cau 5: Cha't nao sau day khbng th^ dung d^ l^m kh6 khi hidro clorua?
3. Cac bai tap tir luyen
^^u 1: Hoa tan hoan toan 20,88 gam m6t oxit sat bang dung djch H2SO4 dac, nong thu duoc dung djch X va 3,248 lit khi SO2 (san phdm khir duy nha 't, b dktc). Co can dung djch X, thu duoc m gam mu6i sunfat khan. Gia trj ciia m la
^•54,0. B.52,2. C.48,4. D. 58,0.
(Trich de thi tuyen sinh Dai hoc, khoi B) Hu&ng ddn gidi
ra: nso2
^qudtrinh nhan electron:
S*' + 2e -
22,4 = 0,145 (mol) S^"
0,29 < 0,145 (mol) 77
+ Tru&ng hop 1: Gia sir oxit sSt 1^ FeO:
Fe-" - le > Fe*' 0,29(mol)<-0,29 > 0,29 (mol)
=> mpeo = 0,29.72 = 20,88 (gam). ' ^ 0 29
Suy ra: m , „ ô = m Fe2(S04)3 = " ^ - ^ ^ = ^^^"^^
+ Tru&ng hap 2: Gia sir oxit sSt la Fe304:
3Fé'^ - 3. 1/ 3e > 3Fế 3. 0,29 < 0,29 (mol)
3 0 29
Suy ra: mFe304 = • 232 = 67,28 (gam) ^ 20,88 (gam)
=^ loai trucmg hop nay.
Dap an dung la D.
Cau 2: Khi hoa tan hidroxit kim loai M(0H)2 bang m6t lucmg vira dii dung djch H2SO, 20% thu dugc dung djch mu6'i trung hoa c6 n6ng d6 27,21%. K i m loai M 1 ^ A . M g . B. Cu. C . Z n . D. Fe.
(Trich de thi tuyen sinh Cao dang khoi Aj Hu&ng ddn gidi
G o i x l a s o m o l M(OH)2.
Phuong trinh phan urng:
M(OH)2 + H2SO4 > MSO4 + H2O X X X
=> m (dd H2SO4) = 98x. 100/ 20 = 490x (gam).
Kh6'i luong dung djch mu6'i thu duoc:
TTIM = m (M(OH)2) + m (dd H2SO4)
=> = ( M + 34). X + 490x = M . x + 524. x (g) Theo bai ra ta c6: ( M + 96). x /[(M + 524).x] = 0,2721
( M + 96) / ( M + 524) = 0,2721 M = 64 (Cu).
Dap an diing la B.
Cau 3: De phan iimg h6't a mol kim loai M cin 1,25a mol H2SO4 va sinh ra khi (san ph^m khir duy nha't). Hoa tan hd't 19,2 gam k i m loai M vao dung di' H2S04tao ra 4,48 lit khi X (san phdm khir duy nha't, dktc). K i m loai M 1^
A . Cu B. M g C. A l D. Fe.
Hu&ng ddn gidi 4 48
Theo bai ra: Ux = —— = 0,2 (mol) 22,4
0 M - n e
fa CO cac qua trinh:
+n M a —ằ na ^ a
S(H2S04) + me- na
m
^2- , o A * n +
+ ( 6 - m )
> s
-na (Bao toan electron)
n S O f + 2 M " ^ ->M2(S04)„ (Bao toan didn tich) an/2 a
V i nguySn t6' liru huynh duoc bao toan ntn ta c6:
na an .
— + — = l,25a m 2
r^—+—=l,25=:>2n + nm = 2,5m \ m 2
Suy ra cap gia t n phii hop: n = 2, m = 8 (H2S) 4M-f
19,2
PTPlT: 4M + 5H2S04->4MS04+H2S + 4H20
M 0,2
19 2 19 2
' -=0,2.4 = 0,8 = > M = - ^ = 2 4 ( M g )
M 0,8 Vay M la k i m loai M g .
Ddp an dung la B. . Chu v: - San pha'm khiJr X (khi) c6 xhi la SO2, H.S.
- Trucmg hop M la Fe c6 the' tao ra sir khac biet ve each giai.
C a u 4: Hoa tan het m6t oxit kim loai ki^m th6 vao dung djch H2SO4 c6 ndng d6 24,4% (vira du), thi thu duoc dung djch mu6i c6 n6ng d6 27,17%. K i m loai ki^m thd la
A.Sr B. Ca C. Ba D. M g Hu&ng ddn gidi
Oxit kim loai la M O (x mol).
PTPIT: M O + H2SO4 MSO4 + H2O
X X X
Gia sir phan iitig da dung h6't 401,64 gam dung djch H2SO4 24,4%.
401,64.24,4
"•a: nH^so4 = — = l(mol) = x Suy
100.98 Kh6'i lugng dung djch sau phan ii'ng:
m,,„ = 401,64 + l . ( M + 16) (gam)
79
Theo b^i ra: 27.17 = ( M + 9 6 ) . 1 0 0 ^ 24 (Mg)
401,64+M + 16 ^ Dap in dung \k D .
Chit v: V i cac k i m loai Ca, & va Ba d6u tao kfi't tua v6i axit sunfuric ntn chi c6 M g tao duoc dung dich mu6i sunfat (chon ngay dap an D).
Dang 2: Xdc eUnh lUffng chat, thanh phdn cac chdt trong hon hap (cac chat tham gia, sdn phdm, cdn lai sau phdn leng)
1. L I tbttyet v&n dung va phuong phap giai:
* Cac phuong phap giai nhanh duoc sijr dung d^ giai cdc hki t&p dang n i y : - Phuomg phap bao toan khoi Iirong.
- Phuong phap bao toan electron.
- Phuong phap su dung cac cong thiirc giai nhanh:
+ Cjing thirc ti'nh kh6i luong mu6i sunfat thu duoc khi hoa tan hd't h6n hop kirti loai bang H,S04 loang: m,^,,, = m,i„^^ + 96n^^^
+ C6ng thiic tinh kh6'i luong muoi sunfat thu duoc khi hoa tan h€t h6n hop oxit k i m loai bang H2SO4 loang:m^„f„ = mt,^,^^ + 80nH2so4
+ C6ng thiic tinh khd'i luong mu6'i sunfat thu duoc khi cho h6n hop cdc kim loai tac dung vdi H2SO4 dac, nong giai phong khi SOj:
"1 muôi = m k i m loai +96.nso2
+ C6ng thiic tinh s6' mol H2SO4 dac, nong cSn dung d^ hoa tan m6t h6n hop k i m loai dira theo san phim khir SO, duy nha't: nn2S04 = ^^soj
+ C6ng thu-c tinh kh6'i luong mu6'i thu duoc khi hoa tan hd't h6n hop g6m Fe, FeO, FcoOj, Fe304 bang H,S04 dac, nong, du giai phong khi SOj:
400 ^ ,^ , ' m „ „ f l i = — (m h6„ hop + l o . n s o j ) - Phuong phap quy doi:
+ Quy d6i la phuong phap bid'n d6i nham dua h6n hop nhi^u chtft phdc tap thanh m6t hay hai cha't don gian, qua do lam don gian hoa bai toan ca mat hod hoc iSn toan hoc.
+ K h i dp dung phuong phap quy d6i cin tuSn thii su bao toan nguydn t6 va bao toan s6' oxi hod.
+ Co nhi^u cdch quy ddi cSn lua chon each quy d6i don gian nha't d^ giai bai tap m6t each chinh xdc va nhanh nha't.
+ Chon hudrng quy d6i thich hop.
+ Dat ^n, lap he phuong trinh dai seằ.
+ Giai h6, tinh cac dai luong theo ydu ciu cm hhi toan.
phUOng phap phuong trinh Ion rut gon:
' + Cac bu6c vie't phuong trinh ion rut gon:
^xtdc 1: Vi^t phuong trinh phan utig ma cac chat tham gia va san ph^m dudi dang phan tir (nhd can b l n g phan ung).
^\jdc 2: Cac cha't difin l i manh duoc vid't duoi dang ion; cac chUt kh6ng tan, khi, dien l i y ^ " ^^'^^ ^'^^ P^^" ^ phuong trinh ion dSy dii.
Bir6c 3: Luoc bo cac ion gi6'ng nhau or hai vd' => phuong trinh ion nit gon.
+ Khi bai todn eo sir tham gia eiia h6n hop nhi^u chat tdc dung vdd nhau nhung c6 cung phuong trinh ion nit gon, d^ giai nhanh c6 th^ siJr dung phuong trinh ion
gon di ti'nh cac ydu ciu cua. bai ra.
+ H6n hop nhilu axit, bazo tac dung voi nhau, phai sir dung phuong trinh ion nit gon:
+ O H ' > H j O d^ giki.
* M6t sdchii y: ; Cac kim loai manh (Na, K, Ca, Ba,...) khi cho vao dung dich mu6i thi cac kim
loai nay tdc dung vdi nu6c trude, ki^m sinh ra tac dung vori mu6'i.
- M6t s6' kim loai c6 hod trj thay d6i khi tdc dung v6i cac chat khdc nhau:
Fe + H2SO4 (loang) -> FeS04 + H2 3Fe + 2 O 3- > Fe304
2Fe + 6H,S04 (dac, nong) Fe2(S04)3 + 380, + 6H2O ,, Sn + 2HC1 -> SnQ^ + H2
Sn + O, SnO, 2. Cac thi du minh hoa:
Thidu 1: Cho 1,37 gam Ba vao 1 lit dung dich CUSO4 0 , 0 I M . Sau khi cdc phan ihig xay ra hoan toan, kh6'i luong ke't tiia thu duoc Id
A. 2,33 gam B. 0,98 gam C. 3,31 gam D. 1,71 gam (Trich de thi tuyen sinh DH khoi A nam 2013) Hudmg dan giai
Theo bai ra: nBa = 0,01 (mol); ncuS04 =0-01 PTHH: Ba + 2 H , 0 ^> Ba(OH)2 + H^l
0,01 - > 0,01 (mol)
Ba(OH)2 + CUSO4 - > BaS04i + Cu(0H)2i 0,01 -> 0,01 -^0,01 ^ 0,01 K^t tiia thu duoc g6m: BaS04 va Cu(OH)2
Kh6'i luong ke't tua: m = mg^so^ + "1CU(OH)2
= 0,01.233 + 0,01.98 = 3,31 (gam) E>dp dn dung Id C .
81
Thidu 2: Cho 25,5 gam h6n hc^p X g6m CuO va AI2O3 tan hoan toan trong dung dich H2SO4 loang, thu diroc dung dich chiia 57,9 gam mu6'i. PMn trftm kh6'i luong cua AljOj trong X 1^
A. 60% B.40% C. 80% , >. D. 20%
(Trich de thi tuyen sink DH khdi A nam 2013) Hu&ng ddn gidi
PTHH: CuO + H2SO4-> CUSO4 + H2O
Ta CO he:
X - > X
AI2O3 + 3H2SO4 ^ Al2(S04)3 + 3H2O y ^ y
80x + 102y = 25,5 fx = 0,255
[I60x + 342y - 57,9 ^ | y = 0.05 •ô(-) + H cr 102.0,05.100%
Vay %mA,203/x = ~ = 20%
Dap an diing D.
Thidu 3: H6n hop X g6m FeO, FcjO, va Fe304. Cho khi CO qua m gam X nung"
nong, sau m6t thcd gian thu du'oc h6n hop cha't rSn Y va hdn hop khi Z. Cho toan b6 Z vao dung dich Ca(OH)2 du, 66n phan ihig hoan toan, thu duoc 4 gam ke't tua. Mat khac, h6a tan hoan to^n Y trong dung dich H2SO4 dac, nong (du), thu duoc 1,008 lit khi SO2 (dktc, san p h ^ khu duy nha't) v^ dung dich chii-a 18 gam mu6'i. Gia trj ctia m la
A. 6,80 B.7,12 C. 13,52 D. 5,68
(Trich de thi tuyen sink DH khdi B nam 2013) Hu&ng ddn gidi
Theo bai ra: Kd't tua la CaCOj: ncacoj = 0,04 (mol) Mu6'i m Fe2(S04)3: np^2(so4)3 = '8/400 = 0,045 (mol) S6' mol khi SO,: nso2 = 1,008/22,4 = 0,045 (mol) PTHH: CO2 + Ca(OH)2 CaCOji + H2O
0,04 < - 0,04
Quy d6i h6n hop X thanh Fe (x mol) v^ O (y mol).
V i nguyen t6' sat duoc bao toan nan:
' '' = "Fe(Fe2(S04)3) = 2.0,045 = 0,09(mol) dc qua trinh nhucmg - nhan electron:
Fe - 3e - > Fe*' 0,09 - > 0,27 (mol)
-2 +2 +4
0 + 2 e - > 0 C - 2 e ^ C
y 2y 0,08 4- 0 , 0 4
+6 .
S + 2e S iii6-rt-6:j !s!H\, i;.'':M"r; . 0,09 <<- 0,045 ri ; ' - ; '
ITieo djnh luat bao toan electron ta c6:
0,27 + 0,08 = 2y + 0,09 => y = 0,13 . . | j . , .
Vay m = + mo = 0,09.56 + 0,13.16 = 7,12 (gam) Dip an diing la B.
'fhfdu4: H6n hop X g6m hai kirn loai k\im \h. m6t kim loai ki^m th6. Hoa tan hoan toan 1,788 gam X vao nude, thu duoc dung djch Y va 537,6 ml khi H2 (dktc). Dung dich Z g6m H2SO4 v^ HCl, trong do s6' mol ciia HCl ga'p hai Mn s6 mol ciia H2SO4. Trung hoa dung dich Y bang dung dich Z tao ra m gam h6n hop mu6i. Gia trj cQa m la ^
A. 4,460 B. 4,656 C. 3,792 D. 2,790
(Trich dethi tuyen sinh DH khdi B nam 2013) Hu&ng ddn gidi
Theo bai ra: n^^^ = 0,024 (mol) A + H 2 0- > A * + 0 H ^ +^ H j
B + 2H3O B-* + 2 0 H - + H , /
=>n _ = 2 n H -2 . 0 , 0 2 4 = 0,048(mol)
OH 2
H* + O H - ^ H2O "
0,048 < - 0,048 (mol)
=>n ^ =0,048 (mol)
Goi s6' mol H2SO4 la X => sd' mol H Q la 2x H
Ta c6:2x + 2x=: 0,048 =>x = 0,012 ' '
Kh6'i luong mu6'i = m i < ô i + m^ts..xi. \
=^ m^utfi = 1,788 + 0,012.96 + 2.0,012.35,5 = 3,792 (gam) E>ap an diing la C.
^hi du 5: Hoa tan hoan toan 14,6 gam h6n hop X g6m A l va Sn bang dung dich Ha (du), thu duoc 5,6 lit khi H , (6 dktc). The' tich khi O. (d dktc) ch\
phan ung hoan toan vdi 14,6 gam h6n hop X la
A. 3,92 lit. B. 1,68 lit. C. 2,80 lit. D. 4,48 lit.
^ (Trich De thi tuyen sinh DH khdi A)
Huong ddn gidi T h e o b i i r a : n u , = = 0,25 (mol)
"2 22,4
Goi X, y ISn lirot la s6' mol cua A I , Sn c6 trong h6n hgfp ban dSu.
Taco: 27x + 119y = 14,6 (1) X + HCl (dir):
A l + 3HC1 > A l Q j + 1 . S H . ! X l,5x (mol) Sn + 2HC1 > SnQ, + H ^ t
y y (mol)
Taco: l,5x + y = 0,25 (2) T\t ( l ) v a ( 2 ) t a d u o c : x = 0 , l ; y = 0 , l .
X + Oj: 4A1 + 3O2 — ^ 2AI2O3 x — X (mol)
4
Sn + O2 ——> SnO, y y (mol)
Suyra: no = - x + y = - . 0 , 1 + 0,1 = 0,175 (mol) ^ 4 4 Vay = 0,175. 22,4 = 3,92 (lit).
Dap an dung la A .
Chii v.- Sn + 2HC1 > SnQ, + H ^ t Sn + O. ) SnO,
Thi du 6: Hoa tan hoan toan 2,43 gam h6n hop g6m Mg va Zn vao m6t luong vira du dung djch H 2 S O 4 loang, sau phan iJng thu duoc 1,12 lit H2 (dktc) va dung djch X. Kh6'i luong muoi trong dung djch X la
A. 7,33 gam. B. 5,83 gam. C. 7,23 gam. D. 4,83 gam.
(Trich de thi tuyen sink DH khoi A) Hu&ng ddn gidi
TheobMra: n ^ ^ =0.05 (mol)
Mg + H2SO4 - ) . MgS04 + H2 i X -> x -> x
Zn + H2SO4 -> ZnS04 + y ^ y ^ y
Tac6: x + y = 0,05 24x+65y = 2,43 X = 0,02; y = 0,03
vay m„u.i = 0,02.120 + 0,03.161 = 7,23 (g) p^p in dung la C.
g ^ ' : Ta biet n 2 - = =^'^5 (mol)
= "iAg.z„ + m^ 2 - => m„„,, = 2,43 + 0,05 . 96 = 7,23 (g).
fiddu?'- Cho cac chat sau: FeCOj, Fe304, FeS, Fe(OH)2. Neu hoa tan ciing s6' mol m6i chS^t vao dung djch H 2 S O 4 dac, nong (du) thi chat tao ra s6' mol khi loll nha't la
A. Fe304 B. Fe(OH) 2 C. FeS D. FeCOj
(Trich de thi tuyen sinh DH khoi B) Hu&ng ddn gidi
Cha't tao s6' mol khi \6n nha't la FeS:
2FeS + 10H2SO4 - ^ F e 2 ( S 0 4) 3 + 9 S O 2 t+ I O H 2 O Dap an dung la C.
ChA y: PTHH cac phan ung xay ra:
2FeC03 + 4 H 2 S O 4 ^ Fe2 (SO4 )3 + 2 C O 2 t +SO2 t -HtHjO 2Fe304 + IOH2SO4 - > 3 F e 2 ( 8 0 4 ) 3 + S O 2 ^^^^2'^
2Fe(OH) 2 + 4 H 2 S O 4 Fe2 ( 8 0 4 ) 3 + ^ + 6 H 2 O
Thi du 8: Khi nhidt phan hoan toan 100 gam m6i chat sau: K C I O 3 (xiic tac MnOj), KMn04, K N O 3 va AgNO,. Chat tao ra luong Oj it nha't la
A. K C I O 3 B.KMn04 C. K N O 3 D.AgN03
(Trich de thi du bi dqi hoc) Hu&ng ddn gidi
2 K C I O 3 > 2 K C l + 302
100 150 50 122,5 122,5 40,8 (mol) 2KMn04 — ^ K 2 M n 0 4 + M n 0 2 + 0 2
100 50
158 158 (mol)
85
2KN03 100 101 2AgN03 100
- ^ 2 K N 0 2 +O2 50 101 (mol) .0 ->2Ag + 2N02 +O2
50 170 ' ,70 ^ " " ' ^
=> Chat tao ra lirong O2 it nhlTt la A g N O j . Dap an dung la D.
Air:
Thidu 9: Nung nong 16,8 gam h6n hop g6m Au, Ag, Cu, Fe, Zn wdi m6t lucm^
du khi O2, den khi cac phan ting xay ra hoan toan, thu duoc 23,2 gam cha't rSn X. Thi tich dung djch H Q 2 M vira dii d^ phan ling vdd chat rSn X Ik
A. 200 ml B. 400 ml C. 800 ml D. 600 ml (Trich detuyen sink Cao dang - Khoi A) , Hu&ng ddn gidi
Soddxayra: O2 — ^ 2 ' o ( o x i t ) ^ ^ ^ ->2H20 Theo djnh luat bao toan khd'i luong:
mo2 =23,2-16,8 = 6,4(g)=>no2 =0.2(mol)
=>n_2 =0,2.2 =0,4 (mol) o
=> n^+ = 0,4.2 = 0,8(mol) - nH^i
=> HCi 2M = ^ = 0,4(1) = 400(ml) Dap an dung la B.
Thidu 10: D6t chay hoan toan 17,4 gam h6n hop Mg va A l trong khi oxi (du) thu duoc 30,2 gam h6n hop oxit. Thd' ti'ch khi oxi (dktc) da tham gia phan ling la A. 4,48 lit. B. 17,92 lit. C. 8,96 lit D. 11,20 lit
(Trich de tuyen sink Cao dang khoi A) Hu&ng ddn gidi
PTHH: 2Mg + O2 —!—> 2MgO 4 A I + 3 O 2 —> . 2 A l 2 0 3
•"Mg.Al +^02 = ' " o x i t Tac6:
=>mo2 = 3 0 . 2 - 1 7 . 4 = I2,8(g)
• ra
" 0 2 = l ^ = 0,4(mol)
=0,4.22,4 = 8,96(1)
ay V02
j p an dung la C.
'p^fdull- Nhiet phan 4,385 gam h6n hop X g6m KCIO3 va KMn04, thu duoc O va m gam chat rSn g6m K2Mn04, MnO, vk KCl. Toan b6 lucmg O2 tac dung hat vdi cacbon nong do, thu duoc 0,896 lit h6n hop khi Y (dktc) c6 t i kh6'i so v6i H2 la 16. Thanh phdn % theo khd'i luong ciia KMnO^ trong X Ik A. 62,76%. B. 74,92%. C. 72,06%. D. 27,94%.
(Trich detuyen sinh Dai hoc, khdi B) Hu&ng ddn gidi
Xic dinh sd'mol O2 sinh ra: ny =0,04(mol) C + O2
X ô
Tac6 ht:
CO, 2C + O2 2CO
y / 2 < - y(mol) x(mol)
x + y = 0 , 0 4
i f ^ _ t ^ = 16 2 = 32 y = 0,03
X + y
=^ô02 = x + -^ = 0,01 + ^ = 0,025 (mol) Xac djnh thanh phSn ciia X:
KCIO, KCI + - O 2 2 ^ a - > l,5a(mol) 2KMn04 K2Mn04 + M n 0 2 +O2
b - > b / 2 ( m o l ) I,5a + b / 2 = 0,025
122,5a + 158b = 4,385 158.0,02.100%
Taco he:
V a y %m ' K M n 0 4
^ap an diing la C.
4,385
a = 0,01; b = 0,02
= 72,06% ^4
••t I F '
'^hidu 12: Cho 6,72 gam Fe vao dung dich chiia 0,3 mol H2SO4 dSc, n6ng (gia thid SO, la san ph&n khir duy nha't). Sau khi phan ting xay ra hoan toan thu duoc A. 0,03 mol Fe2(S04)., va 0,06 mol FeS04.
J5:a02 mol Fe, (804)3 va 0,08 mol FeS04. .
C . 0,12mol F e S 0 4 .
D. 0,05 mol Fe2(S04)3 va 0,02 mol F e du.
(Trich dethi tuyen sink DH - CD khoi B) Hu&ng ddn gidi
6 72 "'•"••''^^^
Theo b^i ra: np^ = -— = 0,12 (mol) 56
PTPU": 2Fe + 6H2S04(,, „,„g, > Fe2(S04)3 + SSO^ + 6H2O Banddu: 0,12 mol 0,3 mol
PhaniJng: 0,1 mol 0,3 mol 0,05 (mol) Con lai: 0,02 mol 0 0,05 (mol) F e( d u ) + Fe2(S04)3 > 3FeS04 Ban diu: 0,02 mol 0,05 (mol)
Phaniing: 0,02 mol 0,02 mol 0,06 mol Con lai: 0 0,03 mol 0,06 mol Vay sau phan ling c6 0,03 mol Fe2(S04)3 va 0,06 mol FeS04.
Dap an diing la A.
Thidu 13: Tr6n 100 ml dung dich (g6m Ba(OH)2 0,1M va NaOH 0,1M) vdi 400 ml dung dich (g6m H2SO4 0,0375 M va HCl 0,0125M), thu duoc dung dich X. Gia tri pH ciia dung djch X la
A. 7. B. 1. C. 6. D . 2 . (Trich de thi tuyen sink DH - CD khoi B) Hu&ng ddn gidi
Theo bai ra: n^^^^^^^^^ = 01.0,1 = 0,01 (mol); HN^QH = 0,1.0,1 = 0,01 (mol) Ba(0H)2 > Ba-* + 2 0 H -
0,01 0,02 (mol) NaOH > Na* + OH"
0,01 0,01 (mol) Suy ra: n^^_ = 0,02 + 0,01 = 0,03 (mol) Theo bai ra: n H2SO4 = 0,4.0,0375 = 0,015 (mol);
nHa = 0,4.0,0125 = 0,005 (mol) H2SO4 > 2H* + SO4'-
0,015 mol 0,03 (mol) H Q —> w + a-
0,005mol 0,005 (mol)
Suy ra: ^ n^^ = 0,03 + 0,005 = 0,035 (mol)
+ OH" > H2O Band^u: 0,035 (mol) 0,03 (mol)
phan ling: 0,03 (mol) 0,03 (mol)
Cdnlai: 0,005 (mol) 0 ? . H M 0,005 0,005 : ; . rtl+l = = =0,01(M) ,,,, M
0,1 + 0,4 0,5
^ p H = - l g O , 0 1 = 2 . Dip an dung la D.
'ff^fdul^' Cho 3,68 gam h6n hop g6m A l va Zn tac dung vdi m6t luong vira dii dung dich H2SO4 10%, thu duoc 2,24 1ft khi H , (or dktc). Kh6'i luong dung dich thu duoc sau phan umg la
A. 101,48 gam. B. 101,68 gam. C. 97,80 gam. D. 88,20 gam.
(Trich de thi tuyen sinh Dai hoc khoi A) Hu&ng ddn gidi
2 24
Theo bai ra: nH2 = r r — =0,1 (mol) ^ nyi^^^ -=n^^^ = 0 , 1 (mol) 0,1.98.100 ,
mddH2S04 = = 98 (gam)
So d6 phan ling: Kim loai (Al, Zn) +H2S04(1) > Mu6'i + HjT
Theo djnh luat bao toan khoi luong: >
3,68 + 98 = m, d n , u ô + 0 , 1 . 2
=>nidd.uôi = 3 , 6 8 + 9 8 - 0 , 1 . 2 = 101,48 (gam).
Dap an dung la A. >!
Thidu 15: Hoa tan hoan toan 3,22 gam h6n hop X gdm Fe, M g va Zn bang m6t liromg viira du dung dich H2SO4 loang, thu duoc 1,344 lit hidro (0 dktc) va dung dich chiia m gam mu6'i. Gia tri ciia m la
A. 10,27. B. 7,25. C. 8,98. D. 9,52.
(Trich de thi tuyen sinh Cao dang khoi A) Hu&ng ddn gidi
- T , 1 344
Theo bai ra: n„= = 0,06 (mol)
"2 22,4
So d6 phan ung: M( Fe, Mg, Zn) + H2SO4 > MSO4 + H2
0,006 0,06 ' Theo djnh luat bao toan khd'i luong, ta c6: ' '
" ^ M +"1H 2 S 0 4 = m + mH2
m = 3,22 + 0,06. 98 - 0,06. 2 = 8,98 (gam).
Dap an diing la C.
Thi du 16: Cho m6t mSu kim loai Na - Ba tac dung vdi nudfc (dir), thu duoc dung djch X va 3,36 lit H , (6 dktc). Th^ tich dung dich axit H2SO4 2 M c^n dung trung hoa dung djch X la > j
A. 30 ml. B. 75 ml. C. 150 ml. D. 60 ml.
(Trich dethi tuyen sink Cao dang khoi/{) Huong ddn gidi
Phirong tnnh phan utig:
2Na + 2 H , 0 > 2NaOH + [ Ba + 2H2O > Ba(OH)2 + H2
=:>n _ = 2 n H - ^ ^ - ^ = 0,3(mol) ' ' H* + O H > H2O
0,3 0,3
H 2 S O 4 ^ 2H" + SO4'-
j 0,15 0,3
: ^ V(dd H 2 S O 4 ) = 2 : 1 ^ ^ = 75 („,1).
Dap an dung la B.
3. C a c bai t a p tir l u y e n :
C a u 1: Tr6n 5,6 gam b6t sat vdri 2,4 gam b6t liru huynh r6i nung nong (trong dieu ki6n khdng c6 khong khQ, thu dugc h6n hop r l n M . Oio M tdc dung voi lucmg dir dung djch H Q , giai phong h6n hop khi X va con lai m6t phSn kh6ng tan G.
dd't chay hoan toan X va G cin vira dii V lit khi O2 (6 dktc). Gia tri cua V la A. 4,48. B.3,36. C. 3,08. D.2,80.'
(Trich de thi tuyen sink Cao dang khoi A)
* Hu&ng ddn gidi Ta quy doi H2S; H , ; S thanh H , va S:
"H2 = "Fe - = 0. Kmol); ns = ^ = 0,075(mol) .56 : " " 32
n
"2
=^Vo2 -22,4.0,125 = 2,80 (lit).
Dap in diing la D.
Vay no2 - Hs + - ^ = 0,075 + 0,05 = 0,125(mol)
2' Cho 13,5 gam h6n hop cac kim loai A l , Cr, Fe tac dung vdi luong dir dung
^^dich H 2 S O 4 loang, nong (trong dieu ki6n khdng c6 kh6ng khO, thu duoc dung dich X va 7,84 lit khi H , {is dktc). Co can dung djch X (trong di^u kifin khong c6 thdng khi) duoc m gam mu6'i khan. Gia trj ciia m la
^ 4 5 5 . B.42,6. C.48,8. D.47, 1 . (Trich de thi tuyen sinh Cao dang khoi A) Hu&ng ddn gidi
= 0.
22,4
Theo bai ra: n^^ = ^ = 0,35 (mol) Matkhac: m„„5i = mkMo,: + n i ^ ^ j -
=^ m„„*i = 13,5 + 0,35.96 = 47,1 (gam) '
Dap an diing la D.
Cau 3: D6't chay hoan toan m gam FeS, bang m6t lucmg O2 vira dii, thu duoc khi X . m'p thu he't X vao 1 lit dung dich chiia Ba(OH)2 0,15M va KOH 0,1M, thu duoc dung djch Y va 21,7 gam ket tiia. Cho Y vao dung dich NaOH, tha'y xuA hitn them kA tua. Gia trj ciia m la
A. 24,0. B. 12,6. C. 23,2 D. 18,0 ' (Trich de thi tuyen sinh Dai hoc khoi B) Hu&ng ddn gidi
Theo bai ra:
2 17
"Ba(OH)2 ^ ^ ' ' 5 ('"ol);nKOH =0-1 (mol); = - L - = Q,\.
Suyra: n^^_ =0,15.2 + 0,1 = 0 , 4 ( m o l ) ; n^2+ =0,15(mol)
PTHH: 4FeS2+1102 )>8S02 t+2Fe203 tô 1^ ~> r?
m/120 -> 2m/120 H i l SO2 + O H " >HS03 0,3 < - ( 0 , 4 - 0 1 )
HSO3 + O H " >S03" + H 2 O 0,1 < 0,1
SOf" + Ba^^ > B a S 0 3 i ' j | 0,1 (0,1<0,15) < - 0,1
2m 120
*^ap an dung la D.
'^aco: 0,3=^ m = 0,3.60 = 18,0(gam)
Q1
C^u 4: Cho 2,8 h't HjS (dktc) tac dung viia du vdd dung dich hoa tan m gam Q Gia tr| cua m la
A. 71,0 B.35,5 C. 14,2 D. 42,6.
Hu&ng dan gidi
2 8 ri- .. . • A...:.'
Theo bai ra: n H , ô =0,125 (mol)
"2S 22,4 . ' PTPlT: H^S + 4 C I 2 + 4 H , 0 ^ H 2 S O 4 + 8Ha
0,125 ->• 0,5 (mol)
=>nci2 =0,125.4 = 0 , 5 ( m o l ) ^ m = 0 , 5 . 7 1 =35,5(gam) Chu v.- H^S +Cl,(khi') -> 2HC1 + S i
H.S + 401. (dd) + 4 H 2 O ^ H 2 S O 4 + 8 H a
Ciku 5: Cdn bao nhidu gam H2SO4.3SO3 d^ pha vao 131 gam dung djch H 2 S O 4 40%
tao oleum c6 ham lugng SO3 la 10%?
A. 178,56 B. 594,14 C. 329,80 D. 364,50 Hu&ng dan gidi
- Tinh s6' mol H^O c6 trong 131 gam dung dich H,S04 40%:
131.40 ^ ^ ^
"iH2SO4(dd40%) = = 52,4 (gam) niH^o = 131 - 52,4 = 78,6 (gam)
78 6
= ^ n H 2 0 = - ^ = 4,37(mol)
Go'i X \k so mol H.SO4.3SO3 cSn dung:
SO3 + H 2 O >H2S04 4,37 < - 4,37
nidd =131 + (98 + 80.3)x = 131 + 338x(g) (3x-4,37).80 10 Theo bai ra: — = — = 0,1
131 + 338X 100
=>240x-349,6 = 13,1+ 33,8x
=> 206,2x = 362,7 => x = 1,7578
Vay mH2S04.3S03 =1.7578.338 = 594.14 Dap an diing la B.
Chu v: Luang SOj (trong H 2 S O 4 . 3 S O 3 ) them vao tham gia 2 thanh pMn:
+ Thu- nha't, tac dung vdri HjO (trong dung djch H 2 S O 4 40% tao H 2 S O 4 .
+ Thu- hai, tao oleum mod (10% SO3).
^. Nhifin lieu ran diing cho ten lua tang t6c cua tau vij tru con thoi la h6n hop
^^amo"' pc'^'°''^^ nh6m. Khi duoc d6't tren ZOO^C amoni peclorat n6 (san ham la N2, O,, CI,, H 2 O ) , gia six tat ca khi oxi sinh ra tac dung he't vdd b6t nhom thi kh6'i lircmg nhdm oxit sinh ra khi tau con thoi tifiu t6'n 587,5 ta'n amoni
tpecloratla
^ 680,0 ta'n B. 340,0 ta'n C. 117,5 ta'n D. 165,0 tSfn Hu&ng dan gidi
^ 587 5 10^
Xheobaira: nNH4Ci04 = ^ j . ^ ' ^ =5.10^(mol)
pTPlT: 2 N H 4 C I O 4 ^"""'^ > N 2 + C l 2 + 2 0 2 + 4 H 2 0 i f >> >
2 (mol) 2 (mol)
5.10' ^ 5.10" (mol)
4A1 + 3O2 > 2 A I 2 O 3 ^
3 (mol) -> 2 (mol) 5.10' ^ 10.1073 (mol)
vay m A,203 = = 340. lO^gam) = 340,0 (ta'n) Dap an diing la B.
Cau 7: Cho 15 gam h6n hop bot cac kim loai Fe, A l , Mg, Zn vao 100ml dung dich h6n hop H 2 S O 4 I M va HCl I M , sau khi cac phan irng x^y ra hoan toan, toan bo khi sinh ra cho qua 6'ng si} dung m gam CuO (du) nung nong. Phan ung xong, trong ong con lai 17,6 gam chat ran. Vay m bang
A. 13,56 B. 16,40 C. 15,60 D. 20,00
Hu&ng dan gidi OIH Theob^ira: n^^+ =2.0,1.1+ 0,1.1 =0,3(mol)
Chirng minh axit hd't, kim loai du:
Gia su h6n hop kim loai chi c6 Zn ( M = 65 ^ 1dm nha't): nz^ = — = 0,23(mol) 65
PTPLT: Zn + 2U* -> Zn"^ + H , 0,23 ^ 0,46 > 0,3
Axit hd't. Do do: nH = - n , = — = 0.15(mol) PTPl/: CuO(,) + H2(,) - ^ C u + H 2 O (,)
0,15 -> 0,15 (mol)
% : m = m,,„„+ mo(„20)
=> m = 17,6 + 0,15.16 = 20,0(gam) Dap dn diing la D.
Chu y: rricuo = m > m => Dap an dung la D
Cau 8: Nung nong 39,2 gam h6n horp X g6m FeO va FcjOs r6i cho lu6ng khi H, ^j.
qua, thu duac h6n hop Y chiira Fe, Fe304. Hoa tan hoan loan Y trong dung djcj,
H2SO4 dac nong, du chi thu duoc 100 gam m6t mu6'i sunfat trung hộ Than], ph^n % \6 kh6'i iuong FeO trong X la
A.36,73 B. 18,37% C. 81,64% D. 33,33%
Hu&ng ddn gidi Theobaira: np^2(S04)3
So d6 phan litig:
2FeO-> 2Fe ^ Fe2 (SO4 )3
X x/2
= 0,25(mol)
Fc^Oy •Fe2(S04)3 y
Tac6: [72x +160y-39,2
[ x / 2 + y=0,25 x = 0 , l ; y = 0,2
Th^nh phSn % khd'i lircmg FeO trong X: %mpgo/x = 0,1.72.100%
39,2 = 18,37%
Dap an diing la B.
c a u 9: X la h6n hop cua SO. va O,, c6 ti kh6'i so vdri H , 1^ 22,4. Nung n6ng X mcM then gian trong binh kin c6 chat xiic tac thich hop, thu dugc h6n hop khi c6 t;
kh6'i so v6i H2 la 26,67. Hieu sua't ciia phan utig t6ng hop SOj la A. 80% B.60% C. 48,03% D. 57,2%
Hu&ng ddn gidi
Goi X, y Mn lugt la s6' mol SO2 va O, trirdc phan ung. Ta c6:
^ ± 3 2 y ^ 2 2 , 4 . 2 = 44,8
^ X + y
=> 64x + 32y = 44,8x + 44,8y =>19,2x = 12,8y z:> x : y = 2 : 3 PTPLT:
Ban diu:
Phan ung:
Con lai:
2SO2
X
2a <-
O2
l,5x a
2SO,
( x - 2 a ) ( l , 5 x - a )
2a 2a Tac6: . ^ ( - 2 a ) ^ 3 2 ( U x - a ) . 8 0 . 2 . ^
( x - 2 a ) + ( l , 5 x - a ) + 2a
112x
' 2 , 5 x - a = 53,34
>ll2.x = 133,35x-53,34a >21,35x=53,34a=:>- = 0,4
r ay hi^u sua't phan ling (tinh theo SOj): X
y = ^ 3^ 0 0 % ^2.0,4.100% = 80%
p^p an dung la A.
Cio 1^' ^^"^ ^ " ^ " " ^ '^^^^ n6ng) thu dugc 8,96 lit khi (dktc). Nfi'u cho m gam h6n hgp do vao dung djch H2SO4
dac d^ ngu6i (du) ddn phan ung hoan toan thi thu dugc 2,24 lit khi SOj (san phim khir duy nha't, dktc). Ph^n tram s6' mol ciia Cr v^ Zn trong m gam h6n hgp ban d^u l^n lugt la
A. 70,59% va 29,41% B.75%va25%
C. 25% va 75% D. 80% va 20%
Hu&ng ddn gidi 8 96 2 24
Theo bai ra: n„ 2 = ^ = 0 , 4 ( m o l ) ; nso2 = ^ = 0 , l ( m o l ) Goi X, y Mn lugt la s6' mol, Cr, Zn trong h6n hgp
1,0 +• US
PTPU": Cr + 2HCI • CrCl2 + H2 Zn + 2HC1
y ^ Tac6: x + y = 0,4 (1)
ZnCl2+H2
Cr + H,S04(d, ngudi) kh6ng xay ra Zn + 2H2S04(d) -^ZnS04+SO2+2H2
y y Taco: y = 0,1.
T i r ( l ) = > x = 0,3
•7
% n c , = 0,3
:.100% = 75%; %nz„ = 1 0 - 7 5 = 25%
Vay
0,1 + 0,3
*^^P an dung la B.
U : Cho 200 ml dung djch X g6m Ba(OH)2 0,5M va NaAlO. (hay Na[Al(OH)4])
•'5M. Them tir tir dung djch H2SO4 0,5M vao dung djch X cho de'n khi k6't tua Jan tro lai m6t phSn, thu dugc ke't tija Y. Dem nung ke't tiia Y nay de'n khd'i '"ong kh6ng ddi thu dugc 24,32 gam cha't ran Z. Th^ tich dung djch H2SO4
da dung la
"^•1'34 lit B. 1,1 lit C. 0,55 lit D. 0,67 lit.
Hu&ng ddn gidi
Hi: bai ra: nB,(OH)2 = ^'""'^^ "NaAi02 =0-3 (mol)