2Al,r, + 3CUSO4 ( d u n g d i c h )
> Al2(S04)3 ( d u n g d i c h ) + 3CU(r)
2. P h a n uTng xay r a :
2NaOH + H2SO4 > Na2S04 + 2H2O
(mol) a b
• Neu a = 2b, phan ijfng trung hba viia du nen moi triidng trung
tinh p H = 7.
• Neu a < 2b, phan ijfng trung hoa dii axit nen moi tn/cfng axit pH < 7.
• Neu a > 2b, phan ijfng trung hoa dtf N a O H nen moi truorng baza => p H > 7.
3. H2SO4 dac l a m kho dugc cac khi: CI2, SO2, CO2, O2, H2, NO2. • P2O5 lam kho difac cac k h i : CI2, SO2, CO2, O2, H2, NO2. • P2O5 lam kho difac cac k h i : CI2, SO2, CO2, O2, H2, NO2.
• C a O l a m kho dugc cac k h i : O2, H2.
• N a O H r a n l a m kho daac cac k h i : O2, H2.
• CaCl2 khan lam klio diigc cac khi: CI2, SO2, CO2, O2, H2, NO2, H2S.
//.
D a n k h i clo vao 2 ong nghiem:
1. O n g nghiem 1 chufa dung dich N a O H :
2NaOH + CI2 > N a C l + N a C l O + H2O
Dung dich tao th^nh c6 tinh sat trung v a tay mau vi taang tu
H C I O , N a C l O l a chat oxi hoa manh. • Ong nghiem 2 chufa dung dich Ca(0H)2:
2Ca(OH)2 + 2CI2 > CaCh + Ca(0Cl)2 + 2H2O
hoac CăOH)2 + Cl2 >CaOCl2 + H20
Dung dich tao thanh c6 tinh t a y mau vk sAt trung v i tijang tii
2 . N h a n biet cdc binh mat nhan chijfa cdc hon hap dung dich
(K2CO3 + Na2S04), (KtlCOa + Na2C03), (KliCOs + NazSÔ,,
(Na2S04 + K2SO4) bang phuang phap hoa hoc.
Trich moi hSn hop dung dich mot i t lam mau thijT.
Cho dung dich H C l Ian liigt vao cac m i u thtf.
- M S U t l i L f khong CO hien tuong sui bot k h i do la (Na2S04 + K2SO,, 1
cac mau con lai deu c6 k h i bay rạ
K2CO3 + 2HC1 > 2KC1 + COat + H2O KHCO3 + HCl > K C l + C 0 2 t + H2O
Na2C03 + 2HC1 > 2NaCl + COat + H2O
- Cho dung dich BăN03)2 vao 3 mau thii con lai thi deu tao ket tiia tr&ig.
K2CO3 + BăN03)2 > B a C O g i + 2KNO3
Na2S04 + BăN03)2 > BaS04i + 2 N a N 0 3
Na2C03 + BăN03)2 > B a C 0 3 l + 2 N a N 0 3
Loc lay ket tiia cua tiing mau cho tac dung v6i dung dich H C L
• Ket tua tan 1 phan la dung dich ( K 2 C O 3 + Na2S04)
• Ket tua tan het la (KHCO3 + Na2C03)
• Ket tua khong tan trong dung dich HCl la (KHCO3 + Na2S04).
Chii y: Bai todn nay con each gidi khdc, xln nhuang ban doc.
3 . Cong thufc cau tao c6 the c6 cua CeH^ va C3H5CI3
• C C H H gom: CH3-CH2—CH2—CH2—CH2—CH3 CHg — C H — CHg ~" CHg — j ^ ^ 3 ~ ~ ~ ^ ^ 2 ~ C H 3 CH3 CH„ CH3 CH3 - C - CH2 - CH3 ; CH3 - C H - C H - CH3 CH3 . CH3 • C3H5CI3 gom: CI CI I I • C H „ - C H , - C - C l ; C H „ - C H - C H CI CI CI CI CI I I C H , - C H , - CH ; CH„ - C - C H , ; CH, - CH - C H , CI CI CI CI CI CI CI f Goi
Xac d i n h k i m loai t r o n g cac t r t f d n g hgp: Goi k i m l o a i k i e m t o n g quat l a R va c6 a m o l . 29,78 X 49,03 T a c6: nuci = 100 X 36,5 = 0,4 m o l P h a n ufng: R + H C l -> R C l + - H 2 2
a) Ba r a n c h i c6 m o t chat, c6 k h o l liicfng 67,4 gam.
Theo gia t h i e t , k i m l o a i t h a m gia p h a n ufng h e t vdi H C l tao t h a n h
d u n g d i c h chufa R C l . R + H C l 1 (gam) R (gam) 53,2 Ta CO t i le: R > R C l + - H 2 2 (R + 35,5) 67,4 R + 35,5 o 67,4R = 53,2R + 1888,6 53,2 67,4 c> 14,2R = 1888,6 o R = 133: Cesi (Cs)
b ) B a r a n chijfa h a i chat, c6 k h o i l i i g n g 99,92 gam.
Theo gia t h i e t , sau p h a n ving t h u diJOc h a i chát sau: R O H va R C l
P h a n u f n g : R + H C l (mol) 0,4 0,4 R + H2O -> R C l + ^ H 2 0,4 - ^ R O H + - H 2 It (mol) (a - 0,4) ^ (a - 0,4)
So m o l ciia R d i i de p h a n ijfng vcJi niidrc l a : (a - 0,4) m o l Theo de, t a c6 he phiiang t r i n h :
j a R = 53,2
" 0 , 4 ( R + 35,5) + ( a - 0 , 4 ) ( R + 17) = 99,92
G i a i he p h u a n g t r i n h t a dugc: a = 2,313
R = 23 : N a t r i (Na)
^J^jl^" V a y k i m l o a i k i e m t r o n g triicfng h o p n a y \h. n a t r i (Na).
c ) Ba r a n t h u dUcfc g o m ba c h a t , c6 k h o i l u g n g 1^ 99,92 gam. Theo g i a t h i e t , c h a t rSn gom: RCl, ROH v a R dụ Theo g i a t h i e t , c h a t rSn gom: RCl, ROH v a R dụ
Goi X la so m o l cua R t h a m gia phan ifng v6i nudfc.
Phan ufng: R + H C l > RCl + - H a d j (mol) 0,4 0,4 0,4
R + HaO > ROH + | H2 ( V
(mol) X • X
TLT (1), (2) => so mol kirn loai R duf la: a - (0,4 + x) (mol).
Dieu kien: a > (0,4 + x) ( , Theo de, ta c6 he phiiOng t r i n h :
aR = 53,2 "0,4(R + 35,5) + x(R + 17) + R ( a - 0 , 4 - x ) = 99,92 aR = 53,2 85,72-53,2 , , = > X = — — = 1,913 m o l 17x + aR = 85,72 17 Tir (*): a > 2,313 =ị R < 23
=> k i m loai kiem trong triJdng hop nay la L i t i (Li).
Cdu IV.
= - 0,03 (mol) va " N a = ^ = 0,2 (mol)
a ) C d c phan lifng xay ra:
Na + HaO ^ N a O H + ^ H a t (1) (mol) 0,2 -> 0,2 0,1
2NaOH + C U S O 4 . — — > Cu(0H)2i + N a 2 S 0 4 (2'
(mol) 0,06 <- 0,03 ^ 0,03 0,03
b ) T i n h nong do phan trSm cac chat c6 trong dung dich (A) vk (C).
* Dung dich (A) chufa NaOH
Tii (1) = > n N a O H = 0,2 (mol) m d u „ g d i c h A = 100 gam
0 9 X 40
Vay C% NaOH = ' x 100 = 8%
100
1 « z Lfll GiAl o f THI HnC SINH GIOI HflA HOC S
* Dung dich (C) thu dUcfc gom: • NaaSÔ : 0,03 mol
NaOH du : 0,1 - 0,06 = 0,04 mol Khoi lugng dung dich (C) thu dugc la:
m j u n g d i c h C = m d u n g d i c h A + ™ d d CuSỌ, " " ^ C u ( O H ) ^ = 50 + 30 - 0,03 X 98 = 77,06 (gam) 0,03x142 Vay: va C% c % NâSÔ N a O H du 77,06 0,04 X 40 77,06 X 100 = 5,537c X 100 = 2,08%.
c ) T i n h khói lugng cua ( X ) tham gia phan ufng v6i H2.
s .
Phan ijrng: Cu(0H)2
(mol) 0,03 -> CuO + H 2 O 0,03 (3) (4) C u O + H 2 > Cu + H 2 O (mol) a a Ta c6: 2,08 = 64a + 80(0,03 - a) =^ a = 0,02 (mol) Vay lugng X tham gia phan tfng vdri H 2 la:
mcuo = 0,02 X 80 = 1,6 (gam).
C ^ w V. Goi a, b, c, d Ian lugt la so mol cua F e 3 0 4 , F e 2 0 3 , F e O , F e va goi
s6 mol cua CuO ban dau la: e mol. a) Tinh khoi lugng muoi khan thu dugc:
^ , 360x 18,25
71? Ta co: nnci = = 1,8 (mol) 100x 36,5 Phan ufng: F e 3 0 4 + 8HC1 (mol) a ^ 8a F e z O a + 6HC1 (mol) b -> 6b ,* F e O + 2HC1 - (mol) c ^ 2c F e + 2HC1 — (mol) d 2d 2 F e C l 3 + F e C l a + 4 H 2 O 2 a a 4 a > 2FeCl3 + 3 H 2 O 2b 3 b FeClz + H 2 O c c F e C l 2 + H a t d d H2 + CuO (mol) d d
iifilGlAl 0 ^ THI HOC SINH HOA HOC 9
-> Cu + HaO d d (1) (2) (3) (4) (5) 137
Theo de: m c o g m m = 16d = 3,2 => d = 0,2 (mol). A p dung dinh luat bao toan khoi lugng ta dugc:
n i h o n hap + niHCl = T^mudi + H^n^o + "^Hj (*)
T a c6: m„_^ = 0,2 x 2 = 0,4 gam
iiHci = 8a + 6b + 2c + 2 d = 1,8
» 8a + 6b + 2c = 1,4 o 4a + 3b + c = 0,7
m „ o = (4a + 3b + c) X 18 = 0,7 x 18 = 12,6 (gam) The vao p h i i a n g t r i n h (*) t a difac:
57,6 + 65,7 = mnu,6-, + 12,6 + 0,4 => m , „ u o i = 110,3 (gam)
b ) T i n h n o n g do p h a n t r a m
T a c6: = 232a + 160b + 72c + 56d = 57,6
232a + 160b + 72c = 46,4 « 29a + 20b + 9c = 5,8 Theo de b a i : b = c o 29a + 29b = 5,8 « a + b = 0,2 = a + c K l i o i l i / g n g dung dich B t h u di/cfc l a :
nidiing djch B = niA + mdung dich H C l " Hlj^^
= 57,6 + 360 - 0,2 X 2 = 417,2 gam
« . c . = ^ ? ^ ^ . i o o . M ^ . i o o = .^^^^
c ) Chufng m i n h (X) t a n het.
Goi X , y , z, t I a n liigt la so m o l cua Fe304, Fe203, FeO, Fẹ
T a c6: 232x + 160y + 72z + 56t = 100 ( ]
H H C I phiin ling = 8 x + 6 y + 2z + 2 t
va nuci ban diu = 2 x 2 = 4 m o l
Tii (1) ^ 25(8x + 6y + 2z + 2t) + 32x + lOy + 22z + 6 t = 100 => 25(8x + 6y + 2z + 2t) < 100
=> 8x + 6y + 2z + 2t < 4. V a y h o n h o p t a n h e t .
LUu y: Bdi nay con c6 nhiiu each chiJCng minh khdc. Xin nhuang ban dó'
• I h i niAi ( I F THi H n n mm n i n i HHA Hnn 9
D E SO 2 9 (Tl/ LUYEN)
THi HOC Sil\iH GIQI HOA HOC 9, HUYEN CU CHI, TP. HCM NAM HOC 2005 - 2006
^ ^ j ^ 2; (1,5 diem) ThUc hien 3 thi iighiem sau:
fhi nghi^m 1: Nliiet phdn muoi NaHCOs. Dan sdn phdm vdo dung
^ich Că0H)2 dụ
Thi nghiem 2: Sue khi CO2 vdo nUac c6 nhugni quy tim sau do dun nhẹ
fhi nghiem 3: Cho 1 vien Na vdo 6ng nghiem dUng dung dieli CuSỘ
Cho bict hien tugng xdy ra a ba thi nghiem tren vd gidi thich, vict phuang trinh phdn ling (neu c6).
Cdu 2: (3 diem)
Co 1 hdn hap gom 4 Jam loqi : Al, Cu, Fe, Mg. Bdng pJiUang phdp hoa hoc hay tdeh rieng tUng kim logi ra khoi hon hap. Viet edc pliuang trinh phdn ling.
Cdu 3: (2,5 diem) Hoa tan hodn todn 27,4 gam hon ligp 2 muoi M2CO3 vd
MHCO3 bdng 400ml dung dich HCl IM, Thodt ra 6,72 lit khi CO2 (dktc).
a) Ten cua kim logi M.
b) Viet phiCang trinh phdn iCiig chuyen hoa:
M2CO3 > MCI > MOH
MHCO3 ^
CO2 > MHCO3 > MCI
Cdu 4: (3 diem) Cho 6 gam hdn hgp X gom Fe, FeO, Fe203 chia thdnh 2 phdn bdng nhau:
Phdn I: Bern nung nong roi cho dong khi H2 di qua sau khi phdn Íing xdy ra hodn todn con 2,52 gam Fẹ
Phdn II: Cho tdc dung vai dung dich CUSO4 da sau khi phdn iJcng xdy ra hodn todn Igc xdy kho can ndng dUgc 3,16 gam mot chat rdn.
Tim khoi lugng cdc chat trong X.
(3 diem) Mot hon hgp gom 2 kim logi, kim logi A (hoa tri H) vd kifn logi B hoa tri HỊ Chia Idm 3 phdn bdng nhaụ
Phdn I: Cho tdc dung vai dung dich HCl dit sau khi phdn ling thu dixgc
1,792 lit khi H2 (dktc).
Phdn II: Cho tdc dung vai dung dich NaOH du thu dUgc 1,344 lit Ih
(dktc) vd khoi lugng chat rdn khong tan bdng — khoi lugng kim logi B.
9
Phdn III: Dot trong oxi dU thu duac 2,84 gam chdt rdn. Xdc dinh A; B.
D E S O 3 0
DE THI HOC SINH GIQI HOA HOC 9 , TJNH QUANG NGAI N A M HOC ZOOS - 2 0 0 6
Cdu 1.
1. Hay sit dung nhang hoa chat: Cu, MgO, NaOH, CaCOs, CuSỘSHjj
dung dich H2SO4 lodng, dung dich H2SO4 đc de Idm thi ?ighie„^
chUng niinh r&ng:
a) Dung dich H2SO4 lodng cd day đ tlnh chat hoa hoc cda axit.
b) Dung dich H2SO4 dgc cd tinh chat hoa hoc rieng.
2. Cd hai nhdni chat:
Nhom A: HCl, NaOH, H2Ọ
Nhum B: CaO, CO2, CuO, CO, SO2, Fe203.
Clio cdc chat d nhdni A Idn lugt tdc dung vol cdc chat d nhom B, uict
cdc philang trinh phdn iCng xdy ra (neu cd).
Cdu IỊ Tic dung dich A chica a mot CUSO4 vd b mot FeS04. Ngudi ta thuc hien cdc thi nghieni sau: