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.
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