04
'\)ang 4: B a i tap ve p h a n ung N a 2 C 0 3 ; N a H C C ) , \(n axit
[sjho tir tir tirng giot den het 30 m l d u n g d i c h H C l I M vao 100 m l d u n g d i c h chira ' NạCỎ 0 , 2 M va N a H C O , 0 , 2 M , sau phan ting t h u dugc so m o l C O , la
A . 0,015 B. 0,010 C. 0,020 D . 0,030 Huong đn gidi Huong đn gidi T h e o bai ra: n n c i = 0 , 0 3 ( m o l ) " N n i C O . , = " ' 0 2 ( m o l ) ; n N , H a ) , = 0 , 0 2 ( m o l ) 1^. PTPÚ: N a j C O , + H C l > N a C l + N a H C O , 0,02 - > 0 , 0 2 ^ 0,02 N a H C O ^ + H C l > N a C l + C O , t + H 2 O 0,01 < - ( 0 , 0 3 - 0 , 0 2 ) ^ 0,01 V a y n(-02 = 0 . 0 1 0 ( m o l ) D a p a n d u n g la B .
2. Cho tir tir d u n g d i c h chira a m o l H C l vao d u n g djch chiia b m o l NâCO, d o n g thai khuay t h u d u g c V 1ft k h i ( a d k t c ) va d u n g d i c h X . K h i cho dir nircfc v o i t r o n g vao khuay t h u d u g c V 1ft k h i ( a d k t c ) va d u n g d i c h X . K h i cho dir nircfc v o i t r o n g vao d u n g d i c h X thay c 6 xuat hien ket tuạ
Bieu thiic l i e n he g i u a V v d i a, b la
A . V = 22,4 .(a + b). B. V = 11,2 .(a - b). C. V = l l , 2 . ( a + b). D . V = 22,4 . ( a - b ) C. V = l l , 2 . ( a + b). D . V = 22,4 . ( a - b )
Huong đn gidi
Co cac P T P l T xay ra:
H C l + N a , C O , > N a H C O j + N a C l Ban d a u : a ( m o l ) b ( m o l )
Phan u n g : b < b ( m o l ) > b ( m o l )
Con l a i : a - b ( m o l ) 0 b ( m o l ) '
( T i n h theo NâCO,, v i H C l con d u , t h a m gia phan úng d\x6\y dá tao ra C O , )
H C l + N a H C O , > N a C l + C O , ! + H^O
Ban dau: (a - b ) m o l b ( m o l ) ^^ Phan l i n g : (a - b) m o l (a - b) m o l
( T i n h theo H C l , v i N a H C O , con d u , do c6 phan iJng:
N a H C O , + C a ( O H ) , > C a C O , i + N a O H + H3O)
V a y V = (a - b). 22,4 ( l i t ) D a p a n diing l a D.
3- D u n g d i c h X chira h 6 n hop g o m N a , C O , 1,5M va K H C O , I M . N h o tir tir tirng
g i o t c h o d e n het 2 0 0 m l d u n g d i c h H C l I M vao 100 m l d u n g d i c h X , s i n h ra V lit k h i ( d d k t c ) . G i a t r i cua V la
A . 4,48. B. 1,12. C. 2,24. , _ ^ D . 3,36.
Phan locii va pfiiịyn,: /'Inir vliii nluinli BĨHda DOc IZ -LU 1 lUlllll 1 om Huong dan gidi
Theo bai ra: riHci = 0,2 . 1 = 0,2 (mol).
HNajco, =0,1 . 1,5 = 0,15 (mol); n^Hcoj =0,1 . 1 = 0,1 (mol). Khi nho tir tfir dung dich HCl (H*) vao X thi: ' * v'.!
Na.CO, + HCl > NaCl + NaHCO,
0,15 > 0,15 )• 0,15(mol)
Sau phan ung nay, lucmg HCl con lai la nHc,,d«, = 0,2-0,15 = 0,05 (mol).
Do do, xay ra qua trinh: .,, ^.ti v\
HCO3 + H^ > C O , t + H . O
0,05 < 0,05 > 0,05 (mol) ,j Vay V = Vco2 = 0,05. 22,4 =1,12 (lit).
Dap an diing la B . (adsbygoogle = window.adsbygoogle || []).push({});
4. Cho h6n hop K 2 C 0 3 v a NaHC03(ti le mol 1:1) vao binh dung dich Ba(HC03)2 thu dugc ket tua X va dung dich Y . Them tiir tiir dung djch HCl 0,5M vao binh den khi khdng con khi thoat ra thi het 560 ml. Biet toan bo Y
phan ling vCra du \6\0 ml dung dich NaOH I M . Khoi lirgng ket tiia X la
Ạ 3,94 gam. B. 11,28 gam. C. 7,88 gam. D. 9,85 gam.
Huong đn gidi
Theo bai ra: nHci = 0,5.0,56=0,28mol; n^ a O H =0,2.1 =0,2mol
Goi s6' mol K2CO3, NaHC03 va BăHC03)2 Mn luot la x, x va y (mol) ^ S6' mol CO3", HCO3 la x va (x + 2y) mol.
* HCl + X + Y
i I ' * CÔ" + 2H^ -> CO2 + H2O
, X ^ 2x
HCO3 + H ^ CO2 + H2O (x + 2y) (x + 2y) (x + 2y) (x + 2y)
^ r ^ 2 x + (x + 2y) = 0,28 =>3x + 2y = 0,28 ( l )
*NaOH + d d Y :
HCOJ + O H ' -> COf + H2O (x + 2y) -> (x + 2y)
=:>x + 2y = 0,2(2) 9 6
Cty TNHH MTV DWH Khang Viet
T i r ( l , 2 ) x = 0,04; y = 0,08
K2CO3 + BăHC03)2 -> BaCOj i +2KHCO3 |
Banđu: 0,04 0,08 '
pU": 0,04 0,04 -> 0,04
=>mB,co3; = ' " x i =0,04.197 = 7,88(g) ; ' ! ! ' .
Dap an dung la C.
Chu y: Khi cho HCl vao binh thi HCl phan ling ca vdi ket tiia X (BaCO,) va
dung djch Y (NaHCO,, KHCO,, BăHC03)2).
Dang 5: Bai tap ve phan urng nhiet nhom
1. Phan dug hoa hoc xay ra trong trudng hop nao duori day khong thu6c loai phan ung nhiet nh6m? ung nhiet nh6m?
Ạ A l tac dung \6\ nung nong. B. A l tac dung vori CuO nung nong.
C. A l tac dung voi FcjOj nung nong. D. A l tac dung vdri H2SO4 dac nong.
Huong đn gidi
Phan dng nhiet nh6m la phan utig giOa bot nhom vdi oxit kim loaị
Do do, A l tac dung vdi H2SO4 dac, nong khong thudc loai phan dug nhidt nh6m. (adsbygoogle = window.adsbygoogle || []).push({});
Dap an dung la D.
2. Day g6m cac oxit deu bj A l khij 6 nhiet d6 cao la
Ạ FeO, MgO, CuO B. FeO, CuO, Cr^Oj C. Fe304, SnO, BaO D. PbO, K^O, SnỌ
Hu&ng đn gidi
Cac oxit bj A l khii b nhiet do cao la cac oxit ciia cac kim loai c6 tinh khif trung binh va yeu, nhu FeO, CuO, Cr.O,:
3FeO + 2 A l — ^ 3 F e + Al203 y
3CuO + 2 A l — i % 3 C u + Al203 < • C r 2 0 3 + 2 A 1 ^ 2 C r + Al203
Dap an dung la B
Chu f: Cac oxit MgO, BaO, K.O kh6ng bi khir bai A l a nhiet d6 cao
3. Nung 21,4 gam h6n hop A g6m bot A l va FcjOj (trong binh kin kh6ng c6 khong khi), thu dugc h6n hgp B g6m Fe, FeO, Fe304, FẹO, va A l (du). Cho B tac dung khi), thu dugc h6n hgp B g6m Fe, FeO, Fe304, FẹO, va A l (du). Cho B tac dung v6i dung djch HCl du dugc dung djch C. Cho C tac dung vdi dung djch NaOH du dugc ket tua D. Nung D trong khdng khi den khoi lugng khong d6i thu dugc
16 gam chat ran. Khd'i lugng A l trong h6n hgp A la
Ạ4,4g. B.5,4g. C. 10,8g. D. 2,7g.
Huong đn gidi
Phcin loai va plnfcfii}^ ;>/UI/J t^idi itlianh Hoa hoc 12 —Cu Thanh Toan
FêO, > Fe,0,, Fe
Fe(OH),, Fe(OH)3 ^ Fé^ Fe-*
Theo dinh luSt bao toan nguydn to, suy ra:
m Fe203 ~ 1 ^ S^f" ^ " ' A I = 21,4 - 16 = 5,4 gam.
Dap an dung la lị
4. Nung nong m gam h6n hop A l va FcjOj (trong m6i truofng khong c6 khong khi) den khi phan iJng xay ra hoan toan, thu duoc h6n hop rSn Ỵ Chia Y thanh hai phdn bSng nhau:
- Fliiin 1 tac dung vdi dung dich H,S04 loang (dir), sinh ra 3,08 lit khi H, (d dktc). - Ph&i 2 tac dung vdi dung djch NaOH (du), sinh ra 0,84 lit khi H , {b dktc).
Gia trj ciia m la
Ạ 29,40. B. 22,75. C. 29,43. D. 21,40.
^ Hu&ng đn gidi
T h e o b a i r a:nH2,i, = = 0,1375(mol); nH2(2) = = 0.0375(mol). PTPU" nhiet nhom: 2Al(d„) + FẹO, — — > 2Fe + AUOj
Trong hop sau phan ling c6: Fe, AKOj, A l du (vi h6n hop sau phan ung tac dung vdi NaOH tao ra khf H,), (FcjÔ phan ling het vi phan ling xay ra hoan toan).
Phan 1: Fe + H,S04 (/) > FeSO, + H , t
X X
* ' " 2A1 + 3H3SO4 (1) > AU(S04)3 + 3 H , t
• y l,5y (adsbygoogle = window.adsbygoogle || []).push({});
Taco: x + l,5y = 0,1375 (1)
Phan 2: 2A1 + 2NaOH + 6Hfi > 2Na[Al(OH)4] + SH^t
y ''5y Taco: l,5y =0,0375 (2)
Tir(1,2) taco: x = 0 , l ; y = 0,25.
Theo PTPir nhiet nhdm: n Fe203 = ^ " P e = ^-x = ^-0,1 = 0,05(mol). => m Fe203 = 0.05.160 = 8(gam).
nAKp/u) = n F e = 0 , l ( m o l ) .
= > niAKbanđu) = mAi(du) + H A I (p/ir) = (0,025 + 0,1). 27 = 3,375 (gam). Vay m = 2. (m p^^o, + ^ M ) = 2. (8 + 3,375) = 22,75g.
, J,. Dap an diing la B.
Cty TNHH MTV DVVH Khans ^ier
5. Nung h6n hop b6t gom A l va FejO, (trong di^u kidn khong c6 oxi), thu duoc h6n hop chat rdn X. Chia X thanh 2 phan bang nhau:
- Cho phan 1 vao dung dich HCl (du) thu duoc 7,84 lit khi H . (dktc); - Cho phdn 2 vao dung djch NaOH (du) thu duoc 3,36 lit khi H2 (dktc).
Bi6't rang cac phan ling d6u xay ra hoan toan. Phin tram khdi luong ciia Fe trong X la
Ạ 66,39%. B. 42,32%. C. 46,47%. D. 33,61%.
Hu&ng đn gidi
Theobai ra: nH2 (Pi) = 0,35mol;n^^2 (P2) = 0,15mol
V I X + NaOH(đ) tao ra khf H . => A l con du (FẹO, het; X gom Fe, Al.Oj va Al du).
; j 2A1 + Fe203 -> AI2O3 + 2Fe
2x <- x X - » 2x
S6' mol A l du (trong X ) la y => so mol A l ban dc1u la (2x + y)mol. * Phdn 1: Fe + 2HC1 ^ FeCl2 + H2 t 2x -> 2x A l + 3HC1 ^ AlCl3 + l,5H2 T y ^ l,5y =>2x + l,5y = 0,35.2 =^2x + l,5y =0,7 ( l ) * Ph^n?.: A l + NaOH + 3 H 2 0 - > N a A l ( O H ) ^ + l,5H2 4 y ^ l,5y =>l,5y = 0,15.2 = 0,3 (2) ' . T i r( l, 2 ) = ^ x = 0 , 2 ; y = 0,2 *
Khdi lirgmg hdn hop X : m ^ = 2.0,2.56 + 0,2.27 + 0,2.102 = 48,2(g)
^, 2.0,2.56.100% ^ , . ^ Vay %mp^/x = — " 46,47% Dap an dung la C .
6. Tron 10,8 gam bot A l vdi 34,8 gam bot Fe304 roi tien hanh phan u^g nhiet nhdm trong di^u kien khong cd khdng khị Hoa tan hoan toan hdn hop rdn sau phan trong di^u kien khong cd khdng khị Hoa tan hoan toan hdn hop rdn sau phan ling bang dung djch H2SO4 loang (du), thu duoc 10,752 lit khi H , (dktc). Hieu
sua't ciJi: phan ting nhidt nhdm la
Ạ 80% B. 70% C. 60% D. 90%
Huong đn gidi
Cdchl: '
PTHH: 2 A l + 3 F e, 0 4 > 9 F e O + AI2O3 X ^ l,5x 8A1 + 3 F e 3 0 4 — ^ 9 F e + 4Al203 v' l : D y 3y/8 >9y/S ^C.^P: F e + H 2 S 0 4 ( ^ ) >FeS04 + H2 9y/8 > 9y/8 2A1 + 3H2S04(^) >Al2(S04)^+3H2 ( 0, 4- x - y ) l , 5 ( 0, 4- x - y ) Taco: ^ + 1,5.(0,4 - x - y ) = 0,48 8 => 9y + 1 2 . ( 0 , 4 - x - y ) = 3,84 => 12x + 3y = 0,96 . ^, . , . , , (l,5x + 3y/8).I00%
Hiei suat phan ung nhiet nh6m (tinh theo F e 3 0 4 ) : H = -^^ — i
0,15 ( I 2 x + 3y).100% 0,96.100% „ „ „ = P n = = = 80% 8.0,15 1,2 Dap an dung la A . Cdch 2:
Coi phan iJng xay ra: (adsbygoogle = window.adsbygoogle || []).push({});
8A1 + 3Fe,04 >9Fe + 4AI2O3
Banđu: 0,4 0,15
P/ir: X ^ 3x/8 9x/8
Con: 0,4 - x 9x/ 8
2A1 + 3H2SO4 >Al2(S04), + 3H2
0, 4- x 1,5(0,4-x)
Fe + H2SO4 > FeS04 + H2
9x/8 9x/8 T a c o : l , 5 . ( 0, 4- x ) + 9x/8 = 0,48 1 2 ( 0, 4- x ) + 9x - 3 , 8 4
3x = 0,96 =>x= 0,32
Vay hieu suat phan u-ng nhiet nh6m: H = ^ a32JI00 ^ ^^^^
0,4 0,4
7, Nung nong m gam h6n hgfp g6m A l va F e 3 0 4 trong di^u kidn kh6ng c6 kh6ng
khị Sau khi phan ling xay r a hoan toan, thu duoc h6n hop r S n X. Cho X tac
dung vdri dung djch NaOH (du) thu duoc dung djch Y, chat ran Z va 3,36 lit khi
H2 (c* dktc). Sue khi CO, (du) vao dung djch Y, thu duoc 39 gam kfét tijạ Gia tn ciia m la
Ạ 36,7. B.48,3. C. 45,6. D. 57,0. '
Hu&ng dan gidi t.'>'^<
T h e o b a i r a : j
" " 2 TTI"^ ^''^ " A I ( O H ) 3 i = ^ = 0,5 (mol). So d6 phan ling xay ra:
A l t" y A ! +NaOH đ NaAl(OH)4 + CO2 du A K O H ) , !
Fe304 Fe di( Fe (rSn)
AI2O3 H , t
A l + NaOH + 3H2O > NaAl(OH)4 + | H2T
0,1 (mol) 0,15
V i nguyen to nhom duoc bao toan nen:
" A I (ban d i u ) = " A K O H ) 3 = " A I ( d u ) + " A I (p/u iihi<l riham)
=> 0,5 = 0,1 + n A i (p/^,„hi«, „ h « m ) = > " A I (p/u,>hi«. nhom) = 0,5 " 0,1 = 0,4 (mol)
8A1 + 3Fe304 — ^ 9Fe + 4AI2O3
0,4 > 0,15 (mol)
Suy ra: n p ^ . ^ ^ ^j^, = 0,15 (mol). > , 1 (adsbygoogle = window.adsbygoogle || []).push({});
V a y m = m^ K b a n d d u ) + m p , p ^ = 0,5. 27 + 0,15 . 232 = 48,3 (gam).
Dap an dung la B. 8. D6't nong h6n hofp gom A l va 16 gam FêOj (trong d i l u kidn khong c6 khong
khi) den khi phan iJng xay ra hoan toan, thu duoc h6n hop ran X . Cho X tac dung vira dii voi V ml dung djch NaOH I M sinh ra 3,36 lit H2 ( a dktc). Gia trj c u a V l a
Ạ 300. B. 100. C. 200. D. 150. „ r
Huong đn gidi •;. A
3,36 22,4 22,4
Trong X (g6m AUOj, Fe va A l du) chi c6 A l tac dung vdri dung djch NaOH: A! + NaOH + 3H2O > N a A l ( 0 H ) 4 + USH, t
0,1 (mol) 0,15 (mol) T h e o b a i r a : n ^ ^ =0,15 (mol)
S u y r a : V = " ' ' j " " " ^ 100 ( m l ) D a p a n diing l a B .
9. H 6 n h o p X g 6 m F e , 0 4 va A l c 6 t i le m o l t u o n g iJng 1:3. Thirc hi€n phan l i n g nhiet n h 6 m X ( k h o n g c 6 k h o n g k h i ) den k h i phan l i n g xay ra hoan toan t h u dugc nhiet n h 6 m X ( k h o n g c 6 k h o n g k h i ) den k h i phan l i n g xay ra hoan toan t h u dugc h 6 n h o p g 6 m > /,
A . A l , F e , F e , 0 4 va A I 2 O 3 . B. A l j O , va Fẹ