X gom: Mg3N2 =0,1 mol; Mg= 0,2 mol
Rieng cac kim loai nhom lA va IIA (trir Be, Mg) se tac dung voi nuoc trong dung dich > Baza, baza tao thanh c6 the tac dung tiep vai muo'i.
dung dich -> Baza, baza tao thanh c6 the tac dung tiep vai muo'i.
dm nang On luy?n thi d^i hgc 18 chuyfin dg H6a hqc - Nguygn Van HSi
Vi du: Cho Na kim loai vao dung dich CuS04 xay ra cac phuong trinh: 2Na + 2H2O > 2NaOH + H2T , , ,
2NaOH + CuS04 > NaiSO* + Cu(OH)2i v i DV M A U
V i du 1: Cho 1,12 gam kim loai M (hoa trj n) tac dyng het vai khi clo, thu du^c 3,25 gam muoi clorua. Kim lo^i M la * < - .r ,
A.Cu. B.Mg. C.Ca. ' D.Fe. ,.r,/._,
man xet: Bao toan khoi lug-ng: fo^^j^ « « i ^ „ v | f c ;
mci2 = 3,25-1,12 = 2,13 gam-> nci2 = ^ = 0,03 mol. • '
71 Phuong trinh hoa hpc: 2M + nCh
2 0,6 1,12 -> n^= —nQ2 =• -> n^= —nQ2 =• -4 2MCln 56n n n 0,06/n 3 -> n = 3 va M = 56 (Fe) Dap an D.
V i dy 2: Cho 7,84 gam hon hgp gom Fe304 va Cu vao dung djch HCl du. Sau khi cac phan ung xay ra hoan toan, thu dugc dung dich X chiia m gam muoi va cha't ran Y. Hoa tan het Y trong dung dich HNOa, sinh ra 0,336
lit khi NCh (san pham khii duy nhat, 6 dktc). Gia tri cua m la * 8 A. 7,78. B.7,62. C. 10,32. D. 14,37. l<
Laigidi: .,a^yi,riui ,,§1;/
Phan ling hoa hpc: Fe304 + 8HC1 Mol: a
2FeCl3 + Cu - Mol: 2a a
-> 2FeCl3 + FeCh + 4H2O
2a a 2FeCl2 + CuCh
2a a
•6v gni/fe 'X
u:
Chat ran Y la Cu du. Phan ung hoa tan Cu bang HNO3 loang:
Cu + 4HN03 > Cu(N03)2 + 2NO2 + 2H2O
Mol: 0,03 <- 0,015 -> 232a + 64(a + 0,03) = 7,84 gam -» a = 0,02 mol.
Vay: m = 127.3.0,02 +135.0,02 = 10,32 gam - > Dap an C.
V i dy 3: Cac chat vua tac dyng vai dung dich HCl, vua tac dyng du(?c vo' dung djch AgNOa la: * ; > ' :
A. MgO, Na, Ca. B. Zn, Ni, Fe. C. Cu, Fe, Ag. D. CuO, Mg, Be.
- . n . , * . rh> - Laigidi:
Nhqn xet: Loai C vi Cu khong tac dyng voi dung djch HCl.
Cty IMHH MTV DWH Khang Vijt
Loai A va D vi MgO va CuO khong tac dung voi dung dich AgNOa
—> Dap an B. Cac phuong trinh hoa hpc (M = Zn, Ni, Fe): ' (adsbygoogle = window.adsbygoogle || []).push({});
M + 2HC1 > MCh + H2 i:,,: ,.
M + 2AgN03 > M(N03)2 + 2Ag v =
Vi dy 4: Hoa tan hoan toan hon hgp X gom Mg va Zn bang mot hrgng vua du
dung djch H2SO410%, thu dugc dung dich Y. Nong dg cua MgS04 trong Y
la 3,89%. Nong dg phan tram ciia ZnS04 trong Y la
A. 10,43%. B. 12,84%. C. 18,21%. D. 15,42%.
Lai gidi:
Phan ung hoa hgc: ^ ^ ^ ^
Mg + H2SO4 >• MgS04 + H2
Zn + H2SO4 > ZnS04 + H2 "go^^ i
Xet voi 100 gam dung dich Y: -> mMgso4 = 3,89 gam. j , , ^
^ mznS04 + n^HjO = 100 - 3,89 = 96,11 gam.
Nhan xet: Lugng H2O trong Y cung chinh la lugng H2O c6 trong dung dich
H2SO4 ban dau.
M§t khac, do nong dg H2SO4 bang 10% -> mHjO = 9mH2S04 aj->£,
mMgS04+ 9mH2S04 =96,11. < - ^KU) £0,0 : i '
Bao toan nguyen to'S: < ' * viS
3,89 1 1 •)•;/;
^ + mznso4-•^mH2S04 m:'
"MgS04 + ^Zx&Oi - "H2SO4
120 161
mznS04 = ^^'^^ C%(ZnS04) = 10,43% Dap an A. ' i ; " '
Vi d\ 5: Cho 2,6 gam hon hgp X gom ba kim loai Mg, Cu va AI 0 dang bgt tac dung hoan toan vai oxi (du) thu dugc 4,2 gam chat ran Y. The tich dung djch HCl 2M vua dii de phan ung het vai Y la:
A. 60ml. B.50ml. C. 80ml. D. 100ml.
Lai gidi:
Nhan xet: Bao toan khoi lugng mo2 = 4,2 - 2,6 = 1,6 gam '
n^2- =0,1 mol.
Vay: no = =0,1 mol
Khi cho oxit kim loai tac dung vai axit, ion O^- trong oxit se ket hgp voi H*
trong axit tao thanh H2O: (adsbygoogle = window.adsbygoogle || []).push({});
02- + 2H^ > H2O ^.
Mol: 0,1 0,2
0,2
Vay: nHci = n^^^ = 0,2 mol -> VHCI = - T - = 0,1 lit = 100ml ^> Dap an D.
Ca'm naiig oil luyeii thi dai hgc 18 chuy6n dg Hlia hpc - Npuyin Van Hai
V i dv 6: H o a tan hoan toan 1,12 gam bgt Fe t r o n g 100ml dung d}ch gom H N 0 3 0,6M va H2SO4 0,5M thu dvtcyc dung djch X. Dung djch X c6 the tac
dung voi toi da m gam bpt C u . Biet khi N O la san pham k h u duy nhat
ciia N*'' trong cac qua trinh tren. Gia tri cua m la ^ , A . 1,92. B.2,56. C . 0,64. D . 1,28. , !
Lot Stat: ; ,
112 * „
npg=-^^—=0,02mol. . d'AJ^tttS s;.s,,;'j iTirnf rr"''-' '
Nhan xet: K h i axit HNO3 c6 mat dong thoi voi axit H2SO4 loang thi lugng trong dung dich la do 2 axit phan li ra —> Giai theo phuong trinh ion.
n „ + = "HNOg + 2nH2S04 = 0,06 + 2.0,05 = 0,16 mol
V t r ^ -= mol; n „ ^ 2 - = 0/05 mol. , , v v r t
Trong X:
Phuong trinh ion rut gQn: 'g(M.mvi&-
Fe + 4H* + N O J >• Fe^* + N O + 2H2O r n 4-
: Mol: 0,02 < - 0 , 0 8 - > 0,02 - > 0,02.. ! t »
Khi cho X tac dung voi C u : * '
3Cu + 8 H - + 2 N O i > 3Cu2^ + 2 N O + 4H2O
Mol: 0,03 <r- 0,08 - > 0,02 - > 0,03
C u + 2Fe3* > Cu^* + 2Fe2+
Mol: 0,01 < - 0,02
- > m = 0,04.64 = 2,56 gam - > Dap an B.
V i dv 7: Hoa tan hoan toan 3,68 gam hon hop gom Fe va kim loai M (hoa tri 2) trong dung dich H2SO4 loang du, thu dugc 2,24 lit khi H2 (dktc). Mat khac, de hoa tan hoan toan 2,4 gam M can dung chua den 250ml dung dich H C l
I M . K i m loai M la '! A. Be. B. C a . C . Z n . D. Mg. ,
.„„ ^, Lcngiai:
Cac phuong trinh phan ung: Fe + H2SO4 > FeS04 + H2
M + H2SO4 > MSO4 + H2 ' " ' -X^'"'
Nhanthay:nFe + n M = n H 2 = a i m o l « x i . . 4 , M ; ' (adsbygoogle = window.adsbygoogle || []).push({});
-> M = — = 36,8 -> MM<36,8< Mpe ^ LoaiB v a C . Mat khac: M x > - ^ = 1 9 , 2 -> Loai A.
0,125
-> Dap an D. '
0
17A
CtyTNnH 1TV DWH Khang Vi?t
yji du 8: Hoa tan hoan toan 3,22 gam hon hop X gom Fe, M g va Z n bang dung dich H2SO4 loang, thu dugc 1,344 lit H2 (dktc) va dung djch chua m gam muoi. Gia tri cua m la ^ ^ , ,, g , ,
A. 9,52. B. 10,27. C . 8,98. ' D . 7,25. ' ' ,r
]slhan xet: n ^ = nR^soA = "^H, = 0,06 mol.
^ ^ ^ ii'i voa. ••'
Mlitkhac:
1 mol kim loai M > MSO4 Khoi lugng tang 96 gam. s ' 0,06 Khoi lugng tang 0,06.96. ^ .j 0,06 Khoi lugng tang 0,06.96. ^ .j
m = 3,22 + 0,06.96 = 8,98 ^ Dap an C . _ 7
Vi du 9: Hon hgp khi X gom clo va oxi. Cho X phan ung v u a het voi hon hgp gom 0,36 gam magie va 0,27 gam nhom tao ra 2,21 gam cha't ran. Phan tram ve the tich cua khi clo trong X la
A. 50%. B.40%. C.80%. D . 60%.
Lai gidi:
HMg ~ = 0,015 mol; " A I = ^ = 0,01 mol.
N/ifln xet Day la bai toan hai chat khu (2 kim loai) va hai chat oxi hoa (2 phi
kim) nen can ap dung dinh luat bao toan electron.
Bao toan khoi lugng: m 2 kh,- = 2,21 - 0,36 - 0,27 = 1,58 gam ^ 71x + 32y = 1,58. Bao toan electron: 2nMg + 3nAi = 2nci2 + 4no2 " "
-> 2.0,015 + 3.0,01 = 2x + 4y ^ X = 0,02 mol; y = 0,005 mol. jjerc.ai'.. 1
-> % V r u = - ^ ^ . 1 0 0 % = 80%-> Dap an C .
Vi d\ 10: H o n hgp X gom Z n va S. D u n nong 3,24 gam X (trong dieu kien khong CO khong khi), thu dugc chat ran Y . Cho Y tac dung voi dung dich H C l du, thu dugc 0,16 gam chat ran va 896ml hon hgp khi (dktc). H i ? u suat cua phan ling giua Z n v a S la -vr > A. 75%. B. 50%. C . 80%. D . 85%.
Ung^dii ,.-^>•':>: ,
Nhanxet: Z n -> Hz ; .8 rife q6<3 •
•Zn -> ZnS H2S • fM nfiJ 6oH,:El:
3,24-0,04.65 ^ — - ^ ' ' ' ' ' ^ ^ . f i - ' V£ly: n = nH2 + H H J S = 0,04 mol; n 5= 0,02 mol.
nsdu= = a 0 0 5 m o l nsph4nung = a015mol.^ ^^^'"^
H = Ml^.100% = 75% -> Dap an A. cn Q « J?I: , , , ,
dm nang On luygn thi d^i hpc 18 chuy8n d6 H6a hgc - Nguyjn van HSi (adsbygoogle = window.adsbygoogle || []).push({});
V i d\ 11: Hoa tan hoan toan 8,8 gam hon hop X gom Zn, Fe, C u trong lugng dij. axit H2SO4 dac, nong, thu dugc dung dich Y va 4,48 lit khi SC2 (san phaVti
khu duy nhat, a dktc). Co c^n Y thu dugc m gam muoi sunfat khan. Gia tri
cua m la
A. 28,4. B.24,0. , ; C. 18,6. D. 28,C.
_ 4 , 4 8 _ . „ , :^mMif.l - 22,4 ~ ' '''-"^^ • ' - '• - ^ ^«Pi maion-
Cach 1: Goi cong thuc chung cua cac kim loai la M .
2M + 2nH2S04(^flc) — ^ M2(S04)n + n S 0 2 + 2nH20,^ ^ g
Mol: 0,4 <- 0,2 -> 0,4,'^^o*„,;:
Bao toan khoi lugng: 8,8 + 98.0,4 = m + Q,1M + 18.a4 jjj fbh gdi I
m = 28 gam -> Dap an D. 3 .g .,^06
NMn xet: n 2- ("^"6') = "soo = 0/2 mol -> m = 8,8 + 0,2.96 = 28 gam. .^^ - ^ D a p a n D . ,., , . ... ..^ ^v'.. : ' V ' . ; , r , , . y; , . > 7 ^
V i d\ 12: Hoa tan hoan toan 2,8 gam hon hgp X gom C u va A g bang dung dich H2SO4 dac nong, toan bg khi tag thanh dugc sue vac nuoc brom (du), thu dugc dung dich Y . Cho dung dich BaCh d u vac Y, thu dugc 4,66 gam ket tua. Thanh phan phan tram khoi lugng ciia C u trong X la
A. 45,71%. B. 22,86%. C . 68,57%. D. 91,43%.^*
Laigiai: t'i -4-
Ggi so'mol: C u = x; A g = y. Ta c6: 64x + 108y = 2,8.
Bao toan nguyen to S theo so do: ^ ' ' •'"'''^ H''"' ^ - ^ ^ "^^