dam dac, ta c6:
Cu + 2H2SO4 (dam dac) > CUSO4 + SO2T + 2H;.('
CuO + H2SO4 > CUSO4 + H2O
Vay k h l Y la SO2 va dung dich X gom: CUSO4, H2SO4 du, H2Ọ - CUSO4/X tac dung vcfi KOH tao T la ket tiia Cu(0H)2 mau xanh
CUSO4 + 2 K 0 H > Cu(0H)2i + K2SO4
- Y (SO2) tac dung vdi dung dich KOH ta diJOc dung dich Z, Z vti 1
tac dung di/cfc vdi dung dich NaOH vCfa tac dung diigc vdi dun-
dich BaCl2 -> Z la hon hop KHSO3 va K2SO3. SO2 + 2 K 0 H > K2SO3 + H2O SO2 + K O H > K H S O 3
2KHSO3 + 2NaOH > Na2S03 + K2SO3 + 2H2O
K2SO3 + BaCl2 > BaSOai + 2KC1
Ldl GIAI Oi THI HOC SINH GIOI HOA HOC 9
u 3.
1. a) Trucfng hgp ca 2 la kim loai deu tan het. Goi a gam la khói lifgng moi la kim loai Goi a gam la khói lifgng moi la kim loai
(mol)
(mol)
56 Fe + 2HC1
a
^ mol, nAi = -r^mol
56 2A1 + 6HC1 a 27 27 > FeCl2 + H2t a 56 -> 2AICI3 + SHat a 18 (1) (2)
Do liigng k h i H2 d binh c6 la nhom thoat ra nhieu hcfn
r a ' a ^
> — I nen sau phan ufng (2), dia can c6 coc chufa la nhom
I18 56,
se nhe hcfn, bong len, dIa can c6 coc chiifa la sSt se tram xuong.
b) Trufdng hgp the tich hidro thoat ra cf hai coc la bSng nhau: Luc
nay khoi liigng 2 coc giam nhuf nhau, nen hai dia can a vi t r i thang bang.
2. a) Nung nong canxi cacbonat, ta thu diigc C02t CaCOa CaO + C02t CaCOa CaO + C02t
Thi nghiem chuTng minh: Cho k h i sinh r a sue qua dung dich niJcfc voi trong c6 dii, thay c6 ket tua trSng xuat hien, chufng to khl sinh r a la CO2.
CO2 + Ca(0H)2 —> CaCOsi + H2O
b) Cho mangan dioxit tac dung vdi dung dich H C l dam dac thi c6 khi CI2 thoat rạ khi CI2 thoat rạ
^ Mn02 + 4HC1 -> MnCla + Cl2t + 2H2O
Thi nghiem chufng minh: Cho khi sinh ra tac dung vdi giay quy tim tham nude, ban dau quy t i m hoa do, sau do mat mau, chufng to k h i sinh ra la CI2.
CI2 + H2O > HCl + HCIO
HCIO cótinh oxi h6a manh nen c6 tinh tay mau manh
c) Cho kem tac dung vdi dung dich H2SO4 loang t h i c6 k h i H;; thoat ra: thoat ra:
Zn + H2SO4 -> ZnS04 + H2T
Thi nghiem chufng minh: Cho k h i sinh ra loi i i i tCf qua CuO (den) t h i c6 Cu (do) xuat hien, chufng to khi 1^ H2.
H2 + CuO —> Cu + H2O
d ) P h a n urng: N a t r i sunfit
NaaSOs + H2SO,, — > Na2S04 + SO2T + II2O T h i n g h i e m chuTng m i n h : Cho k h i s i n h r a sue qua dung die: nUcfc v o i t r o n g c6 dil, t h a y c6 k e t t i i a t r S n g x u a t h i e n , chijfng 1 , k h i s i n h r a l a SO2 hoac l a m m a t m a u n a u do nuldc b r o m .
SO2 + Ca{0n)2 > C a S O g i + H2O hoac: SO2 + Bra + 2H2O > H2SO4 + 2 H B r
Cdu 4.
1. * Cdc phtfong t r i n h p h a n ijfng chay:
C n H 2„ , 2 + [ ^ ^ J O 2 > n C 0 2 + (n + D H s O
• C n l l a n + f — 1 O2 n C 0 2 + nHaO
C n H 2 n - 2 + ( ^ ^ j ^ 2 n C 0 2 + (n - DHaO
C n H 2 n- 6 + O 2 ^ > n C 0 2 + (n - 3)1120 * H a i ntfcfc s i n h r a l a m mc/ n'gon ItTa, n e n chat c6 t i le h o i ntfot
c a n g \6n t h i do sang ngon liia cang k e m . Thi/c te do sang cua
n g o n lufa t a n g d a n theo thiJ t i i sau:
C „ H 2 u + 2 < C n H 2 „ < C n H 2 n ^ 2 < C n H 2 „ - 6, l a do:
T i le m o l I I 2 O : C O 2 cua C n H g n + 2 > C n H 2 n > C n H 2 „ - 2 > C n l i a n - G * M o i quan he ve l o a i hidrocacbon vcfi t i le m o l : H 2 O : C O 2 .
C n H 2 n + 2 <^ > C „ H 2 n <^ - > C„H2a-2n <^ > C „ H 2„ - 6 <->
n n n n - N e u njj^Q > npQ_^ :=> hidrocacbon la C „ H 2 n + 2-
- N e u nj^^Q = n^Q^ => hidrocacbon la CnH2„.
- N e u njj^Q < n^Q^ => hidrocacbon la C n H 2 u - 2; CJH^n - 6-
2. L a y 1 m o l m o i m a u thuf I a n liicft tac d u n g vdfi dung die!
A g N O s / N H s ( A g 2 0 / N H 3 ) d i i , chat tao duac l i j g n g k e t t u a n h i e u hoí
la C H ^ C H va chat tao daqc lugng k e t t u a i t h a n la C I i = C - C H 3 .
C H . C H + A g 2 0 A g N 0 3 / N H 3 , A g - C C - A g i + H 2 O
1 m o l 240g
2 C H - C - C H 3 + A g 2 0 , 2 A g - C - C - C H 3 4 + H 2 O
1 m o l 147g
I l a i chat k h o n g c6 h i e n t u g n g p h a n Cirng l a C H 2= C H 2 ,
C H 2 = C H- C H 2 - C H 3 . L a y 28 gam m o i chat I a n l u g t cho tac d u n g v 6 i dung d i c l i chuTa 1 m o l Br2, cho den k h i p h a n uTng h o a n t o a n , chat nao l a m m a t m a u dung dich Br2 l a CH2=CH2, chat c h i l a m n h a t m a u d u n g d i c h B r a 1^ C H 2 = C H - C H 2- C H 3 . #3 C H 2= C H 2 + B r a > C H 2 B r - C H 2 B r (mol) — = 1 1 28 C H 2 = C H - C H 2 - C H 3 + Br2 > C H 2 B r - C H B r - C H 2- C H 3 g . (mol) ^ = 0,5 1 '* 56 V d i C H 2= C H- C H 2- C H 3 t h i Bra d J n e n Bra c h i n h a t m ^ u . Cdu 5.
a) * A n k e n va a n k i n c6 ciing so nguyen ttf C , n e n t a goi cong thufc cua chung I a n liigt la: C , i H a n (b mol) va C„Han - a (c mol) va só mol H a
la a mol.
Do t i le % k h o n g p h u thuoc vao t o n g só m o l h o n hcfp dau, n e n t a chon h o n hop dau la 100 m o l .
T a c6: a + b + c - 100 (I) K h i n u n g n o n g h 6 n hop t a c6 t h e c6 cac p h a n ufng:
C j i H a n + H a > C „ H a n + a C u H a n - 2 + 2 H 2 > C n H a n + 2
* Co 2 triicfng hop:
' - Tri/ofiig i i ( / p 3: H a tac d u n g h e t k h i s i n h r a gom:
C „ H a n + 2, C„Han du, C , i H a n ^ 2 dtf.
G o i M l a k h o i l i / g n g m o l ciia a n k e n ( C n H a n ) , t h i khói liicfng m o l t r u n g b i n h cua h o n hcfp sau k h i n u n g la M , t a c6:
( M - 2) < M < ( M + 2) => Gia t r i ciia M diicfc x6c d i n h t r o n g k h o a n g sau:
[26 - 30]; [40 - 44]; [54 - 58];... — 20 Theo dau b a i , t a c6: M = 7,8 x 2 x — = 34,6
9
T a t h a y M k h o n g thuoc cac k h o a n g xac d i n h da n o i , chufng to H2 k h o n g tac d u n g het. T a l o a i trifcfng hop naỵ
Triiiing hop 2: H2 d i i
C „ H 2 „ + H 2 > C n H a n . 2
(mol) b b b
C n H 2 n - 2 + 2 H 2 > C n H g n . 2
Khói lugng mol trung b i n h c u a hon hap trudfc phan ufng:
M ' = 2a + Mb + (M - 2) c
= 7,8 X 2 = 1 5 , 6
100
=^ M(b + c) + 2(a - c) = 1560 d , H o n hcfp trirdc v a sau phan urng c6 khói lagng khong doi, nen ta eo
khoi Itfcfng mol trung binh cua hon hcfp sau phan ufng l a :
a - c
20
M(b + c) + 2(a - c) = 15,6 x — ( a - c)
9 (II)
Tii (I) va (II) 15,6 20
9 (a - c) = 1560 (a - c) = 45 M(b + c) = 1470 (a - c) = 45 M(b + c) = 1470 1470 1470 105 (b + c) = V i : a + b + c = 100 a - c = 45 M 14n b + 2c = 55 n V i : (b + 2c) la so mol H 2 phan tfng.
Do vay só mol H 2 ban dau a > b + 2c = 55. V i : (b + c) = 100 - a => (b + c) < 45 Mat kliac: (b + 2c) = 55 ^ 2(b + c) = 55 + b (b + c) = 27,5 + 0,5!. 105^ Do b > 0 27,5 < b + c < 45 --=> n = 3 (n e N*) => vdi b + c = b + c = 3f =^ a = 65; c = 20; b = 1
Cong thufc phan tuf ciia anken va ankin la: CSHQ va 03114.
Cong thufc cau tao cua anken va ankin la: C I l 2 = C H - C H 3 va CH=C-Cri3
b) C a c phiJang trinh phan ufng dieu che 1,3,5-trimetylbenzGn: C H 2 = C H - C H 3 + Br2 > C n 2 B r - C H B r - C H 3 C H 2 = C H - C H 3 + Br2 > C n 2 B r - C H B r - C H 3 C H 2 B r - C H B r - C H 3 K O H / n r o u -> C I I ^ C - C H . + 2 H B r
3 C H = C - C H 3 t " , p . xt
CII3
Lfll GIAI Oi THI HOC SINH GIOI HOA HOC '•
DE SO 28
OE THI HOC SINH GIQI HOA HOC 9, TP. HO CHI MINH NAM HOC 2005 - 2006
Cdu I-
1. Mot nguyen to A c6 tdng so hat trong nguyen tit bang 40, tong so hat
^ niang dien nhieu han so hat khong niang dien bdng 12.
a) Xdc dinh vi tri cila nguyen to A trong bdng he thong tudn hoan,
ten nguyen to Ạ
b) Viet phitang trinh hoa hoc dieu che A tit oxit cua nọ
c) Cho mot day lam bdng nguyen to A vdo dung dich CUSO4. Quaii sat liien tugng, viet phuang trinh phan itng. sat liien tugng, viet phuang trinh phan itng.
Cho a mol dung dich NaOH tdc dung vai b mol dung dich H2SO4. Hay bien lugn cac truang hap xdy ra de dung dich sau phan icng c6 pH^7, <7,> 7.
Cho cac khi CI2, CO2, SO2, O2, Ih, NO2, H2S. Moi khi deu chita hoi dm. Hoi diing mot trong cac chat ndo sau day de lam kho moi khi:
H2SO4 dgc, P2O5, CaO, NaOH rdn, CaCL rdn.
Cau I I .
1. Dan khi do vao trong 2 dug nghiem. Ong 1 chita dung dich NaOH,
dng 2 chita dung dich Că0H)2 (biet cdc phan itng xdy ra vila du). Vict cdc phuang trinh phan itng. Clio biet cdc dung dich tao nen c6 tinh gỉ Tgi saỏ
2. Co 4 binh mat nhdn, moi binh chiia mot hon hgp dung dich nhu sau:
K2CO3 va Na2S04; KHCO3 vci NasCOs; KHCO3 va TVâSOj; Na2S04 va K2SO4. Trinh bay phuang phdp Iioa hoc de nhdn biet 4 binh nay ma chi diing them dung dich HCl vd dung dich BăN03)2.
3. Viet cong tliiic cdu tgo c6 the c6 ciia CeHi4 va C3H5CI3.
Cdu I I I . Cho 49,03 gam dung dich HCl 29,78% vdo binh chiia 53,2 gam nigt kim logi kiem (nhom I). Cho bdc hoi can than dung dich tgo thank,
trong dieu kien khong c6 khong khi, thu dUgc bd rdn. Xdc dinh kim logi trong cdc tFitdng hgp sau:
a) Bd rdn chi c6 mot chat, c6 khoi litgng Id 67,40 gam. b) Bd rdn la hon hgp hai chat, c6 khoi lugng Id 99,92 gam. b) Bd rdn la hon hgp hai chat, c6 khoi lugng Id 99,92 gam. c) Bd rdn la hon hgp ba chat, c6 khoi litgng Id 99,92 gam.
Cho biet: Kim loqi kiem diing truac Mg trong day hoqt dong hoa hoc cua kim
logi, CO hoa tri I vd gom cdc nguyen to: Li (7), Na (23), K (39), lib (85), Cs (133), j i | Fr (223) (so a trong dciu ngoac la nguyen ta khoi).
Cau rV. Cho 4,6 gam natri tdc dung hodn todn vai nitac, sau phdn ling th^
dugc 100 gam dung dich Ạ Dung 50 gam dung dich A cho tdc dung vOi
30 gam dung dich CuSOj 16% thu dugc ket tua B vd dung dich C. a) Viet cdc phiiang trinh phdn itng hoa hgc xdy rạ a) Viet cdc phiiang trinh phdn itng hoa hgc xdy rạ
b) Tinh nong do phdn tram cdc chat c6 trong dung dich A, C.
c) Lgc ket tua B, rvCa sgch dem nung den khoi lugng khong doi thu dugc
chat rdn X. Cho mot luong khi H2 qua X a dieu kien nhiet do cao, sau
phdn ling thu dugc 2,08 gam chat rdn. Tim lugng X tham gia phchi
ling vai H2.
Cdu V. Hoa tan hodn todn 57,6 gam hdn hgp A gom FesÔ, FeoOs, FeO,
trong dung dich HCl thi can diing 360 gam dung dich HCl 18,25% dc
tdc dung vica diị Sau phdn I'ing thu dugc V lit H2 vd dung dich B.
Cho todn bg H2 sinh ra tdc dung het vai CuO du a dieu kien nhiet dn cao, sau phdn iCng thu dugc hon hgp rdn gom Cu vd CuO cd khoi lUgng
nhd han khoi lugng CuO ban đu la 3,2 gam.
a) Neu CO can dung dich B, ta thu dUgc bao nhieu gam miioi khan? b) Neu hon hgp A ban đu c6 tl le mol Fe203 : FeO = 1:1. Tinh n6ni< b) Neu hon hgp A ban đu c6 tl le mol Fe203 : FeO = 1:1. Tinh n6ni<
do phdn tram cdc chat cd trong dung dich B.
c) Hon hgp X cdng chiia Fe304, Fe203, FeO, Fẹ Neu dung 100 gam X cho tdc dung vai 2 lit dung dich HCl 2M. ChvCng minh rdng hon ligp cho tdc dung vai 2 lit dung dich HCl 2M. ChvCng minh rdng hon ligp
X tan het.
LCil GIAI
Cdu I .
1. Goi n, p, e Ian liicft la so natron, proton v a electron cua Ạ
Theo de: n + p + e = 40 (1'
Vi nguyen tijf trung hoa dien nen p = ẹ
(1) . « 2p + n = 40 (•'• Tong so hat mang dien nhieu hon so hat khong mang dien 1^ 12.
=^ 2p - n = 12 (* G i a i (*) va (**) ta diTOc: p = 13, n = 14.
a) X a c dinh v i tri nguyen to A
- A thuoc 6 13 cua bang tuan hoan.
- A thuoc chu ki 3, phan nhom chinh nhom I I => A la nguyen to nhom (Al).
b ) Dieu che A iU oxit cua no
2AI2O3 d'en phan nong chay ^ ^
cnohc ^
c) H i $ n tirang v^ viet phdn ufng:
- Co chat mau do bdm ngoai day nhom. Nhom tan dan, mau xanh cua dung dich nhat dan.