CẨM NĂNG ôn THI 18 chuyên đề hóa học nguyễn văn hải

D i n hon hgp Y qua dung dich brom du, sau khi phan ung xay ra hoan toan, khoi lugng brom tham gia phan ung la •.... Lai gidi: Nhan xet: Theo bai, axit phan ung hoan toar\i baza c6 the

TS NGUYEN VAN HAI (Chu bien) NGUYEN NAM TRUNG - TRAN THE NGA - NGUYEN THj THU HA

Cam naligFluyen thi dai hoc

18 CHUYEN OE ^ HOA HOC

^ He thons cac phUdng phap siai nhanh bai tap hoa hoc

^ Duns cho on tap va thi tot nshiep THPT.,

^ Luyen thi vao Dai hoc va Cao dans

r W VJEWTIWHSINHTHUAN

D m MHA yi lAT RAN RAI HOC QUOC GIA HA NOl

MUC LUC

huyen di 1 C a c phucmg phap giai nhanh 3

huyen de 2 C a c axit v6 ca d i l n hinh 71

huydn de 3 Tinh chat cua cac hidroxit 99

huySn di 4 Tinh chat cua cac mu6'i v6 c a 111

huyen d^ 5 Tinh chat cua cac oxit 131

huyen di 6 C a c nguydn to phi kim di^n hinh 148

huydn d6 7 D a i cufotng v6 kim loai 173

huyen d^ 8 An mon k i m loai - Di^u chS' kim loai 190

huyen de 9 K i m loai kiem - K i e m tho - Nhom 204

huyen de 10 Sat - Cr6m - Dong 231

huyen d l 11 C a c li thuy^t co ban ciia hoa hoc 253

huyen de 12 Su dien li - Axit, bazof, muoi 268

huyen de 13 A n c o l - Phenol 285

huyen 14 Andehit - A x i t cacboxylic 301

huyen de 15 Este - Lipit 325

huyen d6 16 Cacbohidrat 341

huyen d6 17 A m i n - A m i n o axit - Peptit 352

huyen di 18 Polime va vat lieu polime 370

e thi thir dai hoc 378

Khoi luong dung dich sau phan ung = Khoi \ugng dung dich truoc phan

ling - (khoi luong chat ket tua + chat bay hoi)

vi DU M A U

V i du 1: H o a tan hoan toan 1,794 gam kim loai kiem M vao 400ml dung dich

H C l 0,1M C o can dung dich sau phan ving thu du(?c 3,86 chat ran khan Y

K i m logi kiem M la y,,?j,, ,

^ m ^ ^ =3,86-1,794-0,04.35,5 = 0,646 gam ^ n^^^ = 0,038 mol

1 794

- n M n H a - n „ „ = 0,078 ^ M ^ 2 3 ( N a ) ^ ^ a : , - ,

—> Dap an A

V i 2: Cho 100ml dung dich H3PO4 a molA vao 100ml dung dich K O H 2M

thu dugc dung dich Y c6 chua 15,44 gam hon hop muoi Gia tri cua a la

A 0,75 B.1,00 C.0,50 D 0,80

Lai gidi:

Nhan xet: V i Y chua hSn hop muoi - > chac chan c6 chiia muoi axit

- > K O H phan ung het

S a d o p h a n u n g : H3PO4 + K O H > Muoi + H2O , i

Bao toan khoi luong: 0,la.98 + 0,2.56 = 15,44 + 0,2.18 a = 0,80.^•' ' ^^

- > Dap a n D |

dm nang fln luy$n thi d<ii hgc 18 chuy§n H6a hpc - Nguygn Van H5i

Vi 3: Dot chay hoan toan 17,4 gam h6n hop Mg, Zn va Al trong khi O2 (du)

thu du(?c 30,2 gam hon hgp oxit The tich khi O2 (dktc) da tham gia phan

-> = ^ ^ = 0'4 mol V 0 2 = 8,96 lit ^ Dap an B

V i dv 4 (A-08): Cho 2,13 gam hon hgp X gom Mg, Cu va A l 6 dang bot tac

dyng hoan toan voi O2 thu dugc hon hop Y gom cac oxit c6 khoi luong 3,33

gam The tich dung dich HCl 2M vira dii de phan ling het voi Y la

A 60ml B 45ml C 75ml D 80ml

Lai gtat:

Nhan xet: bai nay cac em nhat thiet phai ap dung bao toan khoi lugng de

tinh kho'i lugng oxi tham gia phan ling:

Vi dv 5 (A-12): Hoa tan hoan toan 2,43 gam hon hgp gom M g va Zn vao mpt

, lugng vira dii dung djch H2SO4 loang, sau phan ling thu dugc 1,12 lit H2

(dktc) va dung dich X chiia m gam muoi Gia tri ciia m la ^

A 4,83 gam B 5,83 gam C 7,33 gam D 7,23 gam

Lai giai:

O bai nay, cac em c6 the giai chi tiet dua theo phan ling hoa hgc

Nhan xet: nH2S04 = r'H2 = 0,05 mol

So do phan ling: Kimloai + H2SO4 > Muoisunfat + H2

Bao toan kho'i lugng cho so do tren: ^ 6 I > ' ' •

->• m = 2,43 + 0,05.98 - 0,05.2 = 7,23 gam ' i i £ 0 t O B > ;

- > D a p a n D , , ^'^ ^^-^^^ ' ^

Cty TNHH MTV DWH Khanq Vigt

V i dv 6 (CD-08): Hoa tan het 7,74 g a m h o n h g p bgt Mg, Al b a n g 500 m l d u n g

d i c h h o n h g p HCl I M v a H2SO4 0,28M thu d u g c d u n g d i c h X v a 8,736 lit k h i H2 (dktc) Co c a n d u n g djch X thu d u g c l u g n g m u o i k h a n la

A 38,93 gam B 103,85 gam C 25,95 gam D 77,86 gam

Lai giai:

Nhan xet: Bai nay n e u cac e m giai theo p h u o n g trinh p h a n l i n g se gap k h o k h a n

v i can viet d e n 4 p h u o n g trinh

Truoc het, c a n x e m axit c6 p h a n l i n g het hay khong b a n g each so s a n h so

-V i dv 7: Nung hon hgp bgt gom 15,2 gam Cr203 v a m gam A l a nhi?t d g cao

Sau k h i p h a n u n g h o a n toan, thu d u g c 23,3 g a m h o n h g p r a n X Cho X p h a n

l i n g v o i axit H C l (du) thoat ra V lit k h i H2 (6 dktc) Gia tri ciia V l a

A 7,84 B.4,48 C 3,36 D 10,08

Lai giai:

Nhan xet: Bai nay truoc he't cac e m can tinh d u g c so m o l Al b a n d a u Lim y rang, trong p h a n u n g n h i f t n h o m , t h u o n g a p d u n g d i n h luat bao toan k h o i l u g n g , cia the:

m c ^ 0 3 + i " A l = m x m A i = 23,3 -15,2 = 8,1 g a m -> n ^ r 0,3 m o l

.0 Phuong trinh h o a hgc: Cr203 + 2A1 > 2Cr + Al2C)3 (j^p.^v,-.^

V i d\ 8 (B-09): Cho 100ml d u n g djch KOH 1,5M vao 200ml d u n g d j c h H3PO4

0,5M, thu d u g c d u n g d j c h X Co c^n X thu d u g c k h o i l u g n g chat ran k h a n la

A 15,5 gam. B 18,2 gam. C 12,8 gam. D 16,4 g a m

Lai giai:

" K O H = 0,15 m o l ; nH3P04 = 0,10

Nhan xet: Cac em c6 the xet ti le m o l KOH va H3PO4, sau do viet 2 p h u o n g trinh p h a n u n g va d a t so m o l de giai

Ca'm nang 6n luyfn thi d^i hpc 18 ctiuySn dg H6a hpc - Nguygn Van HSi

Tuy nhien, ne'u ap dung bao toan kho'i lugng cho so do:

H3PO4 + K O H > Muoi + H2O

Taco: mH3P04 + rnKOH= niHjO

mx = 0,1.98+ 0,15.56-0,15.18 = 15,5 gam

—> Dap an A

V i du 9: Dun nong hon hgp khi X gom 0,06 m o l C2H2 va 0,04 mol H2 v o l xiic

tac N i , sau mot thoi gian thu dugc hon hgp khi Y Dan toan bg Y Igi t u t u

qua binh dung dung dich brom (du) thi con lai 0,448 lit h6n hgp khi Z (6

dktc) CO ti kho'i so voi O2 la 0,5 Khoi lugng binh dung dich brom tang la

V - A 1,04 gam B 1,32 g a m C 1,64 gam D 1,20 gam

Nhan xet: Bao toan khoi lugng: mY = mx = 0,06.26 + 0,04.2 = 1,64 gam

0 448 or,, Mat khac: nz= = 0,02 mol va Mz= 0,5.32 = 16 ,

22,4

mz = 0,02.16 = 0,32 gam

Luu y: Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thy

Bao toan khoi lugng, ta c6:

Kho'i lugng binh brom tang = mv - mz = 1,64 - 0,32 = 1,32 gam ,.j j-^,^, •^i^^.j^.^

, Dap an B

V i du 10 (A-10): Dun nong hon hgp khi X gom 0,02 mol C2H2 va 0,03 mol H2

trong mot binh kin (xuc tac Ni), thu dugc hon hgp khi Y Cho Y Igi t u t u

vao binh dung djch brom (du), sau khi ke't thiic cac phan ung, khoi lugng

binh tang m gam va c6 280ml hon hgp khi Z (dktc) thoat ra Ti kho'i aia Z so

voi H2 la ia08. Gia tri ciia m la • ,iaw; : ,

% A.a585 B 0,620 C 0,205 D 0,328

Lbi gidi:

Nhan xet: Bao toan khoi lugng: mv = mx = 0,02.26 + 0,03.2 = 0,58 gam

, Mat khac: nz= ^ = 0,0125 mol va Mz= 10,08.2 = 20,16

22,4 ?

mz = 0,0125.20,16 = 0,252 gam

Nhanthay: 'r^^.^ i i,*!.'.a " ; m»s;c,ci ;,/•> •

Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thu

Bao toan khoi lugng: m = mv - mz = 0,58 - 0,252 = 0,328 gam

—>^ Dap anD

Cty TKi, 1TV DVVH Khang Vi$t

V i du 11 (B-12): Hon hgp X gom 0,15 mol vinylaxetilen va 0,6 mol H 2 Nung nong hon hgp X (xiic tac Ni) mot thoi gian, thu dugc hon hgp Y c6 ti khoi so voi H 2 bang 10 D i n hon hgp Y qua dung dich brom du, sau khi phan ung xay ra hoan toan, khoi lugng brom tham gia phan ung la • ) ; ,

A 0 gam B 24 gam C 8 gam D 16 gam

Lai gidi:

Bao toan khoi lugng: mY = mx = 0,15.52+ 0,6.2 = 9 gam 5 ,

Mat khac: M Y = 10.2 = 20 " 2 = = 0/45 mol " ' ^ •

Nhan xet: vinylaxetilen c6 chua 3 lien ke't 71 —> So' mol lien ke't K ban dau = 3.0,15 = 0,45 mol • ,,^.J^:n

Gia su so mol H2 tham gia phan umg = a mol So mol lien ke't n phan

ling = a mol

Mat khac, so mol khi giam = so mol H2 phan ung = n x - n Y nkfusi'^'^

-> a = 0,75 - 0,45 = 0,3 mol

-> So mol lien ke't 7i du = 0,45 - 0,3 = 0,15 mol = So'mol Br2 phan ung

Khoi lugng brom phan ung = 0,15.160 = 24 gam ,

Bao toan khoi lugng: mY = mx = 0,1.28 + 0,2.26 + 0,5.2 = 9 gam

Mat khac: M Y =10.2 = 20 n z = = 0,45 mol " '

Nhan xet: etilen c6 chiia 1 lien ke't 71, axetilen c6 chiia 2 lien ke't 71 - > So mol lien ke't 7t ban dau = 1.0,1 + 2.0,2 = 0,5 mol

Gia su so mol H2 tham gia phan ung = a mol - > So' mol lien ke't n phan

ling = a mol

Mat khac, so mol khi giam = so mol H2 phan ung = nx - nY ^ ^

- > a = = 0,8 - 0,45 = 0,35 mol

- > So mol lien ke't 7t d u = 0,5 - 0,35 = 0,15 mol = So mol Br2 phan ung

Khoi iugng brom phan ung = 0,15.160 = 24 gam

—> Dap anD '

7

dm nang On luygn thi dgi hpc 18 chuySn dg H6a hpc - Nguygn van HJi

Vi dv 13: Cho 2,1 gam h6n hgp X gom hai amin (no, don chiic, dong dang ke

tiep) phan ling het voi dung dich HCl (du), thu du(?c 3,925 gam hon hgp

muoi Cong thuc cua hai amin trong X la

A CH3NH2 va C2H5NH2 B C2H5NH2 va C3H7NH2 Y^ic

C C3H7NH2 va C4H9NH2 i '> • D C H 3 N H 2 va (CH3)3N S 0 >

Lot gidi:

Nhan xet: Loai D vi hai amin khong dong dang ke tiep Cac phuong an con

lai deu la amin don chuc, bac I

Khi cho X tac dung voi axit HCl:

R- N H 2 + H C l > R - N H 3 C I '

i<-Bao toan kho'i lugng: mamin + mnci = mmuoi • > ' ;

mnci = 3,925 - 2,1 = 1,825 gam -» nnci = 0,05 mol = namm

^ Mamin = 7 ^ = 42 ^ R- N H 2 = 42

0,05 -> R = 26 -» Hai goc hidrocabon la CH3- va C2H5- *

Dap an A

Vi dij 14: Dot chay hoan toan m gam hon hgp X gom 3 ancol (don chuc, thugc

Cling day dong dang), thu dugc 26,4 gam khi CO2 (dktc) va 19,8 gam H2O

Neu thuc hien phan ling ete hoa m gam X (hieu suat 100%) thi tong khoi

lugng ete thu dugc la

A 8,40 gam B 14,80 gam C 10,92 gam D 12,90 gam

Trong phan ung tao ete thi: nx = Zn^jO i^H20 = 0,25 mol

Bao toan kho'i lugng:

"^x = r^ete + nHjO mete = 17,4 - 0,25.18 = 12,90 gam -> Dap an D

Vi du 15 (B-08): Cho 3,6 gam axit cacboxylic X (no, don chiic) tac dung hoan

toan voi 500 ml dung dich gom KOH 0,12M va NaOH 0,12M Co can dung

dich thu dugc 8,28 gam chat ran khan Cong thiic cua X la

A C2H5COOH B.CH3COOH C.HCOOH D.C3H7COOH

Lai gidi:

Nhan xet: Theo bai, axit phan ung hoan toar\i baza c6 the con du, do

vay neu giai dua vao phuong trinh phan ung se rat kho khan

Cty TNHH MTV DWH Khang Vijt

Tuy nhien, bai nay dugc giai nhanh chong khi ap dung bao toan khoi lugng: maxlt + mbazo= mmuol + mnuoc

_^ mH20= 3,6+ 0,06.56+ 0,06.40-8,28 = l,08gam^> nH20= 0,06 mol

3 6 Mat khac: n H , o = Max.t = = 60 Axit la C H 3 C O O H

A 2,65 B.3,70 C 5,15 ' " " D 3,25

Lai gidi:

n 02 = 0,35 mol; n CO2 = 0,3 mol; n = 0,3 mol

Nhan xet: UQQ^ = nH20 ^^^^ este deu no, dan chuc -> Cong thiic cua 2

este la C n H 2 n 0 2 * ' • " ' ••.:

Bao toan nguyen to'oxi: 2n ^^te 2n = 2n CO2 ^ H2O ~ ^ " este 0,1 mol

_> n = ^^^22 = = 3 2 este trong X c6 cong thuc phan tu la C3H6O2

Heste 0,1

-> Cong thuc cau tao 2 este la: HCOOC2H5 va CH3COOCH3

Phan ling hoa hgc:

HCOOC2H5 + NaOH — ^ HCOONa + C2H5OH CH3COOCH3 + NaOH —!—> CHsCOONa + CH3OH '

De tha'y: nNaOH = neste = nz = 0,2 mol ^ Bao toan kho'i lugng: meste+mNaOH = mv + mz

Vi d^ 17: Xa phong hoa hoan toan 0,1 mol este X (dan chiic, mach ho) bang

100 g^m dung dich MOH 11,2% (M la kirn loai kiem) Co can dung dich sau phan ling, thu dugc 15,4 gam chat ran khan, dong thai ngung tu phan hoi bay ra tha'y tao thanh 92 gam chat long Cong thiic cua X la

9

Ca'm nang On luygn thi dgi hpc 18 chuyen dg H6a hpc - Nguygn VSn Hii

-> kho'i l u a n g d u n g m o i (nuac) = 100 - 11,2 = 88,8 gam a r i f / v i •«

g f i ; Phan u n g hoa hoc: iv;y£;f{;-^,';f0<;j; ^

rfsB<.m RCOOR' + M O H )• R C O O M + R ' O H fti fO !d>i Oi

S^'- M o l : 0,1 0,1 0,1 0,1 ^

N/ian f/ifli/: Cha't long sau k h i n g u n g t u g o m nuoc va ancol nen:

mancoi = 93,4 - mnuoc = 92 - 88,8 = 3,2 gam

^ M R O H = 3 2 ^ > R ' = 15(R'lanh6mCH3-)".''*^' ^ „

Bao toan k h o i l u g n g : meste = 15,4 + 3,2 - 11,2 = 7,4 g a m

^ Meste = 7 4 R C O O C H s = 74 ^ R la C H 3

-^ Este la CH3COOCH3 -^ D a p an A

V i dv 18: Cho 700 g a m cha't beo c6 chi so axit la 8 tac d u n g v o i d u n g dich K O H

du, sau phan u n g , khoi l u g n g muo'i t h u dugc la 764,6 gam Kho'i l u g n g

V i 19 (B-08): Xa phong hoa hoan toan 17,24 g a m chat beo can vvra d u

0,06 m o l N a O H Co can d u n g dich sau phan u n g t h u dugc m g a m xa

^ phong Gia t r i ciia m la -'"^^^ *

A 16,68 B 18,24 C 17,80 D 18,38

Cty TIMHH MTV DVVH Khang Vigt

Lmgidi:

Ggi cong thuc cua chat beo la (RCOO)3C3H5 , < 'i-b tfW 1, i ,

K h i cho chat beo tac d u n g v o i N a O H :

(RCOO)3C3Hs + 3 N a O H > 3RCOONa + C3Hs(OH)3

M o l : 0,06 0,02

A p d u n g bao toan kho'i l u g n g : mch.nt beo + mN.nOH = m x a phong (muoi) + mgiixeroi

- > m x i phong = 17,24 + 0,06.40 - 0,02.92 = 17,8 gam D a p an C ' * "• '

V i dv 20: Xa p h o n g hoa hoan toan m gam m o t triglixerit bang K O H t h u dugc

0,92 g a m glixerol va 9,58 gam hon h g p muo'i cua axit linoleic va axit oleic

-> D a p an D

V i du 21: A m i n o a x i t X chua m g t n h o m -NH2 Cho 10,3 g a m X tac d u n g v o i axit

H C l (du), t h u dugc 13,95 gam muo'i khan Cong thuc ca'u tao t h u ggn cua X

A CH3CH2CH(NH2)COOH B H2NCH2CH2COOH

C CH3CH(NH2)COOH D H2NCH2COOH

Laigiai:

Nhan xet: Cac dap an deu cho aminoaxit chua m g t n h o m - C O O H

ggi cong thuc cua X la H2N-R-CC)OH - j||||vi.^^^ ^g,; >

K h i cho X tac d u n g v o i axit H C l : f'^^,

V i du 22: D u n nong m gam hon hgp g o m a m o l tetrapeptit mach h o X va 2a

mol tripeptit mach h o Y v o i 600ml d u n g djch N a O H I M (vua d u ) Sau k h i cac phan u n g ke't thiic, c6 can d u n g dich t h u dugc 72,48 g a m m u o i khan

cua c^c amino axit deu c6 m g t n h o m - C O O H va m g t n h o m - N H 2 trong

phan t u Gia t r i a i a m la

A 51,72. B 54,30 C 66,00 , D 44,48

• • • • • 11

dm nang On luy?n thi dgi hpc 18 chuySn 6i H6a hpc - Nguygn Van HJi

Lai gidi:

Nhan xet: K h i cho cac peptit tac dung voi N a O H thi c6 cac lien ket peptit v a

nhom - C O O H (cua amino axit dau C) tham gia phan ung

Theo bai, a mol tetrapeptit mach h a X c6 a.3 = 3a mol lien ket peptit

2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit Tong so'mol nhom c a c b o x y l - C O O H bang a + 2a =3a

Cac phan ling nit ggn cua lien ke't peptit va nhom - C O O H la: '

'yinit - C O - N H - + N a O H > - C O O N a + H 2 N - ,

• nnMQh u 7a 7a

- C O O H + N a O H > - C O O N a + H 2 O , ^ Mol: 3a 3a 3a M ' - A

Theo bai: nwaOH = 7a + 3a = 10a = 0,6 —> a = 0,06 mol

Bao toan khoi lugng: m + mNaOH = mmuai + m ' i( >.:-i(jin

-> m = 72,48 + 0,06.3.18 - 0,6.40 = 51,72 gam

Dap an A , ,

V i d\ 23: D u n nong m gam hon hop gom a mol dipeptit mach ho X va 2a mol

tripeptit mach ho Y voi 400ml dung djch H C l I M (vira du) Sau khi cac

phan ung ket thuc, c6 can dung dich thu dugc 48,1 gam muoi khan cua cac

amino axit deu c6 mpt nhom - C O O H va mpt nhom - N H 2 trong phan ttr

Gia tri ciia m la

A 33,5 B 29,0 C 30,8 D 28,1

' ' , Lai gidi:

Nhan xet: K h i cho cac peptit tac dung voi H C l thi c6 cac lien ket peptit va

nhom H 2 N - (ciia amino axit dau N) tham gia phan ung

Theo bai, a mol dipeptit mach ho X c6 a.l = a mol lien ket peptit

2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit >

Tong so mol nhom amino H 2 N - bang a + 2a = 3a. ' f f

Cac phan ung riit gpn ciia lien ket peptit va nhom H 2 N - la:

, C O N H + H C l + H 2 O ) C O O H + C I H 3 N

-Mol: 5a 5a 5a

H 2 N + H C l > C I H 3 N

-Mol: 3a 3a

Theo bai: nHci = 5a + 3a = 8a = 0,4 —> a = 0,05 mol V'

Bao toan khoi lupng: m + mHci + m = mmuoi

b, H ? qua Tong so mol nguyen tu ciia mpt nguyen to c6 trong cac chat truoc phan ung

va sau phan ung luon bang nhau. o , ^ , r '

LKU y: C a n xac dinh diing va day dii cac chat c6 chiia nguyen to dang xet 6

truoc va sau phan ling , ,

V I D V M A U

V i du 1 (A-08): Hoa tan hoan toan 0,3 mol hon hpp gom A l va AUCs vao dung dich K O H (du), thu dupe a mol hon hpp khi va dung dich X Sue khi CO2 (du) vao X, lupng ket tua thu dupe la 46,8 gam Gia tri cua a la

V i dv 2: Hoa tan hoan toan hon hpp gom 0,12 mol FeS2 va a mol CU2S vao axit

H N O 3 (vua dii), thu dupe dung djch X (chi chiia hai muoi sunfat) va khi duy nhat N O Gia tri cua a la

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

Lai gidi:

Nhan xet: Cac em luu y la dung djch X chi chua hai muoi sunfat - > sau cac

phan ung, S nam het a dang goc sunfat

Ta CO cac so do chuyen hoa: ; i

^ FeS2 > i F e 2 ( S 0 4 ) 3 i'ouU-,,r!>! •

• Mol: 0,12 0,06 'ih.^1bm^l^^., : ^; ].,>f:v., r v ; v : '

-C^m nang 6n luygn Ihi dgi hpc 18 chuyfin ii Hoa hpc - Nguyin van H i !

rt:

Cu2S > 2CuS04

Bao toan nguyen to S, ta c6: 2 npeSj + "CuzS = "504 - H 1 a

-> 0,12.2 + a = 0,06 3 + 2a -> a = 0,06 -> Dap an C

Vi du 3 (A-12): Cho 18,4 gam hon hop X gom Cu2S, CuS, FeS2 va FeS tac dung

het voi HNO3 (dac nong, du) thu duoc V h't khi chi c6 NO2 (6 dktc, san

pham khu duy nha't) va dung dich Y Cho toan bo Y vao mot lugng du

dung dich BaCl2, thu dugc 46,6 gam ket tiia; con khi cho toan bp Y tac

dung voi dung dich NH3 du thu dugc 10,7 gam ket tua Gia tri cua V la ,^

A 38,08 B 24,64 C 16,8 D 11,2

NMn xet: Dung dich Y chiia cac ion: Fe^ Cu^^ SO4 , va NO3

Khi cho dung dich Y + dung dich BaCh:

Ba2* + SO^" — ^ BaS04i /'1 v >

Bao toan nguyen to'S: ng (X) = nBaS04 ^ mol ,

Khi cho Y + dung djch NHa du:

Fe3* + 3NH3 + 3H2O > Fe(OH)3i

Cu2* + 2NH3 + 2H2O > Cu(OH)2i

Luu y: Cu(OH)2 tan trong NH3 du tao thanh phuc chat:

Cu(OH)2 + 4NH3 — ^ [Cu(NH3)4](OH)2 "

10 7 ' Bao toan nguyen to Fe: npe (X)= nFe(OH)3 =

Bao toan khoi lugng: mx = m^u + mpg + mg

-» mcu = 18,4 - 0,1.56 - 0,2.32 = 6,4 gam -> ncu (X)= — = 0'^

i"ol-Qui doi X ve hon hgp gom cac don chat:

Fe = 0,1 mol; Cu = 0,1 mol va S = 0,2 mol. A iKt/tij:

-» ne (X) = 2ncu + Snpe + 6ns = 1,7 mol n^o^ = (x) = 1,7 mol

-> V = 1,7.22,4 = 38,08 lit ^ Dap an A

Vi dv 4 (B-10): Mot loai phan supephotphat kep c6 chua 69,62% Ca(H2P04)2, con

lai gom cac chat khong chiia photpho Dp dinh duong ciia loai phan Ian nay la

A 48,52% B 39,76% C 42,25% D 45,75%

Loigidi:

6 day, cac em can nho la dp dinh duong cua phan supephotphat dupe tinh

I theo % khoi lupng cua P2O5

•;• NMn xet: 1 mol Ca(H2P04)2 hay 1 mol P2O5 deu chua 2 mol P.,» ,u MM

mm

Cty TNHH MTV D W H Khang Vi$t

Bao toan nguyen to P theo so do:

Ca(H2P04)2 < > P2O5 ^ i, n <ti u i s 1' , ' Khoi lupng mol: 234 gam 142 gam o ' " ' ! ( , ,

A 18,0 B.22,4 C 15,6 D 24,2 ;

npg=0,lmol

Nhan xet: Bai nay ne'u dua theo phuong trinh phan ung se rat dai dong va

ton nhieu thoi gian O day, cac em can su dung so do phan ung:

Fe > X (Fe, FeO, Fe203, Fe304) > Fe(N03)3

va ap dung dinh luat bao toan nguyen to'Fe: ; j ,0

Fe > Fe(N03)3 „ ,

.0 f:5 Mol: 0,1 0,1

m =0,1.242 = 24,2 gam ^ Dap an D

Vi du 7: Cho 31,2 gam hon hpp gom Al, Cu va Ag tac dung vua dii voi 900ml dung dich HNO3 1,5M, thu dupe dung djch chua m gam muoi va 4,48 lit hon hpp khi X (dktc) gom NO va N2O Ti khoi ciia X so voi H2 la 16,75 Gia tri ciia m la ' ' '

A 98,3 B.97,2 C 96,3 D 91,0

Ca'm nang fln luygn thi dji hgc 18 chuy6n dg H6a hpc - Nguyin Van Hai

bai nay, truoc het cac em can t i m so' m o l m o i k h i trong X de t h u dugc ket

qua: nNo= m o l ; n ^ j o " ^ O'OS

i^ol-A l , C u , i^ol-A g ) M u o i n i t r a t + N O + N2O „ /, / '

C h a t o x i h o a : N^^ + 3e > N O ; 2N*5 + ge > N2O

va CO the xay ra ca qua trinh: Z N * ' + 8e > NH4NO3 (a mol)

K h i cho k i m loai + HNOs:

n^^Q-(muoi) = ng trao J6i = 3 nfyio + 8 nivj20 + 8 nfvjH^NOs ~ 0'^^

Bao toan nguyen to N : n H N O a = + " N O + 2 n N 2 0 + 2 n N H 4 N 0 3

0,9.1,5 = 0,85 + 8a + 0,15 + 2.0,05 + 2a ^ a = 0,025 m o l ' iTs-> K ! )

Bao toan khoi lugng: m = m ^ i , cu, Ag + m ^ ^ ^ + m N H 4 N 0 3 '

= 31,2 + (0,85 + 8.0,025).62 + 0,025.80 = 98,3 gam

-> Dap an A

NMn xet: Bai nay da "giau d i " san pham NH4NO3

V i d u 8: Hoa tan het 7,8 gam hon hgp gom A l va AI2O3 bang d u n g dich H C l

(du), thu dugc V lit k h i H2 (dktc) va dung djch X N h o d u n g dich NHs d u

vao X, IQC ket tua va dem nung den kho'i l u g n g khong doi thu dugc

10,2 gam chat ran Gia t r i cua V la

A 2,24 B 3,36 C 5,60 D 4,48 '

G<?i s o m o l : A l = x; AI2O3 = y Ta c6: 27x + 102y = 7,8 ' 'Atii /v

So do phan ung: ^

, A l A l ^ O g )AlCl3 "NH3.H2O > A i ( O H ) 3 _ ' % AI2O3

Bao toan nguyen to A l : x + 2y = 2nAi203 -> x + 2y = 2 ^ ^ = 0,2

X = 0,1; y = 0,05 n H 2 = - H A I3 = 0,15 m o l ^ V = 3,36 lit

2 '

Dap an B ' '

V i d v 9 (B-12): N u n g nong 46,6 gam hon hgp gom A l va Cr203 (trong dieu

J k i f n khong c6 khong khi) den k h i phan u n g xay ra hoan toan Chia hon

I hgp thu dugc sau phan u n g thanh hai phan bang nhau Phan mot phan

i u n g vua d u voi 300 m l dung djch N a O H I M (loang) De hoa tan het phan

hai can vira d u d u n g dich chiia a mol H C l Gia trj cua a la

A 0,9 B 1,3 C.0,5 D 1,5

Cty TNHH MTV DWH Khang Vi§t

Laigidi:

hlhan xet: K h i cho phan 1 tac d y n g v o i N a O H , tat ca A l va AI2O3 deu tac

d y n g va chuyen thanh NaA102

D o vay, bao toan nguyen to A l , ta c6: nAi= njvjaOH = 0/3 niol

_^ Ban dau: nAi = 2.0,3 = 0,6 m o l "€1203= ^^^-r^^^= 0,2 m o l

Cr203 + 2A1 — ^ AI2O3 + 2Cr , ,

Mol: 0,2 ^ a 4 : r , ^ ,R>0,2;; OA : ^ ,

Cac chat trong phan 2: n ^ i = 0,1 m o l ; n c r = 0,2 m o l ; 0^1203 = 0,1 m o l

- > H H C I = 3 n A i +2ncr+6nAi203 = l ' 3 m o l ^ D a p a n B

W d v 10: Hoa tan hoan toan 8,16 gam hon hgp gom Fe304 va FeS2 trong d u n g

dich axit HNO3 (dac, du), thu dugc 4,032 lit k h i NO2 (dktc) va d u n g djch X

Cho X tac d u n g v o i d u n g dich Ba(OH)2 d u , Igc ket tua va n u n g trong khong

k h i den khoi l u g n g khong doi t h u dugc m gam cha't ran Gia t r j cua m la

FeS2 -15e > Fe^^ + 2S^ - li^a.W/

Bao toan electron: '',;t»r,'

" N O z = n F e 3 0 4 + 1 5 " F e S z a + 15b = ai8 - » a = a03; b = aOl '[ _

Bao toan nguyen to Fe va S: U m fiiJ5 ifi BfO > i

Fe304 > ^Fe203 -8

2

M o l : 0,03 0,045 808,£

1 ( \.1.0 » FeS2 > -Fe203 + 2BaS04

2

M o l : 0,01 0,005 0,02

-> m = 160.0,05 + 0,02.233 = 12,66 -> Dap an A

V i 1 1 (A-08): H o n hgp X g o m propan, propen va p r o p i n T i k h o i cua X so

v o i H2 Ja 21,2 K h i dot chay hoan toan 0,1 m o l h o n h g p X, tong khoi lugng cuaCCh va H2O thu dugc la

A 20,40 gam B 18,60 SB^JHUZ^Ji^^^j^^^ff^fpf^^

Ca'm nang On luyjn thi dgi hgc 18 chuyfin dg H6a hgc - Nguygn Van H5i

Laigiai:

NMn xet: Cac e m can nh|in ra cac chat trong h o n h o p X deu c6 chiia 3

nguyen t u cacbon

N h u vay, k h i do't chay 0,1 m o l X thu dugc 0,3 m o l C O 2 - > nc = 0,3 m o l

Mat khac: M x = 21,2.2 = 42,2 -> mx = 42,4.0,1 = 4,24 gam ' " ' '"*V

' Bao toan kho'i l u g n g , ta c6: mx = mc + mn '>rn 6,(J " f' > - ^ 1 /•-(.'" M^' «f^ '

-> mH = 4,24 - 0,3.12 = 0,64 gam ^ nn = 0,64 m o l -> nn^o = 0/32 m o l

Tong khoi l u g n g C O 2 va H 2 O bang 0,3.44 + 0,32.18 = 18,96 gam

-> Dap an C

V i d\ 12: Do't chay hoan toan m p t the tich k h i thien nhien g o m metan, etan,

propan bang o x i khong k h i (trong khong k h i , oxi chiem 20% the tich), t h u

dugc 7,84 l i t k h i C O 2 (a dktc) va 9,9 gam H 2 O The tich k h o n g k h i (a dktc)

.X nho nhat can d i i n g de dot chay hoan toan l u o n g k h i thien nhien tren la

:gfi A 70,0 lit ir sv B B 78,4 lit C 84,0 lit D 56,0 lit. Ij

Lmgidi: > 1 0 0 •JIA «;:.«,) jfW nc02 = - ^ = 0 3 5 m o l ; n H 2 O = ^ = 0 ' 5 5 m o l

Nhan xet: Ban dau, nguyen to' oxi 0 dang O 2 t u do, con sau phan l i n g chay

thi chuyen he't vao C O 2 va H 2 O

Bao toan nguyen to O, ta c6: 2 n o j = 2 TYQQ^ + n^jjo •

= ^'^^-^^^'^^ = 0,625 m o l V Q J = 0,625.22,4 = 14,0 l i t

Vay: Vkhongkhi = 5 V o 2 = 70,0 l i t D a p an A. s a t w ;.::;;;»•*

V i du 13 (A-10): D o t chay hoan toan m gam hon hgp X g o m ba ancol d o n chiic,

thupc cung day d o n g dang, t h u dugc 3,808 l i t k h i C O 2 (dktc) va 5,4 gam

D o X chua cac ancol d a n chiic ^ no = noH = nx no = 0,13 m o l

5' j Bao toan khoi l u g n g trong X, ta c6:

m x = m c + mH + m o = 0,17.12 + 0.60.1 + 0,13.16 = 4,72 g a m

- > ' D a p an C M / » t j ; ; , t'vs'jji'ii l i i ^ ^ i , > , t

-Cty TNHH MTV D W H Khang ViQt

V i 14 (B-10): Do't chay hoan toan m p t l u p n g h o n hgip X gom hai ancol ( deu

no, da chiic, mach h o , c6 cung so n h o m - O H ) can vira d u V l i t k h i O2, t h u

dupe 11,2 l i t k h i C O 2 va 12,6 gam H 2 O (cac the tich k h i d o a dktc) Gia t r j

Theo bai ra, X chiia 2 ancol no - > nx = nH20 " ^COi ^ ^,2 m o l

-> So'nguyen tvr C t r u n g b i n h = = 2,5 - > X chua m p t ancol da chuc c6

so nguyen t u C nho h o n 2,5 -> ancol d o la C2H4(OH)2

D o X chua hai ancol cung so' n h o m chuc (hai chiic) - > no = noH = 2nx

^ no = 0,4 m o l Bao toan nguyen to' O, ta c6: no (OH) + 2 no^ = 2 TXQQ.^ + nyi^^Q

2.0,5 + 0 , 7 - 0 , 4 , ^^i^rW '

n o 2 =— ^ ^ = 0,65 m o l

^ V02 = 0,65.22,4 = 14,56 l i t ^ Dap an A

V i dy 15: H o n h p p X g o m hai axit cacboxylic d o n chuc Do't chay hoan toan

0,1 m o l X can 0,24 m o l O2, t h u dupe C O 2 v a 0,2 m o l H 2 O Cong thiic hai

Ca'm nang 6n luy$n thi dgi hpc 18 chuyen dg H6a hgc - IMguySn VSn H5i

V i dv 16 (A-12): Hon hg-p M gom mpt anken va hai amin no, don chuc, mach

" h6 X va Y la dong dang ke tiep (Mx < MY) Do't chay hoan toan mot lugng

r M can dung 4,536 lit O2 (dktc) thu dugc H 2 O , N2 va 2,24 lit C O 2 (dktc) Chat

, Bao toan nguyen to'O, ta c6:

6: 2 no2 = 2 ncoj + " H J O ^ H J O = 0/205 mol

Nh^n thay, khi dot chay 1 mol anken thu du(?c XXQQ^ = n H 2 0' vol hai

amin no, don chijfc thi n H j O " ' ^ C 0 2 1 ' ^ ph"^°''^8

^""^^'^'^"'^S-C x H 2 x 3 N + O 2 — ^ X C O 2 + ( x + l , 5 ) H 2 0 + 0,5N2 ^ '

0,205-0,1 sot os8 Dovay: nH20"'^C02 =l/5namin - > namin=—^ '— =0,07mol

-> So'nguyen t u C trung binh trong M = — ^ ^ < — ^ 2 2 _ = ~ 1^43 Y

" M I^aniin 0,07

f, -> M chua mQt chat c6 so' nguyen t u C nho hon 1,43 -> do phai la amin

ii C H 3 N H 2 (X) O day cac em can luu y: anken chua tir 2 nguyen tu C tra len!

-> Amin ke tiep la C 2 H 5 N H 2 (Y) Dap an A

V i d\ 17: Hon hgp X gom mpt anken va hai ancol (no, don chiic, mach ho)

Dot chay hoan toan mpt lugng X can vua du V lit khi O2, thu duQC 15,68 lit

'' khi C O 2 va 18 gam H 2 O Cac the tich do 6 dktc Gia tri cua V la • *^' •

A 22,40 B 17,92 C 16,24 D 23,52

n c o 2 = 0,7mol; n H 2 0 = l m o l mt?iV6J 1

NMn xet: Khi dot chay a n k e n thu duQC UQQ^ = XIH^Q , con k h i d o t chay ancol

"ancol = "H2O ~ ^C02 •

Do v^y: nancol = " H Z O - ^coz = 1 - 0,7 = 0,3 m o l , „,

Do ancol d o n c h u c n e n U Q (ancon ^ "ancol =0,3 m o l

Bao toan n g u y e n to O: n o (ancol) + 2no2 = 2nco2 + nH20

2.0,7 + 1-0,3 ,

- > H Q ^ = = 1 , 0 5 mol

2

V 0 2 = 1,05.22,4 = 23,52 lit ^ Dap an D

Cty TNHH MTV DWH Khang Vi$t

V i dv 18: Hon hgp X gom hai amino axit no (chi c6 nhom chuc -COOH va - N H 2 trong phan tu), trong do ti 1§ mo : mN = 80:21 De tac d\ing vvra du voi 3,83

gam hon hgp X can 30ml dung dich HCl I M Mat khac, do't chay hoan toan

3,83 gam hon hgp X can 3,192 lit O2 (dktc) Dan toan bp san pham chay ( C O 2 ,

H 2 O va N2) vao nude voi trong d u thi kho'i lugng ket tiia thu dugc la

A 13 gam " B 20 gam C 15 gam D 10 gam

Laigidi:

H H C I = 0,03 mol; n o 2 = ai425 mol. • C>D < - - 1

^ r n o ^ S O n p _ 80/16 10 '

21 ^ n ^ 21/14 ~ 3 •

M|it khac: n N = nNH2 = " H C I (x)= 0,03 mol -> no(X)= 0,1 mol

^ Ggiso'mol: n^ (x) = a i^ioJ va nH(x)=bmol

Bao toan khoi lugng: mx = m^ + mj^ + mo +T^-H

-» m c + niH = 3,83 - 0,1.16 - 0,03.14 = 1,81

man xet: Ixx ti 1? khoi lugng:

Bao toan nguyen to O: n Q (x) + 2no2 = 2 n c o 2 + ^HjO

12a+ b = 1,81

'fl iiAA US'

^ 0,1 + 2.0,1425 = 2a + 0,5b -> 4a + b = 0,77 ^ a = 0,13 , , Bao toan nguyen to C: n c ( X ) = n c o 2 = '^CaC03 = 0 , 1 3 ^ mcaCOa =13 gam

-> Dap an A

3 PHl/ONG PHAP T A N G - G I A M K H O I L l / p N G , ffl>| {

a Npi dung Khi tham gia phan ling hoa hgc, nguyen tu (nhom nguyen tu) cua chat ban

dau dugc thay the (ho^c cgng hgp) bang nguyen t u (nhom nguyen tit) mai

de tao thanh san pham ^

Do do, khoi lugng cua chat tao thanh c6 the tang len hay giam di do chenh l|ch khoi lugng mol cua cac nguyen t u (nhom nguyen tu)

Dua vao sy tang hay giam nay c6 the xac djnh so mol cac chat trong phuang trinh hoa hgc, t u do c6 the giai nhanh nhieu bai toan

S\f thay the (cpng hgfp) Bien doi khoi lugmg (tinh cho 1 mol)

dm nang fln luygn thi dgi hgc 18 chuySn de H6a hpc - Nguyjn Van Hi\

V I D U M A U

V i d\ 1 (A-08): Cho V lit hon hop khi (dktc) gom CO va H2 phan ung vol mot

lugng du hon hop ran gom CuO va Fe304 nung nong Sau khi cac phan ung

xay ra hoan toan, khoi lugng hon hgp ran giam 0,32 gam Gia tri ciia V la

A 0,448 B 0,112 C 0,224 D 0,560

Lot giai:

1 mol CO —*° > C O 2 Kho'i lugng chat ran giam 16 gam ^^^m

1 mol H 2 — * ° > H 2 O —> Kho'i lugng chat ran giam 16 gam

- > 1 mol (CO va H2) ^ ° > ( C O 2 va H 2 O ) giam 16 gam

Theobai: 0,02 mol < - giam 0,32 gam

- > V = 0,02.22,4 = 0,448 lit

—> Dap an A

V i d\ 2: Hon hgp Y gom FeO, Fe203 va CuO Hoa tan hoan toan 6,8 gam Y

bang dung djch HCI (du), thu dugc dung dich chua 12,3 gam muoi Mat

khac, neu khu hoan toan 6,8 gam Y bang CO (du), thu dugc m gam kim

loai Gia tri cua m la :

A 5,2 B.5,6 C.6,0 D 4,8 ^

Led giai:

Nhan xet: Khi cho Y + HCl thi oxit chuyen thanh muoi clorua va mgt ion

0^~ trong oxit dugc thay the bang hai ion CI"

, 1 mol O^- > 2 mol CI" -> Khoi lugng tang 71 -16 = 55 gam

a mol <— tang 12,3 - 6,8 = 5,5 gam

Khi cho Y tac dung voi CO thi cac oxit deu bi khu thanh kim loai ^

Bao toan khoi lugng: my = m + mo -> m = 6,8 - 0,1.16 = 5,2 gam

-> Dap an A ' ' " "*

Vi dvi 3: Hon hgp X gom CuO va Fe203 Hoa tan hoan toan 22 gam X bang

dung dich H 2 S O 4 , thu dugc 52 gam muoi Mat khac, neu khu het 22 gam X

bang CO (du), dan hon hgp khi thu dugc vao dung dich Ca(OH)2 (du), tao

thanh m gam ket tua Gia trj cua m la

A 45,5 B.37,5 C 40,5 D 50,0

Lai giai:

Nhan xet: Khi cho X + H 2 S O 4 thi oxit chuyen thanh muoi sunfat va mgt ion

0-2 trong oxit dugc thay the bang mgt ion SO 4 "

Cty TNHH MTV DVVH Khang Vi§t

1 mol 0-2 > 1 mol SO l~ Khoi lugng tang 96 - 16 = 80 gam f f

a mol tang 52 - 22 = 30 gam f •

_^ a = — = 0,375mol -> no (X) = 0,375 mol ' " ' '' •' '

80 Khi cho 22 gam X tac dung voi CO thi: no (X) = n^oj = riCaCOa = 0375 mol

-> m =0,375.100 = 37,5 gam

—> Dap an B ' '''t*''-t> i?>'; f vcr-f vi K T * •'•H'>I

Vi dv 4 (A-10): Dot chay hoan toan mgt lugng hidrocacbon X Hap thu het san pham chay vao dung dich Ba(OH)2 du, tao ra 29,55 gam ket tiia, dung dich

sau phan ung c6 khoi lugng giam 19,35 gam so voi ban dau Cong thuc phan tu cua X la

Khoi lugng dung dich giam = Mat - Dugc (r\

= mBaCOs - (mco2 + n^Hoo) = 19,35gam

-> m H 2 0 = 29,55 - 0,15.44 - 19,35 = 3,6 gam -> nH20 = 0,2 mol ' *"

Vay hidrocacbon X c6: So C : so'H = nc : nn = 0,15 : 0,4 = 3 :>8

^ Dap an A „.,,.n ,,a , ,

Vi dy 5 (B-12): Cho 21 gam hon hgp X gom glyxin va axit axetic tac dung vua

dii vol dung djch KOH, thu dugc dung djch Y chiia 32,4 gam muoi Cho Y tac dung voi dung dich HCl du, thu dugc dung djch chua m gam muoi

Gia trj cua m la

A 44,65 B 50,65 C 22,30 ^ D 33,50

Ldigiai:

Glyxin ( H 2 N - C H 2- C O O H ) = a mol va axit axetic ( C H 3 C O O H ) = b mol

Khi cho X tac dung voi dung dich KOH: • •' • :> 1 mol -COOH ^ 1 mo! -COOK -> khoi lugng tang 38 gam

Theobai: 0,3 mol <- tang32,4-21 = 11,4 gam -> a + b = 0,3 mol

Mat khac: 75a + 60b = 21 a = 0,2 mol; b = 0,1 mol

Khi cho Y + HCl du, thu dugc cac muoi: C I H 3 N - C H 2 - C O O H (0,2 mol) va

KCl (0,3 mol) theo cac phan ung: 23

elm nang fln luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hi\

H2N-CH2 -COOK + 2HC1 > ClH3N-CH2-CC)OH + KCl

' CH3COOK + HCl > CHCOOH + KCl •,l.>m^V

Vay: m = 111,5.0,2 + 74,5.0,3 = 44,65 gam -» Dap an A

Luu y: Trong bai nay cac em de quen KCl khi tinh kho'i lugng muo'i va chpn

nham phirang an C!

V i 6 (A-12): Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chat khi a dieu

ki^n thuong) roi dem toan bp san pham chay hap thu het vao binh dvmg

dung dich Ba(OH)2 Sau cac phan ling thu dugc 39,4 gam ket tua va khoi

^j lugng phan dung dich giam bot 19,912 gam Cong thiic phan tu ciia X la

Lim y: Ci day bai toan khong cho dung dich Ba(OH)2 d u nen neu cac em viet

phuang trinh hap thu: Ba(OH)2 + CO2 > BaCOa + H2O

va suy ra ncQ2 = TI^^QQ^ = 0,2 mol thi se bi mac sai lam ngay!

V i di;i 7: Hon hgp Y gom axit axetic, axit acrylic va axit adipic Cho 16,9 gam Y

tac dyng vua dti vai dung djch NaOH, thu dugc 22,4 gam muo'i Mat khac,

16,9 gam X tac dung vai dung dich NaHCOs du, thu dugc V lit khi CO2

(dktc) Gia trj aia V la • - • • p

A 5,60. q B 8,40 - C 4,48. n D 7,84

\

Khi cho Y tac dyng voi dung djch KOH:

1 mol -COOH -> 1 mol -COONa - > Khoi lugng tang 22 gam

Theobai: 0,25 mol tang 22,4-16,9 = 5,5 gam

ncooH =0,25 mol

Khi cho Y tac dyng vai dung djch NaHCOa: nco2 = "cooH = 0/25 mol

^ Vco2 = 0/25.22,4 = 5,6 lit

^ -> Dap an A

Cty TNHH MTV DVVH Khang Vigt

Vi dv 8: Hon hgp X gom hai amino axit (mach ho, moi amino axit deu chiia mgt nhom chiic -NH2 va mgt nhom chuc -COOH) Cho 16,4 gam X tac

dyng vua du vai dung dich KOH, thu dugc 24 gam muo'i M|t khac, 16,4

gam X tac dung vua du V lit dung djch HCl 2M Gia tri ciia V la

A 0,1 B.0,5 C.0,2 D.0,3 f

Loigidi: 8""?' " " U : i' ^ , ,

Khi cho X tac dung vai dung dich KOH: { i > y,i i ,/ <,

1 mol -COOH > 1 mol -COOK khoi lugng tang 38 gam

Theobai: 0,2 mol < - tang24-16,4 = 7,6 gam

nx = ncooH = 0,2 mol Khi cho X tac dung vai dung djch HCl:

nHCl=nNH2 = " x ^ nHci=0,2mol ^ VHCI=0,2 lit

-> Dap an C ^ ,,,,, , ,

V i dy 9: Cho 12 gam hon hgp X gom glyxin va etylamin tac dyng vua du voi

dung dich HCl, thu dugc dung dich Y chiia 19,3 gam muo'i Cho Y tac dung voi dung dich NaOH du, thu dugc dung dich chua m gam muo'i Gia trj cua

m la

A 9,60 B 15,45 C 21,40 D 19,10

Loigidi:

Glyxin (H2N-CH2-COOH) = a mol va etylamin (C2H5NH2) = b mol

Khi cho X tac dung voi dung dich HCl: ^

1 mol -NH2 > 1 mol -NH3CI -> khoi lugng tang 36,5 gam

Theobai: 0,2 mol tang 19,3-12 = 7,3 gam -> a + b = 0,2 mol

Mat khac: 75a + 45b = 12 ^ a = 0,1 mol; b = 0,1 mol

Khi cho Y + NaOH du, thu dugc cac muoi: H2N-CH2-COONa (0,1 mol) va

NaCl (0,2 mol)

Cac phan ung hoa hgc:

CIH3N-CH2 -COOH + 2NaOH > H2N-CH2-COONa + NaCl

Mol: 0,1 0,1 0,1

C2H5NH3CI + NaOH > C2H5NH2 + NaCl

-Vay: m = 97.0,1 + 58,5.0,2 = 21,40 gam ^ Dap an C

Luu y: Neu quen tinh khoi lugng NaCl cac em se chgn nham phuong an A !

25

Ca'm nang 6n luygn thi dgi hgc 18 chuySn dg H6a hoc - Nguyin VSn Hii

4 PHL/ONG PHAP BAO TOAN ELECTRON 7

a Npi dung 4 v si-lk-l ymi t ^'bdn vym

*• ,1 Tong so' mol electron cac chat khu nhuong = Tong so' mol electron cac cha't

oxi hoa nhan ; f-Aji,

b Cach ap di^ng

Cac em can xac djnh dung va day du cac chat khii va chat oxi hoa ciing nhu

su bien doi trang thai oxi hoa cua chiing

Viet cac qua trinh oxi hoa (nhuong electron) va qua trinh khu (nhan

J, electron) de xac dinh so'mol electron trao doi roi ap djmg dinh luat bao toan

electron , ;

VI D U MAU

Vi du 1 (CD-11): Hoa tan hoan toan 13 gam Zn trong dung dich HNOs loang,

du thu dup-c dung dich X chua m gam muoi va 0,448 lit khi Ni (dktc) Gia

Bai toan nay cac em c6 the giai khi viet phuong trinh phan ung

6 day cac em se dugc huong dan giai theo phuong phap bao toan electron

Chatkhu': Zn - 2e > Zn*^ —> ne nhuong = 2nzn = 0,4 mol

Chat oxi hoa: 2N^5 + lOe > N2 ^ nenh,in = lOn^^ = 0,2 mol

" Nhu v$y so mol electron trao doi chua bang nhau—> "chua on" O day,

' mot san pham khu da dugc "gia'u di", do la su tao thanh muoi NH4NO3: -i

Chat oxi hoa: 2N^5 + se > NH4NO3 '

Mol: 0,2 - > 0,025

Vay: m = mzn(N03)2 + mNH4N03 = 0,2.189 + 0,025.80 = 39,8 gam

—> Dap an D

Luu y: 1- Bai toan nay c6 the giai nhanh hon bang each ap dung ngay

phuong trinh bao toan electron: 2 n^g = 3 n-^Q + 8 n[,jH4N03 •

2- Ba kim loai kha manh (Mg, Al, Zn) tac dung voi axit HNO3, c6 the t^io

thanh muoi NH4NO3! ^

Vi dy 2: Hoa tan hoan toan h6n hop gom 0,03 mol FeS2 va a mol CU2S vao axit

HNO3 (vira dii), thu dugc dung djch X (chi chua hai muoi sunfat) va V lit

khi duy nha't NO (dktc) Gia trj ciia V la

A 1,12 B.5,60 C.2,24 D 4,48

Cty TNHH MTV DVVH Khang Vi$t

Lai gidi:

'Sh^n xet: Cac em luu y la dung dich X chi chua hai muoi sunfat sau cac

phan ung, S nam he't 6 dang goc sunfat ,

Ta CO cac so do chuyen hoa: : • ' FeS2 > -Fe2(S04)3 Cu2S > 2CuS04

2 \ Mol: 0,03 0,015 Mol: a -> 2a Bao toan nguyen to S, ta c6: 2 npeSj + ncu2S = " 5 0 4 ^ ' 0,03.2 + a = 0,015.3 + 2a ^ a = 0,015 • ^^''^

Cac phan ung khu:

FeS2-15e > Fe*^ + 25*^ Cu2S-10e > 2Cu^2 + 5-6 Bao toan electron:

3n^o = 15npesj +10ncujS " N O = 0,2 mol V^o = 4,48 lit c

-> m= mMg(N03)2 + mNH4N03 = 0,09.142 + 0,0075.80 = 13,92 gam

- > D a p a n B ^ ^ ^ : ,

Vi d\ 4: Hoa tan hoan toan hon hgp gom 9,75 gam Zn va 2,7 gam Al vao 200 ml dung dich X chua dong thoi H N O 3 I M va H2SO4 1,5M Sau khi phan ung xay ra hoan toan thu dugc khi NO (san pham khu duy nha't) va dung dich

Y (chi gom cac muoi) Khoi lugng muoi c6 trong Y la:

A 41,25 gam B 53,65 gam C 44,05 gam D 49,65 gam

Lot gidi:

9 75 '

n2n=-^—= 0,15 mol; nAl = 0,1 mol : • •

Ca'm nang On luy$n thi dgi hgc 18 c h u y § n de Hoa hqc - Nguyin Van H&i

Nhan xet: Khi axit HNO3 c6 mat dong thai voi axit H2SO4 loang thi lirgng

trong dung dich la do 2 axit phan li ra -> giai theo phuong trinh ion

= '^HNOg + 2nH2S04 = 0,2 + 2.0,3 = 0,8 mol , ,^, ^ Trong X:

-Phuong trinh ion rut gon:

3Zn + 8H* + 2NO; > 3Zn2+ + 2NO + 4H2O

i f r ^ f - 8 A l + 4H* + NO~ > A P + N O + 2 H 2 O

Mol: 0 , 1 ^ 0 , 4 ^ ai • :Mig(m ufifki-^iiO

—> H + v a N O j tham gia phan ling het "+g': 4 - < K ' !

-> Khoi luong muoi trong Y = 9,75 + 2,7 + 0,3.96 = 41,25 gam Dap an A

V i du 5 (B-07): Nung m gam hot sat trong oxi, thu dugc 3 gam hon hg-p chat

ran X Hoa tan het X trong dung dich H N O 3 du, thoat ra 0,56 lit N O (san

pham khu duy nha't 6 dktc) Gia tri cua m la:

A 2,52 B.2,22 C 2,62 D 2,32

Laigidi: •

n N O = — = 0 , 0 2 5 mol .88;;-',A

22,4

Bao toan khoi lugng: m.Q^ = mx- mpg= 3-m

Nhan xet: Neu dua theo phuong trinh phan ung, bai giai se rat dai va kho giai

Cach 1: 6 day, cac em can su dung so do phan ung:

Fe (1) - ^ 1 ° ^ X (2) , Fe^3 (3)

Bao toan electron: 3.npg = 4nQ^ + 3nj^Q

^ 3.E = 1:21 + 0,075 -> m = 2,52 gam : ^v

—> Dap an A

Cach 2: Qui doi X thanh Fe (a mol) va O (b mol) Ta c6: 56a + 16b = 3

Qua trinh oxi hoa: Fe - 3e ^ Fe*^

Qua trinh khu: O + 2e - > O-^; N*^ + 3e - > N O

Bao toan electron: 3a = 2b + 0,075 —> a = 0,045; b = 0,03

- > m = 0,045.56 = 2,52 gam

V i d\ 6: Hoa tan hoan toan 5,84 gam hon hop gom Fe304 va FeS2 trong dung

dich axit H N O 3 (dac, du), thu dugc 3,808 lit khi NO2 (dktc) va dung djch X

Cho X tac dung voi dung djch Ba(OH)2 du, Ipc ket tiia va nung trong khong

khi den kho'i lugng khong doi thu dugc m gam chat ran Gia tri cua m la

nN02= " F e 3 0 4 + 15nFeS2 a + 15b = 0,17 ^ a = 0,02; b = 0,01.^ '

Ta CO cac so do chuyen hoa:

Fe304 , 3Fe(N03)3 "^'^^""'^ > 3Fe(OH)3 1,5 Fe203 Mol: 0,02 0,03

FeS2 > Fe(N03)3 + 2H2SO4 ) Fe(OH)3 + 2BaS04

0,5Fe2O3+2BaSO4 Mol: 0,01 -> 0,005 0,02

Nhan xet: Day la bai toan c6 nhieu chat khu (3 kim loai) va tao ra 2 san

pham khi nen can ap dung dinh luat bao toan electron

Chat khu: Fe - 3e > F e - ; Mg - 2e > Mg^^; Z n - 2e >Zn^^

Chat oxi hoa: N - + 3e > N O ; N - + l e > NO2

Bao toan electron: netraod6i = 3njsjo + l-nN02 = ^''^"^°^- Den day, cac em can luu y la khi cho kim loai tac dung voi HNOs, so mol go'c N O 3 nam trong muoi dugc tinh nhu sau: ^

n , (muoi) = netraod6i = 0,7mol <»i (' 1 j f ' j u ' / j

C^m nang 6n luy^n thi d<ii hpc 18 chuyen ai H6a hpc - IMguygn VSn HSi

V i 8 (CD-12): Hoa tan hoan toan 8,9 gam hon hg-p gom Mg va Z n bang

lugng vua du 500ml dung dich HNOa I M Sau khi cac phan ling ket thiic,

thu dugc 1,008 lit khi N2O (dktc) duy nhat va dung d|ch X chiia m gam

muoi Gia tri ciia m la , ,1

A 31,22 B 34,10 C 33,70 D 34,32 '

Loigidi:

nN20= ^ ^ = 0 , 0 4 5 mol;nHN03 = 0/5 mol i» ,

M g , Z n > Muoi nitrat + N2O ' * '>•

Chat oxi hoa: 2N*^ + Be > N2O "

va CO the xay ra qua trinh: 2N*^ + Be > NH4NO3 (a mol)

Khi cho kim loai + HNO3: r

* i^jsjQ-(muoi) = ng trao doi = 8 njs420 + 8 nNH4N03 ~ (0/36 + 8a) mol

Bao toan nguyen to N : nHNOs = " N O 3 ^ ^ " ^ ^ 2 0 + 2nNH4N03

-> 0,5 = 0,36 + 8a + 2.0,045 + 2a a = 0,005 mol

' ' Bao toan khoi luong: m = mMg, Zn+ m^^^^ + mNH4N03 ,^ ^

= 8,9 + (0,36+B.0,005).62 + 0,005.80 = 34,1 gam

f Dap anB

Nhan xet: Can luu y Mg tac dung vai HNO3 c6 the tao ra muoi NH4NO3

V i dyt 9: Dan luong khi C O di qua hon hgp gom C u O va Fe203 nung nong,

sau mpt thoi gian thu dug-c cha't ran X va khi Y Cho Y hap thy hoan toan

vao dung dich Ba(OH)2 du, thu dugc 29,55 gam ket hia Chat ran X phan

ung voi dung dich H N O 3 d u thu dugc V lit khi N O (san pham khu duy

,s,i nhat, a dktc) Gia tri cua V la

A 2,24 B.4,48 C 6,72 D 3,36

/J < Lffigiai:

"BaC03 = 0,15 mol

6 day, cac em can su dyng so do phan ung:

CuO, Fe203 > X ) Muoi nitrat + N O

k>r> n

CO2 + Ba(OH)2 > BaCOs + H2O

Mol: 0,15 0,15 '

Nhan xet: Khi cho hSn hgp oxit tac dung vdi khi C O thi:

n,, trao doi = IIXQQ M a n^o = ^C02 ~ 0/15 mol -> ng trao doi = 0,3 mol

Bao toan electron: ngtraodoi = 3 n N o ^ T^NO =0,1 mol

V N O =2,24 lit Dap an A

Cty TNHH MTV DVVH Khang Vijt

V i dV 10 (B-10): Nung 2,23 gam hon hgp X gom cac kim loai Fe, Al, Zn, Mg trong khi oxi, sau mot thoi gian thu dugc 2,71 gam hon hgp Y Hoa tan hoan toan Y vao dung dich HNO3 (du), thu dugc 0,672 lit khi N O (san

pham khu duy nhat, a dktc) So mol HNO3 da phan ling la 1,;, v

A 0,12 B 0,14 C 0,16 . j -' D- 0/18 ;,; {

Lodgidi: ' W :,,0,f rrr'''.':

Bao toan khoi lugng: m o 2 = 2,71 - 2,23 = 0,48 gam rni-ii ricirt •

-> no2 = 0/015 mol -> no= 0,03 mol ^ nQ.2 ^^^.jj= 0,03 mol

each 1: Su dung so do: X > Y ^ " ^ " ^ > Muoi nitrat + N O

Ltm y: Y chua ca kim loai (con du) va oxit

+ Khi cho kim loai + HNOs: T^^Q- ( " " " O O = ng trao doi = 3 n ^ o = 0,09 mol

+ Khi cho oxit + HNO3: 1 goc O^^ trong oxit se bi thay the bang 2 go'c

3 i '

N O o de tao muoi nitrat, do do: n = 2n , = 0,06 mol

Bao toan nguyen to N: nnwos = " N O " " ^'^^ " ^'"^^

—> Dap an D

Cach 2: Qui doi Y thanh X va O (0,03 mol)

Bao toan electron: ng(x)=2nQ+3nr^o =0,15 mol - > n ^ ^ =0,15 mol

Bao toan N : nnNOs = ^^Q'^ " ^'^^ ~^ ^' ''

V i d^ 11 (B-12): Cho 29 gam hon hgp Al, C u va A g tac dung vua du vai 950ml

dung dich HNO31,5M, thu dugc dung dich chua m gam muoi va 5,6 lit khi

X (dktc) gom N O va N2O Ti khoi caa X so vai H2 la 16,4 Gia tri ciia m la

A 98,20 B 97,20 C 98,75 D 91,00

Lai gidi:

6 bai nay, truoc he't cac em can tim so mol moi khi trong X de thu dugc ke't

qua: nNo= 0,20 mol; n^^o= 0,05 mol

A l , C u , A g ) Muoi nitrat + N O + N2O

Chat oxi hoa: N^=^ + 3e > N O ; 2N*^ + 8e > N2O

va CO the xay ra qua trinh: 2N*5 + Se > NH4NO3 (a mol) ,

Khi cho kim loai + HNO3:

" N O 3 ^ " " ^ ' " ° ^ ^ ^ ^ " N 2 O ^ 8 riNH4N03 = (1 + 8a) mol ,, ^ j

Bao toan nguyen to N: nHN03 ^^oi ^^'^ ^ " N 2 O 2nNH4N03 '

-> 0,95.1,5 = 1 + 8a + 0,2 + 2.0,05 + 2a -> a = 0,0125 mol

31

elm nang On luyjn thi dgi hgc 18 chuySn dg H6a hpc - Nguyjn Van HSl

Bao toan khoi luqmg: m = m^i, c u , A g + "1,^0^ + m N H 4 N 0 3

= 29 + (1+8.0,0125).62 + 0,0125.80 = 98,2 gam

Dap an A *'

Nhan xet: Can nhan ra bai toan da "giau d i " san pham NH4NO3

V i dv 12: Cho hon hop khi X gom CI2 va O2 tac dung vua du vol hon hgp bpt

gom 10,8 gam A l va 2,4 gam Mg, thu dugc 40,9 gam hon h(?p chat ran Y

Phan tram the tich cua khi CI2 trong X la

A 80% B.40% C.50% , D 60%

, Laigidi: - •',

n^,= M = a 4 m o l ; n M , = ^ = 0 , l m o l X gnyb ig :f d i 5

27 ^ 24 M&ii) Igol^M m mifl:4 y i[ mil

Nhan xet: Day la bai toan hai chat khu (2 kim loai) va hai chat oxi hoa (2 phi

kim) nen can ap dung dinh luat bao toan electron

Chat khu: A l - 3e > Al*^ M g - 2e > Mg*^

Mol: 0,4-^1,2 Mol: 0 , 1 - > 0,2 * f'^^'^

Chat oxi hoa: O2 + 4e > ICf^ CI2 + 2e > lOr

Bao toan electron: 4x + 2y = 1,2 + 0,2 = 1,4 mol

Bao toan kho'i luong: mx + Tn^i + m j ^ g = my

V i dv 13: Nung hon hop X gom a mol Fe va 0,03 mol C u trong khong khi mot

thoi gian, thu dugc 12,64 gam chat ran Y Hoa tan hoan toan Y bang dung

dich HNO3 loang (du), thu dugc 0,896 lit khi N O (san pham k h u duy nhat 6

dktc) Gia tri ciia a la

Laigidi:

nNO= ^ ; r — = 0,04 mol

22,4 ;

Bao toan kho'i lugng: TTIQ^ = my - mx = 12,64 - 56a -1,92 = 10,72 - 56a

Nhan xet: Neu dya theo phuong trinh phan ung se rat dai va kho giai

Cach 1:

6 day, cac em can su diing so do phan ung:

Fe, C u (1) — Y (2) """^^^ ) Fe^^ Cu^^ (3)

Cty TNHH MTV DWH Khang Vijt

Xet su trao doi electron a cac giai doan:

Cach 2: Qui doi Y thanh: Fe (a mol); C u (0,03 mol) va O (b mol) , „ l i t , ;

Bao toan khoi lugng: 56a + 16b = 12,64-0,03.64 = 10,72. ysb n

Bao toan electron: 3a + 2.a03 = 2b + ai2 ^ a = ai4; b = ai8 \ ] (w^ - ,

- ^ D a p a n D ; „ , , i i O ftoi v M l ' o ^ ' i ' :

5 P H l / O N G P H A P T R U N G H O A D I E N ;gnu nerfq 6u'>a

a N p i d u n g Dung dich cac chat dien li luon luon trung hoa ve di^n Q si*: q&CJ <

Tong so'mol di^n tich duong = Tong so mol di^n tich am: 5^! ( ;v

Lieu y: De tinh so'mol di?n tich, cac em la'y so mol ion x di^n tich ion do

V I D V M A U

V i dv 1: Dung dich X chua cac ion: Fe^* (0,1 mol), AP" (0,2 mol), C h (x mol) va S04~ (y m o l ) C o c^n dung djch X thu dugc 46,9 gam chat ran khan Y Gia tri cua X va y Ian lugt la v • nr,^ipf\n\~n f:,

A 0,2 va 0,3 B. 0,1 va 0,2 C 0,4 va 0,4 D 0,3 v a 0,3 >

Laigidi: 3 orbs

Nhan xet: Dung dich da cho chua 4 loai ion, trong do 2 loai ion chvra bie't so

mol, do vay can lap dugc 2 phuong trinh dai so de tim so mol cua chung

+ A p dyng dinh luat trung hoa di?n:

Tong so mol di^n tich duong = Tong so mol di^n tich am '^^

dm nang On luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hai

Luu y: D i e m m a u cho't d bai nay la cac em dua ra p h u a n g t r i n h d^ii so' cua d j n h

lu^t t r u n g hoa dien tich.:t.'e'-f ,ji*i„trt,»sj:fr '^4f^- :>''J*!;/i J.K •:i.«jSJi':'iie|'

V i d y 2 (B-12):, M p t d u n g dich gom: Na* (0,01 mol); Ca^* (0,02 mol); H C O ,

(0,02 mol) va i o n X (a mol) I o n X va gia t n cua a la

••• A O H - va 0,03 B Ch va 0,01 C COs^- va 0,03 D N O ^ va 0,03

Lai gidi:

Goi dien tich ion X la - n "• < f J>0 + eC rrfor; ' , ,('

+ A p d y n g d i n h luat t r u n g hoa di?n: ' ' ' ' 't;'^

1.0,01 + 2.0,02 = 1.0,02 + n.a n.a = 0,03 fJom «> irtmrU Y r S f f r

Den day c6 2 p h u a n g an thoa man la A va D Vay chpn i o n X la i o n O H " hay

ion N O 3 ?

+ N h a n thay ion O H " khong the ton tai cung ion H C O 3 trong d u n g dich ban

d a u d o c o p h a n u n g : , i - , - ' A 6 i- U ) M U * I -

O H - + H C O ; > C 0 ^ - + H 2 0 '-IdU'dfi

—> Dap an D jt^n ife graiQ

V i d^ 3 (B-IG): D u n g dich X chira cac ion: C a ^ Na*, H C O , va CI" (0,1 mol)

Cho 1/2 d u n g dich X phan l i n g v o i d u n g dich N a O H d u , t h u dugc 2 g a m

ket tua C h o 1/2 d u n g dich X con lai phan l i n g v o i d u n g dich Ca(OH)2 d u ,

thu duoc 3 gam ket tua M a t khac, neu d u n soi den c^n d u n g dich X t h i thu

dugc m gam chat ran khan Gia trj cua m la

A 9,21 ; V B.7,47 C 9,26 D 8,79 kUA^gi

fjv \ Lai gidi: Itfbl'

' Gpi so m o l trong 1/2 d u n g djch X: Ca^* (a mol); Na* (b mol), H C O 3 (c mol)

va CI-(0,05 mol)

+ A p d y n g d i n h luat t r i i n g hoa di^n: 2a + b = c + 0,05

Nhqn xet: K h i cho Ca(OH)2 d u vao 1/2 d u n g dich X, toan bp goc H C O 3 se d i

* vao ket tua:

D o v%y, 1/2 cha't ran khan thu dupe chua cac ion: Ca^* (0,02 m o l ) ; Na* (0,04

mol); CO 3 ' (0,015 mol); CI-= 0,05 m o l *^ '• ' '

+ A p d u n g d i n h luat t r u n g hoa dien: 2a + 2b = c '~ '

+ Cac phan l i n g hoa hpc k h i cho V l i t d u n g dich Ca(OH)2 x mol/1 vao coc:

Ca(OH)2 + Ca(HC03)2 > 2CaC03 + 2 H 2 O U , , J , , W

V i d v 5: D u n g dich Y c6 chvia: K^ (0,01 mol), Fe^^ (0,02 mol), N O 3 (0,04 mol)

va SO 4~ (x m o l ) Co can Y t h u dupe m gam m u o i khan Gia trj cua m la

A 3,98 B.6,87 C 5,43 D 4,78

Lai gidi:

O bai nay, d u n g dich da cho chua 4 lo?ii ion, trong d o i o n SO 4 " chua biet so

m o l , do vay can lap dupe 1 p h u o n g t r i n h dai so de t i m so m o l nay

+ A p d y n g d i n h luat t r u n g hoa di^n: * i ^ Tong so'mol d i ^ n tich d u o n g = Tong so'mol dien tich a m

0,01.1 + 0,02.3 = 0,04.1 + x.2^^, x = 0,015 m o l m'^'^'^'^^^'^'

+ Bao toan k h o i l u p n g : m = m^^^ + m^^^ + m^^^ + m^^2 f,

= 39.0,01 + 56.0,02 + 62.0,04 + 96.0,015 = 5,43 gam .,.^j, ^

- » Dap an A

Lieu y: tai nay cac e m can ap d u n g d j n h luat t r u n g hoa di?n de t i m dupe so

m o l cua goc sunfat " t,Oi:<; r ' i , „ i

• 35

Ca'm nang On luyjn thi dgi hgc 18 chuySn dS H6a hpc - Nguygn Van Hi'i

Vi dv 6: Dung djch X c6 chua: Fe^* (0,05 mol), Na* (0,07 mol), CI" (0,03 mol) va

SO 4" Cho dung dich Ba(OH)2 du vao dung dich X thu dugc ket tiia Y

Nung Y a nhi^t dg cao ngoai khong khi den khol lugng khong doi thu dugc

^, m gam chat tin Z (coi BaS04 khong hi nhi?t phan) Gia tri cua m la

+ Phan 1 cho tac dung voi dung dich NaOH du, dun nong, thu dugc 0,58 gam

ket hia va 0,672 lit khi (dktc) .y

+ Phan 2 tac dyng voi dung djch BaCl2 du, thu dugc 4,66 gam ket tiia

Tong kho'i lugng cac chat tan trong Y la

A 3,055 gam B 6,11 gam C 5,35 gam D 7,05 gam

' Lot gtat:

NMn xet: Bai nay cac em giai dua theo cac phuong trinh ion

Phan 2 + dung djch BaCh du:

Ba^^ + sol" , BaS04; f ' - A q « Q ^ - '

^ 'h '"^nt^h 'qsi^^^IKi! -yin ••{in itff ->;; :

~^ "Ba2+ = "BaS04 = ^ = 0/02 mol jf;|„yi j ^ : , ,j

Cty TNHH MTV DWH Khang Vi$t

Ap dung djnh luat trung hoa di^n: 0,01.2 + 0,03.1 = 0,02.2 + n^^ .1

-> n^j_ = 0,01 mol ,'^f^•v:iJiJr^;,'^^ii: ^ ^':^^Otti >

+ Bao toan khoi lugng:

-Khoi lugng cac chat tan trong Y = ^^^^2+ + ' " N H ^ "^cr "^sol" ' •'• '

= 2.(0,01.24 + 0,03.18 + 0,01.35,5 + 0,02.96) = 6,11 gam _> Dap an B

Lim y: Bai nay cac em de chgn nham dap an A (3,055 gam) do chi tinh khoi

lugng chat tan trong mgt nua dung dich Y , , , ,

Vi 8: Dung dich Y c6 chua dong thoi cac ion: Ba^*; Ca^*; CI" (0,01 mol) va

NO J (0,03 mol) Cho V ml dung djch Na2C03 I M vao Y de thu dugc ket tiia Ion nhat Gia trj nho nhat ciia V la

A 30 B.20 C.40 D 70

Ggi so'mol ciia cac ion : Ba^* = X ; Ca^* = y -M.J'jj \t

Ap dyng djnh luat trung hoa di?n: 2x + 2y = 0,01.1+ 0,03.1 = 0,04 mol

Cac phuong trinh phan ung : jii 0 Ba2* + CO3" > BaCOsi H ^ ^ - , B V p : ^ ; - :

C a - + C O r CaC03l ^ ' : Xn^^^2- = X + y = 0,02 mol ,

3 nNa2C03 = 0,02 mol V= = 0,02 lit = 20 ml - > Dap an B.,

Vi dv 9 (A-10): Dung dich X c6 chua: Na^ (0,07 mol); SO^" (0,02 mol) va OH" (x mol) Dung djch Y c6 chua CIO 4, NO 3 va H"" (y mol); tong so mol CIO4

va NO3 la 0,04 Trgn X va Y dugc 100ml dung dich Z Gia tri pH ciia Z la

•*• Ap dung djnh luat trung hoa dien vai dung dich Y: ,^ ( i n i l ) ;

^•"ciOi -^l-^NOS =^-V V =0'04mol • ^

Trgn X vai Y: H* + OH- > H2O

-> H^du = 0,01 mol -» IH^]= ^ = 0,1 = 10"^ ->.pH = l

- > Dap an A. rf.^ o:

-37

Ca'm nang 6n luygn thi dgi hgc 18 chuy§n dg H6a hgc - Nguygn van H&\

V i dv 10 (CD-07): Dung dich Z chiia: Cu^* (0,02 mol), K"" (0,03 mol), C h (x mol)

va S O 4 " (y mol) Tong kho'i lugng cac muo'i tan c6 trong Z la 5,435 gam Gig

tri cua x va y Ian lugt la

A 0,02 va 0,05 B 0,05 va 0,01 C 0,01 va 0,03 D 0,03 va 0,02

LOT gidi:

• O bai nay, dung dich Z chiia 4 loai ion, trong do 2 loai ion chua biet so'mol,

do vay can lap dugc 2 phuang trinh dai so de tim so' mol cua chung

0,02.2 + 0,03.1 = x.l + y.2 X + 2y = 0,07 mol " ' ' ^

W Baotoankhoi lugng: m z = m ^ 2 + m + m^, + m ' N ^ anuQ :8 i#b >;

Bill ;H^i,^>vvi ^ " ^ ^' -Kilofn'eOiO) 'rOVi

^ 64.0,02 + 39.0,03 + 35,5x + 96y = 5,435 S,,

35,5x + 96y = 2,985 ^ x = 0,03 mol; y = 0,02 mol ,

—> Dap an D

V i dvi 11 (CD-08): Chia dung d k h X chiia cac ion: Fe^^ S O | ~ , N H 4 , N O 3

thanh hai phan bang nhau. "t* , ' f • ,-,4 v ' c ! * 1 ; ^ ,r|/

- Phan 1 tac dung voi dung dich N a O H du, dun nong thu dugc 0,672 lit khi

(dktc) v a 1,07 gam ket tiia; , , ;„)- i^^:;; • :';©aji-4):;::.,3-plIiJ's

- Phan 2 tac dung voi dung djch BaCh du, thu dugc 4,66 gam ket tua .^

Co can X thu dugc m gam muo'i khan Gia tri cua m la

A 3,52 gam B 7,04 gam C 8,52 gam D 4,26 gam

Lai gidi:

Nhan xet: bai nay cac em giai dua theo cac phuong trinh ion. 1*'

Phan 1: + Bung dich N a O H du: , ' r i n » * W i t

Dap an C , J:^.Asxu'^'^^^'^^^: i^^^y •

CtyTI\ih.i 1TV DWH Khang Vi^t

V i dxf 1 (CD-07): Trong tu nhien, nguyen to dong c6 hai dong v i la 29 C u va

29 Cu Nguyen tu khoi trung binh cua dong la 63,54 Thanh phan phan tram tong so nguyen tu ciia dong vj 29 C u la

A 27% B 50% C 54% ' D 73%. n :

Lot gidi: " - 'o

Cach 1: Ggi so nguyen h i 29 C u la a va 2^ C u la b

Ap dung cong thuc cua phuong phap duong cheo, ta c6:'

65 - 63,54

I 63-63,54 Dap an D

1,46 0,54

73

27 » % § C u = 73% a

Cach 2: Nhan thay A c u = 63,54 < 63 + 65 dong v i 29 C u chiem u u the hon

- > % 29 C u > 50% Dap an D (Cac dap an khac deu < 50%) '

V i d^;i 2 (CD-07): Cho 4,48 lit khi C O (dktc) tu tir di qua ong sii nung nong dung 8 gam mgt oxit sat den khi phan ling xay ra hoan toan, thu dugc hon hgp khi X c6 ti khoi so voi hidro bang 20 Cong thiic ciia oxit sat va phan tram the tich ciia khi C O 2 trong X la

A F e O ; 7 5 % B FezOs; 75%.' C FeaOs; 65% D.Fe304;75%

Lai gidi:

4 48 Theo bai: M = 20.2 = 40 v a nx = n c o = T T T = 0'2 mol , , ,

Ap dving cong thiic ciia phuong phap duang cheo, ta c6: ' ''

39

Ca'm nang 6n luy$n thi dgi hpc 18 chuyfin dg H6a hpc - Nguygn Van Hi\

T a c o : no(oxit)= "cOz = 0'15 m o l - > 8-0,15.16

56 = 0,1 m o l ion _ ^ : n o = 0,1 : 0,15 = 2:3 -> Fe203 D a p an B \ i ' fin.?'"/-:

D u n n o n g X m o t t h o i gian t r o n g b i n h k i n (c6 Fe l a m xuc tac), t h u d u g c h o n

h g p k h i Y c6 t i k h o i so v o i h e l i bang 2 H i f u sua't ciia p h a n u n g t o n g h g p

Theo bai: M x = 1,8.4 = 7,2; M Y = 2.4 = 8 i ! i:M :

A p d y n g cong t h u c ciia p h u o n g p h a p d u o n g cheo v o i X, ta c6:

suat ciia p h a n u n g h i d r o hoa l a

dm nang On Iuy0n thi dgl hpc 18 chuyfin ai H6a hqc - Nguyin van Hit

A p d y n g cong t h u c ciia p h u a n g p h a p d u o n g cheo v a i X, ta c6:

G o i : ric2H4 = 1 iriol; riH2 = ^ "^ol- ;lado':.;riT

V i d u 1 (CD-11): D e hoa tan hoan toan 6,4 g a m h o n h g p X g o m k i r n loai R (chi c6

hoa t r i 2) va o x i t ciia no can v u a du 400ml d u n g d i c h H C l I M K i m l o ^ i R la

A Ba B Be C M g D Ca 163

:i Laigiai: -A

Ca'm nang 6n luygn thi d^i hpc 18 chuySn 6i H6a hgc - Nguygn Van Hai

V i d\ 2: Cho 1,67 gam hon hgip gom hai kim loai (6 2 chu ky lien tiep thupc

nhom IIA) tac dung het voi dung dich HCl (du), thoat ra 0,672 lit khi H2

(dktc) Hai kim loai do la

A BevaMg B.MgvaCa C S r v a B a D Cava Sr

n H 2 = ^ ^ = 0 , 0 3 m o l ,V

^^'^ _ _ t ! >fr ^

-Phuang trinh phan ung: M + 2HC1 > MCI2 + H2 ,

Nhan thay: nj;^ = n ^ j = 0,03 mol i

-> ^ ^ ^ ^ " ^ 55,67 Hai kim loai la Ca va Sr -> Dap an D

V i dxjL 3 (A-10): Cho 7,1 gam hon hop gom mot kim loai kiem X va mpt kim loai

if^l kiem tho Y tac dung he't vai lugng du dung dich HCl loang, thu dupe 5,6 lit

khi H2 (dktc) Kim loai X, Y la

A Kali va bari B Liti va beri C Natri va magie D Kali va canxi

Lap luan: Mx hoac M Y < M < 28 -» Loai A va D vi hai kim loai deu > 28

Mx hoac M Y > M > 14 -> Loai B vi hai kim loai deu < 14

b'j -> Dap an C

Cach 2: Gpi cong thiic chung aia hai kim loai la M , hoa tri chung la n

Ta c6: n^ = 2nj^^ = 0,5 mol Bao toan electron: ng (j^j = ng= 0,5 mol

^ n.nM = 0,5mol-> - ^ = 0,5 ^ M = 14,2n

M

Mat khac: 1 < n < 2 14,2 < M < 28,4 Lap luan nhu tren -» Dap an C

V i A\y 4 (CD-12): Hoa tan hoan toan 1,1 gam hSn hpp gom mpt kim loai kiem X

va mpt kim loai kiem tho Y (Mx < MY) trong dung dich HCl du, thu dupe

1,12 lit khi H2 (dktc) Kim loai X la

A L i B.Na C Rb D K

Cty TNHH MTV DWH Khang Vi§t

Lot gidi:

Gpi cong thiic chung ciia hai kim loai la M, hoa tri chung la n * *

-Ta c6: n^ = 2n^^^ = 0,1 mol Bao toan electron: n^ " " e " 0,1 mol

_> n n M = 0 / l m o l - ^ = 0,1 ^ M = l l n

M a t k h a c : l < n < 2 -> l l < M < 2 2 V a y M x < M < 2 2 Loai B, C, D

Dap an A

Vi du 5 (B-08): Cho 1,9 gam hon hpp X gom muoi cacbonat va hidrocacbonat ciia

kim loai kiem M tac dung he't voi dung dich HCl (du), sinh ra 0,448 lit khi (a dktc) Kim loai M la

Vi d^ 6: Cho 1,7 gam hon hop gom kim loai X (nhom IIA) va Zn tac dung voi

dung dich HCl du, sinh ra 0,672 lit khi H2 (dktc) Mat khac, khi cho 1,9 gam

X tac dung voi axit H2SO4 loang, du thi the tich khi hidro sinh ra chua deh 1,12 lit (dktc) Kim loai X la

Vi dii 7: Dot chay hoan toan 6,72 lit (dktc) hon hpp M gom hai hidrocacbon X

va Y (Mx < MY), thu dupe 11,2 lit khi CO2 (dktc) va 10,8 gam H2O Cong thiic phan tu cua X la

.C2H4 B.C2H2 C.C2H6 •Q™D.CH4 '

4";

Ca'm nang On luygn thi dgi h9C 18 chuygn H6a hpc - Mguyin Van Hii

V i d\ 8 (B-10): H o n hop k h i X gom m o t ankan va m o t anken T i khoi cua X so

voi H2 bang 11,25 Dot chay hoan toan 4,48 lit X, thu dugc 6,72 l i t CO2 (cac

the tich k h i do 6 dktc) Cong thuc ciia ankan va anken Ian l u g t la

Theo bai: M x =11,25.2 = 22,5 X chua m o t hidrocacbon c6 phan tvr khoi

nho h o n 22,5 - > D o la metan (CH4) Loai B

M x = 2 — ^ ^20,33 ^ 22,5

46

Cty TNHH MlV DWH Khang Vigt

V i dV 9 (B-08): D o t chay hoan toan 1 lit hon hop k h i gom C2H2 va hidrocacbon

X sinh ra 2 lit k h i CO2 va 2 lit hoi H2O (cac the tich k h i va h o i d o 6 ciing dieu kien nhi^t dg, ap suat) Cong thuc phan t u ciia X la

A C 2 H 6 , B.C2H4 C C H 4 D C 3 H 8

Lai giai: j , ,

Xheobai: ' ,: "^*::i*>,^i:^'O^H:,,>«:'''''

-+ So nguyen t u H trung binh = —^^^2-= ^ = 4 _> Hidrocacbon X c6 so

nguyen t u H I o n h o n 4 (vi hon hgp chua C2H2 so nguyen t u H nho h o n 4)

Loai B va C : ; - i

+ So nguyen t u C trung binh = — — = 2 —> Hidrocacbon X c6 so nguyen

t u C bang 2 (vi hon h g p chua C2H2 so nguyen t u C bang 2) j j , j , Dap an A

V i dy 10: D o t chay hoan toan 6 lit hon hgp X gom 2 anken ke tiep nhau trong

day dong d i n g can v u a d u 21 lit O2 (cac the tich k h i d o trong cung dieu ki?n nhi^t do, ap suat) H i d r a t hoa hoan toan X trong dieu k i | n thich h g p thu dugc hon h g p ancol Y, trong d o tong khoi l u g n g cac ancol bac m o t gap 13/6 Ian khoi l u g n g ancol bac hai Phan tram khoi l u g n g ciia ancol bac mot (c6 so cacbon Ian hon) trong Y la

3 Taco: K = - ^ 2a + 3 b ^ 7 ^ ^ ^ ^ b - > Dat a = 2 m o l ; b = 1 m o l

dm nang On luygn thi dji hqc 18 chuySn dg H6a hpc - Nguygn Van Hii

Cach 2: De thay: % mcHgCHzOH = " 60,53% ' • >

M a t khac, theo bai: % mcH3CH(OH)CH3 = -100% = 31,58%

^ % mcH3CH2CH20H = 100 - 60,53 - 31,58 = 7,89%

—> D a p an D ' , , ,,

V i dv 11: H o n h o p X g o m hai axit cacboxylic d o n chiic Do't chay hoan toan

0,1 m o l X can 0,26 m o l O2, thu duoc CO2 va 0,2 m o l H2O Cong thiic hai

V i d\ 12: D o t chay hoan toan hon hop X g o m hai este can d u n g 14 lit k h i O2,

t h u dugc 12,32 l i t k h i CO2 va 9,9 gam H2O N e u cho m gam X tac d u n g

vira d u v o i d u n g d i c h K O H , c6 can d u n g dich sau phan u n g t h i t h u dxxgc

m u o i k h a n cua mQt axit h u u co va m gam hon hgip ancol Y la d o n g d5ng ke

tiep Gia t r i ciia m la

M a t khac, k h i cho X tac d u n g v o i K O H , t h u duQfc m u o i ciia m g t axit h i h i co

va hon hop ancol Y la dong dang ke tiep —> 2 este t r o n g X h o n kem nhau 1 nguyen t u cacbon

—> Cong thiic p h a n t i i 2 este: C2H4O2 va CsHeCh - , j j

- > Cong thuc cau tao: H C O O C H 3 (a mol) va HCOOC2H5 (b mol) ,5

V i dvi 13 (A-10): Cho h o n X g o m ancol metylic v a hai axit cacboxylic (no, d o n

chiic, ke tiep n h a u t r o n g day dong dang) tac diang het v o i Na, giai p h o n g ra 6,72 l i t k h i H2 (dktc) Neu d u n nong X (co H2SO4 dac xiic tac) t h i cac chat

trong h o n hgp p h a n l i n g vtra d i i v o i nhau tao thanh 25 g a m h o n hgp este

(gia thie't phan l i n g este hoa dat h i f u suat 100%) H a i axit t r o n g X la

A. C3H7COOH va C4H9COOH B C H 3 C O O H va C2H5COOH

C. C2H5COOH va C3H7COOH D H C O O H va C H 3 C O O H

Laigidi:

n H , = — = a 3 m o l

' 22,4 _ Gpi cong thuc chung ciia 2 axit la R C O O H 4 v, ,

Dat so m o l : C H 3 O H = a; R C O O H = b Ta co: a + b = 2 n H j = 0,6 m o l

Phan u n g este hoa: R C O O H + C H a O H ; e = ± RCOOCH3 + H2O

49

dm nang On luygn thi dgi hpc 18 chiiy6n 6i H6a hqc - Nguyln Van H5i

Theo bai: cac cha't tham gia phan ung este vua du -> a = b

N h u vay: a = b = 0,3 mol ,,.„;„, ,.uv.v ;; ; a,,v •'• J.Q«

^ M R C O O C H S = — = 83,33 R + 44 + 15 = 83,33 R = 24,33

-> Goc hidrocacbon trong 2 axit la CHs- va C2H5—> Dap an B

V i d\ 14 (A-11): H o n hgp M chua 15,52 gam mpt axit no, don chiic X va mpt

axit no, da chuc Y (nx > nv) Hoa hoi M thu duQC mpt the tich hoi bang the

tich ciia 5,6 gam N2 (do trong cimg dieu ki§n nhift dp, ap suat) Ne'u dot

chay hoan toan M thi thu dupe 10,752 lit CO2 (dktc) Cong thuc cau tao cua

= = ^ = 0'2 mol; n c o j = ^^Y= 0,48 inol

+ So nguyen t u C trung binh trong M : C M = — 2 , 4 ;^-ifYY! i

-> Loai A v i 2 chat deu c6 so'nguyen tvf C < 2

J -> Loai C v i nx > nv ^ X chie'm u u the So nguyen t u C trung binh can

thoaman: C M > ^ ^ ^ = — = 2 , 5

Thii50:50 • vov'¥i<ri s n u b t s f & h an'Sh v /b

+ Phuong an B: CHaCOOH = a mol; H O O C - C H 2 - C H 2 - C O O H = b mol .d

+ b = 0,2; 2a + 4b = 0,48; 60a + 118b = 15,52 rs,eis.:;|, <|v" n&si i^/nni

-» V6 nghiem Loai

+ Phuong an D: CHaCOOH = a mol; HOOC-CH2-COOH = b m o l

a + b = 0,2; 2a + 3b = 0,48; 60a + 104b = 15,52

^ a = 0,12; b = 0,08 Thoa man -> Dap an D

V i dv 15: Hoa hoi 7,76 gam hon hpp M gom axit X (no, dan chuc, mach ho) va

axit Y (no, da chuc Y, mach cacbon ho, khong phan nhdnh) thu dupe mpt

the tich hoi bang the tich cua 2,8 gam N2 (do trong cung dieu k i | n nhift dp,

ap suat) Dot chay hoan toan 7,76 gam M , thu dupe 10,56 gam C O 2 Phan

tram khoi lupng cua X trong M la

f^han xet: Y la axit no, da chuc, c6 mach cacbon ho, khong phan nhanh -> Y

chua 2 nhom chuc (vi ne'u chua tir 3 nhom chuc tro len, mach cacbon se phan nhanh)

Theo bai: n x= n N 2 = 0,1 mol <»

— 0 24

So nguyen t u cacbon trung binh: C(X) = = 2,4

7 Phan t u khoi trung binh: M(X) = —— = 77,6

0,1 Mpt axit chua so' nguyen t u cacbon < 2,4 Chia 2 truong hpp:

+ Truong hop 1: Ne'u so' cacbon cua Y < 2,4 -> Y chua 2C (vi Y da chuc) ->

la HOOC-COOH

Mat khac: M Y = 90 > 77,6 ^ Mx < 77,6 ^ X la H C O O H hoac CHaCOOH V6 l i v i so cacbon cua X > 2,4 , + Truong hap 2: Ne'u so cacbon cua X < 2,4 -> X la H C O O H ho^c CHaCOOH

Gpi cone thiic cua Y la CnH2n-204 (b mol)

va 0,6 mol H2O Thyc hifn phan ung este hoa 12,2 gam hon hpp tren v d i

h i f u suat 60% thu dupe m gam este Gia trj cua m la

51

Ca'm nang an luy$n thi d?i hgc 18 chuy6n dg H6a hpc - Nguygn van Hi\

mo = 12,2 - 0,45.12 -1,2.1 = 5,6 -> no = 0,35 mol

Nhan thay: XXQ =2nx + nY -> nx =0,1 mol =0,25mol

-''^i/P-So nguyen tu cacbon trung binh: C ( M ) = = 1,8. i*w > ••'sW'v

-> M chiia rriQt chat chiia 1 nguyen tvr cacbon

Neu X chiia 1 nguyen tir cacbon -> X la HCOOH = 0,1 mol

V i dyi 17: Cho 4,2 gam hon hop X gom hai amin no, don chiic, ke tiep nhau

trong day dong dang tac dung het vol dung dich HCl (du), thu duac

7,85 gam hon hgp muoi Cong thuc ciia hai amin trong X la

Phuong trinh hoa hpc: R - N H 2 + HCl > R - NH3CI

Bao toan khoi lugng: mamin + mnci = mmuoi

mHci= 7,85 - 4,2 = 3,65 gam -> n^ci = 0,1 mol = rumin ^,

^ M a m i n = = 42 ^ R - N H 2 = 42 ^ R = 26 f MS A

0,1

-> Go'c hidrocacbon ciia 2 amin la C H 3 - va C2H5 > Dap an B

V i dv 18: Dot chay hoan toan 100ml hon hp-p khi X gom trimetylamin va hai

^ hidrocacbon dong dSng ke tiep bang mpt lugng oxi vira dii, thu dupe 750ml

hon hgp Y gom khi va hoi Dan toan bp Y di qua dung dich H2SO4 d^c

(du), the tich khi con lai la 350ml Cac the tich khi va hoi do 6 cimg dieu

ki^n Hai hidrocacbon do la

Cty TNHH MTV DVVH Khang Vi^t

A C 2 H 4 v a C 3 H 6 B C2H6 va C3H8

C C3H6 va C4H8 , „ ^- ^SHS va C4H10 ,

phan ung hoa hoc: „ „„,^ ,.,„., „ ,

(CH3)3N + O2 3C02 + | H 2 0 + - N 2 j^ol: a 3a 0,5a

Nhan xet: Khi dan Y di qua dung dich H2SO4 dac (du), chi c6 hoi nuoc bj

Mat khac, khi dot chay amin va hidrocacbon: V QQ^ > 3a = 6 V JSJ^ "en suy ra

Vc02 > y-350 - 300ml So nguyen tu C trung binh = ^75^ = ^ So'nguyen t u C trung binh cua hai hidrocacbon > 3 -> Lo^ii A va B

Dap an C » MyO M *c,f •

8 PHl/ONG PHAP GIAI THEO PHl/ONG T R I N H ION

a Npi dung Khi cac ion tham gia phan ung do nhieu chat phan l i ra, can giai bai toan

dya theo cac phan ung dang ion thu gpn

b Van dung

+ Phan rnig trao dot ion: 1 - nhieu axit tac dyng voi nhieu bazo; 2- ha'p thu khi cacbonic vao dung dich chua nhieu bazo; 3- trpn hai dung djch chua nhieu chat di^n l i voi nhau

+ Phan ung oxi hoa khu: 1 - Dung djch chua axit H N O 3 va HCl (hoac H2SO4

loang); 2- Dung djch chua muoi nitrat va axit HCl (hoac H2SO4 loang)

Vi DV MAU

Vi dv 1 (B-09): Trpn 100ml dung dich X gom H2SO4 0,05M va HCl 0,1M voi

100ml dung djch Y gom NaOH 0,2M va Ba(OH)2 0,1M, thu dugc dung dich

2 Dung dich Z c6 p H la

A 13,0 B.1,2 C 1,0 D 12,8

53

dm nang On luy$n thi dgi hpc 18 chuyfin 66 H6a hpc - Nguyjn Van HSi

Lot gidi:

Nhan xet: Bai nay chua nhieu chat dien li nen giai d^a theo phuang trinh

ion Do bai toan chi hoi ve pH -> chi quan tarn den va OH"

'r >H+ = HHCI + 2nH2S04 = 0,01 + 2.0,005 = 0,02 mol ''^^ <f' Trong X:

Trong Y:

n = 0,01 mol; n o - = 0,005 mol O '/•^tryf^

" O H - " "NaOH + 2nBa(OH)2 = ^'^^ + 2.0,015 = 0,04 mol

V i dv 2: Cho m gam hSn hop Na va Ba (ti le mol 1:1) tac dyng vai nuoc (du),

thu du(?c dung dich X va 3,36 lit H2 (dktc) Cho 200ml dung dich Al2(S04)3

0,2M vao X, sau khi phan ung hoan toan, thu dugc a gam ket tiia Gia tri

Trong X: " O H - " "NaOH + 2nBa(OH)2 = ^'^ + ^.0,1 = 0,3 mol

V i dV 3= 8^"^ 250ml dung djch X chua hon hgp axit HCl I M va axit H2SO2 0,5M, thu dugc 5,32 lit H2 (dktc) va dung dich Y (coi the tich dung dich khong doi) Dung djch Y c6 pH la _

A 7 B l C 2 , D.6

Lmgtat: ^

nH2= 1^=0-2375 mol

luu y: C* bai nay neu cac em vie't phuong trinh phan tii se gap kho khan vi

can viet deh 4 phuong trinh ^, ^ , ; Trong X: n^+ = nHci + 2x1^^504, = 0,25 + 2.0,125 = 0,5 mol ' '

Truoc het, can xem axit c6 phan ling het hay con du bang each so sanh so mol H* ban dau va so'mol khi H2 bay ra theo ti 1#:

Y (chi gom cac muoi) Khoi lugng muoi c6 trong Y1^: '• W"'- '

A 37,9 gam B 47,5 gam C 35,1 gam D 43,5 gam

' Loigidi: 1 ,.:'t;.iiio,y

nMg= 0,15 mol; n A i = 0,1 mol. ! 0 0

-^Mn xet: Khi axit HNO3 c6 mat dong thoi voi axit H2SO4 loang thi lugng

trong dung dich la do 2 axit phan li ra -> giai theo phuong trinh ion

dm nang On luygn thi dgi hgc 18 chuy6n d6 H6a hgc - Nguygn Van HJi

-> AP* + N O + 2 H 2 O

0,1

.'' - > H"*" v a N 0 3 tham gia phan ling he't ' '

" - > Khoi lu^ng muoi trong Y = 3,6 + 2,7 + 0,3.96 = 35,1 gam D a p an C

V i dv 5 (A-11): Hap thu hoan toan 0,672 lit khi C02^(dktc) vao 1 lit dung dich X

gom N a O H 0,025M va Ca(OH)2 0,0125M, thu du(?c m gam ket tiia Gia trj

a i a m la

A 2,00 B 1,00 " C 1,25 fn ^^-'vf: D 0,75 '

Lot giai:

Nhan xet: Bai nay neu cac em viet phuong trinh phan tu se gap nhieu kho

khan Do vay, nen giai d y a theo cac phuong trinh ion

Trong X: " O H - " "NaOH+ 2nca(OH)2 = ^'^^S + 2.0,0125 = 0,05 mol

- » phan ung tao ra 2 muoi: cacbonat va hidrocacbonat

Cac phuong trinh ion:

Luu y: N e u t?o dong thai 2 goc axit: H C O 3 va CO§~, cac em c6 the ap

d\ing cong thiic tinh nharUi ra so mol C O 3 2-

" c o 2 - " " O H - ' "CO2 = 0/05 - 0,03 = 0,02 mol

Tir do so sanh voi "^,^2+ va suy ra n^gcos = 0,0125 mol ;

V i dvi 6 (A-09): Cho 0,448 lit khi C O 2 (dktc) hap thu he't vao 100ml dung dich Y

chiia hon hgp N a O H 0,1M va Ba(OH)2 0,1M, thu dug-c m gam ket tiia Gia

hidrocacbonat :no,i jfi-;* C K I - ; i t

Ap dung cong thuc tinh nhanh so mol C O 3 " : " »

" c o ^ - " " o H - • " C O 2 = 0,03 - a02 = aOl mol i =- h :)inxyu\, r v

3 Nhan thay: n^^2- = "3^2+ ^ "BaCOs = " c o ^ " " ' ' • '

- > m = 0,01.197 = 1,970 gam ^' '

—> Dap an C

V i dvi 7 (B-12): Sue 4,48 lit khi C O 2 (dktc) vao 1 lit dung d k h Z chua hon hop Ba(OH)2 0,12M va N a O H 0,06M Sau khi cac phan ung xay ra hoan toan thu dug-c m gam ket hia Gia trj cua m la

A 19,70 B 23,64 C 7,88 ' " ' f ' D 13,79

Lai giai:

n _ = nN30H + 2nBa(OH)2 = 0'06 + 2.0,12 = 0,3 m o l ' ^ Trong Z : OH'

Ca'm nang fln luygn thi dgi hgc 18 chuyfin dg H6a hgc - Nguygn Van Hii

Nh^n thay: ^^^^ = = - > 2 OH- du va phan ung chi t^o ra

muoi cacbonat • , ^ ^4^'' •^"'•"^ I

Phuong trinh ion: CO2 + 20H- > CO^" + H2O

Mol: 0,015 0,03 0,015 Dung dich Y gom cac ion: Na^ = 0,02 mol; = 0,02 mol; OH" = 0,01 mol va

c o r =a015mol v , „

Baotoankhoilugng: a= m^^, + mj^, + m^^

^m^^j-= 0,02.23 + 0,02.39 + 0,01.17 + 0,015.60 m ^m^^j-= 2,31 gam

-> Dap an C

Vi dii 9 (B-07): Thuc hien hai thi nghiem:

1) Cho 3,84 gam Cu phan ung voi 80ml dung dich HNO3 I M thoat ra Vi lit NO

2) Cho 3,84 gam Cu phan ung voi 80ml dung dich chua HNO3 I M va H2SO4

0,5 M thoat ra V2 lit NO

Biet NO la san pham khu duy nhat, cac the tich khi do o cung dieu ki^n

QuanhegiOaVi v a V 2 l a •

A V 2 = V i B V 2 = 2 V i C V 2 = 2 , 5 V i D V2 = l,5Vi

Loigidi: ' ' ; ;S§noiT

Thi nghiem 1: nr-u = 0.06 mol: nHMOj = 0,08 mol 'ifl J v

BV Hao&y 3 C u + 8HNO3 > 3Cu(N03)2 + 2 N O + 4H2O ,^4J

Mol: 0,03 <- a08 -> 0,02 " ' I

Vi = a02.22,4 = 0,448 lit .mi xjIgsMifjKl:

Thi nghiem 2: Nhan xet: Su c6 mat axit H2SO4 loang se "dong gop" them luong

H"^ trong dung dich Khi do, lugng H"^ la do 2 axit phan li ra nen can giai

theo phuong trinh ion

Ta c6: n^+ = nHNOj + 2nH2S04 = 0/16 mol; n^^^, = nHNOg = 0/08 mol

3Cu + 8H^ +2NO3 )• 3Cu2^ + 2 N O + 4H2O

^ " M O I : 0,06 <-0,16- 0,04 -> 0,04 » # IV

-> V2 = 0,04.22,4 = 0,896 lit V2 = 2V, ^^^"^ « ^

-> Dap an B *' 'xmih • ^^'^ •

Vi d\ 1 0 ( A - 0 8 ) : Cho 3,2 gam hot Cu tac dyng voi 100ml dung djch X hon h(7p

gom HNO3 0,8M va H2SO4 0,2M Sau khi cac phan ung xay ra hoan toan,

sinh ra V lit khi NO (san pham khu duy nhat, 6 dktc) Gia trj ciia V la

t^han xet: Khi axit HNO3 c6 mat dong thoi voi axit H2SO4 loang thi lugng H*

trong dung dich la do 2 axit phan li ra —> giai theo phuong trinh ion

Vi 1 1 ( A - 1 1 ) : Cho 7,68 gam Cu vao 200ml dung dich X gom HNO3 0,6M va H2SO4 0,5M Sau khi cac phan ung xay ra hoan toan thu dugc khi NO (san pham khu duy nhat la NO) va dung dich Y Co can can than Y thu dugc bao nhieu gam chat ran khan?

A 19,76 B 22,56 C 20,16 " D 19,20

Lai giai: -J

n c „ = ^ = 0 , 1 2 - m o i : ' ' - ' - * ^ ^ j ^ ^ ' '

Nhan xet: Khi axit HNO3 c6 mat dong thoi voi axit H2SO4 loang thi lugng

H* trong dung dich la do 2 axit phan li ra giai theo phuong trinh ion

V " " H N 0 3 + 2nH2S04 = 0,12+ 0,2 = 0,32 mol ,,,,,

Phuong trinh ion riit ggn: gruji;; •

3Cu + 8H* + 2NO3" > 3Cu^* + 2 N O + 4H2O

Hil]-Mol: 0,12 <<-0,32-> 0,08 -> 0,12 0 , 0 8 Dung dich Y: Cu^* = 0,12 mol; N O ; du = 0,04 mol va SO 4" = 0,1 mol

-> Khoi lugng chat ran khan = 64.0,12 + 62.0,04 + 96.0,1 = 19,76 gam

~> Dap an A \'''\frf^ • •/ ,' '>•-.•<

Lmi y: C) bai nay, ngoai vi|c giai theo phuong trinh ion, cac em can nho

tinh den goc sunfat c6 trong dung dich sau phan ung

^1 1 2 (B-10): Cho 0,3 mol bgt Cu va 0,6 mol Fe(N03)2 vao dung dich chua 0,9 morH2S04 (loang) Sau khi cac phan ung xay ra hoan toan, thu dugc V lit khi NO (san pham khu duy nhat, 6 dktc) Gia tri ciia V la

A 6,72 B.8,96 C 4,48 D 10,08

59 Trong X:

ca'm nang 6n luyjn thi dgi hgc 18 chuyfin H(Sa hpc - NguySn Van H3i

Lot gidi:

Cach 1: Khi hoa tan vao dung dich H 2 S O 4 , Fe(N03)2 se phan li thanh cac ion

Do vay, truoc het cac em can tinh so mol cac ion nhu sau: O ' •

Phuong trinh ion rut gon: ' ^''^''/'i •••^'J^/•^ -:> <>^:r'^ ^i ^mrp

3Cu + 8H* + 2NO^ • > 3Cu2- + ' ^ N O + 4H2O

Mol: a3 <- 0,8 0,2 -> 0,3 0,2 c O / ; ,

Cac em luu y, Fe^* cung bi oxi hoa thanh Fe^* va giai phong khi N O :

3Fe2* + 4H* + N O ; > 3¥e^ + N O + 2H2O

Mol; 0,6 ^ 0,8 ^ 0,2 0,6 0,2 ^ ^ _

> V = (0,2 + 0,2).22,4 = 8,96 lit > Dap an B rr,fj{),p« y 4

-Cach 2: Cac chat khu: Cu - 2e ->• Cu^^; Fe^^ - le -> Fe^*

Chat oxi hoa: N*-^ + 3e -> N O

Bao toan electron: 2ncu + lnp^2+ = Sn^o

2.a3+a6 , v r ^

r i N O = z =0,4 mol

^ V = 0,4.22,4 = 8,96 lit ^ Dap an B

Vi d^ 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4

0 ,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch

-I X Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ke't tua Ian nhat

Gia trj nho nhat ciia V la

A 80 B.50 C.60 D 40

Ldi gidi:

Nhan xet: Dung dich chiia muoi NaNOs va H2SO4 loang -> can giai theo

phuong trinh ion ;?> v, t, • 5

ncu =0,06 mol; n^+ = 2.0,2.0,5 = 0,20 mol; n ^ ^ , =0,2.0,2 = 0,04 mol

Phuong trinh ion thu gpn:

3Cu + 8H* + 2NO3- > 3Cu^*+ 2NO + H 2 O

;^Mol: a06 ai6 a04

MjoriX^:-Cu tan het -> X CO chiia: n^ 2+ = 0,06; n + du = 0,04 '

De thu dugc luang ke't tua Idn nhat thi: njs^aOH = n + + 2n 2+ /

H C u

" N a O H = 0,04 + 2.0,06 = 0,16 mol

y _> V = 0,08 lit = 80ml

Cty TNHH MTV D W H Khang Vigt

14: Hoa tan hoan toan 0,1 mol FeS2 trong 200ml dung dich HNO3 4M, san pham thu dugc gom dung dich X va mot chat khi thoat ra Dung dich X

CO the hoa tan toi da m gam Fe Biet trong cac qua trinh tren, san pham khu duy nhat cua N*^ deu la NO Gia tri ciia m la

-Fe3^ = 0,1 mol; H* = 0,4 mol; N O ; = 0,3 mol va SO ]' = 0,2 mol

^ , ^ , , , , „ " WO '-* "'"^ * mi

Cac phan ung hoa tan Fe:

Fe + 4H* + N O ; > Fe3* + N O + 2H2O QkHi Mol: 0,05 4-0,4 - > 0,05 ••idl,0» j„rtS«.^^iT -.'ihtvU/

Fe + 2Fe3^ > 3Fe2- q% lOH dfegfror* ,;>fefii JfiSv)

Mol: 0,075 <- 0,15

—> mpe = 0,125.56 = 7,0 gam - > Dap an B

Vi d^ 15: Iron 200ml dung djch X gom Ba(OH)2 0,1M va NaOH 0,3M voi

100ml dung dich Y gom Al2(S04)3 0,1M va H2SO4 0,1M, thu dugc a gam ke't tua Gia tri cua a la

Al(OH)3 + O H " > AlO; + 2H2O

0,02 <- 0,02 ,Ba2^ + S04~ > BaS04>l 0,02 -> 0,02 ^ 0,02

C&m nang fln luyjn thi dgi hgc 18 chuy6n dg H6a hoc - MguySn Van H5i

V i d v 16: H o a tan hoan toan 9,46 gam hon h g p gom Na, K va Ba vao nuoc, thu

duQC d u n g dich X va 1,792 l i t k h i H2 (dktc) D u n g dich Y g o m H C l I M va

1^ H2SO4 0,5M T r u n g hoa d u n g dich X b o i d u n g dich Y, tong khoi l u g n g cae

V i d\ 17: Cho h o n h o p g o m Ba va A l (ti le m o l 1:1) vao nuac (du), t h u dugc

d u n g djch X va 1,12 lit k h i H2 (dktc) D u n g dich Y gom H C l 0,5M va H2SO4

0,1M Cho t u t u den het 100ml d u n g dich Y vao X, thu dugc m gam ket tiia

no=2ncooH n o = 2nco2

Vi Dv M A U

V i d v 1 (A-09): K h i d o t chay hoan toan m gam h o n h g p X g o m hai ancol no,

d o n chuc, mach h o thu dugc V l i t k h i CO2 (6 dktc) va a g a m H2O Bieu thu-c

lien h? giiia m , a va V la:

dm nang On luygn thi dji hpc 18 chuy6n H6a hpc - IMguySn Van HSi

V i d u 2: D o t chay hoan toan m g a m hon h o p Y g o m ba ancol d o n chuc, thuoc

cung d a y d o n g d i n g , t h u dugc 37,4 g a m k h i C O 2 (dktc) v a 27 g a m H 2 O Gia

V i d y 3: Do't chay hoan toan m o t l u g n g h o n h o p X g o m hai ancol (no, d a chuc,

mach ho, c i i n g so n h o m - O H ) can v u a d i i V l i t k h i O 2 , t h u dugc 5,6 l i t k h i

C O 2 v a 6,3 g a m H 2 O (cac the tich k h i d o 0 dktc) Gia t r j ciia V la a t

A 5,60.' B.3,92 C 7,28 D 1,12

Laigidi: ' r •

5,6 6 3 ' ''^''^.ymm-Mtifbw

yi: ^C02 m o l ; n H 2 0 = 0 , 3 5 m o l ; , ^

Theo bai ra, X chua 2 ancol n o -> nx = n ^ j o " "CO2 " ^ o l •

Cty TNHH MTV DWH Khang Vigt

So nguyen t u C t r u n g binh = "'''"^ = 2,5 X chua m g t ancol d a chuc c6

V i d v 4 (CD-12): D o t chay hoan toan hon hg-p X g o m hai ancol (no, hai chuc,

mach ho) can v u a du V i l i t k h i O2, t h u dugc V 2 l i t k h i C O 2 va a m o l H 2 O

Cac k h i d e u d o 6 dieu kien tieu chuan Bieu thuc lien he giiia cac gia t r i

Do X chua 2 ancol hai chuc > no = noH = 2nx > no = 2a

-Bao toan n g u y e n to O: no (on) + 2 n o 2 = 2 nQQ^ + VXH^Q

2V2 ^ 2 V i _ 2V2

22,4

2 ^ 22,4

-> 2a - + ^=-^ = ^ - ^ + a V i = 2V2 - 11,2a o n

22,4 22,4 22,4 -> Dap an D

Nhan xet: Bai nay cac e m can n h o v o i ancol no t h i : nancoi = n^^Q " "CO2 '

dong t h a i bie't ap d u n g bao toan nguyen to oxi

V i d ^ 5 (B-12): Do't chay hoan toan m gam h o n h g p X g o m hai ancol, t h u dugc

13,44 l i t k h i C O 2 (dktc) va 15,3 gam H 2 O M a t khac, cho m g a m X tac d y n g

v o i N a (du), t h u dugc 4,48 l i t k h i H 2 (dktc) Gia trj cua m la

Ca'm nang 6n luy^n thi d<ii hpc 18 chuy6n 6i H6a hgc - Nguyin Van H^i

Bao toan kho'i l u g n g : m x = m c + m n + m o

= 0,6.12 + 2.0,85.1 + 0,4.16 = 15,3 gam

D a p an B

V i d u 6 (B-12): Cho h o n h o p X g o m metanol, etylen glicol v a glixerol Do't chay

hoan toan m gam X t h u dugc 6,72 l i t k h i C O 2 (dktc) CQng m gam X tren cho

tac d u n g v o i N a d u t h u dugc to'i da V l i t k h i H 2 (dktc) Gia t r i a i a V la

V i d u 7: H o n h o p X g o m hai axit cacboxylic d o n chuc D o t chay hoan toan 0,1

m o l X can 0,24 m o l O 2 , t h u dugc C O 2 va 0,2 m o l H 2 O Cong thuc hai axit la

A H C O O H v a C H 3 C O O H i

B C H 2 = C H C O O H v a C 2 H 5 C O O H i o , rm • -f i

C C H 3 C O O H v a C 2 H 5 C O O H , ;

D C H 3 C O O H v a C H 2 = C H C O O H * 3iwb fefl loarifi dG

N h | n thay X chua 2 axit cacboxylic d o n chiic -> chua 2 nguyen t u oxi

+ So' nguyen t u C t r u n g b i n h = = - 5 ^ = 2,4 -> L o a i A ( v i cac axit chua

0,1

so n g u y e n t u C < 2)

- > D a p an D ^ '••m'r\fl•„,,••

V i d\ 8 (B-11): H o n h g p X g o m v i n y l axetat, m e t y l axetat v a etyl fomat D o t

chay hoan toan 3,08 gam X, t h u dugc 2,16 gam H 2 O Phan t r a m so m o l ciia

N h a n thay, k h i d o t chay 1 m o l m o i chat m e t y l axetat v a etyl fomat deu t h u

dugc n c o 2 = n H 2 0 ' " e n g v o i v i n y l axetat t h i : n^^^ - n^^o = 1- v /

Do v^y: nvinyi axetat = r\QQ^ - n H 2 0 0,13 - 0,12 = 0,01 ' "

- > % nvinyi axetat = ^ ^ 1 0 0 % = 25% - ¥ D a p an D , , b f n i J 3 » , , , A

V i d\ 9 (A-11): H o n h g p X g o m axit axetic, axit fomic v a axit oxalic K h i cho m

gam X tac d u n g v o i NaHCOs (du) t h i t h u d u g c 15,68 l i t k h i C O 2 (dktc) M a t khac, d o t chay hoan toan m gam X can 8,96 l i t k h i O 2 (dktc), t h u dugc 35,2

gam C O 2 va v m o l H 2 O Gia t r i cua v la

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

N/ion xet: K h i cho axit cacboxylic tac d u n g v o i NaHCOa ta l u o n c6:

" C 0 2 = r » C 0 0 H -> no(X) = 2ncooH = 2nco2 " «^

-> no(X) = 2 —'-— = 2.0,7= 1,4 m o l r , y '

22,4 "^^ • ^ Bao toan n g u y e n to O: no(X) + 2 no2 = 2 nco2 '^H20

^ 1 4 + 2 ^ = 2 ^ + n H , o ^ 2,2 = l , 6 + y y = 0 , 6 m o l ' ' 2 2 , 4 44

''•••ft ^

dm nang 6n luygn thi d?! hgc 18 chuy6n d j H6a hpc - IMguygn Van Hai

L o t gidi:

Nhan xet: Khi cho axit cacboxylic tac dyng voi NaHCOa ta luon c6: j , 3

J n c 0 2 = " C O O H ' ' •

1 344 no(x) = 2ncooH = 2 ticoj no(x) = 2 = 2.0,06= 0,12 mol

i'r'f, • 22,4

Cty TNHH MTV DWH Khang Vi^t

Bao toan nguyen to O: no (x) + 2 = 2 ncoj + riHjO

0,12 + 2 2,016 ^ 22,4 44 2 4.84 + "Hjo ^ "Hjo = 0/08 mol

mH20= 0,08.18 = 1,44 gam

—> Dap an B

Vi dy 11: Cho m gam hon hg-p X gom hai ancol tac dung v6i Na (du), thu duoc

) 4,48 lit khi H2 Mat khac, dot chay hoan toan m gam X, thu dugc 13,44 lit khi

C O 2 va 16,2 gam H 2 O Cac the tich do 6 dktc Gia tri ciia m la

A 14,5 B 15,4 C.12,2 D 13,8

( • • ' Ldfigiai:

nH2 = 0,2 mol , " ' • • ' - a w N;'"^-" #

I nco2 = 0,6->nc=0,6; D H J O ^ O ' ^ ->nH=l,8 Y ,1 4 ' r i * r r * M > „ i , v

Dya tren moi quan h^ nguyen to-nhom chiic thi voi ancol: no (X) = noH

-> no(X)= noH" 2nH2 = 0,4 mol ••!«?» m-nmi mod «;i,ib fob ,xtM

Bao toan kho'i lugng: m = m(- + m H + mo

^ m = 0,6.12 + 1,8.1 + 0,4.16 = 15,4 gam

-> Dap an B. • M

Vi dy 12: Hon hgp X gom 2 amino axit no (chi c6 nhom chiic C O O H va

-NH2 trong phan tu), trong do ti 1^ mo : m N = 80:21 De tac dung vua du voi

7,66 gam X can 100 ml dung djch KOH IM Cho 7,66 gam X tac dyng vtra

du voi dung dich HCl, thu dugc dung dich Y Co can Y thu dugc m gam

muoi khan Gia tri ciia m la

m = mx+ m H c i = 7,66 + 0,06.36,5 = 9,85 gam. O r M ^ ' ^" " ,

_> Dap an C

Vi 13: Dun nong m gam hSn hgp X gom cac este voi 350ml dung dich NaOH 2M, thu dugc dung dich Y chiia muol cua mgt axit cacboxylic don chuc va 13,9 gam hon hgp ancol Z Cho Z tac dyng voi Na du, thu dugc 4,48 lit khi H2 (dktc) Co can Y, nung nong chat ran thu dugc voi CaO cho deh khi phan ling xay ra hoan toan, thu dugc 4,8 gam mgt chat khi Gia tri cua m la

A 40,6 B.26,6 C 30,7 D 34,5 ' "

rAv>r^ itT,'.' ".sn-.' Loigidi: .\i life

n = 0,7 mol; n = 0,2 mol H2

Nhan xet: Cac este tao thanh tu cung mgt axit cacboxylic don chuc

Khi ancol tac dung voi Na:

- O H + Na -> - O N a + - H 2 i M v^ j n d a E v

0,2 fa;

Mol: 0,4 noH=2nH2 =0,4 mol

<-M|it khac: n.coo-= " O H = 0,4 mol

"RCOONa=0,4mol

- > nwaOH dir = 0,7 - 0,4 = 0,3 mol

Phan ling voi toi xut (NaOH he't, R C O O N a con du):

R C O O N a + N a O H ) R H + Na2C03 Mol: 0,3 <- 0,3 0,3

Ca'm nang 6n luygn thi dgi hpc 18 chuy§n dg H6a hqc - Nguygn Van Hki

V i dy 14: Do't chay hoan toan m gam hon hgp X gom hai ancol, thu dugc 11,2

h't khi C02 (dktc) va 12,6 gam H2O Mat khac, cho m gam X tac dung voi

N a (du), thu dixgc 4,48 h't khi H2 (dktc) Gia tri cua m la m 0 - ,„

no = noH = 2nH2 -> no = 0,4 mol -> mo = 0,4.16 = 6,4 gam

Bao toan kho'i lugng: mx = m^ + mpi + m o = 6,0 + 1,4 + 6,4 = 13,8 ^ j,^ ^

->Dap an D .,

Cty TN TV DVVH Khang Vi§t

CAC AXIT vo CO mm HIND

1 A X I T C L O H I D R I C : H C l

a, Lithuyet

+ Tfnft dung voi kim loai, bazo, oxit baza, muoi V i du: Si n ? f

Fe + 2HC1 > F e C h + H a t ''^^^ ''' CaC03+ 2HC1 > C a C h + C O a t + H2O '^^-'^^'^-^ "'^

+ Tin?z fc^""- Tac dung voi cac chat oxi hoa manh: Mn02, KMn04, KCIO3, K2Cr207 V i d y : ,.,^„„ „ ;„,,.,,„/, ^

M n 0 2 + 4HC1 - ^ - ^ M n C h + CI2 +2H2O " ry^.^ ^^ ^^^^

2KMn04 + 16HC1 > 2KC1 + 2MnCl2 +5CI2 + 8H2O qfi(}

K2Cr207 + 14HC1 2KC1 + 2CrCl3 + 3Cl2 +7H2O C O H /

b V i d u m a u , ^-.•cr^^' rfi ' • ' i - i - '

V i d u l : Cho cac phan ling sau: ,;.^d Au -^iyti^'^^ '-Xji::;

(a) H C l +Mn02 > M n C h +CI2 +2H2O , ,„,f^eH (b) 2HCl + Fe > F e C h +H2 , v,t X: jfefb g n i i ^ X i V t (c) 6HCl + 2A1 > 2AICI3 + 3H2 " ' -W! M i:i ,i/:}c:: i

(d) 16HCl + 2 K M n a > 2KCl + 2MnCl2 + S C h + 8H2O iM>f § ' ?

-Cac phan ling trong do H C l the hi?n tinh oxi hoa la " s f 8 f i

A.(b),(c) B.(a),(b) C (b), (c), (d) D.(a),(d)

• Laigidi:

Nhan xet: Trong cac phan ung (b) va (c), so' oxi hoa cua hidro giam tu +1

(trong H C l ) xuohgO (khi H2) , ,

- > Dap an A

Vi dv 2: Hoa tan hoan toan 7,6 gam hSn hgp hot FesOA va C u trong 200ml dung

djch H C l 1,2M (loang) Sau khi cac phan ung xay ra hoan toan, thu dugc dung dich X (khong chua axit du) Co can X thu du^c m gam muo'i khan

Gia tri cua m la

A 10,39 B 14,20 C 5,16 D 11,10

Lffigidi:

Gpi so mol: Fe304 = a; C u = b ' ^ • ' j Theo bai: 232a + 64b = 7,6 • ?-•<.}( • ••: '^^j^ X :,,; i

^han xet: chi c6 Fe304 phan ung tryc tiep voi axit f| ,* g j :

Trong X khong con axit d u nen Fe304 phan ung vua dii voi H C l : Cac phan ung hoa hpc:

Caim nang On luy$n Ihi dgi hgc 18 chuy6n dg H6a hgc - Nguygn VSn Hit

i * • Fe304 + 8HC1 > FeCh + 2FeCh + 2H2O

V i 3: Day g o m cac chat deu tac d u n g dugc v o i d u n g dich H C l loang la

A. KNO3, CaC03, Fe(OH)3 ^ B FeS, BaS04, K O H , ^.^^^

C AgNCte, ( N H 4 ) 2 C 0 3 , CuS D NaHCOs, FeS, CuO ' , , ,

Lai gidi:

Loai A v i KNO3 khong tac dung; loai B, C v i BaS04 va CuS khong tan trong

d u n g d i c h H C l loang ^ •

D a p an D Cac p h u a n g trinh hoa hoc: ^''^ ^ ' i O n M > i

NaHCOa + H C l > N a C l + H2O + CO2 ' '

FeS + 2HC1 > FeCh + H2S

C u O + 2HC1 > C u C h + H2O

V i d u 4: Hoa tan hoan toan 8,55 gam hon hop gom Na, K va Ba vao nuoc, thu

duoc d u n g dich X va 1,792 l i t k h i H2 (dktc) D u n g djch Y gom H C l va

H2SO4, t i 1$ m o l t u o n g u n g la 2:1 T r u n g hoa d u n g djch X boi d u n g dich Y,

tong k h o i l u o n g cac muoi dugc tao ra la

A 13,81 gam B 11,39 gam C 15,23 gam D 19,07 gam

Cty TNHH MTV DVVH Khang Vijt

V i dV 5: Cho 2,13 gam hon hgp X g o m M g , Cu va A l a dang bgt tac d y n g

hoan toan v o i O2 t h u dugc hon hgp Y gom cac oxit c6 khoi l u g n g 3,33 gam

"The tich d u n g d i c h H C l 2 M vua d u de phan u n g het v o i Y la

V i d\ 6: Hoa tan hoan toan 7,8 gam K vao 500ml H C l 0,2M, t h u d u g c k h i H2 va

d u n g dich X Co can X t h u dugc m gam chat ran khan Gia trj ciia m la

-> m = m K c i + mKO H = 74,5.0,1 + 56.0,1 = 13,05 gam Dap an B

V i d y 7: Hoa tan hoan toan 8,97 gam k i m loai k i e m M vao SOOnil d u n g djch

H C l 0,2M, t h u d u g c k h i h i d r o va d u n g djch X Co c?n X t h u d u g c 14,73 chat

ran khan Y K i m loai kiem M la

A R b B K C N a D L i

Lai gidi: Jl/ion Ob ^

Cac phan u n g hoa hgc: •< •= , I^^/TU' »• | ,

Ca^m nang 6n luy^n thi dgi hqc 18 chuy6n dg H6a hgc - Nguygn VSn Hai

8 9 7

r i M = n H c i + n Q ^ = 0,23 -> M = - ^ = 3 9 ( K ) ^ Dap an B

V i dv 8: Cho m gam hon hop X gom Cu, M g , Fe tac d u n g v o i axit H C l d u , thu

d u g c d u n g dich Y, 448ml k h i (dktc) v a 0,64 gam chat ran C h o d u n g dich

N a O H d u vao Y, loc ket tiia va nung trong khong k h i t o i khoi l u o n g khong

doi, t h u dugc 1,2 gam chat ran Gia t r i ciia m la ,

Luu f K h i n u n g ngoai k h o n g k h i , Fe(OH)2 chuyen thanh Fe(OH)3 v a b i

phan h i i y thanh Fe203

V i du 9: Hoa tan hoan toan hon hgp X gom Fe va M g bang m o t l u g n g vira d i i

d u n g dich H C l 20%, t h u d u g c d u n g dich Y N o n g d o cua FeCk trong Y la

15,76% N o n g d o phan t r a m cua M g C h trong Y la

Cty TNHH MTV DVVH Khang Vift

' do ^'^^ ^^"^ ^ NaC\a K C l v o i HiS04 dac, d u

jChi thoat ra cho hoa tan vao nuoc t h u dugc d u n g d i c h Y Cho b g t Z n d u vao Y t h u dugc 448ml k h i (dktc) K h o i l u g n g N a C l trong X la

^ 0,585 B 1,170 C 1,755 D 2,340 '

Loigidi:

Goi so m o l : N a C l = x; K C l = y Cac p h u o n g t r i n h phan u n g : jsjaCl + H2SO4 — ^ NaHS04 + H C l t '''' ' jCCl + H2SO4 KHSO4 + H C l t

Z n + 2HC1 > Z n C h + H a t " > ' ' • ' ' ^

Bao toan nguyen to: C I " > H C l > -Hi

Ta c6: mx = 58,5x + 74,5y = 2,66 va n H 2 = 0,5(x + y) = 0,02

_» x = 0,02; y = 0,02 -> mNaci = 0,02.58,5 = 1,17 gam D a p an B

V i du 11: Hoa tan het m gam hon hgp M g va MgCOa trong d u n g dich H C l , t h u

dugc 4,48 l i t h o n hgp k h i X (dktc) T i k h o i cua X so v o i H2 la 11,5 Gia t r i cvia

Theo bai: M = 11,5.2 = 23 va a + b = 0,2 mol , ,^ ,

A p d u n g cong thuc cua p h u o n g phap duong cheo, ta c6: ,

"*" Dwn^djc^ H2SO4 (fflc: T i n h oxi hoa manh Ngoai t i n h axit m a n h , axit sunfuric dac con the h i ^ n t i n h oxi hoa m ^ n h , tac

d i i n g d u g c v o i n h i e u k i m loai, h g p chat, : ff;-!;! » {'"-l

C^m nang 6n luy$n thi dgi hpc 18 chuySn dg H6a hpc - NguySn VSn HJi

Cu + 2H2S04(^flc) — — > CuS04 +SO2 + 2H2O :/>

2Fe + 6H2SO4 (dac) Fe2(S04)3 + 3SO2 + 6H2O ',' '

2FeO + 4H2SO4 (dac) > Fe2(S04)3 + SO2 +4H2O

2Fe304 + IOH2SO4 (dac) > 3Fe2(S04)3 + SO2 + IOH2O

L u u y: Cac kim loai Al, Fe, Cr khong tac dung vdi Ji2S04 dac, nguQi

Dieu che

So do: Quang pirit FeS2 hoac S ) SO2 — ^ SO3 — ^

' • ,.H,.- »,I '• ,

H2S04.nS03 > (n+1) H2SO4

Cac phan ung: S + O2 > SO2

4FeS2 + I I O 2 — ^ 2Fe203+ 8SO2 •• • "ID :6i fiB'lii^H

2SO2 + O2 2S03 , - x m : o : f i i ,

H2S04.nS03+ n H 2 0 > (n+l)H2S04

b Vi dv mau:

Vi dy 1: Cho day cac chat sau: KBr, S, Si02, FeO, Cu va Fe203 So'chat trong day

CO the bi oxi hoa boi dung djch axit H2SO4 (dac, nong) la

A 4 B.5 ' G.3.c|f: D.2

Lai gidi:

Nhan xet: Axit H2SO4 dac, nong the hien tinh oxi hoa khi tac dung vai chat

CO tinh khu (chiia nguyen to' dang 6 muc oxi hoa tha'p)

2KBr + 3H2SO4 2KHS04 + Br2 + SO2 + 2H2O

2 F e O + 4H2SO4 — ^ Fe2(S04)3 + SO2 + 4H2O

Cu + 2H2SO4 CuS04 + SO2 + 2H2O ^ fH',:^::

Dap an A

Luu y: Khi tac dyng voi H2SO4 d^c, FeaOs the hi^n tinh baza:

FeaOs + 3H2SO4 — ^ Fe2(S04)3 + 3 H 2 O

Vi dy 2: Cho oxit ciia kim loai M (hoa tri 2) tac dung vira dii vai dung dich

H2SO4 10% (loang), thu dug-c dung dich muoi c6 nong dp bang 14,45% Kim

Cty TNHH MTV DWH Khang ViSt

Gia thie't dung djch ban dau chiia 1 mol H 2 S O 4 (tiic chiia 98 gam H 2 S O 4 )

100 _> Khoi lupng dung djch H 2 S O 4 = 98 — = 980 gam

^^^^^•^ -=0,1445 - > M = 56 (Fe)-^ Dap an B

980 + M + 16

Vi dy 3: Nung hon hop X gom a mol Fe va 0,015 mol Cu trong khong khi mpt

thoi gian, thu dupe 6,32 gam chat ran Y Hoa tan hoan toan Y bang dung dich H2SO4 dac nong (du), thu dupe 0,672 lit khi SO2 (san pham khu duy nha't 6 dktc) Gia tri ciia a la

A 0,04 B.0,05 C 0,07 D 0,06

Lm gtat: ^

n c r i ^ = — ^ =0,03 mol , , ,, , i ,

Bao toan khoi lupng: mo2 = my " "^x = 6,32 - 56a - 0,96 = 5,36 - 56a

Nhan xet: ne'u dya theo phuong trinh phan ung se rat dai va kho giai

Cach 1:6 day, cac em can su dyng so do phan ung: t y s j j , : Fe,Cu(l) > Y (2) > Fe^3Cu^M3)

Xet su trao doi electron 6 cac giai doan:

(1) -> (3): Fe -3e > Fe*-^ nenhuong = 3 np^ = 3a OsH »• ;

Cu -2e > Cu*^ nenhiKmg=2ncu =0,03 "^s:

(1) (2): O2 +4e > 2Ct^ ^ nenhan = 4 n o 2 = ^ ^ ^ ^ ^ ^ = 0 , 6 7 - 7 a

(2) (3): S** +2e > SO2 nenh,^n=2nso2 =0,06 Bao toan electron: 3a + 0,03 = 0,67 - 7a + 0,06 ^ a = 0,07 mol -> Dap an C

-CachZ:

Qui doi Y thanh: Fe (a mol); Cu (0,015 mol) va O (b mol)

Bao toan khoi lupng: 56a + 16b = 6,32 - 0,015.64 = 5,36

Bao toan electron: 3a + 2.0,015 = 2b + 2.0,03 a = 0,07; b = 0,09 , , ,

im nang 6n luy^n thi dgi hoc 18 chuygn dg H6a hqc - Nguygn Van Hi\

HlS + 3H2SO4 ' > 4SO2 + 4H2O ,

8HI + H2SO4 ^'-^ 4I2 + H2S + 4H2O f-SnO'w! ;oi*:^f '

2Fe304 + IOH2SO4 3Fe2(S04)3 + SO2 + IOH2O , 1/

Dap a n C

Luu y: ^eiOj, AgNOs xay ra phan ung trao doi, Na2SQj khong tr.c dung:

Fe203 + 3H2SO4 Fe2(S04)3 + 3H2O

2AgN03 + H2SO4 Ag2S04i + 2HN03

i d\ 5: Cho 6,08 gam hon hop gom Li, Na v a Ba vao nuoc (du), thu duoc

dung dich X v a 1,344 h't khi H2 (dktc) Dung dich Y gom HCl I M v a H2SO4

0,5M Trung hoa dung dich X boi dung dich Y, tong khoi lugng cac muoi

Nhan xet: n ^ ^ = 2nH2 = 0,12 mol ' ' ' '

Mat khac, nong do HCl gap 2 Ian H2SO4 trong cung mot the ti'ch thi

1 dvi 6: Hoa tan hoan toan 5,28 gam hon hgp bpt Fe304 v a Cu trong 80ml dung

dich H2SO4 I M (loang, vua dii) Sau khi cac phan ung xay ra hoan toan, thu

duoc dung djch X Co can X thu du(?c m gam muoi khan Gia trj ciia m la

A 8,64 B.7,68 C 15,68 D 11,68

Cty TNHH IVITV DVVH Khang Vigt

Lcngiai:

Gpi so mol: Fe304 = a; Cu = b Theo bai: 232a + 64b = 5,28 ; i ;

jV/ian xet: chi CO Fe304 phan ling true Hep voi axit. ji b uAi

Trong X khong con axit du nen Fe304 phan ling vua du voi H2SO4:

Cac phan ung hoa hpc:

Fe304 + 4H2SO4 > FeS04 + Fe2(S04)3 + 4H2O • • '

Vi 7: Cho 3,68 gam hon hop gom A l , Mg va Zn tac dung voi mot luong vua

du dung dich H2SO410%, thu dxxgc dung dich X va 2,24 lit khi H2 (dktc)

Khoi lugng dung dich X la

A 101,68 gam B 88,20 gam C 101,48 gam D 97,80 gam

Lcngiai:

Bao toan nguyen to hidro: nH2S04 = " H 2 = 0,10 mol ''^ " ' ' '

-> Khoi luong dung dich H2SO4 = 0,10.98 ^ = 98 gam

Cu + Fe2(S04)3

M o l : 0,01 - > 0,01

10 Bao toan k h o i l u o n g : 3,68 + 98 = mx + 0,10.2 ^ mx = 101,48 -> Dap a n C

Luu y: Bai n a y cac e m d i q u e n t r u khoi l u o n g k h i H2 b a y ra, v a c h i tinh: mx

= 3,68 + 98 = 101,68 v a se chgn n h a m d a p a n A!

V i 8: Hoa tan hoan toan 6,44 g a m h o n hgp X g o m A l , Fe v a Zn bang mot

l u o n g v u a d u d u n g dich H2SO4 loang, t h u d u g c 2,688 lit H2 (dktc) v a d u n g