Cẩm nang ôn luyện thi đại học rèn luyện giải nhanh các đề thi ba miền bắc trung nam hóa học phần 1

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... Thi du 4: Ddt chay hoan toan 4,64 gam mdt hidrocacbon X chat khi d di^u

CU 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

LOI 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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cty TNHH MTV DWH Khang Vi$t

B A I Xp> HQA 119c 1 r u , PhUCSng phap 1: PHLfONG PHAP BAO TOAN KHOI LUONG

I LITHUYET

- Gia sir CO phan ling: ' Jii i •! •

1-Ta c6: + nig = + triy

- 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: ||; || , (,

|, (Trich de thi thdDai hoc khdi A, B nam 2013) 3

Ca'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)

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

C^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 ! ^'

elm 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

A.C,H(,COOH B.C2H5COOH C C H 3 C O O H D.HCOOH

(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

= > I N ^ O H = 0,025 (mol) = n^„ (don chiJc) ; 1 \ H

Cgm nang 6n luygn thi DH 3 mign BSc - Trung - Nam mOn H6a hpc - CD Thanh To^n

Thi du 14: D e h o a tan hoan toan 14,2 g h6n hop N a H C O , , NH4HCO, v a K H C O ,

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

Thi du 16: X la h6n hop g o m H , v a hoi cua hai andehit ( n o , don churc, mach ho,

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 )

Chu v: So m o l h6n hop k h i g i a m = so m o l H , phan ung = s6' m o l andehit d o n chiic,

Thi du 17: D o t chay hoan toan m g a m h6n hop X g6m ba ancol (don chiJc, thuoc

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 = ^'^^ — ^ ' ^ ~ ^ ' ^ ^ ("^"'^

dm 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

elm 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

Ca'm nang On luygn thi DH 3 mign BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan

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,

M g - 2e > M g r W J VicAJ ItNH i^lNH THUAN

0,09 0,18(mol) 0,09 (mol) , JAiflu JQ

^ I n , , , , = 0,09 2 = 0,18 (mol) WyLlSJl^L^

dm 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 l > AP+ +3e Mg > Mg^+ + 2e a(mol) - > 3a b(mol) - > 2b /* f

19

Ca'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

d 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

ca'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 c6 can chi c6 nu6c bay hai) jom 6« r<i!t

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

cam 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'

dm 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

ca'm nang 6n luyjn thi DH 3 mign B&c - Trung - Nam mOn H6a hpc - CCi Thanh ToSn

Sa 66 p h a n irng:

- C O O H + N a H C 0 3 - > - C O O N a + CO2 t + H 2 O ; 0

0,06 * - 0,06 H.ii!;OJ>rff

=> "-cooH(x) = "CO2 = 0 , 0 6 ( m o l ) ^n^^^^ -0,06.2 = 0,12(mol) <nes

So d6 phan ling: X + O 2 - > CO2 + H2O

V i nguydn t6' oxi dugc bao toan n&n ta c6: nQ^^) + " 0 ( 0 2 ) = " 0 ( 0 0 2 ) " o ( H 2 0 )

=>0,12 + 0,09.2 = 0,11.2 + n H 2 o l =^ n H 2 0 =0-08 ^ a =0,08.18 = l,44(g) ^

Dap an dung la A

Thi du 7: Dot chay hoan toan 4,64 g a m mot hidrocacbon X (cha't khi of di6u kifin

thuofng) roi dem toan b6 san phdm chay ha'p thu h6't vao binh dung dung

djchBa(OH)2 Sau cac phan umg thu dugc 39,4 gam ket tua va khoi lirgng ph&n

dung dich giam bot 19,912 gam C6ng thiic phan tijf cua X la

Thi du 8: Hoa tan het m gam h6n hop FeS^ va CU2S trong dung dich HNO3, sau cac

phan ling hoan toan thu dugc dung dich X chi c6 2 chat tan, \6i tong khoi lugng

cac chat tan la 72 gam Gia tri ciia m la

Cty 1MHH MIV L.'VVH Khang Vigt

V i cac nguyen t6' (Fe, Cu va S) dugc bao toan ntn ta c6 so do:

2FeS2 + CU2S Fe2 ( 5 0 4 ) 3 2CUSO4

2x <- X - > X -> 2x (mol) i ; '

Theo bai ra: 400.x + 160.2x = 72 => 720x = 72 => x = 0,1

v a y m = 2.0,1.120 + 0,1.160 = 40(g) jg m vnios n i o i / '

D a p a n d i i n g l a B ^rB.V> iM 0 : 1

Thidu 9: D6't 5,6 gam Fe trong khong khi, thu dugc h6n hgp cha't ran X Cho toan

b6 X tac dung vdi dung djch HNO3 loang (du), thu dugc khi NO (san ph^m khir duy nha't) va dung djch chiia m gam mu6'i Gia trj cua m la ,j

Thi du 10: D6't chay hoan toan 0,2 mol h6n hgp g6m mdt ancol va m6t axit don

chiic CO cung so nguydn t6' cacbon cin diing 0,45 mol O,, thu dugc 0,4 mol CO,

va 0,5 mol H^O Phdn tram khoi lugng ciia axit trong h6n hgp trdn la:

dm nang On luy^n thi DH 3 mign B&c - Trung - Nam mOn H6a hqc - CCi Thanh Toan

0,1.60 + 0,1.62

y'-Thidu 11: H6n horp X g6m axit axetic, axit fomic va axit oxalic Khi cho m gam X

tac dung vdri N a H C O , (du) thi thu dugc 15,68 lit khi C O , (dktc) Mat khac, d6't

chay hoan toan m gam X cdn 8,96 lit khf O, (dktc), thu duoc 35,2 gam C O , va y

mol H i O Gia tri ciia y la

A 0,3 B 0,8 C.0,2 D 0,6

(Trich de thi thUDqi hoc khoi A, B) Huong ddn gidi

Theobai ra: nco2 ( l ) = 15,68/22,4 = 0,7(mol) ;

no2 = 0,4(mol); nco2 (2) = 35,2/44 = 0,8(mol)

C H 3 C O O H + HCOOH + HOOC - COOH ^ ^ ! ^ y ^ ^ C 0 2

a b c —• (a + b + 2c)(mol)

Suy ra, s6' mol nguyfen tir nguyen to oxi trong X :

"o(x) = 2a + 2b + 4c = 2(a + b + 2c) = 2.nco2

Thi du 12: Hoa tan hoan toan h6n hop g6m 0,12 mol FeS, va a mol CujS vao axit

HNO, (viJra dii), thu duoc dung djch X (chi chiia hai mu6'i sunfat) va khi duy

nha't NO Gia tri ciia a la

A 0,04 B 0,075 C.0,12 D 0,06

(Trich de thi tuyen sinh Dai hoc khoi A) Huong ddn gidi

Vi cac nguyen to (Cu, Fe va S) duoc bao toan ntn ta c6 so 66:

2FeS, + C U 3 S > Fe,(S04)3 + CuS04

Thidu 14: H 6 n hop X g6m CiH, va H , c6 cung s6' mol La'y m6t lugng h6n hop X

cho qua chat xuc tac, nung nong dugc h6n hgp Y g6m C,H4, CiH^ va C,H2, H,

du DSn Y qua nuorc brom tha'y binh nu6c brom tang 10,8 gam va thoat ra 4,48

lit h6n hgp khi (do b dktc), c6 ti khd'i so vdi H , la 8 The ti'ch khi O, (khi do a

dktc) vira dii de dot chay hoan toan h6n hgp Y la

A 33,61ft B 22,4 lit C 26,88 lit D 44,8 lit

(Trich dethi thi Dai hoc khoi A, B) Huong ddn gidi

X (C,H,, H,) ) Y (C,H4, C,H„ C,H,, H,) > C,H„ H,"'^*

- Goi X la so mol C,H, (cung nhu H,) trong h6n hgp X

mx = (26 + 2) x = 28x (gam) (1)

- Theo dinh luat bao toan kh6'i lugng, ta c6: mx = my (2)

- Mat khac, theo bai ra: my = m ,bi„h b^om a„g) + " i (khf hoi n)

• X i

4 48 / ,oMi iish'

my = 1 0 , 8 + ^ — 8 2 = 14 (gam) (3) ,

22,4

- Theo dinh luat bao toan nguyen t6': V i nguy6n t6' C, H dugc bao to^n ntn thi

tich khf O, d^ dot chay hoan toan h6n hgp Y cung bang the' tfch O, de d6't chay hoan toan h6n hgp X :

Ca'm nang On luygn thi B H 3 mfo.f ^ a c - Trung - Nam mOn H6a hi?c - Cu Thanh To^n

=> X"o2 =0,25+1,25 = 1,5 (mol) =i> Vo^ ,dk.c) = 1,5 22,4 = 33,6 (lit)

Thi du 15: H6n hop M g6m andehit X, xeton Y (X, Y c6 ciing s6' nguy6n tir

cacbon) va anken Z Dot chay hoan toan m gam M cdn diing 8,848 lit O2 (dktc)

sinh ra 6,496 lit CO, (dktc) va 5,22 gam HjO Cong thuc ciJa andehit X la

A CH3CHO B.C3H7CHO C C4H,CHO D C2H5CHO

(Trich de thi thi Dai hoc khoi A, B) Huong dan giai

Theo bai ra: no2 = 0,395 (mol); ncoj = 0,29mol; ni^^Q = 0,29mol

V i T\QQ^ = n^^o => X, Y deu no, her, dcrti chiic

VI X, Y CO Cling s6' nguyen tir cacbon n6n diu c6 ciing CTPT la C„H2nO(x mol)

D a t Z l a C^ H 2 ^ (ymol)

So dd phan ling: X, Y , Z + O 2 - > CO2 + H 2 O

V i nguyen t6' oxi dupe bao toan nen: n ^ ^ y ) + "0(02) = "0(002) ^ "O(H20)

Thi du 16: H6n hop X g6m hai axit cacboxylic dcm chiic D6't chay hoan toan 0,1

mol X cin 0,24 mol O,, thu dirac CO2 va 0,2 mol H2O C6ng thiic hai axit la

A HCOOH va C 2 H 5 C O O H

B CH,=CHCOOH va CH2=C(CH3)COOH

C CH3COOH va C2H5COOH

D CH3COOH va CH2=CHC00H

(Trich de thi thicDqi hoc khoi A, B)

Ctv iMMii M i v iJVVH Khang V i § t

Huong ddn giai

V i axit don chiic => phan tir X c6 2 nguySn til oxi

Nguyen to oxi dupe bao toan nen: 0,1.2 + 0,24.2 = 0,2.1 + n^^^^ 2

Nhan xet:

nco > " H , o c6 axit khOng no (loai A, C) >»A H 61 nSv,,, nc =0,24/0,1 = 2,4 (=> loai B) • Dap an diing la C

Thi du 17: Cho h6n hpp khi X gom HCHO va H , di qua 6'ng sii dung b6t N i nung

nong Sau khi phan ling xay ra hoan toan, thu dupe h6n hpp khi Y g6m hai chat h&u CO Dot chay het Y thi thu dupe 11,7 gam H2O va 7,84 lit khi CO, (o dktc)

PhSn tram theo the tich ciia H , trong X la „:.,

A 65,00% B 46,15% C 35,00% D 53,85%

(Trich dethi thu: Dai hoc khoi A, B) Hu&ng ddn giai

7 84 117 Theo bai ra: nco2 = ^ = 0,35 (mol); nH20 " " f ^ = ("'"'^ ' ' '

Ta CO so do cac phan ling xay ra:

HCHO > HCHO > CO, (0,35 mol)

H, CH3OH H , 0 (0,65 mol) (X) (V) HCHO + H2 > CH3OH u

Thi du 18: Dot chay hoan toan m gam h6n hpp g6m n hidrocacbon khac nhau, thu

dupe 11 gam CO, va 9 gam H2O Gia trj ciia m la ,

rr;i

> 1, > •

J I

d m nang On luygn thi DH 3 mi^n BJlc - Trung - Nam mOn H6a hqc - Cu Thanh ToSn

Ạ 4,0 B 6,2 C 8,0 D 13,6

(Trich de thi thuDai hoc khoi A, B) Huong đn gidi

P T P l / : C , H , ) x C O , + ^ H O * r(X>^ '^^

Vi nguydn t6' C, H duoc bao toan, nen:

nictCxHy) + mH(CxHy) = n i c x H y = ITl0(C02) + " 1 H ( H 2 0 )

Rat nhi^u bai toan phiic tap c6 thé giai nhanh bang each t6 hofp cac djnh luat bao

toan (bao toan khoi lirong, bao toan electron, bao toan dien tich va bao toan

nguydn i6)

I I V A N D U N G

Thi du 1: Hoa tan hoan toan 13,5 gam b6t A l vao dung dich H N O , loang du, sau

cac phan u-ng hoan toan thu duoc 5,6 lit h6n hop 2 khf NO va N.O (dktc) Trong

dung djch kh6ng c6 NH4NO, Kh6'i luong dung djch sau phan ling thay d6i so

vdi kh6'i lircfng dung djch H N O , ban đu la

Ạ khong thay doi B tang 3,9 gam

C tang 13,5 gam D giam 9,6 gam

(Trich de thi thvcDai hoc khoi A, B nam 2013) Huong đn gidi

Theo bai ra: n^, = 13,5 / 27 = 0,5 (mol)

So 66 phan ling: A l + HNO3 ^ AKNO,)., + N O t + N p t + H j O ^ '

Theo djnh luat bao toan kh6'i luong: m^i + mđHN03 = "^đsaứ"NỚ"N20

=> 13,5 + m^dHNOj =mđsau +0.1-30 +0,15.44 ^

=^in„ds.„- niđHNOj =13,5 - 3 -6,6 =3,9 (gam) , , ,

Dap an dung la B • '

Thi du 2: Hoa tan hét 52 gam kim loai M trong 739 gam dung djch HNO3, két thiic

phan iJng thu duoc 0,2 mol NO; 0,1 mol N2O va 0,02 mol N , Biét kh6ng c6 phan ling tao mu6'i NH4NO3 va HNO3 da láy du 15% so vdri luong c^n thiét Kim loai M va n6ng d6 phSn tram ciia HNO3 ban dSu Mn luot la

Ca'm nang On luy$n thi DH 3 mign Bjc - Trung - Nam mOn H6a hpc - CCi Thanh Toan

Vay M la Zn va C % ( H N O , ) = 20%

Dap an diing la B

Thi du 3: Cho 18,4 gam h6n horp X g6m Cu2S,CuS,FeS2 va FeS tac dung hd't vdi

HNO3 (dac nong, du) thu duoc V lit khi chi c6 NO2 (a dktc, san ph^m khu duy

nha't) va dung dich Y Cho toan bo Y vao mot luomg d u dung djch BaCi2 , thu

duoc 46,6 gam ket tua; con khi cho toan bo Y tac dung vori dung djch NH3 du

thu duoc 10,7 gam ket tua Gia trj cua V la

Mat khac: z = = ng^^Q^ =0,2 (bao toan nguydn t6' liru huynh)

Ta lai c6: x = np^ = npg^Qp^^^ = 0,1 (bao toan nguydn to sSt)

64

+3 +2 T6ng s6' m o l electron do Fe - > Fe; Cu - > Cu

Thi du 4: Hoa tan hoan toan 8,9 gam h6n hop g6m M g va Z n bang luong vifa du

500 m l dung djch HNO3 I M Sau khi cac phan iJng ket thuc, thu duoc 1,008 lit

khf N2O (dktc) duy nha't va dung dich X chiia m gam mud'i Gia trj ciia m la

2x + 2 y - 8 z = 0,36 (3) ;

( 1 , 2, 3) => X = 0, l ; y = 0 , l ; z = 0,005 fS'= y;: x 'il ahx nh::i'

Vay m = ni^g(N03)2 + '"zn(N03)2 + "INH4NO3 "q ^_ ,., ,

=0,1.148 + 0,1.189 + 0,005.80 = 34,10(g) * x xn •

Dap an dung la A

Thi du 5: De phan ling h6't a m o l kirn loai M cSn 1,25a m o l H 2 S O 4 va sinh ra k h i X (san ph^m khir duy nha't) Hoa tan hd't 19,2 gam k i m loai M vao dung djch H.S04tao ra 4,48 l i t khi X (san ph^m khir duy nha't, dktc) K i m loai M la

(Trich de tuyen sinh dqi hoc khoi A) Hu&ng ddn gidi

Theo bai ra: nx = 4,48/ 22,4 = 0,2(mol) ^^.^^f,

Ta CO cac qua trinh: (^.^ff

19 2 , 9 2 ••••^

= > — = 0 , 2 4 = 0,8 = > M = — = 2 4 ( M g ) -rx

M 0,8 Vay M la k i m loai M g

Dap an diing la B

Chu v: San phdm k h u X (khi) c6 th^ la SO,, HjS

/ 39

Cim nang On luyQn thi DH 3 mien B^c - Trung - Nam mOn H6a hpc - Cii Thanh Toan

Thi du 6: Cho x mol Fe tan hoan toan trong dung dich chura y mol H 2 S O 4 (ti 16 x: y

= 2: 5), thu duoc m6t san phim khir duy nha't va dung djch chi chiia mu6'i

sunfat S6' mol electron do lucmg Fe trSn nhircmg khi bi hoa tan la

Theo dinh luat bao toan electron: nx = ma (3)

Theo dinh luat bao toan difen tich: nx = 2b (4)

Theo djnh luat bao toan nguydn t6': a + b = y (5)

T i r ( 3 , 4 ) = > m a = 2b (6)

+ V<5ri m = 2 (san ph£m khir SO,):

Tir (6): 2a = 2b => a = b

Thay vac (5): 2b = y

=t> nx = 2b = y (thoa man di^u kidn (2))

Vay s6' mol electron do Fe cho la y e

Dap in dung la C

+ Vdri m = 6 (san phdm khir la S):

Tir (6) ^ 6a = 2b 3a = b

Thay vao (5): 4a = y => 6a = 1,5y '

=> nx = 6a = 1,5y = 1,5 2,5x = 3,75x (=> loai do kh6ng thoa man di6u kifen (2))

+ Vdi m = 8 (san phdm khiJr la HjS):

Tuf (6) => 8a = 2b =i> 4a = b

Thay vao (5): 5a = y => 8a = l,6y

=> nx = 8a = 1,6y = 1,6 2,5x = 4x (=> loai)

Thi du 7: D6't chay hoan toan 7,2 gam kim loai M (c6 hoa trj hai khong d6i trong

hgrp chat) trong h6n hop khi v^ O, Sau phan iJng thu duoc 23,0 gam ch&L ran

va th^ tich h6n hop khi da phan ihig la 5,6 lit (6 dktc) Kim loai M la:

A M g B Cu C B e D Ca

(Tnch de tuyen sinh Cao dang khoi A)

Cty TNHH MTV DWH Khang Vijt

7,2n

M = 0,4 + 0,2=0,6 =>7,2n=0,6M = > M = 12n Suyra: n = 2 ^ M = 24(Mg)

vay kim loai M la Mg

Dap an dung la A

Thidu 8: Cho 29 gam h6n horp g6m A l , Cu va Ag tac dung viJra dii vdi 950 ml dung

djch H N O 3 1,5M, thu duoc dung djch chiia m gam mu6'i va 5,6 lit h6n hap khi

X (dktc) g6m NO va NjO TI kh6'i cua X so vdi H2 la 16,4 Gia trj ciia m la

" N 2 0 = 0.05; n^io = 0,2 (sij dung phuang phap duomg cheo hoac giai he)

* Goi s6' mol NO3 tao mu6'i la x; s6' mol NH4 trong mu6'i la y ''^

- Nguydn to nito duoc bao toan ntn:

41

dm nang On luyjn thi DH 3 miSn B&c - Trung - Nam mOn H6a hpc - CD Thanh Toan

Thi du 9: Cho 5,75 gam h6n hop M g , A l va Cu tac dung vdri dung dich H N O ,

loang, du, thu diroc 1,12 1ft h6n hop khi X g6m N O va N j O {b dktc, t i khd'i cua

X so vori H , la 20,6; san ph^m khir khong c6 N H 4 N O 3 ) Kh6'i lirong mu6'i nitrat

khan thu diroc khi c6 can dung djch la

ii> l e (nhan) = 0,32 + 0,03 = 0,35 (mol)

Cty TNHH MTV CVVH Khang ViSt

Theo nguyen tac bao toan electron:

rr> l e (cho) = 0,35 (mol) (do cac kirn loai nhucmg)

Theo djnh luat bao toan dien tich, ta c6: •••• / ^.^mif

(cho) = X^n^^_ = 0,35 (mol). A & jifurb n

Xheo djnh luat bao toan khoi lucmg ta c6:

Of:, ri

= 5,75 + 0,35.62 = 27,45 (gam) Dap an dung la A

Phuong phap 6: P H a O N G PHAP QUY DOI

I L I T H U Y E T

- Quy d6i la phuong phap bien ddi nham dua h6n hop nhifiu chat phiic tap thanh m6t hay hai chat don gian, qua do lam don gian hoa bai toan ca ve mat hoa hoc iSn toan hoc

- K h i ap dung phuong phap quy d6i c^n tuan thu sir bao toan nguydn t6' va bao

toan so o x i hoa

I I V A N D U N G

Thi du 1: Hoa tan hoan toan 2,44 gam h6n hop bot X gom Cu va Fe^Oy bang dung

djch H 2 S O 4 dac, nong, du Sau phan ling thu dugc 0,504 l i t khi SOj (san ph§im khir duy nha't, dktc) va dung djch chiia 6,6 gam h6n hop muoi sunfat Phdn tram khoi luong Cu trong X la

M a t k h a c t a c o : 64a + 56b + 16c = 2,44 ( 2 ) ^! ; :>

CU-S.CUSO4 2Fe->Fe2(504)3 • y ,

^ - > a b - ) • 0,5b : > H X t : : r ;

AT,

dm nang On luygn thi SH 3 mign Bic - Trung - Nam mOn H6a hpc - Cii Thanh ToAn

=> 160a + 400.0,5b = 6,6r:> 160a + 200b = 6,6 (3)

TO(1,2,3) =^a=0,01;b = 0,025;c=0,025 (ori

Vay %mcu/x =26,23% • ' t( rfrtih <v ,'r«

Dap an diing la A

Thi du 2: Hoa tan hoan toan 24,8 gam h6n hop X gom Fe va Fe^O^ bang dung djch

H2SO4 dac, nong, du thu duoc dung dich Y va 4,48 lit khi SO, (san ph^m khii

duy nha't, dktc) Phdn tram khoi lucmg nguyen tooxi trong X la

A 20,97% : B 16,84% C 25,73% D 32,56%

V i V (Ml L ; r n c (Trich de thi thu Dai hoc khoi A, B nam 2013)

Hu&ng ddn gidi

Theo bai ra: H s o j = 0 2 (molO

Quy d6i h6n hop Fe,Fe^Oy => Fe va O

Thidu 3: Cho 18,4 gam h6n hop X gom Cu2S,CuS,FeS2 va FeS tac dung het v6i

HNO3 (dac nong, dir) thu duoc V lit khi chi c6 N O 2 (or dktc, san ph^m khir duy

nha't) va dung djch Y Cho toan b6 Y vao mot luong du dung djch BaClj , thu

duoc 46,6 gam ket tua; con khi cho toan b6 Y tac dung v6i dung dich N H , du

thu duoc 10,7 gam ket tiia Gia tn cua V la

Cty TNHH MTV DWH Khang Vi^t

j^at khac: z = = n^^st-u = 0.2 (bao toan nguyen to liru huynh) ^ ^ ^

Ta lai c6: X = n p, = n ^ ^ ^ ^ ^ = 0,1 (bao toan nguyen to' sat) ^ , ,

18,4-56.0,1-32.0,2 ^ , >

T i r ( l ) = > y = 0,1

+3 2+ J ^ l ^Omd • ,rj,.:r^

T6ng s6'mol electron do F e - > Fe;Cu-> Cu

S-J-s" Ian, (cho) = 0,1 3+0,1.2+0,2 6 = 1,7 mol •.,.i (dt

+.<; +4

=> T6ng s6' mol electron do N + le -> N la 1,7 mol

nN02 = I.7mol=> V^,,^ = 1,7.22,4 = 38,08 (/)

Dap an diing la B

Thi du 4: H6n hop X gom C^H, va c6 cung s6' mol La'y mot luong h6n hop X

cho qua chat xiic tac, nung nong duoc h6n hop Y g6m C2H4, C^H^ va C^H,, H2

du Dan Y qua nudfc brom thay blnh nude brom tang 10,8 gam va thoat ra 4,48 lit h6n hop khi (do o dktc), c6 t i khd'i so vori la 8 The' tich khi O , (khi do o dktc) vvtdi du de dot chay hoan toan h6n hop Y la

A 33,6 lit B 22,4 lit C 26,88 lit D 44,8 lit

(Trich de thi thi Dai hoc khoi A, B) Huong ddn gidi

Ta c6: mx = my => mx = 10,8 + ^ ^ ^ 8 2 = 14(g) ^ I ' l

22,4

=>26a + 2 a = 14 => a = 0,5 Quy h6n hop X thanh cacbon va khi hidro

=> He = 2a = Imol; n^^^ = a + a = 1 (mol) '

Vi C va H duoc bao toan nan nc,x) = H C Y ) = 1 (mol); n H 2 ( X ) = " H 2 ( Y ) = ' m o l

C + O , ^ C O , 2 H , + O , 2 H 7 O ^2 ^^^2 ^ 1 1 2 ^ ^ i i 2 v

l - > l ( m o l ) 1 0,5(mol) or vay: Vo^ = ( I + 0,5).22,4 = 33,6 (/) ^^^T - ' Dap an diing la D '

Thi du 5: Cho mot luong khi CO di qua 6'ng sir dung m gam Fe^O, nung nong Sau

mot thai gian thu duoc 10,44 gam chat rdn X gom Fe, FeO, FcO,, va F e A - Hoa tan het X trong dung dich H N O , d i e , nong thu duoc 4,368 lit N O , (san ph^m

khir duy nha't a d i l u kien chudn) Gia tri ciia m la

A 12 B 24 C 10,8 D 16

(Trich de thi thu: Dai hoc khoi A, B)

Ca'm nang fln iuyen tin Ud 3 [iniin BSC - Trung - Nam mOn H6a hgc - Cu Thanh Toan

Thi du 6: De hoa tan hoan toan 2,32 gam h6n hop gom FeO, Fe304 va Fe203 (trong

do s6' mol FeO bang so mol FcjO,), cin dung vita dii V lit dung djch HCl I M

Gia tri ciia V la

FeO Fe.Oj + 8HC1 > FeCK + 2FeCl3 + 4H2O

Thi du 7: Cho 9,12 gam h6n hop g6m FeO, Fe^Oj va Fe304 tac dung v6i dung djch

HCl (dit) Sau khi cac phan ling xay ra hoan toan, duoc dung djch Y, c6 can Y

thu duoc 7,62 gam F e C l , va m gam FeCl3 Gia trj ciia m la

Theo bai ra: n ^^^^ = = 0,06(mol)

V i Fe304 = FeO.Fe203, ntn c6 the coi h6n hop g6m FeO (x mol) va FejOj (y mol)

Theo bai ra, ta c6: 72 x + 160 y = 9,12 (1)

PTPU-; FeO + 2HC1 > F e d , + H.O

Fe^Oj + 6HC1 > 2FeCl3 + 3H2O

y 2y

nFeCi2 = 0.06 (mol) (2)

Tir ( 1 , 2) ta c6: x = 0,06; y = 0,03 (mol)

v a y m = m p,ci3 = 2 y 162,5 = 2 0,03 162,5 = 9,75 (gam). , ; s ^ Dap an diing la A •

Thi du 8: Cho m gam h6n hop X gom F e O , F e 2 0 3 , F e 3 0 4 vao mdt Iitong vita dii

dung djch HCl 2 M , thu dugc dung djch Y c6 ti \t s6 mol Fe^"" va Fe"'* la 1: 2

Chia Y thanh hai phSn bang nhau C 6 can phSn m6t thu dugc m, gam muoi khan Sue k h i clo (du) vao phSn hai, c6 can dung djch sau phan itng thu dugc m2 gam mud'i khan Biet m2 - m, = 0,71 The ti'ch dung djch HCl da diing la

A 320 m l B 80 ml C 240 m l D 160 m l

(Trich de thi thucDai hoc khoi A, B) Hu&ng ddn gidi

Fe^O^ = FeO.Fe203 => coi h6n hop X chi gom FeO (x mol) va Fe203 (y mol)

FeO + 2HC1 ^ FeCl2 + H 2 O Fe203 + 6HC1 ^ 2FeCl3 + 3 H 2 O

=^\<iHa = 0 , 3 2 / 2 = 0 , 1 6 ( 1 ) - 1 6 0 m l

47

dm nang On luyjn thi BH 3 mi^n B^c - Trung - Nam mOn H6a hpc - CCi Thanh Toan

Thi du 9: Hoa het 2,32 gam h6n hop FeO, Fe,04, Fe.Oi (trong do FeO, Fe.O, c6 s6'

mol bang nhau) trong 80 m l dung djch HCl I M thu duoc dung dich X Cho X

tac dung vdri dung dich AgNO, du thi thu duoc bao nhi^u gam chat kh6ng tan?

=^nFeCi3 =0'02(mol); npgci2 =0,01(mol)

FeCl3 + 3AgN03 >3AgCl i +Fe(N03)^

Thidu 10: Dot chay hoan toan mot h6n hop A (glucozo, andehit fomic, axit axetic)

cdn 2,24 lit O2 (didu ki^n tieu chuSin) Dan san phsim chay qua binh dung dung

djch Ca(0H)2du, tha'y khdi luong binh tang m gam Gia trj ciia m la

axit axetic CH3COOH => ( ^ 2 0 ) ^

Do do, ta quy h6n hop A la CH2O

C H 2 O + 0 2 ^ CO2 + H 2 O

0 , 1 ^ 0,1 - > 0,1 (mol)

Cty TNHH M W OWM KfaWQ Vigt

Khdi luong binh tang chinh bang kh6'i luong san p h ^ chay (C02,H20)

* Cdc hudc viet phifcmg trinh ion nit gpn: ^./i.) i- i

Budc 1: Viet phuang trinh phdn I'fng md cdc chat tham gia vd sdn pham duai dang phdn ti((nh('rcdn hang phdn I'/ng)

- Buac 2: Cdc chat dien li manh duac viet dudi dang ion; cdc chat khong tan, khi, dien li yen diMc viet dudi dang phdn ti( ^phucfng trinh ion day du •

- Bi(c/c 3: LiMc ho cdc ion giong nhau d hai ve phucfng trinh ion rut gpn

* Hdn hajj nhieu axit, baza tdc dung vdi nhau, phdi svt dung phuang trinh ion rut gon W + OH' > H2O degidi

* Di/a vdo pH cua dung dich sau phdn itng => H* hay 0H~ phdn itng het

I I VAN DUNG

Thi du 1: Cho 0,3 mol b6t Cu va 0,6 mol bot Fe(N03), vao dung djch chiia 0,9 mol

H 2 S O 4 (loang) Sau khi cac phan ihig xay ra hoan toan, thu duoc V lit khi NO (

san phim khu duy nha't, d dktc) Gia trj cua V i a '

dm nang On luy$n thi OH 3 mign BJic - Trung - I u i ; l i o i ; t i o a i v u C i ihanh Tocin Cty INHH M i v I W V M Khang Vijt

Thi du 2: La'y 200 m l dung djch A chiia HCl, HNO3, H,S04 c6 ti \t so mol 1: 5 : 1

cho tac dung vdi A g (du), r6i dun nong tha'y thodt ra t6'i da 22,4 m l khi NO^

(duy nha't, do of dktc) Gia trj pH cua dung djch A la

A 1,795 B 2,79 C 2 D 3 "

(TrichdethithuDaihgckhoiAyB ndm2013) Hu&ng dan gidi

Theo bai ra: nN0 2 = 22,4.10"V 22,4 = 0,001 (mol)

Phuong trinh ion rut gon ciia phan ling:

Thidu 3: Thirc hien thi nhgiem sau:

1 Cho 3,84 gam Cu phan ling vdi 80 ml dung djch HNO3 I M thoat ra V , lit NO

2 Cho 3,84 gam Cu phan urng vdi 80 m l dung djch HNO3 I M va H2SO4 0,5 M

Thinghieml: nc„ = 0,06 (mol); n H N O j = 0,08 (mol)

3Cu + 8HNO3 > 3Cu(N03)3 + 2NO + 4H2O

Thi du 4: Hoa tan hoan toan 8,94 gam h6n hop g6m: Na, K va Ba vao niT6c, thu

dugc dung djch X va 2,688 lit khi H , (dktc) Dung djch Y g6m HCl va H,S04 ti

\t mol tuong ling la 4: 1 Trung hoa dung djch X bai dung djch Y , t6ng khd'i

luang cac mu6'i d\sgc tao ra la

A 13,70 gam B 14,62 gam C 18,46 gam D 12,78 gam

(Trich de thi thi Dai hoc khoi A, B) Huong ddn gidi

Theo bai ra: nH^ = 2,688/22,4 = 0,12(mol)

Thi du 5: Trong dung djch X c6: 0,02 mol Ca""; 0,05 mol Mg""; 0,02 mol

HC03;Cr Trong dung djch Y c6: 0,12 mol O H " ; 0,04 mol CF; Cho X vao

Y, sau cac phan ling hoan toan khoi Itrcmg ke't tiia thu duoc Idn nha't la:

A 2,0 gam B 4,2 gam C. 4,9 gam D 6,2 gam

(Trich de thi thii Dai hoc khoi A, B) Hu&ng ddn gidi ,XM ^im a»iti< • ^ PTHH (ion nit gon):

H C O 3 + O H "

0,02 - > 0,02 Mg2+ + 2 0 H '

0 , 0 5 ^ 0 , 1

> coj- + H2O

> 0,02 (mol)

- > M g ( O H ) 2 ^ 0,05 (mol)

Ca^* + C O ^ 0,02 ^ 0,02 -

• C a C O j i 0,02 (mol)

mi

dm nang On luygn thi BH 3 miin BJc - Trung - Nam mOn H6a hpc - Cu Thanh Todn

Khoi lirgng kd't tiia thu duoc Idn nha't la:

m ^ = m c a C 0 3 + m M g ( o H ) 2 = 0,02.100 + 0,05.58 = 4,9(g)

DAp an diing la C

Chu v: KA tiia Mg(OH) c6 d6 tan be hon nhieu so vdi MgCO,

SMg(OH)2 = M 1 0 - ' m o l / l ; S M g c o 3 - 3 , 2 1 0 - ' m o l / l

Thi du 6: Cho 0,87 gam h6n hop gom Fe, Cu va A! vao binh dung 300 ml dung

dich H2SO4 0 , l M Sau khi cac phan ung xay ra hoan toan, thu duoc 0,32 gam

chat ran va c6 448 ml khi (dktc) thoat ra Them tie'p vao binh 0,425 gam NaNO,,

khi cac phan iJng ke't thiic thi the' ti'ch khi NO (dktc, san ph^m khir duy nha't) tao

thanh va khoi luong mu6'i trong dung djch la

A 0,224 lit va 3,750 gam B 0,112 lit va 3,750 gam

C 0,112 lit va 3,865 gam D 0,224 lit va 3,865 gam

(Trich de tuyen sinh Dai hoc khoi A) Hu&ng ddn gidi

Theobaira: n H 2 S 0 4 = 0.03mol; HNaNOj =0,005mol

Khi phan irng vdi NaNOj c6:

n =0,02mol; n^,^_ =0,005mol;nr„ =0,005mol; n =0,005mol

Vay V N O = ( 0 ' 0 1 / 3 + 0,02/12).22,4-0,112(1) , „rD >,

m , „ : , = m p , c u A , - m ^ ^ , + m ^ 2 - =0,87.0,005.23.0,03.96 = 3,865(g)

Dap an diing la C

Chuj: Cu,Fe^^ (FeS04) deu bj oxi h o a boi NaNO, t r o n g m6i trucmg a x i t H 2 S O 4

Thidu 7: Cho 7,68 g a m Cu vao 200 m l dung d i c h g6m HNO3 0,6M va H2SO4 0,5M

Sau khi cac p h a n l i n g xay ra h o a n t o a n (san p h ^ m khir d u y nha't la NO), c6 can

c^n than t o a n bo d u n g d i c h sau p h a n umg thi k h o i l u o n g mu6'i k h a n thu d u o c la

=:>m = 7,68 + 0,04.62 + 0,1.96 = 19,76(gam) ^ ,

Dap an d u n g la B

Thi du 8: Cho 3,2 g a m bot Cu t a c d u n g vdi 100 m l d u n g d j c h hdn hop g d m H N O 3

0,8M va H 2 S O 4 0,2M Sau k h i c a c p h a n l i n g xay ra h o a n t o a n , s i n h ra V lit k h i

NO (san p h d m k h i r d u y nha't, d d k t c ) Gia trj c i i a V la

d m nang 6n luyQn thi DH 3 miin BJc - Trung - Nam mOn H6a hqc - Cu Thanh Toan

Do do: H N O = 0 , 1 2 2 / 8 = 0,03 (mol)

= > V N O = 0,03 22,4 = 0,672 (lit)

Dap an dung la C

Thi du 9: Cho 1,82 gam h6n hop b6t X g6m Cu va A g ( t i Id s6' m o l tuong ling 4:1)

v&o 30 m l dung djch g6m H 2 S O 4 0 , 5 M va H N O , 2 M , sau khi cac phan iJug xay

ra hoan toan, thu duoc a mol k h i N O (san ph£m khir duy nha't ciia N*'') Tr6n a

mol N O tren vdi 0,1 mol O , thu duoc h6n hop khi Y Cho toan b6 Y tac dung

v6i H2O, thu duoc 150 m l dung dich c6 pH = z Gia t n ciia z la

3Ag + 4H+ + N O ; ,

0,005 0,11/3 0,14/3 0,005

Dap an dung la A

Cty TNHH MTV DWH Khang Vi^t

Thi du 10: D u n g dich X c6 p H = 1 chira H C l 0 , 0 2 M va H 2 S O 4 D u n g dich Y c6

pH = 13 chura K O H 0,025 M \h Ba(OH)2 Cho V lit dung djch Y v^o 0,100 lit

dung djch X thu duoc dung djch c6 p H = 12 va m gam kd't ttia Gia trj ciia V va

m l ^

A 0,122; 1,006 B 0,14; 0,932 C 0,122; 0,932 D 0,110; 0,874

(Trich de thi thi Dai hoc khoi A, B)

Trong 0,1 lit dung djch X :

Phan ling:

Con:

H^ O H ' -> H 2 O 0,01 0,1V (mol) 0,01 0,01 (mol) ^ ' ^ ' ^

0 ( 0 , 1 V - 0 , 0 1 )

=> 0,1 V - 0,01 = 0,01 ( V + 0,1) 0,1 V - 0,01 = O.OIV + 0,001 0

= > 0 , 0 9 V = 0,011 = > V = 0,122(1)

P T H H : Ba^^ + SO^" • Ban ddu: 0,0046 0,004 Phan umg: 0,004 < - 0,004

dm nang On luy^n thi DH 3 m\6n BJc - Trung - Nam mOn H6a hoc - Cii Thanh Toan Cty TNHH WTV P W H Khang Viet

Thidu 11: Co V lit dung djch chiia 2 axit HCl a (M) va H,S04 b (M) C^n la'y x lit

dung dich chiia 2 baza la NaOH c (M) va Ba(OH)2 d (M) de trung hoa vCra du V

lit dung djch 2 axit trdn Bie'u thiJc tinh x theo V , a, b, c, d la

Thidu 12: Co hai dung djch A va B Dung dich A chufa H2SO4 0,2M va HCl 0,1 M

Dung djch B chura K O H 0,3M va Ba(0H)2 0,1M Cho dung djch A trung hoa v6i

0,5 lit dung djch B, sau phan ling tha'y c6 m gam ket tiia Gia trj cua m la:

A 46,60 B 5,825 C 11,65 D 23,30

(Trich de thi thu Dai hoc khoi A, B) Huang ddn gidi

Trong 0,5 lit dung djch B c6:

Thi du 13: Hoa tan 19,2 gam Cu vao 500 m l dung djch NaNO, I M , sau do them

vao 500mi dung djch HCl 2 M Kd't thiic phan lifng thu duoc dung djch X v^ khf

NO duy nha't, phai them bao nhieu ml dung djch NaOH I M vao X kd't tiia hfi't

=>n _ = 0 , 2 + 0,6 = 0,8(1) = 800(ml)

OH

E)ap an dung la B

Thi du 14: Cho 4,8 gam b6t Cu.S vao 120 m l dung djch NaNO, I M , sau do them

200 mi dung djch HCl I M vao, ket thuc phan ling thu duoc dung dich X va V lit khi NO (san ph^m khir duy nha't, dktc) Gia tri ciia V la

A 67,2 B.22,4 C 2,24 D 6,72

(Trich de thi thu:Dai hoc khoi A, B)

B&c - Trung - Nam mOn H6a hpc - Cti Thanh Tojin

Huong dan gidi

Thidu 15: Cho 14,4 gam h6n hop M g , Cu, Fe c6 s6' m o l bang nhau vao 0,8 l i t dung

djch H 2 S O 4 I M (loang) C^n phai them it nha't bao nhidu gam N a N O , vao h6n

hop sau phan ling thi khong con khf N O (san ph^m k h u duy nha't) thoat ra?

* Phan ung vdti H 2 S O 4 (1):

M g + H2SO4 > M g S 0 4 + H2 t Fe + H2SO4 > FeS04 +

V i con dir (0,4/3 + 0,8/3 = 0,4 < 1,2) nfen Fe"* va Cu phan iJng he't

Vay mNaN03 = ( 0 , l / 3 + 0,2/3).85 = 8,50(gam)

Dap an diing la D

Cty TNHH MTV D W H Khang Vi^t

-^^^^^^r^^^bpB: PHUONG PHAP GHEP AN

Nhieu ht toan hoc ciia b^i tSp hoa hoc kh6ng giai dirge vi s6' ^n n h i l u hon s6'

phuong trinh, nhung v§n giai quyet dugc cau hoi ciia Ai bai bang phirong phap

ghep (tiJc la khong c6 ket qua tirng an ma chi t i m dugc cum ^ n , cum ^n cung chinh la cau tra Icri ciia cau hoi trong bai tap) r r'" '^e.tX) 1/

Da'u hi^u nhan biet la he s6' c6 so in nhi^u hon so phuong trinh

11 V A N D U N G

Thi du 1 : Hoa tan hoan toan h6n hgp 3 k i m loai Na, Z n va Fe trong dung djch

H^S04 loang, d u thu dugc V lit H2 (dktc) Hoa tan hoan toan h6n hgp g6m k i m loai M (c6 hoa t r i I I ) va Fe trong H2SO4 loang, du cung thu dugc V lit khi (dktc) Bifit Fe c6 kh6'i lucmg nhu nhau trong 2 hdn hgp va kh6'i lugng cua M bang nira t(5ng kho'i lugng ciia Na va Zn trong h6n hgp ban ddu K i m loai M la

A M g B N i C Ca D Ba

(Trich de thi thii Dai hoc khoi A, B nam 2013) Hu&ng dan gidi

P T H H : ' + 2Na + H2SO4 - > Na2S04 + H2 :m) dt.,u =: a

23a + 65b 23a + 32,5.2b

• M =

-a + 2b -a + 2b _^ 23a + 23.2b 32,5a + 32,5.2b ^

a + 2b a + 2b 23{a + 2 b ) 32,5(a + 2 b ) -'^

= > 2 3 < M < 3 2 , 5 = > M = 2 4 ( M g ) Dap an dung la A

CSm nang on luyijii tin UH 3 rnjgn B^c - Trung - Nam mOn H6a hgc - Cu Thanh ToAn

Thi du 2: Cho butan qua xuc tac (a nhiet d6 cao) thu duoc h6n hop X gom C4H,o,

C4H8, C4He, va H , T i kh6'i cua X so v6i butan la 0,4 Nd'u cho 0.6 mol X vac

dung djch brom (du) thi s6' mol brom toi da phan ting la j j ^

A 0,60 mol B 0,36 mol C 0,48 mol D 0,24 mol

(Trich de thi thu Dai hoc khoi A, B) Hu&ng dan giai

Thi du 3: Di hoa tan het h6n hop X gom Cr^O,, CuO, Fe304 cdn viTa dii 550 ml

dung dich HCI 2 M , sau phan ling thu duoc dung dich Y M6t nira dung dich Y

tha'y hoa tan hd't td'i da 2,9 gam bdt N i C6 can nijfa dung djch Y con lai thi thu

duoc bao nhieu gam mu6'i khan?

A 30,525 gam B 30,8 gam C 61,6 gam D 61,05 gam

(Trich dethi thvcDai hoc khoi A, B) Huong dan giai

Theo bai ra: n^^Q =l,l(mol);nf^i =0,05(mol)

Goi X, y, z \in luot la s6' mol cua Cr^Oj, CuO, Fe,04 c6 trong h6n hop X

PTPU: Cr^O, +6HCI -> 2CrCl, +3H2O

do la phuong phap duy nha't

Cac dai luong trung binh thucmg duoc sir dung la:

- Nguyen tii kh6'i trung binh

- Phan tir khoi trung binh

- S6' nguydn tif cacbon trung binh trong phan tir

- So nhom chiic trung binh trong phan tu

n V^kN D U N G

Thidu 1: Di hoa tan hoan toan 6,4 gam h6n hop g6m kim loai R (chi c6 hoa trj II)

oxit ciia no cSn vira du 400ml dung djch HCI I M K i m loai R la

dm nang On luy$n thi DH 3 mign BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan

Oxit cua k i m loai R la RO

Thi du 2: Hoa tan hoan toan 6,645 gam h6n hop mu6'i clorua ciia hai k i m loai k i ^ m

thuoc hai chu k i k6 tiep nhau vao nude dugc dung dich X Cho toan bo dung

djch X tac dung hoan toan vdi dung djch A g N O , (du), thu duoc 18,655 gam ket

tiia Hai k i m loai kiem trdn la

Thi du 3: Hoa tan hoan toan 1,1 gam h6n hop gdm mot k i m loai kiem X va mot

k i m loai kiem th6 Y ( M x < M y ) trong dung djch H C l du, thu dircc 1,12 l i t khi

Thidu 4: H6n hop X chiia dong thdi hai muoi natri ciia hai halogen Wtn tiep trong

bang tuSn hoan La'y m6t luong X cho tac dung vira dii vori 100ml dung dich

A g N O , I M thi thu dugc 15 gam ket tiia Cong thiJc phan tir ciia hai muoi trong

X l a

A NaF va NaCl B NaBr va N a l C NaCl va N a l D NaCl va NaBr

(Trich de thi thu Dai hoc khoi A, B) Hu&ng ddn gidi

Theo bai ra: n^gNOs = 0 , l m o l

* Tririmg hop 1 (xet rieng): Hai mu6'i la NaF va NaCl ' ' '

NaF + A g N O , > khong xay ra irf>S 'i;

-NaCl + A g N O , ^ A g C l i + N a N O , ^ ,

0,1 ^ 0,1 m o l

=> m ^ g Q i =0,1.143,5 = 14,35gam 5'15gam => loai trucmg hop nay : t v r y '

* Truong hop 2: H a i halogen deu tao duoc ket tiia vori Ag*

Dat cong thiic chung ciia muoi la N a X

P T P l / : N a X + A g N O ^ - > A g X i + N a N O j

=^ M ^ g - = i ^ = 1 5 0 =^ 108 + X = 150 = ^ X = 42

Suy ra, hai halogen la Cl(35,5) va 3 r (80)

Vay hai muoi trong X la NaCl va NaBr !;

Dap an dung la D

Thi du 5: Cho dung dich chira 6,03 gam h6n hop gdm hai mu6'i N a X , N a Y ( X , Y la

hai nguyen to c6 trong tir nhien, a hai chu kl lien tie'p thudc nhom V I I A , so hieu nguyen tir Zx < Zy) vao dung djch A g N O , (du), thu dugc 8,61 gam ket tiia Phin tram khoi lugng ciia N a X trong h6n hop ban diu la

A 47,2% B 5 2 , 8 % C 58,2% D 4 1 , 8 %

(Trich de thi tuyen sinh DH - CD khoi B) Hu&ng ddn gidi

* Tru&ng h ffp 1-.Xla no, Y\a do: la H

Goi X, y Idn lugt la s6 m o l cua NaF, NaCl 'i** A'-'

Theo bai ra, ta c6: 42x + 58,5y = 6,03 (1) >: • NaF + A g N O j > kh6ng xay ra

63

cam nang on luygn tin U H 3 inign B&c - Trung - Nam mOn H6a hpc - CD Thanh ToAn

NaCl + AgNO, > AgCl i + NaNO,

* Tru&ng hop 2: Hai mu6'i NaX va NaY d^u tao ket tua vdi AgNOj

Dat c6ng thiic chung cua 2 mudi la N a X

NaX + AgNO, > AgX i + NaNO,

Khi 1 mol NaX > 1 mol AgX => khdi luong tang 85 gam ,

Vay khi tang (8,61 - 6,03) thi so mol NaX la:

C O trong tu nhien ma c6 M > 178 => loai trudng hop nay)

Thi du 6: Dot chay hoan toan h6n hop X g6m hai hidrocacbon kfi' tiep nhau trong

day d6ng dang, thu duoc 2,24 lit khf CO, (dktc) va 3,24 gam H,0 Hai

Thi du 7: H6n hop M g6m m6t andehit va m6t ankin (c6 cung s6' nguydn tir

cacbon) D6't chay hoan toan x mol h6n hop M , thu duoc 3x mol CO, va l,8x

mol H2O PhSn tram s6' mol ciia andehit trong h6n hop M la

Thi du 8: Dot chay hoan toan mot lugng h6n hop X g6m 2 ancol (d^u no, da chiic,

mach hor, c6 cung s6' nhom - O H ) c5n viTa dii V lit khi O,, thu duoc 11,2 lit khf

CO, va 12,6 gam H , 0 (cac the tfch khf do a dktc) Gia tri cua V la , „

A 11,20 B.4,48 C 14,56 D 15,68

(Trich de thi thi Dai hoc khdi A, B) Hu&ng ddn gidi

Theobaira: n ^Oj =0,5 mol; n H 2 o = 0 , 7 m o l

=> S6' mol h6n hop ancol no, da chiic, mach ho bang:

(n + 1)

0,7

M.S \

65

Ca'm nang On luy^n thi DH 3 mign B^c - Trung - Nam mOn H6a hqc - Cii Thanh Toan

Suy ra: rT = = M = 2,5 => Co ancol C , H 4 ( O H ) , a = 2)

0,2.(3n + l - a ) 0 , 2 ( 3 2 , 5 + 1 - 2 ) ^ ^ ,^

" o , = ^ = z - 0 , 6 5 ( m o l ) ,

= 0,65.22,4 = 14,56 (lit)

Dap an dung la C

Thi du 9: H6n hop X gom hai ancol don chiic, mach ho thuoc ciing day dong d i n g

va 3 ete tao ra hai ancol do D6't chay hoan toan m gam X cin vira du V l i t O,

(dktc), thu duoc 0,81 m o l C O , va 0,99 m o l H^O Gia t r i cua m va V l^n lirot la

Chit y: Ancol no, hot, don chiic va ete no, hd, don chiic la d6ng phan nhom chiic cua

nhau, d6u c6 c6ng thiic chung la C„H2„+20 K h i d6't chay thi n^^o > "CO2

^ ^ • uOd to

Thidu 10: D6't chay hoan toan m gam h6n hop X g6m hai ancol, thu duoc 13,44 1ft

khi C02(dktc) va 15,3 gam H 2 O Mat khac, cho m gam X tac dung vdri Na (du),

thu duoc 4,48 l i t k h i H2(dktc) Gia tri ciia m la

A 12,9 B 15,3 C 12,3 D 16,9

(Trich de thi thu Dai hoc khdi A, B) Hu&ng ddn gidi

Theo hki ra: n^Oj = 0,6 mol; rxy^^Q - 0,85 mol; n ^ ^ = 0 , 2 mol

- "H20 > nco2 => ancol no, ho C„H2„^20,

Thi du 11: Dot chay hoan toan 1 li't h6n hop k h i gom C ^ H , va hidrocacbon X sinh

ra 2 h't k h i C O 2 va 2 li't hoi H.O (cac the ti'ch khi va hoi do 6 cung di^u kien

nhiet do, ap sua't) Cong thiic phSn t i i c i i a X la

Thi du 12: Dot chay hoan toan h6n hop n hidrocacbon cung day ddng dang thu

duoc 5,04 l i t C O 2 (do 0 dktc) va 6,75 gam nude Trong h6n hop hidrocacbon

nay phai c6

A etan B metan. C xiclopropan D xiclobutan

(Trich de thi thu: Dai hoc khoi A, B) Hu&ng ddn gidi

Theo b ^ i ra: nco2 = 5,04/ 22,4 = 0,225 (mol); nH20 = 6,75/ 18 = 0,375 (mol)

" H20 > nco2 =^ <^ay <^'^ng dang ankan (C„H2„ ^2) ,^01

dm nang On luy^n thi BH 3 mign BJc - Trung - Nam mOn H6a hpc - Cii Thanh Toan

Chii v: Khi dd't chay hoan toan hOn hop ankan thi:

r ^ i J M i J ; D6't chay hoan toan 6,72 h't (dktc) h6n hop g6m hai hidrocacbon X va Y

(My > Mx), thu duoc 11,2 h't khi CO, (dktc) va 10,8 gam H^O Cong thiic ciia X

V i X (CH4) C O 4 nguyen tu H trong phan tir

=> Y phai C O 4 nguydn tu H trong phan tif

Y m C2H4 hoac C,H4 hoac C4H4

Thi du 14: Cho 4,48 lit hdn hop X (o dktc) gom 2 hidrocacbon mach ho I6i tir tiJf

qua binh chiia 1,4 lit dung dich Br, 0,5M Sau khi phan urng hoan toan, s6' mol

Br, giam di m6t niJfa va khoi luong binh tang them 6,7 gam Cong thiJc phan tir

=> Trong X c6 1 anken va 1 ankin hoac 1 ankin va 1 ankan

(suy ra loai phuong an D)

+ Gia sir X chiJa 1 anken va 1 ankin (chiing d^u bi dung djch Br, du giO lai)

=> mx= 6,7 gam (khd'i luong binh dung dung dich Br, tang)

Suy ra: M x = ^ = ^ - 3 3 , 5

' nx 0.2

Cty TNHH U j V iJWH Khang ViQt

£X) do, trong X phai c6 m6t hidrocacbon khong no (mach hor) c6 M < 33,5 do phai l a C , H , ( M = 26)

Pap an diing la A

+ Gia sir X chiia 1 ankan va 1 ankin => m,,,^;,, = 6,7 gam ^^

n „ = l / 2 - n B , , =0,35/2 = 0,175(mol) „ : ,

Suy ra, trucmg hop nay loai ' • • '

Chuli d\jtgc hidrocacbon thii hai la C4H8 s6 mol m6i

hidrocacbon (0,15 mol CjH, va 0,05 mol C4H8) • • < v-t^ — d '

Thi du 15: DSn 1,68 lit hdn hop khf X gdm hai hidrocacbon vho binh dung dung

dich brom (du) Sau khi phan iJng xay ra hoan toan, c6 4 gam brom da phan ufng

\h con lai 1,12 lit khf Ne'u d6't chay hoan toan 1,68 lit X thi sinh ra 2,8 lit khf

C O 2 C 6 n g thiic phan tir hai hidrocacbon la (hi6t cac the' ti'ch khf d^u do 6 dktc)

A C H 4 v a C , H 4 B.CH4vaC3H4 C.CH4vaC3H6 D Q H ^ va CaH^

(Trich de thi thiiDai hoc khoi A, B) Hu&ng ddn gidi

Theo bai ra:' nx = 1,68/ 22,4 = 0,075 (mol) ^ ' '

nx 0,075

=> Trong h6n hop X c6 CH4 (HCH ^ = 0,05 mol)

n (hidrocacbon c6„ lai) = 0,075 - 0,05 = 0,025 mol = ner 2 (phin tag) 1^

=> hidrocacbon con lai la C„H2„

0,05 > 0,05 (mol) 0,025 > 0,025n (mol) Tac6: 0,05 + 0,025n = 0,125

=^n = 3(C3H<,)

Vay trong X c6 CH4 va CjH*

Dap an diing la C

elm nang On luyOn thi BH 3 mi^n Bjc - Trung - Nam mOn H6a hgc - Cu Thanh Toiln

Phaong phap 10:

PHaONG PHAP TIJ CHON LLrONG CHAT THICH HOP

I L I T H U Y E T

Co m6t so bai tap tirong chirng nhu thifiu dO kien ho&c 66 ra cho duori dang t6ng

quat nhu m gam, n mol, V h't, hoac cho dudi dang ti le (ti le ve khdi luong, so

mol, the tich, ) gay kho khan va be' tSc cho viec tinh toan De giai nhanh cac

bai tap dang nay td't nha't ta tu lira chon mdt luong chat (1 mol, 100 gam, 1 lit,.)

di cho viec giai quyet bai toan trof ntn don gian

I I V A N DUNG

Thi du 1: Khi oxi hoa C 2 H 5 O H bang CuO nung nong, thu duoc h6n hop hoi X chi

gom CH,CHO, H , 0 va C 2 H 5 O H du X c6 khdi luong phan tu trung binh bang 36

dvC Hidu sua't phan dug oxi hoa C^H^OH da xay ra la -OD

Ban dau: 1 mol

Phan ling: x x ^ x (inol)

Con lai: 1 — x x x (mol

Thi du 2: D6't chay hoan toan m gam ancol R, san phdm thu duoc cho di qua binh

dung dung dich nude voi trong du, tha'y khdi luong binh tang them p gam va c61

gam ke't tiia

Xac dinh c6ng thiifc ciia R Biet p = 0,7 It; t = (m + p) /1,02

(Trich de thi thifDgi hoc khoi A, B) Hu&ng ddn giai

Chon t = 100 = mcaco3 => ticacoj = 1 mol

=^ p = 0,71t = 7 1 ; m = 100 1,02 — 7 1 = 3 1

Goi C T T Q cua ancol la C^H^O,

QHyO, ) xCO, + y/2H,0

CO, + Ca(OH)2 > CaCOj + H^O

Vi nguy^n td cacbon duoc bao toan ntn n c ( R , = nco2 = "caCOs - ^ '"o'

Khdi luong binh tang len: p = mcoj + • " H J O

Vay ancol R la C:H4(OH), •

Thi du 3: Cho dung dich axit axetic c6 ndng do x% tac dung viia dij vori dung dich

NaOH C O ndng do 20% thi thu duoc dung djch mudi c6 ndng dd 10,25% Gia trj

Thi du 4: Crackinh C4H10 duoc hdn hop chi gdm 5 hidrocacbon c6 khdi luong mol

la 36,25 gam Hieu sua't ciia phan ihig crackinh la

71

crin! luiny on luyC)iHtji_DH_3_nnie(! H.1r, iuing - Nam mOn H6a hQC - Cu Thanh Toan

Trong phan iJng crackinh, ta tha'y ciJ 1 mol C4H10 phan ihig thi sinh ra 2 mol cua

hai hidrocacbon, do do s6' mol C4H10 tham gia phan ling dung bang s6' mol

hidrocacbon tang sau phan ling

^^T "C4H,o = 1,6 - 1 = 0,6 mol

Hifiu sua't phan ling: H % = ^ 100% = 60%

Dap an diing la D

Thi du 5: Hoa tan het m6t oxit kim loai kiem tho vao dung djch H2SO4 c6 n6ng do

24,4% (vira dii), thi thu duoc dung djch mu6'i c6 n6ng d6 27,17% K i m loai

ki^m th6 la

(Trich de thi thd Dai hoc khoi A, B) : Hu&ng ddn gidi

Oxit kim loai la M O (x mol)

PTPU: M O + H2SO4 ^ MSO4 + H2O

Kh6'i luong dung djch sau phan ling: m^^,, = 401, 64 + 1 ( M + 16) (gam)

Theo bai ra: 27,17 = J M l ^ 00 _ ^ ^ ^ ^4 (Mg)

Co nhi^u bai tap, doc d^ xong ta ra't be tac each giai (do de thieu dir kidn, ne'u

dat dn se rat nhieu in, s6 in nhi6u hon s6' phuong trinh dai s6', ) Ne'u ta xem

xet d^ ra k l luong ta se tha'y c6 nhCng da'u hieu dac biet de' giam s6' ^n va khi do

bai toan 16 ra la m6t bai toan quen thu6c De hieu ro hon va'n de nay chiing ta

nghien ciJu m6t s6' thi du sau

I I VAN DUNG

Thi du 1: Cho 3,36 lit ( a dktc) h6n hop khi X g6m cac hidrocacbon mach her

•^^ C,,H2„, C^Hjn, + 2 , Q, + m + 1H2 („ + ni) + 2 (m < n) CO s6' mol bSng nhau, loi cham qua

binh dung dung djch brom du, ket thiic phan umg tha'y kh6'i luong ciia binh dung

brom tang x gam Gia tri ciia x la

S6'mol m6i hidrocacbon bfing 0,15/3 = 0,05 mol >" " d* : ,•

Theo bai ra: m < n (1) ^ : Cac hidrocacbon khi nen n + m + 1 < 4 n + m < 3 (2) ^ ,

T i r ( l , 2 ) = > m = 1; n = 2 / - s e j

Vay cac hidrocacbon trong X gom: ^ ;

C2H4 (0,05 mol); C H 4 (0,05 mol) ' C4H8 (mach hor => anken; 0,05 mol) '" ' '

PTHH:C2H4 + Br2 ^ C2H4Br2

C4H8 + Br2-> C4HsBr2

0,05 Kh6'i luong binh brom tang chinh bang khd'i luong h6n hop ankan C2H4, C4H8 c6 trong X

vay x = 0,05.28 + 0,05.56 = 4,2 (g) Dap an diing la A

Thi du 2: H6n hop T gdm 3 hidrocacbon khi (6 dktc), mach hor c6 cOng thiic tong

quat la: C^H^,,, C„H2,„ C„^„_|H2„ (n, m c6 ciing gia tri trong ca 3 cha't va m < n)

Khi do 15,12 gam h6n hop T (m6i chat c6 s6' mol bang nhau) qua binh dung dung djch brom du den phan dug hoan toan, thi khoi luong binh brom tang len la

A ll,52gam B.10,08gam C.15,12gam D 7,56 gam

(Trich de thi thi Dai hoc khoi A, B) Hu&ng ddn gidi

- C„H2„ (mach ho): anken

dm nang On luy$n thi BH 3 mign Bcic - Trung - Nam mOn H6a hpc - Cii Thanh Toan

So mol m6i chat trong 15,12 gam T:

=> Khoi lucmg binh brom tang bang khoi liromg ciia C j H g va C4Hft (bj giiJ lai):

mbinhi.ng = mcHfi + m c 4 H 6 = 0'12(42 + 54) = 11,52 (gam)

Dap an diing la A

Thi du 3: H6n hop khi X gom etilen, metan, propin va vinylaxetilen c6 ti kh6'i so

vdi H2 la 17 D6't chay hoan toan 0,05 mol h6n hop X roi ha'p thu toan bo saii

pha'm chiiy vao binh dung djch Ca(OH), (du) thi khd'i luong blnh tang thdm m

gam Gia tri ciia m la

(Trich dethi thu: Dai hpc khoi A, B) Huotig ddn gidi

Cac chat trong X ( C 2 H 4 , C H 4 , C , H 4 , C 4 H 4 ) c6 dac diem chung la c6 ciing so

nguyen tir H trong phan ti!r => Dat cong thiic chung la C^):{^

Theo bai ra: M , = 17.2 = 34 =^ 12n + 4 = 34 => 12n = 30 => n = 2,5 {C.^,\{,)

Thi du 4: Mot h6n hop X gdm C^H^, Q H ^ , C 4 H 6 c6 ti kh6'i so vdi H la 18,6 Dot

chay hoan toan 4,48 lit h6n hop X (dktc), san ph^m chay cho Ian luot qua binh 1

dung dung djch H 2 S O 4 dac (du), binh I I dung dung djch K O H (du) thi khoi

luong binh I , I I tang \in luot la

A 21,6 gam va 26,4 gam B 10,8 gam va 26,4 gam

C 10,8 gam va 22,88 gam D 20,8 gam va 26,4 gam

(Trich dethi thdDgi hpc khoi A, B) } Huong ddn gidi

Theo bai ra: nx = 0,2 (mol)

M x = 18,6.2 = 37,2 (g/mol) ^m^ = 0,2.37,2 = 7,44 (gam)

- V i trong X, cac phan tir c6 6 nguydn tij H => nH = 6nx = 1,2 mol

3:>nH20 = l ' 2 / 2 = 0,6(mol)

= > ' " H 2 0 = 0,6.18 =10,8 (gam) = m (binh I tang)

Cty TNHH MTV DVVH Khang Vi$t

m c ( X , = ' " x - n i H = 7,44-1,2 = 6,24(8)

nc = 6,24/12 = 0,52 (mol) => Uco^ = 0,52 (mol)

mco = 0,52.44 = 22,88 (gam) = m (binh I I tang)

-ay binh I , I I tang lin luot la 10,8g va 22,88g ^ j , ^.^.^j, _ ,

E)ap an dung la C

Thidu 5: H6n hop A gom C 3 H 4 , C j H g , C j H ^ c6 ti kh6'i so vori H 2 bang 21 Dot chay

hoan toan 1,12 lit h6n hop A (do a dktc) roi dSn toan bo san ph^m chay vao

binh dung nu6c v6i trong du Do tang kh6'i luong ciia binh la

A. 9 , 3 gam B 10,6 gam C 14,6 gam D 12,7 gam

(Trich de thi thucDai hpc khoi A, B) Huong ddn gidi

Theo bai ra: n^ = 1,12/22,4 = 0,05 (mol)

m c 3 H 4 = 40, m c 3 H 6 = 42, mc^H^ =44

Do do C O the gia thiet coi h6n hop A chi chiia C j H g (0,05 mol):

C3H5 ) 3 C O 2 + 3 H 2 O

0,05 (mol) > 0,15 (mol) 0,15 (mol)

Do tang khd'i luong binh dung nu6c voi trong du chinh bang tong khoi luong

CO, va H 2 O (san ph^m chay):

Dap an dung la A '

Thi du 6: H6n hop X c6 ti kh6'i so vori H 2 la 21,2 gom propan, propen va propin

Khi d6't chay hoan toan 0,1 mol X, t6ng kh6'i luong C O 2 va H 2 O thu dugc la

A 16,80 g B 18,60 g C 20,40 g D 18,96 g

(Trich de thi thu Dai hpc khoi A, B) Hu&ng ddn gidi ' * '

Dat c6ng thiic chung ciia 3 hidrocacbon nay la CjHy

PT d6't chay: C,H, + ( 3 + | ) 0 2 > ^CO + ^ H.O

Ca'm nang nii i u v e n t h i O H :•, i n i e i i \',Sc \t\uv] K m mOn H6a hpc - Cu Thanh Toan

=5> riH^o = " H C X / 2 = 0,64/ 2 = 0,32 (mol)

Vay tdng kh6'i luomg C O j va H j O thu duoc la:

m = m c o 2 + mH20 = 0,3- 44 + 0,32 18=18,96 (gam)

Dap an diing la D

CM± Su dung djnh luSt bao toan nguyfen t6'd^ giai b ^ i tap tr6n

Thi du 7: D6't chay hoan toan m gam mot hidrocacbon 6 the k h i , nang hon khong

k h i , mach ho thu duoc 7,04 gam CO, K h i sue m gam hidrocacbon nay vao nuoc

brom du, sau k h i phan ling xay ra hoan toan tha'y c6 25,6 gam brom tham gia

phan ling Gia trj cua m la

A 2 B 4 C 10 D 2,08

^ (Trtch de thi thUDai hoc khdi A, B)

Huong ddn gidi

G o i cong thiic cua hidrocacbon nay dang C„H2„ + 2-2k

- V i hidrocacbon d the khi ntn n < 4

- V i hidrocacbon nang hon khong khi nen:

' Suy ra: CTPT ciia hidrocacbon c6 the la:

|- C3H2: khdng CO ca'u tao nao thoa man

- C4H2: H O C - C =CH (VI mach cacbon ho)

X = 0,16/ 4 = 0,04 ( m o l ) =^ m = 0,04 50 = 2 (gam)

Dap an dung la A

Cty TNHH MTV DWH Khang Vi$t

2 C6ng thiic ti'nh s6' d6ng phan ete don chiJc, no, mach hcf C„H2„^.20:

S6d6ng phan ete C ^ H j ^ + z ^ = ^ ( n - 0(" 2) (2 < n < 6)

3 Cong thirc tinh so ete tao boi h6n hop n ancol don chiic (H2SO4 dac, 140°C):

n ( n + l ) S6' ete = X, J , t '/"v •

4 Cong thiJfc tinh so nguyen t i j C trong phan t u (n) cua ancol no, mach ho (don

chilc hoac da chiic) C„H2„+20 dua vao phan ling chay:

S6' nguyen tur C ciia ancol no, mach hcf n = nco2 '

"H20 "CO2 I I I } ! 5'i>(; S;(x.:

Phuong phap 12:

PHJONG PHAP sCr DUNG CONG THLfC GlAl NHANH

1 L I T H U Y E T

Cac c6ng thurc giai nhanh bai tap hoa hoc:

1 C6ng thiic tinh s6' ddng phan ancol don chiic, no, mach ho C„H2„+20:

S6' d6ng phan ancol C„H2„,20 = 2"'- ( l < n < 6 )

76

5 C6ng thurc ti'nh khoi luong ancol don chiJc, no, mach hd (hoac hdn hop ancol don

chirc, no, mach ho) theo khd'i lugng CO2 va khdi luong H2O:

-2 cte rifiit Mill j(n6'~'

9 Cdng thu-c tinh sd ddng phan andehit don ehiic, no, mach hd:

12 Cdng thue tinh sd d i , t r i , tetra n peptit tdi da tao bdi hdn hop gdm x amino

axit khac nhau:

Sd peptit = x" f^OH + , •)( ,(>g,H ''nS'-f if!.-:: v;;,.:

C6ng thuc tinh khdi luong amino axit A (chiia n nhdm NH2 va m nhdm COOH)

•^hi cho amino axit nay vao dung djch ehira a mol H C l , sau do cho dung dich sau phan ung tac dung vira dii vdi b m o l N a O H : m ^ = b - a

m

77

e l m nang 6n luy^n thi DH 3 mjgn BJc - Trung - Nam mOn H6a hpc - Cu inann lodn

14. C 6 n g thiic tinh khoi lugng amino axit A (chiia n nhom N H va m nhom COOH)

khi cho amino axit nay vao dung dich chiia a mol NaOH, sau do cho dung dich

b — a

sau phan ung tac dung vira dii \6i b mol HCl: m ^ =

15. Cong thij-c tinh lugng ket tua xua't hien khi ha'p thu het m6t lugng CO vao dung

dich Ca(OH)2 hoac BaCOH).: n,„„, = n ^ ^ _ - n c o 2

16. Cong thiic tinh lugng C O cfin thiet dugc hap thu het vao mot dung dich

Ca(OH)2 hoac Ba(OH) de thu dugc mot lugng ka't tiia theo yeu cSu:

nco2 =

U r n - , = U — n I

17. Cong thiic tinh s6' mol NaOH cdn cho vao dung dich A l ' ' ' de xua't hien m o t

lugng ket tiia theo yeu cSu:

OH" = 3.n,

_ = 4 n

OH

Al-n _ _ = 4 Al-n - Al-n ^

18. Cong thiic tinh s6' mol H C l cin cho vao dung djch N a j A ^ O H ) ^ ) (hoac

NaA102) di xua't hien mot lugng ket tiia theo ytu cSu:

H+ -I

= 4.n^ - , _ - 3 n ,

19. C6ng thiic tinh s6' mol N a O H cin cho vao dung dich Zn"* de' xua't hien m6t

lugng ket tiia theo yeu cSu:

22. C6ng thiic t i n h kh6'i lugng mu6'i sunfat t h u dugc k h i hoa tan het h6n hgp oxit

k i m loai bang H.SO4 loang: m , „ „ f „ = mh6„ hc^, + 80nH2SO4

23. C6ng thiic t i n h k h o i l u g n g m u o i c l o r u a t h u dugc k h i hoa tan het h6n hgp oxit

k i m loai bang d u n g d i c h HCl: m , | „ u a = mhAnhgp + 27,5nHci

Cty TNHH MTV DVVH Khang Vi$t C6ng thiic t i n h k h o i l u g n g m u o i nitrat k i m loai thu dugc k h i c h o hOn hcyp cac

k i m loai tac d u n g vdi H N O , (san p h i m k h i i khOng c6 N H 4 N O 3 ) :

28. C6ng t h i i c t i n h k h o i l u g n g m u o i t h u d u g c k h i c h o h6n hgp sat v a c a c o x i t sat tac

d u n g vdfi HNO3 du, g i a i p h o n g k h i NO:

m m u 6 i gQ V 111)11 11U|, 'NO2 + 8 n N o J

30 C6ng thiic tinh khoi lugng muoi thu dugc khi hoa tan he't h6n hgp gdm F e , F e O ,

F e O j , Fe304 bang H S O 4 dac, nong, du giai phong khi S O :

400 (m h6n hnp fnutfi ~ + 16.nsoJ