S d G D& D T T / P D A N A N G
K Y T H I T U Y ^ N S I N H V A O L C J P I O T R L ; O N G THPT CHUVeN le QUtDON . rinP
NAM HOC: 2 0 1 2 - 2 0 1 3 Di C H I N H THirC
K Y T H I T U Y ^ N S I N H V A O L C J P I O T R L ; O N G THPT CHUVeN le QUtDON . rinP
NAM HOC: 2 0 1 2 - 2 0 1 3 MON: HOA HOC Th()1 gian lam bai: 150 phut (khong keth&i gian giao de) Cau 1. (3,0 diem)
(1) Viet cac phudig trinh hoa hoc ciJa phan ling xay ra trong qua trinh san xuat thiiy tinh tir cat trdng, s6da, da v6i.
( 2 ) Tron 50 gam dung djch muoi sunfat ciia m6t kirn loai kiem (dung dich A) n6iig d6 26,4% v6i 50 gam dung djch NaHCO, thu dixgc dung djch X c6 khdi lirgng nho hon 1 0 0 gam. Cho 0,1 mol BaQ, vao dung dich X thay vSn con du muoi sunfat. Them tiep vao do 0,02 mol BaQ. thi dung djch thu duoc vSn con Baa2 dir.
Bie't cac phan ling xay ra hoan toan.
(a) Xac djnh cong thiirc hoa hoc ciia mu6'i sunfat ban ddu.
(b) Viet phuong trinh hoa hoc cua phan ling xay ra (neu c6) khi cho ISn lugt cac chat sau tac dung v6i dung djch A: Fe, Fc(OH),, Fe,04, Ag, NaAlO,.
Ckull. {4,50 diem)
(1) Khong dung thuoc thir, biing phuong phiip hoa hoc, hay phftn biet 6 dung dich kh6ng mau: MgCl., NaHCO,, H,S04, BaCI,, NaCl va NaOH dung rieng biet trong cac binh mat nhan. Viet cac phuong trinh hoa hoc minh hoa.
( 2 ) Dun nong 10,8 gam b6t A l trong O, mot thofi gian, thu diroc m gam h6n hop chat rdn A. Hoa tan het A bang mot lirong vua du dung djch h6n hop HCl I M va
H 2 S O 4 0,5M thu duoc V lit H . (dktc) va dung djch B. Co can dung djch B thu diroc (m + 44,34) gam muoi khan. Tinh m, V.
(3) Hoa tan hoan toan 8,4 gam kirn loai M trong dung dich H 3 S O 4 dac, nong (vira dii) tha'y thoat ra 5,04 lit khi SO, (dktc). Dung djch sau phan ung dem c6 can thi thu duoc 42,15 gam chat rdn X. Xac dinh M va X.
cau I I I . {3,0 diem)
(1) Trinh bay phuong phap hoa hoc de' tach rieng biet cac khi tir h6n hop sau: HCl, O,, CO,.
(2) Cho mot binh km dung ti'ch khong doi chu-a 80ml nude va 4 lit khdng khf (xem hinh ve). Phdn khong khi chi cMa N , va O, theo ti le 4 : 1 v^ the' tich.
Bom 0,04 mol h6n hop khf B bao g6m NO, va NO c6 ti kh6'i so vdi H , bang 19 vao binh va lac kT binh tdi khi cac phan ung sau xay ra hoan toan:
2 N O + O, 2NO,
4 N O , + O, + 2 H, 0 -> 4 H N O 3
70
Kliong khi
NmVc
-a d U c K dung dich X. Tinh n6ng d 6 % dung dich X.
Gia sir ap sua't hoi liudc trong binh khong dang ke'; cac the ti'ch khi deu do dktc; khdi luong rieng cua nude Dn^o = ' g/m'-
cau IV. (^,<>
(1) '^'^"0 Ve hinh minh hoa each tien hanh thi nghiem dieu che va thu khf axetilen an toan trong phong thi nghiem. Viet phuong trinh hoa hoc minh hoa.
(2) ( 2 diem) X la hdn hop gdm axetilen va hidro cd ti khdi so vdi heli la l,*-).
Cho toan bo X qua dng sii dung N i , dun nong mot thdi gian, thu duoc hdn hop Y Cho Y tac dung vdi luong du dung djch AgNO, trong N H , thay tao thanh 7,2 gam ke't tiia va hdn hop khi Z. D3n Z qua nude brom du thu duoc hdn hop khi T, ddng thdi thay cd 4,8 gam brom da tham gia phan tag. Dot chay hoan hoan hdn hop T thu duoc 0,896 lit CO, (dktc).
(a) Xac djnh thanh phdn hdn hop Y, Z, T.
(b) Tinh ti khdi cua Y so vdi heli. ^ Cau V. {2,50 diem)
(1) Neu hien tuong vii giai thi'ch:
(a) Nhd vai giot iot vao mat mdi cdt cua cii khoai lang.
(b) Cho vai giot chanh vao cdc sua bd.
(c) Cho mot mdu cao su tu nhien vao xang.
(2) Len men giam 115ml dung djch rugu etylic 10° mdt thdi gian thu duoc dung djch A. Neu cho toan bd dung djch A tac dung vdi luong vua dii Na den khi phan ung xay ra hoan toan thu duoc 251,7 gam cha't rdn khan. Tinh hieu sudt qua trinh len men giam, biet khdi luong rieng DC^H^OH = g/"^'- ^ ^H o = ' g/f"'- • Cau VI. {3,0 diem)
Ddt chay hoan toan 3,74 gam hdn hop X gdm CH,COOH, CH,COOC,H,, C,H,OH thu dugc 3,584 lit CO, (dktc) va 3,42 gam H , 0 . Mat khac, cho 3,74 gam X phan ung het vdi 40ml dung djch NaOH I M , thu duoc dung djch Y va 0,05 mol C.HyOH. Cd can dung djch Y, thu duoc 2,86 gam cha't rdn khan.
(a) Xac djnh c6ng thiJc phan tir ciia ancol C^H^OH.
(b) Tinh % theo khdi luong cac chat trong X.
Cho C = 12, H = 1, N = 14, O = 16, Na = 23, Ba = 137, K = 39, L i = 9, A l = 27, Ag = 108, He = 4, S = 32, CI = 35,5, Fe = 56, Br = 80.
HI/(!JNG DAN GlAl
Cau I. (7,5 rf////i) i^) 1,0 diem
Phuong trinh hda hoc: SiO, + Na,CO, > Na,SiO, + CO, SiO, + CaCO., - 1 ^ - ^ CaSiO, + CO3 (2) 2,0 diem
(a) m.. + m > m mudi sunfat kirn loai kiem la mud'i axit
(MHSO4) do cd S U thoat khi CO, nen lam giam khdi luong dung djch sau khi pha trdn.
-71
2MHS04 + ZNaHCOj -> M^SO^ + Na,S04 + 2CO, + 2H,0 M2SO4 + B a C l , - > B a S 0 4 + 2 MCI
Dat MHSO4: X mol
( M . 9 7 ) x = ^ = 1 3 . 2 ^ x = . ' "
100 M+97 T h e o d a O, l < x< 0 , 1 2 : ^ 13 < M < 35 => M = 23 (Na) la phu hop Vay cong thirc cua muoi sunfat la NaHS04
(b) 2NaHS04 + F e - * Na3S04 + FeS04+ H ,
8NaHS04 + Fe304 ^ 4Na.S04 + Fe,(S04), + FeS04 + 4 H , 0 2NaHS04 + 2Fe(OH)2 -> Na2S04 + 2FeS04 + SH.O NaHS04 + NaAlO, + H.O-* Na.S04 + Al(OH),
Co the': 6NaHS04 + 2 A 1 ( 0 H ) 3A l 3 ( S 0 4) 3 + 3Na,S04 + 6H.O Cdu U. (4,50 diem)
(I) 1,50 diem '^^-f^
Tn'ch cac mSu thu va danh dtfu. Ldn luot cho cac m5u thir tac dung v6i nhau tirng doi m6t, ta c6 bang ket qua sau:
MgCK NaHCO, H.SO4 BaCl. NaCl NaOH
MgCK - - - - - i
NaHCOj - - t - - -
H:S04 - t - i - -
BaCU - - - - -
NaCI - - - - - -
NaOH i - - - - -
MSu thu nao khi cho vao cac mau thir kia tao duoc 1 ket tiia la NaOH, MgCK, BaCU.
MSu thir nao khi cho vao cac mSu thu kia tao 1 chat khi thoat ra: NaHCO, MSu thu nao khi cho vao cac mSu thir kia tao duoc 1 chat khi va 1 ket tiia la H2SO4, chat tuong urng la NaHCOj va BaCl,
Kh6ng C O hien tuong nao la NaCl.
2NaOH + M g C l 2 - > Mg(OH), i + 2Naa
; H:S04 + BaCl, -> BaS04 i + 2HC1
; 2NaHC03 + H,S04 -> Na,S04 + 2H.O +2C0, Phan bidt NaOH, MgCl,:
C6 can va nung nong 1 mSu NaHCOj, hoa muoi thu duoc vao 2 m3u thir nay, neu C O ket tua tao duoc la MgCU, kh6ng c6 hien tuong gl la NaOH. ằ 1
* K : t 2NaHC03 — ^ Na^CO, + CO, + H^O MgCU + Na3C03 -> MgCOj + 2NaCl (2) (2,0 diem)
Dun nong A l trong O,: 4A1 + 3 0 , - -> 2AKO3 72?
S5'niol A! = 10,8:27 = 0,4 mol
phuong trinh hoa hoc: A l + 3HC1 AICI3 + ^ H, t 2A1 + 3H,S04 -> Al2(S04)3 + 3 H, t ' AI2O3 +6HC1 ->2AlCl3 + 3 H 2 O AI3O3 +3H2SO4 ^ A 1: ( S 0 4) 3 + 3 H, 0
oat s6' mol H:S04 la x => s6' mol HCl: 2x '•'0)^:)
X 2x
Iv1u6'i thu duoc la h6n hop: A1:(S04)3: — mol va AICI3: — mol
3 3 2x 2x
Bao loan nguyen to Al ta c6: —- + — = 0,4 => x = 0,3 Kh6'i luong muoi khan:
mAia, + '"Ai2(S04)3 = " I A I + nig^caxi. = 10,8 + 0,3x96 + 0,3x2x35,5= 60,9 Theo d^: m + 44,34 = 60,9 => |m = 16,56 gam|
. m-10,8 16,56-10,8 ^ , „ , Bao toan khoi luong ta co: no2 = ——— = — = 0,18 mol
.S6'mol A l d u t r o n g A l a : 0 , 4 - - . 0 . 1 8 = 0,16 mol
> S6' mol khi H , thoat ra: 3/2. 0,16 = 0,24 mo!
Vay the tich khi H , 6 dktc la: |V= 0,24 x 22,4 = 5,376 lit (3) (1,0 diem)
So mol SO, = 5,04:22,4 = 0,225 (mol)
2M +2nH3S04 — ^ M,(S04)„ + nSO, + 2nH,0 , 0 45
^ ' 0,225 (mol) n
=> = — => M = — n => Nghiem phu hop: n = 3; M = 56 n M 3 • *
So mol Fe,(S04)3 = ^ so mol Fe = 0,075 (mol) in'
Khoi luong Fe3(S04)3 =0,075.400 =30 gam < 42,15 gam cha't rdn.
=> Chat rdn thu duoc la tinh the ngam nu6c: Fe3(S04)3.xH20
— = 0,075 X = 9 . vay chat rdn la: |Fe,(S04),.9H,^
18x
f^^ni. (3,0 diem)
^^^(i,50die)n)
^^n h6n hop khi qua binh dung dung dich Ca(OH)2 du, khi O, kh6ng tham gia phan ling thoat ra, thu la'y O,; HCl va CO, bj hap thu het:
7-*
2HC1 + Ca(OH)2 -> CaCl, + 2H2O CO, + Ca(OH), -> CaCOj + H, 0
- Lpc lay ket tua, hoa tan het trong dung djch HCl du, thu khi CO, thoat ra CaCO, + 2HC1 CaCl, + H^O + CO^t
- Co can dung djch chira CaCU, CaCOH), den khan, sau do cho tac dung v6
H2SO4 dun nong, thu khf HCl thoat ra. '
(2) (1,50 diem)
Ca(OH), + H2SO4 ' ) CaS04 + 2 H , 0 CaCU + H,S04 — ^ CaS04 + 2 H C l t
n,HB= 0,896 : 22,4 = 0,04 (mol) va M R = 19.2 = 38 Sof do duong cheo:
8l M N O2= 4 6
M N O = 3 0
^ M = 3 8 ^ 'NO2 ' N O
= 1 ^ n N 0 2 = " N O = 0 , 0 2 (mol)
Cac PTHH xay ra khi bofm khi B vao nu6c:
2NO + O, ^ 2 N 0 ,
4NO2 + O, + 2 H , 0 -> 4HNO,,
" N0 2( 1 ) = " N O = 0 , 0 2 (mol)
n N 0 2 ( 2 ) = n N 0 2( i ) + n N 0 2 = 0,02 + 0,02 - 0 , 0 4 (mol) (1) (2)
no2 (ban ddu) = = 0,036 (mol) > n o , (p/iing) = 0,03 (mol)
^ " H N O ^ = 0 . 0 4 (mol)
mdd = m„„.i,i„„jj„ + m H N 0 3 - ' " H2 0 ( 2 ) = 80 + 0,04.63 - 0,02.18 = 82,16 (g)
C% (HNO,) = . 100 = 3,07 (%)
OduW. (4,0 diem) K (1) (1,75 diem)
Theo hinh ve 4.12 trang 122 sach giao khoa Hoa hoc 9 CaC, + 2 H , 0 ^ Ca(OH), + C , H , t
(2) (2,25 diem)
Theode: d^, = 2,9 ^ M = 2,9. 4 = 11,6 => = - 'H2 Cac phucmg trinh phan ung xay ra:
C,H, + H , -> CH4 (1)
C2H2 + 2 H , Q H , + AgjO C3H4 + Br,
2 C , H 6 + 70, H , + 1/20,
J ^ i ^ C,Ag, + H , 0
> C,H4Br,
.0 4 C 0 , + 6 H , 0 H, 0
(2) (3) ( 4 ) (5) ( 6 ) Tif dff kien de bai, suy ra thanh phan cac h6n hcrp:
Y: C,H4, C , H 6 , C,H, du, H , du Z: C,H4, CiH^, H ,
T: CMt, H":
S6' mol C,H,,d., = s6' mol C,Ag, = 7,2: 240 = 0,03 (mol) S6' mol C2H4 ô so mol Br, = 0,03 (mol)
S6' mol C , H 6 = ^ s6' mol CO, = 0,02 (mol)
TCr (1) va (2) suy ra so mol C,H, ban dau = 0,03 + 0,02 + 0,03 = 0,08 (mol) So mol H , ban dau = 3/2 so mol C,H, ban ddu = 0,12 (mol)
S6' mol H2 tham gia phan iJng: = 0,03 + 0,02.2 = 0,07 (mol) Ta c6: M y = • m . 26.0,08 + 2.0,12 ^ 2 , 3 2
ny ~ 0,03+ 0,03+ 0,02+ ( 0 , 1 2 - 0 , 0 7 ) " 0,13 = 17,85 17,85
' Y / H e - 4 , 4 6 Cau V. (2,50 diem)
(1) (0,75 diem)
(a) Ch6 nho iot tha'y xufi't hien mau xanh dac trung do trong khoai lang c6 chiJa tinh b6t, tao phan ihig mau v6i iot.
(b) Co hidn tucmg dong von siJa do trong sOa c6 1 so protein d6ng tu khi ti6'p xiic v6i moi truong axit.
(c) M5u cao su tan tao thanh dung dich nhdt do cao su la polime tan dupe trong dung m6i hiru co: xang, benzen, ddu hoa...
i'^) (1,75 diem)
11,5x0,8 '^C2H50H = 115xl0° = l l , 5 m l = > n C2H5OH
46 = 0,2 mol 'H20
( 1 1 5 - l l , 5 ) x l
18 = 5,75 mol
G o i x l a s6'moIC,H50H tham gia phan dng len men:
C 2 H, O H + O2 ''"'"'^"^''"^ )CH3COOH + H2O (moi)
- .at
Cho A tac dung v6i Na:
2CH3COOH + 2 N a - > 2 C H 3 C O O N a + H 2 t ( m o l )
75
2H2O + 2Na 2NaOH + H2
(5,75 + x ) - > (5,75+ x) (mol) 2C2H,OH + 2Na -> 2C2HsONa + H2 t
( 0 , 2 - x ) ^ ( 0 , 2 - X ) (mol) 'if
>m ( r i l l khun) = 82x + 68(0,2 - X ) + 40(5,75 + x) = 251,7 g => x = 0,15 mol 0,15
Hieu sua't qua trinh len men gia'm: H = — — x 100% = |75%
0,20 Cau V I . (3,0 diem)
(a) CTPT cua ancol :
Ta CO so do chay X: X + O, -> CO, + H , 0
nco2 = (3,584 :22,4) - 0,16mol; nn,,) = (3,42 :18) = 0,19 mol TCr cac PTHH ciia phan ling chay:
Ta c6: C H , C O O H va este C H ^ C O O Q H ^ trong do goc Q H ^ boa trj 1 va no khi chay tao thanh so m o l H , 0 b i n g so m o l C O , .
Theo de ra: n^^o > "002 C^HyOH la ancol no don chiic: y = 2x + 1 T C r d o t a c o : n c ^ n ^ ^ ^ j O H = 0 , 1 9 - 0 , 1 6 = 0,03 mol
Sau khi thuy phan thu dugc long s6' mol ancol la 0,05 mol.
Suy ra : so mol este CH3COOC^H2,+| la : 0,05-0,03 = 0,02 mol V a y X g 6 m : CH,COOH (a mol), CH3COOC,H2,+, (0,02 mol)
va C,H2^+,0H (0,03 mol)
PTHH ciJa phan ung v6i dung djch NaOH ciia X:
CH ,COOH + NaOH -> CH ,COONa + H^O
a -> a -> a (mol) CH,COOC,Hy + NaOH ->• CH^COONa + C . H y O H
0,02-> 0,02-> 0 , 0 2 ^ 0,02 (mol) - Neu NaOH phan ling vira dii vdri X, chat rdn la CH ^COONa .
nCH3COONa = " NaOH = 0,04 mol => mcH3CO()Na = "'O^ X 82 = 3,28g > 2,86g - Vay NaOH con du.
Ta c6: m,j„ ,,han = 82(0,02 + a) + 40[0,04 -(0,02 + a)] = 2,86g a = 0,01
=> m x = 0,01X60 + 0,02(60 + 14n) + 0,03(14n +18) = 3,74 =>n = 2
=> CTPT cua ancol: |C,H,OH.
0.01x60 (b) Trong X:
% m
%m
CH3COOC2H5
CH3C00H - ' 0,02x88
3,74
3,74 16,04%
47,06%!; %mc2H50H = 0.03x46
3,74 36,90%
If,
sdGD&OT
HAI DLTONG p ^ T H I N H T H U C
KV THI TUY^N SINN VAO idP 10 TRLTONG THPT CHUYeN NGUYEN TRAI
NAM HOC: 2012-2013
MOH: HOA HOC Thoi gian lam bai: 120 phut (khong kethdigian giao de)
C^l. (2,0 diem)
(1) Cho h6n hop g6m AKO,, Cu, Fe^O, vao dung djch H,S04 loang du thu duoc dung djch X va chat ran Y. Cho X\x tir dung djch NaOH toi du vao dung djch X thu diroc dung djch Z va ket tiia M . Nung ke't tiia M ngoai kh6ng khf tdi kh6'i lircmg kh6ng doi thu duoc chat ran N . Cho khi H , du di qua N nung nong thu ducK; chat ran P. Sue khi CO2 toi du vao dung djch Z thu duoc ket tiia Q.
(a) Xac djnh thanh phdn cac chat c6 trong X, Y, Z , M , N , P, Q. Bia't cac phan ling xay ra hoan toan. ; ' i >
(b) Vid't cac phuong trinh phan utig hoa hoc xay ra.
(2) Cho h6n hop kim loai M g , Fe vao dung djch chiia h6n hop mu6'i Cu(N03)2, AgNOj. Phan ung xay ra hoan toan, thu dugc h6n hofp chat rdn A g6m 3 kim loai va dung djch B chu-a 2 muoi. Trinh bay phuong phap tach riang tirng kim loai ra khoi h6n hop A. Viet phuong trinh hoa hoc.
cau 2. (2,0 diem)
Cho hai hop chat hffu co X , Y chii-a (C,H,0) chi chffa mot loai nhom churc da hoc va c6 khd'i lugng mol phan tir deu bang 46 gam.
(1) Xac djnh c6ng thffc ca'u tao cua X, Y. Biet X, Y d6u phan ling vdri Na, dung djch cOa Y lam quy tim hoa do.
(2) Tir X Viet cac phuong trinh hoa hoc dieu che' Polyvinylclorua (PVC) va Polyetylen (PE). , ; >
cau 1.(2,0 diim)
(1) Hay chon cac chat thich hop va viet cac phuong trinh phan irng hoan thanh so
<J6 bie'n h6a sau:
A - i M -
+ (X) ->B; +(X)+... ± D <-
+(X)+... • N +(Y)
i ( + Y )
^ho biet:
Cac chat A , B, D la hop chat cua Na;
Cac cha't M va N la hop chat cua A l ; Cac chat P, Q, R la hop chat cua Ba;
Cac chat N , Q, R khong tan trong nu6c.
X la chat khf kh6ng miii, lam due dung djch nu6c v6i trong;
muO'i Na, dung dich Y lam do qui tfm. ^
(2) Tir 9 kg t i n h bot c6 the' didu che dirge bao nhiSu h't rugu (ancol) etylic 46°? Biq hieu suat ciia ca qua trinh didu che la 72%, kh6'i luong r i e n g cua ruou etylj^, nguyen chat la 0,8g/ml.
Cau 4. (2,0 diem)
Nung 9,28 gam h5n hop A g6m FeCO, va mot oxit sit trong khong khi den k h d j
lugng khong doi. Sau khi phan lirng xay ra hoan toan, thu dugc 8 gam mot oxj(
sdt duy nha't va khf CO,. Ha'p thu het luong khi CO, vao 300ml dung dic|i Ba(OH)2 0,1M, ket thiic phan ling thu dugc 3,94 gam k€i tiia.
(1) Tim cong thiic hoa hoc ciia oxit sat.
(2) Cho 9,28 gam h6n hop A tac dung vofi dung djch HCl du, sau khi phan lifng ket thiic thu dugc dung dich B. DSn 448ml khi CU (dktc) vao B thu dugc dung dic|, D. Hoi D hoa tan tdi da bao nhieu gam Cu?
Cau 5. (2,0 diem)
Thuy phan hoan toan 19 gam hcfp chat hCru co A (mach his, phan ling dugc vo, Na) thu dugc m, gam cha't B va m , gam chat D chiia hai loai nhom churc.
- Dot chay m, gam chat B cin 9,6 gam khi O, thu dugc 4,48 lit khi CO2 va 5,4 gam nu6c.
- Dot chay m , gam chat D can 19,2 gam khi O, thu dugc 13,44 lit khi CO, va 10,X gam nuorc.
(1) T i m cong thiic phSn tir A, B, D.
(2) Xac djnh cong thiic ca'u tao ciia A , B, D.
C h o : F e = 56; B a = 1 3 7 ; C = 1 2 ; 0 = 1 6 ; H = 1; Na = 23; CI = 35,5; Cu = 64 .5^ Hl/CJNG DAN GIAI
Cau 1. (2,0 diem) 1. (1,0 diem)
(a) Dung dich X : A1,(S04)3, CUSO4, FeS04, H,S04du Chat rdn N : CuO, Fe,03
Chat ran Y : Cu - ; r Chat ran P : C u , Fe
Dung djch Z : NaAlO,, Na,S04, NaOH
Ket tua Q : AKOH)^ ' Ket tua M : Cu(OH),, Fe(OH), ' (b) PTHH: A1,03 + 3H,S04
Fe,03 + 3H,S04 Cu + Fe,(S04)3 2NaOH + H,S04
6NaOH + A1,(S04)3 — N a O H + Al(OH) 3
2NaOH +FeS04
A1,(S04)3 + 3 H , 0 Fe,(S04)3 + 3 H , 0 CUSO4 + 2FeS04 Na,S04 + 2 H , 0 2Al(OH) 3+3Na,S04 NaAIO, + 2 H , 0 F e ( O H ) , 4 + N a , S 0 4
^1/ I mill 111 Y vii njiaiig viae
> Cu(OH), i + Na,S04 -> CuO + H , 0
-> 2Fe,03 + 4 H , 0 Cu + H , 0 - > 2Fe + 3 H , 0
NaHC03
> A1(0H)3 + NaHC03
.1
-> MgCNOj), + 2 A g 2NaOH +CUSO4 —
Cu(OH), — 4Fe(OH), + 0 , — CuO + H , — F e A + 3 H , — CO, + N a O H — C O , + H , 0 + NaAIO, 2. (1,0 diem)
PTHH: M g + 2AgN03 — M g + CuCNOj), — Fe + 2AgN03
Fe + CuCNOj), — Cha't ran A : A g , Cu, Fe
Dung djch B : MgCNO,),, FeCNO,),
Cho chat ran A vao dung dich HCl du. Thu dugc pMn cha't ran la k i m loai Cu, Ag va phdn dung dich FeCl, va HCl
Fe + 2HC1 ). FeCl, + H , t ,^;
Cho NaOH d u vao phdn dung dich, thu dugc ket tiia la Fe(OH), ^ N a O H + HCl > N a C l + H , 0
2NaOH + FeCl, > F e ( O H ) , i + NaCl
Nung ket tua ngoai khong khi den khoi lirgng khong doi. Cho luong khi H , d u di qua cha't rdn, nung nong, thu dugc Fe.
-> MgCNOj), + Cu )-Fe(N03), + 2 A g -> FeCNOj), + Cu
4Fe(OH), + O, Fe,03 + 3 H ,
->2Fe,03 + 4 H , 0 -> 2Fe + 3 H , 0
^ " n g 2 k i m loai trong kh6ng khi toi khoi lugng khong d6i thu dugc cha't rdn 8<5m CuO va A g . Cho cha't rdn sau phan ling vao dung dich HCl du thu dugc k i m J o a i A g .
2Cu + 0 , CuO + 2HC1
-^2CuO -ằCuCl, + H , 0
Cho dung djch NaOH du vao dung dich sau phan iJng thu dugc kd't tua, Igc la'y
•^^'t tua dem nung ngoai khdng khi t6i khoi lugng khdng d6i, cho ludng H , du di 9Ua chat rdn, nung nong thu dugc Cu tinh khia't.
T O
NaOH + HCl CuCU +2NaOH Cu(OH),
CuO +H2 Cau 2. (2,0 diem) 1.(1,0 diem)
> NaCl + H.O
> 2NaCl + Cu(OH )2 - ^ C u O + H,0
' 1 '
r Cu + H,0
Goi cong thiic tdng quat cua X, Y la QH^O, (x, y, zeN*) Tac6:Mx,Y = 46 <=> 12x + y + 16z = 46.
46-(12x + y) z =
z<
16 46-14 16 - 2
Choz= l=:>12x + y = 30(CHJ Choz = 2^12x + y = 14 (CH,)
Vay c6ng thiic phan tir cua X, Y c6 the la QH^O, CH,0,.
Vi Y phan irng \6\, lam do quy tfm, Y c6 nhom -COOH
=^ CTPT Y: CH,0. => CTCT cua Y: H-COOH 2HC00Na + H, PTHH: 2HCOOH + 2Na —
• X phaii irng vol Na, X phai c6 nhom -OH.
o CrPT Y: CH,0 => CTCTcua X : CH,-CH,-OH
PTHH: 2CH,-CH,-OH + 2Na > 2CH3-CH,-ONa + H,
2. (1,0 diem) .iCH Viet PTHH dieu che polyetilen (P.E) ô iUi '<\A ^nilti.
^ .<> . CH,-CH,-OH -il252dc !_^CH, = CH, +H,0
. . . .„ • 1 7 0 "
nCH, -CH2 lCXX)ulni H C H, - C H 2 ^ „
Dieu che |x)lyvinylclorualcn (PVC)
CjH^OH + O2 >CH,COOH + H^O CH,COOH + NaOH > CH,COONa + H,0 CH.,COONa + NaOH - > Na.CO, + CH4
2CH4 >CH^CH + 3H,
it. lib i CH = CH + HCl > CH,=ClICi
nCHj =CH-C1 Xt, 1^ -^-eCH.-CHCl V
•'inn
u5. (2,0 diem) , (1,25 diem)
Khi X khdng miii, lam due dung dich nude v6i trong la CO,