p/idn iXng 1x6a hoc xdy ra (gid svC khong khi cJii gom N2 vd O2).
2. Trgn 190 cm^ klii oxi vai 120 cm^ lion hap kiii X gom N2, H2 vd CH^. Dot c/idy liodn todn hdn hap X, sau do Idm Ign/i, t/iu dugc /ion /igp Y Dot c/idy liodn todn hdn hap X, sau do Idm Ign/i, t/iu dugc /ion /igp Y
CO t/ie tich la 110 cm^. Cho Y qua dung dich NaOH dU, khi t/iodt ra cd
t/ie tich la 40 cm^. Tin/i t/ie tich moi khi trong /iSn /igp X. Biet cdc
k/ii do a Cling dieu kien nhiet do vd dp sudt. Cdu 4. Cdu 4.
1. Clio a gam dung dich H2SO4 10% p/idn ting vai a gam dung dich KOH
20%. Do pH cda dung dich thu dugc sau p/idn dng n/iU t/ie ndỏ
2. Trgn Idn 100 gam dung dich H2SO4 10% vai 200 gam dung dich
H2SO4 C% t/iu dugc dung dich H2SO4 30%. Tinh C% vd trinh bay
cdc/i p/ia trgn.
3. C/io 21,3 gam P2O5 p/idn dng vdi 200 ml dung dich hon /igp KOH IM vd NaOH 0,5M. Tinh khoi lugng muoi t/iu dUgc sau phdn ling. vd NaOH 0,5M. Tinh khoi lugng muoi t/iu dUgc sau phdn ling.
Cdu 5. Ngdm mot thanh dong vdo 250 gam dung dich AgNOs 6,8%. Sau
mot t/iai gian n/idt thanh dong ra, t/iu dUgc dung dich X c6 lilioi iMng
243,92 gam (gid sd bgc sin/i ra bdm het vdo t/ianh dong).
a) Tinh nong do p/idn trdm cdc c/idt trong dung dich X.
b) De p/idn dng /let vai cdc c/idt trong dung dich X p/idi diing 3,25 ga"^
kim logi M c6 /i6a tri /i/i6ng doị Xdc dinh ten /dm logi M.
Cdu 6,
1. Dung dich Că0H)2 bdo /loa a 25°C c6 nong do la 0,027M. C/io 6 g^^"^ canxi p/idn dng vdi 100 gam ndac. Tinh /i/i6i lugng Că0H)2 rdn canxi p/idn dng vdi 100 gam ndac. Tinh /i/i6i lugng Că0H)2 rdn dugc (gid sd the tich dung dich bdng the tich nuac).
2. De m gam sdt ngodi khong k/ii, sau mot t/idi gian t/iu dugc c/idt rdn ^
cd k/idi lugng Id (m + 1,6) gam. Neu c/io todn bg X tdc dung vdi d'^'^^ dich H2SO4 đm dgc, nong, dU thi dUgc 4,48 lit li/ii SO2 (d/itc) d'O dich H2SO4 đm dgc, nong, dU thi dUgc 4,48 lit li/ii SO2 (d/itc) d'O
n/idt t/iodt rạ Viet cdc p/iuang trinh p/idn dng cd t/ie xdy rạ Tin/i-
i r i i R i i i [ I F THI w n n s i w H niQI HÓ*
DE SO 50
OE THI HOC SINH GiOi HOA HOC 9, THANH PHO HO CHJ MINH NAM HOC 2010 - 2011
Cdu 1' (4 diem)
a) C/io sa do bien doi sau:
Ai + B, -> A, NaCl- -> NaCl NaCl- -> NaCl A , _lB3 -> NaCl -> A , -> NaCl -> A ,
C/io biet Ai id /dm logị Tim cdc c/idt Ai, A2, A3, A4, A5, Ag, B^, B2, B 3 , B4 vd /lodn t/idn/i cdc p/iUang trinh /i6a /igc cda sa do tren. B4 vd /lodn t/idn/i cdc p/iUang trinh /i6a /igc cda sa do tren.
b) C/io bgt sdt tdc dung /lodn todn vai lugng du dung dich H2SO4 dgc, nong nong do 78,4% t/iu dugc dung dich A trong do nong do % cua nong nong do 78,4% t/iu dugc dung dich A trong do nong do % cua
Fe2(S04)3 vd cda H2SO4 dU bdng n/iau vd gidi p/idng /du SO2. Tinh nong do % cua Fe2(S04)3 vd H2SO4 dụ nong do % cua Fe2(S04)3 vd H2SO4 dụ
Cdu 2. (4 diem)
a) Cd 5 Ig /diong n/idn, biet 4 Ig dUng dung dich sau cd cung nong do
mol/lit: NaOH, NaCl, NaHS04, BaCh vd 1 Ig dung nudc. C/u dung t/iem t/iudc t/id duy n/idt Id dung dich p/ienolp/italein, /idy neu cdc/i t/iem t/iudc t/id duy n/idt Id dung dich p/ienolp/italein, /idy neu cdc/i
n/idn ra tdng Ig.
b) Hon /igp X gom Fe, FeO, Fe203. C/io mot luong /dii CO di qua ong sd dung m gam hon /igp X nung nong. Sau k/ii ket t/idc t/ii ng/iiem t/iu dung m gam hon /igp X nung nong. Sau k/ii ket t/idc t/ii ng/iiem t/iu dugc 64,0 gam chat rdn A trong ong sd vd 11,2 lit /du B ( d/dc), cd ti khoi /lai so vdi H2 Id 20,4. Tin/i m.
3. (4 diem)
0-) Trong /igp c/idt C4H4CI4 cd 4 nguyen td CI deu a vi tri cdu tgo liodn todn gidng n/iaụ Hay viet cong t/idc cdu tgo cd t/ie cd cda c/idt /idu ca ndỵ gidng n/iaụ Hay viet cong t/idc cdu tgo cd t/ie cd cda c/idt /idu ca ndỵ
b) Cho mot an/ian A a t/ie k/iị De dot c/idy hodn todn 1,2 lit A can dung vda /let 6 lit oxi lay d cung dieu kien. Xdc dinh cong thdc p/idn td cda vda /let 6 lit oxi lay d cung dieu kien. Xdc dinh cong thdc p/idn td cda A, viet cong t/idc cdu tgọ Neu c/io c/idt A tdc dung vdi Clo d 25°C vd
CO dn/i sdng t/ii t/iu dugc boo n/iieu dan xudt monoclo cua Ạ
C a u 4. (4 diem)
Kid dieu die axetilen bdng phuang phdp nhiet phdn nhanh metan thu
dugc lion hap khi A gom axetilen, hidro vd melan chua phdn I'Cng. Ti
kiwi hai cila A so vai hidro bdng 5.
a ) Viet phuang trinh hoa hoc xdy rạ
b) Tinh hieu sudt cila qua trinh chuyen hoa metan thanh axetilen.
c ) Neu ban đu lay 1 mol metan de thuc hien phdn I'ing nhiet phdn uJm
tren de tao thanh hon hap khi A cilng gom 3 chat axetilen, hidro
metan tlii khi dot chdy hodii todn A ta tliu diCgc bao nhicu gam CỌ,
vd bao nJiieu gam H2Ỏ
C a u 5. (4 diem)
Hoa tan hodn. todn 46,4 gam mot oxit kim loai bdng dung dich HoSÔ
đc, nong (viia dil) thu dugc 2,24 lit khi SOo (dktc) vd 120 gam mudị
a) Viet phuang trinli hoa hoc xdy rạ
b) Xdc dinh cong thUc cila oxit kim loaị
c ) Viet phuang trinh hoa hoc cila oxit tren vai dung dich HCl.
L C f l GIAI
C a u 1.
a) 2 N a C l di^n phan nong chay ^ g N a + C ^ t
(Ai) (A4) 2 N a + CI2 > 2 N a C r (Ai) (A4) 2 N a + 21120 > 2 N a O H + H a t (Ai) ( B i ) (A2) CI2 + H2 —> 2HC1 (A4) (B3) (As) N a O H + H C l > N a C l + H2O (A2). (As)
2 N a O I i + CO2 > NaaCOa + H2O (A2) (B2) (A3)
2HC1 + B a S > BaCl2 + H2S
(As) (B4) (Ae)
NaaCOa + BaCl2 > N a C l + B a C O g i
(As) (Ae)
i r l i niAi T U I u n n c i u u nini unA u n r Q
Fe-ASOAh + SSOat + 6H2O
0,5a 1,5a
Goi a l a so m o l Fe va b la so m o l cua H2SO4 trong h o n hop b a n daụ 2Fe + 6 H 2 S 0 4 d a c (mol) a ^ 3a T a c6: m H 2 S 0 4 b a n d d u = 98b ( g a m ) 98b X 100 , = > n i d u n g dich H 2 S O 4 78,4% = z ^ — = 125b ( g a m ; 78,4
= ^ m d u n g dich sau phan .rng = 56a - 125b - 96a = I 2 5 b - 40a (gam)
^ mn^sô du = 98b - 294a (gam); m^e^isô^^ - 200a (gam) Theo d4 b a i , t a c6: C% H ^ S O^ d.r = C% F e s c s o ^ , ,
(98b - 2 9 4 a ) x 100 ^ 200a x 100 ^ b = 5,0408a 1 2 5 b - 4 0 a ~ 1 2 5 b - 4 0 a
I r^cf nr/ 200a X 1 0 0 % _ qo oqoL
a u 2.
a) T r i c h m o i dung dich m o t i t l a m m a u thuf.
Cho p h e n o l p h t h a l e i n Ian liJcft vac cac mSu tb"^ t r e n
+) M a u thuf x u a t h i e n m a u h o n g la dung dich N a O H (nhdm I)
+) Cac m a u con l a i k h o n g c6 h i e n tUdng g i : N a C l , N a H S 0 4 , BaClz
va nifofc (nhdm II).
Cho s a n p h a m d n h o m I (nho vai giot) I a n l^*?*^ vao cac m a u thijf 6 n h o m ( H )
+) M a u thijf nao l a m m a t m a u h o n g la dung d i c h N a H S 0 4 N a O H + " N a H S 0 4 > Na2S04 + H2O
+) M l u thtf k h o n g l a m m a t m a u h 6 n g l a : N a C l , BaCla va nadfc Cho san p h a m v t o t h u diTc^c nho I a n lugt v a o cdc m a u thCf c6n laị
+) M a u thir tao k e t t u a m a u t r S n g l a dung d j c h BaClg
Na2S04 + BaCl2 > B a S 0 4 i + 2 N a C l
+) Mau t h u f khong c6 hien tugng gi la NaCl va n U d c . Den day t a d ,
rat nhieu phuang phdp de nhan ra 2 chat nay, sau day 1^ p h i T o n g
phdp minh hpạ
Phitctng phdp 1: Dung dua thuy tinh nhung vao hai dung dich s a u
do dot tren ngon l u f a den con, m i u thuf chay vdri ngon lufa m a u
vang la NaCl, con lai la niidc
PhUtfng phdp 2: Nho vai giot hai chat len tam kinh sach va dun
nhe tren ngon lufa den con, mau thuf tao can mau trdng la NaCl,
khong CO hien tiictng gi la nUdc.
b) Cac phan ufng khuf sit oxit c6 the c6:
SFeaOg + CO — > 2Fe304 + CO2 ' (1)
Fe304 + CO — > 3FeO + CO2 (2)
FeO + CO Fe + CO2 (3)
Nhir vay chat rSn A c6 the gom 3 chat Fe, FeO, Fe304 hoac i t han,
dieu do khong quan trong va viec can hhng cac phacrng trinh tren
cung khong can thiet cho viec xac dinh dap an, quan trpng 1^ so mol CO phan ufng bao gid cung bang so mol CO2 tao thdnh.
Ta c6: n„ = ^ = 0,5 (mol) ^ 22,4
Goi x la so mol cua CO2, ta c6 phiicfng trinh ve khoi laang cua B:
44x + 28(0,5 - X ) = 0,5 x 20,4 x 2 = 20,4
=> x = 0,4 (mol) va do cung chinh la so mol CO t h a m gia phan u f n g
Ap dung dinh luat bao toan khoi liicfng, ta c6: mx + mco = m A + m^Q^
^ m = 64 + 0,4 X 44 - 0,4 x 28 = 70,4 (gam).
LOI GIAI Oi THI HOC SINH GIOI HOA HOC 9
a) Cac cong thiJc cau at5o c6 teh c6 cua hgp chat C 4 H 4 C I 4 thoa man diou kien de bai la:
CI CI ^CH—C—HC ^CH—C—HC CI 1!H, CI 01^ / C I CH—CH ^CH—CH CI CI CI , CI ^CH—CH=CH—HC^ CI CI CI CI—C CH2- HpC C—CI CI CH3 CH / \ CI—C C—CI CI CI
b) Gpi cong thufc tong quat cua ankan c6 dang: C„H2„ + 2 (n > 1).
Phan úng: C Ji2„ , 2 + '"^^^^ ^ O2 (lit) (lit) Turd) 1 1 , 2 3n + 1 1 _ " 2 ^ 2 3n + 1 2 6 -> nC02 + (n + DllaO ( I j n (n + 1) 6 = 0,6(3n + 1) n = 3 1,2 6
Vay cong thufc phan tuf cua ankan (A) a the khi la: Callg.
Cong thufc cau tao cua C3II8 la: CH3-CII2-CH3
Phan ufng: ^ CH3-CH2-CH2CI + IICl
anil sang
cii3-cri2-cn3 + CI2
CHs-CHCl-ClIa + HCl Vay khi cho Callg tac dung vdri clo khi c6 anh sang lam xuc tac thi tliu diioc 2 dan xuat
C a u 4.
a) P h a n ufng: 2CH4 t.ch nhan.;! iSiluh . . h a n i r - > + 31-12 (1 !
(mol) ' X - > 0,5x l , 5 x
b) T i n h h i e u suat cua qua t r i n h chuyen hoạ
Goi a l a so m o l cua CH4 b a n dau va x l a so m o l cua CH4 t h a m gia p h a n ijfng
TO (1) Hf,;,^ = a - X (mol); n,, „^ = 0,5x (mol) v a n^^^ = l , 5 x (mol)
M a : CIA/H = T T ^ = 5 => M A - 5 X 2 - 10 ( g a m )
1 6 ( a - x ) + 13x + 3 x ^ . , ^ i r , - ^ nr
<=> — = 10 c;> 16a = 10a + l O x => x = — - 0,Ca
( a - x ) + 0,5x + l , 5 x 10
6a
V a y , h i e u suát p h a n ifng l a : H = ™ x 100 = -iS- x 100% - 6 0 %
a a
c ) P h a n l i n g : CH4 + 2O2 CO2 + 2II2O (1)
(mol) 0,4 0,4 0,8 C2H2 + 2,502 2CO2 + H2O (2) (mol) 0,3 0,6 0,3 2H2 + O2 — > 2H2O (3) (mol) 0 , 9 - > 0,9 TCf (1) v a (2) ^ n,,ô = 0,4 + 0,6 = 1 (mol) => m^,)^ = 1 X 44 = 44 (gam)
Tii (1), (2) v a (3) ^ n„^o = 0-8 + 0,9 + 0,3 = 2 (mol) => m „ „ = 2 X 18 = 36 (gam)
C a u 5, G o i cong thufc oxit k i m loai can t i m l a IlxOy v a c6 a m o l p h a n ufng.
a) P h a n ufng:
2RxOy + ( 2 n x - 2y)H2S04 dac —> xR2(S04)„ + ( n x - 2y)S(> ax a ( n x - 2 y ) (mol) a 2 2 + ( 2 n x - 2y)H20 I ril nifli r i F T M I w n n H U H m n i H f i a H n n 9 b) T a c6: n . . , = 2,24 22,4 = 0,1 (mol)
Theo de bai, ta c6 he phúong trinh:
a ( n x - 2 y ) nso, = ^ = 0,1 m„,i^ = a ( R x + 16y) = 46,4 ax - ( 2 R + 9 6 n ) = 120 a n x - 2 a y - 0,2 Rax + 16ay = 46,4 <=j> R a x + 4 8 a n x = 120 R a x + Banx = 4 8 R a x + 4 8 a n x = 120
Nghiem hop l i : n = 3; R = 56: sSt (Fe). a ( n x - 2 y ) a ( n x - 2 y ) 8 a n x - 1 6 a y = 1,6 Rax + 16ay = 46,4 Rax + 4 8 a n x = 120 Rax = 33,6 a n x = 1,8 R - — n 0,2 n x - 2 y <=> 1 M a t khac, _ , ,
ăRx + 16y) 4 6 , 4 R x + 16y 2 3 2 232(nx - 2y) = R x + 16y o 232(3x - 2y) = 56x + 16y o 696x - 464y = 56x + 16y « 640x = 480y =^ - =
y 6 4 0 V a y cong thufc oxit k i m loai can t i m l a : Fe304
c) P h a n ufng cua Fe304 v 6 i dung dich H C l
Fe304 + 8HC1 > FeCl2 + 2FeCl3 + 4H2O
3 4
DE SO 51 (TLT L U Y E N )
2.2. Nguyen to kini loai A c6 hoa tri III trong Jigj) dial vdi oxị Diet rdiig
cil 6,4 gam oxit ciia A tdc dung uiCa dii vol 0,4 lit dung dich IlCl 0,6M. a) Xdc dinh him loai Ạ
b) Viet phUang trinh phdn ÍCng ciia A theo so do bien hoa suu:
ACI2 > Ă0H)2 > A S O j
^ACls > Ă0II)3 > A2O3
Cdu 3. (2 diem)
Cho mot mieng kim loai ndtri vdo dung dich FcofSOJ^ c6 du, ngudi hi thu dilgc chat ket tiiạ Bern nung nong chat ket tila nay cho den khi no
CO kiwi liigng khong đi tin dugc 4,8 gam moat chat rdn mdu do ndu Tinli klioi lagng kim loai natri da dung.
Cdu 4. (4 diem)
4.1. Co nigt dung dich chica dong thai axit HCl vd H2SO4. Cho 200 gun: dung dich do tdc dung vai BaCL c6 du thi tao tlidnh 46,G gam chut dung dich do tdc dung vai BaCL c6 du thi tao tlidnh 46,G gam chut ket tiiạ Lgc ho ket tiiạ Be trung hoa phdn nUac Igc (dung dich thu dugc sou khi tdch bo ket tila b&ng each lgc) ngiidi ta plidi dung SOOnil dung dich NaOH 1,6M.
Tinh nong do phdn tram ciia moi axit trong dung dich axit ban đụ
4.2. Hoa tan kim loai Mg vdo GOOml dung dich HCl ta thu dugc 1 dung
dich muoi A vd chat khi thodt rạ Bat chdy chat khi nay ta thu duuc 1,8 gam niiac.
a) Tinh khoi lugng kim loai Mg.
b) Cho vdo dung dich A mot lugng du AgNOj. Neu muon thu dugc
43,05 gam ket tila trdng thi nong do M ban đu ciia dung dich HCl Ut bao nhieu ? Y dinh nay c6 thuc hien dugc khong ? Vi sao ?
Cdu 5. (6 diem)
5.1. Mot hon hgp X goni Al, Fc, Cu can ngng 68,7 gam. Khi cho X tuc
dung vai HNO3 dgc ngugi c6 26;88 lit khi NO2 (dktc) bay rạ Con khi cho X tdc dung vai V lit dung dich H2SO4 0,5M thi thu dugc 23,52 khi (dktc). Xdc dinh khoi lugng moi kim loai trong hon hgp X vd the . tich V cua dung dich H2SO4 0,5M da diing.
5.2. Hoa tan 8,32 gam Cu vdo 3 lit dung dich HNO3 thu dugc dung did' A vd 4,928 lit hon hgp NO, NOo (dktc). Hoi a (dktc) 1 lit hon hgp - A vd 4,928 lit hon hgp NO, NOo (dktc). Hoi a (dktc) 1 lit hon hgp - khi nay c6 khoi lugng bao nhieu gam.
Diet: H ^ 1; O ^ 16,Cl = 35,5; Fe - 56; S - 32; Na = 23: Ba = 13<' Mg - 24; Ag = 108; N = 14; Cu = 64; Al - 27.)
MUC LUC
E S O I . D E THI H O C S I N H GldlHOA H O C 9, T H A N H P H O H O CHI MINH NAM H O C 1998 - 1999
E S O 2. D E THI CHQN DO) T U Y E N HOC S I N H G161 HOA H O C 9, Q U A N T A N BJNH TP, H O C H I MINH N A M H Q C 1998 - 1999 TP, H O C H I MINH N A M H Q C 1998 - 1999
S O 3. D E THI H O C S I N H G161 H O A H Q C 9, Q U A N 3 T P . H O C H I MINH NAM H O C 1998 - 1999 NAM H O C 1998 - 1999
£ 5 0 4. D E T H ! H O C S I N H G161 HOA H O C 9, T P . H O CHi MINH NAM H O C 1999 - 2000
E S O 5. D E THI H O C S I N H GI61 H O A H O C 9, Q U A N 9 ( V O N G 2) TP.HO C H I MINH NAM H O C 1999 - 2 0 0 0 NAM H O C 1999 - 2 0 0 0
E S O 6. D E THI H Q C S I N H G161 H O A H Q C 9, C A P TP. H O C H I MINH NAM HOC 2000 - 2001 NAM HOC 2000 - 2001
E S O 7. D E THI C H Q N H Q C SINH G161 HOA H O C 9 TP. Q U Y NHON,
TINH BiNH DINH NAM HOC 2000 - 2001
E S O 8. D E THI H O C S I N H G161 H O A H O C 9, T P . H O C H I MINH NAM H O C 2001 - 2 0 0 2 NAM H O C 2001 - 2 0 0 2
S O 9. D E THI C H O N H Q C S I N H G161 HOA H Q C 9, T P HAI P H O N G ( B A N G A) NAM H O C 2001 - 2 0 0 2 NAM H O C 2001 - 2 0 0 2
i S O 10. D E THI C H O N H Q C S I N H G161 HOA H Q C 9, TINH BINH DINH