de chon HSG Hoa 12

S GD&T BC NINH THI CHN HC SINH GII LP 12

TRNG THPT L THI T Nm hc 2014 2015

Mn: Ho hc

Ngy thi: 14/09/2014

Thi gian lm bi 150 pht

Cu 1: (6,0 im).

1. Kh A khng mu c mi c trng, khi chy trong kh oxi to nn kh B khng mu, khng mi. Kh B c th tc dng vi liti kim loi nhit thng to ra cht rn C. Ho tan cht rn C vo nc c kh A. Kh A tc dng axit mnh D to ra mui E. Dung dch mui E khng to kt ta vi bari clorua v bc nitrat. Nung mui E trong bnh kn sau lm lnh bnh thu c kh F v cht lng G. Xc nh cc cht A, B, C, D, E, F, G v vit phng trnh ho hc ca cc phn ng xy ra.

2. a) Cho dung dch H2O2 tc dng vi dung dch KNO2, Ag2O, dung dch KMnO4/H2SO4 long, PbS. Vit phng trnh ho hc ca cc phn ng xy ra.

b) Nu phng php iu ch Si trong cng nghip v trong phng th nghim. Vit phng trnh ho hc ca cc phn ng xy ra.

c) - Tinh ch kh NH3 c ln kh N2, H2.

- Tinh ch NaCl c ln Na2HPO4, Na2SO4

3. A, B, C, D, E, F l cc hp cht c oxi ca nguyn t X v khi cho tc dng vi NaOH u to ra cht Z v H2O. X c tng s ht proton v ntron b hn 35, c tng s oxi ha dng cc i v 2 ln s oxi ha m l -1. Hy lp lun tm cc cht trn v vit phng trnh phn ng. Bit rng dung dch mi cht A, B, C trong dung mi nc lm qu tm ha . Dung dch E, F phn ng c vi dung dch axit mnh v baz mnh.

Cu 2: (6,0 im).

1/ T naphtalen v cc cht v c cn thit, vit phng trnh chuyn ho thnh axit phtalic. Ghi r iu kin nu c.

2/ Oxi ho khng hon ton etilenglicol thu c hn hp 5 hp cht hu c cng s nguyn t cacbon trong phn t. Hy vit cng thc cu to ca 5 cht v sp xp theo th t tng dn nhit si. Gii thch ngn gn.

3/ Hon thnh s chuyn ha sau:

A

0

,

t

xt

B

0

4

,

t

KMnO

C

HCl

D

25

PO

G.

Bit G c cng thc phn t C12O9. A l but-2-in.

4/ Anken A c cng thc phn t l C6H12 c ng phn hnh hc, khi tc dng vi dung dch Brom cho hp cht ibrom B. Cho B tc dng vi KOH trong ancol un nng, thu c ankaien C v mt ankin D. Khi C b oxi ho bi dung dch KMnO4/H2SO4 v un nng thu c axit axetic v CO2

a/ Xc nh cng thc cu to v gi tn A, C, D. Vit phng trnh ho hc ca cc phn ng xy ra.

b/ Vit cc ng phn hnh hc ca C.

Cu 3: (3,0 im).

Cho 3,58 gam hn hp X gm Al, Fe, Cu vo 200 ml dung dch Cu(NO3)2 0,5M. Khi phn ng hon ton c dung dch A v cht rn B. Nung B trong khng kh nhit cao n phn ng hon ton thu c 6,4 gam cht rn. Cho A tc dng dung dch NH3 d, lc kt ta nung trong khng kh n khi lng khng i thu c 2,62 gam cht rn D.

1/ Tnh phn trm khi lng mi cht trong hn hp ban u.

2/ Ho tan hon ton 3,58 gam hn hp X vo 250 ml dung dch HNO3 a (mol/l) c dung dch E v kh NO (sn phm kh duy nht). Dung dch E tc dng va ht vi 0,88 gam bt ng. Tnh a.

Cu 4: (5,0 im).

1. Hp cht hu c A (cha 3 nguyn t C, H, O) ch cha mt loi nhm chc. Cho 0,005 mol cht A tc dng va vi 50 ml dung dch NaOH (khi lng ring 1,2 g/ml) thu c dung dch B. Lm bay hi dung dch B thu c 59,49 gam hi nc v cn li 1,48 gam hn hp cc cht rn khan D. Nu t chy hon ton cht rn D thu c 0,795 gam Na2CO3; 0,952 lt CO2 (ktc) v 0,495 gam H2O. Nu cho hn hp cht rn D tc dng vi dung dch H2SO4 long d, ri chng ct th c 3 cht hu c X, Y, Z ch cha cc nguyn t C, H, O. Bit X, Y l 2 axit hu c n chc. Z tc dng vi dung dch Br2 to ra sn phm Z c khi lng phn t ln hn Z l 237u v MZ < 125 u. Xc nh cng thc cu to ca A, X, Y, Z, Z.

2. Cho hn hp A gm 3 hirocacbon X, Y, Z thuc 3 dy ng ng khc nhau, hn hp B gm O2 v O3. Trn A v B theo t l th tch tng ng l 1,5 : 3,2 ri t chy hon ton thu c hn hp ch gm CO2 v hi H2O theo t l th tch l 1,3 : 1,2. Bit t khi ca kh B i vi hiro l 19. Tnh t khi ca kh A i vi hiro?

----------------------------------------------HT-------------------------------------------------

( thi gm 02 trang)

Cho: C = 12; O = 16; H = 1; Ag = 108; Na = 23; Cl = 35,5; K = 39; N = 14; Br = 80;

Cu = 64; Ca = 40; P = 31; Si = 28; S = 32; Ba = 137; Al = 27; Fe = 56; Zn = 65;

Li = 7; Rb = 85; Cs = 133.

- Hc sinh khng c dng bng HTTH.

- Cn b coi thi khng gii thch g thm.

S GD&T BC NINH HNG DN CHM THI CHN HSG LP 12

TRNG THPT L THI T Nm hc 2014 2015

Mn: Ho hc

Ngy thi: 14/09/2014

Thi gian lm bi 150 pht

Cu

Ni dung cn t

im

Cu 1

(6,0)

1.

2

Lp lun a ra: kh A l NH3. Kh B l N2. Cht rn C l Li3N. Axit D l HNO3. Mui E l NH4NO3. .................................................................

Vit cc phng trnh ho hc xy ra: (Mi pt 0,25x5=1,25 )

4NH3 + 3O2

0

t

N2 + 6H2O.

N2 + Li

Li3N.

Li3N + 3H2O

NH3 + 3LiOH

NH3 + HNO3

NH4NO3.

NH4NO3

N2O + H2O.

0,75

1,25

2

2

a. Phng trnh ho hc xy ra: (Mi phng trnh 0,25 x 4 pt =1,0 )

H2O2 + KNO2

KNO3 + H2O.

H2O2 + Ag2O

2Ag+ O2 + H2O.

5H2O2 + 2KMnO4 + 3H2SO4

5O2 + 2MnSO4 + K2SO4+ 8H2O.

4H2O2 + PbS

PbSO4 + 4H2O.

b. iu ch Si trong CN: dng than cc kh SiO2 trong l in:

SiO2 + 2C

Si + 2CO....................................................................

iu ch Si trong phng th nghim: Nung Mg vi SiO2:

SiO2 + Mg

Si + MgO......................................................................

c. - Dn hn hp (NH3, H2, N2) qua dung dch axit (VD: dd HCl), NH3 b gi li. Tip n cho dung dch baz d (VD dd Ca(OH)2) v un nng nh, kh thot ra cho i qua ng ng CaO d s thu c NH3 kh

NH3 + H+ NH4+

NH4+ + OH- NH3 + H2O

Tinh ch NaCl c ln Na2HPO4 v Na2SO4

Cho hn hp vo dung dch BaCl2 d

Na2HPO4 + BaCl2 2 NaCl + BaHPO4

Na2SO4 + BaCl2 2NaCl + BaSO4

lc b kt ta, dung dch thu c cho vo bnh cha Na2CO3 d

BaCl2 + Na2CO3 2 NaCl + BaCO3

lc b kt ta, thm lng d dung dch HCl vo dung dch thu c, sau c cn ri nung nng nh thu c NaCl khan.

1,0

0,25

0,25

0,25

0,25

3

2

Xc nh X: p+n 0,1 mol =>Cht rn sau khi nung B trong khng kh c khi lng > 0,1.80 = 8(g) (khng ph hp).

Vy Cu2+ d nn Al v Fe ht.

Gi s mol Al ,Fe, Cu trong hn hp X ln lt l: a, b, c.

Phng trnh v khi lng hn hp: 27a + 56b + 64c = 3,58 (I)

Cht rn sau khi nung ch c CuO: 3a/2 + b + c = 0,08 (II)

Dung dch A cha: Al3+, Fe2+, Cu2+ d

Al3+, Fe2+, Cu2+

3

NH

+

d

Fe(OH)2, Al(OH)3

0

,

tkk

Fe2O3, Al2O3.

khi lng cht rn D: 102.a/2 + 160.b/2 = 2,62 (III)

Gii h (I), (II), (III) ta c: a = 0,02; b=0,02, c=0,03.

% khi lng ca mi kim loi trong hn hp l:

Al =15,084%; Fe=31,28%; Cu=53,63%.

1,0

0,5

0,5

2.

1,0

Theo gi thit nhn thy: hn hp X v 0,88 gam Cu ( tc 0,01375 mol) tc dng va vi 250 ml dung dch HNO3 a(mo/l). Theo L bo ton e suy ra s e nhn do HNO3 bng tng s e nhn do hh X v 0,88 gam Cu.

S e nhng =

322

AlFeCu

nnn

++=

0,06+0,04+0,0875=0,1875 (mol)

Qu trnh nhn e: 4H+ + NO

3

-

+3e

NO + 2H2O

0,25 0,1875

S mol HNO3=s mol H+=0,25 (mol)=> a = 1M.

0,5

0,5

Cu 4

5,0

1. 4,0

2. 1,0

p dng bo ton khi lng ta c: mA + mddNaOH = mhi nc + mD

mA = 59,49 + 1,48 50.1,2 = 0,97 (g)=> MA = 0,97/0,005=194 (g)....

Mt khc theo gi thit: D

chy

0,795 gam Na2CO3 + 0,952 lt CO2 (ktc)

0,495 gam H2O.

=>

232

CO

0,0075(); n0,0425()

NaCO

nmolmol

==

p dng LBT nguyn t C ta c:

nC(trong A) =

()()

232

NaCOCO

CC

nn

+

= 0,0075 + 0,0425 = 0,05 (mol)

BT nguyn t H:

22

()( )( dd NaOH)( H) ( D)

HtrongAHtrongNaOHHtrongHOHtrongOH

nnnnn

++=+

ban ucahit chy

nH(trongA) = 0,05 (mol)

Gi cng thc phn t A l CxHyOz. Ta c:

x = nC/nA = 0,05/0,005=10

y = nH/nA = 0,05/0,005 =10 => z = (194-10.12-10)/16 = 4

Vy cng thc phn t A l C10H10O4. .....................................................

Xc nh cng thc cu to ca A:

S mol NaOH phn ng vi A = 2.

23

NaCO

n

=0,015 (mol)

Vy t l:

0,0051

0,0153

A

NaOH

n

n

==

; Trong A c 4 nguyn t O nn A c th cha 2 nhm chc phenol v 1nhm chc este COO- hoc A c 2 nhm chc este COO- trong 1 nhm chc este lin kt vi vng benzen. Nhng theo gi thit A ch c mt loi nhm chc do A ch cha hai chc este (trong mt chc este gn vo vng benzen) => A phi c vng benzen.

Khi A tc dng vi dd kim thu c X, Y l 2 axit hu c n chc.

Z l hp cht hu c thm cha 1 nhm chc phenol v 1 chc ancol

S nguyn t C trong Z 7

Tng s nguyn t C trong X, Y = 3.

Vy 2 axit l CH3COOH v HCOOH.......................................................

Nh vy Z phi l: OH-C6H4-CH2OH (c 3 ng phn v tr o ,m, p)

Khi Z tc dng dd nc brom to ra sn phm Z trong :

'

237

Z

Z

MM

-=

=> 1 mol Z th 3 nguyn t Br. Nh vy v tr m l thun li nht. CTCT ca Z v Z l:

( Xc nh Z, Z mi cht 0,5 )

CH

2

OH

OH

EMBED ChemDraw.Document.6.0

CH

2

OH

OH

BrBrBr

Br

.............................................

CTCT ca A c th l

CH

2

-O-CO-CH

3

O-CO-H

hoc

CH

2

-O-CO-H

O-CO-CH

3

..

2 (1,0 im)

t cng thc cht tng ng ca hn hp A l

xy

CH

= 19.2 = 38

B

M

=> t l s mol O2 v O3 l 5:3

Trn A vi B theo t l th tch 1,5: 3,2.

Chn nB = 3,2 mol => n (O2) = 2 mol; n (O3) = 1,2 mol

nO = 7,6 mol

Khi nA = 1,5 mol. Khi t chy A ta c th coi:

xy

CH

+ (2

x

+

2

y

) O

x

CO2 +

2

y

H2O

Mol 1,5 1,5(2x+

2

y

) 1,5

x

1,5

2

y

Ta c: nO = 1,5(2x+

2

y

) =7,6 (*)

V t l th tch CO2 : H2O = 1,3:1,2 =>

x

:

2

y

= 1,3:1,2 (**)

Gii h (*), (**) ta c:

x

= 26/15;

y

= 16/5 = 3,2

= 12x + y = 24

A

M

=> dA/H2 = 12

0,5

1,5

0,5

1,0

0,5

0,5

0,5

Ghi ch: Hc sinh lm theo phng php khc, nu ng vn cho im ti a ng vi mi phn.