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GII CHI TIT MT S CU 201

Cu 20.Cc dung dch ha tan Cu(OH)2 l cc dung dch axit hoc cc ancol c cc nhm OH k nhau:

mantoz; saccaroz; fructoz; glixerol; axit axeticCu 21.Ta c nFe = 0,05 mol;

3HNOn = 0,3 mol

Fe + 4HNO3Fe(NO3)3 + NO + 2H2O

mol 0,05 0,2 0,05

Vy dung dch Y c:

3+

+

-3

Fe

H

NO

n = 0,05 mol

n = 0,8 mol

n = 0,25 mol

Cc phn ng xy ra khi cho Cu vo dung dch Y:3Cu + 8H+ + 2NO3

- 3Cu2+ + 2NO + 4H2O (1)mol 0,3 0,8 0,2

Cu + 2Fe3+ Cu2+ + 2Fe2+ (2)mol 0,025 0,05

Vy mCu (max) = 64(0,3 + 0,025) = 20,8gam (Chn A)Cu 22.Ch c ZnO v FeO l tha cc d kin cho.

Cu 23.+ Khng th chn A v iu ny ch ng vi axit mnh+ Khng th chn B v theo nguyn l chuyn dch cn bng th cn bng

CH3COOH CH3COO- + H+ s chuyn dch theo chiu nghch tc in li ca axit

axetic trong dung dch s gim khi thm dung dch H2SO4 long vo.

+ Khng th chn C v theo cng thcK

=C

th khi pha long dung dch trn, in li

ca axit axetic trong dung dch phi tng.+ Chn D, th theo ta c

M

10D . C% 10 . 1 . 0,6C = = = 0,1M

M 60

CH3COOH CH3COO + H+

0,1M 0,1MVy pH = -log0,1M = 3 = 0,01 hay 1%Cu 26.

V2 3Na CO

n = 0,1 mol nn nNaOH = 0,2 mol

Nh vy neste n chc X : nNaOH = 1 : 2 nn X l este ca phenolPhn ng xy ra:

RCOOR' + 2NaOH RCOONa + R'ONa + H2Omol 0,1 0,2 0,1

Bo ton khi lng cho MX =19,8 + 0,1 . 18 - 0,2 . 40

= 1360,1

Do X l C8H8O2Cu 27.

Theo dung dch X phi cha x mol Na2CO3 v y mol NaHCO3. Cc phn ng:CO2 + 2NaOH Na2CO3 + H2O

mol x 2x xCO2 + NaOH NaHCO3

mol y y y

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+Phn 1:BaCl2 + Na2CO3 BaCO3 + 2NaCl

mol 0,5x 0,5x+Phn 2:

BaCl2 + Na2CO3 BaCO3 + 2NaClmol 0,5x 0,5x

2NaHCO3 Na2CO3 + CO2 + H2Omol 0,5y 0,25y

BaCl2 + NaHCO3 BaCO3 + NaOH + H2Omol 0,25y 0,25y

Vy

0,5x = 0,05 x = 0,1

0,5x + 0,25y = 0,2 y = 0,6

Do V = 22,4(x + y) = 15,68 lt v a2x + y

a = = 3,20,25

Cu 28.V mi anehit tc dng va vi H2 theo t l mol 1 : 2 nn anehit no cng c 2 trong

phn t. Vy n2anehit =2 2O H O

n - n = 0,425 - 0,25 = 0,175 mol.

Ta ctb

10M = = 57,14

0,175nn MX < 57,4 < MY

Anehit c 2 v c M < 57,14 ch c th l CH2=CH-CHO. Vy X l CH2=CH-CHO

Mt khc s C trung bnh =tb

0,425M = = 2,42

0,175trong CH2=CH-CHO c s C > 2,42 nn

anehit cn li phi c s C < 2,42. Anehit c 2 v c s C < 2,42 ch c th l OHC-CHO. VyY l OHC-CHO

Gi a, b ln lt l c s mol CH2=CH-CHO v OHC-CHO trong 10 gam X, ta c h:

56a + 58b = 10 a = 0,075

a + b = 0,175 b = 0,1. Vy

0,075 . 56 . 100%X = = 42(%)

10

Cu 31.

Loi A v phn m NaNO3 khng lm t b chuaLoi C v phn ln cung cp nguyn t photpho cho cy di dng ion photphat PO43.

Loi D v phn nitrophotka l hn hp gm (NH4)2HPO4 v KNO3Cu 32.

Theo th 55,5 gam hn hp mui khan gm:

2 3 2 2 2 3

NaCl: 0,2mol

55,5 - 58,5 . 0,2CH (NH Cl)CH CH CH CH(NH Cl)COOH: = 0,2 mol

219

Vy nlysin =2 3 2 2 2 3CH (NH Cl)CH CH CH CH(NH Cl)COOH

n = 0,2 mol

tc mlysin = 146 . 0,2 = 29,2 gam

Cu 34.D dng tm c2NO NO

n = n = 0,2 mol . Vy m = 22,4 + 622NO NO

(3n + n ) = 72 gam

Cu 37.Theo %O = 55,18. t cng thc cn tm l CxHyOz, ta c:

41,37 3,45 55,18x : y : z = : : = 1 : 1 : 1

12 1 16. Vy A c cng thc (CHO)n

V A l axit cacboxylic mch khng phn nhnh nn A ch c 2 kh nng+A l axit cacboxylic n chc. Khi A phi c 2 nguyn t oxi trong phn t. Vy cng

thc phn t A phi l C2H2O2 (loi, v khng ph hp ha tr ca mt axit)+A l axit cacboxylic nh chc. Khi A phi c 4 nguyn t oxi trong phn t. Vy cng

thc phn t A phi l C4H4O4. Phn ng xy ra:

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C4H4O4 + 3O2 4CO2 + 2H2Omol 0,1 0,3

Cu 40.X phi l caprolactam C6H11ON

Cu 42.V Y tc dng vi dung dch NaOH d gii phng 0,15 mol H2 nn Y phi c 0,1 mol Al d.

V Y tc dng vi HNO3 long d c 0,4 mol NO nn Y cn cha 0,3 mol FeTheo phn ng 2Al + 3FeO Al2O3 + 3Fe, th Y gm:

2 3

Al: 0,1 mol 2,7 gamFe: 0,3 mol 16,8 gam

Al O : 0,1 mol 10,2 gam

Vy m = mX = mY = 2,7 + 16,8 + 10,2 = 29,7 gamCu 44.

Theo , hn hp cho gm RCOOC2H5 v RCH2CH2COOHRCOOC2H5 + NaOH RCOONa + C2H5OH

mol a a aRCH2CH2COOH + NaOH RCH2CH2COONa + H2O

mol b b b

Ta c phn t lng trung bnh 2 mui = 17,8 17,8= = 71,2a + b 0,25 , chng t phi c mt mui l

HCOONa v mui cn li l C2H5COONa.Vy X, Y u c cng thc phn t l C3H6O2 nn m = 74 . 0,25 = 18,5 gamCu 46.

Ta c:e

I.t 9,65 . 90 . 60n = = = 0,54 mol

96500 96500

rng s in phn dung dch cho ch to OH- catot2H2O + 2e 2OH

- + H2mol 0,54 0,54

M 3+ln = 0,16 mol nn - 3+OH Aln > 3n , tc kt ta tan mt phn. p dng cng thc

- 3+OH Aln = 4n - n ta c

-3+Al OHn = 4n - n = 4 . 0,16 - 0,54 = 0,1 mol tc m = 7,8 gam

Cu 48.- t X trn ngn la n kh cho ngn la mu vng x l hp cht ca natri- Dung dch X lm xanh qu tm- Dung dch X tc dng c vi kh CO2- Dung dch X tc dng vi dung AlCl3 gii phng kh X l Na2CO3

Na2CO3 + CO2 + H2O 2NaHCO33Na2CO3 + 2AlCl3 + 3H2O 6NaCl + 2Al(OH)3 + 3CO2

Cu 50.V x phng ha este n chc E ch thu c cht hu c Y duy nht chng t E l este vng

CxHyO2 + NaOH HOCxHyO2Namol a a

2HOCxHyO2Na ot Na2CO3 + (2x - 1)CO2 + (y + 1)H2O

mol a 0,5a 0,5a(2x + 1) 0,5a(y + 1)

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Theo , ta c h:

5,30,5a = = 0,05

106 a = 0,110,08

0,5a(2x-1) = = 0,45 x = 522,4

y = 88,1

0,5a(y+1) = = 0,4518

Vy m = 100a = 10Cu 55.

Phn ng xy ra RCOOR' 4LiAlH RCH2OH + R'OH

V s C trung bnh ca 2 ancol = 2

2 2

CO

H O CO

n 0,2= = 2

n - n 0,3 - 0,2, m 2 ancol ny c s mol bng nhau nn

chng ch c th l CH3OH v C3H7OH. Vy este l HCOOC3H7Cu 56.

Trong th nghim chun dung dch HCl bng dung dch chun NaOH vi cht ch th mu lphenolphtalein, dung dch HCl (c pha t git phenolphtalein) ban u khng mu. Khi dung dchny va chuyn sang mu hng th th nghim kt thc (im cui). Nhng thi im ny, phnng trung ha HCl bng NaOH xy ra xong (im tng ng) t trc, v ti im tngng, ng vi pH = 7 th cht ch th phenolphtalein vn cha ha hng (phenol phtalein ha hng

pH 8,3). Vy trong th nghim chun dung dch HCl bng dung dch chun NaOH vi chtch th mu l phenolphtalein, th im cui n sau im tng ng.Cu 58.

Ta c phn ng:

NH3 + H2O NH4+ + OH

Ban u 0,1M 0,1M 0in li xM xM xMLc c/bng (0,1-x)M (0,1+x)M xM

Vy 3

+ --5 -5 -54

NH

3

[NH ][OH ] (0,1 + x)xK = = 1,8.10 = 1,8.10 x = 1,8.10

[NH ] (0,1 - x)

Do pH = -log[OH

] = -log(1,8.10-5

) = 4,74 tc pH = 9,24Cu 60.Gi a, b ln lt l s mol CH3OH b oxi ha to anehit v axitGi c l s mol CH3OH cn d trong X, ta c cc phn ng:

2HCH2OH + O2 2HCHO + 2H2Omol a a a

HCH2OH + O2 HCOOH + H2Omol b b b

Vy X gm:

2

3

HCHO: a mol

COOH: b mol

a + bO: mol2

CH OH: c mol

Do ta c h:

26,194a + 2b = = 0,2425

108 a = 0,05125b a + b c 0,672

+ + = = 0,03 b = 0,018752 4 2 22,4

c = 0,006250,42

b = = 0,0187522,4

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Vy %CH3OH b oxi ha =(a + b).100

= 91,8(%)a + b + c