dich cd 1 mudị
a) Tim mdi quan he giUa a, b, c trong tiCng thi nghiem tren.
b) Neu a = 0,2; b = 0,3 vd so mol Mg Id 0,4 mol. Hdy tinh khdi luoni
chdt rdn thu diCgc sau phdn ling. Cdu I I I .
1. Cd 6 lo hoa chdt khdng nhdn chda rieng biet cdc chdt bgt saÚ
Mg(0H)2, Zn(0H)2, Fe(0H)2, BaClz, NasCOs, NaOH. Chi dung niiô
vd mot hoa chdt thong dung nUa (tiC chon), hdy trinh bdy cdch nlu}''
biet cdc chdt tren.
2. Ta cdc chdt: NasO, Fe2(S04)3, H2O, H2SO4 (dgc), Cụ Hdy viet pjiicai-:
trinh phdn iCng dieu che: NaOH, Cu(0H)2; hon hap (Fe(OJl Cu(0H)2); Na2S04, CuỌ
Ldl GIAI THI HOC SINH GIOI HOA HQC ^
TV. Xdc dink cdc chdt A, B, €,..., X,, Xo,... vd uict cdc phuang trinh ;. phdn itng xdy ra trong sa do saụ Biet D Id ihdnh phdn chinh cda quQtng
boxit, E dugc sdn xudt tic qudng nianhetit.
A,ranj > dung dich B ^^"^^^ c i J9,,,,,; - i E i ^
ĂrAn) iiiliJ > E(rin) ^ > P(rdn) ^ duUg dich G
, _ x . ! ^ u i f , , , - J ^ ^ E L (Cu) * dung dich X ' ^ P d d c a m . n g n g . n ^ ^ ^ ^ ^ ^ . ^ ^ X^ + X2 f + X3 f * * * Natri clorat > X5 T + X * Dong + X2 t >X6 * Cacbon + X5 t (du) —> X4 T
Cdu V. Mot hon hap khi A gom 2 ankan X vd Y (vdi Mx < My), đng lien
tiep trong day dong đng. Khi dun ndng vdi chdt xdc tdc xdy ra phdn ling tdch 1 phdn td hidro vdi hieu sudt phdn dug chung bdng 75%. Khi đ thu dugc hon hgp khi B cd d^, = 15.
/ H ,
a) Lap cdng thdc phdn td cda 2 ankan vd tinh thdnh phdn % the tich
cda hon hgp Ạ
b) Tinh thdnh phdn the tich hdn hgp B neu hieu sudt tdch hidro cda 2 ankan bdng nhaụ 2 ankan bdng nhaụ
c) Neu hieu sudt tdch hidro cda X lap gdp 1,2 Idn hieu sudt tdch hidro
cda Y thi đ cd bao nhieu phdn tram moi chdt thani gia phdn dng?
Trong truang hgp ndy hdy tinh thdnh phdn phdn tram the tich cda
" hon hgp khi D con Igi sau khi dan B qua dung dich KMn04.
d) Ve sa do vd viet cdc phuang trinh phdn dug thOc hien sU chuyen hoa tic X
chuyen hoa thdnh Y vd ngicac Igị Biet Y cd mgch cacbon khong nhdnh.
L d l GIAI
Cda I .
1. a) Dung dich H2SO4 loang c6 day du tinh chat hoa hoc cua axit: - Khong tac dung vcJi k i m loai kem hoat dong. - Khong tac dung vcJi k i m loai kem hoat dong.
Cu + H 2 S O 4 khong tac dung.
- Tdc dung vdi oxit bazcf:
MgO + H 2 S O 4 > MgS04 + H2O - Tac dung vdi baza:
2NaOH + H 2 S O 4 > Na2S04 + 2H2O
- Tac dung vdi muói:
CaCOa + H 2 S O 4 > CaS04 + CO2T + H2O
b) Dung dich H2S04 dac c6 tinh chat hoa hoc rieng:
- T a c dung vdi k i m loai k e m hoat dong:
Cu + 2H2SO4 ,dac) > C U S O 4 + S O z t + 2H2O - H a o n L f d c , l a y nifdc cua muói ngSm niidc.
2. * V d i H C l : CuS04.5H20 + H2S04(dac) (mail xanh) 2HC1 + CaO — 2HC1 + CuO — -> CUSO4 + H2SO4.5H2O (mail trdng) > C a C l 2 + H 2 O 6HC1 + FeaOg * V d i N a O H : N a O H + C O 2 - H a y : 2 N a O H + C O 2 N a O H + S O 2 - > C u C l 2 + H 2 O 2FeCl3 + 3 H 2 O -> NaHCOg - > NaaCOg + H 2 O 2 N a O H + S O 2 H 2 O + CaO — -> NaHSOs NaaSOg + H 2 O Că0H)2 H 2 O + C O 2 ^ ^ H 2 C O 3 H 2 O + S O 2 :^ H2SO3 CduII
a ) D u n g d i c h A : C U S O 4 (a mol); FeS04 (b mol).
M g liu t i e n p h a n ijfng vdi CUSO4 trilcfc FeS04.
• Till nghiem 1: A + M g (c m o l ) 3 muói C U S O 4 dii
M g + CUSO4 > Cu + M g S 0 4
(mol) c a
c < a ^ d u n g d i c h t h u diidc c6 3 m u o i : CUSO4 dii, FeS04, MgSÔ
• Thi nghiem 2: A + M g (2c m o l ) - > 2 m u o i M g + CUSO4 > Cu + M g S 0 4 (m.ol) 2c a K h i 2c = a - > d u n g d i c h t h u dugc c6 2 m u o i : FeS04, M g S 0 4 M g + FeS04 > Fe + M g S 0 4 (mol) (2c - a) b
K l i i b > (2c - a) > 0 - > dung dich t h u diidc c6 2 muoi: FeS04 d i i , MgSOj -> a < 2c < a + b.
1 Ldl GIAI THI HOC SINH GIOI HOA HOC ^
• Thi nghiem 3: A + Mg (3c m o l ) --> 1 muói
Mg + C U S O 4 > Cu + M g S o 4 (mol) a a M g + FeS04 > Fe + M g S 0 4 (mol) b b K h i 3c > (a + b) ^ dung d i c h t h u difgc c6 1 m u o i : M g S 0 4 b ) K h i a = 0,2; b = 0,3; n w g = 0,4 m o l M g + CUSO4 > Cu + M g S 0 4 (mol) 0,2 0,2 0,2
Sau p h a n ufng (1), so m o l M g dii: nMg = 0,4 - 0,2 = 0,2 m o l , t i e p tuc tac d u n g v d i FeS04 (0,3 mol).
M g + FeS04 > Fe + M g S 0 4 (2) (mol) 0,2 0,2 0,2
So m o l M g dif sau p h a n ufng (1) = 0,2 < so m o l FeS04 (0,3 mol) => só m o l Fe s i n h r a : 0,2 m o l
=> FeSOi con dK: 0,3 - 0,2 = 0,1 m o l
=> K h d i lirong c h a t r S n t a o t h a n h :
m c u + mpe = 0,2 x 64 + 0,2 x 56 = 24 (gam). / / /
1. H o a chat t i i chon l a H2SO4.
• H o a t a n cac chat vao nifdc, n h o m chat k h o n g t a n gom: Mg(0H) 2 ,
Fe(0H)2, Zn(0H)2. N h o m t a n gom: N a O H , Na2C03, BaCk.
• D u n g d u n g d i c h H2SO4 n h a n r a Na2C03 (c6 k h i C O 2 t h o a t r a ) ;
B a C l 2 t a o két t u a t r S n g ; N a O H c h i t a n b i n h thiJdng.
H 2 S O 4 + Na2C03 > Na2S04 + C 0 2 t + H 2 O H 2 S O 4 + BaCl2 > B a S 0 4 i + 2HC1
• ' * D u n cac chat k h o n g t a n v d i niidc t r o n g k h o n g k h i , chat nao
chuyen t h a n h m a u n a u do l a Fe(0H)2.
4Fe(OH)2 + O2 + 2 H 2 O —> 4Fe(OH)3
• D u n g d u n g d i c h N a O H n h a n b i e t Z n ( 0 H ) 2 : két t u a t a n la Z n ( 0 H ) 2 , két t u a k h o n g t a n l a M g ( 0 H ) 2 .
Z n ( 0 H) 2 + 2 N a O H > NagZnOa + 2 H 2 O
2. Phuang t r i n h phan ijfng dieu che: NaOH, Fe(OH)2, Cu(0H)2.
• Dieu che NaOH: NâO + H 2 O > 2NaOH • Dieu che Cu(0H)2:
Cu + 2 H 2 S O 4 (dac) —> CUSO4 + SOgt + 2 H 2 O
CUSO4 + 2NaOH > Cu(0H)2 + Na2S04 • Dieu che h6n hap (Fe(0H)2, Cu(0H)2):
Fe2(S04)3 + Cu > CUSO4 + 2FeS04
CUSO4 + 2NaOH > Cu(0H)2i + Na2S04 FeS04 + 2NaOH > Fe(0H)2i + Na2S04 • Dieu che NaaSOsi SO2 + 2NaOH > NazSOs + H2O
• Dieu che CuO: Cu(0H)2 —> CuO + H2O
CduIV
Dung dich X: NaCl; Xn NaOH; X z t : CI2; X g t : H2; X4t: CQv,
X s t : O2; Xg: CuClz; A: A l ; B: NaA102; C: AKOII);,*; D:Al203; E: Fe; F: FeClg; G: FeClg; H : Fe(On)„ I : Fe(0H)3; K: FeaOs; L: Cụ C O , t / H , 0 1 U 2 — (A) (B) * Al(rdn) -^^^^S^ đ NaAlO, ) A1(0H)3 i (C)
^ (Tin) ^P"" ) Al(r^n) ^^^^^P"' ) Fe(r^n) (D) . (A) ^ ' (E)
C i t p^cigCr^n) ) đ Feci, ^ 1 ™ ^ Fe(OH).,
(F) ^ (G) ^"'^ \ H )
^ Fe(0H)3 i Fe,03 (ran) Fe
(I) (K) i « i T u i u n r c i i u u n n i u n A u n r 9 2NaCl + 2 H 2 O (X) * 2NaC103 * Cu + CI2 (X2) * C + Oat (di/) ( X 5 ) d p d d C6 , n . n g ngan ^ 2NaOH + C l 2 t + H2T (Xi) (X2) (X3) — > 302t + 2NaCl ( X 5 ) ->CuCl2 (Xe) C02t ( X 4 ) Cdu V.
a) Goi cong thufc phan tijf trung binh cua 2 ankan dufng lien tiep trong day dong d i n g la: C-H^-^
Gia suf ban dau c6 1 mol A ( C - H - „) n 2 n + 2 Ta c6: M A = 14n + 2
C-H - „
n 2 n + 2 -> C-H - + H2
n 2 n Sau phan ufng diJgc 1,75 mol hon hop B: Sau phan ufng diJgc 1,75 mol hon hop B:
0,75 mol H , 1,75 mol B 0,75 mol 0,25 mol M B = 15 X 2 = 30 - 14n + 2 1,75 n = 3,6 Hai ankan ban dau 1^: CsHs va C4H10.
Gia suf trong 1 mol A c6: x mol C3H8 (1 - x)mol C4H10.
n = 3x + 4(1 - x) = 3,6 =^ X = 0,4.
Thanh phan phan trSm so mol va the tich: C3H8: 40%; C4H10: 60%.
b ) Neu hieu suat phan ling cua moi ankan deu bang 75%:
l,75mol B C3H6 : 0,4x0,75= 0,3mol ^ 1 7 , 1 4 % C^Hg : 0,6 X 0,75=0,45mol 25,71% C3H8 : 0,4 X 0,25 = 0 , Imol -> 5,71% C ^ H J O : 0,6 X 0,25 = 0 , 1 5 m o l -> 8,57% H2 : 0,75mol 42,86%