Thi du 4: Ddt chay hoan toan 4,64 gam mdt hidrocacbon X chat khi d di^u kien thudng rdi dem toan b6 san ph^m chay ha'p thu het vao binh dung dung dich Cty TNHH MTV DWH Khang Vigt BaOH2
Trang 1CU THANH TOAN
\iibm nddl duh KI-:' '.,.>'
CAM NANG ON LUYEN THI DAI HOC
• • • R^n luiien ^ nhanh thi
HOA HOC
^^M |Ml^^lilHRI H$thdngcacphuongphapgi^ nhanh bai tap hoa hoc X
Wn chon va gifli thieu cac d6 thi thii nam 2013, cac dfi thi thii ciia cac truong THPT, cac truong Chujen va cac Trung tarn liij^n thi dai hoc uij tm ctia 3 mien B^c - Trung - Nam
cac detf«d& dupe gi^ctitiet,dehfduvatheo cac phuongphapgi^i nhanh X
: i 4
H u u A V I I S ' T D A M TKUR urtD T U A N U DUffHf! H H f M I N H
Trang 2LOI N O I D A U
Cac em hoc sinh than me'n!
Chiing t6i xin trSn trong gidi thiSu vdri cdc em tap sach: Cam nang on
luyen thi Dai hoc 3 mien Bdc - Trung - Nam Hoa hoc
Tap sach la tai lieu cap nhat cho cac em cac bo de thi thir Dai hoc, Cao
dang tren ba mi^n Bac - Trung - Nam de cac em tu ren luyen ky nang lam
nhanh bai thi cho minh Cac de thi deu dugc chiing toi tuyen chon k i cang,
noi dung bam sat chaong trinh thi Dai hoc, Cao dang va theo dung ca'u true
de thi ciia B6 GD - DT M 6 i bai tap d^u dugc giai chi tiet, de hi^u, theo
nhieu each va dac biet c6 cac phuong phap giai nhanh, d6c dao
Tap sach g6m hai phan:
- Ph&i I Cac phuong phap giai nhanh bai tap hoa hoc
- PhSn I I Cac b6 d^ thi thit Dai hoc va hudng dan giai chi tie't
Chiing toi tin tucmg rang tap sach se cap nhat cho cac em day dii cac dang
de thi tuyen sinh Dai hoc, Cao dang theo hudmg ra de thi cua Bg GD - DT,
trang bi dSy du cho cac em toan bg kie'n thiic hoa hoc Trung hgc phd thong
va quan trgng hon la mang lai cho cac em su tu tin trong cac ky thi sap tdi
Dd cuon sach hoan thien hon, rat mong nhan dugc su dong gop y kien
chan thanh ciia cac ban dong nghiep va ciia cac em hgc sinh
X i n tran trgng cam on!
TAG GIA
Nhd Sdch Khang Viet xin trdn trgng gioi thieu toi Quy dgc gid va
xin Idng nghe mgi y kien dong gop de cuon sdch ngdy cdng hay han, bo
ich han.Thu xin gui ve: ^ ,1
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Fax:(08) 39110880 ^
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cty TNHH MTV DWH Khang Vi$t
Phan 1. CAC PHCWUfG P H A P GIAI imAIUI
- Ap dung: Trong mot phan ling, c6 n chat (ke ca cha't phan ling va san phdm), n€u
biet khdi lirong ciia (n - 1) chat thi tinh dugc khoi lirgng ciia cha't con lai
II VAN DUNG ^A
Thi du 1: H6n hop X gom 0,15 mol vinylaxetilen va 0,6 mol H2 Nung nong h6n hop X (xiic tac Ni) mot thcfi gian, thu dugc h6n hop Y c6 ti kh6 'i so voi H, bang
10 DSn h6n hgp Y qua dung dich brom du, sau khi phan ling xay ra hoan toan,
so mol brom tham gia phan ling la
A 0,1 mol B 0,15 mol C 0,05 mol D 0,2 mol
(Trick dethi thu Dai hoc khdi A, B nam 2013}
Hu&ngddngidi
H6n hgp X c6: S6'lien ket 7t = 0,15.3 = 0,45 (mol) Kh6ilugng: m^ =0,15.52 + 0,6.2 = 9(gam) * Sdmol: nx =0,15 + 0,6 = 0,75(mol)
Soddphanurng: X(C4H4,H2)-^Y Theo dinh luat bao toan khoi lugng => my = m^ = 9(g) ,
=onY =9/(10.2) = 0,45(mol)
=>n,,h,ii,. =0,75-0,45 = 0,3(mol) = nH2 (phan umg)
=>n„ (con lai) =0,45-0,3 = 0,15(mol) = nB,^ ^ , Dap an diing la B
Thi du 2: H6n hgp X g6m m6t hidrocacbon o the' khi va H (ti khd'i hoi ciia X so v6i H, bang 4,8) Cho X di qua Ni dun nong den phan iJng hoan toan, thu dugc h6n hgp Y (ti kh6'i hoi ciia Y so v6i CH4 bang 1) C6ng thiic phan tir cua hidrocacbon la: ||; || , (,
Trang 3Ca'm nang fln luy$n thi DH 3 mign BSc - Trmg - Nam mOn H6a hpc - Cu Thanh To^n
Phan ting: x -> kx ^ x (mol)
Thidu 3: Nung nong m6t h6n hop g6m CaCO^ va MgO tori khd'i luong khong ddi,
thi so gam cha't rSn con lai chi bang 2/3 so gam hdn hop trudc khi nung Vay
phdn tram theo khd'i luong ciia CaCO, trong hdn hop ban ddu la
A 75,76% B 24,24% C 66,67% D 33,33%
(Trich de thi thuDai hpc khoi A, B nam 2013) Hu&ng ddn gidi
Gia six ban d^u c6 100 gam hdn hop CaC03,MgO
=> Khd'i luong hdn hop con lai sau khi nung la: 100.2/3 = 200/3(g)
C a C O , — ^ C a O + C02 t
Theo djnh luat bao toan khd'i luong:
mco2 = 100-200/3 = 100/3(gam)^nco2 = 100/132(mol)
^ ncaC03 = 100/132(mol) = nco2 mcaco., = 10000/132(g) |
vay %CaC03 = 10000/132 = 75,76%
Dap an diing la A
Thi du 4: Ddt chay hoan toan 4,64 gam mdt hidrocacbon X (chat khi d di^u kien
thudng) rdi dem toan b6 san ph^m chay ha'p thu het vao binh dung dung dich
Cty TNHH MTV DWH Khang Vigt
Ba(OH)2- Sau cac phan ung thu duoc 39,4 gam ke't tiia va khdi lucmg phSn
dung djch giam bdt 19,912 gam C6ng thurc phan i\i ciia X la: / , ,
A CH4 B C4H,o C C2H4 D C3H4
(Trich de thi thADai hoc khdi A, B ndm 2013) Hu&ng ddn gidi
So dd cac phan ling xay ra:
CO2 ddBa(OH)2 fBaCOj i
=> mo2 = 19,488 - 4,64 = 14,848(g) => no2 = 0,464(mol)
x + - O2 ->xC02+^H20
4
ijjBflfji i r i i fi6b oflc) (}f>'J,b
4,64 12x + y > 0,464
Dap an dung la D
Thi du 5: Hdn hop X gdm 0,15 mol CH^C—CH=CH, va 0,6 mol hidro Nung
nong hdn hop X (xiic tac Ni) mot thdi gian, thu dugc hdn hop Y cd ti khd'i so vdi khi hidro bang 10 DSn hdn hop Y qua dung dich brom du, sau khi phan ung xay
ra hoan toan, khd'i luong brom tham gia phan ufng la
A Ogam B 24 gam C 8 gam D 16gam •
(Trich de thi thi Dai hoc khoi A, B nam 2013) Hu&ng ddn gidi ,j »,)»
Xet hdn hop X c6: n^^^) =0,15.3 = 0,45(mol) /, ,J,H,h nk qpO
Trang 4C^m nang On luy?n thi BH 3 mign B&c - Trung - Nam mfln H6a hpc - Cu Thanh ToAn
m x = 0,15.52 +0,6.2 = 9 ( g a m ) ; n x =0,75(mol) ut.c
Sodo: X - ^ Y I K>:i'IJA.;^ ri;;!o j{r»,/>
Theo djnh luat bao toan khdi lucfng ta c6: m y = = 9(gam) j^^"' '
n^ (conlai) =11^^^ (phan ling)
Thi du 6: Dun nong m gam h6n hop X g6m cac chat c6 cung m6t loai nhom chiic
v6i 600 ml dung dich NaOH 1,15M, thu duoc dung dich Y chiia mu6'i cua mot
axit cacboxylic don chiic va 15,4 gam hoi Z gom cac ancol Cho toan bo Z tac
dung vdri Na du, thu duoc 5,04 lit khi (dktc) Co can dung dich Y, nung nong
.cha't ran thu dugc vai CaO cho den khi phan u-ng xay ra hoan toan, thu dugc 7,2
gam m6t chat khi Gia tri ciia m la
A 40,60 B 22,60 C 34,30 D 34,51
, (Trich de thi tuyen sinh DH khdi B)
Huong dan gidi
Theo bai ra: nf^^Q^ =0,69mol;nH2 =0,225mol
So d6: X + NaOH -> Y + Z
"NaOH (phanurng) =2.nH2 =0,45mol
=^nNoOH ('li') = 0 - 6 9 - 0 , 4 5 = 0,24mol
RCOONa + NaOH > R H T +Na2C0,
Cty TMHH MIV JVVH Khang Vigt
Thi du 7: Dot chay hoan toan x gam h6n hop g6m hai axit cacboxylic hai chiic,
mach ho va deu c6 mot lien ket d6i C=C trong phan tir, thu dugc V lit khi CO2
(dktc) va y mol H , 0 Biau thurc lien he giua cac gia tri x, y va V la
Theo dinh luat bao toan khoi lugng, ta c6:
x + l,5y.32 = — 4 4 + 18y z=> x + 30y = 44V/22,4
22,4 22,4.(x + 30y) 28 • : ' _ ^ V = iZ = _ ( x + 30y)
44 55 ^ ^ Dap an dung la C ,
Thi du 8: Cho 3,68 gam h6n hgp g6m A l va Zn tac dung vdri m6t lirgng vijfa dii
dung dich H:S04 10%, thu dugc 2,24 lit khi H (6 dktc) Khdi lugng dung dich thu dugc sau phan ling la
A 101,48 gam B 101,68 gam C 97,80 gam D 88,20 gam
(Trich de thi tuyen sinh DH - CD khdi A) Hu&ng ddn gidi
2 24 Theo bai ra: = =0.1 (mol) => n H 2 S 0 4 = " H 2 = ^ ' 1 ('"O')
0,1.98.100
^dd H2SO4 = ^ = 98 (gam)
So do phan urng: K i m loai ( A l , Zn) + H2S04(1) > Mu6i + Hjt 4
Theo dinh luat bao toan khdi lugng:
mkin, loai + m ^ d H2SO4 = "^ddmM + m H 2
3,68 + 98 = m , , „ „ , i + 0,1.2
= > m , , , „ , i =3,68 + 9 8 - 0 , 1 2 = 101,48 (gam). ^ c ^ ; ^ '
Dap an diing la A ! ^'
Trang 5elm nang On luy^n thi DH 3 mign B^c - Trung - Nam mOn H(5a hpc - Cu Thanh Toan
Thi du 9: Nung nong 16,8 gam h6n hop gom Au, Ag, Cu, Fe, Zn vdi m6t luong du
khi O2, de'p khi cac phan ling xay ra hoan toan, thu duoc 23,2 gam chat ran X
Th^ tich dung djch HCl 2M vCra dia de phan umg vdi chat rSn X la:
A 200 ml B 400 ml C 800 ml D 600 ml
(Trich detuyen sink Cao dang khoiA) Hu&ng ddn gidi
Soddxayra: - ^ 2 0 ( o x i t ) - ^ ^ 2 H , 0 m^-^-^
Theo djnh luSt bao toan khoi luong:
mo2 =23,2-16,8 = 6,4g=>no2 =0'2(mo!)
=>n_2 =0,2.2 = 0,4(mol) =:>n ^ =0,4.2 =0,8(mol) = n^ci
o "
=>VddHCi2M = 0 , 8 / 2 = 0,4(1) = 400(ml)
Dap an diing la B
Thi du 10: Khi d6't chay hoan toan m gam h6n hop hai ancol no, don churc, mach
ha thu duoc V lit khi CO (a dktc) va a gam H,0
Bieu thirc lien he gii?a m, a va V la:
Ancol no, don churc, mach ho: C„H2„+20
Phuong trinh hoa hoc d6't chay:
Thidu 11: Xa phong hoa 8,8 gam etyl axetat bang 200 ml dung djch NaOH 2M
Sau khi phan irng xay ra hoan toan, c6 can dung dich thu duoc chat ran khan c6
kh6'i luong la •
A 8 , 2 g B 8,56g C 20,2 g D 10,4g
f' (Trich de tuyen sinh Dai hoc khoi A)
Cty TNHH MTV DWH Khang Vi$t
Hu&ng ddn gidi
8 8 Theo bai ra: n^aOH = 0.2 2 = 0,4 mol ; ncHjCOOCaH^ = ^ "
PTPLT: CH3COOC2H5(l) + NaOH(r) CHjCOONaCr) + C2H,OH t
Thidu 12: Cho 13,8 gam axit A tac dung vdi 16,8 gam KOH, c6 can dung djch sau
phan ling thu duoc 26,46 gam chat ran C6ng thiic ca'u tao thu gon ciia A la
(Trich de thi du bi tuyen sinh Dai hpc khoi A) Hu&ng ddn gidi
Gia thie't axit tac dung het v6i KOH " -S"** " "
Sodo: Axit X + K O H ^ Chat ran (mudi, KOH du) + H20 ,, j, Theo djnh luat bao toan khd'i luong, ta c6: + mj^QH - " ' ( r )
Thi du 13: Xa phong hoa hoan toan 1,99 gam h6n hop hai este bang dung dich
NaOH thu duoc 2,05 gam mu6'i ciia mot axit cacboxylic va 0,94 gam h6n hop hai ancol la d6ng dang ke tiep nhau Cong thiic ciia hai este do la
A. HCOOCH3 va HCOOC2H5 B C2H5COOCH3 va C2H5COOC2H5
C CH3COOC2H5 va CH3COOC3H7 D CH3COOCH3 va CH3COOCH5
(Trich de thi thu:Dai hoc khoi A, B) Hu&ng ddn gidi
Sodd phan ung: Este + NaOH > mu6'i + ancol , ^ Theo djnh luat bao toan khd'i luong: m„,e + mN^oH " mmu«i + ancol
=> 1,99 + m^aOH = 2,05 + 0,94 ^ m ^^oH = 1 (gam) , ,
= > I N ^ O H = 0,025 (mol) = n^„ (don chiJc) ; 1 \ H
Trang 6Cgm nang 6n luygn thi DH 3 mign BSc - Trung - Nam mOn H6a hpc - CD Thanh To^n
cdn vira d u 54,75 g d u n g d i c h H C l 1 0 % Kh6'i lironig h 6 n h o p m u o i clorua t h u
duoc sau phan l i n g la
Suy ra: mH20 O^o ra) = 0,15 18 = 2,7 g
mco2 (bay ra) = 0 , 1 5 4 4 = 6 , 6 g
T h e o d i n h luat bao toan kh6'i l u o n g :
"^hh + " ^ H C l muoi clorua + "^C02 + " 1 H 2O (>ao ra)
m muoi clorua = 14,2 + 5 , 4 7 5 - 2 , 7 - 6 , 6 = 10,375g
D a p an d i i n g la D
Thidu 15: D e t r u n g hoa 25,6 g a m h6n h o p 2 axit c a c b o x y l i c d a chuc cSn d u n g 1 l i t
d u n g d i c h h 6 n h o p N a O H 0 , 2 M v a Ba(0H)2 0 , 1 M Sau phan l i n g c 6 can thu
phan tir ddu c 6 so nguyen t u C nho hon 4 ) , c 6 t i k h o i so v6\i l a 4,7 D u n
n o n g 2 m o l X (xuc tac N i ) , duoc h6n hop Y c 6 t i k h o i hoi so voi heli la 9,4 T h u
la'y toan bo cac ancol t r o n g Y r o i cho tac d u n g vdri N a (du), duoc V l i t H , ( d k t c )
G i a t r i Ion nhat ciia V la
—^ 1(aiidehil phan lillg) ~ Idiiilro ph,iu I'nig) ~ 1 ( m o l )
no, phan u n g = s6' m o l ancol don chiic tao ra ' ^
c u n g d a y d6ng dang), thu duoc 8,96 l i t k h i C O , (dktc) v a 11,7 g a m H O M a t
khac, neu d u n n o n g m g a m X v 6 i H2SO4 dac t h i t6ng k h o i l u o n g ete t6'i da thu
duoc la
A 5,60 gam B 6,50 gam C 7,85 gam D 7,40 gam
(Trich dethi thu Dai hpc khoi A, B) Huong ddn gidi
T h e o bai ra: n^o^ = 0 , 4 m o l ; nH20 = ^'^5 m o l
" ^ ' "H2O >"co2 ^ -dncol nay no, mach h o (don churc) dang Cj^H2n+|OH
va s6 m o l h6n hop ancol n^h = ny^^Q - nco2 = ^'^^ — ^ ' ^ ~ ^ ' ^ ^ ("^"'^
Trang 7dm nang On luygn thi DH 3 m\in B&c - Trung - Nam mOn H6a hpc - Cii Thanh Togin
0,25 'hh
Scrddtaoete: ROH + R ' O H "^'^^ > R - O - R ' + H.O
1 0,25, Theo s 0 do: n H 2 0 = - l a n c o i = — (mol): ^HjO = ^ = 2,25 (gam) 0,25.18 ^
Theo djnh luat bao toan kh6'i luong: m„„„| = + m H 2 0
=> 0,25(14n + 18) = m + 2,25 => 0,25(14.1,6 + 18) = m + 2,25
=> m = 7,85 (gam)
Dap an dung la C
Thi du 18: Cho mot lucmg bot Zn vao dung dich X g6m FeCl va CuCl Kh6'i
lugng chat rSn sau khi cac phan ting xay ra hoan toan nho hcfn khdi luong bdt
Zn ban dSu la 0,5 gam C6 can pMn dung dich sau phan ling thu duoc 13,6 gam
mu6'i khan T6ng khoi lugng cac mu6'i trong X la
A 13,1 gam B 17,0 gam C 19,5 gam D 14,1 gam
( Trich de thi thu: Dai hpc khoi A, B) Huong ddn gidi
Ta CO so 66 phan ling: FeCl2
CuCl, + Zn- -> ZnClj +
Fe
Cu Theo djnh luat bao toan khdi lugng, ta c6:
r"hh(FeCl2,CuCl2) " ^ Z " ~ ^ ZnCI 2 '^hh(Fe, Cu)
m hh(FeCl2,CuCl2 ) +mz„ = 13,6 +(mz„ - 0 , 5 )
m , ' h h ( F e a 2 , c u c i 2 ) = '3,6 - 0,5 =13,1 (gam)
Vay tdng khdi lugng cac mudi trong X bang 13,1 gam
Dap an dung la A
Thi du 19: Dun nong hdn hgp khi gdm 0,06 mol C,H2 va 0,04 mol H vdi xiic tac
N i , sau mot thdi gian thu dugc hdn hgp khi Y DSn toan b6 h6n hgp Y Idi tCr ttr
qua binh dung dung dich brom (du) thi con lai 0,448 lit hdn hgp khi Z (d dktc)
CO ti khdi so vdi Oj la 0,5 Khdi lugng binh dung dung dich brom tang la
A 1,64 gam B 1,20 gam C 1,04 gam D 1,32 gam
( Trich de thi thii Dai hoc khoi A, B)
! • Hu&ng ddn gidi
Nhan xet: Khdi lugng binh dung dung dich brom tang chinh bang khdi lugng
etilen, axetilen cd trong hdn hgp Y
Mz = 32 0,5 = 16; n^ = 0,448/ 22,4 = 0,02 (mol)
Theo djnh luat bao toan khdi lugng, ta c6:
Cty TNHH MTV D W H Khang Vi$t
^ 0,06.26 + 0,04.2 = 0,02 16 + m H ™
=> ni(bi„h brom ,a„g) =1,32 (gam). M ' i •;
-Dap an dung la D ium) JBAU ^.^^ v-.K'
Phaong phap 2: PHl/ONG PHAP BAO TOAN ELECTRON
I L I T H U Y E T
- Trong phan ling oxi hoa - khu, xay ra ddng thdi qua trinh oxi hoa va qua trinh
- Djnh luat bao toan electron: Trong phan ilng oxi hoa khir, tdng sd electron do chat khu cho phai diing bang tdng sd electron do chat oxi hoa nhan:
2 : e ( c h o ) = Se(nhan)
- Da'u hieu de nhan ra bai tap c6 thd' su dung phuong phap bao toan electron de giai la cac bai tap cd phan ling oxi hoa - khu
11 V A N D U N G
Thidu 1: Cho hoi nude di qua than nong do thu dugc 15,68 lit (dktc) hdn hgp khi A
gdm C O , C O o va Ho Cho toan b6 A tac dung het vdi hdn hgp MgO, CuO du, nung nong thu dugc hdn hgp chat rdn B Hoa tan toan bg B bang HNO3 dac, ndng, du dugc 26,88 lit N O , (san pham khu duy nhat, dktc) Sd mol khi H , trong A la
A 0,4 B 0 , 2 C 0 , 3 D.'o,5
(Trich dethi thu: Dai hpc khoi A, B nam 2013) Hu&ng ddn gidi
Theo bai ra: n ^ = 0,7(mol);nNO2 " ' ' ^ ( m o l ) ^ " •< rv?
C 4- H2O ^ CO + H2 ,^ ^ C + 2H2O ^ CO2 + 2H2
x - ^ x y _> 2y : ^ x + x + y + 2y = 0 , 7 ^ 2 x + 3y =0,7 ( l )
+2 +4 +1
C - 2e ^ C H2 - 2e ^ 2 H
X - > 2 x (x + 2 y ) ^ 2(x + 2y)
N + le N 1,2<- 1,2 Theo djnh luat bao toan electron, ta cd: 2x + 2(x + 2y) = 1,2 => 4x + 4y = 1,2
= > X + y = 0,3 (2) ^ , y , , v , , ,
T i l f ( l , 2 ) = > x = 0 , 2 ; y = 0 , l ' ' - M u
Vay sd mol khi H2 la: x + 2y =0,2 + 2.0,1 =0,4 (mol) ltr,ubo'.,n
Dap an dung la A
Trang 8elm nang On luy^n thi DH 3 mjgn BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan ^
Thi du 2: Hoa tan hoan toan 2,44 gam h6n hop hot X gom Cu va Fe.Oy bang dung
dich H 2 S O 4 dac, nong, du Sau phan urng thu duoc 0,504 h't khf S O , (san ph^m
khir duy nhat, dktc) va dung dich chiia 6,6 gam h6n hop muoi sunfat Phan tram
khoi luong Cu trong X la
Theo nguydn tac bao toan electron ta c6: 2a + 3b = 2c + 0,045 (1)
Matkhac, taco: 64a + 56b + 16c = 2,44 (2)
Thi du 3: Hoa tan hoan toan 24,8 gam h6n hop X gom Fe va Fe.Oy bang dung djch
H 2 S O 4 dac, nong, du thu duoc dung dich Y va 4,48 lit khi S O 2 (san phim khir
duy nhat, dktc) Phan tram khd'i luong nguyen to oxi trong X la
A 20,97% B 16,84% C 25,73% D 32,56%
( Trich de thi thu Dai hoc khoi A, B nam 2013) Huong ddn gidi
Theo bai ra: n^Q^ =0,2 (mol)
Quy d6i h6n hop Fe.Fe^Oy => Fe va O
Thi du 4: Dot chay h6n hop g6m 1,92 gam Mg va 4,48 gam Fe vdi h6n hop khf X
g6m clo va oxi, sau phan dng chi thu duoc h6n hofp Y g6m cac oxit va muoi
clorua (khong con khf du) Hoa tan Y bang mot luong vira dii 120 ml dung dich HCl 2M, thu duoc dung dich Z Cho AgNOy du vao dung dich Z, thu duoc 56,69 gam ket tua Phan tram the tfch ciia clo trong h6n hop X la
A 51,72% B 76,70% C 53,85% D 56,36%
(Trich de thi tuyen sink DH khoi B) Huong ddn gidi ^
Theo bai ra: n^, =0,08mol;np^ =0,08mol;np.p| =0,24mol
Goi x,y,z iSn luot la s6'mol ciia Cl2,02 va Ag"^ (tao ra Ag)
Theo nguyen tdc bao toan electron, ta c6: 0,08.2 + 0,08.3 = 2x + 4y + Iz
=:>2x + 4y + z = 0,4 (1) • - ' ] ( •
0,24->0,12 0,06 <- 0,12
=>y = 0,06 (2) • Ag+ + l e - > A g i Ag+ + Cr -> A g C U
z ^ z ( 2 x + 0 , 2 4 ) - > (2x +0,24)
^ 108.Z + 143,5.(2x + 0,24) = 56,69
^ 1 0 8 z + 287x = 22,25 (3) Tir (1, 2, 3) => x = 0,07;y = 0,06;z = 0,02
Dap an diing la C "
Thi du 5: Cho h6n hop g6m 6,72 gam Mg va 0,8 gam MgO tac dung het v6i luong
du dung djch HNO3 Sau khi cac phan ling xay ra hoan toan, thu duoc 0,896 lit m6t khf X (dktc) va dung djch Y Lam bay hoi dung djch Y thu duoc 46 gam mu6'i khan Khf X la
Trang 9Ca'm nang On luygn thi DH 3 mign BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan
+5
N + 8e 0,16
-3 -3 NH4NO3
Dap an diing la C
Thi du 6: Cho 24,15 gam h6n hop khi A gom clo va oxi phan ung vtra het vdi mot
h6n hop gdm 4,8 gam magie va 8,1 gam nh6m tao ra h6n hop cac mudi clorua va
oxit ciia hai k i m loai Thanh ph^n % theo khdi luong ciia o x i trong hdn hop A la
A 26,5% B 32,0% C 20,8% D 16,8%
(Trich de thi thu Dai hoc khoi A, B) Hu&ng dan gidi
Dat x, y Idn luot la sd mol C I , O, trong hdn hop A
So do: hh A + hh CU, O , - > hh mudi clorua + oxit
Thi du 7: Cho 9,1 gam k i m loai M tan hS't vao dung dich H N O , loang, d u thu duoc
0,06 mol hdn hop 2 khi X va Y , cd khdi lirong 2,08 gam vdi M^/My = 1,467
Bi6't trong dung dich thu duoc khdng cd mudi NH4NO3, k i m loai M la
Vay n = 2, M = 65 ( Z n ) Dap an dung la A
Thi du 8: Cho 2,16 gam M g tac dung vdi dung dich H N O , (dir) Sau khi phan umg
xay ra hoan toan thu duoc 0,896 1ft khi N O (0 dktc) va dung djch X Khdi luong
mudi khan thu duoc k h i lam bay hoi dung dich X la
A 8,88 gam B 13,92 gam C 6,52 gam D 13,32 gam
(Trich de thi thtiDai hoc khoi A, B) Hu&ng ddn gidi
24 22,4 Xay ra cac qua trinh:
+2 I
Theo bai ra: n^^ = = 0,09 (mol); n^o = -ir^ = 0,04 ( m o l ) ,v,
0,09 0,18(mol) 0,09 (mol) , JAiflu JQ
^ I n , , , , = 0,09 2 = 0,18 (mol) WyLlSJl^L^
Trang 10dm nang On luyjn thi BH 3 niien BJc imny - Nam mOn H6a hgc - Cu Thanh Tocin
Thi du 9: Hoa tan het 0,03 mol m6t oxit sat c6 c6ng thiic Fe.Oy vao dung dich
HNO3 loang du thu duoc 0,01 mol m6t oxit nito c6 c6ng thiic Nfi, (san ph^m
khir duy nha't) M6'i quan he giila x, y, z, t la
Thi du 10: Nung 1,92 gam h6n hop bot X gom Fe va S trong binh km khong c6
kh6ng khi, sau mot thoi gian thu duoc chat ran Y Hoa tan het Y trong dung
djch H N O , dac, nong, du thu duoc dung djch Z va V lit khi thoat ra (dktc) Cho
Z tac dung vdri dung dich BaCK du thu duoc 5,825 gam ket tiia Gia tri ciia V la
Theo bai ra: n^^^^ = = 0,025(mol) " }•
Cty TNHH MTV DWH Khang Vijt
=> ms(x) = 0,025.32 = 0,8(gam) => mp^^^) = 1,92 - 0 , 8 = 1,12(gam>
=^np,(x) = ' ' 1 2 / 5 6 = 0,02(mol) ^ Qua trinh cho - nhan electron: ; d M u T l
Fe - 3e vFe'^ S - 6 e — > S ^ ( s o 2 - )
0,02 -> 0,06 0,025 - > 0 , 1 5 ^_,5Q
N + le > N(N02)
V V 22,4 22,4
V Theo nguyen tac bao toan electron, ta c6: 0,06 + 0,15 = => V = 4,704 (lit)
22,4 Dap an diing la A
Thi du 11: Hoa tan hoan toan 8,862 gam h6n hop g6m A l va Mg vao dung djch
HNO3 loang, thu duoc dung dich X va 3,136 lit (0 dktc) h6n hop Y g6m hai khi
khong mau, trong do c6 m6t khi hoa nau trong khdng khi Kh6'i luong ciia Y la 5,18 gam Cho dung djch NaOH (du) vao X va dun nong, khong c6 khi mui khai
thoat ra Phin tram khoi luong ciia A l trong h6n hop ban 6iu \k
A 19,53% B 10,52% C 12,80% D 15,25%
(Trich de thi thvcDai hgc khoi A, B) Hu&ng ddn gidi
Theobaira: n^ =3,136/22,4 = 0,14mol Trong Y c6 N O (khi bj hoa nau trong khong khi:
2NO + 0 2 ^ 2NO2 do nau) •
Khf con lai trong Y c6 the la N , hoac N , 0 (deu kh6ng c6 mau) Ta c6:
V i = 30 < 37 =^ khi con lai c6 M > 37, vay khi do la N O ( M = 4 4 ) , ,
Xac dinh s6 mol N O (x mol), NoO (y mol) trong Y
x + y - 0 , 1 4
Ta c6: <
30x + 44y = 5,18 Giai he ta duoc: x = 0,07; y = 0,07
Cac qua trinh oxi hoa - khij xay ra: c<• • ' '
0 0
a(mol) - > 3a b(mol) - > 2b /* f
19
Trang 11Ca'm nang Qn luyQn thi DH 3 mign B j c - Trung - Nam mOn H6a hgc - Cu Thanh To^n
Thidu 12: Cho 10,8 gam b6t Al tan hoan toan trong dung djch H N O 3 tha'y thoat ra
3 khi Nn, NO va N2O c6 ti le mol tuong l i n g la 1: 2: 1 Trong dung dich thu duoc
khdng CO NH4NO3. The tich 3 khi tren (dktc) la
A 2,24 lit B 3,36 lit C 4,48 lit D 6,72 lit
( Trich de thi tkuDqi hoc khoi A, B)
f Hu&ngddngidi
; Theo bai ra n ^ , =0,4(mol)
' Goi X la s6'mol N , => nNo=2x; nN20=x(moI)
Cac qua trinh nhucmg - nhan electron
Thidu 13: Cho 24,3 gam m6t kim loai M (c6 hoa trj n duy nha't) tac dung v6i 5,04
lit khi O (6 dktc) thu duoc chat rSn A Cho A tac dung het v6i dung dich HCl
tha'y CO 1,8 gam khi thoat ra Kim loai M la
A Mg B.Zn C A l D Ca
( Trich dethi thi Dai hoc khoi A, B) Hit&ng dan gidi
5 ^ ^ ^ ^ ^ ^ , , 1,8 22,4
Theo bai ra: no^ = - ^ = 0,225(mol); n^^ = — = 0,9(mol)
Theo dinh luat bao toan electron, ta c6: = 0,9 + 1,8 = 2,7
Thi du 14: Chia m gam A l thanh hai phSn bang nhau:
- PhSn m6t tac dung vori luong du dung dich NaOH, sinh ra x mol khi H2;
- PhSn hai tac dung v6\g du dung djch HNO3 loang, sinh ra y mol khi N^O (san phaim khir duy nha't)
Thi du 1: Dung dich X c6 chiia: 0,07 mol Na"; 0,02 mol S O 4 " va x mol OH"
Dung djch Y c6 chiia C104,N03 va y mol t6ng s6' mol CIO4 va NO3 la 0,04. Tr6n X va Y duoc 100 ml dung djch Z Dung djch Z c6 pH ( bo qua sir
dien l i cua H^O) la
A 1 B.2 C 12 D 13 ' '
(Trich de thi thi Dai hoc khoi A, B nam 2013)
21
Trang 12d m nang On luy$n thi DH 3 mign Bjc - Trung - Nam mOn H6a hpc - Cii Thanh Toan
Huong ddn gidi
Ap dung dinh luat bao loan dien tich, ta c6:
Dung dich X: 0,07.1 = 0,02.2+ x l =>x=0,03(mol)
Dung dich Y: 0,04.1 = y 1 => y = 0,04 (mol)
Thi du 2: Hoa tan h6n hop mu6'i vao nu6c duoc dung dich X chiia 3 ion trong s6'
cac ion sau:
Ion Na" Ba^^ cr NO3- Ag^
SO4 S6'mol 0 , 2 0,1 0,3 0,4 0,1 0,2
Dung djch X chiia 3 ion la
A Na^ A g ^ N O , - B Na\^ C f
(Trich de thi thii Dai hoc khoi A, B nam 2013) Huong ddn gidi
Phuong phap di giai nhanh bai nay la thir cac phuong an:
+ Cac ion c6 ciing t6n tai trong mot dung djch duoc khong (dieu kiSn 1)
+ Dung dich c6 trung hoa dien kh6ng (dieu kien 2)
Phuong an A, B, D thoa man di^u kien 1
Phuong an C khdng thoa man diSu kitn 1 (loai)
Ap dung dinh luat bao toan didn tich:
- dphuomgan A: |0,2.1(+) + 0,l.l(+)|7i|0,4.1(-)| (loai)
- 6 phuong an B: |0,2.I(+) + 0,1.2(+)|^|0,3.1(-)| (loai)
- d p h u o n g a n D |0,2.1(+) + 0,1.2(+)| = |0,4.1(-)| (thoaman)
Vay dap an dung la D
Thi du 3: M6t dung dich chiJa 0,02 mol Cu"*, 0,03 mol K*, x mol CP va y mol
504-^ Tdng kh6'i luong cac mu6'i tan c6 trong dung dich la 5,435 gam Gia trj
ciia X va y lin luot la
A 0,01 \k 0,03 B 0,05 va 0,01 C 0,03 v i 0,02 D 0,02 va 0,05
^ (Trich de thi thu! Dai hoc khoi A, B)
Cty TNHIi MIV DWH Khang Vi^t
Huong ddn gidi
Theo djnh luat bao toan diSn tich, ta c6: 0,02.2 + 0,03.1 = x 1 + y 2 ,
=^ x + 2y = 0,07 (1) Mat khac: T(5ng khd'i luong muoi bang t6ng khO'i luong cac ion: ,
0,02 64 + 0,03 39 + 35,5 x + 96 y = 5,435 '
T i r ( l , 2 ) : x = 0,03; y = 0,02 , Dap an dung la C , , ( ' ' " ! •
Thi du 4: Dung djch E g6m x mol Ca'*, y mol Ba'*, z mol HCO3 Qio tif tiT dung djch Ca(OH)2 nong d6 a mol/1 vao dung djch E den khi thu duoc luong k6't tiia
\6n nha't thi vira het V lit dung dich Ca(0H)2 Bieu thiic lien he giira cac gia tri
Theo djnh luat bao toan didn tich ta c6: z = 2x + 2y ( l )
HCO3 + OH" -> CO^" + H2O I
z - > z
=>z = 2.V.a (2) i:.V;<, •
T i r ( l , 2 ) =>2x + 2y = 2.Va=>x + y = V.a ^ V = ^ ^
a Dap an diing la A
Chu v: Luong ket tiia lom nha't khi toan bo luong HCO3 chuye'n het thanh CO3"
Thidu 5: Dung dich X g6m 0,1 mol H", z mol A l ' " , t mol NO3 va 0,02 mol SO4"
Cho 120 ml dung dich Y g6m KOH 1,2M va Ba(0H)2 0,1M vao X, sau khi cac
phan ling ket thiic, thu duoc 3,732 gam ke't tua Gia trj ciia z, t Mn luot la
Trang 13ca'm nang On luy^n thi DH 3 m'lin BSc - Trung - Nam mOn H6a hpc - CO Thanh Toan
Thi du 6: Dung dich X chiia cac ion: Ca'"^, Na*, HCO3 va CI", trong do so nnol cua
ion Cr la 0,1 Cho 1/2 dung djch X phan ihig vdi dung dich NaOH (du), thu
ducKc 2 gam ket tiia Cho 1/2 dung dich X con lai phan ling v6i dung dich
Ca(OH)2 (du), thu dugc 3 gam ket tua Mat khac, neu dun s6i de'n can dung djch
X thi thu dugc m gam chat rSn khan Gia tri ciia m la
Chiing to trong phan ling (1) c6 Ca"^ het (con du H C O 3 )
^ "Ca2+ " "CaC03 = 0,02(mol)
94
Cty TNHII MIV UVVII Kliaiig Vi$t
Theo (2): n^^^_ = nc,co3 = 0,03(mol)
vay trong X c6: 0,1 mol CP; 0,04 mol Ca-^ 0,06mol HCO3 ,,,.ui>; ,
Theo djnh luat bao toan dien tich, ta tinh dugc s6' mol Na* : t • i v
n ^ = 0 , l l + 0 , 0 6 1 - 0 , 0 4 2 = 0,08(mol). ,0 •
* Dun soi dung djch X:
2HCO3 >C02 T +co^- + H2O
vay dun can dung djch X thu dugc CaCO,, CaCU, NaCl (chat ran),
vay m = m ^^2+ + r"^n2- + "^r,- + cot c r Na"'
=> m = 0,04.40 + 0,03.60 + 0,1.35,5 + 0,08.23 m = 8,79 (gam)
Dap an diing la D
Thi du 7: Dung djch X chiia cac ion: Fe'", SO4-", N H / , CP Chia dung djch X
thanh hai phdn bang nhau:
- Phan mot tac dung vori lugng du dung dich NaOH, dun nong thu dugc 0,672 lit khi (0 dktc) va 1,07 gam ket tua;
- Phdn hai tac dung vai lugng du dung djch BaCU, thu dugc 4,66 gam ket tua T6ng khoi lugng cac mu6'i khan thu dugc khi c6 can dung djch X la (qua trinh
A 3,52 gam B 3,73 gam C 7,04 gam D 7,46 gam
( Trich de thi thu:Dai hoc khoi A, B) Hu&ng ddn gidi
Phuong trinh phan ling xay ra:
Fe'^ + 3 0 H - •
N H ; + O H
-SO^- +Ba-^ > BaS04 i
-> Fe(OH), NHi t + H , 0
Trang 14cam rang 6n luygn thi DH 3 mjgn Bac - Trung - Nam mfln H6a hgc - Cu Thanh Toan
Khoi lircmg muoi bang long kh6'i luong cac ion:
m„„«i = 0,02 56 + 0,06 18 + 0,04 96 + 0,04 35,5= 7,46 (gam)
Dap an diing la D
Thi du 8: Hoa tan h6n hop mud'i vao nuofc duoc dung djch chiia 0,1 mol N H / ; 0,2
mol k?*\3 mol NO", 0,1 mol CF va 0,15 mol ion X Di6n tich ciia ion X la
Dat dien tich cua ion X la a ( - )
Ap dung dinh luat bao toan dien tich ta c6:
0,1.1(+) + 0,2.3(+)| = |0,3 l ( - ) + 0,1 l ( - ) + 0,15.a(-)| =^ a = 2
Dap an diing la C
PHirONG PHAP 4: PHlTONG PHAP BAO TOAN N G U Y E N T O
I L I T H U Y E T
- Trong phan ting hoa hoc, cac nguyen to duoc bao toan Dieu nay c6 nghla la
t6ng s6' mol nguyen tir ciia cac nguy6n t6' trudc va sau phan ling lu6n bang nhau
- C&n chu y, han che viet PTHH ma thay vao do nen viet so do phan ihig (c6 he
s6') d^ bieu didn cac bien doi co ban ciia cac nguyen t6' quan tam
I I VAN DUNG
Thi du 1: H6n hop X g6m hai axit cacboxylic don chiic Dot chay hoan toan 0,1
mol X cSn 0,24 mol O, thu duoc COj va 0,2 mol H^O Cong thiic hai axit la
Dat cong thiic 2 axit don chiic trong X la C^HyOj £
Sor d6 dot chay: C,Hy02 + O2 -> CO2 + HjO
0,1 0,24 0,2(mol)
V i nguyen to oxi duoc bao toan ntn ta co:
Cty TNHH MTV DWH Khang Vigt
= n 0 ( C 0 2 ) + " 0 ( H 2 0 )
no(x) +"0(02)
=^0,1.2 + 0,24.2 = nco2-2 + 0,2.1 nco2 =0'24{mol) i ^ ( i i ( > ) f A
S6 nguyfin tir cacbon trung blnh trong phan tir: x = _ "CO2 _ 0,24
'hh 0,1 ""^'^.^'^''V'
V i nco2 >"H20 • X khdng phai la 2 axit no, mach ho, don chiic
=> Loai A va C
Mat khac, vl nc = x = 2,4 => loai D
Vay 2 axit trong X co the la CH3COOH va CHj = CH - COOH Dap an diing la B
Thi du 2: D6't chay hoan toan 20 cm' hoi hop chat hOu co X (chi g6m C, H, O) cSn
vira du 110 cm^ khi O2, thu duoc 160 cm' h6n hop Y g6m khi va hoi Dfin Y qua dung djch H2SO4 dac (du), con lai 80 cm' khi Z Bife't cac th^ tich khi va hoi do 6 C l i n g di^u kien Cong thiic phan tir ciia X la
A CjHgO B C4Hg02 C C4H,oO D. C4H8O ^ ' ^ ' "
(Trich dethi thu Dai hgc khoi A, B nam 2013) Huong ddn gidi ' I •
Thi du 3: Hoa tan hoan toan 0,3 mol h6n hop gdm A l va AI4C3 vao dung djch
KOH (du), thu duoc a mol h6n hop k h f va dung djch X Sue k h i CO2 (du) vao
dung djch X, lugng ket tiia thu duofc la 46,8 gam Gia trj ciia a la:
A 0,55 B 0,60 C.0,40 D 0,45
(Trich de thi thucDgi hgc khoi A, B nam 2013)
So d6 cac phan ling xay ra:
Huong ddn gidi
Al [A14C3'
Trang 15dm nana On luyQn thi BH 3 m\6n Bic - Trunp - Nam mOn H6a hqc - Cii Thanh Toan Cty TNHH MIV iiVVH Khang Vi$t
Ke't tiia thu duoc la A l ( O H ) , ; n ^ | , O H ) = 4 6 , 8 / 7 8 = 0 , 6 ( m o l )
Thi du 4: Nung nong m gam h6n hop gom A l va Fe304 trong d i ^ u kifen khong c6
khong k h i Sau khi phan ung xay ra hoan toan, thu duoc h6n hop rSn X Cho X
tac dung v6i dung djch N a O H (du) thu duoc dung dich Y , chat ran Z va 3,36 lit
khi H2 (a dktc) Sue khi C O j (du) vao dung djch Y , thu duoc 39 gam kd't tua
Gia t r i ciia m la:
2 H 2 0 + 2 e ^ H , + 2 0 H " ^
0 , 3 < - 0 , 1 5 Theo djnh luat bao toan electron ta c6: 1,5 = 8x + 0 , 3 =>8x = 1,2 => x = 0 , 1 5
=>mpe3O4=0''5-232 = 34,8(g)
v a y m = 13,5 + 3 4 , 8 = 4 8 , 3 ( g )
Dap an diing la A
Chuji: P T H H cac phan iJng xay ra:
8A1 + 3Fe,04 9Fe + 4AI2O3 2A1 + 2 N a O H + 6 H 2 O 2 N a A l (OH)^ + 3H2 t
N a A l ( O H ) ^ + C O 2 - > NaHCO^ + A1(0H)3 i
Thi du 5: D6't chay hoan toan h6n hop X gom hai ancol no, hai chuc, mach ho can
vua du V | li't khi O,, thu duoc V , Ii't khi C O va a mol H j O Cac khi diu do a
dieu kien tieu chuan Bieu thiic lien he giua cac gia trj V | , V 2 , a la
22A,
= > V , = 2 V 2 - 1 1 , 2 a Dap an dung la B
Thidu 6: H6n hop X gom axit fomic, axit acrylic, axit oxalic va axit axetic Cho m
gam X phan ling he't voti dung dich N a H C 0 3 t h u duoc 1,344 lit C 0 2 ( d k t c ) Dot
chay hoan toan m gam X cin 2,016 lit 0 2 ( d k t c ) , thu duoc 4,84 gam C O 2 va a
gam H 2 O Gia trj ciia a la
A 1,44 B 1,62 C.3,60 D 1,80
(Trich de thi tuyen sinh DH khdi A nam 2012) Huong ddn gidi •
Theo bai ra: nco2 = 1,344/22,4 = 0,06mol;
=0,09mol;nco2 = 4,84/44 = 0,11 mol
29