PP2_Tang Giam Khoi Luong

8/4/2019 PP2_Tang Giam Khoi Luong

1/16

16

RO = 31 R = 15 (CH3) X l CH3OH

Phng php 3: Tng gim khi lng

Phng php 3

Phng php tng gimI. PHNG PHP GII

1. Ni dung phng php

- Mi s bin i ha hc (c m t bng phng trnh phn ng) u c lin quan n s tnghoc gim khi lng ca cc cht.

+ Da vo s tng hoc gim khi lng khi chuyn 1 mol cht X thnh 1 hoc nhiu mol cht Y(c th qua cc giai on trung gian) ta d dng tnh c s mol ca cc cht v ngc li, t s molhoc quan h v s mol ca 1 cc cht m tas bit c s tng hay gim khi lng ca cc cht X, Y.

+ Mu cht ca phng php l: * Xc nh ng mi lin h t l mi gia cc cht bit (cht X)vi cht cn xc nh (cht Y) (c th khng cn thit phi vit phng trnh phn ng, m ch cn lp s chuyn ha gia 2 cht ny, nhng phi da vo LBT nguyn t xc nh t l mi gia chng).

*Xem xt khi chuyn t cht X thnh Y (hoc ngc li) th khilng tng ln hay gim i theo t l phn ng v theo cho.

* Sau cng, da vo quy tc tam sut, lp phng trnh ton hc

gii.2. Cc dng bi ton thng gp

Bi ton 1: Bi ton kim loi + axit (hoc hp cht c nhm OH linh ng) mui + H2

2M + 2nHX 2MXn+ nH2 (l)

2M + nH2SO4M2(SO4)n+ nH2 (2)

2R(OH)n+ 2nNa 2R(ONa)n+ nH2 (3)

T (l), (2) ta thy: khi lng kim loi gim v tan vo dung dch di dng ion, nhng nu c cndung dch sau phn ng th khi lng cht rn thu c s tng ln so vi khi lng kim loi ban u,nguyn nhn l do c anion gc axit thm vo.

T (3) ta thy: khi chuyn 1 mt Na vo trong mui s gii phng 0,5 mol H2 tng ng vi s tng khi

lng l m = MRO.Do , khi bit s mol H2 v m => R. Th d: Cho m gam ancol n chc X vo bnh ng Na d, sau phn ng c 0,1 mol H2 v khilng bnh tng 6,2gam. Xc nh CTPT ca X.

Hng dn gii

Theo (3), vi n = 1 : 1 mol Na 1 mol R- ONa

0,5 mol H2: m = MRO

0,1 mol H2: m = 6,2gam

Bi ton 2: Bi ton nhit luyn

Oxit (X) + CO (hoc H2) rn (Y) + CO2 (hoc H2O)

Ta thy: d khng xc nh c Y gm nhng cht g nhng ta lun c v oxi b tch ra khi oxit v

thm vo CO (hoc H2) to CO2 hoc H2O

m = mX - mY = mO nO =16

m

V= nCO = n 2

CO (hoc =2H

n = n 2H )


2/16

16


Bi ton 3: Bi ton kim loi + dung dch mui: nA + mBn+nAm+ + mB

Ta thy: tng (gim) khi lng ca kim loi chnh l gim (tng) khi lng ca mui (vmanion = const) .

* Ch : Coi nh ton b kim loi thot ra l bm ht ln thanh kim loi nhng vo dung dch mui.

Bi ton 4: Bi ton chuyn ha mui ny thnh mui khc.Khi lng mui thu c c th tng hoc gim, do s thay th anion gc axit ny bng anion gc

axit khc, s thay th ny lun tun theo quy tc ha tr (nu ha tr ca nguyn t kim loi khng thayi).

* T 1 mol CaCO3 CaCl2: m = 71 - 60 = 11

( c 1 mol CO32ha tr 2 phi c thay th bng 2 mol Cl ha tr 1)

* T 1 mol CaBr2 2 mol AgBr: m = 2. 108 - 40 = 176

( c 1 mol Ca2+ ha tr 2 phi c thay th bng 2 mol Ag+ ha tr 1)

Bi ton 5: Bi tonchuyn oxit thnh mui:

MxOy MxCl2y (c 1 mol O-2

c thay th bng 2 mol Cl

)MxOy Mx(SO4)y (c 1 mol O-2 c thay th bng 1 mol SO42

)

* Ch : Cc iu ny ch ng khi kim loi khng thay i ha tr.

Bi ton 6: Bi tonphn ng este ha:

RCOOH + HO R RCOOR + H2O

- meste < m : m tng = m - meste

- meste > m : m gim = meste m

Bi ton 7: Bi tonphn ng trung ha: - OHaxit, phenol + kim

- OH(axit, phenol) + NaOH - ONa + H2O

(c 1 mol axit (phenol) mui: m = 23 1 = 22)

3. nh gi phng php tng gim khi lng- Phng php tng gim khi lng cho php gii nhanh c nhiu bi ton khi bit quan h v

khi lng v t l mi ca cc cht trc v sau phn ng.- c bit, khi cha bit r phn ng xy ra l hon ton hay khng hon ton th vic s dng

phng php ny cng gip n gin ha bi ton hn.- Cc bi ton gii bng phng php tng gim khi lng u c th gii c theo phng php

bo ton khi lng, v vy c th ni phng php tng gim khi lng v bo ton khi lng l 2 anhem sinh i. Tuy nhin, ty tng bi tp m phng php ny hay phng php kia s l u vit hn.

- Phng php tng gim khi lng thng c s dng trong cc bi ton hn hp nhiu cht.4. Cc bc gii.

- Xc nh ng mt quan h t l mi gia cht cn tm v cht bit (nh vn dng LBTNL).- Lp s chuyn ho ca 2 cht ny.- Xem xt s tng hoc gim ca M v m theo phng trnh phn ng v theo d kin bi ton- Lp phng trnh ton hc gii.

II.TH D MINH HA

V d 1: Khi oxi ho hon ton 2,2 gam mt anehit n chc thu c 3 gam axit tng ng. Cng thc

anehit l

mui mui

mui mui


3/16

16


A. HCHO. B. C2H3CHO. C. C2H5CHO. D. CH3CHO.

Gii:

RCHO [ O ] RCOOH

x mol x mol

m tng= 16x = 3 2,2 x = 0,05

Manehit = (R+29) = == CHOCH15R440,05

2,23 p n D

V d 2 : Oxi ho m gam X gm CH3CHO, C2H3CHO, C2H5CHO bng oxi c xc tc, sn phm thu c

sau phn ng gm 3 axit c khi lng (m + 3,2) gam. Cho m gam X tc dng vi lng d dung dch

AgNO3/NH3 th thu c x gam kt ta. Gi tr ca x l

A. 10,8 gam B. 21,6 gam C. 32,4 gam D. 43,2 gam

Gii

2 +0txt,

2OCHOR 2 COOOHR

Khi lng tng 3,2 gam l khi lng ca oxi tham gia phn ng

nx = 2 2On = 2 x 0,2(mol)32

3,2=

V cc anehit l n chc (khng c HCHO) nAg= 2nx= 2. 0,2 = 0,4 (mol)

mAg = x = 0,4. 108 = 43,2 gam p n D

V d 3 : Cho 3,74 gam hn hp 4 axit, n chc tc dng vi dung dch Na2CO3 thu c V lt kh CO2

(ktc) v dung dch mui. C cn dung dch th thu c 5,06 gam mui. Gi tr ca V lt l:

A. 0,224 B. 0,448. C. 1,344. D. 0,672

Gii:

OHCOCOONaR2NaCOCOOHR 223 +++

a mol a mol 0,5a mol

m tng = (23 - 1)a = 5,06 3,74 a = 0,06 mol

2CO

V = 0,06. 0,5. 22,4 = 0,672 lt p n D

V d 4: Cho 2,02 gam hn hp hai ancol n chc, ng ng k tip tc dng va vi Na c 3,12

gam mui khan. Cng thc phn t ca hai ancol l :A. CH3OH, C2H5OH. B. C2H5OH, C3H7OH.

C. C3H7OH, C4H9OH. D. C4H9OH, C5H11OH.

Gii:


4/16

16


2

1ROH Na RONa H

2+ +

a mol a mol

mtng = 22a = 3,12 2,02 a = 0,05 mol

M 2 ru = M R+17 =


5/16

16


Zn + CuSO4 ZnSO4 +Cu (1)

x x

mgim = (65 - 64)x = x

Fe + CuSO4 FeSO4 +Cu (2)

y y

m tng = (64 - 56)y = 8y

V khi lng hn hp rn trc v sau phn ng i mgim = mtng x = 8y

%Zn = =+

90,28%100%x56y65x

65xp n A

V d 8: Cho 4,48 lt CO (ktc) tc dng vi FeO nhit cao mt thi gian, sau phn ng thu c

cht rn X c khi lng b hn 1,6gam so vi khi lng FeO ban u. Khi lng Fe thu c v %

th tch CO2 trong hn hp kh sau phn ng ln lt l:A. 5,6gam; 40% B. 2,8gam; 25%

C. 5,6gam; 50% C. 11,2gam; 60%

Gii:

FeO + CO2

tCOFe

0

+

mgim = mO(oxit phn ng )= 0,1(mol)16

1,6=

Fen = 2COn = 0,1 (mol) mFe = 0,1.56 = 5,6gam (*)

Theo bo ton nguyn t: n hn hp kh sau phn ng = nCO(ban u) = 0,2 (mol)

% th tch kh CO2 = 50%(**)x100%0,2

0,1=

T (*) v (**)p n C

V d 9 : Tin hnh 2 th nghim :

- TN 1 : Cho m gam bt Fe d vo V1 (lt) dung dch Cu(NO3)2 1M.

- TN2 : Cho m gam bt Fe d vo V2 (lt) dung dch AgNO3 0,1M.

Sau khi cc phim ng xy ra hon ton, khi lng cht rn thu c 2 th nghim u bng nhau. Gi

tr ca Vl so vi V2 lA. V1 = V2 B. Vl = l0V2 C. Vl = 5V2 D. Vl = 2V2

Gii:

Fe + Cu2+Fe2+ + Cu

V1 mol V1 mol

m tng= 64V1 56V1 = 8V1 gam


6/16

16


Fe + 2Ag+ Fe2+ + 2Ag

0,05V2 mol 0,1V2 mol

mtng = 108.0,1V2 56.0,05V2 = 8V2 gam

Theo mrn(TN1) = mrn(TN2) 8V1= 8V2V1 = V2 p n A

V d 10 : Nung 1 hn hp rn gm a mol FeCO3 v b mol FeS2 trong bnh kn cha khng kh d. Sau

khi cc phn ng xy ra hon ton, a bnh v nhit ban u thu c cht rn duy nht l Fe 2O3 v

hn hp kh. Bit p sut kh trong bnh trc v sau phn ng bng nhau v sau cc phn ng lu hunh

mc oxi ho +4, th tch cc cht rn l khng ng k. Mi lin h gia a v b l

A. a = 0,5b. B. a = b. C. a = 4b. D. a = 2b.

Gii:

2FeCO3+ 32t

2OFeO

2

1 0 +2CO2

a4

aa

Phn ng lm tng 1 lng kh l (a -4

a)= mol

4

3a

2FeS2 + 232t

2 4SOOFeO2

11 0+

b4

11b2b

Phn ng lm gim mt lng kh l: mol4

3b2b4

11b

=

V ptrc = psau == ba4

3b

4

3ap n B

V d 11: Cho 5,90 gam amin n chc X tc dng va vi dung dch HCl sau khi phn ng xy ra

hon ton thu c dung dch Y. Lm bay hi dung dch Y c 9,55 gam mui khan. S cng thc cu

to ng vi cng thc phn t ca X l:

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

Gii:

RNH2 + HCl RNH3Cl

x mol x mol x mol

m tng = 36,5x = 9,55 5,9 x = 0,1

Mamin = MR+16 =1,0

9,5=59 MR= 43 X: C3H7NH2


7/16

16


CH3 CH2 CH2 NH2 ; (CH3)2CHNH2; CH3NHCH3CH2; (CH3)3N p n B

V d 12: Trong phn t amino axit X c 1 nhm amino v 1 nhm cacboxyl. Cho 15,0 gam X tc dng

va vi dung dch NaOH. C cn dung dch sau phn ng thu c 19,4 gam mui khan. Cng thc

ca X l

A. H2NC3H6COOH. B. H2NCH2COOH.

C. H2NC2H4COOH. D. H2NC4H8COOH.

Gii:

H2 NRCOOH + NaOH H2NRCOONa + H2O

x mol x mol

mtng = 22x = 19,4 15,0 x = 0,2 mol

Mx = MR+61 = 75 MR= 14 X: H2NCH2COOH p n B

V d 13: t chy hon ton 4,40 gam cht hu c X n chc thu c sn phm chy gm 4,48 lt

CO2 (ktc) v 3,60 gam H2O. Nu cho 4,40 gam X tc dng vi dung dch NaOH va n khi phn

ng hon ton c 4,80 gam mui ca axit hu c Y v cht hu c Z. Tn ca X l

A. etyl propionat. B. metyl propionat

C. isopropyl axetat. D. etyl axetat.

Gii :

2COn = OH2n = 0,2mol X l este no n

CnH2nO2 + ( )2

13n O2

0t

nCO2 + nH2O

moln

0,20,2 mol

mX = (14n + 32) n0,2

= 4,4 n = 4 X: C4H8O2 v nX = 4

0,2= 0,05 mol

RCOOR + NaOH RCOONa + ROH

0,05 mol 0,05 mol

mX < mmui mtng = (23-R) 0,05 = 4,8 4,4 = 0,4 R= 15

Cng thc cu to ca X l: C2H5OHCOOCH3

p n BV d 14: Hn hp X gm HCOOH v CH3COOH (t l mol 1:1). Ly 5,30 gam hn hp X tc dng vi

5,75 gam C2H5OH (xc tc H2SO4 c) thu c m gam este (hiu sut ca cc phn ng este ho u

bng 80%). Gi tr ca m l:

A. 10,12 gam. B. 6,48 gam.

C. 16,20 gam. D. 8,10 gam.


8/16

16


Gii:

x mol x mol x mol

532x60x46xMX =+=

nX = 5,3: 53 = 0,1 mol < OHHC 52n = 0,125 mol khi lng este tnh theo s mol ca axit

mtng = (29-1)x = m -5,3 m = 8,1 gam

Khi lng este thc t thu c l 6,48gam100%

8,1.80%=

p n B

V d 15: Dn t t hn hp kh CO v H2 qua ng s ng 55,4 gam hn hp bt CuO, MgO, ZnO,

Fe3O4 un nng. Sau khi phn ng xy ra hon ton thu c 10,08 lt (ktc) hn hp kh v hi ch cha

CO2 v H2O, trong ng s cn li mt lng cht rn c khi lng l

A. 48,2 gam. B. 36,5 gam. C. 27,9 gam D. 40,2 gam

Gii:

Bn cht ca cc phn ng CO, H2+[O] CO2 , H2O

nO = 2COn + OH2n = nCO+ 2Hn = 0,45mol

m rn = moxit mO = 55,4 0,45.16 = 48,2 gam p n A

V d 16: Nung 47,40 gam kali pemanganat mt thi gian thy cn li 44,04 gam cht rn. % khi lngkali pemanganat b nhit phn l

A. 50%. B. 70%. C. 80%. D. 65%.

Gii:

2KMnO4 0

t

K2MnO4 + MnO2 + O2

gim khi lng ca cht rn =2O

m = 47,4 44,04 = 3,36gam

2O

n = 3,36: 32 = 0,105 mol 4KMnO

m tham gia = 0,105.2 = 0,21 mol

% 4KMnOm phn ng = 4,47158.21,0

.100%= 70% p n B

V d 17 : Nhit phn a gam Zn(NO3)2 sau 1 thi gian dng li lm ngui v em cn thy khi lng

gim i 2,700 gam (hiu sut phn ng l 60%). Gi tr a l

A. 4,725 gam. B. 2,835 gam. C. 7,785 gam. D. 7.875 gam.

Gii:


9/16

16


Zn(NO)2 0

t

ZnO + 2NO2 + 2

1O2

xmol 2xmol 0,5xmol

m rn gim = 2NOm + 2Om = 92x + 16x = 2,7 x = 0,025mol

H = 7,875gama60%.100%a

189x== p n C

V d 18 : Cho 3,06 gam hn hp K2CO3 v MgCO3 tc dng vi dung dch HCl thu c V lt kh (ktc)

v dung dch X. C cn dung dch X c 3,39 gam mui khan. Gi tr V (lt) l:

A. 0,224 B. 0,448 C. 0,336 D. 0,672.

Gii:

mtng = 11 2COn = 3,39 3,06 2COn = 0,03 mol 2COV = 0,672 lt

p n D

V d 19 : Ho tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500ml dung dch H2SO4

0,1M va . Sau phn ng hn hp mui sunfat khan thu c khi c cn dung dch c khi lng l

A. 7,71 gam. B. 6,91 gam. C. 7,61 gam. D. 6,81 gam.

Gii:

O2-(trong oxit)-2

4SO

Khi lng tng: 0,05 (96 -16) = 4,0 gam

mmui = moxit + mmui = 2,81 + 4 = 6,81 gam

p n DIII. BI TP T LUYN

Cu 1: Dn 130 cm3 hn hp X gm 2 hirocacbon mch h qua dung dch Br2 d kh thot ra khi bnh

c th tch l 100cm3, bit dx/He = 5,5 v phn ng xy ra hon ton. Hai hirocacbon cn tm l

A. metan, propen. B. metan, axetilen.

C. etan, propen. D. metan, xiclopropan.

Cu 2 : un nng 1,77 gam X vi 1 lng va 1,68 gam KOH c 2,49 gam mui ca axit hu c Y

v 1 ancol Z vi s mol Z gp 2 ln s mol Y (bit phn ng xy ra hon ton). X lA. CH2(COOCH3)2 B. (COOCH3)2

C. HCOOC2H5 D. C2H4(COOCH3)2

Cu 3: Trung ho 5,48 gam hn hp axit axetic, phenol v axit benzoic cn dng 600ml dung dch NaOH

0,1M. C cn dung dch sau phn ng c hn hp cht rn khan c khi lng l



10/16

16


Cu 4: Cho 5,76 gam axit hu c X n chc mch h tc dng ht vi CaCO 3 c 7,28 gam mui ca

axit hu c. Cng thc cu to thu gn ca X l:

A. CH2=CH-COOH B. CH3COOH

C. CH C-COOH D. CH3-CH2-COOH

Cu 5: Ho tan hon ton 2,1 gam mui cacbonat ca kim loi ho tr II trong dung dch H 2SO4 long

c 3 gam cht rn khan. Cng thc mui cacbonat ca kim loi ho tri II l:

A. CaCO3 B. Na2CO3 C. FeCO3 D. MgCO3

Cu 6: Cho ancol X tc dng vi Na d thy s mol khi bay ra bng s mol X phn ng. Mt khc, X tc

dng vi lng d CuO nung nng n phn ng hon ton thy lng rn gim 1,2 gam v c

2,7 gam cht hu c a chc Y. Cng thc cu to thu gn ca Y l:

A. OHC-CH2-CH2-CHO B. OHC-CH2-CHO

C. CH3-CO-CO-CH3 D. OHC-CO-CH3

Cu 7: Cho 26,80 gam hn hp KHCO3 v NaHCO3 tc dng ht vi dung dch HCl d c 6,72 lt kh(ktc). Sau phn ng c cn c a gam mui khan. Gi tr ca a gam l:

A. 34,45. B. 20,15. C. 19,15. D. 19,45.

Cu 8: Dn V lt (ktc) hn hp gm CO v H2 qua ng s nung nng cha hn hp FeO, Al2O3 (cc

phn ng xy ra hon ton) c hn hp kh v hi nng hn hn hp kh ban u 2 gam. Gi tr ca V

lt l

A. 2,80. B. 5,60. C. 0,28. D. 0,56

Cu 9: Nung hn hp rn gm FeCO3 v FeS2 (t l mol 1 : 1) trong 1 bnh kn cha khng kh d vi p

sut l p1 atm. Sau khi cc phn ng xy ra hon ton a bnh v nhit ban u thu c cht rn duynht l Fe2O3 v p sut kh trong bnh lc ny l p 2 atm (th tch cc cht rn khng ng k v sau cc

phn ng lu hunh mc oxi ho + 4). Mi lin h gia p l v p2 l:

A. pl = p2 B. pl = 2p2 C. 2pl = p2 D. pl = 3p2

Cu 10: Dn kh CO i qua ng s nung nng cha 0,02 mol hn hp X gm FeO v Fe2O3 phn ng

xy ra hon ton thu c 1,96 gam cht rn Y, kh i ra khi ng s hp th hon ton vo dung dch

Ca(OH)2 d th thy khi lng bnh tng 2,20 gam. Hn hp X c:

A. 50%FeO v 50% Fe2O3 B. 13,04%FeO v 86,96% Fe2O3

C. 20%FeO v 80% Fe2O3 D. 82%FeO v 18%Fe2O3

Cu 11: Ho tan ht 1,625 gam kim loi M vo dung dch Ca(OH) 2 thy khi lng dung dch sau phn

ng tng 1,575 gam. M l

A. Al. B. Be. C. Zn. D. Cr.

Cu 12: Dn V lt kh CO2 (ktc) hp th hon ton vo 750ml dung dch Ba(OH) 2 0,1M, sau phn ng

khi lng dung dch gim 5,45 gam v c hn hp 2 mui. Gi tr V lt l


11/16

16


A. l,68. B. 2,24. C. 1,12. D. 3,36.

Cu 13: Cho 1,825 gam amin X tc dng va vi dung dch HCl, sau khi phn ng xy ra hon ton

thu c dung dch Y. Lm bay hi dung dch Y c 2,7375 gam mui RNH 3Cl. X c tng s ng

phn cu to amin bc 1 l:

A. 4. B. 6. C. 7. D. 8.

Cu 14: Cho a gam hn hp gm metanol v propan-2-ol qua bnh ng CuO d, nung nng. Sau khi

phn ng xy ra hon ton a hn hp kh v hi c khi lng l (a + 0,56) gam. Khi lng CuO

tham gia phn ng l

A. 0,56 gam. B. 2,80 gam C. 0,28 gam. D. 5,60 gam.

Cu 15: Cho a gam hn hp cc ankanol qua bnh ng CuO d, nung nng. Sau khi phn ng xy ra

hon ton c hn hp kh v hi c khi lng l (a + 1,20) gam v c t khi hi i vi H 2 l 15. Gi

tr ca a gam l

A. 1,05 gam. B. 3,30 gam. C. 1,35 gam. D. 2,70 gam.Cu 16: Cho amino axit X tc dng va vi Na thy s mol kh to ra bng s mol X phn ng.

Ly a gam X tc dng vi dung dch HCl d c (a + 0,9125) gam Y. un ton b lng Y thu c

vi 200ml dung dch NaOH thu c dung dch Z. Bit X lm qu tm ho . Nng mol ca dung

dch NaOH phn ng l

A. 0,2500M. B. 0,1250M. C. 0,3750M. D. 0,4750M.

Cu 17: Cho amino axit X tc dng va vi Na thy s mol kh to ra bng s mol X phn ng.

Ly a gam X tc dng vi dung dch HCl d c (a + 0,9125) gam Y. em ton b lng Y tc dng

va vi dung dch NaOH un nng c dung dch Z. C cn Z c 5,8875 gam mui khan. Bit Xlm qu tm ho . Gi tr a gam l

A. 3,325. B. 6,325. C. 3,875. D. 5,875.

Cu 18: Cho amino axit X tc dng va vi Na thy s mol kh to ra bng s mol X phn ng.

Ly a gam X tc dng vi dung dch HCl d c (a + 0,9125) gam Y. em ton b lng Y tc dng

va vi dung dch NaOH un nng c dung dch Z. C cn Z c 5,8875 gam mui khan. Bit X

lm qu tm ho . Cng thc cu to ca X l

A.HOOC-CH(NH2)-COOH

B. HOOC-CH2CH(NH2)CH2-COOHC. HOOC-CH2CH2CH2NH2

D. HOOC-CH2CH(NH2)-COOH

Cu 19: Cho amino axit x tc dng va vi Na thy s mol kh to ra bng s mol X phn ng. Ly

a gam X tc dng vi dung dch HCl d c (a + 0,9125) gam Y. em ton b lng Y tc dng va


12/16

16


vi dung dch NaOH un nng c dung dch Z. C cn Z c 1 lng mui khan. Bit X lm qu tm

ho . Khi lng mui khan thu c so vi khi lng ca Y s

A. tng 1,65 gam. B. gim 1,65 gam.

C. tng 1,10 gam. D. gim 1,10 gam.

Cu 20: t chy hon ton 3,72 gam hp cht hu c X (bit2X/H

d < 70), dn ton b sn phm chy

thu c qua bnh ng dung dch Ba(OH)2 d thy to ra 41,37 gam kt ta ng thi khi lng dung

dch gim 29,97 gam. Bit s mol NaOH cn dng phn ng ht vi X bng s mol kh hiro sinh ra

khi cho X tc dng vi Na d. Cng thc cu to thu gn ca X l:

A. CH3-C6H4(OH)2 B. C6H7COOH.

C. C5H6(COOH)2 D. HO-C6H4-CH2OH.

Cu 21: Th tch oxi phn ng l bao nhiu nu chuyn 1 th tch oxi thnh ozon thy th tch gim i

7,0 cm3 (th tch cc kh o cng iu kin)

A. 21,0 dm3 B. 7,0 cm3 C. 21,0 cm3 D. 4,7 cm3

Cu 22: Trong 1 bnh kn dung tch khng i cha 0,2 mo1 CO v 1 lng hn hp X gm Fe3O4 v

FeCO3 (t l mol 1 : l). Nung bnh nhit cao cc phn ng xy ra hon ton v a bnh v nhit

ban u (th tch cc cht rn khng ng k) thy p sut trong bnh tng 2 ln so vi ban u. Tng

s mol ca Fe3O4 v FeCO3 l:

A 0,4 B. 0,3. C. 0,2. D. 0,1.

Cu 23: t chy hon ton 16,8 gam mui sunfua ca kim loi ho tri II khng i thu c cht rn X

v kh B. Ho tan ht X bng 1 lng va dung dch H 2SO4 35% c dung dch mui c nng

44,44%. Ly dung dch mui ny lm lnh xung nhit thp thy tch ra 25 gam tinh th ngm nc Y

v dung dch bo ho c nng 31,58%. Y c cng thc l

A. CuSO4.3H2O. B. MgSO4.2H2O.

C. CuSO4.5H2O. D. CuSO4.2H2O.

Cu 24: Thu phn hon ton 1,76 gam X n chc bng 1 lng va dung dch NaOH un nng

c 1,64 gam mui Y v m gam ancol Z. Ly m gam Z tc dng vi lng d CuO nung nng n phn

ng hon ton thy lng cht rn gim 0,32 gam. Tn gi ca X l

A. etyl fomat. B. etyl propionat.

C. etyl axetat. D. metyl axetat.

Cu 25: Cho hn hp X gm 2 axit ng ng k tip nhau tc dng vi Na d thy s mol H 2 bay ra

bng2

1mol X. un 20,75 gam X vi 1 lng d C2H5OH (xc tc H2SO4 c) c 18,75 gam hn hp


13/16

16


este (hiu sut ca cc phn ng este ho u bng 60%). % theo khi lng cc cht c trong hn hp X

l:

A. 27,71% HCOOH v 72,29% CH3COOH.

B. 27,71 % CH3COOH v 72,29% C2H5COOH.

C. 40% C2H5COOH v 60% C3H7COOH.

D. 50% HCOOH v 50% CH3COOH.

Cu 26: Ho tan 5,4 gam Al vo 0.5 lt dung dch X gm AgNO3 v Cu(NO3)2 c 42 gam rn Y khng

tc dng vi dung dch H2SO4 long v dung dch Z. Ly ton b dung dch Z cho tc dng vi dung dch

NaOH d th c 14,7 gam kt ta (cho phn ng xy ra hon ton). Nng mi ca AgNO 3 v

Cu(NO3)2 trong dung dch X ln lt l:

A. 0,6M v 0,3M. B. 0,6M v 0,6M.

C. 0,3M v 0,6M. D. 0,3M v 0,3M.

Cu 27: Nhng m gam kim loi M ho tr II vo dung dch CuSO4 sau 1 thi gian ly thanh kim loi thykhi lng gim 0,075%. Mt khc, khi nhng m gam thanh kim loi trn vo dung dch Pb(NO 3)2 sau 1

thi gian ly thanh kim loi thy khi lng thanh kim loi tng 10,65% (bit s mol ca CuSO4 v

Pb(NO3)2 tham gia 2 trng hp l nh nhau). M l

A. Mg. B. Zn. C. Mn. D. Ag.

Cu 28: Nhng 1 thanh Al v 1 thanh Fe vo dung dch Cu(NO3)2 sau 1 thi gian ly 2 thanh kim loi ra

thy dung dch cn li cha Al(NO3)3 v Fe(NO3)2 vi t l mol 3 : 2 v khi lng dung dch gim

2,23 gam (cc phn ng xy ra hon ton). Khi lng Cu bm vo thanh Al v Fe l:

A. 4,16 gam. B. 2,88 gam. C. 1,28 gam. D. 2,56 gam.Cu 29 : Cho 32,50 gam Zn vo 1 dung dch cha 5,64 gam Cu(NO 3)2 v 3,40 gam AgNO3 (cc phn

ng xy ra hon ton v tt c kim loi thot ra u bm vo thanh kim loi). Khi lng sau cng ca

thanh kim loi l


Cu 30: in phn l00ml dung dch M(NO3)n. Vi in cc tr cho n khi b mt catot xut hin bt kh

th ngng in phn. Phi dng 25ml dung dch KOH 2M trung ho dung dch sau khi in phn. Mt

khc, nu ngm 20 gam Mg vo 100ml dung dch M(NO3)n. Sau mt thi gian ly thanh Mg ra, sy kh

v cn li thy khi lng tng thm 24% so vi lng ban u. Bit cc phn ng xy ra hon ton.Cng thc ho hc ca M(NO3)n l

A. Cu(NO3)2 B. Ni(NO3)2 C. Pb(NO3)2 D. AgNO3

Cu 31: Nung 46,7 gam hn hp Na2CO3 v NaNO3 n khi lng khng i thu c 41,9 gam cht

rn. Khi lng Na2CO3 trong hn hp u l



14/16

16


Cu 32: Nung 104,1 gam hn hp K2CO3 v NaHCO3 cho n khi khi lng khng i thu c

88,6 gam cht rn % khi lng ca cc cht trong hn hp u l

A. 20% v 80%. B. 45,5% v 54,5%.

C. 40,35% v 59,65%. D. 35% v 65%.

Cu 33: Dn kh CO qua ng s cha 7,6 gam hn hp gm FeO v CuO nung nng, sau 1 thi gian

c hn hp kh X v 6,8 gam rn Y. Cho hn hp kh X hp th hon ton vo dung dch Ca(OH) 2 d

thy c kt ta. Khi lng kt ta

A. 5 gam. B. 10 gam. C. 15 gam. D. 20 gam.

Cu 34: t chy hon ton m gam hai kim loi Mg, Fe trong khng kh, thu c (m + 0,8) gam hai

oxit. hon tan ht lng oxit trn th khi lng dung dch H2SO4 20% ti thiu phi dng l

A. 32,6 gam. B. 32 gam. C. 28,5 gam. D. 24,5 gam.

Cu 35: Ly 2,98 gam hn hp X gm Zn v Fe cho vo 200ml dung dch HCl 1M, sau khi phn ng

hon ton ta c cn (trong iu kin khng c oxi) th c 6,53 gam cht rn. Th tch kh H 2 bay ra(ktc) l

A. 0,56 lt. B. 1,12 lt. C. 2,24 lt. D. 4,48 lt.

Cu 36: em nung nng m gam Cu(NO3)2 mt thi gian ri dng li, lm ngui v em cn thy khi

lng gim 0,54 gam so vi ban u. Khi lng mui Cu(NO3)2 b nhit phn l


Cu 37: trung ho 7,4 gam hn hp 2 axit hu c n chc cn 200ml dung dch NaOH 0,5M. Khi

lng mui thu c khi c cn dung dch l

A. 9,6 gam. B. 6,9 gam. C. 11,4 gam. D. 5,2 gam.Cu 38: Cho 5,615 gam hn hp gm ZnO, Fe2O3, MgO tc dng va vi 100ml dung dch

H2SO4 1M th khi lng mui sunfat thu c l


Cu 39: t chy hon ton 33,4 gam hn hp X gm Al, Fe, Cu ngoi khng kh thu c 41,4 gam

hn hp Y gm ba oxit. Th tch ti thiu dung dch H 2SO4 20% (D =1,14 g/ml) cn dng ho tan ht

hn hp Y l:

A. 215ml. B. 8,6ml. C. 245ml. D. 430ml.

Cu 40: X l mt -aminoaxit ch cha 1 nhm -NH2 v 1 nhm -COOH. Cho 0,445 gam X phn ngva vi NaOH to ra 0,555 gam mui. Cng thc cu to ca X c th l

A. H2N-CH2-COOH. B. CH3-CH(NH2)-COOH.

C. H2N-CH2-CH2-COOH. D. H2N-CH=CH-COOH.

Cu 41: Cho hn hp X gm NaCl v NaBr tc dng vi dung dch AgNO3 d th lng kt ta thu c

sau phn ng bng khi lng AgNO3 tham gia phn ng. Thnh phn % khi lng NaCl trong X l


15/16

16


A. 27,88%. B. 13,44%. C. 15,20%. D. 24,50%.

Cu 42: Cho 1,52 gam hn hp hai ancol n chc l ng ng k tip nhau tc dg vi Na va , sau

phn ng thu c 2,18 gam cht rn. Cng thc phn t ca hai ancol v th tch kh thu c sau phn

ng ktc ln lt l:

A. CH3OH; C2H5OH v 0,336 lt. B. C2H5OH; C3H7OH v 0,336 lt

C. C3H5OH; C4H7OH v 0,168 lt. D. C2H5OH; C3H7OH v 0,672 lt.

Cu 43: Hn hp X c khi lng 25,1 gam gm ba cht l axit axetic, axit acrylic v phenol. Lng hn

hp X trn c trung ho va bng 100ml dung dch NaOH 3,5M. Tnh khi lng ba mui thu c

sau phn ng trung ho l


Cu 44: Ngm mt inh st sch trong 200ml dung dch CuSO4 n khi dung dch ht mu xanh, ly inh

st ra khi dung dch, ra sch, sy kh, cn thy inh st tng 0,8 gam. Nng mi ca dung dch

CuSO4 lA. 0,5M. B. 5M. C. 0,05M. D. 0,1M

Cu 45: Nung l00 gam hn hp gm Na2CO3 v NaHCO3 cho n khi khi lng hn hp khng i

c 69 gam cht rn. Xc nh phn trm khi lng ca mi cht trong hn hp ln lt l:

A. 16% v 84%. B. 84% v 16%.

C. 26% v 74%. D. 74% v 26%.

Cu 46: Ly 2,98 gam hn hp X gm Zn v Fe cho vo 200ml dung dch HCl 1M, sau khi phn ng

hon ton ta c cn (trong iu kin khng c oxi) th c 6,53 gam cht rn. Th tch kh H 2 bay ra

(ktc) lA. 0,56 lt. B. 1,12 lt. C. 2,24 lt. D. 4,48 lt.

Cu 47: Cho mt anken X tc dng ht vi H 2O (H+, t0) c cht hu c Y, ng thi khi lng bnh

ng nc ban u tng 4,2 gam. Cng cho mt lng X nh trn tc dng vi HBr va , thu c

cht Z, thy khi lng Y, Z thu c khc nhau 9,45 gam (gi s cc phn ng xy ra hon ton). Cng

thc phn t ca X l:

A. C2H4 B. C3H6 C. C4H8 D. C5H10

P N

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

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

21C 22A 23C 24C 25A 26B 27B 28A 29B 30D

31A 32C 33A 34D 35B 36D 37A 38B 39A 40B


16/16

16


41A 42B 43A 44A 45A 46B 47A

* Xem thm bi ging trc tuyn ca: PGS.TS o Hu Vinh trong phn video th nghim ha hc(video: Phng php tng gim khi lng)

Documents

PP2_Tang Giam Khoi Luong