Lcdgidi: ; Nhan xet: Dung dich Y lam qui tim hoa xanh -> Loai C va D vi cac dung
djch muoi BaCl2 va K2SO4 c6 moi truong trung tinh.
Trong cac phuong an con lai, Y la Na2C03. Mat khac, trgn X voi Y thu dugc ket tua Loai Ạ
Dap an B.
nang On luygn thi dgi hpc 18 chuy8n 66 H6a h(?c - NguySn Van HSi
dv 2 (B-09): Hon hap M gom hai muoi NaX va NaY (X, Y la hai nguyen to c6 trong tu nhien, a hai chu ki lien tiep thuQC nhom VILA, Zx < ZY). H o a tan het 6,03 gam M vao nuac roi cho tac dung voi dung dich AgNOs (du), thu
dugc 8,61 gam ket tuạ Phan tram khoi lugng cua NaX trong hon hgp M la
Ạ 58,2%. B.41,8%. C . 52,8%. D . 47,2%.
Ldigidi:
Neu X ^ V: GQI cong thuc chung ciia hai muói la N a X.
Nhan xet: 1 mol N a X > 1 mol A g X -> khoi luQ-ng tang 108-23 = 85 gam.
Theobai: n ^ < - ' " ' " 8,61-6,03 = 2,58 gam.
= ± ^ ^ M N a X = — - 8 5 = 198,67 ^ X = 175,67 -> Mpt hguyen to c6 85 2,58
nguyen tu khoi > 175,67 —> Loai vi khong c6 trong ty rJiien. * • • Vay X = F - > Y = CỊ Ggi so mol NaF = a; N a C l = b mol. .
Ta CO : 42a + 58,5b = 6,03. ^^^'^ " Phuang trinh hoa hoc: N a C l + AgNOs >• A g C l + NaNOa
8 61 ''''''
-> nAgci = —— = 0,06 mol -> b = 0,06 mol -> a = 0,06 mol W))n\
- > % m N a F = 4 1 , 8 % - ^ D a p a n B .
Luu y: NaF khong tac dung voi AgNOs do AgF la muói tan. ' f'--^^A / i dy 3 (CD-11): H o a tan hoan toan 6,645 gam hon hgp muói clorua cua hai
kim loai kiem thuoc hai chu ki ke tiep nhau vao nuac dugc dung dich X. Cho X tac dung voi dung djch A g N 0 3 (du), thu dugc 18,655 gam két hiạ Hai kim loai kiem tren la
A . R b v a C s . B. N a v a K . C . L i v a N a . D . K v a R b .
Lot giai:
_ 18,655 _ - . - , ' ' - ' Mb iV "^sci - - "'^^ '"ó- i^f^ ri:,ifo ^mm iU-m Xvb
Ggi cong thuc chung ciia hai muói la MCỊ Phan ung hoa hgc : ^
MC\ AgNOs > MNO3 + A g C U ''^'omX
Mol: 0,13 „„,.. • 0,13 •'"•"-"•SJH'S
-> MCI = =51,11 -> M =15,61 -> H a i k i m l o ^ i l a L i v a N a .
';.>niii> ' . f i' v f ' ' V fJ',4 I'ly'i »i,Fit> mt'> V tHrh vtms'O -X':-! mv>"'/:
-> Dap an C .
Vi d\ 4: Cho tung gigt den het 30ml dung dich H C l I M vao 250ml dung dich
NazCOs 0,1M, thay giai phong V ml khi CO2 6 dktc. Gia trj ciia V la Ạ56. , B.112. r D.224. D. 3 3 6 . ' ' '
Cty Jh: 'TV DWH Khang Vi$t
Lot gidi:
nHCi=0'03-l mol; nNa2C03 = 0,1.0,25 = 0,025 mol. ' .V'
t^han xet: Khi nho tung giot H C l thi Na2C03 cQng nhan H C l mgt each "tu tix" theo 2 giai doan:
NazCOs + H C l > NaHCOs + N a C l + H 2 O
Mol: 0,025 0,025 0,025
A tlfj
nna = 0,03 - 0,025 = 0,005 mol; nNaHCOa = 0^025 mol.
N a H C O s + H C l > N a C l + CO2T + H2O
Mol: 0,005 ' ' " ' ' • 0,005 _> Vco2 = 0,005.22,4 = 0,112 lit = 112ml Dap an B.
Luu y: Bai nay cac em de nham khi viet ngay phan ung: , r
Na2C03 + 2HC1 > 2NaCl + C 0 2 t + H2O 3 H S Mol: 0,015 0,03 0,015
Vco2 = 0,015.22,4 = 0,336 lit = 336 ml -> Chgn dap an sai D.
V i  5: Cho 100ml dung dJch X gom K2CO3 0,1M va NaHCOa 0,1M. Nho tung gigt den hét 25ml dung dich H C l I M vao X, thu dugc V ml khi CO2 (dktc). Gia tri ciia V la
Ạ 448. 'M-:m- B.112. C. 224. D. 336.
Ldigidi /
n^^2- = 0,01 mol; n^^^_ = 0,01 mol; n^+ = 0,025 mol.
HCO3
Nhan xet: K h i nho tung gigt axit (H^^) vao X, phan ung theo 2 giai doan:
CO3" + H * > HCO3'
Mol: 0,01 0,01 0,01 "'^
-> di, = 0,015 mol. Xay ra phan ung kétiép:
H C O ; + H^ > CO2 + H2O
Mol: 0,015 < - 0,015 - > 0,015
.iOUíM .1*-"
nco2 = 0,015 mol ^ V c o 2 = 0,336 lit = 336ml -> Dap an D.
V i 6: Hoa tan hoan toan hon hgp gom 0,02 mol FeCh va 0,02 mol N a C l vao mgt lugng nuac (du), thu dugc dung dich X. Cho dung dich AgNOs (du) vao, t h u d u g c m gam chat ran. Gia tri cua m la
Ạ 8,61. B.5,74. C . 10,77. D.6,82. :,
>v Lot gidi:
Nhan xet: X gom: Na" = 0,02 mol; Fê* = 0,02 mol; C h = 0,06 mol.
Cac phan ung hoa hgc:
elm nang 6n luyjn thi dgi hpc 18 chuy6n 66 H6a hpc - Nguygn Van HSi
Ag^ + CI- > A g C l ^,^.,,^„ ạ!, , : . M o l : 0,06 - > 0,06 „': "
Ag^ + Fê^ >Fe3*+ A g _ . . > ụ-n^
M o l : 0,02 0,02
m = mAgci + mAg = 0,06.143,5 + 0,02.108 = 10,77 gam - > Dap an C.
Lieuy: Fê*c6 the k h u A g * thanh A g . , , y :,rn r . .
V i d\ 7: Cho 8,95 g a m hon hop X gom m u o i cacbonat va hidrocacbonat cua
k i m loai k i e m M tac d y n g het v o i d u n g dich H C l (du), sinh ra 2,24 lit khi (dktc). K i m loai M la
Ạ L i . B . N a . C. K. ' " ' - f ^ ' D . Rb. '
Lcri gidi: n<r r> h rrít •<
M2CO3 + 2HC1 > 2MC1 + CCht + H2O ~03iBl/i
M H C O 3 + H C l *• M C I + CCht + H 2 O
Nhan xet: nx = n ^ o , = 0/1 m o l . . , 0 ' I1SvO*„';oV-4-
M x = - ^ = 89,5 M H C O 3 < 89,5 < M 2 C O 3 . , , , K.;,
-> M + 61< 89,5 < 2 M + 60 ^ 14,75 < M < 27,5 M la Nạ ' —> Dap an B.
2. T A C D V N G V 6 I K I M L O A I x:-i- ^h^^^^-^-im-.
ạ L i thuyet
+ Cac k i m loai m ^ n h h o n c6 the " d a y " dugc k i m loai yeu h o n ra khoi d u n g d j c h m u o i : '^E .^•••f-• .•'••{7.
Fe + CuS04 > FeS04 + Cu f t )0 |(!j) ^ , M C u + 2 A g N 0 3 > Cu(N03)2 + 2 A g J,: r; f
+ Rieng k i m loai n h o m l A va I I A ( t r u Be, M g ) se tac d u n g v o i nuoc trong d u n g dich m u o i bazo, bazo sau do c6 the tac d u n g tiep v o i m u o i .
Vi du: cho N a k i m loai vao d u n g d}ch CuS04 xay ra cac p h u o n g t r i n h :
2Na + 2 H 2 O > 2 N a O H + H 2
2 N a O H + CuS04 > Na2S04 + Cu(OH) 2 4' '•^^ *:
b. V i d v m a u . ^ ^hi
V i dy 1: N h i i n g m p t la k i m loai M (hoa trj 2) c6 khoi luqmg 5 gam vao 200ml
d u n g dich A g N 0 3 0,1M cho deh k h i phan l i n g hoan toan t h u duoc chat ran X va d u n g djch Y chua 1,88 gam m u o i . K i m loai M la
^ A . M g B . C u C.Fe D . Z n
Lai gidi: Nhan xet: • :.
Trong cac dap an, k i m loai c6 k h o i lug-ng m o l Ion nhat la Z n
Cty TNHH MTV DWH Khang Vi^t
^ A M > n A g N 0 3 ^ AgNOa p h a n ung het. ^ . ,^ Phan u n g hoa h<?c: I-MHI GJ-) .CRS^: SVỤV: :»t.«.:?,:. rre:j nui^ -iv,,.
M + 2AgN03 > M(N03)2 + 2Ag -an.^'; - r ! mol: 0,01 < - 0,02 ^ 0,01
M M(N0 3 ) 2 = ^ = 188 ^ M = 64 (Cu) ^ Dap an B.
V i dv 2: Cho hon hgp gom Fe va Z n vao d u n g dich A g N O den k h i cac phan
l i n g xay ra hoan toan, thu duoc d u n g dich X gom hai m u o i va chat ran Y gom hai k i m loaị H a i m u o i trong X la ^. , ,
Ạ Fe(N03)3 va Zn(N03)2. ^ B. Zn(N03)2 va Fe(N03)2. V !
C A g N 0 3 vaZn(N03)2. ' D . Fe(N03)2 va AgNOs. '
Lai gidi:
Nhan xet: Y gom 2 k i m loai —> Y chua Z n hoac Fe mpt k i m loai con d u —> AgNOs het Loai C va D.
Tinh k h u Z n > Fe Z n phan ung truoc roi mod den Fẹ
Gia su Z n con d u Fe con nguyen va Cu b i " d a y " ra het chát ran X gom 3 k i m loai dieu gia su la saị ! 0 „ U ) : =
Z n phan u n g het va Fe phan u n g mot phan - > Y gom Fe d u va Cụ
-» H a i m u o i trong X la Zn(N03)2 va Fe(N03)2 Dap an B.
V i dv 3 (CD-09): Cho m gam A l vao 100ml d u n g dich g o m Cu(N03)2 0,3M va
AgN03 0,3M. Sau k h i cac phan u n g xay ra hoan toan, t h u dupe chat ran X.
Cho X tac d u n g v o i d u n g djch H C l (du), t h u dupe 0,336 l i t k h i (dktc). Gia t r i c u a m l a :
Ạ 8,10. B. 1,08. C. 0,54. D . 2,7C. \
Lai gidi: ,
" A g N 0 3 = 0,03 m o l ; ncu(N03)2 = 0,03 m o l ; n H j = 0,015.
C a c h l : Nhan xet: Cho X + H C l giai phong H 2 ^ X chua A l -> A l con d u .
A l + 3AgN03 > A1(N03)3 + 3 A g
2A1 + 3Cu(N03)2 > 2A1(N03)3 + 3Cu
2 A 1 + 6HC1 > 2 A I C I 3 + 3 H 2
_ " A g N 0 3 ^ ^"^"(N03)2 ^ ^ ^ 0 04 = 1^08 gam 3 3 3
-> Dap an B.
Cach 2:
Nhan xet: A l tac d u n g v o i 3 chat oxi hoa la Ag*, Cu^* va H * .
Bao toan electron: 3 n A i = In^^^ + 2n^^2+ + 1"^+ =^ ^ ^ ^'^^ "
- > nAi = 0,04 m o l m A i = 1,08 gam Dap an B. *