Bai Tap Hoa 10Phi Kim
-
Upload
virustestcheck -
Category
Documents
-
view
223 -
download
0
Transcript of Bai Tap Hoa 10Phi Kim
-
8/8/2019 Bai Tap Hoa 10Phi Kim
1/76
Chng I: Cc halogen
A. Tm tt l thuyt:
Nhm halogen gm flo (F), clo (Cl), brom (Br) v iot (I). c im chung ca nhm l
v tr nhm VIIA trong bng tun hon, c cu hnh electron lp ngoi cng l ns2np5. Cc
halogen thiu mt electron na l bo ha lp electron ngoi cng, do chng c xu hng
nhn electron, th hin tnh oxi ha mnh. Tr flo, cc nguyn t halogen khc u c cc
obitan d trng, iu ny gip gii thch cc s oxi ha +1, +3, + 5, +7 ca cc halogen.Nguyn t in hnh, c nhiu ng dng nht ca nhm VIIA l clo.
I- Clo
a. Tnh cht vt l L cht kh mu vng lc, t tan trong nc.
b. Tnh cht ho hc: Clo l mt cht oxi ho mnh th hin cc phn ng sau:
1- Tc dng vi kim loi Kim loi mnh: 2Na + Cl2 2NaCl
Kim loi trung bnh: 2Fe + 3Cl2 2FeCl3
Kim loi yu: Cu + Cl2 CuCl2
2- Tc dng vi phi kim Cl2 + H2 as 2HCl
3- Tc dng vi nc Cl2 + H2O HCl + HClONu dung dch nc clo ngoi nh sng, HClO khng bn phn hu theo phng trnh:
HClO HCl + OS to thnh oxi nguyn t lm cho nc clo c tnh ty mu v dit trng.
4- Tc dng vi dung dch kim: Cl2 + 2KOH0t th-ng KCl + KClO + H2O
3Cl2 + 6KOH075 C> 5KCl + KClO3 + 3H2O
2Cl2 + 2Ca(OH)2 long CaCl2 + Ca(OCl)2 + 2H2O Cl2 + Ca(OH)2 huyn ph CaOCl2 +H2O
5- Tc dng vi dung dch mui ca halogen ng sau:
Cl2 + 2NaBr 2NaCl + Br2 Cl2 + 2NaI 2NaCl + I2
6- Tc dng vi hp cht: 2FeCl2 + Cl2 2FeCl3 6FeSO4 + 3Cl2 2Fe2(SO4)3 + 2FeCl3
SO2 + Cl2 + 2H2O H2SO4+ 2HCl H2S + 4Cl2 + 4H2O H2SO4+ 8HCl
c. iu ch Nguyn tc: Oxi ho 2ClCl2bng cc cht oxi ho mnh, chng hn nh:
MnO2 + 4HCl c0t MnCl2 + Cl2 + 2H2O
2KMnO4 + 16HCl 2KCl + 2MnCl2 + 5Cl2 + 8H2O
2NaCl + 2H2Opddmnx
2NaOH + Cl2 + H2
II- Axit HCl
1- Tc dng vi kim loi (ng trc H): 2Al + 6HCl 2AlCl3 +3 H2 Fe + 2HCl FeCl2 +
H2
2- Tc dng vi baz: HCl + NaOH NaCl + H2O 2HCl + Mg(OH)2 MgCl2 + H2O
3- Tc dng vi oxit baz Al2O3 + 6HCl 2AlCl3 + 3H2O CuO + 2HCl CuCl2 + H2O
4- Tc dng vi mui (to kt ta hoc cht bay hi) CaCO3 + 2HCl CaCl2 + CO2+H2O
FeS + 2HCl FeCl2 + H2S Na2SO3 + 2HCl 2NaCl + SO2+ H2O
AgNO3 + HCl AgCl + HNO3
5- iu ch H2 + Cl2 as 2HCl NaCl tinh th + H2SO4 c0t NaHSO4 + HCl
(hoc 2NaCl tinh th + H2SO4 c0t 2Na2SO4 + HCl)
3
-
8/8/2019 Bai Tap Hoa 10Phi Kim
2/76
III. Nc Giaven Cl2 + 2KOH KCl + KClO + H2O Cl2 + 2NaOH NaCl +NaClO + H2O (Dung dch KCl + KClO + H2O hoc NaCl + NaClO+ H2O -
c gi l nc Giaven)
IV. Clorua vI iu ch: Cl2 + Ca(OH)2 sa viCaOCl2 + 2H2O
(Hp cht CaOCl2 c gi l clorua vi)
B. Bi tp c li gii:
bi
1. Gy n hn hp gm ba kh trong bnh kn. Mt kh c iu ch bng cch cho axit
clohiric c d tc dng vi 21,45g Zn. Kh th hai thu c khi phn hu 25,5g natri nitrat, ph
ng trnh phn ng:
2NaNO3 0
t 2NaNO2 + O2
Kh th ba thu c do axit clohiric c, c d tc dng vi 2,61g mangan ioxit.
Tnh nng phn trm (%) ca cht trong dung dch thu c sau khi gy ra n.
2. Khi cho 20m3 khng kh c cha kh clo i qua mt ng ng mui KBr, khi lng ca
mui gim bt 178 mg. Xc nh hm lng ca kh clo (mg/m3) trong khng kh.
3. Hn hp A gm hai kim loi Mg v Zn. Dung dch B l dung dch HCl nng a mol/lt.Th nghim 1: Cho 8,9g hn hp A vo 2 lt dung dch B, kt thc phn ng thu c 4,48ltH2 (ktc).
Th nghim 2: Cho 8,9g hn hp A vo 3 lt dung dch B, kt thc phn ng cng thu c4,48lt H2 (ktc).
Tnh a v phn trm khi lng mi kim loi trong A? Cho: Mg = 24, Zn = 65, H = 1, Cl = 35,5.
4. Hn hp A gm KClO3, Ca(ClO3)2, Ca(ClO)2, CaCl2 v KCl nng 83,68 gam. Nhit phn hon
ton A ta thu c cht rn B gm CaCl2, KCl v mt th tch O2 va oxi ho SO2 thnh
SO3 iu ch 191,1 gam dung dch H2SO4 80%. Cho cht rn B tc dng vi 360 ml dung
dch K2CO3 0,5M (va ) thu c kt ta C v dung dch D. Lng KCl trong dung dch Dnhiu gp 22/3 ln lng KCl c trong A.
Tnh khi lng kt ta A. Tnh % khi lng ca KClO3 trong A.5. Ho tan 1,74g MnO2 trong 200ml axit clohiric 2M. Tnh nng (mol/l) ca HCl v
MnCl2 trong dung dch sau khi phn ng kt thc. Gi thit kh clo thot hon ton khi
dung dch v th tch ca dung dch khng bin i.
6. Khi un nng mui kali clorat, khng c xc tc, th mui ny b phn hu ng thi
theo hai phng trnh ha hc sau: 2 KClO32 KCl + 3 O2 (a) 4 KClO33KClO4 + KCl (b)
Hy tnh:
Phn trm khi lng KClO3 b phn hu theo (a)? Phn trm khi lng KClO3 b phn hu
theo (b)?
Bit rng khi phn hu hon ton 73,5g kali clorat th thu c 33,5g kali clorua.
7. Hon thnh s bin ho sau : FAKClOLCvMLOHG
CGA
FEDCSOHMnOA
BAKClO
3
t
2
pnc
422
t3
0
0
+++++
+
+++++
+
8. Cho axit clohiric, thu c khi ch ha 200g mui n cng nghip (cn cha mt lng
ng k tp cht), tc dng vi d MnO2 c mt lng kh clo phn ng vi 22,4g st
kim loi.
4
-
8/8/2019 Bai Tap Hoa 10Phi Kim
3/76
Xc nh hm lng % ca NaCl trong mui n cng nghip.
9. Cn bao nhiu gam KMnO4 v bao nhiu ml dung dch axit clohiric 1M c kh clo
tc dng vi st to nn 16,25g FeCl3?
10. Nung mA gam hn hp A gm KMnO4 v KClO3 ta thu c cht rn A1 v kh O2. Bit
KClO3 b phn hu hon ton theo phn ng : 2KClO32KCl + 3O2
(1)
cn KMnO4 b phn hu mt phn theo phn ng : 2KMnO4 K2MnO4 + MnO2 + O2 (2)
Trong A1 c 0,894 gam KCl chim 8,132% khi lng. Trn lng O2 thu c trn vi khng kh
theo t l th tch V 2o : Vkk = 1:3 trong mt bnh kn ta c hn hp kh A2.Cho vo bnh 0,528 gam cacbon ri t chy ht cacbon thu c hn hp kh A3 gm ba
kh, trong CO2 chim 22,92% th tch. a. Tnh khi lng mA. b. Tnh % khi lng ca
cc cht trong hn hp A.
Cho bit: Khng kh cha 80% N2 v 20% O2 v th tch.
11. in phn nng chy a gam mui A to bi kim loi M v halogen X ta thu c 0,96g kim
loi M catt v 0,896 lt kh (ktc) ant. Mt khc ho tan a gam mui A vo nc, sau
cho tc dng vi AgNO3 d th thu c 11,48 gam kt ta.
1. Hi X l halogen no ?
2. Trn 0,96 gam kim loi M vi 2,242 gam kim loi M c cng ho tr duy nht, ri t
ht hn hp bng oxi th thu c 4,162 gam hn hp hai oxit. ho tan hon ton hn hp
oxit ny cn 500ml dung dch H2SO4 nng C (mol/l).
a. Tnh % s mol ca cc oxit trong hn hp ca chng.
b. Tnh t l khi lng nguyn t ca M v M.
c. Tnh C (nng dung dch H2SO4). Cho: F = 19; Cl = 35,5 ; Br = 80 ; I = 127 ; Ag =
108 ; O = 16.12. A, B l cc dung dch HCl c nng khc nhau. Ly V lt dung dch A cho tc dng vi
AgNO3 d th to thnh 35,875 gam kt ta. trung ho V lt dung dch B cn dng 500 ml
dung dch NaOH 0,3 M.
1. Trn V lt dung dch A vi V lt dung dch B ta c 2 lt dung dch C (cho V + V = 2
lt). Tnh nng mol/l ca dung dch C.
2. Ly 100 ml dung dch A v 100 ml dung dch B cho tc dng ht vi Fe th lng H2thot ra t hai dung dch chnh lch nhau 0,448 lt (ktc). Tnh nng mol/l ca cc dung
dch A, B.
Cho: Cl = 35,5 ; Ag = 108.13. Cho kh HI vo mt bnh kn ri un nng n nhit xc nh xy ra phn
ng sau: 2 HI (k) H2 (k) + I2 (k) H = 52 kJ.
1. Tnh nng lng lin kt H I, bit rng nng lng lin kt H H v I I tng ng bng
435,9 kJ/mol v 151 kJ/mol.
2. Tnh phn trm s mol HI b phn ly thnh H2 v I2 khi phn ng t ti trng thi
cn bng, bit rng tc phn ng thun (vt) v nghch (vn) c tnh theo cng thc: vt= kt [HI]2 v vn = kn [H2][I2] v kn = 64 kt.
3. Nu lng HI cho vo ban u l 0,5 mol v dung tch bnh phn ng l 5 lt th khi
trng thi cn bng nng mol/l ca cc cht trong phn ng l bao nhiu?4. Nhit , p sut v cht xc tc c nh hng nh th no n s chuyn dch cn
bng ca phn ng trn ? Da vo nguyn l Lsatlie hy gii thch ?
5
-
8/8/2019 Bai Tap Hoa 10Phi Kim
4/76
14. un 8,601 gam hn hp A ca natri clorua, kali clorua v amoni clorua n khi lng
khng i. Cht rn cn li nng 7,561 gam, c ho tan trong nc thnh mt lt dung
dch. Ngi ta thy 2 ml dung dch phn ng va vi 15,11 ml dung dch bc nitrat 0,2 M.
Tnh % khi lng ca Na, K, N, H v Cl trong hn hp.
15. 1. Ngi ta c th iu ch Cl2 bng cch cho HCl c, d tc dng vi m1 gam MnO2, m2gam KMnO4, m3 gam KClO3, m4 gam K2Cr2O7.
a. Vit phng trnh phn ng xy ra.
b. lng Cl2 thu c cc trng hp u bng nhau th t l: m1 : m2 : m3 : m4 s phinh th no ?.
c. Nu m1 = m2 = m3 = m4 th trng hp no thu c nhiu Cl2 nht, trng hp no thu
c Cl2 t nht (khng cn tnh ton, s dung kt qu cu b).
2. Nn dng amoniac hay nc vi trong loi kh c Cl2 trong phng th nghim, ti
sao ?
Hng dn gii
1. Zn + 2 HCl ZnCl2 + H2
65g 1 mol
21,45g x = 0,33mol2 NaNO3
0t 2 NaNO2 + O2
2.85g 1mol
25,5g y = 0,15mol
MnO2 + 4 HCl MnCl2 + Cl2+ 2 H2O
87g 1mol
2,61g 0,03mol
Phn ng xy ra gia cc kh theo phng trnh phn ng :
2H2 + O2 2H2O
0,3mol 0,15mol 0,3mol
H2 + Cl2 2HCl
0,03mol 0,03mol 0,06mol
Nh vy, cc kh tc dng vi nhau va , phn ng to thnh 0,3mol nc hay 0,3 . 18 = 5,4
(g) nc ; 0,06mol hiro clorua, hay 0,06 . 36,5 = 2,19 (g) HCl. Kh HCl tan trong nc to thnh
axit clohiric
%85,28%100.19,24,5
19,2%C HCl =
+=
2. Cl2 + 2KBr 2KCl + Br2Sau khi phn ng, mui KBr gim khi lng l v clo thay th brom. Mt mol Br 2 c
khi lng ln hn mt mol Cl2 l: 160g 71g = 89g. S mol Cl2 phn ng l:
(mol002,089
178,0=
Lng kh clo c trong 20m3
khng kh l : 71g 0,002 = 0,0142g hay 14,2 mg
Hm lng ca kh clo trong khng kh l :
3m/mg1,720mg2,14
=
3. T d kin ca bi ton nhn thy lng HCl dng trong th nghim 1 l va hotan ht hn hp kim loi.
6
-
8/8/2019 Bai Tap Hoa 10Phi Kim
5/76
Nn s mol HCl c trong 2 lt dung dch B l:4,48
.2 0,422,4
= (mol) nng HCl trong dung
dch B l: a = 0,2 (mol/l).
Gi s mol Mg, Zn trong 8,9 gam hn hp ln lt l x v y. Ta c h phng trnh ton hc:
24x 65y 8,9
x y 0,2
+ = + =
(0,2 l tng s mol H2 thot ra)
Gii ra ta c x = 0,1 v y = 0,1. Vy %mMg = 0,1.24.100% 26,97%8,9
= v %mZn = 100%
26,97% = 73,03%.
4. Theo nh lut bo ton khi lng, tng s mol KCl trong B = x + y =
52,05,74
111.18,032.78,068,83=
=
(trong 32 v 111 l KLPT ca O2 v ca CaCl2). Mt
khc :
y
3
222.18,0yx =++
Gii h phng trnh, ta c: x = 0,4%55,58
68,83100.5,122.4,0
KClO%Vy 3 ==
5. MnO2 + 4HCl MnCl2 + Cl2+ 2H2O
1 mol 4 mol 1 mol
0,02mol 0,08 mol 0,03mol
S mol MnO2 c ha tan trong axit clohiric l :)mol(02,0
8774,1
=
)mol(4,010002002
:ldchdungtrongcHClmolS =
Nhn vo phng trnh phn ng, ta thy 1 mol MnO2 tc dng vi 4 mol HCl to nn 1 mol
MnCl2. Vy 0,02 mol MnO2 tc dng vi 0,08 mol HCl to nn 0,02 mol MnCl2.
S mol HCl cn li trong dung dch l : 0,4 mol 0,08mol = 0,32 mol
Nng ca HCl cn li trong dung dch l :)l/mol(6,1
200100032,0
=
Nng ca MnCl2 trong dung dch l :(mol/l1,0
200100002,0
=
6. Gi x l s mol KClO3, b phn hu thnh O2 y l s mol KClO3, b phn hu
thnh KClO4
2KClO3 2KCl + 3O2 (a)x x
4KClO3 3KClO4 + KCl (b)y y/ 4
Theo bi ra :2,0y
4,0x
45,05,745,33
4yx
6,05,1225,73
yx
=
=
==+
==+
7
-
8/8/2019 Bai Tap Hoa 10Phi Kim
6/76
-
8/8/2019 Bai Tap Hoa 10Phi Kim
7/76
= 1,6n. V s mol cacbon = =0,528
0,04412
, v v theo iu kin bi ton, sau khi t chy
thu c hn hp 3 kh, nn ta c 2 trng hp:
Trng hp 1: Nu oxi d, tc 1,6n > 0,044, th cacbon ch chy theo phn ng
C + O2 CO2 (3)
192,092,22100.044,0
bngngnphsaukhmolstngnylc =
Cc kh gm:oxi d + nit + CO2 (1,6 n 0,044) + 2,4n + 0,044 = 0,192
Khi lng mA = khi lng cht rn cn li + khi lng oxi thot ra.
)g(53,12048,0.32132,8100.894,0
mA =+=
Trng hp 2: Nu oxi thiu, tc 1,6 < 0,044, th cacbon chy theo 2 cch:
C + O2 CO2 (3)
2C + O2 2CO (4)
Cc kh trong hn hp c N2 (2,4n), CO2 (n') v CO (0,044 n'). Nh vy tng s mol kh = 2,4n+ 0,044. Theo cc phn ng (3,4) th s mol O2 bng:
2)'n044,0(
'nn6,1
+=
)044,0n4,2(10092,22
044,0n2,3'n +==
Gii ra c n = 0,0204
)g(647,1132.0204,0132,8100.894,0
'mVy A =+=
b. Tnh % khi lng cc cht trong A.
)g(47,1012,0.5,122n:(1)ngnphTheo3KClO
==
i vi trng hp a) : %3,887,11100KMnO%
%7,1153,12100.47,1
KClO%
4
3
==
==
i vi trng hp b)%4,876,12100KMnO%
%6,12647,11100.47,1
KClO%
4
==
==
11. 1.Phng trnh phn ng:
2 MXn pnc 2 M + n X2 (1) (n l ho tr ca kim loi M)
MXn + n AgNO3n AgX+ M(NO3)n (2)
S mol X2 =0,896
0,0422,4
= , do s mol X = 0,08.
Theo (2)11,48
0,08108 X
=+
. Suy ra X = 35,5. Vy X l clo.
2. n gin, k hiu cng thc phn t ca cc oxit l M2On v M2On:
2 M +n2
O2 M2On (3) 2 M +n2
O2 M2On (4)
V clo ho tr I, cn oxiho tr II, do 0,96 gam kim loi M ho hp vi 0,08 mol Cl hoc 0,04
mol O, tc l 0,04 . 16 = 0,64 (g) oxi.9
-
8/8/2019 Bai Tap Hoa 10Phi Kim
8/76
Vy khi lng oxi trong M2On= 4,1620,96 2,242 0,64= 0,32(g), tc l 0,02 mol O.
Gi x, y l s mol ca M2On v M2On ta c:
nx 0,04
ny 0,02
= =
x = 2y, tc M2On chim 66,7% v M2On chim 33,3%.
3. Theo khi lng cc kim loi c:2x.M 0,96
2y.M ' 2,242
= =
v v x = 2y nn:M'
4,66M
= .
4. Cc phn ng:
M2On + n H2SO4M2(SO4)n + n H2O (5)
M2On + n H2SO4M2(SO4)n + n H2O (6)
Thy oxi ho tr II v gc SO42 cng c ha tr II.
Do s mol SO42 = s mol O = s mol H2SO4 = 0,04 + 0,02 = 0,06.
Vy nng H2SO4 = 0,06 : 0,5 = 0,12 (mol/l).
12. 1. Gi n, P v m, Q l ho tr v KLNT ca kim loi X v Y. Cc phng trnh ha hc:
2X + n Cu(NO3)22X(NO3)n + n Cu (1)
2Y + m Pb(NO3)22Y(NO3)m + m Pb (2)
2X + 2n HCl XCln + n H2 (3)Y2Om + 2m HCl 2YClm + m H2O (4)
Gi a l khi lng ban u ca mi thanh kim loi v x l s mol ca mi kim loi tham gia
phn ng (1) v (2).
i vi thanh kim loi X, c: (P n
.642
).x =1.a100
(5)
i vi thanh kim loi Y, c: (m 152.a
.207 Q).x2 100
= (6)
T (5) v (6) c: 152.(2P 64n) = 207m 2Q (7)
Theo phn ng (3), c t l:
2 n3,9 1,344P 22,4
=2P = 65n (8)
Theo phn ng (4), c t l:2HCl H
1 2m 2m 2m4,25 n 2n 2.0,06
2Q 16m
= = =
+
Suy ra 2Q = 55m. (9)
T cc phng trnh (7), (8), (9) ta c n = m, ngha l X v Y cng ho tr.
2. V n = m v v s mol 2 kimloi X, Y tham gia phn ng nh nhau, nn s mol Cu(NO3)2v Pb(NO3)2 gim nhng lng nh nhau.
13. 1. Phn ng: 2 HI (k) H2 (k) + I2 (k) H = 52 kJ
Nng lng ph v lin kt cht tham gia phn ng l 2E (H I) . Nng lng to ra khi to thnh
lin kt trong H2 v trong I2 l: 435,9 + 151 = 586,9 (kJ).
Phn ng trn to nhit, ngha l: 586,9 2E (H I) = 52, suy ra E (H I) = 267,45 (kJ/mol).
2. Phn ng: 2 HI (k) H2 (k) + I2 (k)
Ban u: a mol/l 0 0
Khi CB: (a 2x) x xNn: vt = kt (a 2x)2 v vn = kn x2. Khi trng thi cn bng, c vt = vn :
kt (a 2x)2 = kn x22
t2
n
kx 1(a 2x) k 64
= =
10
-
8/8/2019 Bai Tap Hoa 10Phi Kim
9/76
v vy:x 1 a
x(a 2x) 8 10
= =
2x = 20%.a
Vy trang thi cn bng 20% HI b phn hy.
3. C a =0,5
0,1(mol / l)5
= x = 0,01 (mol/l)
trng thi cn bng, c: [HI] = 0,1 0,02 = 0,08 (mol/l)
[H2] = [I2] = 0,01 (mol/l).
4. L phn ng to nhit, nn khi tng nhit , cn bng s chuyn dch sang phathu nhit (sang pha to ra HI), v ngc li.
p sut khng nh hng n s chuyn dch cn bng v phn ng ny s mol
cc phn t kh khng thay i.
Cht xc tc nh hng nh nhau n tc phn ng thun v nghch m khng
lm chuyn dch cn bng,
14. Phng trnh phn ng: NH4Cl0t NH3+ HCl Ag+ + ClAgCl
Lng amoni clorua l: 8,601 7,561 = 1,04 (g)
t lng NaCl l x, th lng KCl l 7,561 x.
C s mol clorua trong 1 lt dung dch l:0,2.15,11
25= 0,1208 0,121 (mol)
C phng trnh:x 7,561 x
0,12158,5 74,5
= =
Gii ra c: x = 5,32 (g) = mNaCl Lng KCl = 7,561 5,32 = 2,24 (g)
T cc lng mui bit, da vo hm lng ca tng nguyn t theo cc cng thc phn t s
tnh c khi lng ca tng nguyn t trong hn hp.
15. 1. a. Cc phn ng:
MnO2 + 4 HCl MnCl2 + Cl2+ 2 H2O (1)2 KMnO4 + 16 HCl 2 KCl + 2 MnCl2 + 5 Cl2+ 8 H2O (2)
KClO3 + 6 HCl KCl + 3 Cl2+ 3 H2O (3)
K2Cr2O7 + 14 HCl 2 KCl + 2 CrCl3 + 3 Cl2+ 7 H2O (4)
b. Tnh khi lng phn t:
2MnOM = M1 = 87 ; M 4KMnOM = M2 = 158
M3KClO
M = M3 = 122,5 ; M 2 2 7K Cr OM = M4 = 294
Gi s trong cc trng hp u c 1 mol Cl2 thot ra, ta c t l:
m1 : m2 : m3 : m4 = M1 : 25
M2 : 13
M3 : 13
M4
= 87 :25
.158 :13
.122,5 :13
.294 = 87 : 63,2 : 40,83 : 97,67.
c. Nu m1 = m2 = m3 = m4 th trng hp KClO3 cho nhiu Cl2 nht v K2Cr2O7 cho t Cl2 nht.
2. Mc d Cl2 tc dng c vi dung dch Ca(OH)2 theo phn ng:
2 Cl2 + 2 Ca(OH)2CaCl2 + Ca(ClO)2 + 2 H2ONhng phn ng xy ra gia cht kh v cht lng s khng th trit bng phn ng gia
hai cht kh vi nhau. Hn na, kh amoniac phn ng vi kh clo sinh ra sn phm khng
c: N2 v NH4Cl.
Phn ng l: 3 Cl2 + 2 NH3N2 + 6 HCl v HCl + NH3NH4Cl
11
-
8/8/2019 Bai Tap Hoa 10Phi Kim
10/76
C. Bi tp t gii:
16. a. Ho tan ht 12 gam hn hp A gm Fe v kim loi R (ha tr 2 khng i) vo 200ml
dung dch HCl 3,5M thu c 6,72 lt kh ( ktc) v dung dch B.
Mt khc nu cho 3,6 gam kim loi R tan ht vo 400 ml dung dch H2SO4 1M th H2SO4cn d.
Xc nh : Kim loi R v thnh phn phn trm theo khi lng ca Fe, R trong hn hp A.
b. Cho ton b dung dch B trn tc dng vi 300ml dung dch NaOH 4M th thu c
kt ta C v dung dch D. Nung kt ta C ngoi khng kh n khi lng khng i ccht rn E.
Tnh : Khi lng cht rn E, nng mol/l ca cc cht trong dung dch D.
Bit : Cc phn ng xy ra hon ton, th tch dung dch thu c sau phn ng bng
tng th tch hai dung dch ban u, th tch cht rn khng ng k.Cho: Be = 9 ; Ca =
40 ; Fe = 56 ; Mg = 24 ; Na = 23.
p s: a. R l Mg ; %mFe = 70% ; %mMg = 30%
b. Cht rn E gm Fe2O3 v MgO c khi lng l mE = 18 gam ;
CM (NaCl) = 1,4 M ; CM (NaOH) = 1 M
17. Mt hn hp A gm ba mui BaCl2, KCl, MgCl2. Cho 54,7 gam hn hp A tc dng vi 600ml
dung dch AgNO3 2M sau khi phn ng kt thc thu c dung dch D v kt ta B. Lc ly
kt ta B, cho 22,4 gam bt st vo dung dch D, sau khi phn ng kt thc thu c cht
rn F v dung dch E. Cho F vo dung dch HCl d thu c 4,48 lt kh H2. Cho NaOH d vo
dung dch E thu c kt ta, nung kt ta trong khng kh nhit cao thu c 24 gam
cht rn.a. Tnh thnh phn % khi lng cc cht trong hn hp A ?
b. Vit phng trnh phn ng, tnh lng kt ta B, cht rn F. ( Fe + AgNO3 to ra
Fe(NO3)2)
p s: a. Sau khi cho st vo dung dch D thu c cht rn F, nn dung dch D
cn d Ag+.
%mBaCl2 = 38,03% ; %mKCl = 27,24% ; %mMgCl2 = 34,73%
b. mB = 14,8 gam ; mF = 54,4 gam
18. Cho 1,52 gam hn hp gm st v mt kim loi A thuc nhm IIA ha tan hon ton trong
dung dch HCl d thy to ra 0,672 lt kh (o ktc). Mt khc 0,95 gam kim loi A ni trn
khng kh ht 2 gam CuO nhit cao.
a. Hy xc nh kim loi A.
b. Tnh thnh phn phn trm khi lng mi kim loi trong hn hp.
Cho Mg = 24 ; Ca = 40 ; Zn = 65 ; Sr = 88 ; Ba = 137.
p s: a. A l canxi b. %mFe = 73,68% ; %mCa = 26,32%
19. kh hon ton 8 gam oxit ca mt kim loi cn dng ht 3,36 lt hiro. Ha tan ht l
ng kim loi thu c vo dung dch axit clohiric thy thot ra 2,24 lt kh hiro (cc kh
u o ktc).
Hy xc nh cng thc phn t ca oxit kim loi ni trn.
p s: Gi cng thc ca oxit cn tm l MxOy, kim loi c ho tr k khi tc dng vi axitHCl. Da vo cc d
kin ca bi ton tm c khi lng mol nguyn t ca M bng 56 M l Fe cng thc caoxit l Fe2O3
12
-
8/8/2019 Bai Tap Hoa 10Phi Kim
11/76
20. Cho 45 gam CaCO3 tc dng vi dung dch HCl d. Ton b lng kh sinh ra c hp th
trong mt cc c cha 500ml dung dch NaOH 1,5M to thnh dung dch X.
a. Tnh khi lng tng mui c trong dung dch X ?
b. Tnh th tch dung dch H2SO4 1M cn thit tc dng vi cc cht c trong dung
dch X to ra cc mui trung ho.
p s: a. Trong dung dch X c 31,8 gam Na2CO3 v 12,6 gam NaHCO3.
b. Th tch dung dch axit cn dng l 375 ml.
21. Ho tan hon ton 4,82 gam hn hp ba mui NaF, NaCl, NaBr trong nc c dung dch A.
Sc kh clo d vo dung dch A ri c cn dung dch sau phn ng thu c 3,93 gam mui
khan. Ly mt na lng mui khan ny ho tan vo nc ri cho phn ng vi dung dch AgNO 3d th thu c 4,305 gam kt ta. Vit cc phng trnh phn ng xy ra v tnh thnh phn
phn trm khi lng mi mui trong hn hp ban u.
p s: %mNaF = 8,71% ; %mNaCl = 48,55% ; %mNaBr = 42,74%
22. Cho 31,84 gam hn hp NaX, NaY (X, Y l hai halogen hai chu k lin tip) vo dung
dch AgNO3 d, thu c 57,34 gam kt ta. Tm cng thc ca NaX, NaY v tnh khi lng ca
mi mui.p s: Hai mui l NaBr v NaI ; %mNaBr = 90,58% ; %mNaI = 9,42%
23. Ho tan 3,28 gam hn hp X gm Al v Fe trong 500 ml dung dch HCl 1M c dung dch
Y. Thm 200 gam dung dch NaOH 12% vo dung dch Y, phn ng xong em thu ly kt
ta, lm kh ri em nung ngoi khng kh n khi lng khng i th c 1,6 gam cht
rn (cc phn ng u xy ra hon ton). Hy tnh thnh phn phn trm theo khi lng mi
kim loi c trong 3,28 gam hn hp X.p s: %mAl = 65,85% ; %mFe = 34,15%
24. A v B l hai kim loi thuc nhm IIA. Ho tan hon ton 15,05 gam hn hp X gm 2 mui
clorua ca A v B vo nc thu c 100gam dung dch Y. kt ta ht ion Cl
c trong 40 gamdung dch Y phi dng va 77,22 gam dung dch AgNO3, thu c 17,22 gam kt ta v dung
dch Z.
a. C cn dung dch Z th thu c bao nhiu gam mui khan?
b. Xc nh tn hai kim loi A v B. Bit t s khi lng nguyn t ca A v B l 5/3 v trong
hn hp X s mol mui clorua ca B gp i s mol mui clorua ca A.
c. Tnh nng % cc mui trong dung dch Y v dung dch Z.
p s: a. mZ = 9,2 gam b. A l Canxi ; B l Magie.
c. Trong dung dch Y: C% (CaCl2) = 5,55% ; C% (MgCl2) = 9,5%
Trong dung dch Z: C% (Ca(NO3)2) = 3,28% ; C% (Mg(NO3)2) = 5,92%
25. Nu cho 18 gam hn hp A gm Mg, Al v Al2O3 tc dng vi dung dch NaOH d th sinh ra
3,36 lt kh H2 ( ktc). Nu cng cho cng lng hn hp nh trn tc dng vi dung dch HCl d
th sinh ra 13,44 lt kh H2 ( ktc).
a. Vit cc phng trnh phn ng xy ra.
b. Tnh thnh phn phn trm khi lng ca tng cht trong hn hp ban u. Mg = 24 ;
Al = 27 ; O = 16.
p s: a. Lu : Mg khng phn ng vi dd NaOH
b. %mAl = 15% ; %mMg = 60% ; %mAl2O3 = 25%
13
-
8/8/2019 Bai Tap Hoa 10Phi Kim
12/76
26. Cho 500ml dung dch A (gm BaCl2 v MgCl2 trong nc) phn ng vi 120ml dung dch
Na2SO4 0,5M (d), th thu c 11,65 gam kt ta. em phn dung dch c cn th thu c
16,77 gam hn hp mui khan. Xc nh nng mol/lt ca cc cht trong dung dch A.
p s: CM(BaCl2) = 0,1M v CM (MgCl2) = 0,2M
27. Ha tan hon ton 4,24 gam Na2CO3 vo nc thu c dung dch A. Cho t t tng git n
ht 20 gam dung dch HCl nng 9,125% vo A v khuy mnh. Tip theo cho thm vo
dung dch cha 0,02 mol Ca(OH)2.
1. Hy cho bit nhng cht g c hnh thnh v lng cc cht .
2. Nu cho t t tng git dung dich A vo 20,00 gam dung dch HCl nng 9,125% v
khuy mnh, sau cho thm dung dch cha 0,02 mol Ca(OH)2 vo dung dch trn. Hy gii
thch hin tng xy ra v tnh khi lng cc cht to thnh sau phn ng. Gi thit cc phn ng
xy ra hon ton. Cho Ca = 40 ; O = 16 ; H = 1 ; Cl = 35,5 ; Na = 23 ; C = 12.
p s: 1. Thiu H+ nn ban u to ra HCO3; 0,02mol CaCO3, trong dung dch c:0,01mol NaOH, 0,01mol NaCl v 0,01mol Na2CO3.
2. D H+ nn kh CO2 thot ra ngay t u; 0,015mol CaCO3, trong dung dch c: 0,03mol
NaOH, 0,05mol NaCl v 0,005mol Ca(OH)2.28. 4,875 gam km tc dng va vi 75 gam dung dch HCl thu c dung dch A v khH2. Tnh nng phn trm ca dung dch HCl v dung dch A. p s: C% (dd HCl) =7,3% ; C% (dd A) 12,82%
29. Cho 33,55g hn hp AClOx v AClOy vo bnh kn c th tch 5,6 lt. Nung bnh chophn ng xy ra hon ton thu c cht rn B (ch c mui ACl) v mt kh duy nht, saukhi a v 00C th P = 3 atm.
Ho tan ht B vo nc c dung dch C. Cho dung dch C tc dng vi dung dch AgNO 3 d toc 43,05g kt ta.
Xc nh kim loi A . p s: Kim loi A l Na30. Hn hp A gm NaI, NaCl t vo ng s ri t nng. Cho mt lung hi brom i qua
ng mt thi gian c hn hp mui B, trong khi lng mui clorua nng gp 3,9 ln khi
lng mui ioua. Cho tip mt lung kh clo d qua ng n phn ng hon ton c cht
rn C. Nu thay Cl2 bng F2 d c cht rn D, khi lng D gim 2 ln so vi khi lng C gim
(i chiu vi hn hp B). Vit cc phng trnh phn ng v tnh phn trm khi lng hn hp
A. p s: %mNaI = 67,57% ; %mNaCl = 32,43%
31. Mt hn hp X gm ba mui halogenua ca natri, trong xc nh c hai mui l
NaBr, NaI. Ha tan hon ton 6,23g trong nc c dung dch A. Sc kh clo d vo dung dch A
ri c cn hon ton dung dch sau phn ng c 3,0525g mui khan B. Ly mt na lng
mui ny ha tan vo nc ri cho phn ng vi dung dch AgNO3 d th thu c 3,22875g kt
ta. Tm cng thc ca mui cn li v tnh % theo khi lng mi mui trong X.
p s:Tng s mol Cl c trong B = 2. 0,0225 = 0,045 khi lng mui NaCl c trong
B l 2,6325 gam trong B c 0,42 gam NaF (y cng l lng c trong X). Kt
hp vi cc d kin khc ca bi ton %mNaF = 6,74% ; %mNaBr = 33,07% ; %mNaI= 60,19%.
32. Hn hp A gm : Al, Mg, Fe . Nu cho 18,2 gam A tc dng ht vi dung dch NaOH d th
thu c 6,72l H2 ( ktc). Nu cho 18,2 gam A tc dung ht vi 4,6 l dung dch HCl th thuc dung dch B v 15,68 lt H2 (ktc). Phn ng xy ra hon ton. 1. Tnh
khi lng mi kim loi trong hn hp A.
2. Chia dung dch B thnh hai phn bng nhau.
14
-
8/8/2019 Bai Tap Hoa 10Phi Kim
13/76
-
8/8/2019 Bai Tap Hoa 10Phi Kim
14/76
p s: 1. a. CM (HCl) = 0,32M
b. Khi lng st khng tan sau khi cho phn ng vi dung dch HCl l m = 4,76
gam. Khi lng m tng thm 0,024 gam chnh l khi lng oxi trong oxit st t c
to thnh mFe b oxi ho =0,024 3
. .56 0,06316 4
= (gam) %mFe khng b oxi ho =
4,76 0,063.100% 98,68%
4,76
= .
2. Gi s khi lng ring ca st l d. Vin bi dng cu v ng u mi im
V = 34
. .r3
. Da vo d kin ca bi ton 0
r.100% 56,30%
r= (r0 l bn knh vin bi
ban u, r l bn knh vin bi cn li).
36. Cho vo nc d 3 gam oxit ca mt kim loi ha tr 1, ta c dung dch kim, chia dung
dch ny thnh 2 phn bng nhau :
Phn I cho tc dng vi 90 ml dung dch HCl 1M, sau phn ng dung dch lm qu tm
xanh.
Phn II cho tc dng vi V(ml) dung dch HCl 1M sau phn ng dung dch khng lm i
mu giy qu.
a. Tm cng thc phn t oxt ? b. Tnh th tch V ? p s: a. Li2O b.
V = 100ml
37. 3,28g hn hp 3 kim loi X, Y, Z c t s nguyn t X : Y : Z l 4 : 3 : 2, t s nguyn t lng
l 3 : 5 :7. Ho tan hon ton hn hp trong axit clohiric th thu c 2,0161t kh ktc v
dung dch (A).
a. Xc nh 3 kim loi , bit rng khi chng tc dng vi axit u cho mui kim loi ha tr
2.
b. Cho dung dch xt d vo dd(A), un nng trong khng kh cho phn ng xy ra hon
ton. Tnh lng kt ta thu c, bit rng ch 50% mui ca kim loi Y kt ta vi xt.
p s: a. X l Mg; Y l Ca v Z l Fe
b. m = 0,04 mol Mg(OH)2 + 0,015mol Ca(OH)2 + 0,02mol Fe(OH)3 = 5,57gam.
38. Hn hp A gm KClO3, Ca(ClO3)2, Ca(ClO)2, CaCl2, KCl nng 83,68g. Nhit phn hon
ton A ta thu c cht rn B gm CaCl2, KCl v mt th tch oxi va oxi ha SO2 thnh
SO3 iu ch 191,1g dung dch H2SO4 80%. Cho cht rn B tc dng vi 360ml dung
dch K2CO3 0,5M (va ) thu c kt ta C v dung dch D. Lng KCl trong dung dch D
nhiu gp 22/3 ln lng KCl c trong A.
a. Tnh lng kt ta C. b. Tnh % khi lng ca KClO3 trong A. C = 12 ; O = 16 ; Cl =
35,5 ; K = 39 ; Ca = 40
p s: a. mC= 0,36 x 0,5 x 100 = 18 gam b. %m (KClO3 trong A) = 58,56%
39. Trn V1 (lt) dung dch HCl (A) cha 9,125g v V2 (lt) dung dch HCl (B) cha 5,475g c
dung dch HCl (C) 0,2M. a. Tnh nng CM ca dung dch A v dung dch B ? Bit
rng hiu s ca hai nng l 0,4 mol/lt.
b. Ly 1/10 dung dch C cho tc dng vi AgNO3(d) tnh lng kt ta thu c ?
p s: a. CM (A) = 0,5M ; CM (B) = 0,1M b. Khi lng kt ta = 5,74 gam40. Ha tan 43,71g hn hp mui cacbonat, hirocacbonat v clorua ca kim loi kim vi
mt th tch dung dch HCl 10,52% (d = 1.05) ly d, thu c dung dch A v 8,96 lt kh B
(ktc). Chia A thnh hai phn bng nhau :
16
-
8/8/2019 Bai Tap Hoa 10Phi Kim
15/76
Phn 1 : Tc dng vi dung dch AgNO3 (d) c 68,88g kt ta.
Phn 2 : Dng 125ml dung dch KOH 0,8M trung ha va .
Sau phn ng, c cn thu c 29,68g hn hp mui khan.
a. Xc nh cng thc cc mui trong hn hp. b. Tnh thnh phn % hn hp.
c. Xc nh th tch dung dch HCl dng. p s: a. Na2CO3 ; NaHCO3 ; NaCl
b. %mNa2CO3 = 72,7% ; %mNaHCO3 = 19,2% ; %mNaCl = 8,1% c. VddHCl =
297,4 ml
Chng II: Oxi Lu hunh
A. Tm Tt l thuyt:
Nhm VIA gm oxi (O), lu hunh (S), selen (Se) v telu (Te). Cu hnh electron lp ngoi cng
l ns2np4, thiu hai electron na l bo ha. Oxi v lu hunh u th hin tnh oxi ha
mnh, tnh oxi ha gim dn t oxi n telu. Trong nhm VIA hai nguyn t oxi v lu hunh
c nhiu ng dng nht trong cng nghip v i sng con ngi.
I- Oxi ozon:
1- Tc dng vi kim loi oxit
2Mg + O2 2MgO3Fe + 2O2khng kh Fe3O4
2Cu + O2 2CuO
2- Tc dng vi phi kim oxit
Tc dng vi hidro:
2H2 + O2 2H2O
Tc dng vi cacbon:
C + O2 CO2
2C + O2 2CO
Tc dng vi lu hunh:
S + O2 SO23- Tc dng vi hp cht:
2H2S + 3O22SO2 + 2H2O
2CO + O22CO24- iu ch oxi trong PTN:
Nhit phn cc hp cht giu oxi v km bn nhit.
Th d: 2KClO3 20MnO
t 2KCl + 3O2
5. Ozon: Tnh oxiho mnh Tc dng vi dung dch KI:
O3 + 2KI + H2O O2 + 2KOH + I2I2 to thnh lm xanh h tinh bt, phn ng trn dng nhn bit O3.
II- Lu hunh v hp cht:
1- Tc dng vi kim loi mui sunfua
Fe + S0t FeS
Zn + S0t ZnS
i vi ring thy ngn, phn ng c th xy ra ngay nhit phng: Hg + S HgS.
V vy, ngi ta c th dng bt lu hunh x l thy ngn ri vi.
17
-
8/8/2019 Bai Tap Hoa 10Phi Kim
16/76
2- Tc dng vi phi kim:
Tc dng vi hiro: H2 + S0t H2S
Tc dng vi oxi: S + O20t SO2
Vi cc phi kim khc, phn ng xy ra kh khn hn.
III. Hirosunfua:
1- Tnh axit yu: Tc dng vi dung dch kim:
H2S + 2NaOH Na2S + 2H2O
H2S + NaOH NaHS + H2O
Tc dng vi dung dch mui (phn ng nhn bit kh H2S)
H2S + Pb(NO3)2PbS en + 2HNO3
H2S + Cu(NO3)2 CuS en + 2HNO3
2- Tnh kh mnh
Tc dng vi oxi: 2 H2S + 3 O20t 2 SO2 + 2 H2O
2 H2S + O2 oxi ho chm0t 2 S + 2 H2O
Tc dng dung dch nc Cl2:
H2S + 4Cl2 + 4H2O H2SO4 + 8HCl
3- iu ch
FeS + 2HCl FeCl2 + H2S
ZnS + H2SO4 long
ZnSO4 + H2S
IV- Lu hunh ioxit (kh sunfur)
1- Tnh oxit axit
Tc dng vi nc axit sunfur:
SO2 + H2O H2SO3
Tc dng vi dung dch baz Mui + H2O:
SO2 + 2NaOH Na2SO3 + H2O
SO2 + NaOH NaHSO3
Nu 2n
n
2SO
NaOH : To mui Na2SO3
Nu 2n
n1
2SO
NaOH
-
8/8/2019 Bai Tap Hoa 10Phi Kim
17/76
Tc dng vi dung dch nc clo, brom:
SO2 + Cl2 + 2H2O H2SO4 + 2HCl
SO2 + Br2 + 2H2O H2SO4 + 2HBr (phn ng lm mt mu dung dch brom)
3- Tnh oxi ha
Tc dng vi H2S: SO2 + 2H2S 3S + 2H2O
4- iu ch:
t qung sunfua:
2FeS2 + 11O22Fe2O3 + 8SO2
2ZnS + 3O2 2ZnO + 3SO2 Cho mui sunfit, hidrosunfit tc dng vi dung dch axit mnh:
Na2SO3 + H2SO4 Na2SO4 + SO2+ H2O
t chy lu hunh: S + O20t SO2
Cho kim loi tc dng vi dung dch H2SO4 c, nng:Cu + 2H2SO4 c
0t CuSO4 + SO2 + 2H2O
V. Lu hunh trioxit:
1- Tnh oxit axit:
- Tc dng vi nc axit sunfuric:
SO2 + H2O H2SO4
- Tc dng vi dung dch baz Mui + H2O:
SO3 + 2NaOH Na2SO4 + H2O
SO3 + NaOH NaHSO4- Tc dng vi oxit baz tan mui sunfat
Na2O + SO3Na2SO4
BaO + SO3BaSO42- iu ch:
SO2 + O2 2SO3
VI. Axit Sunfuric:
a- Dung dch H2SO4 long (th hin tnh axit mnh)
1- Tc dng vi kim loi (ng trc H) Mui + H2:
Fe + H2SO4FeSO4+ H2
2Al + 3H2SO4Al2(SO4)3 + 3H2
2- Tc dng vi baz (tan v khng tan) Mui + H2O
H2SO4 + 2NaOH Na2SO4 + 2H2O
H2SO4 + Mg(OH)2MgSO4 + 2H2O
3- Tc dng vi oxit baz Mui + H2O
Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O
CuO + H2SO4CuSO4 + H2O
4- Tc dng vi mui (to kt ta hoc cht bay hi)19
V2O
5,to
2 0 +4+6
-
8/8/2019 Bai Tap Hoa 10Phi Kim
18/76
MgCO3 + H2SO4 MgSO4 + CO2+ H2O
Na2CO3 + H2SO4 Na2SO4 + CO2+ H2O
FeS + H2SO4 FeSO4 + H2S
K2SO3 + H2SO4 K2SO4 + SO2+ H2O
BaCl2 + H2SO4 BaSO4 + 2HCl
b- Dung dch H2SO4 c:
1- Tnh axit mnh
Tc dng vi hidroxit (tan v khng tan) Mui + H2O
H2SO4 c + NaOH Na2SO4 + H2O
H2SO4 c + Mg(OH)2 MgSO4 + H2O
Tc dng vi oxit baz Mui + H2O
Al2O3 + 3H2SO4 cAl2(SO4)3 + 3H2O
CuO + H2SO4 c CuSO4 + H2O
y cc axit d bay hi ra khi mui
H2SO4 c + NaCl tinh th NaHSO4 + HCl
H2SO4 c + CaF2 tinh th CaSO4 + 2HF
H2SO4 c + NaNO3 tinh thNaHSO4 + HNO3
2- Tnh oxi ho mnh
Tc dng vi nhiu kim loi, k c mt s kim loi ng sau H nh Cu, Ag:
2Fe + 6H2SO4 c0t Fe2(SO4)3 + 3SO2 + 6H2O
Cu + 2H2SO4 c0t CuSO4 + SO2 + H2O
2Ag + 2H2SO4 c0t Ag2SO4 + SO2 + 2H2O
Mt s kim loi mnh nh Mg, Zn c th kh H2SO4 c n S hoc H2S:
3Zn + 4H2SO4 c0t 3ZnSO4 + S + 4H2O
4Zn + 5H2SO4 c0t 4ZnSO4 + H2S + 4H2O
Cc kim loi Al, Fe khng tan trong dung dch H2SO4 c ngui!
Tc dng vi phi kim:
C + 2H2SO4 c CO2 + 2SO2 + 2H2O
S + 2H2SO4 c0t
3SO2 + 2H2O Tc dng vi hp cht c tnh kh ( trng thi oxi ho thp)
2FeO + 4H2SO4c Fe2(SO4)3 + SO2 + 4H2O
2FeCO3 + 4H2SO4c Fe2(SO4)3 + SO2 + 2CO2 + 4H2O
20
2 0 +4+6
-
8/8/2019 Bai Tap Hoa 10Phi Kim
19/76
2Fe3O4 + 10H2SO4c 3Fe2(SO4)3 + SO2 + 10H2O
2FeSO4 + 2H2SO4c Fe2(SO4)3 + SO2 + 2H2O
c- iu ch H2SO4
S iu ch:
Qung prit st FeS2 hoc S SO2 SO3 H2SO4.
d. Nhn bit: Gc SO42 c nhn bit bng ion Ba2+, v to kt ta trng BaSO4 khng tantrong cc axit HNO3, HCl.
B. Bi tp c li gii:
bi
41. T 800 tn qung pirit st (FeS2) cha 25% tp cht khng chy, c th sn xut c
bao nhiu m3 dung dch H2SO4 93% (d = 1,83) ? Gi thit t l hao ht l 5%.
42. Oleum l g ? C hin tng g xy ra khi pha long oleum ? Cng thc ca oleum l
H2SO4.nSO3. Hy vit cng thc ca axit c trong oleum ng vi gi tr n = 1.
43. Lm th no nhn bit tng kh H2, H2S, CO2, CO trong hn hp ca chng bng ph
ng php ho hc.
44. Tnh lng FeS2 cn dng iu ch mt lng SO3 tan vo 100g H2SO4 91%thnh oleum cha 12,5% SO3. Gi thit cc phn ng c thc hin hon ton.
45. Cho ba kh A', B', C'. t chy 1V kh A' to ra 1V kh B' v 2V kh C'. Phn t A' khng
cha oxi. Kh C' l sn phm khi un nng lu hunh vi H2SO4 c. Kh B' l oxit trong
khi lng oxi gp 2,67 ln khi lng ca nguyn t to oxit.
Vit cc phng trnh phn ng khi :
t chy hn hp ba kh trn trong khng kh.
t chy hon ton A' v cho sn phm qua dung dch NaOH, H 2SO4 c nng, HNO3 cnng.
Cho B', C' tng kh qua dung dch Na2CO3(bit rng axit tng ng ca SO2 mnh hn axit t-ng ng ca CO2).
46. Hai bnh kn A, B u c dung tch khng i 9,96 lt cha khng kh (21% oxi v
79% nit v th tch) 27,30C v 752,4 mmHg. Cho vo c 2 bnh nhng lng nh nhau hn
hp ZnS v FeS2. Trong bnh B cn thm mt t bt lu hunh (khng d). Sau khi nung bnh
t chy ht hn hp sunfua v lu hunh, a nhit bnh v 136,50C, lc trong
bnh A p sut l pA v oxi chim 3,68% th tch, trong bnh B p sut l pB v nit chim
83,16% th tch.
1. Tnh % th tch cc kh trong bnh A.
2. Nu lng lu hunh trong bnh B thay i th % th tch cc kh trong bnh B thay
i nh th no ?
3. p sut pA v pB.
4. Tnh khi lng hn hp ZnS v FeS2 cho vo trong mi bnh.Cho: O = 16, S = 32, Zn
= 65, Fe = 56.
47. Trn m gam bt st vi p gam bt lu hunh ri nung nhit cao (khng c mt oxi)
thu c hn hp A. Ho tan hn hp A bng dung dch HCl d ta thu c 0,8 gam cht rn B,
dung dch C v kh D. Cho kh D (c t khi so vi H2 bng 9) sc rt t t qua dung dch
CuCl2(d) thy to thnh 9,6 gam kt ta en.
1. Tnh khi lng m, p.
2. Cho dung dch C tc dng vi NaOH d trong khng kh ri ly kt ta nung nhit
cao ti khi lng khng i th thu c bao nhiu gam cht rn ?
21
-
8/8/2019 Bai Tap Hoa 10Phi Kim
20/76
3. Nu ly hn hp A cho vo bnh kn dung tch khng i, cha O 2 d t0C v nung
bnh nhit cao cho ti khi cht rn trong bnh l mt oxit st duy nht, sau lm
ngui bnh ti t0C ban u th thy p sut trong bnh ch bng 95% p sut ban u.
Bit rng th tch ca cht rn l khng ng k. Tnh s mol oxi ban u trong bnh.
48. Nung m gam hn hp A gm FeS v FeS2 trong mt bnh kn cha khng kh (gm 20%
th tch oxi v 80% th tch nit) n khi phn ng xy ra hon ton, thu c cht rn B
v hn hp kh C c thnh phn th tch N2 = 84,77%; SO2 = 10,6% cn li l oxi.
Ho tan cht rn B bng dung dch H2SO4 va , dung dch thu c cho tc dng viBa(OH)2 d. Lc ly kt ta, lm kh, nung nhit cao n khi lng khng i, thu c
12,885 gam cht rn.
1. Tnh % khi lng cc cht trong A.
2. Tnh m.
3. Gi s dung tch ca bnh l 1,232 lt nhit v p sut ban u l 27,3 0C v
1 atm, sau khi nung cht A t0 cao, a bnh v nhit ban u, p sut trong bnh l p.
Tnh p sut gy ra trong bnh bi mi kh c trong hn hp C.
49. Axit H2SO4 100% hp th SO3 to ra oleum theo phng trnh:H2SO4 + nSO3 H2SO4.nSO3
Ho tan 6,76 gam oleum vo nc thnh 200 ml dung dch H2SO4 ; 10 ml dung dch ny trung
ho va ht 16 ml dung dch NaOH 0,5 M. 1. Tnh n. 2. Tnh hm lng % ca SO3 c
trong olum trn.
3. Cn bao nhiu gam olum c hm lng SO3 nh trn pha vo 100 ml H2SO4 40%
(d= 1,31 g/ml) to ra olum c hm lng SO3 l 10%.
50. Hn hp A gm KClO3, Ca(ClO3)2, Ca(ClO)2, CaCl2 v KCl nng 83,68 gam. Nhit phn
hon ton A ta thu c cht rn B gm CaCl 2, KCl v mt th tch oxi va oxiho SO 2thnh SO3 iu ch 191,1 gam dung dch H2SO4 80%. Cho cht rn B tc dng vi 360
ml dung dch K2CO3 0,5M (va ) thu c kt ta C v dung dch D. Lng KCl trong dung
dch D nhiu gp223
ln lng KCl c trong A.
a. Tnh khi lng kt ta C. b. Tnh % khi lng ca KClO3 trong A.
Hng dn gii
41. Phn ng t chy pirit st: 4 Fe + 11 O2 2 Fe2O3 + 8 SO24 mol (4.120g) 8 mol
Cc phn ng chuyn SO2
thnh H2SO4: 2 SO
2
+ O2
2 SO3 SO3 + H2O H2SO4
Lng FeS c trong 800 tn qung: 800 (800 0,25) = 600 (tn)
S kilomol FeS2 = =600.000
5.000120
(kmol)
S kilomol FeS2 thc t chuyn thnh SO2: 5000 (5000 0,05) = 4750 (kmol)
S kilomol SO2 v l s kilomol H2SO4 c to thnh: 4750 2 = 9500 (kmol)
Lng H2SO4 c to thnh : 98 9500 = 931.000 (kg)
Th tch dung dch H2SO4 93% l:931000
5471,83.0,93
= (m3)
42.a) Oleum l sn phm ca phn ng khi cho SO3 tan trong H2SO4 100%:
H2SO4 + nSO3 H2SO4 . nSO3.
Khi ho tan oleum trong nc c hin tng pht nhit mnh.
b) H2SO4 . nSO3 khi n = 1 c axit H2S2O7 .22
-
8/8/2019 Bai Tap Hoa 10Phi Kim
21/76
43.Cho hn hp kh sc t t qua dung dch Pb(NO3)2 hoc Cu(NO3)2 d: H2S + Pb(NO3)2
PbS + HNO3hn hp kh cn li cho qua nc vi trong d: CO2 + Ca(OH)2CaCO3+ H2OKh cn li (gm H2, CO v c ln hi H2O) cho qua H2SO4 c (hoc P2O5) loi ht hi
H2O. t chy hn hp kh H2 v CO, lm lnh hi nc ngng t v li cho CO2 to thnh qua
nc vi trong.
2CO + O22CO2 2H2 + O22H2O
44.Cc phn ng trong qu trnh iu ch H2SO4 t FeS2:
4FeS2 + 11O20t 2Fe2O3 + 8SO2 (1)
4 mol (4.120g) 8 mol
2SO2 + O20t 2SO3 (2)
2 mol 2 mol
SO3 + H2O H2SO4 (3)1 mol (80g) 1 mol (18g) 1 mol (98g)
Trong 100g H2SO4 91% c 91g H2SO4 v (100 91)g H2O, tc l 0,5 mol H2O. chuyn 100g
H2SO4 91% thnh H2SO4 100% cn dng 0,5 mol SO3, tc l 80 0,5 = 40g SO3 v lng H2SO4100% c to thnh l 100 + 40 = 140g
Oleum l dung dch SO3 trong axit sunfuric khan (100%). Trong oleum 12,5% c 12,5% SO3v 87,5% H2SO4. Vy lng SO3 cn dng ho tan vo 140g H2SO4 thnh oleum 12,5% l:
g205,871405,12
=
Lng SO3 cn dng ho tan vo 100g H2SO4 91% thnh oleum 12,5%:
mol75,0
80
60hay60g2040 ==+
Nhn vo cc phn ng (1) v (2), ta thy 1 mol FeS2 s to nn 2 mol SO3. Vy lng FeS2 cn
dng to nn 60g SO3 l:
(g)45275,0120
=
245. S + 2H2SO40t 3SO2+ 2H2O (C l SO2)
t B' l oxit c dng X2Om trong 16m = 2,67 . 2x x = 3m . X2Om l cht kh nnn l oxit phi kim.
Ta c: 16m = 2,67 . 2x x = 3mKhi m = 1,2 ....8 th x = 3,6...24, trong ch c gi tr m = 4, x = 12 l ph hp vi khi l
ng nguyn t ca C. Vy B' l CO2 .
Khi t chy A' : 2SOCO2O2A0t ++
1V 1V 2V
Vy A' l CS2.
a) Khi t chy hn hp:
22t
22 2SOCO3OCS0
++
CO2 + O2 khng phn ng SO2 + O2 khngphn ng
b) Sn phm t chy A' l CO2 v SO2.
23
-
8/8/2019 Bai Tap Hoa 10Phi Kim
22/76
Vi NaOH: CO2 + NaOH NaHCO3 CO2 + 2NaOH Na2CO3 +H2O
SO2 + NaOH NaHSO3 SO2 + 2NaOH Na2SO3 +H2O
Vi H2SO4 c nng: CO2 + H2SO4 khng phn ng. SO2 + H2SO4 khng phn
ng.
Vi HNO3 c nng: CO2 + HNO3 khng phn ng. SO2 + 2HNO3 0t H2SO4
+ 2NO2
c) Khi cho CO2, SO2 qua dung dch Na2CO3 : CO2 + H2O + Na2CO3 2NaHCO3. SO2 + Na2CO3Na2CO3 + CO2
46.
Cc phn ng: 2ZnS + 3 O20t 2ZnO + 2SO2 (1)
4FeS2 + 11 O20t 2Fe2O3 + 8SO2 (2) S + O2
0t SO2 (3)
1. Theo (3) c 1mol O2 (k) mt i li sinh ra 1mol SO2 (k), ngha l tng s mol kh trong hai
bnh nh nhau, do %V ca N2 trong 2 bnh nh nhau = 83,16% v % SO2 = 100% 83,16% 3,68% = 13,16%.
2. Do tng s mol kh khng i, nn % N2 lun bng 83,16%, cn tu thuc vo lng S m
%O2 hoc bng trong bnh A (nu khng c S) hoc ht (nu nhiu S), tc 0% %O2
3,68%, cn % SO2 th hoc bng trong bnh A (nu khng c S) hoc thm SO2 do t S;
tc l: 13,16% % SO2 13,16 + 3,68 = 16,84%.
3. Th tch v nhit nh nhau, tng s mol bng nhau, nn PA = PB.
Gi tng s mol khng kh ban u l n0, c:
0
760.n .22,4 752,4.9,96
273 273 27,3= + n0 = 0,4
trong c: 0,4 . 21% = 0,084 (mol) O2 v 0,4 . 79% = 0,316 (mol) N2. V %V t l vi s
mol kh nn ta c:
2
2
s mol SO x 13,16
s mol N 0,316 83,16= = x = 0,05 2
2
s mol O y 3,68s mol N 0,316 83,16
= = y = 0,014
Tng s mol kh trong A = 0,316 + 0,014 + 0,05 = 0,38 Vy: AP .9,961.0,38.22,4
273 273 136,5=
+ PA
= 1,282 (atm) = PB
4. S mol O2 tham gia phn ng (1) v (2) = 0,084 0,014 = 0,07.
Gi s mol ZnS v FeS2 ln lt l: a v b, ta c:
S mol SO2 = a + 2b = 0,05 v s mol O2 phn ng =3 11
.a .b 0,072 4
+ = .
Gii ra c: a = 0,01 v b = 0,02 Vy khi lng hn hp = 97 . 0,01 + 120 . 0,02 = 3,37 (g).
47.
1. Cc phn ng: Fe + S0t FeS (1) FeS + 2HCl FeCl2 + H2S (2)
Fe + 2HCl FeCl2 + H2 (3) S + HCl khng phn ngH2S + CuCl2 CuS+ 2HCl (4) FeCl2 + 2 NaOH Fe(OH)2+ 2NaCl
(5)
24
-
8/8/2019 Bai Tap Hoa 10Phi Kim
23/76
4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3 (6) 2Fe(OH)30t Fe2O3 + 3H2O
(7)
Theo (1), (2) v (4) nFeS = 2H Sn = nCuS =9,6
0,196
= (mol)
i vi kh D: gi x l %V ca H2S, ta c:
DM = 9 . 2 = 18 = 34x + 2(1x) x = 0,5 = 50%. Do 2Hn = 2H Sn = 0,1 = nFe cn li.
Vy tng khi lng Fe ban u l m = (0,1 + 0,1) . 56 = 11,2 (g) v khi lng S ban u p =
0,8 + 0,1 . 32 = 4 (g).
2. Theo cc phn ng t (1) n (7) ta c:2 3Fe O
1n
2= nFe ban u =
0,2
2= 0,1.
Khi lng Fe2O3 = 0,1 . 160 = 16 (g).
3. Theo bi ra c cc phn ng: S + O20t SO2 (8) 2 Fe +
3
2O2
0t Fe2O3 (9)
2 FeS +7
2O2
0t Fe2O3 + 2 SO2 (10)
Tng s mol O2 tham gia phn ng (8), (9), (10) l:
2O S Fe
3 4 3 1,1n n .n .0,2
4 32 4 4= + = + =
S mol SO2 to ra: nS + nFeS = 0,025 + 0,1 = 0,125 (mol)
Gi s mol oxi cn li l n2O , c t l s mol trc v sau phn ng:
2
2
O
O
1,1n' 1004
n' 95
+= n 2O = 2,725.
Vy s mol O2 ban u = 2,725 + 0,275 = 3 (mol).
48. t x, y ln lt l s mol ca FeS v FeS2 trong A. t a l s mol ca kh trong bnh tr
c khi nung.
Khi nung, cc phn ng: 2FeS + 27
O2
0t Fe2O3 + 2SO2 4FeS2 + 11 O2
0t 2Fe2O3 +
8SO2
Trc khi nung, s mol kh l: N2 = 0,8a v O2 = 0,2a
Sau khi nung, s mol kh l: N2 = 0,8a v SO2 = (x + 2y)
Vy s mol O2 d l: 0,2a 1,75x 2,75y tng s mol kh = a 0,75.(x+y).
C: %2N
0,8a 84,77V
a 0,75.(x y) 100= =
+a = 13,33. (x+y)
%2SO
x 2y 10,6V
a 0,75.(x y) 100+
= = +
a = 10,184x + 19,618y
T hai gi tr trn ca a suy ra:x 2
y 1=
1. T l v khi lng:
V t l v s mol x 2y 1
= , nn t l v khi lng s l:
%FeS =2.88
.100 59,46%(2.88 1.120)
=+
25
-
8/8/2019 Bai Tap Hoa 10Phi Kim
24/76
%FeS2 = 100% 59,46% = 40,54%
2. Cht rn B l Fe2O3 cha 0,5.(x + y) mol.
Cc phn ng xy ra: Fe2O3 + 3 H2SO4 Fe2(SO4)3 + 3 H2O
Fe2(SO4)3 + 3 Ba(OH)2 2 Fe(OH)3+ 3 BaSO4Khi nung kt ta:
BaSO40t khng thay i
2 Fe(OH)30t Fe2O3 + 3 H2O
T cc phn ng trn c phng trnh: 233 . 1,5 .(x + y) + 160 .0,5 .(x + y) = 12,885
Hay: x + y = 0,03
Mt khc c:x 2y 1
= , nn x = 0,02 v y = 0,01
Vy m = 88 . 0,02 + 120 . 0,01 = 2,96 (g).
3. S mol kh trc phn ng (a mol):
0V .11,232.1273 27,3 273
=+
V0 = 1,12 (lt) a =1,1222,4
= 0,05 (mol)
Th tch hn hp C tnh theo iu kin tiu chun = 22,4 .[0,05 0,75 .(x + y)] = 0,616(V x + y = 0,03)
p sut gy ra bi hn hp C:1,232.p 0,616.1
273 27,3 273=
+p = 0,55 (atm).
V t l th tch ca cc kh trong hn hp cng chnh bng t l v p sut ca chng,
nn:
pN2 = 84,77% . 0,55 = 0,466 (atm)
pSO2 = 10,6% . 0,55 = 0,058 (atm)
pO2 = 4,63% . 0,55 = 0,025 (atm)
49. Khi ho tan oleum vo nc, c phn ng: H2SO4.nSO3 + n H2O (n + 1) H2SO4
s mol H2SO4 to thnh l: x =6,76.(n 1)
98 80n
+
+
(1)
T phn ng trung ho: H2SO4 + 2 NaOH Na2SO4 + 2H2O
suy ra s mol axit H2SO4 c trong 10 ml dung dch l:0,5.0,016
0,0042
= (mol), vy s mol
H2SO4 c to ra t oleum l: x =0,004.200
0,0810
= (mol) (2)
T (1) v (2) suy ra n = 3 cng thc ca oleum: H2SO4.3SO32. Hm lng SO3 t do trong oleum l:
%SO3 =240.100
71%338
=
3. t y l s gam H2SO4.3H2O cn ho tan: Trong 131 g H2SO4 40% c 52,4 g H2SO4 v
78,6 g H2O.
Vy khi ho tan: SO3 + H2O H2SO4y1 78,6 y2
y1 = 78,6.80 349,318 = (g) SO3
y2 =78,6.98
427,918
= (g) H2SO4
26
-
8/8/2019 Bai Tap Hoa 10Phi Kim
25/76
V trong oleum c 10% l SO3, nn:
3
2 4
Khi l- ng SO 10Khi l- ng H SO 90
=
khi lng SO3 d =240y
349,3 (0,71y 349,3)338
= (g)
V khi lng H2SO4 = 427,9 + 52,4 +98y
(480,3 0,29y)338
= + (g)
Da vo t l v khi lng gia SO3 v H2SO4 trn suy ra y = 594,1 (gam).50.
Cc phn ng: 2KClO30t 2KCl + 3 O2 (1) Ca(ClO3)2
0t CaCl2 + 3O2 (2)
Ca(ClO)20t CaCl2 + O2 (3) (CaCl2 v KCl khng b nhit phn iu
kin ny)
2SO2 + O22SO3 (4) SO3 + H2O H2SO4 (5) CaCl2 + K2CO3 CaCO3+ 2KCl (6)
Vy s mol cc cht l:
3 2 4SO H SO
191,1.80n n 1,56
100.98= = =(mol)
2O
1n .1,56 0,78
2= =(mol)
2 2 3 3CaCl K CO CaCOn n n 0,36.0,5 0,18= = = = (mol)
a. Khi lng kt ta C l: 0,18 . 100 = 18 (g)
b. t x v y ln lt l s mol ca KClO3 v KCl c trong A, theo nh lut bo ton khi lng,
c:
Tng s mol KCl trong B = x + y =83,68 0,78.32 0,18.111
0,5274,5
= .
Mt khc: x + y + 0,18 . 2 =22
.y
3Gii h phng trnh c x = 0,4.
Vy %KClO3 =0,4.122,5.100
58,55%83,68
= C. Bi tp t gii :
51. Nung mA gam hn hp KClO3 v KMnO4 thu c cht B v kh O2(lc KClO3 b phn
hy hon ton cn KMnO4 b phn hu khng hon ton). Trong B c 0,894g KCl chim
8,312% v khi lng. Trn lng oxi thu c trn vi khng kh (ch cha O2 v N2) theo t l
th tch 1: 3 trong mt bnh kn thu c hn hp kh C. Cho vo bnh 0,528g cacbon ri
t chy ht cacbon thu c hn hp kh D gm 3 kh, trong CO2 chim 22,92% v th
tch. a. Tnh mA ?
b. Tnh % khi lng mi cht trong hn hp A ?
Cho bit: Khng kh cha 80% nit v 20% oxi v th tch.
p s: a. Trng hp 1: Nu d oxi: Ba kh l O2, N2 v CO2mA = 12,53 gam.
Trng hp 2: Nu thiu oxi: Ba kh l N2, CO2 v CO mA = 11,647 gam.
b. %m (KClO3) = 12,6%; %m (KMnO4) = 87,4%
52. Hy xc nh nng % ca dung dch H2SO4 . Bit rng khi ly mt lng dung dch
cho tc dng vi natri d th lng kh hiro thot ra bng 5% khi lng dung dch H 2SO4 .p
s: C% (H2SO4) 67,38%53. Ha tan hon ton hn hp A gm Mg, Cu vo mt lng va dung dch H2SO4 70%
(c, nng), thu c 1,12 lt kh SO2 (o iu kin tiu chun) v dung dch B. Cho
dung dch B tc dng vi NaOH d, c kt ta C; nung C n khi lng khng i, c hn
27
-
8/8/2019 Bai Tap Hoa 10Phi Kim
26/76
hp cht rn E. Cho E tc dng vi lng d H2(nung nng) thu c 2,72g hn hp cht rn F.
a. Tnh s gam Mg, Cu c trong hn hp A.
b. Cho thm 6,8g nc vo dung dch B c dung dch B'. Tnh nng % cc cht trong
B' (xem nh lng nc bay hi khng ng k). Cho: Cu = 64, Mg = 24, H = 1, O = 16, p s: a.
mMg = 0,48 gam; mCu = 1,92 gam.
b. Dung dch B c: 6 + 6,8 = 12,8 (gam) H 2O, m(MgSO4) = 0,02 x 120 = 2,4
(gam), m(CuSO4) = 0,03 x 160 = 4,8 (gam) C% (MgSO4) = 12% v C% (CuSO4) =
24%.
54. Na2SO4 c dng trong sn xut giy, thu tinh, cht ty ra. Trong cng nghip n c
sn xut bng cch un H2SO4 vi NaCl. Ngi ta dng mt lng H2SO4 khng d nng 75%
un vi NaCl. Sau phn ng thu c hn hp rn cha 91,48% Na2SO4 ; 4,79% NaHSO4 ;
1,98% NaCl ; 1,35% H2O v 0,40% HCl.
1. Vit phn ng ha hc xy ra.
2. Tnh t l % NaCl chuyn ha thnh Na2SO4.
3. Tnh khi lng hn hp rn thu c nu dng mt tn NaCl.4. Khi lng kh v hi thot ra khi sn xut c 1 tn hn hp rn.
p s: 2. %m ca NaCl chuyn ho thnh Na2SO4 = 94,58%.
3. m hn hp rn = 1,343 tn.
4. mHCl = 0,2457 tn; mH2O = 0,2098 tn.
55. Chia 59,2 gam hn hp gm kim loi M, oxit v mui sunfat ca cng kim loi M (c ha tr
2 khng i) thnh hai phn bng nhau :
Phn 1 ha tan ht trong dung dch H2SO4 long thu c dung dch A v kh B. Lng
kh B ny tc dng va vi 32 gam CuO. Cho tip dung dch KOH (d) vo dung dch A, khi
phn ng kt thc lc ly kt ta, nung n khi lng khng i c 28 gam cht rn.
Phn 2 cho tc dng vi 500ml dung dch CuSO4 1,2M, sau khi phn ng kt thc lc b
cht rn, em phn dung dch c cn, lm kh thu c 92 gam cht rn.
a. Vit cc phng trnh phn ng xy ra, xc nh M ?
b. Tnh thnh phn phn trm theo khi lng ca cc cht trong hn hp ban u ? Bit
cc phn ng xy ra hon ton.
p s: a. MM = 24 M l Mg.
b. %mMg = 32,43% ; %mMgO = 27,03% ; %mMgSO4 = 40,54%
56. t chy trong oxi 8,4 gam hn hp A gm FeS 2 v Cu2S thu c kh X v cht rn B gm
Fe2O3 v Cu2O. Lng kh X ny lm mt mu va ht dung dch cha 14,4gam brom. Cho
cht rn B tc dng vi 600ml dung dch H 2SO4 0,15M n khi phn ng kt thc thu c m
gam cht rn v dung dch C. Pha long dung dch C bng nc c 3 lt dung dch D.
Bit rng khi ha tan Cu2O vo H2SO4 long thu c CuSO4, Cu v H2O.
1. Tnh thnh phn % khi lng mi cht trong hn hp A ? 2. Tnh m ? 3. Tnh pH ca
dung dch D ?
p s: 1. %mFeS2 = 42,86% ; %mCu2S = 57,14%
2. Trong B c 0,015mol Fe2O3 v 0,03mol Cu2O lng axit H2SO4 d sau khi phnng vi B = 0,09 (0,045 + 0,03) = 0,015 (mol). Cht rn C l Cu vi m = 1,92
gam.
28
-
8/8/2019 Bai Tap Hoa 10Phi Kim
27/76
3. Dung dch D c pH = 2.
57. Cho 3,0 gam hn hp A (gm Al v Mg) ha tan hon ton bng H 2SO4 long, gii phng
3,36 lt kh H2 ktc v dung dch B. Cho B vo NaOH d, ly kt ta sch nung ti khi lng
khng i c m gam cht rn. Cho 1,5 gam A tc dng vi dung dch CuSO4 d, cui cng
thu cht rn to thnh cho tc dng vi HNO3 c gii phng V lt kh mu nu ktc.
1. Vit cc phng trnh phn ng xy ra.
2. Tnh m v V. Tnh thnh phn % (theo khi lng) mi cht trong A.
p s: 2. m = 2 gam; V = 3,36 lt ; %mAl = 60% v %mMg = 40%
58. Cho 1,68 gam hp kim AgCu tc dng vi dung dch H2SO4 c, nng. Kh thu c tc
dng vi nc clo d, phn ng xy ra theo phng trnh; SO2 + Cl2 + 2 H2O =
2 HCl + H2SO4
Dung dch thu c sau khi phn ng vi clo cho tc dng ht vi dung dch BaCl2 0,15M
thu c 2,796 gam kt ta. a. Tnh th tch dung dch BaCl2 cn dng. b. Tnh thnh
phn %m ca hp kim.
p s: a. Vdd (BaCl2) = 0,8 ltb. %mAg = 77% ; %mCu = 23%
59. X l hn hp hai kim loi Mg v Zn. Y l dung dch H2SO4 long cha r nng .
Th nghim 1 : Cho 24,3 gam X vo 2 lt Y, sinh ra 8,96 lt kh H2.
Th nghim 2 : Cho 24,3 gam X vo 3 lt Y, sinh ra 11,2 lt kh H2.
Bit rng: trong th nghim 1, X cha tan ht ; trong th nghim 2, X tan ht.
Tnh nng mol/l ca dung dch Y v khi lng mi kim loi trong X.(Th tch kh c o
ktc)
p s: CM (dd Y) = 0,1M ; mMg = 0,2 x 24 = 4,8(gam) v mZn = 0,3 x 65 = 19,5(gam)
60. T khi ca hn hp X gm CO2 v SO2 so vi kh nit bng 2. Cho 0,112 lit ( iu kin
tiu chun) ca X li chm qua 500ml dung dch Ba(OH) 2. Sau th nghim phi dng 25,00ml
HCl 0,200 M trung ho lng Ba(OH)2 tha. a. Tnh % s mol ca mi kh trong hn
hp X.
b. Tnh nng dung dch Ba(OH)2 trc th nghim.
c. Hy tm cch nhn bit mi kh c trong hn hp X, vit cc phng trnh phn ng.
p s: a. %nCO2 = 40% ; %nSO2 = 60%
b. CM dd Ba(OH)2 = 0,015M.c. Sc hn hp kh qua nc Brom d, SO2 s lm mt mu Brom. Kh cn li sc qua
nc vi trong, CO2 lm vn c.
61. Ho tan 88,2 gam hn hp A gm Cu, Al, FeCO3 trong 250 ml dung dch H2SO4 98% (d =
1,84 g/ml) khi un nng c dung dch B v hn hp kh. Cho hn hp kh ny i qua dung
dch brom (d) sau phn ng c dung dch C. Kh thot ra khi bnh nc brom cho hp th
hon ton vo bnh ng dung dch Ba(OH)2 c 39,4 gam kt ta ; lc tch kt ta ri
thm dung dch NaOH d vo li thu c 19,7 gam kt ta. Cho dung dch BaCl2 d vo dung
dch C c 349,5 gam kt ta.
1. Tnh khi lng tng cht c trong hn hp A.
2. Tnh th tch dung dch NaOH 2M cn cho vo dung dch B tch ring ion Al3+ ra
khi cc ion kim loi khc.
29
-
8/8/2019 Bai Tap Hoa 10Phi Kim
28/76
p s: 1. mCu = 25,6 gam ; mAl = 16,2 gam ; mFeCO3 = 46,4 gam.
2. VddNaOH = 2,05 lt
62. Mt nguyn t phi kim R to vi oxi hai loi oxit R aOx v RbOy vi a 1 v b 2. T s
phn t khi ca hai oxit l 1,25 v t s %m ca oxi trong hai oxt l 1,2. Gi s x > y.
a. Xc nh nguyn t R.
b. Ha tan mt lng oxt RaOx vo H2O, c dung dch D. Cho D tc dng va vi
1,76g oxt M2Oz ca kim loi M, thu c 1 lt dung dch E c nng mol/l ca cht tan l
0,011M. Xc nh nguyn t M ?p s: a. MR = 32 A l S b. MM = 56 M l Fe
63. Trong bnh kn dung tch khng i cha 35,2x(g) oxi v 160x(g) kh SO2, 136,5C cxc tc V2O5. un nng bnh mt thi gian, a v nhit ban u, p sut bnh l P'.
Bit p sut bnh ban u l 4,5 atm v hiu sut phn ng l H%.
a. Lp biu thc tnh p sut sau phn ng P' v t khi hi d ca hn hp kh sau phn
ng so vi khng kh theo H (coi M kk= 28,8).
b. Tm khong xc nh P', d ?
c. Tnh dung tch bnh trong trng hp x = 0,25 ?
p s: a. Bnh kn, nhit khng i, nn0 0
P' n' 3,6x 1,1x.H%P n 3,6x
= = Khi P0 = 4,5
atm, th P = 4,5 1,375. H% (atm). T khi dhh sau P/kk =195,2
(3,6 1,1.H%).28,8.
b. Khong xc nh: 3,125 P 4,5 ; 1,88 d 2,71.
c. T d kin ca P0V
26,88x
= khi x = 0,25 th V = 6,72 lt.
64. Cho hn hp A gm Al, Zn v S di dng bt mn. Sau khi nung 33,02 gam hn hp A(khng c khng kh) mt thi gian nhn c hn hp B. Nu thm 8,296 gam bt Zn vo B
th hm lng n cht Zn trong hn hp ny bng2
1hm lng Zn trong A.
Ly2
1lng hn hp B ha tan trong dung dch H2SO4 long d, sau khi phn ng kt
thc, thu c 0,48 gam cht rn nguyn cht.
Ly2
1lng hn hp B, thm mt th tch khng kh thch hp. Sau khi t chy hon
ton c hn hp kh C. Trong hn hp kh C, nit chim 85,5% th tch v cht rn D. Cho
hn hp kh C qua dung dch NaOH m c, dng d th th tch gim i 5,04 lt ( iu
kin tiu chun).
1. Vit cc phng trnh phn ng.
2.Tnh th tch khng kh dng.
3.Tnh thnh phn % theo khi lng cc cht trong hn hp B.
65. A l dung dch H2SO4, B l dung dch NaOH. Trn 0,3 lt B vi 0,2 lt A ta c 0,5 lt dung
dch C. Ly 20 ml dung dch C, thm mt t qu tm vo thy c mu xanh. Sau thm
t t dung dch HCl 0,05M ti khi qu tm i thnh mu tm thy ht 40ml axit.
Trn 0,2 lt B vi 0,3 lt A ta c 0,5 lt D. Ly 20 ml dung dch D, thm 1 t qu tm vo
thy c mu . Sau thm t t dung dch NaOH 0,1M ti khi qu tm i thnh mu
tm thy ht 80ml xt.
30
-
8/8/2019 Bai Tap Hoa 10Phi Kim
29/76
1. Tnh nng mol ca cc dung dch A v B.
2.Trn VB lt NaOH vo VA lt H2SO4 trn ta thu c dung dch E. Ly V mol dung dch
cho tc dng vi 100ml dung dch BaCl2 0,15M c kt ta F. Mt khc, ly V ml dung dch E
cho tc dng vi 100ml dung dch AlCl3 1M c kt ta G. Nung E hoc G nhit cao
n khi lng khng i th u thu c 3,262 gam cht rn. Tnh t l VB : VA.
31
-
8/8/2019 Bai Tap Hoa 10Phi Kim
30/76
Chng III: Nit Photpho
A. Tm Tt l thuyt:
Nit v photpho thuc nhm VA ca bng tun hon. Cu hnh electron lp
ngoi cng ca chng l ns2np3. Mc d nit c tnh cht phi kim mnh hn
photpho, tuy nhin, n cht photpho hot ng ha hc vi oxi mnh hn nit.
Tnh cht km hot ng ha hc ca nit c l gii bi lin kt ba bn vng
gia hai nguyn t nit: N N . Nit chim khong 78% th tch khng kh, khngc, nhng khng duy tr s sng. Nguyn t N c vai tr rt quan trng trong cuc
sng, l thnh phn ha hc khng th thiu c ca cc cht protit.
I- Nit:1- Tc dng vi hidro:
N2 + 3H2 2NH32- Tc dng vi oxi:
N2 + O2 2NO
3- iu ch: Trong phng th nghim: NH4NO20t N2 + 2H2O
Trong cng nghip: Chng ct phn on khng kh lng thu c N2 v O2.
II- Amoniac: a- Kh amoniac
1- Tnh baz: NH3 + HCl NH4Cl 2 NH3 + H2SO4 (NH4)2SO4
2- Tnh kh: Tc dng vi oxi: 4NH3 + 3O20t 2N2 + 6H2O 4NH3 + 5O2
0850 CPt
4NO + 6H2O
Tc dng vi clo: 2NH3 + 3Cl2N2 + 6HCl
Kh mt s oxit kim loi: 3CuO + 2NH3 3Cu + N2 + 3H2O
b- Dung dch amoniac
1- Tc dng ca NH3 vi H2O: NH3 + H2O NH4+ + OH
2- Tnh cht ca dung dch NH3: Tnh baz: tc dng vi axit to ra mui amoni NH3 + H+
NH4+
Lm i mu ch th: qu tm xanh ; phenolphtalein hng.
Tc dng vi dung dch mui hiroxit kt ta, th d: AlCl3 + 3NH3 + 3H2O Al(OH)3+ 3NH4Cl
Hay: Al3+ + 3NH3 + 3H2O Al(OH)3+ 3NH4+
Phn ng cng xy ra tng t vi cc dung dch mui FeCl3 ; FeSO4
Kh nng to phc (Th hin tnh baz theo Liuyt): Amoniac c kh nng to phc vi nhiu
cation kim loi, c bit cation ca cc nguyn t nhm ph. Chng hn:Cu(OH)2+ 4 NH3 (dd) [Cu(NH3)4]2+ (dd) + 2OH (dd)
Hoc: AgCl+ 2 NH3 (dd) [Ag(NH3)2]+ (dd) + Cl (dd)
c- iu ch amoniac: * Trong phng th nghim: NH4+ + OH Kim(rn) NH3+H2O
Hay 2NH4Cl (r) + CaO0t 2NH3 + CaCl2
* Trong cng nghip: - Nguyn liu: N2 c iu ch bng phng php chng ct phn
on khng kh lng.
H2 c iu ch bng cch nhit phn metan khng c khng kh: CH40t C + 2H2
- Phn ng tng hp: N2 + 3H20450 500( C)
300 1000(atm),Fe
2NH3
(Xc tc Fe c hot ho bi hn hp oxit Al2O3 v K2O)
32
xt, to
3000oC
-
8/8/2019 Bai Tap Hoa 10Phi Kim
31/76
III- Mui amoni: 1- Phn ng trao i ion:
NH4Cl + NaOH NaCl + NH3+ H2O (phn ng nhn bit mui amoni)
Hay: NH4+ + OH NH3+ H2O2- Phn ng phn hu (th hin tnh km bn nhit):
Phn ng tng qut: (NH4)nX NH3+ HnX (trong X l gc axit c ho tr n)
Th d: NH 4Cl0t NH3+ HCl NH4HCO3
0t NH3+ CO2+ H2O
Nhng vi mui to bi axit c tnh oxi ho th: Do NH 3 th hin tnh kh mnh, nn sn
phm ca phn ng s khng dng li giai on trn.
Th d: NH4NO20t N2 + 2 H2O Hoc: NH4NO3
0t N2O + 2 H2O
IV- Axit nitric: 1- Tnh axit mnh Tc dng vi hidroxit (tan v khng tan)
Mui + H2O
HNO3 + NaOH NaNO3 + H2O 2HNO3 + Mg(OH)2 Mg(NO3)2 + 2H2O
Tc dng vi oxit baz Mui + H2O Fe2O3 + 6 HNO3 2 Fe(NO3)3 + 3 H2O
CuO + 2 HNO3Cu(NO3)2 + H2O
2- Tnh oxi ho mnh:
Tc dng vi hu ht kim loi, k c mt s kim loi ng sau H nh Cu, Ag:
Fe + 6HNO3 c0t Fe(NO3)3 + 3NO2+ 3H2O Fe + 4HNO3 long Fe(NO3)3 + NO+
2H2O
Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O Ag + 2HNO3 AgNO3 + NO2 + H2O
Lu : + Sn phm ca phn ng th thuc vo:
Bn cht kim loi; Nng axit: axit c, ch yu
NO2 ; axit long, ch yu
NO; Nhit phn ng.
+ Mt kim loi tc dng vi dung dch HNO3 to ra nhiu sn phm kh, mi sn phm
vit 1 phng trnh phn ng, th d: 10Al + 36HNO3 10Al(NO3)3 + 3N2+ 18H2O
8Al + 30HNO3 8Al(NO3)3 + 3N2O+ 15H2O
+ Cc kim loi mnh c th kh HNO3 thnh NH3 v sau NH3 + HNO3 NH4NO3, c
ngha l trong dung dch tn ti NH4+ v NO3. Chng hn nh: 4Mg + 10HNO3 4Mg(NO3)2 + NH4NO3 + 3H2O
+ Cc kim loi Al, Fe b th ng trong dung dch HNO3 c ngui!
+ Dung dch cha mui nitrat (KNO3) trong mi trng axit cng c tnh cht tng t nh
dung dch HNO3, v trong dung dch tn ti H+ v NO3. Cch gii:
Vit cc phng trnh in li ca mui nitrat v axit. Vit phng trnh dng ion: M + H+ +
NO3 sn phm
Th d: Cho Cu vo dung dch cha KNO3 v H2SO4 long:
Phng trnh in li: KNO3K+ + NO3 v H2SO4 2H+ + SO42
Phng trnh phn ng: 3Cu + 2NO3 + 8H+3Cu2+ + 2NO+ 4H2O
Tc dng vi phi kim: C + 4HNO3 CO2 + 4NO2 + 2H2O S + 6HNO3 H2SO4 +
6NO2 + 2H2O Tc dng vi hp cht c tnh kh ( trng thi oxi ho thp): 3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O
33
3 0 +1 +2 +4+5
NH4NO
3N
2N
2O NO NO
2
HNO3
-
8/8/2019 Bai Tap Hoa 10Phi Kim
32/76
Fe3O4 + 10HNO3 3Fe(NO3)3 + NO2 + 5H2O FeCO3 + 4HNO3 Fe(NO3)3 + NO2 + CO2+ 2H2O
3Fe2+ + NO3 + 4H+ 3Fe3+ + NO + 2H2O FeS2 + 18HNO3 Fe(NO3)3 +2H2SO4 + 15NO2 + 7H2O
3- iu ch Trong PTN: NaNO3tinh th + H2SO4 c NaHSO4 + HNO3
Trong cng nghip: S iu ch: Khng kh N2 NH3 NO NO2 HNO3.
4NH3 + 5O2 0850 CPt 4 NO + 6H2O 2NO + O2 2NO2 4NO2 + O2 + 2H2O 4HNO3V- Mui nitrat 1- Tnh tan:Tt c cc mui nitrat u tan trong nc.
2- Phn ng nhit phn (th hin tnh km bn nhit):
Mui nitrat ca kim loi hot ng mnh (thng l cc kim loi t Mg tr v trc trong dy hot
ng ho hc) b phn hu bi nhit tao ra mui nitrit v oxi: Th d: 2KNO30t
2KNO2 + O2 Mui nitrat ca cc kim loi hot ng trung bnh (sau Mg n Cu) b phn hu bi nhit
to ra oxit, nit ioxit v oxi: Th d: 2Pb(NO3)2 0t 2PbO + 4NO2 + O2 2Cu(NO3)20t 2CuO + 4NO2 + O2
Mui nitrat ca cc kim loi km hot ng (sau Cu) b phn hu bi nhit to ra kim loi,
nit ioxit v oxi.
Th d: 2AgNO 30t 2Ag + 2NO2 + O2
B. Bi tp c li gii:
bi
66. Cho 1,5 lt NH3 (o ktc) i qua ng ng 16g CuO nung nng, thu c mt chtrn X.
1. Vit phng trnh phn ng gia NH3 v CuO, bit rng trong phn ng oxi ho ca nit
tng ln bng 0.
2. Tnh khi lng CuO b kh. 3. Tnh th tch dung dch HCl 2M tc dng
vi X.
67. Dn 1,344 lt NH3 vo bnh c cha 0,672 lt Cl2(th tch cc kh c o ktc)
1. Tnh thnh phn % theo th tch ca hn hp kh sau phn ng.
2. Tnh khi lng ca mui NH4Cl c to ra.
68. Hn hp A gm ba kh NH3, N2 v H2. Dn A vo bnh c nhit cao. Sau phn ngphn hu NH3(coi nh hon ton) thu c hn hp B c th tch tng 25% so vi A. Dn B i
qua ng ng CuO nung nng sau loi nc th ch cn li mt cht kh c th tch gim
75% so vi B.
Tnh thnh phn % theo th tch ca cc kh trong hn hp A.
69. Bit rng c 9,03.1022 phn t H2 tham gia phn ng vi 3,01.1022 phn t N2 (s
Avogaro bng 6,02.1023). Lng amoniac to thnh c ho tan vo mt lng nc va 0,4
lt dung dch (khi lng ring c coi bng d = 1g/ml) 1. Tnh s mol, s gam v s phn
t NH3 to thnh.
2. Tnh nng % v nng mol/l ca dung dch amoniac.70. Cho dung dch NH3 n d vo 20ml dung dch Al2(SO4)3. Lc ly cht kt ta v cho vo
10 ml dung dch NaOH 2M th kt ta va tan ht.
1. Vit phng trnh phn t v phng trnh ion rt gn ca cc phn ng xy ra.
34
-
8/8/2019 Bai Tap Hoa 10Phi Kim
33/76
2. Tnh nng mol/l ca dung dch Al2(SO4)3.
71. Cho dung dch Ba(OH)2 n d vo 50 ml dung dch A c cha cc ion NH4+, SO42 v
NO3. C 11,65g mt cht kt ta c to ra v un nng th c 4,48 lt (o ktc) mt
cht kh bay ra.
1. Vit phng trnh phn t v phng trnh ion ca cc phn ng xy ra.
2. Tnh nng mol/l ca mi mui trong dung dch A.
72. un nng hn hp gm 200g NH4Cl v 200g CaO. T lng kh NH3 to ra, iu ch c 224
ml dung dch NH3 30% (khi lng ring d = 0,892 g/ml). Tnh hiu sut ca phn ng.73. Trong bnh phn ng c 100 mol N2 v H2 theo t l 1 : 3. p sut ca hn hp kh lc
u l 300 atm v ca hn hp kh sau phn ng l 285 atm. Nhit trong bnh c gi
khng i.
1. Tnh s mol cc kh trong hn hp sau phn ng. 2. Tnh hiu sut ca phn ng
tng hp.
74. Trong bnh phn ng c 40 mol N2 v 160 mol H2. p sut ca hn hp kh lc u l
400 atm, nhit trong bnh c gi khng i. Bit rng khi phn ng t ti trng thi
cn bng th t l N2 phn ng l 25% (hiu sut ca phn ng tng hp). 1. Tnh
s mol cc kh trong hn hp sau phn ng.2. Tnh p sut ca hn hp kh sau phn ng.
75. Mt hn hp N2 v H2 c ly vo bnh phn ng c nhit c gi khng i. Sau
thi gian phn ng, p sut ca cc kh trong bnh gim 5% so vi p sut lc u. Bit rng
t l s mol N2 phn ng l 10%.
Tnh thnh phn % s mol N2 v H2 trong hn hp u.
76. Cho bit A l mt hp cht v c :
1. Hon thnh s bin ha sau, vit cc phng trnh phn ng :
2. Cho CO2 tc dng vi dung dch A thu c hn hp gm 2 mui X v Y. un nng hn hp X, Y
phn hu ht mui, thu c hn hp kh v hi H2O, trong CO2 chim 30% th tch.
Tnh t l s mol ca X v Y trong hn hp.
77. Khi nung hn hp mui nitrat ca ch v bc, thu c 12,32 lt (iu kin tiu chun) hn
hp hai kh. Hn hp kh khi c lm lnh bng hn hp nc v mui n cn li 3,36 lt
(iu kin tiu chun).
Xc nh thnh phn phn trm v khi lng ca hn hp mui.
78. Dung dch A cha hai axit HCl v HNO3 c nng tng ng l a mol/l v b mol/l.
1. trung ho 20 ml dung dch A cn dng 300 ml dung dch NaOH 0,1M. Mt khc ly 20ml dung dch A cho tc dng vi AgNO3 d thy to thnh 2,87 gam kt ta. Tnh cc gi tr
ca a v b.
2. Thm t t Mg kim loi vo 100 ml dung dch A cho ti khi kh ngng thot ra, thu c
dung dch B (th tch vn 100 ml) ch cha cc mui ca Mg v 0,963 lt hn hp D gm ba
kh khng mu cn nng 0,772 gam. Trn kh D vi 1 lt O2, sau khi phn ng hon ton,
cho kh cn li i t t qua dung dch NaOH d th th tch hn hp kh cn li 1,291 lt.
a. Hi hn hp kh D gm cc kh g ? Bit rng trong kh D c hai kh chim % th tch
nh nhau, cc th tch kh o ktc.
b. Vit phng trnh phn ng ha tan Mg di dng ion.c. Tnh nng cc ion trong dung dch B v tnh khi lng Mg b tan.
79. t chy a gam photpho ta c cht A, cho a tc dng vi dung dch cha b gam NaOH.
Hi thu c nhng cht g ? Bao nhiu mol ?
35
-
8/8/2019 Bai Tap Hoa 10Phi Kim
34/76
80. Ho tan 20g hn hp gm bari sunfat, canxi photphat, natri photphat v canxi cacbonat
vo nc. Phn khng tan c khi lng bng 18g c lc ring v cho vo dung dch HCl ly d
th tan c 15g v c 2,24 lt (o ktc) mt cht kh bay ra. Tnh khi lng ca mi mui
trong hn hp.
81. t chy hon ton 6,8 g mt hp cht ca photpho thu c 14,2g P 2O5 v 5,4g nc. Cho
cc sn phm vo 50g dung dch NaOH 32%
1. Xc nh cng thc ho hc ca hp cht. 2. Tnh nng % ca dung dch mui thu c.
82. Cn ly bao nhiu tn qung photphorit loi c cha 65% Ca 3(PO4)2 iu ch c150 kg photpho, bit rng lng photpho hao ht trong qu trnh sn xut l 3%.
83. Ho tan 22 gam hn hp A (Fe, FeCO3, Fe3O4) vo 0,896 lt dung dch HNO3 1 M thu c
dung dch B v hn hp kh C (gm CO2 v NO). Lng HNO3 d trong B phn ng va vi
5,516 gam BaCO3. C mt bnh kn dung tch 8,96 lt cha khng kh (ch gm N2 v O2theo t l th tch l 4:1) c p sut 0,375 atm, nhit 00C. Np hn hp kh C vo bnh
gi nhit 00C th trong bnh khng cn O2 v p sut trong bnh cui cng l 0,6 atm.
1. Tnh % khi lng cc cht trong hn hp A. 2.Tnh %V hn hp kh C
84. Ho tan hon ton 0,368 gam hn hp Al v Zn cn va 25 lt dung dch HNO3 cpH=3. Sau phn ng thu c dung dch A cha 3 mui (khng c kh thot ra).
1. Vit phng trnh phn ng xy ra. 2. Tnh khi lng ca mi kim loi trong
hn hp.
3. Thm vo dung dch A mt lng d dung dch NH3. Tnh khi lng kt ta thu c.
85. Cho x mol Fe tc dng vi dung dch cha y mol HNO3 thu c kh NO duy nht v dung
dch B. Dung dch B tn ti nhng ion no? Bao nhiu mol? Bin lun quan h gia x v y
trong dd B tn ti cc ion .
Hng dn gii
66. 1. 3 CuO + 2 NH3N2+ 3 Cu
(l)1,5x(g)
(l)2.22,4(g)3.80
2. Khi lng CuO b kh:
)g(84,22.280.3.5,1
x =
Khi lng CuO cn tha: )g(8816mCuO ==
3. X gm c CuO v Cu. Ch CuO c tc dng vi dung dch HCl. Phng trnh phn ng:
CuO + 2 HCl CuCl2 + H2OTh tch dung dch HCl cn ly: V dd HCl = 0,2 : 2 = 0,1 (lt)
67.
)mol(03,04,22:672,0n
)mol(06,04,22:344,1n
2
3
Cl
NH
==
==
Phng trnh phn ng: 2 NH3 + 3 Cl26 HCl + N2
)mol(04,002,006,0nd-:cnNH3NH3
==
Do : NH3 + HCl NH4Cl
)mol(02,004,006,0n:licnHClKh HCl ==
1. Hn hp kh sau phn ng gm 0,01 mol N2 v 0,02 mol HCl. Thnh phn % theo th tch
ca hai kh bng thnh phn % theo s mol v bng.
36
-
8/8/2019 Bai Tap Hoa 10Phi Kim
35/76
(%)7,66100.03,0
02,0HCl%
(%)3,33100.03,0
01,0N% 2
==
==
2. Khi lng ca NH4Cl bng:)g(14,25,53.04,0m ClNH4 ==
68. Gi x, y v z l thnh phn % theo th tch ca ba kh NH 3, H2 v N2 trong hn hp A. Ta
c: x + y + z = 1 (1)
Phng trnh phn ng phn hu NH3: 2 NH3 N2 + 3 H2
xx
2
3x
2
Sau phn ng hn hp B gm (y + 1,5x) hiro v (z + 0,5x) nit. B c th tch tng 25% so
vi A tc l bng 125/ 100, do :
( ) ( ) (2)25,1100125
zyx2x5,0zx5,1y ==++=+++
T (1) v (2), rt ra : x = 0,25Khi hn hp B i qua ng ng CuO nung nng th H2 b oxi ho : CuO + H2Cu + H2O
Loi nc th cht kh cn li l N2. Th tch gim 75% so vi B tc l cn bng 25% ca B, do
:
312,0165
100125.
10025
x5,0z ===+
Rt ra: z = 0,3125 (0,5 . 0,25) = 0,1875
Thay cc gi tr ca x v z vo (1), ta c: 0,25 + y + 0,1875 = 1, rt ra y = 0,5625
Thnh phn % theo th tch ca cc kh trong hn hp A:
%75,18N%
%25,56H%
%25NH%
2
2
3
==
=
69.1.)mol(15,0)10.02,6(:)10.03,9(n 2322H2 ==
)mol(05,0)10.02,6(:)10.01,3(n 2322N2 ==
N2 + 3 H2 2 NH3
0,05mol 0,15mol 0,1mol
)g(7,117.1,0m),mol(1,0n 33 NHNH === 2223
3 10.02,610.02,6.1,0NHtnaphS ==
2. Nng % ca dung dch NH3:(%)42,0
7,1400100.7,1
%C )ddNH( 3 =+=
Nng mol/l ca dung dch NH3:)M(25,04,0:1,0C )ddNH(M 3 ==
70. 1. Phng trnh phn ng:
6 NH3 + 6 H2O + Al2(SO4)32 Al(OH)3+ 3 (NH4)2SO4
3 NH3 + 3 H2O + Al3+ Al(OH)3+ 3 NH4+
xmol x molAl(OH)3 + NaOH NaAlO2 + 2 H2O
Al(OH)3 + OH AlO2 + 2 H2O
xmol x mol
37
-
8/8/2019 Bai Tap Hoa 10Phi Kim
36/76
2. Trong 10ml dung dch NaOH 2M c: nNaOH = 0,01.2 = 0,02 (mol), do :
)mol(02,0nOH
=
)mol(02,0nx:bngAlmolSOH
3 == +
S mol Al2(SO4)3 bng:
)mol(01,002,0.5,0n21
n 3342 Al)SO(Al === +
Nng mol/l ca dung dch Al2(SO4)3:)M(5,002,0:01,0C
342 )SO(AlM==
Ch thch: Al(OH)3 khng tan trong dung dch NH3 d v y l dung dch baz yu.
71. 1. Ba2+ + SO42 BaSO4
molxmolx
NH4+ + OHNH3 + H2O
molymoly
2. S mol SO42 c trong dung dch:
(mol0,05233:11,65x ==
S mol NH4+ c trong dung dch: y = 4,48 : 22,4 = 0,2 (mol). S mol mui (NH4)2SO4 bng
s mol : SO42 = 0,05 mol
Do :
)M(105,0:05,0C424 SO)NH(M
==S mol NH4+ ca mui nitrat: 0,2 (0,05 . 2) = 0,1 (mol)
Do :mol1,0n
34NONH=
)M(205,0:1,0C:v )NONH(M 34 ==
72.
Phng trnh phn ng:2 NH4Cl + CaO 2 NH3+ CaCl2 + H2O
Theo phn ng: 2. 53,5g 56g 2. 17g
Theo bi ra: 200g 200g xg
Lng CaO c d v NH4Cl c th phn ng ht, nu hiu sut 100% th s thu c:
)g(55,635,53.217.2.200
xm3NH
===
Khi lng NH3 thc t thu c:(%)59,940,3.0,892.224 =
Hiu sut ca phn ng:(%)3,94100.
55,63
94,59=
73. 1. Tng s mol cc kh trong hn hp sau phn ng:
)mol(95300100.285
n.pp
n 11
22 ===
2.2 2N H
n 100:4 25(mol);n 100 25 75(mol)= = = =
V cc kh ly theo ng t l trong phn ng nn theo phng trnh phn ng:
N2 + 3H2 2NH3
mo2xmol3xmolxmol)x25(n:licnNmolS
2N2=
mol)x375(n:licnHmolS
2H2=
38
-
8/8/2019 Bai Tap Hoa 10Phi Kim
37/76
)mol(x2n:thnhtoNHmolS3NH3
=Tng s mol cc kh: (25 x) + (75 3x) + 2x = 95
(mol)
Gii ra c: x = 2,5 mol Hiu sut ca phn ng tng hp:(%)10100.
255,2
=
74. T l mol lc u:4:1160:40n:n
22 HN==
H2 ly c d, do theo phng trnh phn ng:
N2 + 3H2 2NH3
molymolxmol40.0,25
mol2mol3mol1
1. S mol N2 cn li l: 40 (40.0,25) = 30 (mol) S mol H2 phn ng: x = 3. 40. 0,25
= 30 (mol)
)mol(13030160n:licnHmolS2H2
==S mol NH3 to thnh: y = 2. 40. 0,25 = 20 (mol)
Tng s mol cc kh trong hn hp sau phn ng. )mol(1802013030n2 =++=
2. Khi nhit khng i th p sut ca cc kh trong bnh kn t l thun vi s mol
cht kh:
2
1
2
1
n
n
p
p=
p sut ca hn hp kh sau phn ng bng: p2 =2
11
n 180.400.p 360
n 40 160= =
+ (atm)
75. Gi x l s mol N2 v y l s mol H2 c lc u. S mol N2 phn ng l 0,1x. Theo
phng trnh phn ng: 0,1x mol N2 tc dng vi 0,3x mol H2 to ra 0,2x mol NH3.
Sau phn ng cn li: (1 0,1x) mol N2, (y 0,3x) mol H2
Tng s mol cc kh trong hn hp sau phn ng: x8,0yx2,0)x3,0y(x9,0 +=++p sut ca cc kh trong bnh gim 5% so vi p sut lc u, tc l bng 95% lc u, do
ta c:
95,010095
yxx8,0y
PP
1
2 ==+
+=
hay l: y + 0,8x = 0,95(x + y), rt ra x : y = 1 : 3 Thnh phn % s mol trong hn hp
u:
(%)75100.
4
3H%;(%)25100.
4
1N% 22 ====
76. 1. Hon thnh s : A ch c th l NH3, do vy:
NH3 + HCl NH4Cl (1) NH4Cl + NaOH NH3 + NaCl + H2O(2)
NH3 + HNO3 NH4NO3 (3) NH4NO30t N2O + 2 H2O (4)
2. CO2 + NH3 + H2O NH4HCO3 (5) CO2 + 2NH3 + H2O (NH4)2CO3 (6)
NH4HCO30t CO2 + H2O + NH3 (7) (NH4)2CO3
0t CO2 + H2O + 2 NH3
(8)Gi x v y l s mol NH4HCO3 v (NH4)2CO3, theo (7), (8) ta c:
nCO2 = x + y = 0,3 nhh = 3x + 4y = 1 T rt ra : x = 2y
77. Mui Pb(NO3)2 v mui AgNO3 b nhit phn theo cc phn ng:
39
-
8/8/2019 Bai Tap Hoa 10Phi Kim
38/76
2Pb(NO3)20t 2PbO + 4NO2 + O2 (1)
2mol (2.232g) 4mol (4.22,4l) 1mol (22,4l)
2AgNO30t 2Ag + 2NO2 + O2 (2)
2mol (2.170g) 2mol (2.22,4l) 1mol (22,4l)
Hn hp kh thu c l NO2 v O2, khi c lm lnh NO2 ho lng cn li O2.
(mol55,04,2232,12
khhphnmolS =
(mol15,04,22
36,3licnOmolS 2 =
Gi x l s mol Pb(NO3)2 v y l s mol AgNO3 c trong hn hp mui.
Theo phn ng (1), x mol Pb(NO3)2 phn hu to nn 2x mol NO2 v 0,5x mol O2.
Theo phn ng (2), y mol AgNO3 phn hu to nn y mol NO2 v 0,5y mol O2. Ta c cc phng
trnh:
2,5x + 1,5y = 0,55 0,5x + 0,5y = 0,15
Gii h phng trnh, tm c x = 0,1 v y = 0,2.
Thnh phn ca hn hp mui nung:
332 0,1 = 33,2g Pb(NO3)2 v 170 0,2 = 34g AgNO3
Thnh phn phn trm ca hn hp mui:;%5,49100
342,332,33
=+
%5,50100342,33
34=
+
78. 1. Cc phn ng: HCl + NaOH NaCl + H2O (1) HNO3 + NaOH NaNO3 + H2O
(2)
HCl + AgNO3AgCl + HNO3 (3)
02,05,143
87,202,0.ann:(3)Theo
3AgNOHCl====
a = 1
Theo (1) v (2) nHCl + nHNO3 = nNaOH = 0,02. (1 + b) = 0,3 . 0,1 b = 0,52. a. Khi ha tan Mg vo A c th cho cc kh khng mu l : H2, NO, N2O, N2.
mol043,04,22
963,0n:cTa D ==
95,17
043,0
772,0MiV D ==
Do mt kh phi l H2. Khi trn kh D vi O2, th tch kh b ht do c cc phn ng:
2NO + O2 2NO2 (4)
v 2NO2 + 2NaOH NaNO2 + NaNO3 + H2O (5)
Vy kh th hai l NO.
Th tch kh b ht = 0,963 + 1 1,291 = 0,672 lt
02,04,22
672,0.
32
n32
n:(4)theovyNh- kh htNO ===
Gi M l KLPT ca kh th 3 Xt 3 trng hp:
Th1. H2 v NO c th tch nh nhau: 2 . 0,02 + 30 . 0,02 + M(0,043 0,04) = 0,772
M = 44 l kh N2O
Th2. NO v kh th 3 c th tch nh nhau : 30 . 0,02 + M . 0,02 + 2 . 0,003 = 0,772
M = 8,3 : loiTh3. H2 v kh th 3 c th tch nh nhau:
77,02
02,0043,0.M
202,0043,0
.20,02.30 =
+
+ = 12,95: loi
40
-
8/8/2019 Bai Tap Hoa 10Phi Kim
39/76
b. Mg + 2H+Mg2+ + H2 (6)
3Mg + 8H+ + 2NO3 3Mg2+ + 2NO+ 4H2O (7)
4Mg + 10H+ + 2NO34Mg2+ + N2O+ 5H2O (8)
c. Ion Cl khng tham gia phn ng nn [Cl] = 1 mol/l
Theo (7), (8) [NO3] cn li =0,5.0,1 0,02 0,003.2
0,24(mol / l)0,1
=
V dung dch trung ha in nn:
[Mg2+] = 3Cl NO 1 0,24
0,62(mol / l)2 2
+ + = =
Khi lng Mg tan vo dung dch: 0,62 . 0,1.24 = 1,488 (g)
(c th tnh s mol Mg2+
(x) theo s bo ton electron: 2x = 0,02 . 3 + 0,02 . 2 + 0,003 . 8
x = 0,062)
79. Cc phn ng: 2 P + 25
O2
P2O5 (1)
P2O5 + 3 H2O 2 H3PO4 (2) H3PO4 + NaOH NaH2PO4 + H2O (3)NaH2PO4 + NaOH Na2HPO4 + H2O (4) Na2HPO4 + NaOH Na3PO4 + H2O (5)
Theo (1, 2)
mol'b40b
n
mol'amol31a
nn
NaOH
pPOH 43
==
===
Ta xt cc trng hp:
b' < a' : phn ng (3) cha ht, ta c b' mol NaH2PO4 v cn (a' b') mol H3PO4b' = a' : va ht phn ng (3) Ta c: a' = b' mol NaH2PO4 a' < b' < 2a' : phn ng (3) kt
thc v c mt phn phn ng (4) Ta c: (2a' b) mol NaH2PO4 v (b' a') mol Na2HPO4 b' =
2a' : va ht phn ng (3) v (4) Ta c: a' mol Na2HPO4 2a' < b' < 3a' : phn ng (3, 4)
kt thc v mt phn phn ng (5) Ta c:
(3a' b) mol Na2HPO4 v (b'2a') mol Na3PO4 b' = 3a': va kt thc c 3 phn ng (3, 4, 5) Ta c : a' mol Na3PO4
b' > 3a' : sau phn ng (5) cn d NaOH Ta c: a' mol Na3PO4 v (b' 3a') mol NaOH.
80. Ch c Na3PO4 tan trong nc, khi lng ca Na3PO4 = 20 18 = 2 (g). Bari sunfat khng tantrong dung dch HCl, c khi lng bng 18 15 = 3 (g). Cc mui Ca3(PO4)2 v CaCO3 tan trong
dung dch HCl, phng trnh phn ng: Ca3(PO4)2 + 6 HCl 3 CaCl2 + 2 H3PO4 (1)
CaCO3 + 2 HCl CaCl2 + CO2+ H2O (2)
Theo phn ng (2), khi lng CaCO3 = nCaCO3 . 100 = nCO2 . 100 = 10 (g) Khi lng Ca3(PO4)2 =
15 10 = 5 (g)
81. 1. Tng khi lng ca P v H: 6,2 + 0,6 = 6,8 (g) bng khi lng ca cht b t chy,
hp cht khng c O.
Cng thc c dng PxHy, ta c t l:3:16,0:2,0
1
6,0:
31
2,6y:x ===
Ly t l n gin nht ta c cng thc ho hc ca hp cht: PH3
2. S mol P2O5 thu c:)mol(1,0142:2,14n
52OP==
41
-
8/8/2019 Bai Tap Hoa 10Phi Kim
40/76
S mol NaOH c trong dung dch:)mol(4,0
40.10032.50
nNaOH ==
Cc phn ng c th xy ra gia P2O5 v NaOH: P2O5 + 2 NaOH + H2O 2 NaH2PO4(1)
P2O5 + 4 NaOH 2 Na2HPO4 + H2O (2) P2O5 + 6 NaOH 2 Na3PO4 + 3 H2O(3)
Theo (2), t 1 mol P2O5 v 4 mol NaOH to ra 2 mol Na2HPO4.
Vy: t 0,1 mol P2O5 v 0,4 mol NaOH to ra 0,2 mol Na2HPO4.
Khi lng ca Na2HPO4:)g(4,28142.2,0m
42HPONa==
Khi lng ca dung dch: )g(6,694,52,1450mdd =++=
Nng % ca dung dch:(%)41
6.69100.4,28
)HPONa(%C 42 ==
82. Theo s hp thc: P2)PO(Ca 243
kg150kg750x
g31.2g310
=
Khi lng Ca3(PO4)2 thc t cn c:)kg(773
65100.750
Khi lng qung photphorit:)tn(189,1)kg(1189
65100.773
==
83.1. Gi s mol Fe, FeCO3, Fe3O4 trong hn hp A l x, y, z mol. Ta c cc phng trnh phn
ng:
Fe + 4 HNO3 Fe(NO3)3 + NO + 2 H2O (1) 3 FeCO3 + 10 HNO3 3 Fe(NO3)3 + NO + 5
H2O + 3 CO2 (2)3 Fe3O4 + 28 HNO3 9 HNO3 + NO+ 14 H2O (3) BaCO3 + 2 HNO3 Ba(NO3)2 + H2O
(4)
Theo cc phng trnh phn ng (1), (2), (3) ta c: S mol NO: a = x + y/3 + z/3 mol
S mol CO2: y mol Theo cc phng trnh phn ng (1), (2), (3), (4) ta c:
S mol HNO3: 4x + 10y/3 + 28z/3 + 5,516/197 = 0,896.1 = 0,896 (mol) 12x + 10y +
28z = 2,52 (I)
Khi lng hn hp rn ban u: 56x + 116y + 232z = 22 gam (II)
S mol kh c sn trong bnh: n = PV/RT = 0,15 (mol)
Trong N2 c 0,12 mol, O2 c 0,03 mol. Khi np NO v CO2 vo c phn ng: 2 NO +
O22 NO2 (5)Theo bi ra ta thy ht O2 nn NO c th d, v theo phng trnh phn ng (5) th s mol
NO2 bng hai ln s mol O2, mt khc tng s mol NO v NO2 vn bng s mol NO ban u.
Do :
mol0,24
RTPV0,12
3z
34yx0,12yannnn
22 NCONOhh==+++=++=++=
4x + 4y + z = 0,36 mol (III)
Gii h phng trnh: (I), (II), (III) thu c: x = 0,02 mol; y = 0,06 mol; z = 0,06 mol;
42
-
8/8/2019 Bai Tap Hoa 10Phi Kim
41/76
63,273
31,636
5,091%
===
===
===
22%100.06,0.232
22%100.z232
m%
22%100.06,0.116
22%100.y116
m%
22%100.02,0.56
22%100.x56
m%
43
3
OFe
FeCO
Fe
2. S mol NO sinh ra: a = x +y
3
+z
3
= 0,06 mol
S mol NO2 sinh ra: b = 2.nO2 = 0,06 mol Trong hn hp cui cng khng cn NO, ch cn:
NO2 0,06 mol, N2 0,12 mol, CO2 0,06 mol. Tng s mol hn hp: 0,24 mol. % s mol bng
% th tch hn hp:
25%
50%
5%
==
==
==
24,0
%100.06,0
%24,0
%100.12,0%
224,0
%100.06,0%
2
2
2
CO
N
NO
V
V
V
84. 1. Al, Zn tc dng dung dch HNO3 khng c kh thot ra, dung dch cha ba mui snphm c NH4NO3.
Phng trnh phn ng: 8 Al + 30 HNO38 Al(NO3)3 + 3 NH4NO3 + 9 H2O (1)
4 Zn + 10 HNO34 Zn(NO3)2 + NH4NO3 + 3 H2O (2)
2. Dung dch HNO3: nng M1010][HHNO3pH
3 ===+
s mol HNO3 = CM.V = 25.103
= 0,025 (mol)
Gi s mol Al, Zn trong 0,368 gam hn hp ln lt l x, y mol.
Theo cc phng trnh phn ng (1) v (2) ta c: 27x + 65y = 0,368 v 30x/8 + 10y/4= 0,025
ga0,26065ymgam,0,10827xm
mol4.10ymol,4.10x
ZnAl
33
====
==
3. Dung dch A cha Al(NO3)3; Zn(NO3)2 cho tc dng vi dung dch NH3 d:
Al(NO3)3 + 3 NH3 + 3 H2O Al(OH)3+ 3 NH4NO3 (3)
Zn(NO3)2 + 2 NH3 + 2 H2O Zn(OH)2+ 2 NH4NO3 (4)
Zn(OH)2 + 4 NH3Zn(NH3)42+ + 2 OH (5)
Khi lng kt ta thu cm = 78x =0,312 (gam)85. Phng trnh phn ng: Fe + 4 HNO3Fe(NO3)3 + NO+ 2 H2O (1)
Nu d Fe s xy ra phn ng: Fe + Fe(NO3)3Fe(NO3)2 (2)Gi s mol Fe l x, s mol axit HNO3 l y. Trong dung dch c th c cc ion: H+, NO3, Fe2+,
Fe3+, c th nh sau:
Trng hp 1: Nu d hoc va HNO3, tc l: y 4x phn ng (1) xy ra ht Fe nnkhng c phn ng (2).
Dung dch cha HNO3 (y 4x) mol v Fe(NO3)3 x mol: H+
(y 4x) mol; Fe3+ x mol; NO3
(y x)
molTrng hp 2: thiu HNO3, tc l: y < 4x phn ng (1) xy ra ht HNO3, d Fe nn c phn
ng 2 xy ra. S mol Fe(NO3)3 sinh ra t phn ng (1) l y/4 mol lng Fe cn li sau phn ng
(1) l (x y/4).
43
-
8/8/2019 Bai Tap Hoa 10Phi Kim
42/76
Nu: (x y/4) y/8 th phn ng (2) xy ra cn d Fe(NO3)3 th dung dch cha:
Fe(NO3)3 : y/4 2(x y/4) = (3y/4 2x) mol v Fe(NO3)2 : 3(x y/4) mol;
Fe3+ : (3y/4 2x) mol; Fe2+ : (3x 3y/4) mol; NO3
: 3y/4 molNu: (x y
4 ) >y
8 phn ng
(2) xy ra ht Fe(NO3)3 v Fe d. Dung dch cha: Fe(NO3)2c (y
4+
y
8) =
3y
8mol.
Dung dch cha Fe2+ : 3y8
mol, NO3 : 3y4
mol.
C. Bi tp t gii:86. a. C hai kh A v B: Nu trn cng s mol A, B th thu c hn hp X c dX/He = 7,5
Nu trn cng khi lng A, B th thu c hn hp Y c dY/O2 = 11/15 Tm khi lng mol ca A v
B.
b. Mt hn hp kh gm N2 v H2 c t khi i vi H2 bng 3,6. Sau khi nung nng mt thi gian
t h cn bng th t khi ca hn hp sau phn ng i vi H2 bng 4,5.
Xc nh % theo th tch ca cc hn hp trc v sau phn ng. Tnh hiu sut phn ng.
p s: a. MA = 44 ; MB = 16 b. %VN2 = 12,5% ; %VH2 = 62,5% ; %VNH3 = 25% ; Hiu sut
= 50%
87. a.Vit cc phng trnh phn ng iu ch NH3 trong phng th nghim v trong cng
nghip.
b.Vit cc phng trnh phn ng iu ch axit nitric t amoniac.
c. Cho 3 mnh kim loi Al, Fe, Cu vo ba cc ng dung dch axit nitric c nng khc
nhau v thy:
Cc c Al: Khng c kh thot ra, nhng nu ly dung dch sau phn ng tc dng vi
dung dch NaOH thy kh c mi khai bay ra.
Cc c Fe: C kh khng mu bay ra v ho nu trong khng kh.
Cc c Cu: C kh mu nu bay ra.
Hy vit cc phng trnh phn ng xy ra.
p s: a. NH4+ + OH0t NH3+ H2O
hoc dng NaOH rn ht nc ca dung dch NH3 m c thu kh NH3.
b. SGK
c. Cc c Al: to ra mui amoni. Cc c Fe: to ra kh NO. Cc c Cu: to ra kh
NO2.
88. a. Baz Bronstet l g ? Nhng baz no c gi l kim ?
b. Hy gii thch ti sao amoniac v anilin u c tnh baz ?
c. Dung dch NH3 1M c = 0,43 %. Tnh hng s KB v pH ca dung dch .
d. Nu nhn xt khi qut v s phn li ca baz trong dung dch nc.
p s: a. SGK
b. C hai u l baz v u c kh nng nhn H+ (do trn nguyn t nit
u cn i electron cha tham gia lin kt). c. Kb = 1, 86 .105 ; pH =
11,633
44
-
8/8/2019 Bai Tap Hoa 10Phi Kim
43/76
89.a. Nhm v magie tc dng vi HNO3 long, nng u sinh ra NO, N2O v NH4NO3 .
Vit phng trnh ion thu gn ca cc phn ng xy ra.
b. Khi ha tan hon ton 1,575 gam hn hp A gm bt nhm v magie trong HNO3 th c
60% A phn ng to ra 0,728 lt kh NO (ktc).
Tnh phn trm khi lng ca Mg v Al trong hn hp.
p s: a. Vit 3 PTP cho mi cp.
b. %mMg = 71,43% ; %mAl = 28,57%.90. a. Hy vit cc phng trnh phn ng ho hc thc hin dy bin ho sau:
A1 0t N2 2O A2 2O A3 OH2 A4 C u A5
0t A3
b. Ch dng qu tm, hy phn bit cc dung dch ng trong cc l mt nhn sau:
BaCl2 ; NH4Cl ; (NH4)2SO4 ; NaOH ; Na2CO3.
p s: a. A1 l NH4NO2 ; A2 l NO ; A3 l NO2 ; A4 l HNO3 ; A5 l Cu(NO3)2.
b. Lu NH4+ l mt axit yu lm i mu qu tm thnh hng.
91. Vit cc phn ng nhit phn mui amoni sau y:
NH4Cl, NH4HCO3, (NH4)2CO3, NH4NO2, NH4NO3, (NH4)3PO4, (NH4)2SO4, (NH4)2Cr2O7
p s: a. NH4Cl0t NH3 + HCl NH4HCO3
0t NH3 + CO2 + H2O
(NH4)2CO30t 2 NH3 + CO2 + H2O NH4NO2
0t N2 + 2 H2O
NH4NO30t N2O + 2 H2O (NH4)3PO4
0t 3 NH3 + H3PO4
(NH4)2SO40t 2 NH3 +H2SO4 (nu tip tc nung nng th cui cng s thu c N2 + SO2 +
H2O)
(NH4)2Cr2O70t
N2 + Cr2O3 + 4 H2O
92. a. Cc cht: NO, NO2, SO2, H2O2 va c tnh oxi ha va c tnh kh. Vit cc phn ng
chng minh mi tnh cht cho mi cht nu trn m khng c cho cc cht tc
dng ln nhau (tt c 8 phn ng).
b. T qung photphorit v cc cht khc, vit cc phn ng iu ch photpho,
supephotphat n, supephotphat kp.
(Lu : iu ch pht pho: 2 Ca3(PO4)2 + 6 SiO20t 6 CaSiO3 + P4O10 P4O10 + 10 C
0t
P4 + 10 CO)
93. Hp cht MX2 kh ph bin trong t nhin. Ha tan MX2 bng dung dch HNO3 cnng, d ta thu c dung dch A. Cho A tc dng vi BaCl2 thy to thnh kt ta trng, cn
khi cho A tc dng vi NH3 d thy to thnh kt ta nu . a. Hi MX2 l cht g ? Gi
tn n. Vit cc phng trnh phn ng xy ra.
b. Vit cu hnh electron ca M v ca cc ion thng gp ca kim loi M.
p s: a. MX2 l FeS2 (Qung pirit st) b. Fe: 1s22s22p63s23p63d64s2; Fe2+:
1s22s22p63s23p63d6 ;
Fe3+: 1s22s22p63s23p63d5
94. a. Vit cc phng trnh phn ng xy ra trong quy trnh iu ch axit nitric t kh NH 3v oxi khng kh.
b. Tnh th tch dung dch axit nitric 50% (c d = 1,31 g/ml) to thnh khi dng ht 1m 3
kh NH3 ( iu kin tiu chun). Bit rng ch c 98,56% NH3 chuyn thnh axit nitric.
p s: b. V = 4,232 lt dung dch HNO345
t0
-
8/8/2019 Bai Tap Hoa 10Phi Kim
44/76
95. Cho bt Zn vo dung dch hn hp NaNO3 v NaOH, kt thc phn ng thu c 8,96 lt
(ktc) hn hp kh NH3 v H2. Cho hn hp ny vo mt bnh kn nung nng nhit phn
NH3, sau phn ng thu c V lt hn hp kh trong N2 chim12
1
th tch V, H2 chim12
8
th tch V, cn li l NH3.
a. Tnh th tch V ktc.
b. Tnh t khi ca hn hp trc v sau phn ng so vi H2, gii thch s thay i ca t khi
ny.c.Sc hn hp kh A vo dung dch HCl va . Sau phn ng cho thm nc ct vo thnh 1
lt dung dch. Tnh pH ca dung dch mui to thnh, bit 3NHpK = 4,75.
p s: a. V = 10,752 lt
b. dhh trcP/H2 = 4,75 v dhh sau P/H2 =9524
3,96 ; Do khi lng hn hp kh khng
i, nhng tng s mol kh sau phn ng tng.
c. pH = 5,09
96. Bnh kn c V = 0,5 lt cha 0,5 mol H2 v 0,5 mol N2 ( toC) khi t n trng thi cn
bng c 0,2 mol NH3 to thnh.
a. Tnh hng s cn bng Kc ( toC)
b. Tnh hiu sut to thnh NH3. Mun hiu sut t 90% cn phi thm vo bnh bao nhiu
mol N2 ?
c. Nu thm vo bnh 1 mol H2 v 2 mol NH3 th cn bng s chuyn dch v pha no ? Ti
sao ?
d. Nu thm vo bnh 1 mol heli, cn bng s chuyn dch v pha no ? Ti sao ?
p s: a. Kc = 3,125b. Hiu sut = 60% ; t hiu sut 90% th cn thm vo bnh 57,25 mol
N2.
c. Cn bng s chuyn dch theo chiu nghch (Da vo gi tr ca Kc trn
xt)
d. Theo chiu thun, phn ng lm gim s mol kh, tc l lm gim p sut
chung ca h (do phn ng thc hin trong bnh kn). V vy, khi thm kh He
s lm cho p sut chung ca h tng, nn cn bng b dch chuyn theo chiu
thun (nguyn l Lsatlie).
97. Mt hn hp X gm 2 kim loi Al v Cu. Cho 18,20 gam X vo 100ml dung dch Y chaH2SO4 12M v HNO3 2M, un nng cho ra dung dch Z v 8,96 lt (ktc) hn hp kh J gm
NO v kh D khng mu. Bit hn hp kh J c t khi i vi H2 = 23,5.
a. Tnh s mol kh D v kh NO trong hn hp kh J.
b. Tnh khi lng mi kim loi trong hn hp u. Tnh khi lng mi mui trong dung dch Z.
c. Tnh th tch dung dch NaOH 2M thm vo dung dch Z bt u c kt ta, kt ta