Trong phan ung nhiet nh6m, vi A l dir ndn n^, > 2np^203 = 2.0,1 = 0,2 (mol)
= > m A , > 0 , 2 . 2 7 = 5,4(g)
Trong 4 phircmg an, chi c6 phuong an D (7,02 g) cd m > 5,4 (g).
Thidu 2: Nung h6n hop b6t g6m A l va FcjOj (trong diiu kifin khdng cd oxi), thu duoc hdn hop chat ran X. Chia X thanh 2 phSn bang nhau:
- Cho pMn 1 vao dung djch HCl (du) thu duoc 7,84 lit khf (dktc);
- Cho phdn 2 vao dung djch NaOH (du) thu duoc 3,361ft khf H , (dktc).
Biet rang cac phan ung deu xay ra hoan toan. Ph^n tram khd'i luong ciia Fe trong X la
A. 66,39%. B. 42,32%. C. 46,47%. D. 33,61%.
(Trich de tuyen sinh Cao dang khoi A) Hu&ng ddn gidi
Theobaira: (P,) = 0,35 (mol);nH2 (P2) = 0,15 (mol)
^> X + NaOH(dd) tao ra khf H2 =:> A l c6n du (FcoOj he't; X gdm Fe, AI2O3 va Al du).
2A1 + FejOj - > AI2O3 + 2Fe '
2x < - X - > X -> 2x
^ m o l A l du (trong X ) la y s6' mol A l ban d^u 1^ (2x + y)mol.
^ ^ I : Fe + 2 H C l - > F e C l 2 + H . t
2x 2x
323
A l + 3HCI A i a 3+ 1, 5 H 2 t
=>2x+ l,5y = 0,35.2 i:>2x + l, 5 y -0 , 7 ( l )
* P h S n l l : A l + NaOH + 3 H 2 0- * N a A l ( O H) 4 + l, 5 H 2 i
y l,5y
=>l,5y = 0,15.2 =0,3 ( 2 ) T i r ( l , 2 ) = > x = 0 , 2 ; y = 0,2
Kh6'i lucmg h6n hop X: nix =2.0,2.56+ 0,2.27+ 0,2.102 = 48,2(g)
Dap an diing la C.
Thi du 3: H6n hop X gdm Fe304 va A l c6 ti 16 mol tuong ling 1: 3. Thuc hien phan ting nhifet nh6m X (khdng c6 kh6ng khO d6^n khi phan umg xay ra hoan toan thu duoc h6n hop gdm
A. A l , Fe, Fe304 va A I 2 O 3 B. A I 2 O 3 va Fe.
C. A l , Fe va A l j O j . D. A I 2 O 3, Fe va Fe304.
(Trich dethi tuyen sink DHkhoiA) Hu&ngddngidi
PTHH:
3Fe304 + 8A1 -> Fe + 4 A I 2 O 3
3 ->• 8
Gia sir CO 3 mol Fe304 => nAi diu) = 9 mol ma n^i ang) = 8 mol => A l du Vay sau phan ling c6: Fe, AI2O3 va A l du.
Dap an dung la C.
Thi du 4: Cho 1,56 gam h6n hop gdm A l v^ AI2O3 phan ling he't v6i dung dich HCl (du), thu duoc V lit khi H . (dktc) va dung djch X. Nho tir tir dung dich NH3 den du vao dung djch X thu duoc ke't tiia, loc he't luong ke't tiia, nung den kh<5i luong kh6ng d6i thu duoc 2,04 gam ch^ft rdn. Gia trj ciia V la
A. 0,448 B. 0,224 C. 1,344 D. 0,672
(Trich de tuyen sinh Cao dangkhoiAl Hu&ng ddn gidi
2 04
Theo bai ra: n . 1, 0 . = = 0'02 (mol) So d6 phan ung:
A l , AI2O3 - ^ ^ A i a 3 - ^ " 3 ^ " 2 " >Al(OH)3 ^ A I 2 O 3
Taco: X = 0,02; y =0,01 2A1 >Al203
X — > X / 2 (mol) '27x + 102y = l,56 _
x/ 2 + y = 0,02 ^
2A1 + 6HC1 > 2Aia3 + 3H2 t 0,02 -> 0,03 V = 0,03.22,4 = 0,672 (lit)
pap an dung la D.
A I 2 O 3 > A I 2 O 3
y -> y (mol)
Thidu 5: Nung h6n hop g6m 10,8 gam A l va 16,0 gam FejOa (trong di^u kidn khong C O khdng khQ, sau khi phan ting xay ra hoan toan thu duoc chat rSn Y.
Kh6'i luong kim loai trong Y la
A. 16,6 gam. B. 11,2 gam. C. 22,4 gam. D. 5,6 gam.
(Trich de tuyen sinh Cao dang khoi A) Hu&ng ddn gidi
Theo bai ra: n^, = 0 , 4 ( m o l ) ; np^^Oj = 0 , l ( m o l ) 2 A l + FejOj — ^ 2Fe + A I 2 O 3
PhaniJng: 0,2<e-0,l -> 0,2(mol) Sau p/u: 0,2 (mol) 0,2 (mol)
vay m^™ = m^i + mp. = 0,2.27 + 0,2 + 56 = 16,6(g) Dap an dung la A.
Thidu 6: Day gdm cac oxit deu bi A l khir o nhiet d6 cao la A. FeO, MgO, CuO B. FeO, CuO, C r A - C. Fe304, SnO, BaO D. PbO, K2O, SnO.
(Trich de tuyen sinh Cao dang khd'i A) Hu&ng ddn gidi
oxit bj A l khir o nhidt d6 cao la cac oxit ciia cac kim loai c6 ti'nh khuf trung
|nh va ye'u, nhu FeO, CuO, Cr^Oj:
. 0
->3Fe + Al203
^ 3 C U + Al203 „ „,,ff|,,
Cr203 + 2A1 ^ 2Cr + A I 2 O 3 " ; "ifc*'';;
5dp an diing la B. ' ^ 3FeO + 2Al- '
3CuO + 2Al-
Qhu v.- Cac oxit MgO, BaO, KjO kh6ng bi khiJr bcri A l b nhiet d6 cao
325
Thi du 7: Thuc hidn phan uôg nhifet nhdm h6n hop g6m m gam A l 4,56 g-,^
C F J O J (trong difiu kidn khOng c6 O2), sau khi phan ling ke't thiic, thu dugc hft,, hop X. Cho toan b6 X vao m6t luong du dung dich HCl (loang, nong), khi cac phan ling xay ra hoan toan, thu duoc 2,016 lit H2 (dktc). Con neu ci^j toan bo X vao mOt luong du dung djch NaOH (die, nong), sau khi cac ph^^
ihig V.6\. thiic thi s6' mol NaOH da phan iJng la
A. 0,16mol. B. 0,06 mol. C. 0,14 mol. D. 0,08 mol.
(Trich detuyen sink Dai hoc khojH)
Theo bai ra: n
Hu&ng ddn gidi
^ ' ^ ^ = 0,03 (mol); nH2 = = 0,09(mol) Trong X con c6 A l du.
2A1 + CrjOj AI2O3 + 2Cr
22,4
V i n H 2 > 2. n c ^ o 3
0,06 < - 0,03 Cr + 2HC1 0,06
2A1 + 6HC1 0,02
0,03 0,06 CrCl2 + H 2
0,06 2AICI3 + 3H2
(0,09 - 0,06) X + dd NaOH (dac, n o n g , du):
A l + NaOH + 3H2O -> Na 0,02->0,02
AI2O3 + 2NaOH + 3H2O 0,03 -> 0,06
Vay nNaOH =0,02 +0,06 = 0 , 0 8( m o l )
Dap an d i i n g la D.
A l ( O H ) ^
A l ( O H ) , 2Na
Thi du 8: Nung nong m gam h6n hop A l va FcjOj (trong m6i truomg kh6ng co kh6ng khi) de'n khi phan umg xay ra hoan toan, thu duoc h6n hop ran Y. Chu>
Y thanh hai phSn bang nhau:
- Phdn 1 tac dung v6i dung djch H3SO4 loang (du), sinh ra 3,08 lit khi H, (6 dktc).
- Phdn 2 tac dung v&i dung djch NaOH (du), sinh ra 0,84 lit khi H , (6t dktc).
Gia tri cia m la
A. 29,40. B. 22,75. C. 29,43. D. 21,40.
(Trich De thi tuyen sinh DH - CD khoij^
Hu&ng ddn gidi
Xheobaira: n H 2 ( , ) = ^ ^ = 0,1375(mol); n H 2 ( 2 ) =5^ = 0,0375(mol) 22,4
pTPLT nhiet nhom: 2Al,d„) + FejOj
22,4 2Fe + A l A
Trong hop sau phan ihig c6: Fe, Al.Oj, A l du (vi hOn hop sau phan ung tac dung vdi NaOH tao ra khf H^), (FcjOj phan ihig he't vi phan ihig xay ra hoan toan).
phan I : Fe + H, S 0 4 (1) —
X
2A1 + 3 H, S 0 4 ( ] ) y
^ FeSO. + H , t Al2(S04)3 + 3H2T
l,5y
(1) 2Na[Al(OH)4] + 3H2T
l,5y (2) Taco: x + l,5y =0,1375
Phan I I : 2A1 + 2NaOH + 6H2O y
Taco: l,5y =0,0375 T i r ( l , 2 ) t a c 6 : x = 0 , l ; y = 0,25.
Theo PTPLT nhiet nhOm:
^ m Fe203 = 0.05 . 160 = 8(gam) ' '
" A I (p/u) = " F C = 0,1 (mol)
" I A I (ban d i u ) = niAi + m^i , p / „ = (0,025 + 0,1). 27 = 3,375 (gam) Vay m = 2. (m p , 2 0 3 + H I A, ) = 2. (8 + 3,375) = 22,75g. • W- Dap an diing la B.
Thidu 9: Nung h6n hop b6t gom 15,2 gam CrjOj vdi m gam A l 0 nhiet d6 cao.
Sau khi phan ihig hoan toan, thu dugc 23,3 gam h6n hop ran X. Cho toan b6 h6n hop X phan ihig vdri axit HCl (du) thoat ra V lit khi H , ( 0 dktc). Gia tri ctia V la
A. 3,36. B. 4,48. C. 7,84. D. 10,08.
(Trich Delhi tuyen sinh DH - CD khoi B)
I'M:
Hu&ng ddn gidi
£>e }>iai hai nay can si( dung dinh luat hdo toan khd'i lucfng vd chu y kirn loai Cr phan Ung vdi dung dich HCl tao ra muoi crom (II) clorua.
S d m o l C r A v n c , 2 0 3 = ^ = 0,l(mol)
Trong phan ung nhidt nh6m: m^, + m m c^ 0 3 = ' " h ô n h ô , r i „ ô u p w„ a„ g
^ m ^ , = 23,3 - 15,2 = 8,l(gam) = 0,3 (mol)
PTPLT: 2A1 + CrjOj ——> 2Cr + AUOj Saupu: 0,1 (mol) 0,2 (mol)
Al + 3 H a > Aia, + USH^t 0,1 (mol) 0,15 (mol) Cr + 2Ha > CrQj + H^t 0,2 (mol) 0,2 (mol)
=> I n =0,15 + 0,2 = 0,35 (mol)
=> = 0,35. 22,4 = 7,84 (lit).
Dap an dung la C.
Thidu 10: D6't nong h6n hop g6m Al va 16 gam FcjO, (trong di^u kifin kh6ng CO khdng khi) dd'n khi phan ling xay ra hoan toan, thu duoc h6n hop ran X.
Cho X tac dung vCfa du vdi V ml dung djch NaOH I M sinh ra 3,36 lit (6 dktc). Gia tri ciia V la
A. 300. B. 100. C.200. D. 150.
(Trich De thi tuyen sinh Cao dangkhdlA) Theobaira: iii^^ =0,15 (mol)
Hu&ng ddn gidi 3,36
22,4
Trong X (g6m AUO,, Fe va Al du) chi c6 Al tac dung vdi dung dich NaOH:
Al + NaOH + 3 H, 0 > NaAl(OH)4 + ^SH^T 0,1 (mol) 0,15 (mol) Suyra:V= ^^1^1^ = 100 (ml)
Dap an dung la B.
Thidu 11: Tr6n 10,8 gam b6t Al vdi 34,8 gam b6t Fe304 rdi tid'n hanh phan ihig nhiet nhom trong didu kidn khdng c6 kh6ng khi. Hoa tan hoan toan h6n hap ran sau phan ihig bang dung dich H2SO4 loang (du), thu duoc 10,752 lit khi H2 (dktc). Hieu sua't ciia phan ling nhiet nh6m la
A. 80% B. 70% C.60% D. 90%
(Trich de thi tuyen sinh Dai hoc khdlB)^
Hu&ng ddn gidi Cdch 1:
Theobaira: n^i =0,4(mol);npgp^ =0,15(mol); n^^ =0,48(mol) PTHH: 2Al + 3Fe304—^9FeO + Al203
X -> l,5x 328
8A1 + 3 F e 3 0 4 — ^ 9 F e + 4Al203 t y 3y/8 >9y/8
Fe + H2S04(1) ).FeS04+H2 9y/8 > 9y/8 2A1 + 3H2S04(1) >Al2(S04)3+3H
( 0 , 4 - x - y ) 1,5(0,4 Taco: ^ + 1,5.(0,4-x-y) = 0,48
o
=>9y + 1 2 . ( 0 , 4 - x - y ) = 3,84 =^ 12x + 3y= 0,96 Hidu sua't phan ihig nhidt nh6m (tinh theo Fe304):
^ _ ( l . 5 x + 3y/8).100%
0,15
_ ^ _ ( l 2 x + 3y).100% 0,96.100%
8A15 - — - j ^ - 8 0 % Dap an dung la A.
Cdch 2:
Coi phan ung xay ra:
8A1 + 3Fe304 >9Fe + 4AI2O3 BandSu: 0,4 0,15
IVu: X ^ 3x/8 ^ 9x/8 Con: 0,4 - x 9x/ 8
2AI + 3H2SO4 >Al2 ( 8 0 4 ) 3 + 3H2
0 . 4- X 1,5(0,4-x)
Fe + H2SO4 > FeS04 + H j
9x/8 9x/8 Taco: 1,5.(0,4-x) + — = 0,48
8
=^ 12(0,4-x) + 9x = 3,84 i:>3x = 0,96 =>x=0,32
^ay hieu sua't phan ung nhiet nh6m:
H = ^ ^ = 2 : ^ . 8 0 % 0,4 0,4
3. Cac bai tap tir luyen:
Cau 1: Thirc hien phan umg nhiet nh6m giOa A l va CrjOj trong di6u kidn kh6ng ^ kh6ng khi, sau phan umg hoan toan thu duoc h6n hop X c6 kh6'i luong 43,9 gy^
Chia X lam 2 ph^n bang nhau. Cho phdn 1 tac dung vol dung dich NaOH (H thu duoc 1,68 lit khi (dktc). PhSn 2 phan ung vira du vdri V lit dung dich HCj (loang, nong). Gia tri ciia V la
A. 0,65 . B. 1,05 C. 1,15 D. 1,00
Khi thoat ra la H,: n^^ = - O,075(mol) 22,4
Trong X con c6 A l dir:
2Al + 2NaOH + 6H20 •2Na Al(OH)^
K •
+ 3H2 T
0,05 < - 0,075(mol)
=> Trong X (ca 2 phdn) c6 0,05. 2 = 0,1 mol A l du 2A1 + C r 2 0 3A I 2 O 3+ 2 C r
2x < - X - > X ^ 2x (mol) Taco: (2x + 0,l).27 + 152x-43,9
=>206x = 4 3 , 9-2, 7 = 41,2 = > x = 0,2 Do do trong m6i phdn c6: 0,05 mol A l du
0,2 mol Cr sinh ra (x mol Cr) 0,1 mol AKOj (x/2 mol AUOj) 'k^': A l + 3HC1->A1C13 + 1,5H2
0 , 0 5 ^ 0,15(mol)
AI2O3 + 6HC1 ^ 2AICI3 +3H2O 0,1 - > 0,6(mol)
0,15
Cr + 2HC1 -> 2CrCl2 + H . 0,2 -> 0,4(mol)
=>nHCi =
1 = 1,15(1) Dap an diing la D.
Cau 2: Tron 0,25 mol b6t A l voi 0,15 mol b6t Fe.Oj r6i ti^'n hanh phan ling nh'^' nhom trong di^u ki^n khong c6 khdng khi (gia sir chi c6 phan umg khir Fe.Oi ^ Fe) thu duoc h6n hop rSn X. Cho X tac dung voi dung dich NaOH (du) thu dung dich Y , m gam chat ran khan Z va 0,15 mol Hj. Hi6u suS't phan ling ' i ' ^ ' -
nh6m va gia trj ciia m I5n luot la
A. 60% va 20,40 B. 60% va 30,75.
C. 50% va 30,75. D. 50% va 40,80.
Cty TNHH MTV DVVHjn^n^^^
PTHH xay ra:
Ban ddu:
Phan umg:
Con:
Hudfng ddn gidi 2A1 + FejOj ->
0,25 0,15
2x X
((/,25-2x) ( 0 , 1 5- X )
2Fe + A I 2 O 3
0 2x:
2x!
0
X X
2A1 + 2NaOH + 6H2O 2 N a [ A l ( O H ) ^ ] + 3H2 t
( 0 , 2 5 - 2 x ) - ) . 1,5(0,25-2x) AI2O3 + 2NaOH + 3 H 2 0 - > - 2 N a [ A l ( O H ) ^ "
=> Chat ran Z g6m: Fe^O, va Fe
Theo bai ra, ta c6: 1,5.(0,25-2x)=0,15 =:> x =0,075 Vay:
O..A • - tinhtheoAl):
+ Hieu suat cua phan iJng nhiet nh6m ( ^ < 2 l H
„ ^ 0,075.2.100%
H % = = 60%
1 0,25
•
Kh6'i luong ch^t ran Z:
m- m F e 2 0 3 +r"Fe = (0,15 - 0,075). 160 + 2.0,075.56 = 20,40(gam) Dap an diing la A.
Cau 3: Tr6n 5,4 gam b6t A l vdi 14 gam Fe^Oj r6i tid'n hanh phan ung nhidt nh6m (kh6ng CO oxi, FcjOj, bi khir ve Fe). Sau khi kd't thiic phan umg, lam ngu6i h6n hop va hoa tan h6n hop nay bang luong dung dich NaOH (du), cho den phan ling hoan toan thu duoc 1,68 lit khi (dktc). Hieu sua't phan umg nhiet nh6m la A. 66/S7<K, D - . . ^
A. 66,67% B. 85,71% C. 92,68%
Hudfng ddn gidi Theo bai ra: n^, =0,2 (mol); np^^^^ =0.0875 (mol)
1,68
D. 75%
PTPLT:
nH2 = ^ = 0 , 0 7 5 (mol) 2A1 +Fe203 -^Al203 +2Fe
A l + NaOH + 3 H 2 0 ^ N a [ A l ( O H ) 4 ] + | H 2 t 0,05
^ n A i ( P U) = 0 , 2 -0,05=0,15(mol)
"^"Fejoacpu) =~- =0,075(mol).
0,075
Vay hieu sua't phan umg nhidt nh6m: H % = 0,075.100%
=85,71' 0,0875
Dap an dung la B.
C^u 4: Nung m gam h6n hop X g6m A l va Fe304 trong binh kin kh6ng c6 kh6n ' khi dfi'n phan ung hoan toan thu duoc h6n hop ran Y. Chia Y thanh hai p^^
bang nhau. PhSn 1 cho tac dung vdi dung djch H N O 3 loang, du thu duoc 0 1 ]
mol NO (san ph^m khu duy nha't). Cho ph^n 2 tac dung vdi dung djch NaOH du thu duoc 0,03 mol H,. Gia trj cOa m la
A. 19,59 B. 19,32 C. 9,93 D. 9,66 Hu&ng ddn gidi
Vi Y + NaOH (dd) ^ khi => Al con du, Fe304 he't.
Trong Y c6 Fe, AKO, va Al du.
* 1/2 Y + ddNaOH (du):
•Na A l ( O H) J + - H , T
' J 2
• 2Na Al(OH)^
A l + NaOH + 3H20 0,02 (mol) <-
AI2O3 + 2NaOH + 3H20 1/2 Y + ddHNO, loang, du:
A l + 4 H N O 3 - > A l ( N O 3 )3 + NO + 2 H 2 O
0,02 0,02 (mol) Fe + 4 H N 0 3 - > F e ( N 0 3) 3 + NO + 2H20 0,09 (mol) < - (0,11 - 0 , 0 2 )
A I 2 O 3 + 6HN03->2A1(N03)3 + 3 H 2 O
PTPU' nhidt nh6m:
8A1 + 3Fe304 — ^ 9Fe + 4 A I 2 O 3
0,08 < - 0,03 < - 0,09 (mol) v a y m = mA, + mFe304
0,03
•m = 2. (0,02 + 0,08).27 + 0,03.232] ii> m = 19,32(gam) Dap an dung la B.
Cau 5: Tien hanh phan ling nhiat nh6m vori Fe304 trong d i l u kifin kh6ng c6 kh6ng khi, cho bie't phan ihig xay ra hoan toan va Fe304 chi bi khiJ thanh Fe. Chia h^"
hop thu duoc sau phan ung thanh hai phSn bang nhau.
Cho 1 phSn tac dung hS't vdi dung djch NaOH (du) thu duoc 2,52 lit H , (dk"^';
Hoa tan he't phSn 2 bang dung djch HNO3 dac, nong (du) tha'y c6 11,76 lit k^^'
bay ra (dktc). Kh6'i luong Fe sinh ra sau phan umg nhifit nh6m la
A. 5,6 gam B. 8,4 gam C. 11.2 gam D. 16,8 gam
Hu&ng ddn gidi
TTieobai ra: =0,1125(mol); nN02 = ^ ^ = 0,525(mo!)
Tir ke't qua thi nghidm ta tha'y sau phan umg nhifit nhOm thu duoc h6n hop g6m:'
AI2O3; Fe (X mol) va A l du (y mol)
^ Ph^n I + NaOH (dd, du):
A l + NaOH + 3 H 2 O ^ N a [ A l ( O H) J + ^ H 2 t
0,5y 0,75y(mol) + ph^n I I + HNO3 (dac, nong, du):
A l + 6 H N O 3 - ) . A1(N03)3 + 3 N O 2 t+ 3 H 2 O
0,5x - > l,5x fO,75y = 0,1125
Taco:^ => x = 0,2; y = 0,15 . [l,5x + l,5y =0,525 ^
v a y mpe =0,2.56 = 1 l,2(gam) Dip an dung la C.
cau 6: Tr6n 24 g Fe203 vdi 10,8 g A l r6i nung 6 nhifit d6 cao (khdng c6 khdng khO. H6n hop thu duoc sau phan umg dem hoa tan vao dung djch NaOH du thu duoc 5,376 lit khf (dktc). Hieu sua't ciia phan umg nhiet nh6m \k
A. 12,5% B.60% C. 80% D. 90%
1): Hu&ng ddn gidi 24 10 8
Stfmol FejOj = — = 0,15 (mol); s6'mol A l = — = 0,4 (mol)
tlii^ 160 27
2A1 + FcjOj > AI2O3 + 2Fe
2A1 + 2 N a O H + 2 H 2 O > 2NaA102 + S H j t 0,16 mol < 0,24 mol
"Aidu = 0,16 (mol) =>nA,phi„tag = 0,4 - 0,16 = 0,24 (mol)
0,24
ô203 phan ling = = 0,12 (mol) .u suaft phan umg: H = — . 1 0 0 % = 80%
h ^ 0,15 u:
^ p a n d i i n g l a C .
Dang 3: Nhom phdn itng v&i dung dich kiem, tinh lUffng tinh cua hap chat nhom
1, L i thuyet van dung va phirong phap giai:
* Kim loai nlidm tac dung v6i dung dicli kiem:
2A1 + 2NaOH + 6 H 2 O -> 2Na[Al(OH)4] + m^
2A1 + Ca(OH)2 + 6H3O -> Ca[Al(OH)4]2 + 3H,
Do do khi cho h6n hop kim loai manh (Na, K, Ca, Ba, ...) nh6m v^o niroc thi nh6m cung bj tan mot phdn hoac tan hoan toan:
2Na + 2H,0 2NaOH + H,
2A1 + 2NaOH + 6 H 2 O -> 2Na[Al(OH)4] + BH^t
* Nhom oxit AI2O3 va nhom hidroxit AI(OH)3 la cac chat lurong tinh:
, , A l A + 6HC1 > 2AICI3 + 3H,0
A l , 0 3 + 2NaOH + SH^O — ) > 2Na[Al(OH)4]
AICI3 + 3ao
Na[Al(OH)4]
Ai(OH)3+3HCI Al(OH)3 + NaOH
natri aluminat
* Mot so P T H H can chu y:
NaAl(OH)^ + HCl A ^ O H ) ^ i +NaCl + H2O NaAl(OH)^ + 4HC1 ->• NaCl + AICI3 + 4H2O AI2 (504)3 +3Ba(OH)2 ^ 3BaS04 i +2A1(0H)3 4.
Al2(S04)3+4Ba(OH)2->-3BaS04 i + B a [ A l ( O H ) ^ ] ^ Na [ A l (OH)^ ] + CO2 ^ Al (OH)3 i + NaHCOj
KAl (SO4 )^ + 2Ba(OH)^ -> 2BaS04 i +A1 ( 0 H ) 3 i +KOH KA1(S04)2 + 2Ba(OH)2 ^2BaS04 i + K r A l ( O H ) ^ ]
* Cac cong thurc giai nhanh bai tap dang nay:
- C6ng thiic tinh s6' mol NaOH cho vao dung djch A l ' * d^ xuaft hifin ni6t
lirong kfe't tua theo ydu ciu: n =3.ni ''•'
OH
n =4.n -,, - H I
OH~ Ar+
- C6ng thiic tinh s6' mcl HQ cin cho vao dung dich Na(Al(OH)^) (hoac NaAlOj) de xua't hiSn m6t lugng ke't tia theo yfiu cdu:
n . =4.n,
Al{OH)4 . - - 3 . n j 334
c thi du minh hoar
'ifiTdu 1: H6n hop X g6m Ba va A l . Cho m gam X vao nu6c du, sau khi cac phan irng xay ra hoan toan, thu duoc 8,96 lit khi H , (dktc). Mat khac, hoa tan hoan toan m gam X bang dung djch NaOH, thu duoc 15,68 lit khi H , (dktc).
Giatricuamla
^. 16!4 B.29,9 C.24,5 D. 19,1
(Trich dethi tuyen sink DH khoi A nam 2013) Hu&ng ddn giai
8 96 15 68
Theo bai ra: n^^^^^^ = ^ =0,4 (mol); n^^^j) = = (mol) Vi thi nghiem (2) thu duoc lucmg H , nhifeu hom thi nghidm (1) => thi nghidm (1) Al con du.
TNI: Ba + 2A1 + 4H,0 ^ Ba(AI02)2 + 4H2 ,. . 0,1 <- 0,2 <- 0,4 TN2: 2A1 + 2NaOH + 2H2O -ằ 2NaA102 + 3H,
0,2 <- (0,7 - 0,4)
Vay m = 0,1. 137 + (0,2 + 0,2). 27 = 24,5 (gam) . Dap an diing la C.
Thi du 2: The' tich dung dich NaOH 0,25M cin cho vao 15 ml dung dich A1,(S04)3 0,5M de thu duoc lucmg ke't tiia Idn nha't la
A. 210 ml B. 60 ml C. 90 ml D. 180 ml (Trich dethi tuyen sink DH khoi B nam 2013) Hu&ng ddn giai t, a
Sd'mol A1,(S04)3: 11^,2(504)^ =0,015.0,5 = 0,0075(mol)
Khi chi xay ra phan iitig tao kd't tiia thi thu duoc luong kd't tiia 1dm nha't:
A1,(S04)3 + 6NaOH-> 2A1(0H)3>1 + 3Na,S04 0,0075 -> 0,045 (mol)
% V , , ^ , O H 0.25M = ^ . 1 0 0 0 = 180(ml)
^ p an dung la D
</ô 3; H6n hop X g6m Na, A l va Fe (vdi ti Id s6' mol giOa Na va A l tuong
•^g la 2 : 1). Cho X tac dung vdi H,© (du) thu duoc chat ran Y va V lit khi.
Cho toan b6 Y tac dung voi dung djch H2SO4 loang (du) thu duoc 0,25V lit Khi. Biet cac khi do a ciing didu kidn, cac phan litig d^u xay ra hoan toan. Ti H s6 mol cua Fe va A l trong X tuong ihig la
'•5:16. B. 1:2. C. 16:5. ^ D. 5 :8.
(Trich de tuyen sinh Cao dang khd'i A)
Hu&ng ddn gidi
Goi s6 mol Na, Al va Fe trong X Mn luot la 2x, x y.
Na + HjO NaOH + O.SHj t 2x -> 2x -> X
Al + NaOH + 3H2O- ằ NaAl(OH)4 + l.SHj t x ^ x -> l,5x
V
=>x + l,5x = 2,5x = V=:>x = — 2,5 Chat ran Y la y mol Fe.
* Y + H2SO4 (loang, dir)
Fe + H2SO4 (1) ^ FeS04 + H2 t
y ^ y
=>y = 0,25V
Vay n p , : n A , = y : x = 0 , 2 5 V : ^ = 5:8 Dap an dung la D.
Thidu 4: Cho 500 ml dung djch Ba(0H)2 0,1M vao V ml dung djch AI2 (804)^
0,1M; sau khi cac phan umg k6't thiic thu duoc 12,045 gam kd't tua. Gia trj cua V l k
A. 75. B. 150. C.200. D. 300.
(Trich de thi tuyen sink DH khoi A) Hudng ddn gidi
= 0,5.0,
TheobMra: n^^Q^^)'2 =0,5.0,1 = 0,05 (mol)
* Tru&ng hap 1: Chi xay ra phan irng:
3Ba(OH)2 + A I 2 (504)3 -*3BaS04 ^ +2A1(0H)3 i 0,05 0,05 ^ 0,1/3
(Di chi xay ra phan ling trdn, tiic la chua c6 phan utig ho^ tan hot kd't tiia A1(0H)3 thi B a ( O H)2 phai he't).
1 N
• m(ke'ttiia) =0,05.233 + 2ll
. 3 , .78 = 14,25(g)
(;^12,045g => loai). j
* Tru&ng hffp 2: Xay ra 2 phan ting:
AI2 (SO4 )3 + 3Ba(OH)2 -> 3BaS04 i +2Al(OH)3 i
X -> 3x ^ 3x 2x :
Al2{S04)^ • 4Ba(OH)2 ^3BaS04 i + B a [ A l ( O H ) / y 4y —ằ• 3y
fheo bai a, ta c6: 3x + 4y = 0,05 ; j (3x + 3y).233 + 2x.78 = 12,045 t.
-:i> X = 0,0 l ; y - 0 , 0 0 5
^S6'mol Al2(S04)3 =x + y = 0,015 (mol) n.1000 0,015.1000
vay V = •
- M 0,1 • = 150(ml) Dap an diing la B .
Chu y: - Ket tua gom AKOH), va BaS04.
- B a ( O H ) 2 + 2 A l ( O H ) 3 ^ B a r A l ( O H ) '4J2
Thi du 5: Hoa tan ho^n to^n m gam h6n hop g6m NajO va AUOj v^o nudc thu dirge dung djch X trong su6't. Thdm i\x tit dung djch HCl I M vao X , khi hd't
100 ml thi bat ddu xufl't hidn kd't tiia; khi hd't 300 ml hoac 700 ml thi d^u thu dugc a gam ket tua. Gia trj cua a va m \in lugt la
A. 23,4 va 56,3. B. 23,4 va 35,9. C. 15,6 va 27,7. D. 15,6 va 55,4.
(Trich dethi tuyen sink DH khoi A) Ilit&ng ddn gidi
Theo bai ra: n^^Q^ =0,1 (mol);nHci -0,3 (mol); n^ci =0.7 (mol).
Dat s6' mol Na,0; AUOj Ian lugt la x, y (mol). ' Na20 + H2O > 2NaOH
X > 2x
2NaOH + AI2O3 + 3H2O 2Na[ A l ( O H ) J
=:> Dung djch X chi c6 NarAl(OH)^] hoac c6 d6ng then NarAl(OH)4 NaOH du.
Theo bai ra, khi cho HCl vao X chua tao ra kd't tiia ngay => X c6 NaOH Na[Al(OH)^]).