Pp13 Khao Sat Ti Le So Mol Co2 Va h2o

Phng php 13: Kho st t l s mol CO2 v H2O

1

Ph-ng php kho st t l s mol CO2 v H2O I. C S CA PHNG PHP

Cc hp cht hu c khi t chy thng cho sn phm CO2 v H2O. Da vo t l c bit ca

2

2

CO

H O

n

n hoc 2

2

CO

H O

V

V trong cc bi ton t chy xc nh dy ng ng, cng thc phn t hoc tnh

ton lng cht.

1. Vi hydrocacbon

Gi cng thc tng qut ca hidrocacbon l CnH2n+2-2k (k: Tng s lin kt v vng)

CnH2n+2-2k + 3n 1 k

2

O2 nCO2 + (n + 1 k) H2O

Ta c: 2

2

H O

CO

n n 1 k 1 k1

n n n

2

2

H O

CO

n1

n (

2 2H O COn n ) k = 0 hyrocacbon l ankan (paraffin)

Cng thc tng qut l CnH2n+2

2

2

H O

CO

n1

n (

2 2H O COn n ) k = 1

hyrocacbon l anken (olefin) hoc xicliankan Cng thc tng qut l CnH2n

2

2

H O

CO

n1

n k < 1 hyrocacbon c tng s lin kt v vng 2

* Mt s ch :

a, Vi ankan (paraffin): ankann = 2H On - 2COn

b, Vi ankin hoc ankaien): ankinn = 2COn - 2H On

1. Vi cc hp cht c cha nhm chc

a, Ancol, ete

Gi cng thc ca ancol l : CnH2n + 2 2k m(OH)m hay CmH2n + 2 2kOm

CnH2n+2-2kOm + 3n 1 k m

2

O2 nCO2 + (n + 1 k) H2O

2

2

H O

CO

n n 1 k 1 k1 1

n n n

khi v ch khi k = 0


2

Ancol no, mch h, c cng thc tng qut CnH2n+2Om v ancoln = 2H On - 2COn

b, Anhit, xeton

Gi cng thc ca anehit l : CnH2n + 2 2k m(CHO)m

Ta c phng trnh t chy

CnH2n + 2 2k m(CHO)m + 3n 1 k m

2

O2 (n + m)CO2 + (n + 1 k)H2O

2

2

H O

CO

n n 1 k n 1 k

n n m n m n m

2

2

H O

CO

n1

n (

2 2H O COn n ) khi v ch khi k = 0 v m = 1 anehit no, n chc, mch h, cng

thc tng qut l: CnH2n + 1CHO hay CxH2xO (x 1)

Tng t ta c: 2

2

H O

CO

n1

n (

2 2H O COn n ) xeton no, n chc, mch h

c, Axit, este

Gi cng thc ca axit l: CnH2n + 2 2k m(COOH)m

Ta c phng trnh t chy

CnH2n + 2 2k m(COOH)m + 3n 1 k

2

O2 (n + m)CO2 + (n + 1 k)H2O

2

2

H O

CO

n n 1 k n 1 k

n n m n m n m

2

2

H O

CO

n1

n (

2 2H O COn n ) khi v ch khi k = 0 v m = 1 axit no, n chc, mch h, cng thc

tng qut l: CnH2n + 1COOH hay CxH2xO2 (x 1)

Nhn thy: Cng thc tng qut ca axit v este trng nhau, nn: 2

2

H O

CO

n1

n (

2 2H O COn n ) este

no, n chc, mch h, c cng thc tng qut l: CnH2n + 1COOH hay CxH2xO2 (x 2)


3

II. CC DNG BI TP THNG GP

Dng 1: Kho st t l s mol H2O v CO2 cho tng loi hirocacbon:

V d 1. t chy hon ton mt hirocacbon X thu c 0,11 mol CO2 v 0,132 mol H2O. Khi X tc

dng vi kh Clo (theo t l s mol 1 : l) thu c mt sn phm hu c duy nht. Tn gi ca X l

A. 2-metylbutan B. 2-metylpropan

C. 2,2-imetylpropan D. etan

Gii:

OH2n > 2COn

X l ankan, c cng thc tng qut CnH2n+2.

nankan = OH2n - 2COn = 0,022 mol

S nguyn t cacbon = 125HC5

0,022

0,11

Mt khc, do tc dng vi kh Clo (theo t l s mol 1 : 1) thu c mt sn phm hu c duy nht nn

cng thc cu to ca X l :

CH3

CH3 C CH3

CH3

p n C.

V d 2. t chy hon ton hirocacbon mch h X bng O2 va . Dn ton b sn phm chy qua

bnh ng H2SO4 c d, th th tch sn phm gim i mt na. X thuc dy ng ng

A. anken. B. ankan.

C. ankin. D. xicloankan.

Gii:

Sn phm chy l CO2 v H2O, khi i bnh ng H2SO4 c d, th th tch sn phm gim i mt na,

2CO

V OH2V

X l anken hoc xicloankan.

Do X c mch h

X l anken

p n A


4

V d 3: Chia hn hp 2 ankin thnh 2 phn bng nhau:

-Phn 1: em t chy hon ton thu c 1,76 gam CO2 v 0,54 gam H2O.

-Phn 2: Dn qua dung dch Br2 d. Khi lng Br2 phn ng l:

A. 2,8 gam B. 3,2 gam C. 6,4 gam D. 1,4 gam

Gii:

S mol ankin = 2CO

n - OH2n =1,76 : 44 0,54 : 18 = 0,01 mol.

S mol Br2 phn ng = 2nankin = 0,02 mol.

Khi lng Br2 phn ng = 0,02.160 = 3,2 gam

p n B.

Dng 2: Kho st t l s mol H2O v CO2 cho hn hp hirocacbon:

V d 4. t chy hon ton 2,24 lt (ktc) hn hp kh X gm: CH4, C2H4, C2H6, C3H8 v C4H10 thu

c 6,16 gam CO2 v 4,14 gam H2O. S mol C2H4 trong hn hp X l

A. 0,09. B. 0,01. C. 0,08. D. 0,02.

Gii:

Hn hp kh X gm anken (C2H4) v cc ankan, khi t chy ring tng loi hidrocacbon, ta c:

Ankan: OH2n - 2COn = nankan

Anken: OH2n - 2COn = 0

S mol Ankan (X) = OH2n - 2COn = 4,14 : 18 - 6,16 : 44 = 0,09 mol

S mol C2H4 = nX nankan = 2,24 : 22,4 0,09 = 0,01

p n B.

Dng 3: Kho st t l s mol H2O v CO2 cho tng loi dn xut hirocacbon:

V d 5. t chy hon ton m gam mt ru X thu c 1,344 lt CO2 (ktc) v 1,44 gam H2O. X tc

dng vi Na d cho kh H2 c s mol bng s mol ca X. Cng thc phn t ca X v gi tr m ln lt l

A. C3H8O2 v 1,52. B. C4H10O2 v 7,28.

C. C3H8O2 v 7,28. D. C3H8O3 v 1,52.

Gii:

Ta c: 2CO

n = 1,344 : 22,4 = 0,06 mol; OH2n = 1,44 : 18 = 0,08 mol.

2CO

n < OH2n


5

X l ru no, c cng thc tng qut CnH2n+2Om

nX = OH2n - 2COn = 0,02

S nguyn t cacbon = 302,0

06,0

n

n

X

CO2

V s mol kh H2 thu c bng ca X X cha 2 nhm -OH

Cng thc phn t: C3H8O2 v m = 0,02. 76 = 1,52 gam

p n A.

V d 6. Hn hp X gm 2 cht hu c thuc cng dy ng ng. Phn t ca chng ch c mt loi

nhm chc. Chia X lm 2 phn bng nhau.

- Phn 1: em t chy hon ton ri cho ton b sn phm chy (ch c CO2 v hi H2O) ln lt qua

bnh (l) ng dung dch H2SO4 c, bnh (2) ng dung dch Ca(OH)2 d, thy khi lng bnh (l) tng

2,16 gam, bnh (2) c 7 gam kt ta.

- Phn 2: Cho tc dng ht vi Na d th th tch kh H2(ktc) thu c l bao nhiu?

A. 2,24 lt. B. 0,224 lt. C. 0,56 lt. D. 1,12 lt

Gii:

V X tc dng vi Na gii phng H2 X l ru hoc axit.

OH2n = 0,12 > OH2n = 0,07

X gm 2 ru no. t cng thc tng qut 2 ru l CnH2n+2Om

nX = OH2n - 2COn = 0,05 mol

S nguyn t cacbon = 1,40,05

0,07

n

n

X

CO2

Ru th nht l: CH3OH

X l 2 ru no n chc

mol 0,025n2

1n XH2

V = 0,56 lt

p n C.


6

V d 7. t chy hon ton 1,46 gam hn hp 2 anehit mch h ng ng k tip thu c l,568 lt

CO2 (ktc) v 1,26 gam H2O. Cng thc phn t ca hai anehit ln lt l

A. HCHO v CH3CHO B. CH3CHO v C2H5CHO

C. C2H5CHO v C3H7CHO D. C2H4CHO v C3H6CHO

Gii:

Ta thy:

2COn = 1,568 : 22,4 = 0,07 mol.

OH2n = 1,26 : 18 = 0,07 mol.

V 2COn : OH2n = 1 : 1 nn 2 andehit l no n chc mch h.

Gi cng thc chung ca 2 andehit l CHOHC 1n2n

O1)Hn(1)COn(O

2

1n3 CHOHC 2221n2n

a (n+1)a (n+1)a

Do : 4/3n

0,071)na(

1,4630)na(14

p n B.

V d 8. t chy hon ton 0,1 mol hn hp X gm hai cht hu c ng ng lin tip, thu c

3,36 lt CO2 (ktc) v 2.7 gam H2O. S mol ca mi axit ln lt l:

A. 0,04 v 0,06. B. 0,08 v 0,02.

C. 0,05 v 0,05. D. 0,045 v 0,055.

Gii:

(mol) 0,1522,4

3,36n

18

2,7n

22 COOH X l hn hp hai axit no, n chc, mch h, c cng thc

tng qut CnH2nO2

S nguyn t cacbon trung bnh 5,11,0

15,0

n

n

X

CO2 hai axit ln lt l HCOOH (a mol) v

CH3COOH (b mol)

0,05molba

0,152ba

0,1ba p n C

V d 9. t chy hon ton m gam hn hp X gm cc este no, n chc, mch h. Sn phm chy

c dn vo bnh ng dung dch nc vi trong thy khi lng bnh tng 6,2 gam. S mol CO2 v

H2O sinh ra ln lt l


7

A. 0,1 v 0,1. B. 0,01 v 0,1.

C. 0,1 v 0,01. D. 0,01 v 0,01.

Gii:

Khi lng bnh ng dung dch nc vi trung tng = OH2m + 2COm

Mt khc X l hn hp este no, n chc, mch h 2CO

n = OH2n = x

x(44+18) = 6,2 x = 0,1 p n A.

Dng 4: Kho st t l s mol H2O v CO2 cho hn hp dn xut hirocacbon

V d 10. Hn hp X gm cc axit hu c no, n chc, mch h v este no, n chc, mch h. t

chy hon ton m gam hn hp X bng mt lng oxi va . Ton b sn phm chy c dn chm

qua dung dch H2SO4 c d thy khi lng bnh ng axit tng m gam v c 13,44 lt kh (ktc) thot

ra. Gi tr ca m l

A. 5,4 gam B. 7,2 gam. C. 10.8 gam. D. 14,4 gam.

Gii:

- Sn phm chy gm CO2 v H2O khi lng bnh ng dung dch H2SO4 c tng chnh l khi

lng ca H2O b gi li

- V X gm cc axit hu c no, n chc, mch h v este no, n chc, mch h.

X c cng thc tng qut chung l CnH2nO2 v 2CO

n = OH2n = 13,44 : 22,4 = 0,6 mol

m= 0,6. 18 - 10,8 gam p n C.

V d 11: Chia m gam X gm : CH3CHO, CH3COOH v CH3COOCH3 thnh hai phn bng nhau :

- t chy hon ton phn 1 cn ti thiu 5,04 lt O2 (ktc), thu c 5,4 gam H2O.

- Cho phn 2 tc dng ht vi H2 d (Ni, to ) c hn hp Y. t chy hon ton hn hp Y, thu c V

lt CO2 (ktc).

Gi tr ca m v V ln lt l

A. 22,8 v 1,12. B. 22,8 v 6,72.

C. 11,4 v 16,8. D. 11,4 v 6,72.

Gii:

- 3 cht trong X u l no, n chc, mch h, cng thc tng qut : CnH2nOm

Khi t chy: 2CO

n = OH2n = 5,4 : 18 = 0,3 mol

lt 6,72 22,4 . 0,3V2CO

X + O2 CO2 + H2O


8

p dng nh lut bo ton khi lng:

mX (mt phn) = 0,3(44 + 18) 5,04 : 22,4. 32 = 11,4 gam

mX = 22,8 gam

p n B.

Dng 5: Kt hp kho st t l v mi lin h gia cc hp cht

V d 12. Cho hn hp X gm hai anehit l ng ng k tip tc dng ht vi H2 d (Ni, to) thu c

hn hp hai ancol n chc. t chy hon ton hn hp hai ancol ny thu c 11 gam CO2 v 6,3 gam

H2O. Cng thc ca hai anehit l

A. C2H3CHO, C3H5CHO B. C2H5CHO, C3H7CHO

C. C3H5CHO, C4H7CHO D. CH3CHO, C2H5CHO

Gii:

Khi t chy ancol cho 0,3518

6,3n

2CO25,0

44

11n OH2

2 ru l no, mch h

nX = OH2n - 2COn = 0,1 S nguyn t cacbon = 5,21,0

25,0

n

n

X

CO2

hai ru l C2H5OH v C3H7OH hai anehit tng ng l CH3CHO v C2H5CHO p n D.

V d 13. Hn hp X gm CH3COOH v C3H7OH vi t l mi 1: l. Chia X thnh hai phn:

- t chy hon ton phn 1 thu c 2,24 lt kh CO2 (ktc).

- em este ho hon ton phn 2 thu c este Y (gi s hiu sut phn ng t 100%). t chy hon

ton Y th khi lng nc thu c l

A. 1,8 gam. B. 2,7 gam. C. 3,6 gam. D. 0,9 gam.

Gii:

CH3COOH + C2H5OH CH3COOC2H5 + H2O

- Tng s mol cacbon trong hn hp X bng tng s mol cacbon c trong Y (Xem thm phng php bo

ton nguyn t)

Mt khc Y l este no, n chc, mch h, nn:

khi t chy OH2n = 2COn = 2,24 : 22,4 = 0,1 mol OH2m = 1,8 gam

p n A.


9

III. BI TP T LUYN

Cu 1 : t chy hai hirocacbon l ng ng lin tip c nhau ta thu c 7,02 gam H2O v 10,56 gam

CO2. Cng thc phn t ca hai hirocacbon l

A. C2H4 v C3H6 B. CH4 v C2H6.

C. C2H6 v C3H8 D. C2H2 v C3H4

Cu 2 : t chy hon ton mt hirocacbon X bng mt lng va oxi. Dn hn hp sn phm chy

qua H2SO4 c th th tch kh gim hn mt na. X thuc dy ng ng

A. ankan. B. anken. C. ankin. D. ankaien.

Cu 3 : t chy hon ton hn hp hai hirocacbon c khi lng phn t hn km nhau 28 vC thu

c 4,48 l CO2 (ktc) v 5,4 gam H2O. Cng thc phn t ca hai hirocacbon ln lt l

A. C2H4 v C4H8 B. C2H2 v C4H6

C. C3H4 v C5H8. D. CH4 v C3H8.

Cu 4 : t chy hon ton hn hp X gm 2 hirocacbon mch h thu c 16,8 lt kh CO2 (ktc) v

8,1 gam H2O. Hai hirocacbon trong hn hp X thuc dy ng ng no di y ?

A. ankaien. B. ankin. C. aren. D. ankan.

Cu 5 : t chy hon ton m gam hn hp gm mt ankan v mt anken. Cho sn phm chy ln lt i

qua bnh (l) ng P2O5 d v bnh (2) ng KOH rn, d, sau th nghim thy khi lng bnh (1) tng

4,14 gam, bnh (2) tng 6,16 gam. S mol ankan c trong hn hp l

A. 0,06 mol. B. 0,09 mol. C. 0,03 mol. D. 0,045 mol.

Cu 6 : Chia hn hp X gm hai ru n chc, thuc cng dy ng ng thnh hai phn bng nhau:

- Phn 1 : em t chy hon ton thu c 2,24 lt CO2 (ktc)

- Phn 2 : Thc hin phn ng tch nc hon ton vi H2SO4 c, 180oC thu c hn hp Y gm

hai anken. t chy hon ton hn hp Y ri cho ton b sn phm chy i chm qua bnh ng dung

dch nc vi trng d, kt thc th nghim thy khi lng bnh tng ln m gam. Gi tr ca m l

A. 4,4. B. 1,8. C. 6,2. D. 10.

Cu 7 : t chy hon ton hn hp hai ru n chc k tip trong dy ng ng thu c CO2 v hi

nc c t l th tch 2CO

V : OH2V = 7 : 10. Cng thc phn t ca hai ru ln rt l

A. CH3OH v C2H5OH. B. C3H7OH v C4H9OH

C. C2H5OH v C3H7OH D. C3H5OH v C4H7OH.

Cu 8 : Khi thc hin phn ng tnh nc i vi ancol X, ch thu c mt anken duy nht. Oxi ho

hon ton mt lng cht X thu c 5,6 lt CO2 (ktc) v 5,4 gam H2O. S cng thc cu to ph hp

vi X l

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


10

Cu 9 : t chy hon ton m gam hn hp hai ancol n chc, thuc cng dy ng ng thu c

70,4 gam CO2 v 39,6 gam H2O. Gi tr ca m l

A. 3,32 gam B. 33,2 gam. C. 16,6 gam. D. 24,9 gam.

Cu 10 : t chy hon ton mt ru X thu c CO2 v H2O c t l s mol tng ng l 3 : 4. Th

tch O2 cn dng t chy X bng 1,5 ln th tch kh CO2 thu c ( cng iu kin). Cng thc

phn t ca X l

A. C3H8O. B. C3H8O3 C. C3H4O. D. C3H8O2

Cu 11 : Hn hp M gm 2 cht hu c X, Y k tip nhau trong cng dy ng ng phn t ca chng

ch c mt loi nhm chc. t chy hon ton 1,29 gam hn hp M, cho ton b sn phm chy (ch c

CO2 v H2O) vo bnh nc vi trong d thy khi lng bnh tng 4,17 gam v to ra 6,0 gam cht kt

ta. Cng thc cu to ca X, Y ln lt l

A. C2H5OH v C3H7OH. B. CH3COOH v C2H5COOH.

C. CH3CHO v C2H5CHO. D. C2H4(OH)2 v C3H6(OH)2

Cu 12 : Tch nc hon ton t hn hp X gm 2 ancol thu c c hn hp Y gm cc olefin. Nu

t chy hon ton X th thu c 1,76 gam CO2. Khi t chy hon ton Y th tng khi lng nc v

CO2 sinh ra l

A. 1,76 gam. B. 2,76 gam. C. 2,48 gam. D. 2,94 gam.

Cu 13 : t chy hon ton hn hp M gm hai ru (ancol) X v Y l ng ng k tip ca nhau, thu

c 0,3 mol CO2 v 0,425 mol H2O. Mt khc, cho 0,25 mol hn hp M tc dng vi Na (d) thu c

cha n 0,15 mol H2. Cng thc phn t ca X, Y ln lt l

A. C3H6O, C4H8O B. C2H6O, C3H8O

C. C2H6O2, C3H8O2 D. C2H6O, CH4O

Cu 14 : Khi t chy hon ton anehit no, n chc, mch h bng oxi th t l sn phm chy thu

c l

A. 1n

n

2

2

CO

OH B. 1

n

n

2

2

CO

OH C. 1

n

n

2

2

CO

OH D.

2

1

n

n

2

2

CO

OH

Cu 15 : t chy hn hp X gm cc ng ng ca anehit, thu c s mol CO2 bng s mol H2O. X

l dy ng ng ca

A. anehit no, n chc, mch h.

B. anehit no, n chc, mch vng.

C. anehit hai chc no, mch h.

D. anehit cha no (c mt lin kt i), n chc.


11

Cu 16 : t chy hon ton m gam hn hp X gm hai anehit no, n chc, mch h thu c

0,4 mol CO2. Mt khc hiro ho hon ton m gam X cn va 0,2 mol H2 (Ni, to), sau phn ng thu

c hn hp hai ancol. t chy hon ton hn hn hai ancol ny th s mol H2O thu c l bao nhiu

A. 0,3 mol B. 0,4 mol C. 0,6 mol D. 0,8 mol

Cu 17 : t chy hon ton 0,44 gam mt axit hu c, sn phm chy c hp thu hon ton vo bnh

1 ng P2O5 v bnh 2 ng dung dch KOH. Sau th nghim thy khi lng bnh 1 tng 0,36 gam v

bnh 2 tng 0,88 gam. Xc nh cng thc phn t ca axit.

A. C2H4O2 B. C3H6O2 C. C5H10O2 D. C4H8O2

Cu 18 : t chy hon ton hn hp 2 axit cacboxylic thu c 3,36 lt CO2 (ktc) v 2,7 gam H2O. Hai

axit trn thuc loi no trong nhng loi sau ?

A. No, n chc, mch h B. Khng no, n chc

C. No, a chc D. Thm, n chc.

Cu 19 : t chy hon ton 1,76 gam mt axit hu c X mch thng c 1,792 lt kh CO2 (ktc) v

1,44 gam H2O. Cng thc cu to ca X l

A. CH3CH2CH2COOH. B. C2H5COOH.

C. CH3CH=CHCOOH. D. HOOCCH2COOH.

Cu 20 : t chy hon ton mt lng hn hp 2 este, cho sn phm phn ng nhy qua bnh ng P2O5

d, khi lng bnh tng thm 6,21 gam, sau cho qua tip dung dch Ca(OH)2 d, thu c 34,5 gam

kt ta. Cc este trn thuc loi g ? (n chc hay a chc, no hay khng no).

A. Este thuc loi no

B. Este thuc loi no, n chc, mch h

C. Este thuc loi khng no

D. Este thuc loi khng no a chc

Cu 21 : Khi t chy hon ton mt este X cho 2CO

n = OH2n . Thu phn hon ton 6,0 gam este X cn

va dung dch cha 0,1 mol NaOH. Cng thc phn t ca este l

A. C2H4O2 B. C3H6O2 C. C4H8O2 D. C5H8O4

Cu 22 : t chy hon ton 3,7 gam cht hu c X cn dng va 3,92 lt O2 (ktc), thu c CO2 v

H2O c l s mol l 1: 1. X tc dng vi KOH to ra hai cht hu c. S ng phn cu to ca X tho

mn iu kin trn l

A. 1. B. 2. C. 3 . D. 4.

Cu 23 : t chy hon ton a gam hn hp hai este no, n chc mch h. Sn phm chy c dn

vo bnh ng dung dch Ca(OH)2 d thy khi lng bnh tng 12,4 gam v to ra c m gam kt ta.

Gi tr ca m l:

A. 12,4. B. 10. C. 20. D. 28,18.


12

Cu 24 : Khi t chy 4,4 gam hu c X n chc thu c sn phm chy ch gm 4,48 lt CO2 (ktc)

v 3,6 gam H2O. Nu cho 4,4 gam X tc dng vi NaOH va c 4,8 gam mui ca axit hu c Y v

cht hu c Z. Tn ca X l

A. etyl axetat. B. etyl propionat.

C. isopropyl axetat. D. metyl propionat.

Cu 25 : X phng ho hon ton 1,48 gam hn hp hai este A, B l ng phn ca nhau cn dng ht

20ml dung dch NaOH 1M. Mt khc khi t chy hon ton hn hp hai este th thu c kh CO2 v

H2O vi th tch bng nhau ( cng iu kin). Cng thc cu to hai este l

A. CH3COOCH3 v HCOOC2H5

B. CH3COOC2H5 v C2H5COOCH3

C. HCOOCH2H2CH3 v HCOOCH(CH3)2

D. CH3COOCH=CH2 v CH2=CHCOOCH3

Cu 26 : t chy hn hp hai este no, n chc ta thu c 1,8 gam H2O. Thu phn hon ton hn hp

2 este trn ta thu c hn hp X gm ru v axit. Nu t chy hon ton mt na hn hp X th th

tch CO2 thu c l bao nhiu ?

A. 1,12 lt B. 2,24 lt C. 3.36 lt D. 4,48 lt

Cu 27 : C cc loi hp cht sau: anken; xicloankan; anehit no, n chc, mch h; este no, n chc

mch h; ru no, n chc, mch h; axit no, hai chc, mch h. C bao nhiu loi hp cht trn khi

t chy hon ton cho s mol H2O bng mol CO2.

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

P N

1B 2A 3D 4B 5B 6C 7C 8C 9B 10A

11A 12C 13B 14A 15A 16C 17D 18A 19A 20B

21A 22B 23C 24D 25A 26A 27C

Pp13 Khao Sat Ti Le So Mol Co2 Va h2o

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