hoa_hoc_tlbg-thu_thuat_huu_co

8/6/2019 Hoa_hoc_TLBG-Thu_thuat_Huu_co

1/3

Kha hcLTH m bo mn HaThy Sn Th thut lm bi trc nghim phn ha hu c

Hocmai.vnNgi trng chung ca hc tr Vit Tng i t vn: 1900 58-58-12 - Trang | 1 -

BI 2. TH THUT LM BI TRC NGHIMHU CTI LIU BI GING

Cu 1: Cho hn hp kh X gm HCHO v H2i qua ng sng bt Ni nung nng. Sau khiphn ng xy ra hon

ton, thu c hn hp kh Y gm hai cht hu c. t chy ht Y th thu c 11,7 gam H2O v 7,84 lt kh CO2(

ktc). Phn trm theo th tch ca H2trong X l

A. 46,15%. B. 35,00%. C. 53,85%. D. 65,00%.

Hng dn: Khi t chy Y (gm CH3OH v HCHO d) cng chnh l t chy X (HCHO v H2), khi t chyHCHO th s mol CO2 = H2O. Do s mol H2c tnh bng cngthc :

2 2 2H H O CO11,7 7,84

n n n 0,3(mol)18 22, 4

.

Theo phn ng t chy th2HCHO CO

n n 0,35 mol .

2H0,3

%V .100 46,15%0,35 0,3

Cu 2: Cho hn hp X gm hai axit cacboxylic no, mch khng phn nhnh. t chy honton 0,3 mol hn hp X, thu c 11,2 lt kh CO2 (ktc). Nu trung ha 0,3 mol X th cndng 500 ml dung dch NaOH 1M. Hai axit l

A. HCOOH, HOOC-COOH. B. HCOOH, HOOC-CH2-COOH.

C. HCOOH, C2H5COOH. D. HCOOH, CH3COOH.

Hng dn: S nguyn t cacbon trung bnh ca hn hp hai axit l :

2CO

hh

n 0,5n 1,67

n 0,3. Vy trong X c HCOOH.

T l mol X : NaOH = 0,3 : 0,5 = 3 : 5 vy X gm mt axit n chc v mt axit hai chc R(COOH)2. Theo phn ng vi

NaOH tnh c s mol cc cht HCOOH = 0,1 mol v R(COOH)2 = 0,2 mol0,1.1 0,2.nn 1,67 n 2

0,3: HOOC-COOH.

Cu 3: t chy 1,6 gam mt este E n chc c 3,52g CO2 v 1,152g H2O. Nu cho 10 gam E tc dng vi150ml dung dch NaOH 1M, c cn dung dch sau phn ng thu c 16 gam cht rn khan. Vy cng thc ca axitto nn este trn c th l

A. CH2=CH-COOH. B. CH2=C(CH3)-COOH.C. HOOC[CH2]3CH2OH. D. HOOC-CH2-CH(OH)-CH2CH3.

Hng dn:S mol CO2 = 0,08; H2O = 0,064 Vy este ko no.S mol este = 0,08 0,064 = 0,016. S nguyn t C = 0,08/0,016 = 5: C5H8O2

mE + mNaOH = mcht rn: chng t este vng.Cu 4:t chy hon ton m gam hn hp 3 ancol n chc, thuc cng dy ng ng, thu c 3,808 lt kh CO2(ktc) v 5,4 gam H2O. Gi tr ca m l

A. 5,42. B. 5,72. C. 4,72 D. 7,42.

Hng dn:

S mol ancol = H2O - CO2 = 0,3 - 0,17 = 0,13 molm = 12.0,17 + 2.0,3 + 16.0,13 = 4,72

Cu 15Hn hp kh X gm mt ankan v mt anken. T khi ca X so vi H 2bng 11,25. t chy hon ton 4,48

lt X, thu c 6,72 lt CO2(cc th tch kh o ktc). Cng thc ca ankan v anken ln lt l A. CH4 v C2H4 B. C2H6 v C2H4 C. CH4 v C3H6. D. CH4 v C4H8.

Hng dn:

S nguyn t C trung bnh n = 1,5. M = 22,5. Anken khng th l C2H4.

Nu anken l C3H6th vi n = 1,5 ta tnh c s mol CH4 = 3.C3H6. Khi M = 22,4. Ph hp vi bi.

Cu 6:Cho hn hp M gm anehit X (no, n chc, mch h) v hirocacbon Y, c tng s mol l 0,2 (s mol ca


2/3



X nh hn ca Y). t chy hon ton M, thu c 8,96 lt kh CO2(ktc) v 7,2 gam H2O. Hirocacbon Y lA. CH4. B. C2H2. C. C3H6. D. C2H4.

Hng dn:S mol CO2 = 0,4; H2O = 0,4. Vy Y l anken. S nguyn t C trung bnh = 2. S mol Y X th s nguyn t C 2.Ch c th l C3H6.

Cu 7:t chy hon ton mt lng hn hp X gm 2 ancol (u no, a chc, mch h, c cng s nhm -OH)

cn va V lt kh O2, thu c 11,2 lt kh CO2 v 12,6 gam H2O (cc th tch kh o ktc). Gi tr ca V l A. 14,56. B. 15,68 C. 11,20. D. 4,48.

Bo ton mol nguyn t:

2 2 2ancol O CO H Oa.n 2n 2.n n

2 2ancol H O COn n n 0,7 0,5 0,2

S nguyn t C = 2,5 vy phi l 2 chc. a = 2 Tnh c s mol O2 = 0,5 + 0,350,2 = 0,65. V = 14,56

Cu 8:t chy hon ton 6,72 lt (ktc) hn hp gm hai hirocacbon X v Y (MY > MX), thuc 11,2 lt kh CO2(ktc) v 10,8 gam H2O. Cng thc ca X l

A. C2H4. B. CH4. C. C2H6. D. C2H2.

Cu 9: Hn hp kh X gm anken M v ankin N c cng s nguyn t cacbon trong phn t. Hn hp X c khilng 12,4 gam v th tch 6,72 lt (ktc). S mol, cng thc phn tca M v N ln lt l

A. 0,1 mol C2H4; 0,2 mol C2H2. B. 0,2 mol C2H4; 0,1 mol C2H2.

C. 0,1 mol C3H6; 0,2 mol C3H4. D. 0,2 mol C3H6; 0,1 mol C3H4.

Hng dn: Gi cng thc chung ca M v N l x yC H :

12x + y =12,4

41,330,3

x 3, y 5, 3 ; hai cht l C3H6 v C3H4.

Ta thy y 5, 3 nn s mol C3H6 > C3H4. Vy phng n D l hp l.

Cu 10: Cho 0,1 mol hn hp X gm hai anehit no, n chc, mch h, k tip nhau trong dy ng ng tc dng vi lngd dung dch AgNO3 trong NH3, un nng thu c 32,4 gam Ag. Hai anehit trong X l

A. CH3CHO v C2H5CHO. B. CH3CHO v C3H5CHO.C.HCHO v CH3CHO. D. HCHO v C2H5CHO.

T lmol Ag : anehit = 3. Vy c HCHO v CH3CHO

Cu 11: Hnhp kh X gm H2 v mt anken c khnng cng HBr cho snphm hucduy nht. Tkhi ca

X so vi H2 bng 9,1. un nng X c xc tc Ni, sau khi phn ng xy ra hon ton, thu c hn hp kh Y

khng lm mt mu nc brom; tkhi ca Y so vi H2 bng 13. Cng thccu to ca anken l

A. CH2

=C(CH3

)2

. B. CH2

=CH2

.

C. CH2=CHCH2CH3. D. CH3CH=CHCH3.

Hng dn:

262.13MY , chng t trong Y cn H2d.

CnH2n + H2Ni

CnH2n+2

Trc phn ng X gm CnH2n (x mol) v H2 (y mol) ; n1 = x + ySau phn ng Y gm C2nH2n+2 (x mol) v H2d (y x mol), n2 = x + y x = y mol.

7,013

1,9

yx

yhay

n

n

M

McTa.

n

mM;

n

mM

1

2

2

1

22

11 hay 7x = 3y.

Chn x = 3, y = 7 ta c :

84HC:56M2.1,910

2.7M.3M


3/3



C4H8tc dng vi HBr cho mt sn phm nn c CTCT CH 3CH=CHCH3.

Cu 12: Cho hirocacbon X phnng vi brom (trong dung dch) theo t l mol 1 : 1, thu c cht hu cY(cha 74,08% Br vkhi lng). Khi Xphnng vi HBr th thu c hai sn phm hu c khc nhau. Tn gica X l

A.but1en. B. xiclopropan. C.but2en. D. propilen.

Hng dn: Gi cng thc ca sn phm l RBr2 th

84Br HC:5608,74100.160R

160m%

Khi tc dng vi HBr, X cho hai sn phm khc nhau, vy X l but1en.

Cu 13: (2010) Cho 0,15 mol H2NC3H5(COOH)2(axit glutamic) vo 175 ml dung dch HCl 2M, thu c dung dch

X. Cho NaOH d vo dung dch X. Sau khi cc phn ng xy ra hon ton, s mol NaOH phn ng lA. 0,70. B. 0,50. C. 0,65 D. 0,55

Hng dn:S mol NaOH phn ng = 2.glu + HCl = 2.0.15 + 0,175.2 = 0,65.

Cu 11:Cho 13,35 g hn hp X gm CH2NH2CH2COOH v CH3CHNH2COOH tc dng vi V ml dung dchNaOH 1M thu c dung dch Y. Bit dung dch Y tc dng va vi 250 ml dung dch HCl 1M. Gi tr ca V l

A. 100 ml B. 150 ml C. 200 ml D. 250 ml

Hng dn :S mol X + NaOH = HCl13,53/89 + 0,01.V = 0,25=>V = 100ml

Gio vin: Phm Ngc Sn

Ngun: Hocmai.vn
http://hocmai.vn/http://hocmai.vn/http://hocmai.vn/

hoa_hoc_tlbg-thu_thuat_huu_co

Documents