Cach 1: Khi hoa tan vao dung dich H2SO4 , Fe(N03)2 se phan li thanh cac ion. Do vay, truoc het cac em can tinh so mol cac ion nhu sau: O • Phuong trinh ion rut gon: i •••J•..:> :r imrp Mol: a3 < 0,8 0,2 > 0,3 0,2 cO ; , Cac em luu y, Fe cung bi oxi hoa thanh Fe va giai phong khi NO : Mol; 0,6 0,8 0,2 0,6 0,2 _ > V = (0,2 + 0,2).22,4 = 8,96 lit > Dap an B. rr,fj{),p« y 4 Cach 2: Cac chat khu: Cu 2e >• Cu; Fe le > Fe. Chat oxi hoa: N + 3e > N O Bao toan electron: 2ncu + lnp2+ = Sno riNO = z =0,4 mol V = 0,4.22,4 = 8,96 lit Dap an B. Vi d 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4 0,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch I X. Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ket tua Ian nhat. . Gia trj nho nhat ciia V la A. 80. B.50. C.60. D. 40. 3Cu + 8H + 2NO • > 3Cu2 + N O + 4H2O 3Fe2 + 4H + NO ; > 3¥e + N O + 2H2O 2.a3+a6 , v r Ldi gidi:
Trang 1ca'm nang 6n luyjn thi dgi hgc 18 chuyfin H(Sa hpc - NguySn Van H3i
Lot gidi:
Cach 1: Khi hoa tan vao dung dich H 2 S O 4 , Fe(N03)2 se phan li thanh cac ion
Do vay, truoc het cac em can tinh so mol cac ion nhu sau: O ' •
Phuong trinh ion rut gon: ' ^''^''/'i •••^'J^/•^ -:> <>^:r'^ ^i ^mrp
3Cu + 8H* + 2NO^ • > 3Cu2- + ' ^ N O + 4H2O
Mol: a3 <- 0,8 0,2 -> 0,3 0,2 c O / ; ,
Cac em luu y, Fe^* cung bi oxi hoa thanh Fe^* va giai phong khi N O :
3Fe2* + 4H* + N O ; > 3¥e^ + N O + 2H2O
Mol; 0,6 ^ 0,8 ^ 0,2 0,6 0,2 ^ ^ _
> V = (0,2 + 0,2).22,4 = 8,96 lit > Dap an B rr,fj{),p« y 4
-Cach 2: Cac chat khu: Cu - 2e ->• Cu^^; Fe^^ - le -> Fe^*
Chat oxi hoa: N*-^ + 3e -> N O
Bao toan electron: 2ncu + lnp^2+ = Sn^o
2.a3+a6 , v r ^
r i N O = z =0,4 mol
^ V = 0,4.22,4 = 8,96 lit ^ Dap an B
Vi d^ 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4
0 ,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch
-I X Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ke't tua Ian nhat
Gia trj nho nhat ciia V la
A 80 B.50 C.60 D 40
Ldi gidi:
Nhan xet: Dung dich chiia muoi NaNOs va H2SO4 loang -> can giai theo
phuong trinh ion ;?> v, t, • 5
ncu =0,06 mol; n^+ = 2.0,2.0,5 = 0,20 mol; n ^ ^ , =0,2.0,2 = 0,04 mol
Phuong trinh ion thu gpn:
3Cu + 8H* + 2NO3- > 3Cu^*+ 2NO + H 2 O
;^Mol: a06 ai6 a04
MjoriX^:-Cu tan het -> X CO chiia: n^ 2+ = 0,06; n + du = 0,04 '
De thu dugc luang ke't tua Idn nhat thi: njs^aOH = n + + 2n 2+ /
CO the hoa tan toi da m gam Fe Biet trong cac qua trinh tren, san pham khu duy nhat cua N*^ deu la NO Gia tri ciia m la
-Fe3^ = 0,1 mol; H* = 0,4 mol; N O ; = 0,3 mol va SO ]' = 0,2 mol
^ , ^ , , , , „ " WO '-* "'"^ * mi
Cac phan ung hoa tan Fe:
Fe + 4H* + N O ; > Fe3* + N O + 2H2O QkHi Mol: 0,05 4-0,4 - > 0,05 ••idl,0» j„rtS«.^^iT -.'ihtvU/
Fe + 2Fe3^ > 3Fe2- q% lOH dfegfror* ,;>fefii JfiSv)
Mol: 0,075 <- 0,15
—> mpe = 0,125.56 = 7,0 gam - > Dap an B
Vi d^ 15: Iron 200ml dung djch X gom Ba(OH)2 0,1M va NaOH 0,3M voi
100ml dung dich Y gom Al2(S04)3 0,1M va H2SO4 0,1M, thu dugc a gam ke't tua Gia tri cua a la
Al(OH)3 + O H " > AlO; + 2H2O
0,02 <- 0,02 ,Ba2^ + S04~ > BaS04>l 0,02 -> 0,02 ^ 0,02
Trang 2C&m nang fln luyjn thi dgi hgc 18 chuy6n dg H6a hoc - MguySn Van H5i
V i d v 16: H o a tan hoan toan 9,46 gam hon h g p gom Na, K va Ba vao nuoc, thu
duQC d u n g dich X va 1,792 l i t k h i H2 (dktc) D u n g dich Y g o m H C l I M va
1^ H2SO4 0,5M T r u n g hoa d u n g dich X b o i d u n g dich Y, tong khoi l u g n g cae
V i d\ 17: Cho h o n h o p g o m Ba va A l (ti le m o l 1:1) vao nuac (du), t h u dugc
d u n g djch X va 1,12 lit k h i H2 (dktc) D u n g dich Y gom H C l 0,5M va H2SO4
0,1M Cho t u t u den het 100ml d u n g dich Y vao X, thu dugc m gam ket tiia
no=2ncooH n o = 2nco2
Vi Dv M A U
V i d v 1 (A-09): K h i d o t chay hoan toan m gam h o n h g p X g o m hai ancol no,
d o n chuc, mach h o thu dugc V l i t k h i CO2 (6 dktc) va a g a m H2O Bieu thu-c
lien h? giiia m , a va V la:
Trang 3dm nang On luygn thi dji hpc 18 chuy6n H6a hpc - IMguySn Van HSi
V i d u 2: D o t chay hoan toan m g a m hon h o p Y g o m ba ancol d o n chuc, thuoc
cung d a y d o n g d i n g , t h u dugc 37,4 g a m k h i C O 2 (dktc) v a 27 g a m H 2 O Gia
V i d y 3: Do't chay hoan toan m o t l u g n g h o n h o p X g o m hai ancol (no, d a chuc,
mach ho, c i i n g so n h o m - O H ) can v u a d i i V l i t k h i O 2 , t h u dugc 5,6 l i t k h i
C O 2 v a 6,3 g a m H 2 O (cac the tich k h i d o 0 dktc) Gia t r j ciia V la a t
A 5,60.' B.3,92 C 7,28 D 1,12
Laigidi: ' r •
5,6 6 3 ' ''^''^.ymm-Mtifbw
yi: ^C02 m o l ; n H 2 0 = 0 , 3 5 m o l ; , ^
Theo bai ra, X chua 2 ancol n o -> nx = n ^ j o " "CO2 " ^ o l •
Cty TNHH MTV DWH Khang Vigt
So nguyen t u C t r u n g binh = "'''"^ = 2,5 X chua m g t ancol d a chuc c6
V i d v 4 (CD-12): D o t chay hoan toan hon hg-p X g o m hai ancol (no, hai chuc,
mach ho) can v u a du V i l i t k h i O2, t h u dugc V 2 l i t k h i C O 2 va a m o l H 2 O
Cac k h i d e u d o 6 dieu kien tieu chuan Bieu thuc lien he giiia cac gia t r i
Do X chua 2 ancol hai chuc > no = noH = 2nx > no = 2a
-Bao toan n g u y e n to O: no (on) + 2 n o 2 = 2 nQQ^ + VXH^Q
2V2 ^ 2 V i _ 2V2
22,4
2 ^ 22,4
-> 2a - + ^=-^ = ^ - ^ + a V i = 2V2 - 11,2a o n
22,4 22,4 22,4 -> Dap an D
Nhan xet: Bai nay cac e m can n h o v o i ancol no t h i : nancoi = n^^Q " "CO2 '
dong t h a i bie't ap d u n g bao toan nguyen to oxi
V i d ^ 5 (B-12): Do't chay hoan toan m gam h o n h g p X g o m hai ancol, t h u dugc
13,44 l i t k h i C O 2 (dktc) va 15,3 gam H 2 O M a t khac, cho m g a m X tac d y n g
v o i N a (du), t h u dugc 4,48 l i t k h i H 2 (dktc) Gia trj cua m la
Trang 4Ca'm nang 6n luy^n thi d<ii hpc 18 chuy6n 6i H6a hgc - Nguyin Van H^i
Bao toan kho'i l u g n g : m x = m c + m n + m o
= 0,6.12 + 2.0,85.1 + 0,4.16 = 15,3 gam
D a p an B
V i d u 6 (B-12): Cho h o n h o p X g o m metanol, etylen glicol v a glixerol Do't chay
hoan toan m gam X t h u dugc 6,72 l i t k h i C O 2 (dktc) CQng m gam X tren cho
tac d u n g v o i N a d u t h u dugc to'i da V l i t k h i H 2 (dktc) Gia t r i a i a V la
V i d u 7: H o n h o p X g o m hai axit cacboxylic d o n chuc D o t chay hoan toan 0,1
m o l X can 0,24 m o l O 2 , t h u dugc C O 2 va 0,2 m o l H 2 O Cong thuc hai axit la
A H C O O H v a C H 3 C O O H i
B C H 2 = C H C O O H v a C 2 H 5 C O O H i o , rm • -f i
C C H 3 C O O H v a C 2 H 5 C O O H , ;
D C H 3 C O O H v a C H 2 = C H C O O H * 3iwb fefl loarifi dG
N h | n thay X chua 2 axit cacboxylic d o n chiic -> chua 2 nguyen t u oxi
+ So' nguyen t u C t r u n g b i n h = = - 5 ^ = 2,4 -> L o a i A ( v i cac axit chua
0,1
so n g u y e n t u C < 2)
- > D a p an D ^ '••m'r\fl•„,,••
V i d\ 8 (B-11): H o n h g p X g o m v i n y l axetat, m e t y l axetat v a etyl fomat D o t
chay hoan toan 3,08 gam X, t h u dugc 2,16 gam H 2 O Phan t r a m so m o l ciia
N h a n thay, k h i d o t chay 1 m o l m o i chat m e t y l axetat v a etyl fomat deu t h u
dugc n c o 2 = n H 2 0 ' " e n g v o i v i n y l axetat t h i : n^^^ - n^^o = 1- v /
Do v^y: nvinyi axetat = r\QQ^ - n H 2 0 0,13 - 0,12 = 0,01 ' "
- > % nvinyi axetat = ^ ^ 1 0 0 % = 25% - ¥ D a p an D , , b f n i J 3 » , , , A
V i d\ 9 (A-11): H o n h g p X g o m axit axetic, axit fomic v a axit oxalic K h i cho m
gam X tac d u n g v o i NaHCOs (du) t h i t h u d u g c 15,68 l i t k h i C O 2 (dktc) M a t khac, d o t chay hoan toan m gam X can 8,96 l i t k h i O 2 (dktc), t h u dugc 35,2
gam C O 2 va v m o l H 2 O Gia t r i cua v la
A 0,2 B.0,3 C.0,6 D.0,8
N/ion xet: K h i cho axit cacboxylic tac d u n g v o i NaHCOa ta l u o n c6:
" C 0 2 = r » C 0 0 H -> no(X) = 2ncooH = 2nco2 " «^
-> no(X) = 2 —'-— = 2.0,7= 1,4 m o l r , y '
22,4 "^^ • ^ Bao toan n g u y e n to O: no(X) + 2 no2 = 2 nco2 '^H20
^ 1 4 + 2 ^ = 2 ^ + n H , o ^ 2,2 = l , 6 + y y = 0 , 6 m o l ' ' 2 2 , 4 44
''•••ft ^
Trang 5dm nang 6n luygn thi d?! hgc 18 chuy6n d j H6a hpc - IMguygn Van Hai
L o t gidi:
Nhan xet: Khi cho axit cacboxylic tac dyng voi NaHCOa ta luon c6: j , 3
J n c 0 2 = " C O O H ' ' •
1 344 no(x) = 2ncooH = 2 ticoj no(x) = 2 = 2.0,06= 0,12 mol
i'r'f, • 22,4
Cty TNHH MTV DWH Khang Vi^t
Bao toan nguyen to O: no (x) + 2 = 2 ncoj + riHjO
0,12 + 2 2,016 ^ 22,4 44 2 4.84 + "Hjo ^ "Hjo = 0/08 mol
mH20= 0,08.18 = 1,44 gam
—> Dap an B
Vi dy 11: Cho m gam hon hg-p X gom hai ancol tac dung v6i Na (du), thu duoc
) 4,48 lit khi H2 Mat khac, dot chay hoan toan m gam X, thu dugc 13,44 lit khi
C O 2 va 16,2 gam H 2 O Cac the tich do 6 dktc Gia tri ciia m la
A 14,5 B 15,4 C.12,2 D 13,8
( • • ' Ldfigiai:
nH2 = 0,2 mol , " ' • • ' - a w N;'"^-" #
I nco2 = 0,6->nc=0,6; D H J O ^ O ' ^ ->nH=l,8 Y ,1 4 ' r i * r r * M > „ i , v
Dya tren moi quan h^ nguyen to-nhom chiic thi voi ancol: no (X) = noH
-> no(X)= noH" 2nH2 = 0,4 mol ••!«?» m-nmi mod «;i,ib fob ,xtM
Bao toan kho'i lugng: m = m(- + m H + mo
^ m = 0,6.12 + 1,8.1 + 0,4.16 = 15,4 gam
Vi dy 12: Hon hgp X gom 2 amino axit no (chi c6 nhom chiic C O O H va
-NH2 trong phan tu), trong do ti 1^ mo : m N = 80:21 De tac dung vua du voi
7,66 gam X can 100 ml dung djch KOH IM Cho 7,66 gam X tac dyng vtra
du voi dung dich HCl, thu dugc dung dich Y Co can Y thu dugc m gam
muoi khan Gia tri ciia m la
A 40,6 B.26,6 C 30,7 D 34,5 ' "
rAv>r^ itT,'.' ".sn-.' Loigidi: .\i life
n = 0,7 mol; n = 0,2 mol H2
Nhan xet: Cac este tao thanh tu cung mgt axit cacboxylic don chuc
Khi ancol tac dung voi Na:
- O H + Na -> - O N a + - H 2 i M v^ j n d a E v
0,2 fa;
Mol: 0,4 noH=2nH2 =0,4 mol
<-M|it khac: n.coo-= " O H = 0,4 mol
"RCOONa=0,4mol
- > nwaOH dir = 0,7 - 0,4 = 0,3 mol
Phan ling voi toi xut (NaOH he't, R C O O N a con du):
R C O O N a + N a O H ) R H + Na2C03 Mol: 0,3 <- 0,3 0,3
Trang 6Ca'm nang 6n luygn thi dgi hpc 18 chuy§n dg H6a hqc - Nguygn Van Hki
V i dy 14: Do't chay hoan toan m gam hon hgp X gom hai ancol, thu dugc 11,2
h't khi C02 (dktc) va 12,6 gam H2O Mat khac, cho m gam X tac dung voi
N a (du), thu dixgc 4,48 h't khi H2 (dktc) Gia tri cua m la m 0 - ,„
no = noH = 2nH2 -> no = 0,4 mol -> mo = 0,4.16 = 6,4 gam
Bao toan kho'i lugng: mx = m^ + mpi + m o = 6,0 + 1,4 + 6,4 = 13,8 ^ j,^ ^
->Dap an D .,
Cty TN TV DVVH Khang Vi§t
CAC AXIT vo CO mm HIND
1 A X I T C L O H I D R I C : H C l
a, Lithuyet
+ Tfnft dung voi kim loai, bazo, oxit baza, muoi V i du: Si n ? f
Fe + 2HC1 > F e C h + H a t ''^^^ ''' CaC03+ 2HC1 > C a C h + C O a t + H2O '^^-'^^'^-^ "'^
+ Tin?z fc^""- Tac dung voi cac chat oxi hoa manh: Mn02, KMn04, KCIO3, K2Cr207 V i d y : ,.,^„„ „ ;„,,.,,„/, ^
M n 0 2 + 4HC1 - ^ - ^ M n C h + CI2 +2H2O " ry^.^ ^^ ^^^^
2KMn04 + 16HC1 > 2KC1 + 2MnCl2 +5CI2 + 8H2O qfi(}
K2Cr207 + 14HC1 2KC1 + 2CrCl3 + 3Cl2 +7H2O C O H /
b V i d u m a u , ^-.•cr^^' rfi ' • ' i - i - '
V i d u l : Cho cac phan ling sau: ,;.^d Au -^iyti^'^^ '-Xji::;
(a) H C l +Mn02 > M n C h +CI2 +2H2O , ,„,f^eH (b) 2HCl + Fe > F e C h +H2 , v,t X: jfefb g n i i ^ X i V t (c) 6HCl + 2A1 > 2AICI3 + 3H2 " ' -W! M i:i ,i/:}c:: i
(d) 16HCl + 2 K M n a > 2KCl + 2MnCl2 + S C h + 8H2O iM>f § ' ?
-Cac phan ling trong do H C l the hi?n tinh oxi hoa la " s f 8 f i
A.(b),(c) B.(a),(b) C (b), (c), (d) D.(a),(d)
• Laigidi:
Nhan xet: Trong cac phan ung (b) va (c), so' oxi hoa cua hidro giam tu +1
(trong H C l ) xuohgO (khi H2) , ,
- > Dap an A
Vi dv 2: Hoa tan hoan toan 7,6 gam hSn hgp hot FesOA va C u trong 200ml dung
djch H C l 1,2M (loang) Sau khi cac phan ung xay ra hoan toan, thu dugc dung dich X (khong chua axit du) Co can X thu du^c m gam muo'i khan
Gia tri cua m la
A 10,39 B 14,20 C 5,16 D 11,10
Lffigidi:
Gpi so mol: Fe304 = a; C u = b ' ^ • ' j Theo bai: 232a + 64b = 7,6 • ?-•<.}( • ••: '^^j^ X :,,; i
^han xet: chi c6 Fe304 phan ung tryc tiep voi axit f| ,* g j :
Trong X khong con axit d u nen Fe304 phan ung vua dii voi H C l : Cac phan ung hoa hpc:
Trang 7Caim nang On luy$n Ihi dgi hgc 18 chuy6n dg H6a hgc - Nguygn VSn Hit
i * • Fe304 + 8HC1 > FeCh + 2FeCh + 2H2O
V i 3: Day g o m cac chat deu tac d u n g dugc v o i d u n g dich H C l loang la
A. KNO3, CaC03, Fe(OH)3 ^ B FeS, BaS04, K O H , ^.^^^
C AgNCte, ( N H 4 ) 2 C 0 3 , CuS D NaHCOs, FeS, CuO ' , , ,
Lai gidi:
Loai A v i KNO3 khong tac dung; loai B, C v i BaS04 va CuS khong tan trong
d u n g d i c h H C l loang ^ •
D a p an D Cac p h u a n g trinh hoa hoc: ^''^ ^ ' i O n M > i
NaHCOa + H C l > N a C l + H2O + CO2 ' '
FeS + 2HC1 > FeCh + H2S
C u O + 2HC1 > C u C h + H2O
V i d u 4: Hoa tan hoan toan 8,55 gam hon hop gom Na, K va Ba vao nuoc, thu
duoc d u n g dich X va 1,792 l i t k h i H2 (dktc) D u n g djch Y gom H C l va
H2SO4, t i 1$ m o l t u o n g u n g la 2:1 T r u n g hoa d u n g djch X boi d u n g dich Y,
tong k h o i l u o n g cac muoi dugc tao ra la
A 13,81 gam B 11,39 gam C 15,23 gam D 19,07 gam
Cty TNHH MTV DVVH Khang Vijt
V i dV 5: Cho 2,13 gam hon hgp X g o m M g , Cu va A l a dang bgt tac d y n g
hoan toan v o i O2 t h u dugc hon hgp Y gom cac oxit c6 khoi l u g n g 3,33 gam
"The tich d u n g d i c h H C l 2 M vua d u de phan u n g het v o i Y la
V i d\ 6: Hoa tan hoan toan 7,8 gam K vao 500ml H C l 0,2M, t h u d u g c k h i H2 va
d u n g dich X Co can X t h u dugc m gam chat ran khan Gia trj ciia m la
't^»^'-K + H2O > K O H + i H 2 t
2 • O H £ ' •
-M o l : • 0,1 0,1 r<^r ,
-> m = m K c i + mKO H = 74,5.0,1 + 56.0,1 = 13,05 gam Dap an B
V i d y 7: Hoa tan hoan toan 8,97 gam k i m loai k i e m M vao SOOnil d u n g djch
H C l 0,2M, t h u d u g c k h i h i d r o va d u n g djch X Co c?n X t h u d u g c 14,73 chat
ran khan Y K i m loai kiem M la
A R b B K C N a D L i
Lai gidi: Jl/ion Ob ^
Cac phan u n g hoa hgc: •< •= , I^^/TU' »• | ,
Trang 8Ca^m nang 6n luy^n thi dgi hqc 18 chuy6n dg H6a hgc - Nguygn VSn Hai
8 9 7
r i M = n H c i + n Q ^ = 0,23 -> M = - ^ = 3 9 ( K ) ^ Dap an B
V i dv 8: Cho m gam hon hop X gom Cu, M g , Fe tac d u n g v o i axit H C l d u , thu
d u g c d u n g dich Y, 448ml k h i (dktc) v a 0,64 gam chat ran C h o d u n g dich
N a O H d u vao Y, loc ket tiia va nung trong khong k h i t o i khoi l u o n g khong
doi, t h u dugc 1,2 gam chat ran Gia t r i ciia m la ,
Luu f K h i n u n g ngoai k h o n g k h i , Fe(OH)2 chuyen thanh Fe(OH)3 v a b i
phan h i i y thanh Fe203
V i du 9: Hoa tan hoan toan hon hgp X gom Fe va M g bang m o t l u g n g vira d i i
d u n g dich H C l 20%, t h u d u g c d u n g dich Y N o n g d o cua FeCk trong Y la
15,76% N o n g d o phan t r a m cua M g C h trong Y la
Cty TNHH MTV DVVH Khang Vift
' do ^'^^ ^^"^ ^ NaC\a K C l v o i HiS04 dac, d u
jChi thoat ra cho hoa tan vao nuoc t h u dugc d u n g d i c h Y Cho b g t Z n d u vao Y t h u dugc 448ml k h i (dktc) K h o i l u g n g N a C l trong X la
^ 0,585 B 1,170 C 1,755 D 2,340 '
Loigidi:
Goi so m o l : N a C l = x; K C l = y Cac p h u o n g t r i n h phan u n g : jsjaCl + H2SO4 — ^ NaHS04 + H C l t '''' ' jCCl + H2SO4 KHSO4 + H C l t
Z n + 2HC1 > Z n C h + H a t " > ' ' • ' ' ^
Bao toan nguyen to: C I " > H C l > -Hi
Ta c6: mx = 58,5x + 74,5y = 2,66 va n H 2 = 0,5(x + y) = 0,02
_» x = 0,02; y = 0,02 -> mNaci = 0,02.58,5 = 1,17 gam D a p an B
V i du 11: Hoa tan het m gam hon hgp M g va MgCOa trong d u n g dich H C l , t h u
dugc 4,48 l i t h o n hgp k h i X (dktc) T i k h o i cua X so v o i H2 la 11,5 Gia t r i cvia
Theo bai: M = 11,5.2 = 23 va a + b = 0,2 mol , ,^ ,
A p d u n g cong thuc cua p h u o n g phap duong cheo, ta c6: ,
"*" Dwn^djc^ H2SO4 (fflc: T i n h oxi hoa manh Ngoai t i n h axit m a n h , axit sunfuric dac con the h i ^ n t i n h oxi hoa m ^ n h , tac
d i i n g d u g c v o i n h i e u k i m loai, h g p chat, : ff;-!;! » {'"-l
Trang 9C^m nang 6n luy$n thi dgi hpc 18 chuySn dg H6a hpc - NguySn VSn HJi
Cu + 2H2S04(^flc) — — > CuS04 +SO2 + 2H2O :/>
2Fe + 6H2SO4 (dac) Fe2(S04)3 + 3SO2 + 6H2O ',' '
2FeO + 4H2SO4 (dac) > Fe2(S04)3 + SO2 +4H2O
2Fe304 + IOH2SO4 (dac) > 3Fe2(S04)3 + SO2 + IOH2O
L u u y: Cac kim loai Al, Fe, Cr khong tac dung vdi Ji2S04 dac, nguQi
Dieu che
So do: Quang pirit FeS2 hoac S ) SO2 — ^ SO3 — ^
' • ,.H,.- »,I '• ,
H2S04.nS03 > (n+1) H2SO4
Cac phan ung: S + O2 > SO2
4FeS2 + I I O 2 — ^ 2Fe203+ 8SO2 •• • "ID :6i fiB'lii^H
2SO2 + O2 2S03 , - x m : o : f i i ,
H2SO4 + nSOs > H2S04.nS03 (Oleum) ^
H2S04.nS03+ n H 2 0 > (n+l)H2S04
b Vi dv mau:
Vi dy 1: Cho day cac chat sau: KBr, S, Si02, FeO, Cu va Fe203 So'chat trong day
CO the bi oxi hoa boi dung djch axit H2SO4 (dac, nong) la
A 4 B.5 ' G.3.c|f: D.2
Lai gidi:
Nhan xet: Axit H2SO4 dac, nong the hien tinh oxi hoa khi tac dung vai chat
CO tinh khu (chiia nguyen to' dang 6 muc oxi hoa tha'p)
2KBr + 3H2SO4 2KHS04 + Br2 + SO2 + 2H2O
S + 2H2SO4 — ^ 3SO2 + 2H2O : '
2 F e O + 4H2SO4 — ^ Fe2(S04)3 + SO2 + 4H2O
Cu + 2H2SO4 CuS04 + SO2 + 2H2O ^ fH',:^::
Dap an A
Luu y: Khi tac dyng voi H2SO4 d^c, FeaOs the hi^n tinh baza:
FeaOs + 3H2SO4 — ^ Fe2(S04)3 + 3 H 2 O
Vi dy 2: Cho oxit ciia kim loai M (hoa tri 2) tac dung vira dii vai dung dich
H2SO4 10% (loang), thu dug-c dung dich muoi c6 nong dp bang 14,45% Kim
Cty TNHH MTV DWH Khang ViSt
Gia thie't dung djch ban dau chiia 1 mol H 2 S O 4 (tiic chiia 98 gam H 2 S O 4 )
100 _> Khoi lupng dung djch H 2 S O 4 = 98 — = 980 gam
^^^^^•^ -=0,1445 - > M = 56 (Fe)-^ Dap an B
980 + M + 16
Vi dy 3: Nung hon hop X gom a mol Fe va 0,015 mol Cu trong khong khi mpt
thoi gian, thu dupe 6,32 gam chat ran Y Hoa tan hoan toan Y bang dung dich H2SO4 dac nong (du), thu dupe 0,672 lit khi SO2 (san pham khu duy nha't 6 dktc) Gia tri ciia a la
A 0,04 B.0,05 C 0,07 D 0,06
Lm gtat: ^
n c r i ^ = — ^ =0,03 mol , , ,, , i , Bao toan khoi lupng: mo2 = my " "^x = 6,32 - 56a - 0,96 = 5,36 - 56a
Nhan xet: ne'u dya theo phuong trinh phan ung se rat dai va kho giai
Cach 1:6 day, cac em can su dyng so do phan ung: t y s j j , : Fe,Cu(l) > Y (2) > Fe^3Cu^M3)
Xet su trao doi electron 6 cac giai doan:
(1) -> (3): Fe -3e > Fe*-^ nenhuong = 3 np^ = 3a OsH »• ;
Cu -2e > Cu*^ nenhiKmg=2ncu =0,03 "^s:
(1) (2): O2 +4e > 2Ct^ ^ nenhan = 4 n o 2 = ^ ^ ^ ^ ^ ^ = 0 , 6 7 - 7 a
(2) (3): S** +2e > SO2 nenh,^n=2nso2 =0,06 Bao toan electron: 3a + 0,03 = 0,67 - 7a + 0,06 ^ a = 0,07 mol -> Dap an C
-CachZ:
Qui doi Y thanh: Fe (a mol); Cu (0,015 mol) va O (b mol)
Bao toan khoi lupng: 56a + 16b = 6,32 - 0,015.64 = 5,36
Bao toan electron: 3a + 2.0,015 = 2b + 2.0,03 a = 0,07; b = 0,09 , , ,
Trang 10im nang 6n luy^n thi dgi hoc 18 chuygn dg H6a hqc - Nguygn Van Hi\
HlS + 3H2SO4 ' > 4SO2 + 4H2O ,
8HI + H2SO4 ^'-^ 4I2 + H2S + 4H2O f-SnO'w! ;oi*:^f '
2Fe304 + IOH2SO4 3Fe2(S04)3 + SO2 + IOH2O , 1/
Dap a n C
Luu y: ^eiOj, AgNOs xay ra phan ung trao doi, Na2SQj khong tr.c dung:
Fe203 + 3H2SO4 Fe2(S04)3 + 3H2O
2AgN03 + H2SO4 Ag2S04i + 2HN03
i d\ 5: Cho 6,08 gam hon hop gom Li, Na v a Ba vao nuoc (du), thu duoc
dung dich X v a 1,344 h't khi H2 (dktc) Dung dich Y gom HCl I M v a H2SO4
0,5M Trung hoa dung dich X boi dung dich Y, tong khoi lugng cac muoi
Nhan xet: n ^ ^ = 2nH2 = 0,12 mol ' ' ' '
Mat khac, nong do HCl gap 2 Ian H2SO4 trong cung mot the ti'ch thi
1 dvi 6: Hoa tan hoan toan 5,28 gam hon hgp bpt Fe304 v a Cu trong 80ml dung
dich H2SO4 I M (loang, vua dii) Sau khi cac phan ung xay ra hoan toan, thu
duoc dung djch X Co can X thu du(?c m gam muoi khan Gia trj ciia m la
A 8,64 B.7,68 C 15,68 D 11,68
Cty TNHH IVITV DVVH Khang Vigt
Lcngiai:
Gpi so mol: Fe304 = a; Cu = b Theo bai: 232a + 64b = 5,28 ; i ;
jV/ian xet: chi CO Fe304 phan ling true Hep voi axit. ji b uAi
Trong X khong con axit du nen Fe304 phan ling vua du voi H2SO4:
Cac phan ung hoa hpc:
Fe304 + 4H2SO4 > FeS04 + Fe2(S04)3 + 4H2O • • '
Vi 7: Cho 3,68 gam hon hop gom A l , Mg va Zn tac dung voi mot luong vua
du dung dich H2SO410%, thu dxxgc dung dich X va 2,24 lit khi H2 (dktc)
Khoi lugng dung dich X la
A 101,68 gam B 88,20 gam C 101,48 gam D 97,80 gam
Lcngiai:
Bao toan nguyen to hidro: nH2S04 = " H 2 = 0,10 mol ''^ " ' ' '
-> Khoi luong dung dich H2SO4 = 0,10.98 ^ = 98 gam
Cu + Fe2(S04)3
M o l : 0,01 - > 0,01
10 Bao toan k h o i l u o n g : 3,68 + 98 = mx + 0,10.2 ^ mx = 101,48 -> Dap a n C
Luu y: Bai n a y cac e m d i q u e n t r u khoi l u o n g k h i H2 b a y ra, v a c h i tinh: mx
= 3,68 + 98 = 101,68 v a se chgn n h a m d a p a n A!
V i 8: Hoa tan hoan toan 6,44 g a m h o n hgp X g o m A l , Fe v a Zn bang mot
l u o n g v u a d u d u n g dich H2SO4 loang, t h u d u g c 2,688 lit H2 (dktc) v a d u n g
Trang 11C^m nang 6n luygn thi dai hgc 18 chuy6n H6a hpc - Nguygn Van H&\
V i du 9: Hoa tan hoan toan 2,81 gam hon hop X gom Fe203, FeO, CuO can
50ml axit H2SO4IM (loang) Khir hoan toan 2,81 gam X bang khi CO (nung
nong) thu dugc m gam kim loai Gia tri ciia m la
A 3,24 B.2,65 C 3,06 D.2,41. a o f l
Nhan xet: Khi cho oxit kim loai tac dung voi axit/.ion 0^~ trong oxit se ke't
hqp voi H* trong axit tao thanh H2O theo phuong trinh: ^ , ^ Ijjjy,
0 2 - + 2H^ > H2O ^ ia,,j^ u S },.0-'K
Mol: 0,025 0,05
Vay: m o = 0,025.16 = 0,4 gam
-> m kim:o?i = 2,81 - 0,40 = 2,41 gam -> Dap an D s^iBi J O A I flj 4-»f
Vi d\ 10: Hoa tan het 2,088 gam FexOy bang dung dich H2SO4 dac, nong (du),
thu dugc dung dich X va 324,8ml khi SO2 (san pham khu duy nhat, 6 dktc)
Co can dung dich X, thu dugc m gam muo'i sunfat khan Gia tri ciia m la
A 5,22 B.5,40 C 5,80 D 4,84
•HV •• ' ' Laigidi:
Nhan xet: Oxit sat FexOy tac dung voi H2SO4 dac, nong -> SO2 thi oxit la FeO
hoac Fe304
Cac em luu y rang 1 mol FeO hoac Fe304 deu chua 1 mol Fe''- nen deu c6
kha nang nhuong 1 mol electron, do do:
Bao toan nguyen to'Fe: 2FeO > Fe2(S04)3 ; p'
mpe2(so4)3= 0/0145.400 = 5,80 gam-> Dap an C , '
3 AXIT NITRIC: HNO3
A L i thuyet
+ Tinh axit manh: Tac dung voi kim loai, bazo, oxit baza, muoi Vi du:
MgO + 2HNO3 > Mg(N03)2 + H2O
CaC03 + 2HN03 > Ca(N03)2 + C O z t + H2O
+ Tinh oxi hoa manh
+ Tac dung voi kim loai: Axit nitric tac dung dugc voi hau het cac kim loai
(tru Au, Pt), ke ca kim loai c6 tinh khu ye'u nhu Cu, Ag,
2 Al, Fe, Cr khong tan trong HNO3 d$c ngupi ' ' , " ' "
Tac dung vai hap chat
3FeO + 10HNO3 > 3Fe(N03)3 + NO + 5H2O '•'^^ rl-y:
3Fe2* + N O ; + 4H* > 3Fe3* + N O + 2H2O £n,;,
FeC03 + 4HN03 > Fe(N03)3 + NO2 + CO2 + 2H2O * c i i j j
FeS2 + I8HNO3 > Fe(N03)3 + 2H2SO4 + I5NO2 + 7H2O
Dieu che
So do: N2 > NH3 > NO > NO2 -4 HNO3
N2 + 3H2 < > 2NH3 , i 4NH3 + 5O2 '""^•'^ > 4 N O + 6H2O 2NO + O2 > 2NO2
4NO2 + O2 + 2H2O > 4HN03
< 'bCl
4y
Vi du 1: Cho 3,024 gam mpt kim loai M tan het trong dung dich HNO3 loang,
thu dugc 940,8 ml khi NxOy (san pham khu duy nhat, a dktc) c6 ti kho'i doi
voi H2 bang 22 Khi NxOy va kim loai M la
Al, Zn) -> Loai D -> Dap an C
Vi dv 2: Hoa tan hoan toan hon hgp gom 0,04 mol FeS2 va a mol CU2S vao axit HlSf03 (vua dii), thu dugc dung dich X (chi chua hai muoi sunfat) va V lit khi duy nhat NO2 (dktc) Gia tri ciia V la
A 13,44 B 17,92 C 20,16 D 22,40
Trang 12im nang 6n luy^n thi d?i hpc 18 chuySn dS H6a hpc - Nguyin Van HSi
Lai gidi:
Nhan xet: cac em l i m y la d u n g dich X chi chiia hai m u o i sunfat —> sau cac
phan ling, S nam he't 6 dang goc sunfat
Ta CO cac so do chuyen hoa:
1 1' t t- tiVK'
FeS2 > -Fe2(S04)3
Mol: 0,04 0,02 ' ' • *
• Cu2S > 2CuS04 ,;<:J3+, sOl-'';; *• ( O M b U ' - >^ Wlh > ,(fWi
Mol: a - > 2a ^aicHS: + ( O v ^ j r ! < > }A}W\ i r
Bao toan nguyen to S, ta c6:2 npeSz + i^Cu2S = " S O 4 A' • >
-> 0,04.2 + a = 0,02.3 + 2a ^ a = 0,02 , , * / i n ur
Cac phan l i n g k h u : v
FeS2 -15e > Fe*3 + 2S^
Cu2S - lOe > 2Cu^2 +
Bao toan electron: L n ^ O j = 15r>FgS2 ^'^'^Cu2S
nN02 = 0 ' 8 m o l - > V ^ o =17,92 lit - > D a p a n B
/ i 3: Hoa tan hoan toan m p t hon hgp g o m hai k i m lo?ii Fe va Cu bang
d u n g djch HNO3 dac nong t h i thu dugc 2,24 l i t k h i m a u nau (dktc) N e u
thay axit H N O 3 bang axit H2SO4 dac, nong, d u t h i the tich k h i SO2 (dktc)
thu dugc sau phan u n g la bao nhieu?
A 3,36 lit B 5,60 lit C 2,24 lit ' D 1,12 lit
Lot gidi:
nN02 = 0/1 m o l J i * V ; • • • M ,
C h a t k h u : Fe - 2e > Fe*3 va Cu - 2e > Cu^^
Chat oxi hoa: N^^ + i e > NO2 hoac +2e > SO2
Bao toan electron: n g = n f j 0 2 = 2nso2 rcitjirf »s" ='
"502 = ^ = 0,05 m o l Vso2 = 0,05.22,4 = 1,12 lit ^ Dap an D
4: H o a tan hoan toan 2,08 g a m hon hop g o m FeS va FeS2 trong d u n g
djch axit HNO3 (dac, d u ) , t h u dupe 5,376 l i t k h i NO2 (dktc) va d u n g dich X
Cho X tac d u n g v a i d u n g djch Ba(OH)2 d u , Ipc ket tiia va n u n g trong khong
khi den khoi l u p n g khong d o i thu dupe m gam chat ran Gia t r i ciia m la
A 8,43 B 8,59 C 10,19.7 D- 6,19
Lai gidi: , > ,
n N 0 2 = 1 ^=0,24 m o l
Gpi so mol: FeS (a mol) va FeS2 (b mol) Ta c6: 88a + 120b = 2,08
Cty TNHH MTV DWH Khang Vi$t
Cac phan l i n g k h u :
peS_9e > Fe*3 + FeS2-15e > Fe^3 + 25"*
Bao toan electron:
nN02 = ^ "FeS +15 npeS2 -> 9a + 15b = 0,24 ^ a = 0,01; b = 0,01
Bao toan nguyen to Fe va S:
FeS Mol: 0,01 FeS2 Mol: 0,01
-> - F e 2 0 3 + BaS04
2 0,005 0,01
'I it ti
y -> -Fe2C)3 + 2BaS04
2 0,005 0,02
t I I I II U ' I I
-> m = 160.0,01 + 0,03.233 = 8,59 -> Dap an B ^.4 >yf > *'l) -a /'
V i d v 5: Cho 1,92 gam Cu vao 60 gam d u n g dich H N O 3 2 1 % , thu dupe V m l k h i
N O (san p h a m khvr d u y nhat) va d u n g dich Y The tich d u n g dich N a O H
I M can thiet de ket tua he't ion Cu^* trong d u n g dich Y la
- > VNaOH = 0,18 lit = 180ml ^ Dap an D
mi y: Cac em de quen phan u n g trung hoa axit ( H * + O H ) va chi tinh:
= 2n^^2+ - 0/06 m o l —> VNaOH = 0,6 lit = 60ml se chpn n h a m A !
V i d\ 6: Cho 19,2 gam k i m loai M (hoa t r i n) tan hoan toan t r o n g d u n g dich
H N O 3 , t h u dupe 4,48 l i t k h i N O (san pham k h u d u y nhat, 6 dktc) K i m loai
n
n = 2 va M = 64 (Cu) Dap an C
83
Trang 13a'm nang an luy^n thi d^i hoc 18 chuyen H6a hgc - Mguyen van Hi'i
/ i d u 7: Nung 2,94 gam hon hgp X gom cac kim loai A l , Zn, Mg trong khi oxi,
sau mot thai gian thu duoc 3,42 gam hSn hap Y Hoa tan hoan toan Y vao
dung dich HNOs (du), thu duoc 2,016 lit khi NO2 (san pham khu duy nhat,
a dktc) So mol HNO3 da phan ung la
A 0,20 B.0,24 C.0,16 D 0,14
' Lotgtat:
Bao toan k h ^ luc^ng: mo^ = 3,42 - 2,94 = 0,48 gam '^^^^ "^'^^^^^
-> no2 = 0,015 mol no= 0,03 mol n ^ 2 ^^^.j^= 0,03 mol ' ' ^
C a c h l : S u d u n g s a d 6 : X > Y ^ Muoinitrat + NO2
Lm y: Y chua ca kim loai (con du) va oxitsgS + s O r l
4—"-+ Khi cho kim loai 4—"-+ HNO3: nj^^_ (muoi) — ngtraodoi— fij^jQ^ — 0,09 mol ^
+ Khi cho oxit + HNOa: 1 goc trong oxit se bi thay the bang 2 goc N O 3 de
tao muoi nitrat, do do: n =2n , = 0,06 mol " «li 1
Bao toan nguyen to N: n H N O s = " N O 3 "^^o " " ^'^'^
-> Dap an B
:ach 2: Qui doi Y thanh X va O (0,03 mol) ' ' *
Bao toan electron: n^^xj^^ng+nj^Q^ =0,15 mol —> n =0,15 mol
3
Bao toan N : n n N O a " N O 3 ^ ^'^^ ~ ^ ^'
n du 8: De nhan biet ba axit dac, nguoi: H C l , H2SO4, HNOs dung rieng bi$t
trong ba lo bi ma't nhan, ta dung thuo'c thu la
A A l B.Fe C C u O " D C u
Laigidi:
Nhan xet: A l , Fe khong tac dung voi H2SO4 va HNO3 dac ngupi -> Loai A
va B Loai C v i C u O tac dung voi c a 3 axit -> dung dich muoi mau xanh
Dap an D. - - n f i ^ - t „,.,'i;.j c ,, y.'
Giai t h i c h : HNOs hoa tan C u v a c6 khi NO2 mau n a u bay ra:
CU + 4HN03 > Cu(N03)2 + 2NO2 + 2H2O
Dung dich H2SO4 d ^ c hoa tan C u va c6 khi khong mau, miii xoc thoat ra:
C u + 2H2S04d,ic > CuS04 + SO2 + 2H2O
1 du 9: Cho 3,6 gam Mg tac dung het voi dung dich HNO3 (du), sinh ra 0,84 lit
khi X (san p h a m khu duy nha't, a dktc) Khi X la
A.N2 B.N2O CNC)2 D N O
Laigidi:
img = 0,15 mol; n x = 0,0375 mol
Cty TNHH MTV D W H Khang Vi?t
each 1: Cac phan u n g : * J tl ^ >' '
8A1 + 3OHNO3 > 8A1(N03)3 + 3N2O + 15H2O ' ^ ' '
A l + 4HN03 > A1(N03)3 + N O + 2H2O • '' ' uu^.x^ •
Dua theo cac phan l i n g , nhan thay: n H N 0 3 = 1 0 n N 2 O + ^^NO '
_ > U H N O S = 10.0,015 + 4.0,01 = 0,19 m o l - > Dap an D Cach 2: Bao toan electron:
3nAi = 8nN20 + 3 n N o = 8.0,015 + 3.0,01 = 0,15 m o l - > nAi = 0,05 m o l
Bao toan nguyen to N : n n N O a 3nAi(N03)3 + 2 n N 2 0 + ^^NO 1
n H N 0 3 = 0,05.3 + 0,015.2 + 0,01 = 0,19 m o l - > Dap an D
V i d^ 11: Cho 27,45 gam hon hgp gom A l , Z n va C u tac d u n g vvra d u v o i
950ml d u n g dich HNO3 2 M , thu dugc dung djch chua m gam m u o i va 4,48 lit hon hgp k h i X (dktc) g o m N O va N2O Ti khoi ciia X so v a i H2 la 18,5 Gia
A\,Zn,Cu ) Muoinitrat + N O + N2O ''^^^
Chat oxi hoa: N^^ + 3e > N O ; 2N^-'* + 8e > N2O
Trang 14:im nang On luygn thi dgi hgc 18 chuyfin 66 H6a hgc - Nguygn Van HSi
—> Dap an B
Nhan xet: Can nhan ra bai toan da "giau d i " san p h a m NH4NO3
Vi dyi 12: Cho manh Cu phan u n g he't v o i d u n g dich HNOa, t h u dugc 0,896 h't
(dktc) hon hgp k h i X (gom N O va NO2) c6 khoi l u g n g bang 1,52 gam Co
can d u n g djch sau phan l i n g thu dugc a gam m u o i khan Gia trj cua a la
A 5,64 B 7,52 C 9,4 , D 15,04
Lai gidi:
Truoc he't cac em can t i m so m o l m o i khi: n ^ o = nN02 ^ ^'^^
Bao toan electron: • '''V:/'- ' "••'••H' "^•1\::^.,iriyP4^-^,' j>fy'{^
2ncu = Sn^o + " N O Z = 0,02.3 + 0.02 = 0,08 m o l - > ncu = 0,04 m o l , f
Bao toan nguyen to Cu: ncu(N03)2 = " c u = 0,04 m o l ; + | A
^ a = 0,04.188 = 7,52 gam Dap an B
du 13: Hoa tan hoan toan 8,4 gam M g bSng d u n g dich HNO3 vua d u , thu
dugc 0,672 lit N2 bay ra 6 dktc va d u n g dich X K h o i l u g n g m u o i khan t h u
dugc k h i c6 can d u n g dich X la
A 51,8 gam B 30,1 gam C 55,8 gam D 15,04 gam
; s ' Lai gidi: '''
HMg = 0,35 m o l ; txj^^ = • ^ ^ = 0,03 m o l i *
Bai toan nay cac e m c6 the giai khi vie't phuong trinh phan ling
Tuy nhien, giai theo phuong phap bao toan electron se nhanh hem
C h a t k h u : M g - 2e Mg^2 - > nenhucmg = 2nMg = 0,7mol ,
Chat oxi hoa: 2N*'' + lOe > N2 — ^ nenh?n = 10n[vj2 = 0,3mol
N h u vay so mol electron trao doi chua bang nhau —> "chua on" 6 day, mot
san pham k h u da dugc "giau d i " , do la su tao thanh m u o i NH4NO3:
Chat oxi hoa: 2N*s + se > NH4NO3 "••'^***W;'yi'«^
.111 '-•••"'Mol:'.-'' 0,4 - > a05 ^'''^''l;:,
Vay: m = mMg(N03)2 + mNH4N03 = 0,35.148 + 0,05.80 = 55,8 gam Dap an C
Lu-u y: N e n ap dung ngay bao toan electron: 2 n^g = 3 n ^ o + 8 nNH4N03 •
i du 14: Hoa tan hoan toan 12,42 gam A l bang d u n g dich HNOa loang (du),
thu d u g c d u n g dich X va 1,344 lit (a dktc) hon hgp k h i Y gom N2O v a N2
Ti khoi cua Y so v o i hidro la 18 Co can X, thu dugc m gam chat ran khan
Gia tri cua m la
Chatkhu: A l - 3e > A P
Chat oxi hoa: 2N*5 + Se > N2O
va CO the xayra qua trinh: 2N*^ + 8e
V i dv 15: H o n hgp X g o m M g , A l va Z n Cho m gam X tac d u n g VvJ-i d u n g dich
H C l d u , t h u dugc 36,45 gam m u o i clorua M a t khac, hoa tan he't m gam X
trong d u n g dich HNO3 d u , t h u dugc 55 gam m u o i nitrat k i m loai Gia t r i
cua m la
A 5,8 B.8,7 C.11,6 D 14,5 ; :
Lai gidi: • , V - \ T - > ' • • • T
Nhan xet: Do goc Or va N O 3 deu c6 dien tich 1 - nen k l i i ket hgp v o i cung
mot l u g n g ion k i m loai giong nhau thi: n^|_ = " N O 3 " ^ ("^^O • * • ^ '
H3PO4 + 2 N a O H H3PO4 + 3 N a O H
mol 1:1 mol 1:2
-> NaH2P04 + H2O -> Na2HP04 + 2H2O mol 1:3
Trang 15Cim nang On luy^n thi dgi hoc 18 chuy6n dg H(5a hpc - Nguygn Van HSi
—> Dap an B •* r^^hnvl • ~ -M -* '•n4i; ;finri:i ©up jvi v?,?: dvlt 65
Lieu y: Cac em can chuyen P2OS thanh H3PO4 va xet ti I f m o l n h u tren
V i dv 2 (B-09): Cho 100ml d u n g djch K O H 1,5M vao 200ml d u n g dich H3PO4
0,5M, t h u dugc d u n g dich X Co can d u n g dich X thu dugc m gam m u o i
khan Gia t r i ciia m la
A 15,5 gam B 18,2 gam C 12,8 gam D 16,4 gam
V i dv 3: Cho 100ml d u n g dich H3PO4 a mol/1 vao 100ml d u n g dich K O H 2 M
thu dugc d u n g dich Y c6 chua 14,95 gam hon hgp m u o i Gia t r i cua a la
Cty TNHH MTV D W H Khang Vi?t
V i dv 4= Cho 21,3 gam P2O5 vao d u n g djch c6 chua a gam N a O H , thu dugc
d u n g dich c6 chua 28,4 gam Na2HP04 va b gam Na3P04 Gia t r i ciia a, b Ian
P2OS + 3H2O > 2H3PO4 nH3P04 = 2np205 = 0,3mol
Bao toan nguyen to P: nH3P04 = nNa2HP04 + nNa3P04 0,3 = 0,2 + —
- > b = 16,4 gam • ,
So do phan u n g : H3PO4 + K O H > M u o i + H2O
Bao toan khoi l u g n g : 0,3.98 + a = 28,4 + b a = 28 gam Dap an C
V i d u 5: Cho 0,05 m o l P2O5 vao dung dich chua 0,4 m o l N a O H Sau k h i phan ung xay ra hoan toan, d u n g dich thu dugc c6 cac chat la
A.Na3P04vaNa2HP04 B Na2HP04, NaH2P04
C Na3P04 va N a O H D H3PO4 va NaH2P04
i,t:.,dMte- Lai gidi: ,,f,„: '
Nhanxet: P2O5 + 3H2O > 2H3PO4 , i Sij, v«x ;^fi!j ut;.a(,j 'MO ul.-' >
Bai 1: Hoa tan het 1,04 gam hon hop X gom M g , Fe bang 40 gam d u n g dich
H C l 7,3% t h u dugc d u n g dich Y va 0,672 lit k h i H2 (dktc) N o n g dp % ciia
H C l trong Y la
A 2,43% B.2,18% C 1,83% D 1,78%
Bai 2: Hoa tan hoan toan h6n hgp X gom Fe va Z n bang mgt l u g n g vua d i i
d u n g dich H C l C%, thu dugc d u n g dich Y N o n g do phan tram ciia FeCh va
Z n C h trong Y Ian l u g t la 8,05% va 8,63% Gia tri cua C la
A 5 B.15 C.20 D 10
Bai 3: Hoa tan hoan toan 7,8 gam h6n hgp gom M g va A l bang d u n g dich H C l
d u Sau phan u n g thu dugc d u n g djch c6 khoi l u g n g tang them 7,0 gam so voi ban dau K h o i l u g n g ciia A l trong hon hgp ban dau la
A 5,40 gam B 2,70 gam C 1,35 gam D 4,05 gam