh la m c o , => "^C02 ^0,56 mi

Khoi lifdng cac nguyen to :

^ x l O O % = 58.54% ^,^^^,„ £, s^fM)ixSk^^^^^

x 100% = 4.065%

% N = ^ X 100% = 11,18% - ^ ^ ' ^ - ' ^ ^ o ^ ' ' ^ ^ ' '^^^

r:> % 0 = 100% - %C - % H - % N = 26.215%

Bai 5 Hai chat hffu cd A va B ciing chtJa cac nguyen to C, H , Ọ K h i dot chay moi

chat deu phai dung mOt lifdng oxi b^ng 8 Ian lifdng o x i c6 trong m o i chat va thu difdc lifdng k h i CO2 va H2O theo ti le khoi lifdng CO2 so vdi khoi lifdng nuTdc =

— Xac dinh cong thiJc đn gian nhat cua m o i chat ? ' ' ' " " '"

-• ''' CUM '

Theo dieu kỉn de bai thi A, B la hai chat dong phan cua nhaụ •

Dat C T T Q cua A, B la CxHyỘ i ,, ,, , "Mrirả- ^'.-YV ^'^'^ „

Phifdng trinh dot c h a y : , 1 , i

1

Phan djing va phuong ph^p giai H6a hpc 11 HOu co - Dfi Xuan Hung

B a i 6 Vitamin A (retinol) c6 c6ng thiJc phan tuf C2()H3()0, vitamin C c6 cong thiJc

phan tur la CfiHsOfi

a) Vie't cong thiJc ddn gian nhat ciia moi chat 'li,, ,

b) T i n h t i 1? % ve kho'i li/dng va t i ie % so' nguyen tur cua cac nguyen to d vitamin

A va vitamin C

Gidi , "

a) Cong thiJc ddn gian nhat cua cdc cha't: ^

Vitamin A : C2()H3(,0; Vitamin C : C3H4O3 ,0 ' • ' 7 ,

b) * T i 1Q % ve khoi liTdng cdc nguyen to :

* T i 1? so nguyen tuf cdc nguyen to: , / j

+ Vitamin A : C2(iH3„0 (51 nguyen tuf)

B k i 7 Chat A chiJa C, H O c6 t i 1$ khoi li/dng mc : mo = 3 : 2 khi dot chdy h6't A thu

diTdc CO2 va hdi niTdc theo u' I9 the tich Vco2 : V H J O = 4 : 3 (cdc the tich do d cung

dieu k i ^ n nhi§t dp, ip suat) T i m cong thiJc ddn gian nha't cua A ^,

B a i 8 Hay thict lap cong thiJc ddn gian nhat tCf cac so lieu phan tich sau : a) 70,94%C; 6,4%H; 6,9%N; con lai la o x i i r i o n ' r ; o 5 ' b) 65,92%C; 7,75%H; t , ; , c o n l a i la o x i l f : > H : \ t P k v j , ; t v | : : , ' , '

.,!,;•! X^:,, ,,,, Gidi •i^il i o r f ' ' a n f giiW'utkj I.'

a) Thanh phan % cua oxi • 'jiinod:?^'^ ni'^q di'

12 1 16 14

= 5 , 9 1 : 6 , 4 : 0 , 9 8 5 : 0 , 4 9 = 1 2 : 1 3 : 2 : 1 Vay cong Ihu'c ddn gian nhat la C12H13O2N , „ ;

b) Thanh phan % cOa o x i : H , , ,

Bki 9 D o t chdy ho^n loan 4,6g mpt hdp chat huTu cd A thu diTdc 3,6g H2O Dan khi CO2 sinh ra vao dung dich ni/dc voi trong 0,1M thi difdc 8g ket tua v^ dung dich

Y nho tiep dung dich N a O H v^o dung dich Y thi difdc 3,5g ket tua nffa Xac dinh cong Ihu'c ddn gian nhat cua X

Phan djing va phudng phip giSi H6a hpc 11 HQu co - D8 Xuan Hifng

2CO2 + Ca(OH)2 > Ca(HC03)2 ^

Vay cong thtfc ddn gian nhat cua X la CsHgOs •> iah-) ~iSi i i l r f i y \M J

Bai 10 K M cho 5,3 gam hon hdp gom etanol C2H5OH ;H''i?i',fl! ;Civ^,ii' / j ;

va propan-l-ol C H 3 C H 2 C H 2 O H tac dung vdi natri drfthu diTdc 1,12 lit khi (dktc)

a) Vie't phifdng trinh hoa hoc cua cac phan iJng xay ra

b) Tinh thanh phan % khoi lifdng cua m o i chat trong hon hdp S' d i n M T {a

% C 2 H 3 0 H = Q ' " ^ ^ ^ ^ 100% = 43.39% :

=>%CH3CH2CH20H = 56,61%.""" !:>tU^ J l f l n ft? ,, ^ ^ , .fltfc^ :

t.;.,,iV - : 7 - , S ' * j i m ' V ; - & r f i T

(DoMig 3 <£Afi eS^ntf, thi'te fthdit tilt :

- <T)t/a ocLO^ cang, thiie fteiti Qtatt nhdi i <- '

- <T)ita oda khai liMtiq hoae % khoi lii&iig ede nqiujjen t»

- ^i/a i9dn the tieh (phiMuj fthdfi khi nhien ke)

B A I T A P M A U - v ^ ' '

Bai 1 Dot chay ho^n toan 5,6 lit chat hiJu cd A d t h i tich k h i thu difdc 16,8 l i t CO2

va 13,5g, H2O Xac dinh cong thiJc phan tuT cua A biet t i khoi so vdi hidro la 21

Vay trong A khong CO o x i , • 1

Dat C T T Q cua A : CJiy Ta c6 ti I9 : x : y = — : — = 0,75:1,5 = 1:2

, i • 12 1

^ Cong thu-c A c6 dang (CH2)„ , l „, ^ „ ,,

T a c o : 14n = 42 ^ n = 3 V $ y CTPT A : C3H '^''-^'^^^-^^-^ > ^ Bai 2 Limonen la mot chat c6 m d i thdm diu difdc tdch tilf tinh dau chanh Ke't qua

phSn tich nguyen to cho thay limonen diTdc cau tao tiJ hai nguyen to C va H , trong do C chiem 88,235% ve khoi lifdng T i khoi hdi cua limonen so v d i khong khi gan b^ng 4,69 Lap cong thiJc phan tuT cua hmonen , '

V$y CTPT cua limonen m CjpHii

THLT VIEN TIMH BSNH THUAN

-zr::^" ' "A 9 17

Phan dgng va phuong phap giSi H6a hoc 11 Hau co - P i Xuan Hung

B a i 3 Dot chay 100ml hdi cha't hffu cd A chtfa ba nguyen to C, H , O v d i 450ml O 2

la'y dff trong k h i nhien ke Sau phan ffng the tich hon hdp khi va hdi thu difdc

650ml Sau k h i cho hdi nffdc ngffng tu hon hdp chi con 350ml va sau khi cho Igi

tiep qua N a O H chi con 50ml Xac dinh cong thiJc phan tuf cua A

Vay CTPT cua A la CjHfiO ,• ; _ , ^

B a i 4 Hay thie't lap cong thiJc phan tuf hdp chat hffu cd trong m o i trifclng hdp sau :

a) Dot chdy ho^n toan lO.OOmg hdp chat hCu cd Y sinh ra 33,85mg CO2 6,95mg

H2O T i khoi hdi cua hdp chat Ad doi vdi khong khi la 2,69

b) D o t chay hoan toan 28,2mg hdp chat hffu cd Z vk cho cdc san pham sinh ra Ian

Iffdt d i qua cac binh dtfng CaClz khan va K O H dff thi thay binh CaCl2 tSng them

19,4mg con binh K O H tSng them SOmg M a t khac, k h i dot 18,6mg chat do sinh

ra 2,24ml khi nitd (dktc) Biet rkng phan tuf chat d6 chi chffa mpt nguyen tijf nitd

,:;; ; - = 6 : 7 : 1 => C T cd dang (CeBilS^/ Jri^ f)^?'

V I trong phan tuf chi chffa mgt nguydn tuf nitd =>-m=^U.''' ' "P '^^ "'^^i •!

Vay CTPT chaft hffu cd m C 6 H 7 N -"5' < « ^

B a i 5 Phan tich chat hffu cd X chffa C , H , O ta c6 :

mc : mH : mo = 2,24 : 0 , 3 5 7 ^ ^ ' - ' ' ' ' ' ' ' ' " ' a) L a p cong thffc ddn gian nhat cua X t '

b) Xdc dinh cong thffc phSn tff ciia X biet 1 gam J i ^ l a m bay hdi c6 t h ^ t i c h

1,2108 lit a 0"Cva 0,25 atm '

Kh6i Iffdng m o l cua X : M x = — ^ - — = 74 , < ,

0,0135 (^.- ,

T a c d : (3 x 12 + 6 + 2 x 16)n = 74 =i> n = 1 V§y cong thffc phan tff cua X la C3H6O2 r ' f f ) c ! >

19

Phan d?ng va phUdng phAp Q\i\a hgc 11 Hi?u co - D5 XuSn Hung

Bai 6 TO tinh dau hoi, ngi^di ta tach difdc anetol, mpt chaft thdm du"dc diing de san

xuat keo cao su Anetol c6 khoi lifdng mol phan tuf bang 148g/mol Phan tich

nguyen to cho thay anetol c6 %C = 81,08%; % H = 8,1%; con lai la oxi Llipcong

thuTc ddn gian nhat va cong thiJc phan tuf cua anetol

V|iy cong thiJc phan tiJf cua anetol la CioHuO ( t j >l 3 K

Bai 7 Dot chay 200ml hdi 1 chat Mu cd A chiJa C, H, O trong 900ml O2, the tich

h6n hdp khi thu di/dc 1^ 1,3 lit Sau khi cho hdi nifdc ngiftig tu, chi con 700ml

Tiep theo cho qua dung dich K O H dac chi con 100ml (cac the tich do d cUng

dieu ki^n) T i m CTPT cua A

Bai 8 Trpn 4 0 0 cm^ hon hdp hdp chat Mu cd A va nitd vdi 9 0 0 cm' oxi duf roi dot

The tich hon hdp sau phan iJng la 1,4 lit Sau khi cho hdi nufdc ngung tu thi con

8 0 0 cm^ tiep tuc cho qua dung dich KOH thi con 4 0 0 cm"\T cua A la:

r c o 2 ^

H 2 0 N2

^ 0 2 dir 400cm^ 900cm^ 1400cm

Dtfa vao sd do ta tinh difdc: s, j

ngutig tu

- H 2 O

C O 2 N2

Viet lai phi/dng tnnh phan u-ng: 2 C x H y + 7 O 2 -A- 4 C O 2 + 6 H 2 O

Ap dung dinh luat bao toan nguyen to' C va H ta cd: 2x = 4

2y = 6.2'

x = 2

y = 6

=> C T P T c u a A l a :C2H6

Bai 9 Cho 5 cm' C,Hy d the khi vdi 30 cm' O2 (lay diT) vao khi nhien ke Sau khi

bat tia lufa di^n va lam lanh, trong khi nhien ke con 20 c m ' ma 15 c m ' bi hap thu bdi dung dich K O H Phan con lai bi hap thu bdi photpho T i m CTPT cua hidrocacbon , 1

Theo de bai: Vco^= 15 c m ' =^

= > phan ifng = 30 - 5 = 25 c m '

Gidi ,

Vo d ^ = 2 0 - 15 = 5 c m '

2 1

Ptiftn^ang^ phicnq phap giai Hfa^pc 11 Hi?u co - P8 Xuan HiMg

• ^ u a n g trinhphan tfngchdy:

Q H y + (x + ^,)02 ->-xC02 + ^ H z O

=>x = 3 v a y = 8 = > G T P T la CjHs

B ^ O : Cho vao khi nhiSnske 10 cm^ ch^it hffu cd A (chiJa C H , N), 25 cm^ H 2 va 40

tsdl^Oz Bat tia liJa di$^nn:ho hon hcfp no Chuyen hSn hcJp khi nhan diTdc ve dieu

aki^Ii'ban dau, H2O ngiteg tu he't, thu diTdc 20 cm^ hSn hdp khi, trong 66 c6 10

••Wff bi NaOH hap thu vi'S cm^ hi photpho hap thu C T P T cua A la: i^^' '

Photpho hafp thu O 2 = > VQ^ j i , = 5 cm'

© o i n g d i c h N a O H hap thu C O 2 = > VQO = 10 cm' v>v'

20 cm'

=> = 2 0 - ( 5 + 10) = 5 cm'

25 3 Wolchayhkiro) = —- = 12,5 Cm

'•,m ,ix^ '.},>'i i(6o;«iT

BAI TAP NANG CAO

B a i 1 Mot hdp chat hilu cd A chuTa 54,8%C; 4,8%H; 9,3%N con lai la O Cho biet

ph§n tuf khoi cua nd la 1 5 3 X a c dinh c6ng thiJc phan tuf cua hdp chat V i sao phSn tuf khoi cua c&c hdp chat chtfa C , H, O la s6' ch£n ma phan tuf khoi ciia A

lai la so' le (khong ke phan thap phan) ? • u • *, |,

* Phan tuf khoi cua A la so le vl nguyen to nitd c6 h6a tri 16 => nguyen tuT H la so

B^i 2 Cho 400ml mOt hidro g6m N 2 C^Hy v^o 900ml O 2 dif roi dot The tich hon hdp khi thu dtfdc sau khi dot la 1400ml Sau khi cho hdi niTdc ngifng tu thi con 800ml hon hdp ngi/di ta cho loi qua dung dich K O H tha'y con 400ml khi X a c

dinh cong thiJc phan tuf cua hidrocacbon tren, biS't the tich cic khi deu do d ciing

d i l u ki^n nhi$t dp va ap sua't ^^\,

= 4 0 0 - a Theo de bai ta c6 :

ax

< v j H \' ; :Mit iirtJ }flA c^,,-!

VH20 = 1400-800 = 600ml = « c # Vco2 = 800 - 400 = 400ml = ax ^, j, |,

Phan dgng va phuong ph^p giai H6a hpc 11 HQu eo - B5 Xufln Hung

V a y t a c d : ay = 1200 (1)

• V i v i ax = 400 (2)

G l a i ( l ) , (2), (3)

a = 9 0 0 - a (3) ' ' • 'i«i#iffc

• • - • , -itiVj i'kih '•'pli ri/!;y ,fy§^

a = 2 0 0 m l fSsirj '^Iv; Vjnv'i/i) bl u, ii lli

• X = 2 ,:ut}

y = 6 •= "''r;i' m.P 'fli s.i' •>]::

Vay CTPT cua hidrocacbon la C2H6

Bai 3 Dot chay hodn toan 3,6g chat huTu cd A (C, H , O) b^ng 4,48 lit O2 (dktc) thu

diTdc hon hdp k h i va hdi trong do V c o , = 3 V o ; ^ ^ ^ ^ = 11 T i m cong thiJc

phan tuTcua A biet d the hdi; l,8g chat A chiem the tich bang the tich cua 0,8g oxi

cung dieu ki?n , _ ^ , , •

O2 — ^ XCO2+ ^ H 2 0 •« iTt'rf aftd U^;^(m

mi&jB'jof{):,ii ~.tciq atfrti ;ii'n>;!-> idofsi

H5n hdp k h i thu difdc :

CO2 H2O O2 dir

Vay CTPT cua A \k C3H4O2

Bai 4 Dot chay hoan toan l,37g chat hufu cd A thu diTdc 3,08g CO2; 0,63g H2O va

0,126 l i t N2 (30"C va 75 cmHg) Xac dinh cong thiJc phan tuT A Biet l,37g A cho

bay hdi d 100"C va 1 atm thi the tich thu diTdc la 306ml •,.(;./

Vay cong thu-c A cd dang (C7H702N)„

T a c d : (7 x 12 + 7 + 32 + 14)n = 137 n = 1 ^ > -^'''-^ "

Vay C T P T cua A m C7H7O2N AX'kD s ^ H O k J ^

B a i 5 Tri^dc kia, "pham do" dung de nhupm do cho^ng cho cac Hong y gido chu

diTdc tdch chie't tijf mpt l o ^ i oc bien Do la mpt hdp cha't cd th^nh phan nguyen to nhiTsau :

C : 45,7%>; H : 1,9%; O : 7,6%; N : 6,7%; Br : 38,1%

a) Hay xac dinh cong thi^c ddn gian nhat cua "pham do"

b) Phi/dng phdp phd khdi liTdng cho biet trong dd phan tuT "pham do" cd chiJa hai

nguyen tur brom Hay xac dinh cong thufc phan tiJf cua nd

'hSn dgng vk phtwng ph^p giSi H6a hpc 11 HOu CO - D8 Xuan Hung

Vay cong IhiJc ddn gian nhat cua pham do la C K H 4 0 N B r

j) V i phan tuf chiJa hai nguyen tuf Br nen A c6 dang (CsH40NBr)2

Vay C T C T cua pham do la CfiHxOsNsBra

Bai 6 Dot chdy hoan loan hon hdp X gom hai hidrocacbon A B, mach hd cilng

day dong dang Cho toan bp san pham chay vao 4,5 lit dung dich Ca(OH)2

0,02M, thu difdc ke't tua va khoi liTdng dung dich tang 3,78g Cho dung dich

Ba(OH)2 (in vao dung dich thu difdc, ket tua lai tSng them, tong khoi liTdng ket

tua hai Ian la 18,85g T i khoi hdi cua hon hdp X so v d i hidro < 20 So m o l cua A

b^ng 60% tong so m o l cua A va B trong hon hdp X Cdc phan iJng xay ra hoan

toan Hay xac dinh cong thiJc phan tuT cua A va B '

C O 2 + Ca(OH)2 > CaCOs + H 2 O ,/!.,(:), 1-?,":) ; ; l ; (3) yJ:M

b m o l b m o l b m o L j i ffiourU, -lib gauf; "nf> frartq" jnJ; •JSuil 2 }i:H

2 C O 2 + Ca(OH)2 )• Ca(HC03)2 l v : / M & n > n f (4) ,1-^Ij^A

Theo (3) va ( 4 ) : mcafOH), = b + c = 0,09 (mol)

M a t k h a c : 18,85 = mcacoaO) + mcaC03 (4) + mcacogrs) „, , ,

hay 18,85 = 100b + 100c + 197c = 100b + 297c

Ta c5 he phifdng trinh :

b + c = 0,09 fb = 0,04 (mol) 100b + 297c = 18,85 [c = 0,05 (mol)

=> nco2 = b + 2c = 0,14(mol)=> m c o j (3 4) = ( b + 2c).44 = 0,14.44 = 6,16(g)

Ttf (6)=i> m H j O = 3.78 +100.0,04 - 6,16 = 1,62 (g)

1,62

n i - = 0,09 (mol) Theo phifdng trtnh (1), (2) ta c6: n c o j = 3ax + 2a(x + n) = 0,14 1 r i 1

3ay -i-2a y + 2n

2 hay nco2 = 5ax + 2an = 0,14

Bai 7 D o t chay ho^n toan m gam chat hiJu cd X chtfa ba nguyen to C , H , O thu

difdc a gam C O 2 va b gam H 2 O , biet a = va b = Hay x i c dinh cong

thtfc phan tvl cua X Biet 3,6g hdi X cd the tich b^ng the tich cua l,76g C O 2 ciJng

Phan dgng phuong ph&p g\k\a hgc 11 HOu cO- D8 XuSn Hang

V$y CTPT cua X m CjHfiOs v * / r v , - , i v * -' f ,.t •

Bai 8 Oxi hoa hoan loan 18,6g chat hffu cd A thu diTdc 52,8g CO2 va 12,6g H2O

Mat khac k h i phan tich liTdng cha't hffu c d A thu difdc k h i NH3 Dan toan bp k h i

nay v a o 125ml dung dich H2SO4 2 M thi phan axit duT difdc trung hoa vilfa he't bdi

100ml dung djch NaOH 3M

a) Xac d i n h thanh phan % N cd trong A u> ' , ^,

b) L $ p c o n g thiJc phan tuT A biet dA/kk < 3,25

T A 52,8 ^ 12,6 , , , , s f o l T O x J l ^

-T a c d : m c = 1 2 = 14,4(g); mn = 2 — ^ = l , 4 ( g ) ^

=> mo = 1 8 , 6 - ( 1 4 , 4 + 1 , 4 + 2,8) = 0 „

VSy trong A khong cd oxi :

^l- Dat CTTQ cua A : C ^ H y N , rf:^='''^'M ^-^^ ''"^f"^'* - f*-^^' {!•&;)•&• n.^ij^ ptjrfl ^ •

Bai 9 Dot chdy 560 cm^ hon hdp khi (dktc) gom 2 hidrocacbon cd cilng so nguyen

tiJ cacbon ta thu duTdc 4,4g CO2 va 1,9125g hdi niTdc • '

a) Xdc dinh CTPT cac chat hiJu cd ,.,, , ,, j ^y j : •

b) Tinh % khoi lifdng cac cha't

c) Neu cho lifdng CO2 tren vao 100 ml dd K O H 1,3M; Tinh CM muoi tao thanh

Giai

6 bki nay, ta dilng phifdng phap so nguyen tuf H trung bmh ke't hdp vdi phiTdng

phap bi^n luan de giai ij(|,;,,,|v) d d

= ^ y ' = 1 0 r = > C T P T B : C 4 H , o tentO,0 TiWng tir bi?n luan tim CTPT A : y < 8,5, y chSn >|.ii^||''ai '

Bi^n luHn tim CTPT B :

PhSn d^ing phUOng phAp gi^i H6a hgc 11 HOu co - iJu Xuan

c) Tinh CM cdc muol tao thanh r ' i, ,

Bai 10 Mot hidrocacbon A d the khi c6 the tich gap 4 Ian the tich cda lUU huynh

dioxit CO khoi li/dng tifdng difdng trong cung dieu ki?n San pham chay cua A

dan qua binh diTng nifdc voi trong diT thi c6 Ig ket tua dong thdi khoi li/dng binh

tang 0,8g Tim CTPT A

* Tim MA : ' ^\ ^:^ «f«'n-i"fOt^::r:r Ji^

1VA = 4.VSO2 (dcungdieukien)=>nA = 4.nso2 , ("y 1

=> = 4"^^"^ => - i - = (A SO2 c6 khoi liTdng ti^dng diTdng nhau)

MA MSO2 M A MSO2

: ^ M A = ^ = ^ = 16 • ^.'^-<^V ^'^>^^-^.0:>=^j

D 3 t A : Q H y v ' , , - e & r i ( J , u •^:^::rG ^- , ; v : , , n |

Binh dtfng Ca(OH)2 hap thu CO2 va H2O ^, ^, , ,^ •

Ca(0H)2 + CO2 -> CaCOai + H2O r '

<T)cum 4 j£4Lfi emtQ^ thi'f£ pliikn lit: •'•'A -.fid i ' j; ri'fj;

- ^i/a ixda phiiiAig phdfL bifii liiati ' '

- <T)ifii oa» dftih liiai bao^ ioAti khoi ''•'''y'': '

BAr TAP M A U V A BAI TAP N A N G C A O Bai 1 Dot chdy hoan toan ba hidrocacbon A, B, C c6 ciing so nguyen tuT cacbon ta thu diTdc ti le CO2 va H2O c6 gid tri tiTdng iJng la 0,8 : 1 : 2 Xdc dinh cong thiJc phan

=> B la anken hoSc xicloanken: CnH2n /

Vi A, B ciing so nguyen tuT cacbon => B la C4H&. ' f f * I :

* Chat C c6 : = 2 => nco2 = 2nH20 hay nH20 < "coz

-Phan dgng vji phuong pUip giSi H6a hi?c 11 HOu ca - P5 Xuan Hang

Bai 2 Dot chay 3,7g chat hQ^a c d X ( C , H, O) dung vUa du 6,72 h't oxi (dktc) va thu diTdc

0,25 mol H 2 O Xac dinh cong thiJc phan tuT X bie't 70 < Mx < 83

Gidi -i

T a c d : = — = 0,3(mol) ''''^^

'v^;-X + O2 >• C O 2 + H 2 O

Ap dung dinh luat bao toan khoi liWng : • ^idvM •'iimb» '^/.i^,

Ta CO : mx + = mco2 ^^lO ' ^.i^TavAicit 'fw^io n # J '

B a i 3 Cho bon hdp cha't hiJu cd A, B, C, D deu ben, mach cacbon lien tuc, kho'i

lifdng phan tijf cua chiing lap thanh mot cap so cpng Khi do't chdy mot lifdng bat

ki moi chat deu chi thu difdc C O 2 va H 2 O ma khoi liTdng C O 2 bang 1,8333 Ian

khoi liTdng H 2 O Xdc dinh cong thiJc phan tuf cua A, B, C, D

Vay cong thiJc cua bon chat c6 dang CSHRO,,

V i khoi liTdng phan tuf cua chiing lap thanh mpt cap so cgng nen ta c6 z = 0; 1; 2; 3

Vay C T P T cua A , B , C, D Ian iMdl Ik : CaHg, CjHgO, CjHsOz. C J H H O J

32

B^i 4 Dot chay hoan to^n 3,24 gam hon hdp X gom hai chat hiJu cd A va B khac

day dong d i n g , trong dd A hdn B mot nguyen tuT cacbon, ngu'di ta chi thu diTdc

H 2 O va 9,24g C O 2 Bie't ti khoi hdi cua X doi vdi H2 b^ng 13.5 T i m cong thiJc

ciia A, B va tinh % khoi lifdng cua m6i chat trong X 1 i '

„ / s C t : , W ' ! li ^ h M f (Trich TS DHQG TP HCM, dat 2) '•'•'ts,,.,, Gidi ( ',lin^ nj' M> r in 4

= > M A < 36 => A khong c6 oxi va trong A c6 2 nguyen tuT cacbon •

Nen A c6 the la C 2 H 2 , C 2 H 4 hoac C 2 H 6

+ A la C 2 H 2 ( M A = 26) =^ M B = 30 ( C H 2 0 ) ^ * ' " * " ' ' ' ^ ' ^ ' ^ " '''' ' + Ne\ A la C 2 H 4 ( M A = 28) ^ M B = 24 (loai) '^''"""^ + A m C 2 H 6 ( M A = 30) =^ M B = 18 (loai) ^ ' ^ ' ' ^ « ' • V$y CTPT cua A la C 2 H 2 , B la HCHO ( C H 2 O )

Thanh phan % khoi liTdng moi chat trong X: % C 2 H 2 = - — ^ 1 0 0 % = 72,22%

, 3,24

=> %HCHO = 27,78%

Bai 5 MQt hidrocacbon X cl the khi c6 ti khoi hdi so vdi hidro la 15 Lap cong thiJc

phSn tuf ci5a hidrocacbon X

Phan d jng vS phuong pMp giSi H6a hpc 11 HOu cd - B5 Xuan Himg

;/i;,;'Neu x = l => y = 18 (loai) ( K A q n o i p i ^ y n s n ^

x = 2 = > y = 6 (nhan) ivs,t5|vvw;(fa H n>^.n A , ^ 'i'M:

V$y CTPT cua X : C 2 H 6

B a i 6 M p t hon hdp khi A gom hai hidrocacbon thupc ciing day dong dSng Dot

chay 2 lit khi A can diing 7,2 lit khi oxi thu diTdc 4.8 lit CO2 mgt lUdng hdi nifdc

(cdc khi do 3 C l i n g dieu ki#n) s b ©

a) T i n h the tich hdi nUdc thu dUdc , , , , , ,^ ^.^

b) Xdc dinh day dong d i n g cua hai hidrocacbon trong A

c) Lap cong thiJc phan tuf tinh thanh phan % theo t h i tich cac chat trong A Biet

hai chat dtfng cdch nhau mOt chat trong day dong d i n g

c) V I nc = 2,4 = X la anken nen n, = 2 n : = 4 (hai chat c^ch nhau rngt chlft)

V a y CTPT ciia hai hidrocacbon la C2H4 C4Hg ,, , , "^r , ^ j ^ ^ ^

Dat a la nc2H4 trong 1 m o l hon hdp A (1 - a) la sd'mol C 4 H 8 ^ ^.-^.j^ ^

.^^^ Ta c6 : nc = " " " ' ^ " ^ = 2 , 4 ^ a = 0.8 (mol) '^•^^^•^•^ ^1A v T I T O v £ V

Thanh phan % theo t h i tich cdc k h i : ^

B a i 7 Cho ba hidrocacbon A B C 5 thg khi trong d i l u k i ^ n thiTdng A B , C c6 ph^i

; la dong d i n g cua nhau hay khong ? B i ^ t rJlng khi phan hOy deu tao ra cacbon va

j ! hidro the tich hidro gap ba Ian \hi tich hidrocacbon ban d i u (5 cilng d i l u ki§n)

vay A , B , C Ian liTdt c6 CTPT la : C2H6, CjHfi, C 4 H 6

B S i 8 K h i phan tich chat hilu cd A chi chtfa C, H , O thi c6 mc + mw = 3,5.mo

a) T i m cong thrfc ddn g i i n cua A / b) L a y hai nfdu ddn chi?c X , Y dem dun nong v d i H 2 S O 4 dam dac d nhipt dO thich hdp thl thu difdc A Xac dinh cong thiJc cau tao mach hcl cua A , X , Y B i e t r i n g

I wfif,, ,'.f:ft ijOHb

y 44 (loai)

32 (loai)

20 (loai)

8 nh$n

- 4 (loai)

=> A cd CTPT : C4H8O. ' • ^ • ^ ^ w ^ , / - liw.^'^*^;*^^^^ ^^^ iCr ; + N ^ u z = n=> C T cua A c6 dang (C4HgO)„ t •

Vay cong thtJc ddn gian cua A la C4H8O. cn^ii '

Phan djng va phiiOng ph^p g\i\a hpc 11 HOu CO - D5 Xuan Hung

.Ma n nguyen =^ n = 5 =^ CTPT cua A la CsH.ọ / ' ' ' ^ ^ ^ ' " ^

B a i 10 Oxi hoa hoan loan mpt chat hu^ cd Y can l,28g oxi, chi tao thanh 0,66g

^ ^H20 va l,76g CO2 Cho loan bp san pham sau phdn iJng sue vao 250ml dung

dich Ba(0H)2 nong dp a mol/lit thu diTdc b gam két tua, Ipc két tua roi dun nong

dung dich loc dufdc c gam két tua nffạ Cho biet b + c = 5,91 Tinh a, b, c va tim

cong thiJc đn gian nhát cua Y ? ' *''

Cong thiJc đn gian nhat cua Y : Ci2H220n ' i / ^ v ^ ;

Bai 11 Bot chdy hoan toan 0,1 mol hdp chat hCu cd A chi chiJa C, H, O vdi oxi theo ti 1? mol 1 : 2 Toan bp san pham chay diTdc cho qua binh 1 difng dung dich

pdCl2 dii roi qua binh 2 di/ng dung dich CăOH)2 dir Sau thi nghiem, binh 1 tang

0,4g va xuát hi^n 21,2g ket tua, c6n binh 2 cd 30g ket tuạ T i m cong thiJc phan tuTcua Ạ

Binh 1 du-ng PdCb hut CO va H2O, giai phdng CO2 j , , , ; ,^ „ ,^,,

CO + PdCl2 + H2O ^ Pd4 + C O j t + 2HC1 (1)

C02d phufdng trinh (1) va CO2 d phan iJng chay vao dung dich CăOH)2

CO2 + CăOH)2 -> CaCOal + H2O

30

100

Snc02 = n c a C 0 3 = — = 0,3 mol i ( I - m 0< "

2 1 2 Theo ptpỉ (1), ta cd: n c o j s i n h r a = "co = n i p d = — ^ = 0,2 mol •

=> So mol CO2 do phan tfng chay la: 0 , 3 - 0 , 2 = 0,1 mol , ,

Khoi lifdng binh PdCh tSng la: mco + m H 2 0 " i " C 0 2 = ^,4 j

Gpi cong thuTc phan til c i l a A 1^ C^HyỘ , „ , , „ , , » r i , A J S j •

T a c d : m c = 12.(Ucoj + nco) = 12.(0,1 + 0.2) = 3,6 (g) 7{ioffl? b / ,

Vaiy: CTPT cua A la C3H4O2 n * ! ' ' - ^ ^ / ' • '

11

Phan dgnp va phudng phap giai H6a hgc 11 Hgu co - B8 Xufln Hung

B a i 12 Mpt hon hdp gom niQt so hidrocacbon hen tie'p trong day d6ng d i n g c6 khoi

Itfdng phan tit trung binh ( M ) = 64 6 100"C thi hon hdp nay 6 the khi, l a m lanh

den nhift dp phong thi mpt so chat bi ngiftig tu C&c chat khi c6 khoi Itfdng phan

tuf trung binh bkng 54 Cic chat long c6 bKng 74 T o n g khoi lifdng c ^ c chat trong

hon hdp dau la 252 Diet khoi liTdng phan ti3f chat nang nhat g^p d6i chat nhe

nh^t T i m C T P T cic chat % the tich cdc chat trong hon hdp '

- - - K ' ^ - ^ Giai

G o i aj, 3 2 , a n 1^ k h ^ i lifdng phan tijf cua c ^ c hidrocacbon trdn

* Ap dung tilth chott toan hgc :

C d c hidrocacbon lien tiep thupc ciing mpt day dong dSng se tao nen mpt cap so

cOng c6 cong sai d = 14

V§y A la C3H6,1^ hidrocacbon dau tidn trong cap cpng tren

C d c dong dSng k e Uep cua n6 \k C4H8, C5H,,,, C^Hu ( M = 84) ^ ,<'

* T-inA % rfcA cdc chitttroneh3n h<n>: '^'''^ ^"^^ '

Gpi a, b, c, d (mol) Ian li/dt 1^ so mol cdc hidrocacbon tifdng tfng: C 3 H 6 , C 4 H S ,

B a i 13: Dot chdy hoan toan m(g) hdp cha't hiJu c d A chi Ihu diTdc a(g) C O 2 va b(g)

H 2 O B i e t 3a = 1 l b v a 7 m = 3(a + b) T i m C T P T c u a A (biet l i khoi hdi c u a A d5i vdi khong khi nho hdn 3) ' •

Phan dgng va phuong phap giSi H6a hpc 11 HOu co - B5 Xuan Htfng

Bai 14 Dot c h a y hdp chat hiJU c d A (chuTa C, H, O) p h a i d u n g 1 li/dng o x i b a n g 8 Ian lUOng

o x i c6 trong A v a thu dUdc lu-dng C O 2 v a H 2 O theo d le k h o i lUOng

m c o 2 : mHjo = 22 : 9 Biet ti khoi hdi c u a X so vdi H 2 la 29 Tun C T P T c u a X

Gia suf chat hffu c d A chiy sinh ra 22g CO2 v a 9g HzO:'-''-* * '

, = > m c = 1 2 n c o 2 = 12.—= 6 ( g )

44

m H = 2 n H 2 0 = 2 - j ^ = l ( g )

Theo djnh luat hio tohn nguyen to la cO:

modrongA) + mophintfng = mO(trongC02) "'OltrongH2O)

, :„ „, ^,,,, , v;,: = (22 - 6) + (9 - 1) = 24 (g) J^'i' ::-j-f,->V;!S5

• " O phaniJng' ' ' O v - •S'.'C.i., '4^&.:J'9

c a u 1 Mpt chat heru cd X khi d trang thai hdi c6 ti khoi hdi so vdi chat hd^i cd Y la

2 Biet 2,2 gam chat Y c6 the tich b^ng the tich ciia 1,6 gam oxi trong ciang dieu

ki?n nhi?t dp, ap suat Vay ti khoi cua X doi vdi C H 4 la:

A 11 B 10 C.6 D.5,5

C a u 2 Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chdt khi a d'lhu kien

thuong) r6i dem toan bp san pham chay hip thy h^t vao binh dung dung djch

Ba(0H)2 Sau cac phan irng thu dugc 39,4 gam k^t tiia va khoi lugng phin dung

djch giam bat 19,912 gam Cong thirc phan tCr cua X la

i , , ' " Trich di thi tuyin sinh Dai hgc khoi A nam 2012"

c a u 3 Nicotin cd trong thuoc la la mot hdp chat rat doc, c6 the gay ung thiT phoi

Dot chdy 16,2 gam nicotin b^ng oxi v^a dii thu difdc 44 gam CO2, 12,6 gam

nifdc va 2,24 lit N2 (dktc) Cho 85 < M„i,.,ii„ < 230 Cong thiJc phan tuT dung cua

nicotin 1^:

A C 5 H 7 N O B C 5 H 7 N O 2 C C , O H M N 2 D C H , H , 4 N 3 '

Cfiu 4 Do't chdy hoan toan 5,2 gam hdp chat huTu cd X roi cho san pham chdy Ian

lifdt qua binh (1) di/ng H2SO4 dam dac, binh (2) chu"a nifdc voi trong diT, thay blnh (1) tang ISn 1,8 gam, binh (2) thu diTdc 15 gam ket tua Khi hoa hdi 10,4 gam A thu di/dc mot the tich dung bang the tich cua 3,2 gam oxi trong cung dieu

kien nhiet dp dp suat V$yc6ng thu'c phan tuTcua A 1^: 'i 'i/i • i •*

A C5H12O2 i B C 3 H 4 O 4 C C7H4O6 D C ; H S 0 3

c a u 5 Khi d^t chdy hoan toan 100ml hdi cha't B can 250ml oxi, tao ra 200ml C O 2

va 200ml hdi niTdc (cac the tich do d cung dieu ki?n) Vay cong thu'c phan lij cua B la:

A. C 2 H 4 O 2 B C 3 H 4 O 4 ' • C C 2 H 4 O D C 3 H 6 O

c a u 6 Do't chay hoan toSn 3,5 gam mpt hidrocacbon thu difdc 10,68 gam khi

cacbonic va 5,25 gam niTdc Khoi liTdng oxi can diing de dot chdy la:

A 6,21gam,; M B 11,04 gam C 12,43 gam D 12,73 gam

c a u 7 Dot chdy hoan toan 0,366 gam chat hiJu cd X, thu du-pc 0,792 gam CO2 v^

0,234 gam H2O Mat khdc phan buy 0,549 gam X do thu difdc 37,42cm^ nitd (do

d 27°C va 750 mmHg), biet trong phan tuf cua X chi chiJa 1 nguyen tijf nitd Vay cong thu'c phan tuf cua X la:

A C 9 H 1 3 O 3 N B.C9H15O2N C.CyHi303N2 D C K H 1 2 O 3 N

c a u 8 Dot chay 400 ml mpt hon hdp gom nitd va mpt hidrocacbon Y 3 the khi

bing 900 ml khi oxi (dif) The tich hon hdp thu dUdc sau khi dot la 1,4 lit Sau khi ngu-ng tu hdi niTdc thi the tich hon hdp con 800 ml, tiep tuc dan qua dung dich KOH thay con 400 ml khi Biet rkng cac khi do d cilng dieu ki?n nhi?t dp,

dp suat Vay cong thu'c cua Y Id:

A. C 2 H 6 B C 3 H 4 r i i c i C C 2 H 4 D C4HX

Cau9.Sod6ngphancuaC5H,2la: , •

c a u 10 Do't chdy hodn toan mpt liTdng hidrocacbon X Hap thu toan bp san

pham chay vao dung dich Ba(0H)2 (dU) tao ra 29,55 gam ket tua, dung djch sau phan iJng c6 khoi liTdng giam 19,35 gam so vdi dung dich Ba(0H)2 ban

dau Cong thiJc phan tur cua X Id ' A.C3H4 B.C2H6 '^^-'''^^^ C C 3 H 6 : ' ^ ' - ^ ^ ' ^ ^ ^ ^ ^

' - ''^iOslli * „ y^^^ ^ ^ ^ ^ ^.^^ f.^.^^ 2070"

c a u 11 Dot chdy hodn todn 0,9 gam hdp cha't hiJu cd A chuTa C,H,0 thu diTdc 1,32

gam CO2 vd 0,54 gam H2O Ti khoi hdi cua A so vdi hidro la 90 Vay A cd cong

thu'c phan tuf Id:

A.CeHnOft B C,oH,203 C C»H2o04 jin • , D CTHifiO,

Phan dgng va phuong ph^p giai H6a hpc 11 HQu cd - D8 Xufln Hung

Cfiu 12 K h i dot cMy 1 lit khi X can 5 lit khi oxi, sau phan tfng thu dUdc 3 lit CO2 va

4 l i t hdi nifdc Biet cdc k h i diTdc do d cdng dieu ki#n ve nhi^t dp va dp suS't V§y

cong thiJc phan tuT cua X la:

A. C 3 H 6 B CjHx ' C C J H K O D C3H6O2

CSu 13 Trong mot binh kin chiJa hdi chat hOu cd X (c6 dang C„H2n02) mach hcl

O2 (s6 mol O2 gap doi so mol can cho phan uTng chdy) 6 139,9"C, dp suat trong

binh la 0,8 atm Dot chdy hoan toan X sau do diTa ve nhi?t dp ban dau, ap suat

trong binh luc nay la 0,95 atm X cd cong thuTc phan tijf la:

A.C2H4O2 B.CH2O2 C.C4H8O2 D.C3H6O2

(Trich de thi tuyen sink dai hoc khoi B)

CSu 14 D o t chay hoan toan 100 m l hdi chat A , can diing 250 m l o x i , chi tao ra 200

m l CO2 va 200 m l hdi nifdc (cac the tich khi do d cOng dieu ki?n) Xic dinh

cong thtfc phan tur ciia A ,^ , , ^

A.C2H4 B.C2H6O C C 2 H 4 O D.CjHfiO

CSu 15 Dot chay hoan toan 1 lit khi X can 5 h't khi oxi, sau phan uTng thu diTdc 3 lit CO2

v^ 4 lit hdi nufdc B i e t cac k h i dUOc do d ciing dieu k i ? n ve nhi^t dp va ap suat

Cong thifc phan tiy cua X la: f

A C 3 H 6 B.C3H8 , C C 3 H S O D C j H f i O z ^ " ' " '

Cau 16 Co 3 chat hOU cd A, B, C ma phan t& ciia chijng I§p t h ^ 1 cap so cpng Bat ci? chat

n^o khi chay cung chi tao CO2 v^ H2O, trong do n^o^: n^^^o = 2 : 3 CTPT ciia A, B,

ClanlUdtla: , ,

A C2H4, C2H4O, C2H4O2 I i/ :> B C2H4 CzHfiO CjHfiOj \„J'

c. C3HS, C3HS0, C3HXO2 ?! r, D CZH^ CZHSO, C2H6O2 'I'^f ,

Cfiu 2 : So m o l BaCOs = 0,2 m o l so mol C O j = 0,2 m o l - > khdi liTdng = 8,8 gam

K h o i liA;Jng dung dich giam = khoi liMng k e l tua - (CO2 + H2O) , 0 ; •

T d n g k h o i ItfdngCO2 + H2O = 19,488 =>44.nco2 + 18.nH20= 19,488 (1)

Dot X thu diidc CO2 va H2O sfjri ^irf;,, rp ^ jsiLiJ*^ OO'nik^ ' '

A p dung D L B T K L ta c6 : mx + m o j = m c o j + ^ ^ 2 0 ;4i j! r n a ^ q

=> mo2 = nico2 + ^ » 2 0 ~ ' " x = 14,848 g => IXQ^ =0,464 mol V f l i j ' ) A

Bao toSn nguyen to oxi: 2. n c o 2 + " H 2 O ~ 2 n o 2 = 0,464 'f'

TO (1) (2) n c o 2 = 0,348mol; nH20 = 0'232 mol ' "

= > n c : nH = 0 , 3 4 8 : 0 , 4 6 4 = 3 :4=>XmankinC3H4 r -< ,

C f l u 3 , C h o n C ,„,„„.•• „: ,1, ; T •

Nicotin c6 cong thtfc cau tao nhiT sau:

Cfiu 4, Ta c6 phiMng trinh phan iJng: ' ' : = = c >' CO2 + Ca(OH)2 -> CaC03i + H2O , (,;00f _mCaC03^ 15 Oud

Vay C T P T la C3H4O4 => Chon B

Cfiu 5. Goi C T P T cua B Ik CxHyO,

Phan djing phuang ph^p giSi H6a hpc 11 HOu co - D5 Xuan Hiflig

^ m„ = 0,028(g) ; mc = 0,216(g) ; I T I H = 0,026(g); = 0,096(g)

0,216 0,026 0,096 0,028 ,^ ^ ,

x : y : z : t = - ^ :— :— :— = 9 13 : 3 • 1

12 1 16 1

=> Cong thiJc nguyen (C9Hi303N)n *

Do X C O mot nguyen tu" N n6n CTPT la C9H13O3N => Chon A

Ma mco2 = 0,15.44 = 6,6 g=i> mH20= 3.6 g=> nH20 = 0.2 mol - „

Ta thay: n^jo > "002 ^ hidrocacbon X la ankan va:

Cau 12 Goi CTPT cua X la CxHyO,

n02 phan iJng - 1 mOl

=> Tong so mol cua X va O2 luc dau la: 3n - 2 + 1 = (3n - 1) mol

Sau phan iJng chdy va d 139,9°C thi H2O dang d the hdi nen tong so mol khi va

hdi sau phan iJng la:

Skhi = nco2 + nH20 + " 0 2 J" = n + n + ^"^ ^ = (3,5n - 1) mol

Nhiet dO bmh tnfdc vd sau khong doi, gia suf t h i tich binh khong doi, ta c6:

'sau

0,8 3 n - l 0,95 3 , 5 n - l => n = 3 V|iy CTPT cua X Id C 3 H 6 O 2

=> Dap an D

Caul4 Tac6sdd6sau:

C,HyO, + 2,502 -> 2CO2 + 2H2O

Cdn ciJ vdo h0 so phan iJng vd dp dung dinh luat bao todn nguyen to, de dang c6 AldC2H40

Dap an C

45

Phan djng \A phudng phip giil H6a hpc 11 HiTu co - Pg Xuan Hung

CSu 15: Goi cong thiJc phan ttif cua X \k : C,H„0, (z cd the bkng 0)

=> A, B, C deu no, deu c6 cilng so nguyen tuT C va H NhiT vay de phan tuT khoi

cua chiing l|ip th^nh 1 cap so cgng thl chiing khdc nhau \i so nguyen tiJ oxi

trong phan tuf

I A N K A N (parafin): C„H2„+2 (n ^ 1)

* C H 4 C 2 H 6 , C j H g C „ H 2 „ + 2 1 $ P thanh day dong d^ng ciia ankan

* TCf C4 tr3 di cd dong phan cau tao j ^,

* Danhphdp: ^^^j.,.,,, ,.1^:).::.^}^:) < i g i i

+ Ankan khdng phan nhdnh : , Ten ankan = ten mach chinh + an ,, + Ankan phan nhdnh :

T6n ankan = s6 chi vi tri nhdnh + ten nhdnh + ten mach chinh + an

* Tinh cha't hda hoc : phan tfng the', tdch, 0x1 hda ' ,1 ' + Phan itng the': C H 4 + CI2 CH3CI + HCi

-> phan tfng halogen hda j ^ j (

+ PhaniJngtich: , \, d 1 ' i h i ^ l ' ^ r

CaH2a+2 + ChH2b ,

H f( iS'.J)".'!-!:,.) : M.:i sy?

ankan thap anken phan iJng crackinh . ^ r ^ , jy,^.^ ,q ^^.m &MiH^\V I A O J i-'

Khi cd xdc tdc thich hdp: C H 4 + O2 HCHO + H 2 O

R-COONa + NaOH R - H + NajCOj (R Ik C„H2„+,-)

- Mach chinh la mach v6ng

- Ddnh s^ sao cho tdng c&c s6 chi v i tri cdc mach nhdnh Ik nh6 nhSt

* Tinh ch^t hda hoc : ^

41

Phan dgng vS phuong ph^p gl5i H6a hgc 11 HOu c o - B6 XuSn Hung

+ Phan u-ng the : , , » , ^

Xiclopentan brom xiclopentan

+ Phan u-ng cpng md vong : THIYUHF i i '(M f4C: fi

Xiclobutan r f i x - f U 1 f butan i { iin Au > M i ^ i i ) / i - 0 i^n fl'Vi

+ Phan u-ng tdch : / r! u r , # 4 * - f i »'{.'ii* i ^ f i i Sihi\ ^^ , JIMI T I t

Metylxiclohexan metylbenzen (toluen) ri <^ > i^ir " i ri\ +

+ Phan u-ng o x i h6a : C„H2„ + — O g nCO^ + n H 2 0 * V '

* D i e u c h e : C H 3 - ( C H 2 )4-CH3 f j + H 2 H*^ : ;

B P H A N L O A I V A P H O O N G P H A P G I A I C A C D A N G B A I T A P ' '

<ViA eAng thi'te emi t^a im gnl llii atihati^ adehmnkati , ,«

BAI TAP M A U V A BAI TAP N A N G C A O "^^^^ I S ' ' * ^

B a i 1 Viet cong thiJc phan tuf cua ankan va goc hidrocacbon tifcfng i?ng •

b) Viet c6ng thuTc cau tao cua c^c ankan sau : pentan, 2-metylbutan, isobutan. Cic chat

tren c 6 ten g o i nao khdc khong ?

b) Hay vie't c6ng thiJc cau tao thu gon va thu gon nhat cua cac chat sau : ,

* isopentan, neopentan, hexan

* 2,3-dimetylbutan, 3-etyl-2-metylheptan, 3,3-dietylpentan

C H 3 - C - C H 3

I

C H ,

H e x a n : 2,3-dimetylbutan :

C H 3 - C H 2 - C H 2 - C H 2 - C H 2 - C H s i , ;

C H 3 - C H - C H - C H 3 ,'

I

C H , CH3 3-etyl-2-metylheptan: C H 3 - C H - C H - C H 2 - C H 2 - C H 2 - C H 3

Phan dgng phuong phAp giii H6a hpc 11 HOu cO - D8 Xufln Hung

Bai 4 Difa vao khdi ni§m dong d i n g hoac difa tren so electron h6a tri cua cacbon

va hidro Hay chuTng minh cong thtfc tdng qu^t cua ankan Ik C„H2„+2

Gidi

* Di/a v^o khai ni$m dong d^ng :

Dong d i n g cua metan : CH*, CaHft, CsHx, C4H10

=>CH4(CH2)x « Q + , H 2 , ^ ; i f ) ,^rl^ v V ' , ;

D a t n = x + x = n - 1 ' J-i':'-'.;f':^M-,/ J ; ' -

N e n s o n g u y e n t u f H : 2x + 4 = 2 ( n - l ) + 4 = 2n + 2

Vay C T T Q cua ankan m C„H2„+2 •;,;vyi>i'r'' /; il'J-.H'

* Difa v^o so electron h6a t r i : < • ,

Dat so nguyen tuT C 1^ n => So electron h6a tri ciia nC 1^ 4n

So lien ket ddn giffa C - C m (n - 1)

Do do so e h6a tri da suT dung Ik 2(n - 1) (vi m o i li6n ke't ddn diing 2e h6a tri)

=> So electron h6a tri c6n lai la 4n - 2(n - 1) = 2n + 2 dCing g6p chung vdi

electron h6a tri cua H tao thanh lien ket cong h6a tri C - H

=j> So nguyen tuf H la 2n + 2 „ ^:, , , ,

V$y C T T Q cua ankan m C„H2„+2 ?v

Bai 5 Hay viet cong thiJc cau tao va goi ten theo l U P A C c^c ankan c6 cong thiJc

I

C H 3 - C - C H 2 - C H 3 :2,2-dimetylbutan

CH3 -<.m.;:i?M/'vy:J,:' ' /H',

B i i 6 Viet cong thiJc cau tao cdc chat c6 tdn sau :

a) Xiclopentan \''''^ -^''''yi^S'ti b) 3,3-dibrom-5,5,6-trimetyloctan

c) 3-metyl-3-etylpentan d) ,1,4-dimetylxiclohexan e) l-clo-3-xiclopropylpentan f) etylxiclohexan g) l-clo-3-etyl-2-nitroxiclopentan

a) Xiclopentan: !' '

I I ^ b) 3,3-dibrom-5,5,6-trimetyloctan : C H 3- C H 2 - C - C H 2 - ( p — J : H - C H 2 - C H 3

* ':X, \: » ' , ' " r ' - r j : r - ; r ' B r CH3CH3

CH3

I '^-^ • c) 3-metyl-3-etylpentan : C H 3 - C H 2 - C - C H 2 - C H 3

PhSn djng vS phUOng ph^p giSi H6a hpc 11 HOu co - D5 Xufln Hung

i - (Viet ehitdi ftlian ihig

- (Jiiih ehut hdu h^e eua aitktut, aeSdtuitikatt '

B A I T A P M A U V A B A I T A P N A N G C A O

B^i 1 Viet phiTdng trinh hda hoc cua cdc phan rfng sau :

a) Propan tac dung vdi clo (theo ti le mol 1 : 1) khi chie'u sang

b) Tdeh mpt phan tuf hidro tilf phan tuf propan

c) Dd'tchiy hexan

d) Khi syc khi xiclopropan vao dung dich brom ;

e) Dan hon hdp xiclopropan, xiclopentan va hidro di vSo trong 6^ng c6 bpt niken,

b) CH3-CH2-CH3 '"^'^°> CH3-CH=CH2 + Ha;::^ ifv rs&is liz'

G) C6H,4 + —-O2 6CO2 + 7H2O •• •

d) A

+ Br2(dd) > CH2-CH2-CH2 (1,3-dibrompropan)

Xiclopropan ;, Br '

c) Khi dan hSn hdp xiclopropan, xiclopentan va H2 di vao trong ong c6 bgt Ni,

nung nong thi chi c6 xiclopropan phan iJng cpng vdi hidro,j:> H.) -f J1 , / \ H2 CH3-CH2-CH3 ,H:,)

f) + Br2 ^ ^ + HBr « •^^;fi:>~li:|^ f

Xiclohexan ,

B^i 2. Hai chat A, B c6 cung cong thi?c phan tu" C5H12 tdc dung vdi clo theo ti le mol

1 : 1 thi tao ra mot dan xuaft duy nhat con B thi cho bon dan xuat Yie't cong tiiifc

cS'u tao cua A, B, cac dan xuat clo cua chiing va goi ten j

PMn dgng \A phijgng pMp qai H6a hoc 11 HOu co - Dfi Xuan Himg

=> Cic dan xuat mpt Ian the do :

Bai 3 Vi^t phUtfng trlnh gpi ten phan iJng cQa isobutan trong cdc tnfdng hdp sau:

a) Lay 1 mol isobutan cho tac dung vdi 1 mol clo c6 chieu sang ,, i^ ,

b) Lay 1 mol isobutan dun nong vdi 1 mol brom

c) Nupg nong isobutan vdi xQc tac Cr203 de tao thanh C4HX (isobutilen)

d) Dot isobutan trong khong khi '

J'K' '^i.y.U ofe-f Cl'jix ,rufq'/;qokux Qtd t'Aii lUih JiDs i:":

a) CH3-CH-CH3 +CI2 CH3-C-CH3 +HC1 => phan tfng the

CH3 ^ CH3 ' " v :

-h) CH3-CH-CH3 + Br: C H 3 - C - C H 3 + HBr => phan iJng thg" (1

c) CH3-CH-CH3 CH3-C=CH2 + H2 iniiv => ph5n i?ng tach

,s,,;- 'jCHy''I-'- • CH3 'iM'> '•'»^'^ ' • v ^ i ' A s i 'fi;*; * •

13 n ob iimr '^.I'it 'jk> <H;J mh

d) C4H,„ + y 0 2 ).4C02 + 5H20 ^ ,

Bai4 w,-.,

a) Trinh b^y phi/dng phap hda hoc phan bipt hai khi khong m^u propan va

xiclopropan di/ng trong cdc blnh rieng bi?t '^ i^O-iU'^'^J^O-rlf^

b) Nhan biet cdc khi sau : CO2 CH4, H2

a) Dan hai khi di qua dung dich brom c6 m^u v^ng da cam, khi n^o l^m mat

m^u dung dich brom 1^ xiclopropan Con lai la propan ^ ^ ^ ^ ^

PhiTdng tr\nh phan iJng :

' / \ Br2(dd) > C H 2 - < : H 2 - C H 2 '

: V Br Br b) Dan ba khi Ian li/dt di qua dung dich niTdc v6i trong (diT), khi nao lam due nUdc

voi trong l£k CO2, con lai CH4 va H2 ,

CO2 + Ca(OH)2 > CaCOj + H2O sf? • H

Bot chiy hai khi con lai sau dd dan san pham Ian liTdt di qua nifdc v6i trong diT,

Bai 6 6ng nghiem A chiJa dung dich KOH, ong nghiem B chiJa dung dich H2SO4,

ong nghiem C chi?a dung djch KMn04, ong nghiem D chiJa nifdc brom, cho vao moi ong nghiem dd 1ml octan l^c deu roi de yen Dif doan hi§n tiTdng xay ra d moi ong nghiem va giai thich ?

Gidi '

Khi cho 1ml octan Ian lifdt vao moi ong nghiem thi khong thay hi^n tUdng gi xay ra trong tCfng d'ng nghi§m vi: ' ? ,

- d ong A : octan khong cd phan iJng vdi dung dich KOH

- 6 ong B : octan khong phan itng vdi axit H2SO4

- 6 ong C : octan khong cd phan iJng oxi hda vdi dung dich KMn04

- d 6'ng D : octan khong cd phan tJng cgng vdi nifdc brom

B^i 7 Viet ciic phUOng trlnh phan uTng hda hoc dudi dang cong thiJc cau tao rut gon cua

c5c chat theo cdc day chuyen hda sau (cho biet A2 va A3 la cac san pham chinh)

nu xt,t° , A / > Cao su buna

^ > C 4 H 7 B r { A 2 , A 3 )

55

H6a hgc 11 HOu co - P 8 Xuan Hung

Theo sd do ta c6 cdc phifOng trinh phan iJng sau : '' " J ^ - ^ •'

* CnH2n+2 : C4H10 1'A- N.i' '•'.;>'''•• /SI

l|^#a'••^vlyu'i Vi^^^^^^^^^ 3 j - * C H 2 - C H - C = C H 2 j ^ , } ,5,,^ j , ^ ; ^

Br Br C H 3 (Av);,^-,,,,,,,

a) Cong thi?c ciia mpt hidrocacbon A c6 dang (CnHzn+i)^ Vay A thupc day dong

d i n g n^o ?

b) Viet cdc phifdng trinh phan ufng : C„H2„+2 tac dyng vdi CI2 (askt), crackinh b^ng

nhiet Cho vi du ? i.'v/xj ;-:,,;•.••!, ^••••r' • dd ^\iiv:-^krk

a) (CnH2„+i)x hay CnxHjnx+x. • , X : : , ' L : , * : L , „ " '

V i so nguyen tuT H trong hidrocacbon luon 1^ s6' chS'n nen

* Khi X = 1 ==> CnH2„+i : la goc hidrocacbon c6 hda tri 1 (loai) vi A la mpt hdp chat

56

* Khi X = 2 => C2nH4n+2 hay C2nH2.2nf2 la mOt ankan (nhan)

* Khi X = 3 => C3nH6„+3 (loai) vi H c6 so le

* Khi X = 4 => C4nHsn+4 (loai) vi khong phu hdp vdi hda tri cua cacbon V i du :

B a i 9 Xac dinh cong thiJc cafu tao ciia CfiHu, biet rang khi tac dung vdi eld theo ti

1$ mol 1 : 1 chi thu diTdc hai dong phan Gpi ten hai dong phan do

Cong thiJccCu tao CfiHM : f ^ ^ ^ ^ " ' ' " * '

Bai 10 Cho biet C T C T cua pentan trong cac t r i ^ n g hdp sau:

a) Tac dung v d i CI2 (askt) ti le 1:1 cho 4 san pham b) K h i cracking cho 2 san pham

Gidi

Lfng v d i pentan C5H12 c6 cac dang khung C s a u " t i - i i > 1 Si •

••'lii'iV, v.^ii-tVV:/''^ i.^^'i r ' i i ' V • Q ' /'''i^''

(I) C (2) ^ C (3) '•••-•',e',:

a) K h i thiTc hien phan tfng the^:

(1) C O 3 v i t r i the ( C 1 , C2, C3) tao 3 san pham (loai) ' ' I t " ' ' (2) C O 4 v i t r i the ( C I , C2, C3, C4) ^ tao 4 san pham (nhan) (3) C O 1 v i t r i the ( C I hoSc C3) -> tao mpt san pham (Ipai)

V a y C T C T cua pentan la ( 2 ) : 2-metylbutan (isopentan) '

P t p i ? : ^' ^

Phfln d?ng va phuong phap giai H6a hpc 11 HQu cd - O i XuSn Htmg

TMngM CTCT cua pentan 1^ (3): 2,2-dimetylpropan (neopentan), ichi cracking chi cho

2 san pham : ^ •.>.;•,;.•• •V':;?ỉ'ai;V)-.v ruvb JUX

r-C H 3 Srlvj ;lf'f''?i:5 ri,f<iit ;.;;;,) /vyfV';, ( i / =,» ;,.> 4,?L ;;r;;, ta;, j • ; UHV

C H 3 - C - C H 3 cracking,t _

P H > - C H 2 - C - C H 3 + CH4 , „ , ^ ^ , M r ; , n 5 : >

.,H fv:> ,}; C H 3 "•'"^

a) Viet c6ng thil-c cau tao va goi t6n cdc d6ng phan cua CjHsBrj

b) Nh$n bi^t cdc châ sau : C 2 H 6 , SO2, N H 3 , Ộ , ^ , j ,.,M:>. fs':> , H 5 >

- Dan bon khi l^n IiTdt di vao dung djch brom, khi n^o lam mat mau dung dich brom

SO2, con lai la C2H6, O2, NH3

S02 + Br2 + 2H20 >2HBr + H2S04

58

J g W A l g f t VWOT Cho quy tim am Ian Itfdt vao ba khi c6n lai, khi n^o lam quy tim am h6a xanh la

NH3, c6n lai la CzHg, O2

Cho que đm c6n t^n 66 \ko hai khi c6n lai, khi l^m cho que d6m c6n tSn do

bilng chdy m O2, c6n lai m C2H6 , ; , 1 ; , ^ ;

(patui 3 : x;:'.',::,'"' ,'

~ ^Sdi oe ankaitf xielotuikan

- grinh Htdnh fihdit %

- £gft e&ug Iltite (utkaitf xieiemtiuui

BAI TAP MAU

Bai 1 CT đn gian nhat cua 1 ankan \k (C2H5)„ Hay bi^n luan de tim CTPT cua

chat tr6n

Cdch 1: CT đn gian tren la 1 goc ankan h6a tri 1 ttfc c6 kha nSng ket hdp them

vdi 1 goc nhir vay nO-a => n = 2 CTPT ankan C4H,,, - •

GicA 2: CTPT cua ankan t r e n : (C2H5)„ = CxH2x+2 '^:f^€ : ^my^\i '%'%H

Bai 1 Xicloankan đn v6ng X c6 ti khoi so vdi nitd bJlng 2 L§p cong thiJc phan tuT

cua X Viét phiTdng trinh h6a hoc (d dang cong thtfc cau tao), minh hoa tinh chat

h6a hoc cua X, biét rJlng X t i c dung vdi H2 (xuc tdc Ni) chi tao ra mpt san

V i X + H2 (Ni, t") -> mpt sin pham duy nhát nen X c6 CTCT - [ ^

* PhiTdng tfinh hoa hoc minh hpa tinh chat hda hpc ciia X :

• 59

Phan dgng phuang ph^p giai H6a hgc 11 HOu c o - D8 Xufln Hung

- Phan u-ng the :

+ Br2

-Br + H B r ' - P h a n iJng c o n g m d v o n g :

Bai 2 N g i r d i ta thiTc h i ? n phan iJng no m p t liTdng metan tao ra C O 2 va H 2 O , v d i mot

the t i c h k h o n g k h i viifa d u trong b i n h thep k i n , sau p h a n iJng diTa v e n h i e t d o ban

d a u ( h d i niTdc ngifng t u v a a p suat h d i nufdc k h o n g d a n g k e ) HSy xac d i n h % the

U'ch cac k h i t r o n g b i n h sau p h a n ufng T i m t i le a p suat tnfdc v a sau p h a n iJng ?

^G«ir' .>

, Phifdng t r i n h p h a n iJng ' ' • aif;} iMn l i s i j l T"'} 1 liSi

CH4 + 2O2 > C O 2 + 2H2O „ :a^m.ihh

x m o l 2 x m o l x m o l 2x m o l

;H.; J a CO tnA3c p h a n ufng :, -jij) \ ^itiMss :><;t$ i M nM nfeig i*>lj T j J

* T o n g so m o l k h i trirdc p h a n iJng : , • i;t;:i^i*#feiiigi>|i::^^^^^^^

•r n i = ncH4 +no2 +nN2 = x + 2x + 8x = l l x ^.^^^^i^,^ (j^n'WD '& A A

(trong k h o n g k h i t i le the tich O2 va N2 ^ 1 : 4 => UQ^ = 2X = > n^^^ = 8x) '^-l

* T o n g so m o l k h i sau p h a n iJng : n^ = TIQQ^ + nN2 = x + 8x = 9 x ( m o l )

V a y % the t i c h cac k h i sau p h a n iJng :

Bai 3 D o t chdy hoan toan 4 0 0 m l m g t ankan A thu difdc 8 0 0 m l k h i C O 2 (cung dieu

k i $ n ) T i m c o n g thuTc phan tuT cua A V i e t cong thiJc cau tao A ' " ""

p^i 4 A n k a n Y m a c h k h o n g phan nhanh c6 cong thuTc ddn g i d n nhat la C2H5

T i m c o n g thiJc p h a n tuf V i e t c o n g thiJc c a u tao v a g o i t e n chat Y

5) ViS't phifdng t r i n h h d a h g c cua phan Ung giQa Y v d i c l o k h i chie'u sang, c h i ro s^n p h a m c h i n h ciia p h a n u-ng ^:>m m':'^ J ni^'y; <::K<W i f.^'- '

6 1

Phan dang vS phuong phjp giSi H6a hpc 11 HPu cd - D5 Xifln Hung

Thanh phan % khoi liTcJng moi ankan: %mc3Hg = ' ^ ^ g ^ 100% = 43,14%

=> %mc4H|o = 5 6 , 8 6 % <^.iM ^f'lij' :^.v:Vri-i<:'.nf.\i

B a i 6 Dot chay h o i n to^n 3,36 l i t hon hdp k h i A gom metan etan thu dtfdc 4,48 lit

khi cacbon Cdc the tich k h i dtfdc do'd dieu ki$n tieu c h u l n T i n h th^nh p h i n %

the tich cua m o i k h i trong hSn hdp A

B a i 7 Cho 8,8g ankan A phdn tfng v d i clo trong dieu k i ^ n dnh sdng thu dtfdc 15,7g

d i n xuat mpt Ian the clo Xic dinh cong thtfc phSn tuT ciia A dan xuS't mpt l l n

^ C T P T d S n x u r t m 9 t l l n t h e - c l o : C 3 H , C l ' " ' J

BAI TAP N A N G C A O

B a i 1 D ^ t c h i y hoan to^n mOt mau hidrocacbon ngtfdi ta thS'y the tich hdi ntfdc sinh

ra gS'p 1,2 l l n the tich k h i CO2 (do trong cilng d i ^ u k i ^ n ) B i ^ t r^ng hidrocacbon

do chi tao th^nh mOt dan xuat monoclo duy nhat HSy xic dinh c6ng thtfc du tao

chdng m 132 Xic dinh cong thtfc phan tiJ c i a X , Y , Z

D^t M la khoi Itfdng m o i phan tuf trung binh ciia ba ankan lien t i ^ p nhau ( C I l j ^ j )

_ — 112

-T a c 6 : 14n + 2 = M = o 14n + 2 = 4 4 :

=> Ba ankan l l n Itfdt la : X : C2H6; Y : CjHg; Z : C4H,o , •

B i i 3 K h i 1 gam metan chdy t6a ra 55,6kJ Can d^t bao nhifiu l i t k h i metan (dktc)

de Itfdng nhi?t sinh ra du dun 1 l i t ntfdc ( D = I g / c m ' ) ttf 25''C len 100°C Biet

r i n g mudn nang 1 gam ntfdc len 1"C cSn tieu t ^ n 4,18J gid sd nhi?t sinh ra chi dilng de l ^ m tSng nhi^t d$ ciia ntfdc

• Gi&i •' K

Khi nang 1 gam ntfdc len T C c i n tieu ton 4,18J i f •

Vaiy k h i nang 1 gam ntfdc tilf 25''C len 100"C can tiSu ton m0t nhi?t Itfdng la : ( 1 0 0 - 2 5 ) x 4 , 1 8 = 313,5kJ

=> N h i e t Itfdng can tieu ton cho 1 lit ntfdc ttf 25''C len 100"C 1^ :

313,5 X 1000 = 313500J = 313,5kJ

Ta l a i c6 I g CH4 chdy toa ra 55,6kJ ' '

x g? C H 4 c h a y t6ara313,5kJ , ,

3 1 3 , 5 x 1 55,6

Phan dgng vji phuong phap giSi H6a hpc 11 HOu c o - Pg Xuan Hung

=> The' l i c h k h i CH4 (dktc) : V C H 4 = 0,3525 x 2 2 , 4 = 7 , 8 9 6 (Ht)

B a i 4 H a i x i c l o a n k a n A va B d e u c6 t i k h o i h d i so v d i m e t a n b&ng 5,25 K h i

m o n o c l o hoa (cd chie'u sang) t h i A cho bo'n san p h a m , B c h i cho m o t san p h a m

d u y n h a t X a c d i n h c o n g thiJc cafu tao ciia A va B

M A h o a c B = 5,25 X 16 = 84 w ' : ; o , ',

V i A hoac B la h a i x i c l o a n k a n c d khd'i li/dng m o l p h a n tuT b ^ n g n h a u

=> A va B la d o n g p h a n c u a n h a u ,

D a t C T T Q cua A hoac B : C„H2n ' * •

T a c d : 14n = 8 4 => n = 6 C T P T ciia A hoSc B Ih CfiHiz-"''' ^'"''^

K h i m o n o c l o h o a : « v.s?fc i.*30ftoin ikii nib iOm liixiui! o ^ i trf:j n o d ^ ^ ^ f o i x s i V

+ A c h o b o n san p h a m =i> C T C T cua A la H 3 C - C H - C H 3 hoac ( _ )

2\ u U ' , / r 3 c r

+ B c h o m 6 t san p h a m duy n h a t => C T C T cua B : 1 ^

B a i 5 K h i o x i hda hoan toan 7 m g hdp cha't A thu diTdc 11,2ml k h i CO2 (dktc) v a 9 m g

H2O T i k h o i h d i cua A so v d i N2 b i n g 2,5 X a c d i n h c o n g thiJc c a u tao cua A

ne'u k h i c l o hda no t h i c h i thu difdc m g t dan xua't m o n o c l o d u y nhaft '

;'• : '•" Kd ,t!t» l i r i ^ ' i o i i j j & k j loos J^Ci

B a i 6. Do't c h a y m p t hidrocacbon A v d i li/dng o x i vCfa d u T o a n bo san p h a m chay

diTdc d a n q u a b i n h diTng CaCl2 khan cd diT t h i the tich g i a m d i hdn m g t nu:a.,Xdc

d i n h cong thufc cau tao cua A ? B i e t rkng trong A cacbon c h i e m 8 0 % ve k h o i lifdng

V i san p h a m chay d a n qua CaCl2 k h a n , (do C a C l j giu" H2O) n e n the tich g i d m la

thi tic h c u a H2O

T h e o phuWng t r i n h the tich H2O > the tich CO2 j ,

T a c d : n + 1 - a > n => a < 1 => a = 0 ' '

= > C T T Q cua A : C„H2„+2 : ankan i r : v v

T h e o d e : % C = 1 0 0 % = 8 0 % =i> n = 2 ,^

14n + 2 ' ' • • • vay : C T P T cua A la C2H6; C T C T cua A : CH3-CH3: etan '

B^i 7- Tilf m e t a n , ngiTdi ta c d the d i e u che H2 (va C O ) theo h a i cdch sau : _ C d c h 1 : C H 4 + - O 2 " " " " ^ > C O + 2H2 ( H = 8 0 % )

_ C ^ c h 2 : CH4 + H2O ""J^"*^ > C O + 3H2 ( H = 7 5 % ) -i, ,

a) H d i each n a o t h u difdc n h i e u h i d r o hdn ? b) N 6 u theo each thu diTdc nhieu hidro hdn t h i tiif 10* l i t CH4 (0°C; 22,4 atm) c d the

d i e u che difdc bao n h i e u l i t H2 (dktc)

c) V i e t phuTdng trinh d i e u che metan tif n h o m cacbua, natri axetat, C v a H2, iH pentan

Phan d?ng phmng ph^p gi^i H6a hpc 11 HOu co - D5 Xuan Hang

e + 2 H 2 ^ C H 4

B a i 8 Dot chay hoan loan 30,6g hon hdp gom hai ankan d the khi sinh ra 92,4g C Q ^

a) Tinh tong so mol hai ankan

b) Tinh the tich o x i can de do't chay 10,2g hon hdp tren

c) Xac dmh cong thiJc phan tur hai ankan bie't the tich hai ankan trong hon hdp bang

nhau va d the k h i 1 t J i f , ,, ,^<) I ' >

i.'A,'^i^>% i J M f t ' Giai I ' , , )

'C„H2n+2 : a mol ^ • , C n , H 2 m + 2 : b mol '

Bai 9 MOt hdp chat hiJu cd X c6 % khoi luTdng cua C, H , C I Ian lUdt la 14,28%;

J 19%; 84,53% Hay ISp luan de tim cong thiJc phan tur cua X , viet cac cong thiJc ca'u tao c6 the cd cua X

Giai n. ' ' / » n : 4 < « W f ! ( " t U 4 ^ rt Dat C T T Q cua X : Q H y C l , " "

Bai 10 Dot chay ho^n to^n 0,384g hon hdp X gom hai ankan la d6ng d i n g ke tiep San pham sinh ra cho vao 520ml dung dich N a O H 0,1M thu di/dc m u o i trung

, hoa , itsx»;:ic::,

a) Xac dinh cong thiJc phan tuf m o i ankan :,fitdol;)Lx!vJ'.:iffi!-; ^

b) T i n h % khoi lifdng m o i ankan trong hon hdp dau f yfcii) i * ' ' ' '

Giai ^.i,.' ,.„v,4;,,,,,\

Dat C T T Q cua hai ankan dong dang ke tiep : C- H 2 - ^ 2 fjf,,

Phifdng trinh phan iJng :

PhSn dang va phudng phdp gi^i Hoa hpc 11 HQu c d - D5 XuSn Hi /ng

Thanh phan % k h ^ i lUdng m o i ankan :

^^"^ 0,384

I-=> %rnc^Hn =68,75% <'',M'

^ B A I T A P T R A C N G H I | : M

C&u 1 D6't chdy hoan toan mot m l u hidrocacbon X ngifdi ta thay the tich hdi niTdc

sinh ra gap 1,2 Ian the tich k h i cacbonic trong ciing mot dieu ki?n Bie't rang

hidrocacbon do chi tao thanh mpt dan xuat monoclo duy nhS't V a y X c6 ten la:

A 2,2-dimetylpropan B isobutan

C 2-metylbutan D etan

C a u 2 M o t xicloankan X c6 t i k h ^ i so v d i o x i bkng 2,625 B i e t k h i cho X tac dung

v d i clo ngo^i i n h s^ng thi chi tao di/dc dan xuaft monoclo duy nhat V $ y A Ik c6

A xiclohexan , B metylxiclopentan

C 1,2-dimetylxiclobutan """^ D l , 3 - < l i m e t y l x i c l o b u t a n ' ' ' ^ ' ^ - ^

CSu 3 Do't chay hoan to^n m6t Wdng hidrocacbon X Hap thu t o l n bp san

pham chay vao dung dich Ba(OH)2 (diT) tao ra 29,55 gam ket tua, dung dich

sau phan iJng c6 khoi Itfdng giam 19,35 gam so v d i dung dich Ba(0H)2 ban

dau Cong thtfc phan tuT cua X la

„ , ,; ' " Trich de thi tuyen sinh Dai hoc khoi A ndm 2010"

CSu 4 D6't c h i y ho^n t o i n 6,72 l i t (dktc) hon hdp g6m hai hidrocacbon X Y

( M Y > M x ) , thu diTdc 11,2 l i t CO2 (dktc) va 10,8 gam H2O Cong thiJc cua X IS

A C2H6 B C2H4 C CH4 D C2H2

" Trich de thi tuyen sinh Cao ddng ndm 2010"

CSu 5 So lien k e t a (xich ma) c6 trong m o i phan tijr: etilen; axetilen; buta-l,3-dien

l a n l i f d t m , • &gfW"' •••.•vc''

A 3; 5; 9 > toi r B 5 ; 3 ; 9 C 4; 2; 6 D 4; 3; 6

" Trich de thi tuyi'n sinh Cao ddng ndm 2010"

c a u 6 Khi crackinh ho^n to^n mOt the tich ankan X thu diTdc ba the tich hon hdp Y

(cdc the tich k h i do d chng dieu k i ^ n nhi?t dO va dp suat); t i khoi cua Y so vdi

H2 bkng 12 Cong thtfc phan tuf cua X la

A C 6 H , 4 B C j H x C.C4H,o D C5H ,2

" Trich de thi tuyi'n sinh Dai hoc liho'i A nam 2008"

Cflu 7- Cho iso-pentan ike dung vdi C12 theo ti 1$ so mol 1 : 1, so' san pham

monoclo t ^ i da thu difcJc la \nirrM iii\-^''f ':yki!<^ij ••!'X<^}. '^.^,.1 ^.w.,)

A 2 B 3 ; C 5 D 4

" Trich de thi tuyen sinh Dai hoc Ichoi A ndm 2008"

C&tt 8 Hidrocacbon mach hd X trong phan tuT chi chiJa lien ket a va c6 hai nguyen

tvl cacbon b$c ba trong mpt phan tuT D o t cMy hoan toan 1 the tich X sinh ra 6 thi tich CO2 id cung dieu kien nhiet dp, suat) Khi cho X tac dung vdi Clj

(theo ti 1# so m o l 1 : 1), so dan xuat monoclo tdi da sinh ra IS ^' '

A 3 B 4 -•.^ ••^^ C 2 ' D 5 '-^'^ '

H | | , Trich de thi tuyen sinh Dai hoc Ichoi B ndm 2008''

Cfiu 9 Cong thiJc ddn gian nha't cua mpt hidrocacbon la CnH2n+l- Hidrocacbon

dd thuQC day dong dang cua

A ankan B ankin C ankadien D anken

• " Trich de thi tuyen sinh Cao ddng ndm 2008"

c a u 10, Dot chay ho^n toan mpt hidrocacbon X thu difdc 0,11 mol CO2 va 0,132

mol H2O Khi X tac dung vdi khi clo (theo ti 1? so mol 1:1) thu duTdc mpt san

pham hOu cd duy nhat T e n goi cua X la , ^ , , •

A 2-Metylbutan B etan

C 2,2-Dimetylpropan D 2-Metylpropan

" Trich de thi tuyen sinh Cao ddng ndm 2008"

c a u 11 Khi brom hda mot ankan chi thu diTdc mgt dan xua't monobrom duy nhat c6

ti khoi hdi doi v d i hidro la 75,5 Ten cua ankan do la ' ,

A 3,3-dimetylhecxan B 2,2-dimetylpropan

C isopentan , , , , , D 2,2,3-trimetylpentan , ,

, ^,^ :, ,^ , " Trich di thi tuyen sinh Dai hoc khoi B ndm 2007"

cau 12 Dot chay hoan toan mpt the U'ch khi thien nhien gom metan, etan, propan bkng

oxi khong khi (trong khong khi, oxi chiem 20% the tich), thu diTdc 7,84 h't khi CO2 (cf

dktc) va 9,9 gam nifdc The U'ch khong khi (d dktc) nho nhat can dung de dot chay hoan toan lUdng khi thien nhien tren la:

A 70,0 lit B 78,4 lit C 84,0 lit D 56,0 lit

Cau 1 3 , K h i cho ankan X (trong phan tiJf cd phan trSm khoi li/dng cacbon b^ng

83,72%) iic dung v d i clo theo t i 1^ so mol 1:1 (trong dieu k i ^ n chieu sang) chi

thu dirdc 2 dan xuat monoclo dong phan cua nhau Ten cua X la ,

A 2-metylpropan B 2,3-dimetylbutan

C butan D 3-metylpentan '

69

Phan dgng phiiong ph^p giSi H6a hpc 11 HOu c o - D8 Xufln Himg

c a u 14 C h o 2,2g C j H s tdc d u n g \6i 3,55g C I 2 t h u diTdc 2 s5n p h a m the m o n o c l o X

va d i c l o Y v d i k h o i liTdng m x = l,3894mY Sau k h i cho h o n h d p k h i c 6 n l a i sau

p h a n tfng ( k h o n g chiJa X , Y ) d i qua d u n g d i c h N a O H dH, c o n l a i 0,448 l i t k h i

thoat ra (dktc) K h o i lUdng cua X , Y Ian lUdt l a : * i aSfirf* aril s f t iGj 1 4 * H i , i i

A l , 2 7 g v a l , 1 3 g - J B 1,13 g 1,27 g ' S: A

C. 1,13 g v a 1,57 g D 1,57 g v a 1.13 g

C&u 15 C h o m(g ) hidrocacbon A thuoc day d o n g d i n g cua m e t a n tac d u n g v d i clo

C O c h i e u sang, c h i thu dufdc m o t dan x u a t clo d u y nha't B v d i k h o i liTdng 8,52g

D e t r u n g h6a het k h i H C l sinh ra, can vuTa du 80 m l d u n g d i c h K O H I M C T P T

cua A , B Ian lufdt l a : ::^,i^t:^' j^btdma^i^^ I • ] kmr'^y? •)! 1) cMi)

A C s H n va C s H n C l ' ' , ^j: D ' " B C s H ^ v^ CsHjoClj

C C 4 H , , , va C4HyCl D C 4 H H , va C 4 H X C I 2

c a u 16 D o t chay hoan toan hon hdp 2 hidrocacbon dong d i n g lien tiep can 1,35 m o l O 2

tao t h a n h 0,8 m o l C O 2 C T P T cua 2 hidrocacbon l a :

^ • ^ ^ " - ^ ^ " ^ B C 2 H 2 C 3 H 4

C C H 4 , C 2 H 6 , , D C 2 H 6 , C j H s

C a u 17 D o t chay h o a n toan hon h d p 2 hidrocacbon (phan tuT k h o i h d n k e m nhau 14

d v C ) , t h u diTdc 5 m (g) C O 2 va 3 m (g) H 2 O C T P T cua 2 hidrocacbon t r e n l a :

A ankan , B anken C a n k i n D aren ,

C a u 21. K h i do't chdy h o a n toan cdc ankan t h i ta dUdc T = UQQ^ : U H J O • j

T b i e n d o i trong k h o a n g : !' i<i s i ^ 1 1 • - r

A 1 < ' T < 1 , 5 B 0 , 7 5 < T < 1 C 0 , 5 < T < 1 D 1 , 5 < T < 2 „;>

c a u 22 TrOn 2 the tich b^ng n h a u ciia CsHx va O 2 r o i d e m dot Sau k h i cho bay h d i

nufdc ngufng t u , dufa t r d v.e d i e u k i ? n ban d a u Mnihfu iF

N h a n x6t d u n g ve Vh<i„ h,.,p UM pa ( V j ) va Vh<',„

h(.(p sau p\

A V d : V , = 3 : l B V d < V s C V d > V , D V , : V , = 10 : 7

K i i A i g f l i n i g r

(^au 23 K h i dot c h a y hoan toan cijng m p t lUOng hidrocacbon X b a n g m p t liTdng vtfa

O 2 hoac C I 2 , n g i r d i ta thay t i 1§ the U'ch O 2 va C I 2 b^ng 1,25 ( x e t d c i i n g d i e u

k i $ n ) X l a : - - r , ,

A C 3 H 6 * B C J H H C C 4 H f i H D C 4 H S

c a u 24 Thifc h i e n p h a n tfng c r a c k i n h 11,2 l i t h d i isopentan (dktc), t h u du'dc hon hdp

A c h i g o m cac a n k a n va a n k e n T r o n g h o n h d p A c6 chiJa 7,2 g a m m o t cha't X

tnh k h i d o t chay t h i thu dufdc 11,2 li t C O 2 (dktc) va 10,8 g a m H 2 O H i e u suat

phan iJng c r a c k i n h isopentan l a :

A 3 3 , 3 3 % B 5 0 , 0 0 % C 6 6 , 6 7 % D 25,00%

" Trich de thi tuyen sinh Cao ddn^ nam 2012"

c a u 27 C h o butan qua x u c tac ( d n h i ^ t dp cao) thu diTdc h o n h d p X g o m C4H|(i,

C4Hg, C 4 H 6 , H 2 T i k h o i cua X so v d i butan la 0,4 N e u cho 0,6 m o l X v a o dung

djch b r o m (dxi) t h i so m o l b r o m t o i da phan uTng la

A 0,48 m o l B 0,36 m o l C 0,60 m o l D 0.24 m o l

i f c i f i "fit!, o r " ^"y^" '^'"^ ^'-'^ ^ "'•'"^ 2011"

c a u 28 K h i c r a c k i n h h o i n toan m p t the tich ankan X t h u difdc ba the tich h o n hdp

Y (cdc the t i c h k h i do d c i l n g d i e u ki§n n h i e t do va ap suat); t i k h o i cua Y so v d i H2 b^ng 12 C o n g thu-c p h a n tuf cua X l a :

A C 6 H , 4 B C 3 H 8 C C 4 H , o D C H u

c a u 29. N h i ^ t p h a n m e t a n t h u diTdc hon h d p X g o m C 2 H 2 , C H 4 va H 2 T i k h o i hdi

cOia X so v d i H 2 b k n g 5 H i ? u sua't qud trinh nhiQt phan l a : •

A 5 0 % V B. 6 0 % C. 7 0 % , , 5 D 80% ,

Cfiu 30 C r a c k i n h 0,1 m o l pentan diTdc h o n h d p X D o t c h a y het X r o i h a p thu san

p h a m c h a y b d i nufdc v o i trong dU' H d i k h o i li/dng d u n g d i c h c u o i c u n g t h u difdc tSng hay g i a m bao n h i e u gam?

A g i d m 17,2g B tang 32,8g C tSng 10,8g D tang 22g

Cau 31. C r a c k i n h V l i t butan thu difdc 35 l i t hon hdp A g o m H 2 , C H 4 , C 2 H 4 , e 3 H 6 ,

C 4 H 8 va 1 p h a n butan chtfa b i c r a c k i n h Cho hon h d p A l o i rat tiT tiT qua b i n h

dUng b r o m duT t h a y the t i c h c o n l a i 20 l i t % butan da p h a n iJng l a :

A 2 5 % B 6 0 % C 7 5 % D 8 5 % (iiiiiiii 71

Phan djina va phuang pMp g i i i H6a hgc 11 HOu co - Pi Xuan Hung

C a u 32 Nhi^t phan 8,8g CjHs ta thu diTdc hon hop khi A gom C H 4 , C 2 H 4 , C3H6, H ,

va C3H8 chUa bi nhi#t phan Biet c6 90% C s H g bi nhi^t phan T i khoi hOi cua hoti

h d p A s o v d i H z l a : ;

A 11,58 , B 15,58 C 11,85 D 18,55 ;

C a u 33 Hon hcJp khi gom etan propan c6 ti khoi so vdi hidro 1^ 20,25 diTdc nung

trong binh vdi cha't xiic tac de thifc hi?n phan ufng de hidro hda Sau mgt thdi

; gian thu difdc hon hdp khi c6 ti kho'i so vdi hidro la 16,2 gom cd cac ankan,

anken v^ hidro Tinh hieu sua't phan iJng de hidro hda biet rang toe dp phan iJng

ciia etan va propan Ih nhiT nhau^ , j

Cong thiJc cua X la: CnHjn = 84 => n = 6

=> Chi cd xiclohexan cho 1 dan xuat => Chon C

Cflu 6 Goi cong thiJc phan tuf cua ankan X la C„H2„ + 2

Trong C l i n g dieu kien ve nhi^t dp va ap suat thi ti I9 ve the tich cung chinh la t i 1^ ve so mol 1 mol X c r a c k i n h

D a p a n D

C H ,

73

Phan dgng va pnijcing ph^p giai H6a hpc 11 HQu ca - D5 Xuan Hung

C S u 9 T r o n g hidrocacbon so' n g u y e n luf H luon la so c h a n nen:

C T D G N cua hidrocacbon (CnH2n^i) => C T P T (C„H2„+,)2 hay C2nH2(2„+i) ' '^'v '

D a t 2 n = m => CTPT: Cn,H2m+2 => hidrocacbon d o thuQC d a y d o n g d a n g ankan

=> Vkhongkhi = ( 0 , 6 2 5 2 2 , 4 ) — = 7 0 l i t r : > D a p a n A w ) f = i i f

CSu 13 G o i c o n g thiJc p h a n tuT cua ankan X \h CnH2„ + 2

T a c d : % C = 1 0 0 % = 8 3 , 7 2 % n = 6 = > a n k a n X Ih C e H u

C S H M — " ^ " 2 > 2 d a n xu2ft m o n o c l o d o n g p h a n cua n h a u = > X c d 2 hifdng the

=> C T C T d u n g cua X : ' " ' ^ ^ftsaSnfi « l nodaipo-'^^c; :

C H 3 - C H - C H - C H 3 : 2,3 - d i m e t y l b u t a n : * r ! 5

C H 3 C H 3 j t > i

=> Dap an B • ;# ^eSCkf ^5;S;,>.i.*t'?l :i , ' ,.,:.'

Cau 14 T a c d : ncjHx bandJu = ^ = 0,05 m o l ; nci2 = ^ = 0,05 m o l

44

C j H s + C I 2 C 3 H 7 C I + H C l

7S

Phan dgng va phuong ph^p giSi H6a h(?c 11 HQu CO - Bfi Xufln HUng

b 2b b 2b

T h e o d e b ^ i : mx = l,3894mY <=> 78,5a = 1,3894.113b = 157b hay a = 2b (1)

Khi con lai di ra khoi dung dich NaOH dif la C 3 H 8 Knilx'sa i f pi? 7 ; • - :

riQMn dir = = 0,02 mol '

^ "C3H}j phan ilng = "C3H){ bandSu - "C3Hj{ dir = 0,05 - 0,02 = 0,03 Hiol

o 7n + 1 = 36x =i> nghi$m phQ hdp m x = 1 va n = 5 ^ mr'xiy'f" ™'

o CT cua A B Ian lifdt la: CsHu C 5 H 1 1 C I

^ Dap an A siifaA*?^*;i»#4op:^^

C a u l 6 T a c o : nojphantfng = nco2 + - n H 2 0 ; T ; i f I , ,

• ^ n H 2 0 = 2(no2 phan J n g - nco2 ) = 2(1,35 - 0,8) = 1,1 > nco2 = 0.8

=> 2 hidrocacbon la ankan -•-'^••^ ' ':l !>>kyr:^&rTSr:)

Goi cong thut long qudt cua A m C - H 2 - ^ 2 ^ ' ' " ' ' ^ '

Cau 18 Cong thiJc chung cua hon hdp A la CnH2„ + 2: a mol

Cong thiJc chung cua hon hdp B la Cn,H2m: b mol / ' C„H2n + 2 n C 0 2 + ( n + l ) H 2 0 ,+-0 4 /- ^

5-=i> n H j o - n c 0 2 = a = 0,2 mol => b = 0.35 - 0.2 = 0,15 mol=> Dap an C

Cflu 19 Goi c6ng tht?c chung cua hon hdp X la C - H j - ^ j

11 Hau CO - P5 Xuan Hung

= > V h h khi sau phan tfng = V ^ Q , + H„ dlT = — " + (a - - ) = 1,4a l i t

Vj^u = a + a = 2a l i t j^^;;,, /.js^,,;. ^ f i r j , ; , [ ; • : ; smuiu 'XXtiil :ssn63 J:|,if

= > V s a u : Vj^u = 1.4a : 2a = 0.7 hay V j ^ u : V^au = 10 : 7 D a p d n D gn&O

C f l u 2 3 Q H y + ( x + ^ ) 0 2 - > X C O 2 + ^ H 2 0 f

a a ( x + -^) • o^fiirt-rV ^ ODK I ,„^H,„:'*

y (ilai • o m • " d

CxHy + — C I 2 ^ x C + y H C l (J 4 (.,^.j.,fij s.;n j.(,->n ;6o B'T

: ' , ; \ i ^ n & f ; - ^ 'Ti?:r,() = £;tf^ ':?!t,Oacl <fc l o r n £,:().= £ ;'^j03*' O j k "

Dt/a vko sd <i6 ta th^y: C4H10 bi de hidro h6a ra hSn h d p hidrocacbon khdng no

yk H2, h5n hdp hidrocacbon khong no cgng dung dich brom lai tao ra hon h d p

s i n ph^m no

_ ff «'i-:.,;.( k,> • ' •

—^ " B l 2 p h i n iJng - sinh ra

Trong p h i n »?ng 6i hidro h6a s6 mol hon h d p sau phan tfng tang len chinh 1^ so

mol H2 tao ra

V § y : = 1 - 0.4 = 0,6 mol = H B r j phin iTng

T a c 6 : 1 mol hSn hdp X can 0,6 mol Br2

V $ y 0.6 mol h6n hqJp X c i n 0.36 mol Br2

79