Môn Hóa học - Tuyển chọn bài thi trắc nghiệm theo cấu trúc đề thi: Phần 2

Nối tiếp nội dung phần 1 tài liệu Tuyển chọn bài thi trắc nghiệm theo cấu trúc đề thi môn Hóa học, phần 2 giới thiệu các hướng dẫn chọn đáp án cho các đề thi ở phần 1. Mời các bạn cùng tham khảo nội dung chi tiết.

Trang 1

Theo ( I ) va (2) : n,^ = x + 2y = 0,2 + 2.0,05 = 0,3 (mol)

7 2 KhO'i lugng mol cua kim loai M la : ~ = 24 (gam)

Vay kim loai M la magie (Mg)

5 1 1 Dap an dung la C

So mol cac cha't:

ncuci, = 0,1.0,5 = 0,05 (mol) ; n^^.a = 0,5.0,5 = 0,25 (mol)

Cac phuong trinh diSn phan:

CUCI2 ""^^ > C u l + C l 2 t (1) 2NaCl + 2H2O -JS!^ C L t + H z t + 2NaOH (2)

Phan ling (1) xay ra hoan loan, sau do phan ling (2) tiep tuc xay ra

Dua vao edng thiic Faraday :

A It

m = —.—

n F Tfnh dugc thdi gian dien phan hoan toan 0,05 mol CuCij la 1930 giay

Con lai 3860 - 1930 = 1930 giay dien phan mot phan NaCl va tao ra 0,1

mol NaOH

• Dung dich thu dugc chiia 0,1 mol NaOH Hoa tan A l :

, 2A1 + 2 N a O H + 6H2O 2Na[Al(OH)4] + 3 H 2 t 0,1 mol 0,1 mol

S6' kmol CO2 trong hdn hgp X la 3.20% = 0,6 kmol

Dat X la s6' kmol O2 c6 trong 3kmol h6n hgp X

Cac hgp kim: Cu-Fe(I) ; Fe-C(III) va Sn-Fe(IV) chiia Fe la kim loai

boat d6ng han kim loai va phi kim con lai nSn khi tie'p xiic vdi dung

dich chat dien l i , Fe bi an mon

5 1 4 Dap an diing la A

- Khi dien phan dung dich CuClj (vdi dien cue tra), d cue am (catoi)

xay ra phan ling: "

Cu^* + 2e->Cu (1)

- Phan irng an mon dien hoa xay ra cr cue am khi nhiing hgp kim

Zn-Cu vao dung dich HCl:

Z n - > Z n ' * + 2e (2)

Phan ling (1): CO su tham gia cua ion kim loai (Cu^"^)

Phan ling (2): c6 su tham gia cua kim loai (Zn)

5 1 5 Dap an dung la C

Ggi X la s6' mol Zn c6 trong 19,3 gam h6n hgp

S6' mol Cu la 2x mol

Ta c6: 65x + 64.2X = 19,3 - > x = 0,1

Trang 2

Cac phuang trinh phan ling:

3Zn + Fe2(S04)3 ^ 3ZnS04 + 2Fe

0,1 mol 0,1 mol 0,2 mol

Fe + Fe2(S04)3 -> 3FeS04

0,2 mol 0,2 mol

1 mol CUSO4 -> 64 gam Cu va 16 gam O2 kh6'i lucfng giam 80 gam

0,1 mol •<- 6,4gam <- 1,6 gam <-kh6'i luong giam 8 gam

Sau thcfi gian difin phan, dung dich Y con C U S O 4 va H2SO4 (0,1 mol)

Cho b6t Fe vao dung dich Y xay ra cac phan ling:

5,6 gam 0,1 mol

Theo phan iJng (3):

1 mol Fe phan ling, tao ra 1 mol Cu -> khd'i lucmg tang 8 gam

X mol Cu <r- Kh6'i lugng tang 1,2 gam

1.1,2

- > X =

8 = 0,15 (mol)

S6 mol CUSO4 C O trong 200 ml dung dich dien phan la:

ncuso4 = O'l + = 0,25 (mol)

- Zn khur ion Fe^^: Zn + Fe'^-> Zn'^ + Fe

- Fe^^ khijr ion Ag^: Fe^^ + Ag^ ^ Fe'^ + Ag

6 K I M LOAI K I E M , K I M LOAI KIEM THO, NHOM, SAT 6.1 Dap an diing la B

Cac phuong trinh phan iJng :

2Na + 2H20 -» 2Na+ + 20H~ + H21 (1)

Ba + 2H20 ^ Ba^^ + 20H~ + H j t (2) Theo (1) va (2) : D Q ^ - = 2 H H , = 2 = 0,3 (mol)

De trung hoa 0,3 mol OH" cSn 0,3 mol H^" hay 0,15 mol H2SO4

0,15

= 0,075 (lit) hay 75 ml

VddH2S04 6.2 Dap an diing la B

-Cac phuong trinh phan iJng : 2K + 2H2O -> 2K+ +20H'" + H 2 t (1)

• Al^+ + 3 OH" -> A1(0H)3 i (2)

I S6' mol Al^+ = 2.0,2.0,1 = 0,04 (mol)

Di lugng ke't tua Y 16n nha't cin 0,04.3 = 0,12 (mol) OH"

S6' mol OH" da c6 trong 300 ml dung dich la ; 0,3.(2.0,1 +0,1) = 0,09 (mol)

S6 mol OH" d phan ling (1) la : 0,12 - 0,09 = 0,03 (mol)

Theo (1) : S6' mol K = s6' mol OH" = 0,03 mol

- > mK = 0,03.39 = 1,17 (gam)

Trang 3

Gia six hoa tan 1 m o l h6n hop X (trong do c6 x m o l Fe va (1 - x ) mol

M g ) vao dung dich H C l :

Fe + 2HC1 - > FeCl2 + H21 (1)

M g + 2HC1 MgCl2 + H21 (2)

Theo (1) va ( 2 ) : S6' mol HCl phai dung la 2 mol

73 100

- > m H c i = 36,5.2 = 73 (gam) m^juci = = 365 (gam)

Sau k h i hoan tan vijfa du t h i khd'i lirang dung dich ]k :

Theo (1) : CO phan ting tao thanh CO2, do do the' tich vSn la 4,48 lit

(dktc), tu-c la bang 0,2 mol

Goi a la s6' mol CO c6 trong h6n hop sau phan Hag - > s6' mol CO2 la (0,2 - a)

166

- % V r o = — 1 0 0 % = 75%

^ 2 0,2 Theo (1) : (56x + 16y) gam Fe^Oy phan ling vdri y mol CO

8 gam Fe^Oy phan ling vdi 0,05 mol CO

- > - - = - CTPT cua sat oxit la Fe203

y 3 6.7 Dap an dung la A

mco2 = 13,4 - 6,8 = 6,6 (gam) nco^ = ^ = ^'15 (mol)

S6' mol NaOH = 0,075 1 = 0,075 mol < ncoj chi xay ra phan ling (2)

va CO2 vSn con du Kh6'i luong mu6'i NaHCOs dugc ti'nh theo NaOH :

Trang 4

Phucfng trinh phan dng :

2Fe + 6 H 2 S O 4 - > F e 2 ( S 0 4 ) 3 + S S O j t + 6 H 2 O (1)

0,1 mol 0,3 mol 0,05 mol

Sau phan ling (1), Fe du 0,02 mol ntn xay ra phan ling :

F e + F e 2 ( S 0 4 ) 3 -~>3FeS04 (2)

Nhu vay sau (2) thu dugc : 0,05 - 0,02 = 0,03 (mol) Fej(804)3 va

0,06 mol FeS04

6.11. Dap an dung la C

3Fe + 202 ^ Fe304 (Fe203 FeO) (1)

H6n hop X gdm FeO, Fe203 va Fe du

Dat X la s6' mol Fe du, s6' mol Fe304 la y Ta c6 :

56x + 232y = 3 (gam) (I)

Trong h6n hop X chi c6 Fe va FeO tac dung vdi dung dich H N O 3 ,

2A1 + 2NaOH + 6H2O -> 2Na[Al(OH)4] + 3 H 2 1 (2)

• Goi X va y la s6' mol Na va A I c6 trong m gam X

- K h i cho X vao nude, Na tan he't, A l tan khOng he't (vi thie'u NaOH)

Thi tich H2 thoat ra tijr hai phan ling (1) va (2) dugc tinh theo Na :

Theo (1) : X mol Na lam thoat ra ^ l i t H2

Theo (2) : x mol NaOH lam thoat ra 3 ^ lit

mAi 2.27 54 23 + 54

6.13 Dap an dung la A

Cac phuong trinh phan ling :

Al(OH)3 + NaOH -> N a [ A l ( O H ) J (2)

Theo (1) : H A K O H ) , = " A i C b = 1'5 0,2 = 0,3 (mol)

S6' mol Al(OH)3 tham gia phan ling (2) = 0,3 - ^ = 0,1 (mol)

Theo (1) va ( 2 ) : S6' mol NaOH da phan ling = 0,9 + 0,1 = 1 (mol) Gia tri Idn nha't cua V la : V = = 2 (lit)

6.14 Dap an dung la D

Bai tap nay gidi theo phuang phdp can bang dien tich :

Khi tan vao dung dich H N O 3 vCra dii (chi tao thanh muO'i sunfat va

khi N O t ) 0,12 mol FeSz ^ 0,12 mol Fe^+va 0,12.2 = 0,24 mol SO^"

a mol CU2S -)• 2a mol Cu^"*^ va a mol SO4"

-> S6' mol dien tich duong = 3.0,12 + 2.2a = (0,36 + 4a) mol SO' mol dien tich am = 2.0,24 + 2a = (0,48 + 2a) mol

Trong dung dich : 0,36 + 4a = 0,48 + 2a - > a = 0,06 mol

<»-lS Dap an diing la D

Khi trOn dung dich NaOH vdi dung dich A I C I 3 c6 the' xay ra cac PTHH:

A l C l 3 + 3 N a O H ^ Al(OH)3 ^ + 3NaCl (1)

t Al(OH)3+NaOHdu ^ N a [ A l ( 0 H ) 4 ] (tan) (2)

Trang 5

Gia six c6 1 mol AlCl, Di phan iJng xay ra ho^n toan (khdng con

tua) thi phai c&n 4 mol NaOH Do vay, c6 ke't tua thi s6' mc

NaOH phai nho hon 4 mol NghTa la : a : b > 1 : 4

Fe >(Fe, FeO, Fe^Oj, F e j O J - i ^ ^ ^ ^ F e C N O , ) , + N 0 t (1)

Cac k i m loai M g , Cu, A l tac dung hoan tokn v6i O2 tao ra cac oxit

MgO, CuO, AI2O3

Cac oxit nay tac dung vQ\:

Vj.Hci = 0 , 1 5 : 2 = 0,075 lit hay 75 m l

6.20. Dap so diing la D

Cac phuong trinh phan ling:

FeO + 2HC1 FeCl2 + H 2 O (1)

Fe304 (FeO. F e 2 0 3 ) + 8HC1 FeClj + 2 FeCl, + 4 H 2 O (3) ft

Dat s6' mol cua FeO la x - » s6' mol Fe20 3 la x;

Dat s6' mol ciia Fe304 la y ^ so mol FeO la y va so mol F e 2 0 3 la y

Ta c6: 72(x + y) + 160(x + y) = 232(x + y) = 2 , 3 2 x + y = 0,01

Theo (1,2,3): nnci = 8(x + y) = 8.0,01 = 0,08 mol

- > V j j „ c , = 0,08 : 1 =0,08 lit

Cdch khdc: V i s6' mol FeO = s6' mol FejO, nen c6 the' coi 2,32g h6n

hop la chi ciia F e 3 0 4

Fe,0, + 8HC1 ^ FeCl, + 2FeCl3 + 4 H 2 O 0,01 -> 0,08

=>V,,„„=M! = 0,08 (lit) ^

171

Trang 6

6.21 Ddp s6'dung la A

Cac phuong trinh phan ling:

2NaOH + 2 A l + 6H20 -> 2Na[Al(OH)4] + S H j t (2)

Dat s6' mol Na la x mol -> s.6 mol A l la 2x mol

Theo (1,2): S6' mol = - + - x = 2x = — = 0,4 (mol)

^ 2 2 22,4 Theo (2): S6' mol A l da phan ling = so mol Na = 0,2 (mol)

-> S6' mol A l du = 0,4 - 0,2 = 0,2 (mol)

-> m = 0,2.27 = 5,4 (gam)

6.22 Dap so dung la B

7 8 S6' mol Al(OH);, >^ = — = 0,1 (mol)

78 Cac phuong trinh phan ling:

AI2 ( 8 0 4 ) 3 + 6NaOH 2 AKOHX ^xl + 3 Na2S04 (1)

S6 mol cac chat: Hp,. = n^,, = 0,1 m o l ; OAgNo, = 0,55 mol

A l + 3AgN03 -» A1(N03 )3 + 3 A g i

0,1 0,3 0,3 mol

Fe + 2AgN03 Fe(N03 )2 + 2Ag>l

0,1 0,2 0,2 mol

Sau phan ling (1) va (2), AgN03 con du 0,55 - (0,3 + 0,2) = 0,05 (mol) ^

do do xay ra phan ling:

Fe(N03 )2 + AgNO, -4 Fe(N03)3 + Agi

0,05 0,05 mol Vay: m = 108(0,3 + 0,2 + 0,05) = 59,4 (gam)

6.24 Dap s6' diing la A

Phuong trinh phan img:

H6n hop Y tac dung vdi dung dich NaOH sinh ra khi Hj, chung to trong

Y CO AI2O3, Fe va A l du (Fe203 da phan ling he't)

Cac phuong trinh phan ling cua h6n hop Y

| p ' -dphan I: 2A1 + 3H2SO4 ^ Al2(S04), + 3H2T (2)

AI2O3 + 3H2SO4 ^ Al2(S04)3 + 3H2O (4)

- dphan 2: 2A1 + 6H2O + 2NaOH ^ 2Na[Al(OH)4] + 3H2t (5)

S6' mol FcjO, phan ling = ^ s6' mol Al phan ling = 0,05 mol

-> m = 2 [(0,025 + 0,1).27 + 0,05.160] = 22,75 (gam)

6.25 Dap an dung la B

Cac phuong trinh phan ung:

a mol — mol a mol

4 / 4FeS2 + I I O 2 -> 2Fe203 + 8S02t (2)

FeO + 2HC1 FeCl2 + HjO (1)

Fe304 + HCl: xay ra nhu (1) va (2)

Trang 7

Theo (1): ma-o = ^ • 72 = 4,32 (gam)

S6' mol CO, = - ^ ^ ^ = 0,02 (mol) 22,4

Cac phuang trinh phan img: '

MHCO3 + HCl ^ MCI + CO2T + H2O (1)

Theo (1) va (2): S6' mol h6n hop muoi = s6' mol COj

(M + 61 + 2M + 60).0,02 , „

2 Glai phuong trinh diroc M = 23

Vay kim loai M la natri (Na)

6.28 Dap an dung la C

Benzen khong phan irng dirgfc v6i nude brom

Phenol (do anh hudng cua nhom -OH) phan iJng d6 dang vdi nu6c

brom (the Br vao cac vi tri ortho va para)

Cac phuong trinh phan ung:

2A1 + 2NaOH + 6H2O ^ 2Na[Al(OH)4] + 3H2 1 (1)

Dat a la so mol Al tham gia m6i phan ting:

6.31 Dap so diing la Ạ

- S6' mol H^ = 0,5.1 + 0,5.0,28.2 = 0,78 mol (0,39 mol H2)

S6' mol H2 = = 22,4 0,39 mol

W cua axit da bay khoi dung dich du6i dang H2

- S6' mol CI = 0,5.1 = 0,5 mol -> m^, = 17,75 gam

S6 mol SỐ = 0,5.0,28 = 0,14 mol -> m^^- = 1,3,44 gam

Kh6'i luong mu6'i khan la:

m = 7,74 + 17,75 + 13,44 = 38,93 (gam)

Dap s6' diing la Ạ

2A1 + FejO, — > AI2O3 + 2Fe 0,1 mol 0,1 mol

H6n hop ran X gom 0,1 mol AI2O3, 0,2 mol Fe va Al dụ

H6n hop X tac dung v6i dung dich NaOH:

2AI + 6H2O + 2NaOH -> 2Na[Al(OH)4] + 3H2t

S6' mol cac chát: n^gNo, = HCUCNO,), =0,3.0,1 = 0,03mol

Cac phuong trinh phan iJng:

Al + 3AgN03->AI(NO,)3+3Ag>l

0,01 mol 0,03mol 0,03mol

2A1 + 3Cu(N03)2-^2Al(N03)3+3Cu>l

0,02 mol 0,03 mol 0,03 mol

Al du tac dung \6i dung dich HC\:

(1) (2) (3)

Trang 8

Cac phuang tnnh phan irng:

NajO + HjO >2NaOH

AI2O3 +2NaOH + 3H20 >2Na[Al(OH)4]

Chat tan duy nha't la Na[Al(OH)4] c6 s6' mol la 0,5.0,2 = 0,1 (mol)

Th6i khi CO2 vao dung dich Na[Al(0H)4] :

So mol Ba(OH)2 = 34,2 = 0,2 (mol) 171

Phuong tnnh phan irng;

(NH4)2C03 + Ba(0II)2 ^ BaCO,! + 2NH3t + 2H2O

0,1 mol 0,1 mol

^ m = 0,1.197 = 19,7 (gam)

Cac phirang trinh phan irng :

FeO + 2HC1-> FeClj + H2O (1)

Fc^O, + 6HC1 2FeCl, + SHjO (2)

Fe,04 CO c6ng thirc FeO.Fe203 cung phan irng nhu tren va cung thoa man:

so mol Fe^^ : s6' mol Fe'^ = 1 : 2

Goi X la s6' mol ciia m6i oxit trong nira h6n hop X Ta c6 :

Theo (1) va (2) : m, = 127x + 325x = 452x

Sue khf Clj vao phan hai, xay ra phan irng :

2FeCl2 + CI2 -> 2FeCl3 (3) Theo (1), (2) va (3) : m2 = 3x.l62,5 = 487,5x

Theo (2) : n^g = n F e = ^ = 0,06 (mol)

m^g = (0,06 + 0,06).24 = 2,88 (gam)

Dap an dung la B

3 136 H6n hop khf Y g6m NO va NjO ; = 0,14(mol)

Goi X la s6 mol NO c6 trong Y Ta c6 :

: Cho 3a mol electron

hMg^"" : Cho 2b mol electron

S6' mol cac chat: n„t,^„ = — = 0,1 mol ' 474

nBa(OH), =0,2.1 =0,2 m o l

-nsoj- =0,1.2 = 0,2 mol

n^u =0,2 mol

Trang 9

Phuang trinh phan ling :

Ba^^ + SO^- BaSO.^

" B a s o , = 0.2 mol -> mBaso, = 0.2 233 = 46,6 (gam)

T h e o ( 2 ) : = 0,02 - 0,01 = 0,01 mol

Taco: 0,01M + 0,01(M + 16) = 2,9

-> M = 137 K i m loai M la bari

Oxit sat tac dung vdi dung dich H2SO4 dac, nong cho khi SO2 bay ra

la FeO hoac Fe304 (FeO.Fe263)

Phuang trinh phan ling:

2FeO + 4H2SO4 -> Fe2(S04)3 + S02t + 4H2O (1)

T h e o d ) : np^o = 2nso: = 2 ^ ^ = 0,29 (mol)

Dung dich X g6m Al2(S04)3 va FeS04, tac dung v6i Ba(0H)2 dir:

Al2(S04)3 + 3Ba(OH)2 ^ 3BaS04^ + 2 A l ( O H ) 3 i (3) 2Al(OH)3 + Ba(OH)2 du -> Ba[Al(OH)4]2 + 4H2O (4) FeS04 + Ba(OH)2 ^ BaS04^ + Fe(OH)2i (5)

78

Ke't tua Y g6m BaS04 va Fe(OH)2

Nung ke't tua Y den khdi luang khdng doi dugc chat ran Z

Chat ran Z g6m BaS04 va Fe203

6.43 Dap an diing la A

3,36

22,4

8A1 + 3Fe304 — ^ 4AI2O3 + 9Fe H6n hop ran X g6m AI2O3, Fe va A l du

Cho X tac dung vdri dung dich NaOH, xay ra cac phan ling :

AI2O3 + 2NaOH + 3H2O ^ 2Na[Al(OH)4]

2A1 + 2NaOH + 6H2O -> 2Na[Al(OH)4] + 3H2t Dung dich Y la dung dich Na[Al(OH)4], cha't ran Z la Fe

Sue CO2 vao dung dich Y, xay ra phan ling:

Na[Al(OH)4] + CO2 -> Al(OH)34 + NaHC03

39

"Nal A I ( O H ) , 1 = " A K O H ) , " ^ " ^' ^ (mol)

Theo(4) Theo (3) n A i d u = | n H , ==|-0,15 = 0 , 1 (mol)

Theo (2) : n^i^o, =-(0,5-0,1) = 0,2 (mol)

Theo (1) : n^, p., = 2n^i^o, = 2-0,2 = 0,4 (mol)

(1)

(2) (3)

(4)

179

Trang 10

Phuang trinh plian ling:

18A1 + 66HNO3-> ISAKNO,), + 3N2O + 3N2 + 33H2O (1)

0,18 mol 0,18 mol 0,03 0,03 mol

Con 0,46 - 0,18 = 0,28 mol A l phan ling vdfi H N O , tao ra NH4NO3:

8A1 + 30HNO., ^ SAKNOj), + 3NH4NO, + 9H2O (2)

Nhu vay trong dung dich X gom c6 0,46 mol A1(N03)3 va - 0,28

SO' mol cac cha't:

n A , = ^ = 0,4 (mol); n,,^o, = | ^ = 0,15 (mol) '

n H 2 = ^ ^ = 0,48 (mol)

Phuang trinh phan ilng nhiet nh6m:

8A1 + 3 F e 3 0 4 - » 4AI2O3 + 9Fe ( i )

Cac phuong trinh phan img hoa tan h6n hop ran cho bay ra:

Fe + H2SO4 (loang) FeS04 + H2 i (2) 2AI + 3H2SO4 ^ Al2(S04)3 + 3H2t (3)

Goi X la s6' mol A l tham gia phan iJug (1), ta c6:

Hieu sua't phan ling: h = -—.100% = 80%

Taco: 27x + 102y = 1,56 (I)

1 9 04

Theo (1, 2, 3, 4): 0^,^03 = - x + y = — = 0,02mol (II)

^ 2 102

Trang 11

Giai he phuong trinh (I), (II) dugc: x = 0,02

T h e o ( l ) : =|nA, =|.0,02 = 0,03mol

_> = 0,03.22,4 = 0,672 h't

Khi chuyen tur muoi hidrocacbonat thanh mudi sunfat da c6 1 g6'c

S04~ thay the 2 gdc HCO: , khoi luong 1 mol mu6'i giam:

m = 2.61 - 9 6 = 26 (gam)

9 125-7 5 -> s6' mol mu6'i hidrocacbonat = — — = 0,0625 (mol)

26 9,125

^ " ^ = o : 5 6 ^ = ' ' '

-= 146 - (2.61) -= 24 ^ kirn loai Mg

Cong thiJc cua mu6'i hidrocacbonat: Mg(HC03)2

So mol cac cha't: n^g = 0,28 mol, nM^-o = 0,02 mol, nx = 0,04 mol

S6' mol mu6'i Mg(N03)2 = s6' mol M g + s6' mol MgO

Tinh theo phuong phap can bang electron:

0,28 mol Mg cho 0,28.2 = 0,56 mol e (Mg Mg^^ + 2e)

NajCOj + BaCl2 BaC03 i + 2NaCl

Trong dung dich chi con mudi NaCl

Trong dung djch X vSn con ion C03~, chiing to HCl phan ling het, Na2C03 con du Do vfiy, dap an A phu hgp

7.7 Dap an diing la B

C) day luu y phan ling : C2H4 + Br2 -> C2H4Br2

Co ten ggi la phan ling cdng hgp brom, nhung c6 ban cha't la phan ling oxi hda - khuf, v i trong phan ling nay so oxi hda cua brom va cacbon thay ddi

7.8 Dap an diing la D

Khi nung Fe(N03)2, Fe(0H)3, FeC03 0 nhiet do cao chiing deu phan hiiy va bi oxi hda thanh oxit Fe203

183

Trang 12

Ha'p thu khi CO2 vao dung dich Ba(OH)2 c6 thi xay ra cac phan ling:

CO2 + Ba(OH)2 BaC03>l.+ H 2 0 (1) Nd'udu CO2 : BaCOj + CO2 + H2O -> Ba(HC03)2 (2)

Goi n la s6' inol Ba(OH)2 tham gia phan ling (1) -> ni3aco,(i) =

Goi X la s6' mol BaCO^ tham gia phan irng (2) Ta c6 :

S6 mol BaCOg con lai = n - x = = 0,08 mol (I)

^ 22,4

Giai he (I), (II) duoc : n = 0,1 a = - - = 0,04 mol//

2,5

4 chat CO tinh luong Ifnh:

Ca(HC03 )2 , Al(OH)3 , Zn(OH)2 , (NH4 )2C03 „

Fe^^ khdng oxi hoa dugc Cu

7.12 Dap an dung la C ,

X la Fe, tac dung dugc vdi dung dich H2SO4 loang

Y la Cu, tac dung dugc vdi dung dich Fe(N03)3

Cu + 2Fe(N03)3 Cu(N03)2 + 2Fe(N03)2

Cac phirong trinh phan ling:

CuO + H2 (CO) — ^ Cu + H2O (CO2 t ) (1) Fe304 + 4 H2 (4C0) 3Fe + 4 HjO (4 CO, t ) (2)

Theo (1,2): So mol nguyen tir oxi tach ra khoi oxi = so mol h6n hgp khi = 0,32 : 16 = 0,02 (mol) -> V = 0,02.22,4 = 0,448 (lit)

Cac phucrng trinh phan ting:

CaCO, — ^ CaO + CO2 (X) (X,) CaO + H2O -> Ca(0H)2

Ca(OH)2 + NaHC03 CaC03 + NaOH + H2O

(Y) (X) (Y,)

•Ca(OH)2 +2NaHC03 -> CaCO, + Na^CO, + 2 H 2 O

(Y,) 7.17 Dap an dung la C

2 phan irng trong do HCI the hien tinh oxi hoa la (2) va (4)

(2H+ +2e -> H 2 t ) 7.18 Dap s6'dung la B

Phirong trinh phan irng:

Trang 13

7.19 Dapan diing la D j :

Chi CO phan img: O, ~» + O khOng phai la phan iJng oxi hoa - khu

Cac phan ling tao ra don chat la:

(1) O-, + 2KI + H2O 2KOH + O2 + I2

NaX + AgNO,, AgX + NaNO, (D NaY + AgNO, -> AgY + NaNO., (2)

- Xet truorng hop AgX va AgY deu la chat ke't tua Khi do s6' mol hai

muoi la:

8,61-6,03^ (mol) 108-23

- > Mmuoi - = 201

0,03 KhO'i lugng mol cua X va Y nam trong khoang:

X<201 - 2 3 = 178 < Y

186

lot la halogen c6 khO'i luong mol nguyfin tir 16n nha't la: I = 127

Do do, khong c6 halogen Y > 178: loai trudng hop nay

- Xet trucfng hop c6 mOt muoi bac khong ke't tiia Do la AgF Mu6'i AgY la AgClk

H nABCi = 0,06 mol = UN^CI

Phuong trinh ion nit gon chung la :

Ba^^ + SO4- -> BaS044

^ B , Cap nghiem duy nha't phu hop la: x = 2, y = 1

H Vay CTPT cua khi N,Oy la: N2O

^B, nN,o = 0'042 mol -> nN = 0,082 mol

Khi phan ling: N +4e->N -> So mol electron N thu la : 4.0,084 = 0,336 (mol)

• Vay kim loai M la Al (nhom)

Trang 14

7.33 Dap so diing la C

Cac phuang trinh phan iJng:

CaOCl^ + 2HC1 CI, + CaCl, + H , 0

2KMn04 + 16HC1 -> 5C1, + 2MnCl, + 2KC1 + 8 H , 0 (2) (1)

(3) (4)

K2Cr207 + 14HC1 SClj + 2CrCl3 + m.O

M n O j + 4HC1 CI2 + MnCl2 + 2H2O Tir 4 phircfng trinh phan ung ta tháy: Neu dung 1 mol cac chát thl 1 nio]

K2Cr207 cho lugng khi CI2 lorn nhat (3 mol CI2)

Fe304 = FeỌFe203 Cac phuang trinh phan ting:

1) FeO + H N O 3 - > FeCNO,), + N,Oy + H2O

hay (5x-2y)Fe203 + (30x-12y)HNO3 (l0x-4y)Fe(NO3)3 + (15x-6y)H2(i

Cong phan irng ( I ) va (2) thi he s6' cua HNO3 la 46x - 18ỵ

7.35 Dap an dung la Ạ

7.36 Dap an diing la Ạ

Co 4 h6n hop ran:

1 NajO va AI2O3: Tan hoan loan trong HjO di(

2 Cu va FeCl3: Khong tan hoan toan trong HjO dụ

3 BaClj va CUSO4: Khi hoa tan trong mrdrc, tao két tua BaS04

4 Ba va NaHC03: Khi hoa tan trong nude tqo ket tua BaCỘ

TrUc/c phan ifng: Goi x va y la s6' mol va H j c6 trong X Ta c6:

( X - a) mol N2, (y - 3a) mol H2 va 2a mol NH3 Ta c6:

28x + 2y , _

Sail phan i(ng:

» N2 + 3H2 < » 2NH3 Gia su CO a mol phan ling Trong h6n hcfp Y c6:

_ 2 8 ( x - a ) + 2(y-3a) + 34a _ ^ (2)

4(x + y-2a) Thay y = 4x vao (2) va tfnh dugfc - = ^

-> meu suat la 25%

Nam chat tac dung duoc vdi dung dich NaOH loang la:

NaHC03, Al(OH)3, HF, CI2 va N H 4 C I 7.43 Dap an diing la B

Phuong trinh phan ling: NaHS + HCl -> HjSt + NaCl

Phan ling nay loai bo dugc HCl, thu dugc HjS tinh khiét

7.44 Dap an diing la C

So mol cac chát:

17,92 ^ „ 20,16

n 22,4 22,4 = 0,8 (mol); = ^ d - r = ^'^ (^ol)

y = 0,8 mol

2M(Cr, Fe) + 6H2SO4 -> M2(S04)3 + 3SO2 + 6H2O (2)

So mol M = X = -nso, =|.0,9= 0,6 mol

- > X : y = 0,6 : 0,8 = 3 : 4

CTPT ciia M,Oy la Fe304

Trang 15

Hai trudng hop gay ra an mon difin hoa la: Ni - dd CUSO4 vii

Ni - dd AgNO, (Ni phai la kim loai manh hon kim loai tao mu6'i)

So d6 chuyS'n hoa:

P2O, > K 3 P O 4 > KH2PO4 > K 2 H P O 4

S6' mol cac chat:

nBa(OH)2 = 1-0,15 = 0,15 (mol); n^oH = 1.0,1 = O,l(mol)

4 F e S 2 + 1 1 0 2 ^ 2 F e 2 0 3 + 8S02 (1) SO2 + K 0 H ^ K H S 0 3 (2)

1 mol 1 mol

SO2 + Ba(OH)2-> BaS03xl + H2O (3)

0,1 mol 0,1 mol 0,1 mol

2SO2 + Ba(OH)2 ^ Ba(HS03)2 (4)

0,1 mol 0,05 mol

Theo (2, 3, 4): s6' mol SO2 = 0,1 + 0,1 + 0,1 = 0,3 (mol)

Theo (1): s6 mol FeSj = - s6' mol SO2 = 0,15 mol

-> mp,s, = 0,15.120 =18,0 (gam)

Khi chuy^n tijr mu6'i hidrocacbonat thanh mu6'i sunfat da c6 1 g^*^

S04~ thay the' 2 g6'c H C O 3 , khC'i lugng 1 mol mu6'i giam:

m = 2.61 - 96 = 26 gam

9 125-7 5

-> s6' mol mu6i hidrocacbonat = — — = 0,0625 (mol)

26 9,125

M„„,:: = = 146

0,0625

Akin, u,ai = 146 - (2.61) = 24 ^ kim loai Mg

C6ng thiJc cua mu6'i hidrocacbonat: Mg(HC03)2

7.53 Dap an diing la B

CaO + 2 H C 1 C a C l j + H2O

(X) CaClz + 2AgN03 -> Ca(N03)2 + 2 A g C l 4 '

(Y) ' • Ca(N03)2 + (NH4)2C03 ^ CaC034' + 2NH4NO3

"^°1-^•2 Dap an diing la B

Benzen khOng lam mat mau nirdc brom

Stiren lam mat mau nude brom

Anilift lam mat mau nude brom va tao ra ke't tiia trang

Dua vao cac hien tuong tren phan biet dugc 3 lo chat

191

Trang 16

22,4

S6' mol Br2 da phan ling : n^^^ = ' ^ ' = 0,35 (mol) 1,40,5 2

Vi 0,2 < 0,35 < 0,2.2 -> Trong hOn hop X c6 mOt anken va rnoi

ankin (hoac ankadien)

Dat CTPT ciia 2 hidrocacbon la : CnH2n (n > 2) va CmH2m_2 (m > 2)

Phuong trinh phan ling cua X vdi dung dich Br2 :

n, m > 2 va nguySn nen cap nghifim duy nha't thfch hop la : n = 4, m = 2

CTPT cua hai hidrocacbon trong X la : C 4 H 8, C2H2

Ca(0H)2 + CO2 CaC03>l + 3H20 (2)

Theo (1) va (2): ncaco., = n c o 2 = 3 P C J H^ = 0,1.3 = 0,3 (mol)

-> mcaco3 = 0,3.100 = 30 (gam)

Ddp an dung 1& C

Phuong trinh phan ling dO't chay hidrocacbon C^Hy :

C^Hy + (X + ^ ) 0 2 - > X C O 2 + I H 2 O

H6n hop khi Y g6m khi CO2 , hoi H2O va O2 du Qua H 2 S O 4 dac, hoi

H2O bi giil lai, hOn hop khi Z g6m khi CO2 va O2 du

Gia sit CO 1 mol C^Hy chay Ta c6 :

- S6' mol CO2 dugfc tao thanh la x mol

- S6' mol O2 du la : 10 - (x + ^ ) mol = (10 - x - ^ ) mol

Dap an diing la D

Phuong trinh phan ling: CjHj + 2H2

H6n hop khi Z kh6ng phan ling voi dung djch Brj chi g6m CjHf, va H2 dir

phan drng vdi dung dich Br2 la 0,06 - 0,01 = 0,05 mol

KhO'i lircmg binh Br2 tang la: 0,05.26 = 1,3 (gam)

COng thiJc dung la C

S6' mol cac cha't: nx = — = 0,075 (mol)

22,4

1,12 SO' mol hidrocacbon no C„H2n+2 = — — = 0,05 (mol)

22,4

so' mol hidrocacbon kh6ng no C^H^ = 0,025 (mol)

.SO' mol Br2 phan img = = 0,025 (mol)

Trang 17

- > C ^ H y la hidrocacbon khdng no loai anken (C^H2J

So d6 phan ling chay: C„H2„+2 """^ > n C O 2

C,H2, — X CO2

So mol CO2 = 0,05n + 0,025x = — = 0,125

22,4

Vi X > 2 nen cap nghiem duy nha't phu hop la: n = 1, x = 3

CTPT cua hai hidrocacbon la: C H 4 va C^H^

X la hidrocacbon, phan tir c6 6 nguyen tur C

C T C T ^ua X ia; VIH, - C H - C H- C H 3

I i

C H 3 CH, Khi X tac dung vdi CI2 (theo ti le mol 1 : 1), s6' dSn xuat monoclo la 2 (CI

the vao C, hoac C2)

Khi so mol CO2 = so mol H2O thi a = b

-> Thanh pMn % ve s6' mol cua X va Y la 50% va 50%

C,H,2 + C I 2 > m6t san phim huu ca

CH3

I

CTCT cua X la: C H 3 - C - CH3 : 2,2 -dimetyl propan

C H 3

12 Dap diing Ih A

Phuang trinh phan iJng : CH2 = C H - C = C H + 3H2 C H 3 - C H 2 - C H 2 - C H 3 ( l )

Goi X la s6' mol CH2 = CH - C = C H da phan ling

me,, = 0,0999.160= 16 gam 8.13 Dap an diing la B

Goi X, y va z la s6' mol CH4, C2H4 va C2H2 c6 trong 8,6 gam h6n hop

X T a c 6 : 16x + 28y + 26z = 8,6 (I) Cac phuang trinh phan ling cua h6ri hop X v6i dung dich brom:

C2H4 + Br2 ^ C2H4Br2 (1) C2H2 + 2Br2 -> C2H2Br4 (2)

48

T h e o ( l ) v a ( 2 ) : n g , , - y + 2z = — = 0,3 (II)

160

Phuong trinh phan ling cua h6n hap X vdi dung dich AgNO, trong NH,:

C H ^ C H + 2[Ag(NH3)2]OH ^ AgC=CAg^ + 4 N H 3 t + 2H2O (3)

Trang 18

G i a i he p h u a n g t r i n h (I), ( I I ) va ( I I I ) dugfc : x = 0,3, y = 0,7

Thay gia t r i cua x va y vao ( I I ) dugc : n = 4

C T P T cua anken la : C 4 H H

A n k e n cong H B r cho san phdm hiiu ccf duy nha't D o do, c6ng thirc

ca'u tao cua anken la : C H 3 - C H = C H - C H 3

nx = ^ = 0,2 (mol); n^o, = ^ = 0'3 (mol)

TCr s6' m o l X , s6' m o l CO2 va t i kh6'i cua X so v d i H2 b l n g 11,2 xac

d i n h dugc ankan la C H 4 va anken la C^Hfi

- nHjo >nco2 ~^ P^^' ^ hidrocacbon la hidrocacbon no

- D o t 0,3 m o l hidrocacbon, tao thanh 0,5 m o l CO2 - > so nguyfin tir cacbon t r u n g b i n h t r o n g phan t ^ hai hidrocacbon la 1,667

^ C T P T ciia X la C H 4

Trang 19

Chi CO ankin CjHj ( M - 26) thoa man

9 DAN XUAT HALOGEN, ANCOL, PHENOL

D6't chay 0,05 mol X cho 0,15 mol CO2 -> Phan tir X c6 3 nguyfin tir C

X la ancol no, mach hd nen c6 CTPT la CjH ^O,

Phuong trinh phan ling :

Theo (1) : — — — - — = 15,5 - » n = 2

(0,02 + 0,02).2 CTPT ciia ancol la : C2H5OH

23

Theo di bai va theo (1): (12x + y + 17)n = 15,6 (I)

(12x + y +39)n + ( ^ - n ) 23 = 24,5 (11) Giai he phuang trinh (I), (II) duocn = 0,3 mol -> Mhh = = 52 12x + y + 17 = 52 ^ X = 2,5 ; y = 5

CTPT cua 2 ancol phu hop la : C2H5OH va C3H7OH

Hop chat X la ancol bac 2 c6 nhom OH dinh vao C s6' 2

Ba dong phan anken la : (1) CH3-CH = CH-CH3 ; (2) C H 2 - C H - C H 2 - CH3

Trong do (1) CO 2 dong phan hinh hoc

9.7 Dap s6'dung la B

Dat CTPT trung binh cua hai ancol la: C -H2-^,ai20H

Phuong trinh phan ting:

C-Hj-^iCHjOH +CuO — - > C -H2-CHO + H2O +Cu (1)

199

Trang 20

H6n hap hai Y g6m hai andehit C-H^-CHO va hai H 2 O , v6i s6' mol

Phuotig trinh phan ung loai nu6c:

CH3 - ( : H ( C H 3 ) - CH(OH) -CH3 —^ 2 C H 3 - C(CH3) = CH - f

(3-metylbutan-2-ol) (3-iBetylbut-2-en)

9.9 Dap s6'dung la B

CH3OH + CuO — H C H O + H2O + Cu (1)

H C H O + 4[Ag(NH3 )2]OH ^ C O j t + SH^O + 4 A g ^ + 8NH3 (2)

Theo ( 1 2 ) : s6' mol CH3OH = - so mol A g = - = 0,3 (mol)

4 ^ 4 108 Khd'i luong CH3OH da phan ling la: 32.0,3 = 9,6 (gam)

Hieu suat phan ling la: h = ^ 100% = 80%

9.10 C6ng thu-c dung la B

V i M x > M y - > Y la anken (khdng th^ la ete)

Dat CTPT cua ancol X la C.H^O Ta c6: C.H^O > C,Hy.2(Y)

1 2 x ^ y + 1 6 12x + y - 2 cap nghiem duy nha't thich hgp la: x = 2, y = 6

CTPT cua X la CjH^O (CH3 - CH2 - OH) 9.11 Dap an dung la D

Ddt chay hai ancol X , Y cho s6' m o l H j O = s6' m o l CO2 -> X , Y la

hai ancol no

Cho 0,25 m o l h6n hgp M tac dung v6i Na (du), thu dugc < 0,15 m o l

H j - > X , Y la hai ancol dan chiJc

Vie't phuang trinh phan iJng chay cua X , Y va xac dinh dugc:

- Ancol X la C2H5OH

- Ancol Y la C3H7OH

Ancol X chay cho s6' mol H j O > s6' m o l CO2 - > X 1^ ancol no da

cMc C O cong thiJc phan tur la C2H6O2

"co " H o pMn lit ete dem d6't chay cd 1 noi doi

CTPT cua ete do la : C„H2nO

D6't (14n + 16) gam ete, tao ra 44n gam CO2

fl|i D6't 7,2 gam ete, tao ra 44.0,4 gam COj

Trang 21

nH,o > n c o , Ancol no dan chtic c6 c6ng thiic chung C„H2„+20

Phirong trinh phan ling chay:

C„H2„,20 + | ^ 0 2 - > C 0 2 + ( n + l ) H 2 0 (1^

X mol 0 , 1 7 mol 3n

^ = ^ X = 0,255 mol

X 0 , 1 7 Theo dinh lu^t bao loan kh6'i lucmg:

= 7,48 + 5,4 - 8 , 1 6 = 4,72 (gam)

9.16 Dap an dung la B

M (C„H2„,20, C „ H , 0 2 ) C O 2 + H^O 0,5 mol - > 1,5 mol 1,4 mol

Nhdn xet: - Phan tir ancol X va axit Y c6 3 nguyfin tir C

- Hco, > HH^O ^xit Y la axit khong no: CTPT ciia ancol X la C-,HxO

va cua axit Y la C3H4O2. Este tao thanh la Q H K A ( M = 1 1 4 )

V i s6' mol Y > s6' mol X -> So mol este < 0 , 2 5

80

Do do, khO'i lugng este < 0 , 2 5 1 1 4 = 2 2 , 8 (gam)

100 -> Dap an B phii hgp

9.17 Dap an diing la D

( 1 ) sai V I phenol kh6ng phan ling v6\g dich HCl, do do kh6ng tan

trong dung dich HCl

9.18 Dap an dung la B

9.19 Dap an diing la A

S6' mol cac cha't:

nco2 = ^ = (mol); HH^O = ^ = ^'^ (mol)

Ancol no C - H , - , 0 , > nCOj + (n + 1 ) H A ' n 2n+2 X

0,5 ^ n = 2,5 -> X = 2

10 ANDEHIT, XETON, AXIT CACBOXYLIC 10.1 Dap an dung la A

Khi Y g6m : ,

12

1 CH ^ CH -> AgC s CAg I : H C H ^ C H = ^ =

("^°^^-2. C H 2 = C H 2 - > C H 2 B r - C H 2 B r 4 : ncH2=CH2 = nBr2 = ^ = ("^Ol) "H2(2) = (^ol)

^10.2 Dap an dung la C

Goi CTPT cua andehit la CxHy(CHO)a

C,Hy(CHO), • ^ A g N 0 3 / N H 3 ^ 2aAgi

203

Trang 22

Dat CTPT cua axit dan chij-c la : C^HyOs

Phuang tnnh phan iJng chay :

Dat CTPT cua andehit X la C^Hy - CHO Cac phuang tnnh phan ling :

C;^Hy-CHO + AgjO ^"-^ > C^Hy-COOH + 2 A g ^ (1)

4 H N O 3 + 3Ag ^ 3AgN03 + NO t + 2 H 2 O (2)

S6'molNO= ^ =0,1 (mol)

Theo (1) va (2) : S6' mol Ag = 3 s6' mol NO = 3.0,1 = 0,3 (mol)

S6' mol C^Hy - CHO = s6' mol Ag = 0,15 mol

hoac X la H-CHO

S6' mol Na = — = 0,2 (mol)

23 0,1 mol andehit X > 0,1 mol ancol Y

0,1 mol ancol Y tac dung vdri 0,2 mol Na Y c6 2 nhom OH

X phai la andehit hai chiic (OHC-CHO)

10.7 Dap an dung la B

Cac phuang trinh phan ling : HCOOH + C 2 H 5 O H H,so4d ^ H C O O C 2 H 5 + H 2 O (1)

C H 3 C O O H + C 2 H 5 O H H,S04d ^ C H 3 C O O C 2 H 5 + H 2 O (2)

Goi X la s6' mol HCOOH trong h6n hgp X S6' mol C H 3 C O O H

IPI' cung la X mol

Ta C O : (46 + 60)x = 5,3 x = 0,05 mol

-> So mol h6n hop X la : 0,05.2 = 0,1 (mol)

S6 mol C 2 H 5 O H = — = 0,125 (mol) ^ Du C 2 H S O H

46 Theo (1,2) : m H c o o c 2 H 5 = 0-05.74 = 3.7 (gam)

Axit huu ca Y CO 2 nguyen tir C va c6 2 nhom -COOH

vay CTCT ciia axit Y la : HOOC-COOH

Theo d l bai ta bie't:

1 - Chat Z la ancol 2 chiic (phan tur c6 2 nhom chiic -OH)

-> Andehit X cung c6 2 chiic (phan tix c6 2 nhom -CHO)

Trang 23

- I V andehit phan ling v 6 i 2 V - > I V ancol, con d u i V H j

- > G6'c hidrocacbon cua X la g6c nọ

V a y chat X la andehit no, hai chiic

10.10 C6ng thirc dung la A

A x i t cacboxylic no, mach h d c6 C T C T chung la: C„H2„+2-ăCOOHX

T\x c6ng thiic thiic nghiem (C^U^Of,)^

K h i m = 2 - > Q H O f i - > CjHsCCOOH),: phu hop vofi C T C T chung

v a y CTPT cua X la: CfiHsOfị

10.11 C6ng thirc dung la B

roKOH + mNaOH = (56 + 40).0,5.0,12 = 5,76 ( g a m )

mx + niKOH + mNaOH = 3,6 + 5,76 = 9,36 ( g a m )

N h u vay k h i xay ra phan ung trung hoa da c6 9,36 - 8,28 = 1,08 (gam)

H2O dugc tao thanh

1 08

" x = n H , o = = 0'06 ( m o l )

M x = = 6 0 0,06

A x i t no, dan cMc C„H2„+,C00H = 60

14n + 46 = 6 0 ^ n = 1

v a y CTPT cua axit: CH3COOH

10.12 Dap s6' dung la D

Cac phuang t n n h phan iJiig:

H C H O + 4[Ag((NH3)2]OH ^ C 0 2 t + 4Ag^ + 8NH3t + 3H2O (1)

H C O O H + 2[Ag((NH3)2]OH - > C O j t + 2Ag4 + 4 N H 3 t + 2H2O (2)

- > Trong h 6 n hap X phai c6 H C H O va d6ng dSng ké tiép la

CH3CHỌ Chon dap an B

Dat CTPT cua axit X (dcrn chiic) la: C H y C O O H

Cac phuang trinh phan ling:

H C O O H + N a O H - ^ H C O O N a + H20 ( D C^HyCOOH + NaOH->C^HyCOONa + H 2 0 (2)

3,6 0,05 12x + y + 45 = 72 - > X = 2, y = 3

C6ng thiic cáu tao cua axit X la: CH2 = C H - C O O H : axit acrylic

Dat CTPT ciia : A x i t Y la C,Hy02, axit Z la C.H.Ộ

G o i a va b la s6' m o l cua axit Y va axit Z c6 trong - h6n hgp X

Ta CO cac sa 66 phan ling :

2 C , H y 0 2 + 2 N a ^ H2t

C H ^ j + 2Na H2t

n =^ + b = ^ = 0 , 2 (mol) ( I )

' 2 22,4

K h i d6't chay hoan toan ^ h6n hgp X:

1 m o l X chay, tao thanh x m o l C O 2

26 4

(a+b) m o l chay — = 0,6 m o l C O 2

44

- > x ( a + b) = 0,6 (11) Giai he phuang trinh ( I ) va ( I I ) dugc : b = 0,4 - —

Trang 24

S6' mol Ag = — = 0,5 (mol)

^ 1 0 8

Theo di bai:

0,25 mol andehit X lam thoat ra 0,5 mol Ag -> Andehit don chiic

0,125 mol andehit X phan ling vdri 0,25 mol Hj

-> Phan tur andehit c6 1 lien ket d6i

Do vay c6ng thiJc phan tur cua andehit X la: C„H2„_iCHO (n > 2),

S6 mol CuO = = 0,06 (mol) = s6' mol ancol

80 SO' mol andehit = s6' mol ancol = 0,06 mol

S6' mol Ag = = 0,22 (mol)

108 S6' mol Ag > 2.s6' mol andehit -> Trong h6n hop andehit tao thanh

phai CO HCHO va do do trong h6n hop hai ancol phai c6 CH^O!'

(CH,OH -> HCHO)

Dua vao cac dap an ta bifit hai ancol trong h6n hop la hai ancol no,

don chiic, trong do c6 CH^OH:

Ancol thir nha't la CH^OH c6 x mol (andehit tuong irng la HCHO)

Ancol thii hai la C„H2„+iOH c6 y mol (andehit tuong ling la C„H2„0)

Khi hai andehit phan ting v6i dung dich AgN03 trong NH3:

SO' mol cac cha't:

So mol h6n hop Z = = 0,15 (mol)

23-1

So mol Ag = — = 0,2 (mol)

108 Trong h6n hop Z, axit Y la axit fomic (HCOOH) va X la axit C,HyCOOH:

HHCOOH = ^n^g = 0,1 mol, mHcooH = 04-46 = 4,6 (gam)

n c,H,cooH = 0'15 - 0,1 = 0,05 (mol)

^c^HyCooH = 8,2 - 4,6 = 3,6 (gam)

^ Mc^HyCooH " • ^ " ' ^ 2 12x + y + 45 = 72

Cap nghiem duy nha't phu hop la: x = 2, y = 3

CTPT cua X la C 2 H 3 C O O H

^mc^HjCOOH = | | - 1 0 0 % = 43,90%

S6' mol cac cha't:

nco2 = ^ = 0'4 (mol); n„^o =7^= 0,4 (mol)

Nhqn xet: - X la andehit no, don chiic, mach hd C„H2nO

- D6't 0,2 mol h6n hop M cho 0,4 mol CO2 trong phan

tir X va Y CO 2 nguyen tu C

Trang 25

Suy ra: Hidrocacbon Y phai la anken trong phan tir c6 hai nguyfin tir C

Dap an D phu hop

Andehit no mach hb c6 cOng thiic chung la:

C„H2„+2-.2aOa-Dap an C phu hgp: Vdi n = 4, a = 2, ta c6 C4H6O2 (OHC-CH2-CH2-CHO)

10.22 Dap an d i i n g la C

- Cha't X phan l i n g dugc vdi Na va tham gia phan irng trang bac -> X c6 1

chiic ancol va 1 chiic andehit

- Cha't Y phan iJng dugc vdi Na va hoa tan dugc CaCO, -> Y la axit

-> Dap an C phii hgp

10.23 Dap an dui^g la C '

Theo cac dap an, hOn hgp X g6m 2 axit cacboxylic don chiic (phan tiJr

CO hai nguySn tijr oxi)

Dat CTPT C h u n g ciia 2 axit la C-H-COOH

MOH (KOH va NaOH) + C-H-COOH ->C-H-COOM + H2O

X y X y

Ta c6: m,,i, + n i N o H + "IKOH = ^^rumr, + ^^20

16,4 gam + 8 gam + 11,2 gam = 31,1 gam + x l 8 gam

X , = 1 ^ CTPT cua axit thu: nha't la C H 3 C O O H (C2H4O2)

X2 = 2 ^ CTPT cua axit thii hai la C2H,COOH (C^HfiOj)

10.24 Dap a n dung ia C

l l E S T E , LIPIT

Mx = 3,125 32 = 100 Este X c6 gd'c ancol kh6ng no

- Este X don chirc, c6 CTPT la CnH2n-202 = 100

Chi so' axit cua chat beo la so miligam KOH cSn thie't di trung hoa

axit tu do CO trong 1 gam chat beo

S6' miligam KOH can dimg la : 0,015.0,1.1000.56 = 84 (mg)

84 Chi s6' axit cua chat beo la : — =60 1 4

So mol N , = 0,7 = 0,025 mol Mv = = 74 dvC

11.4

28 ' 0,025 Cap chat este D c6 phan tii khd'i la 74 : Phu hgp

R - C O O - R ' +NaOH R-COO-Na + R'OH 0,025 mol 0,025 mol

2,05 0,025 = 82 R = 15 ( - C H 3 )

Trang 26

Phucfng trinh phan ling thuy phan lipit :

(C-H-COO )3C3H5 + 3 H 2 O - > 3C-H-COOH + C3H5(OH)3

46

S6' mol glixerol = — = 0,5 (mol) = s6 mol lipit

- > M H , , = ^ - 888 dvC

3(12x + y + 44) + 41 = 888 x = 17, y = 34

Dua vao 4 dap an cho ta tha'y :

X | = X2 = 17 la p h : : hop (vi khong c6 axit beo C,6 va C|x)

CH3COOC2H5 + NaOH -> CHsCOONa + C2H5OH

Sau khi c6 can, chat ran con lai la CHjCOONa

V i n N , o H < n6n tinh khd'i luong CHsCOONa theo NaOH :

mcHjCOONa = 0,04.82 = 3,28 (gam)

Theo di bai, este X la metyl fomat (HCOOCH3), thuy phan trong mdi

trudng axit: HCOOCH3 + HjO — ^ HCOOH + CH3OH

( Y ) ( Z )

- Axit HCOOH tan v6 han trong H2O •

- Dun CH3OH vdi dung dich H2SO4 dac d 170 "C khdng thu diroc anken,

do vay dap an D sai

11.9 Cong thiic diing la C

g

0,1 mol X tac dung vofi 0,2 mol NaOH -> h6n hop mu6'i

-> X la este cua 2 axit khac nhau vdi 1 ancol 2 chiic Cong thiic 6 dap

an C thoa man

Phuang trinh phan ling:

CH.COOCHjCHpOCC^H, + 2NaOH ^ 0,1 mol

CH.COONa + C2H,COONa + C2H4(OH)

0,1 mol 0,1 mol Kh6'i luong h6n hgp mu6'i:

Theo (1): 3Na da thay the nhom C^H, khi tao thanh xa phong

Khoi lugng xa phong la:

m = 17,24 + 23 0,06 - 41 = 17,80 (gam)

11.11 Dap an diing la A

Este no, don chiic C O CTPT Chung la C„H2„02

Phan ling chay:

C„H2„02 + ' n ,^ n+ - - 1 2 O, nCO, + n H , 0

Ta c6: n + — - 1 = n n = 2

2 CTCT cua este la: HCOOCH, (metyl fomat)

11.12 Dap s6'dung la A

Este HCOOC2H., va CH.COOCH, deu c6 CTPT la C3H6O2 ( M = 74)

n,„ = ^ = 0,3 (mol)

Xa phong hoa 0,3 mol h6n hgp este cAn 0,3 mol NaOH

vay: VjjNaOH = "Y = 0,3 (lit) hay 300 ml

1.13 Dap s6'diing la B

Phuong trinh phan ting:

C H 3 C 0 0 H + C2H5OH < ' ' > CH3C00C2H5 + H2O (1)

Trang 27

Du C 2 H 5 O H nen khd'i lugng este dugc tao thanh tinh theo C H 3 C O O H

Theo (1): S6 mol C H 3 C O O H = s6' mol CH.COOCjH, = 0,1 mol

Hieu sua't phan iJng dat 50% nSn khd'i lugng este tao thanh 1^:

Dat c6ng thiic phan tur cua este la : C^HyCOOQHb

Phuong trinh phan ling:

C.HyCOOC^Hb + NaOH >C,HyCOONa + QHhOH

0,2mol 0,2mol 0,2mol S6' mol NaOH con du la : 0,3 - 0,2 = 0,1 (mol)

Ta c6: 0,2(12x + y + 44 + 23) +0,1 40 = 23,2

12x + y = 29 Cap nghiem duy nha't thich hop la: x = 2, y = 5

Cong thiic phan tir cua este X la: CjHsCOOC.Hb

12a + b + 73 = 100 a = 2, b = 3 Cong thirc cau tao cua este X la: C2H5COOCH=CH2

Phutfng trinh phan irng chay:

C - H , - 0 2 + ^ ^ 0 2 - > n C 0 2 + nH^O (D

3 n - 2 , - , mol nmol

2 0,1775 0,145

214

3n-2 -> — 2 — = — >n = 3,625

0,1775 0,145

n, = 3 CTPT cua este thii nha't la C3H6O2

n2 = 4 ^ CTPT cua este thiir hai la C4H8O2

Dat CTPT cua hai este la C.HyCOO C-H^

Phuong trinh phan iJng xa phong hoa:

C,H,COO C-Hr + NaOH ^ C,H,COONa + C-Hr OH

1,99 gam x gam 2,05 gam 0,94 gam

-^x = 2,05 + 0,94 - 1,99 = 1 gam

S6' mol cac cha't tham gia phan ling va tao thanh deu bang s5' mol

NaOH va bang: nf,,oH =^= 0,025 9mol)

C6ng thiic ca'u tao cua axit tao este la: C H 3 C O O H

Tinh tuang tu cho hai ancol c6 CTPT chung C-Hj^OH, ta dugc ancol

thir nha't la C H 3 O H va ancol thii hai la C 2 H ,0H

11.18 Dap s6'diing la B

Cac phuang trinh phan ling:

H C O O C 2 H , + NaOH -> HCOONa + C2H5OH (1)

C H 3 C O O C H 3 + NaOH -» C H 3 C O O N a + C H 3 O H (2)

2C2H.,OH > C 2 H , , - 0 - C 2 H , + H2O (3) 2CH3OH : ^g^*^ ) CH3 -O-CH3 + H2O (4)

H C O O C 2 H , va CH3COOCH3 la hai d6ng phan, c6 CTFF la C3H6O

-> So mol este = — = 0,9 mol 74 Theo (1, 2, 3, 4): n„o ^^n,,,, = ^ = 0,45(mol)

Trang 28

Khd'i luong cua HjO thu duac la:

Theo cac dap an, khi xa phong hoa phan tit C,„H|405 tao thanh 3 phan

tix muoi chiia 3 lien ket TI (C = O), con lai 1 lien ke't n (C = C)

Mu6'i CO kien ke't TT (C = C) nhung kh6ng c6 ddng phan hlnh hoc Do

do, chi CO dap an A phu hcfp

11.20 Dap ling dung la B

S6' mol NaOH = = 0,6 (mol)

S6' mol NaOH = 3 s6' mol este E

100.40 ' 0,6^

0.2, Este E dirac tao thanh tu ancol ba chvic va hai axit cacboxylic no, don chiic

Dat c6ng thiic chung cua hai mu6'i dirge tao th^nh la C-H2-^,COONa

Ta c6: S6' mol mu6'i = s6 mol NaOH = 0,6 mol

14n + 4 5 + 23 = 43,6

0,6~ = 72,67 -> n = 0,3336

n, = 0 ^ CTPT cua axit la HCOOH

n2 = 1 -> CTPT cua axit la CH3COOH

Vay hai axit do la: HCOOH va CH3COOH

San ph^m sau phan ling la dung dich Y g6m: C H 3 C O O K , C H 3 O H va KOH

du Co can Y thu dugc chat ran g6m CH3COOK va KOH (12,88 gam)

Goi x la s6' mol X tham gia phan ling (2) Ta c6:

ncH,cooK = X mol, nKondu = 0,2.0,7 - x = 0,14 - x

^ 96x + 56(0,1 - X) = 12,88 -> X = 0,12

vay m = 0,12.74 = 8,88 gam

+ Khi X la HCOOC2H,: Ke't qua tim dugc khdng phu hgp vdri cac dap an: loai

- TrKdng hap 2: Khi phan tir X c6 2 lien ke't 7t

-» G6'c hidrocacbon cua X c6 1 lien ke't d6i C6ng thiic phan tii ciia X

la C„H2„ 2O2

Phuong trinh phan ling chay:

C„H2„_202 + 3 ( n - l ) 0 2 ^ n C 0 2 + ( n - l ) H 2 0 Tfnh ra dugc n = 4,5: loai

(D6'i v6i bai tap trSc nghiem, khi tim dugc dap an dung, kh6ng can phai xet cac trudng hgp khac)

X va Y la goc ancol va goc axit trong este Z Muoi chuyen thanh X thanh Y thi s6' nguyen tii C trong X va trong Y phai bang nhau Trong este Z (CH3CH2COOCH3) kh6ng thoa man dieu kien tren

U.24 Dap an diing la D

Ggi CTPT ciia amin don chiic la : C^H^N

Phuong trinh phan ling : C.H^N + HCl C^H^NHCl Theo (1) : 86' mol X = s6' mol HCl = 0,1.1 = 0,1 mol

(1)

217

Trang 29

-> Mx = ^ = 31 -> 12x + y + 14 = 31 x = 1 ; y = 5

CTPT cua amin X la CH5N hay CH3 - N H 2

- Amino axit X + (HCl, NaOH) Vi du : H2N -CH2 -COOH

- Mud'i amoni cua cacboxylic Vi du : CH3 - C O O N H 4

C H 3 - C O O N H 4 +HC1 CH3-COOH + NH4CI

CH3 - C O O N H 4 + NaOH -> CH3 -COONa + NH3 t + H2O

- Este cua amino axit T Vi du : H2N - CH2 - COOC2H5

Dat CTPT cua a - amino axit la : H2N-CxHy - COOH

Phuong trinh phan irng :

H2N-CxHy-C00H +HC1 -> ClHjN-C^Hy-COOH

1 mol X tac dung v6i HCl khd'i lirong tang 36,5 gam

a mol X tac dung vdi HCl kh6'i luong tang 13,95 - 10,3 = 3,65 gam

-> a = 0,1 mol ^ Mv = ^ = 103 dvC

0,1 12x + y + 16 + 45 = 103 ^ x = 3 , y = 6

3 dung dich c6 pH < 7 la:

C6H5NH3CI, CIH3N - CH2 - COOH, HOOC - (CH, )2 - CH(NH2) - COOH

Phirong trinh phan ling:

H2N - CH2 - CONH - CH(CH3) - CONH - CH2 - COOH + 3HC1 ->

2H3N-CH2-COOHCI + H3N-CH(CH3)-COOHCl

12.8 C6ng thiic dung la D

M = 89 ^ n, = — = 0,1 mol

nwaoH = 0,1 1,5 = 0,15 mol

Chat X phan iJng v6i NaOH nen c6 th^ la:

- Amino axit c6 CTPT la: H2NCH2CH2COOH

- Este cua amino axit c6 CTPT la: H2NCH2COOCH3

0,1 mol X + 0,1 mol NaOH 0,1 mol muC'i Y

Con du 0,15 - 0,1 = 0,05 mol NaOH

^ my = 11,7-(0,05.40) = 9,7 (gam)

My = — = 97, ling CTPT H2NCH2COONa

0,1 Vay CTCT thu gon ciia X la: H2NCH2COOCH3

CH3NH2 CO tinh bazo (tinh baza manh hon NH,)

3 CH3NH2 + 3 H2O + FeCl3 Fe(OH)3 i + 3 CH3NH3CI

* 12.10. Dap an diing la C

Tit CTPT C2Hi ,03N2 va tinh chat cua X ta tim dugc CTCT ciia X la:

[ C H 3 - C H 2 - N H 3 r N 0 3

Phuorng trinh phan ling vdfi NaOH:

CH3CH2NH3NO3 + NaOH -> CH3CH2NH2 + NaNO, + H2O

Trang 30

0,1 12x + y + 16 = 5 9 - > X = 3, y = 7

CTPT cua amin X la: C3H7NH2

Co 4 c6ng thu-c CTCT ling vdi CTPT C3H7NH2 D6 la:

- CTCT cua 2 amin bac I

- CTCT cua 1 amin bac I I

- CTCT cua 1 amin bac I I I

(Ban doc tu vie't 4 CTCT tren)

12.12. Dap an dung la D

Dat CTPT ctia amino axit X la: H2N - C.H^ ~ COOH

Phucfng trinh phan ting:

H2N - C,H, - COOH + NaOH ^ H2N - C.H^ - COONa + H2O

X la muoi amoni, trong do g6'c axit c6 li6n k6t d6i

- » chon dap an D: CHj = CH - COONH4 (C3H7O2N)

12.15. Dap an diing la D

12.16. Dap an diing la D

C6ng thuc ca'u tao cua X va Y la:

X : H2N-CH2-COO-CH3 ; Y : CH2=CHCOONH4 H2N-CH2-COO -CH3 + NaOH -> H2N-CH2COONa + CH3OH (Z)

CH2=CHCOONH4 + NaOH -> CH2=CHCOONa + NH3(T) + H2O

12.17. Dap an diing la A

0,02 mol X tac dung v6i 0,02 mol HCl -> X c6 1 nhom chiic amin

0,02 mol X tac dung vdi 0,04 mol NaOH -> X c6 2 nhom chiic axit

CTPT cua amino axit la : H2N-C,Hy(COOH)2

Phuang trinh phan ling ciia X vori HCl:

H2N-C,Hy(COOH)2 + HCl ^ ClH3N-C,H,(C()OiI)2'

0,02 mol 0,02 mol

12x + y + 52,5 + 90 = 183,5

cap nghiem duy nhat phii hop la : x = 3, y = 5

C6ng thiic phan ttr ciia X la : H2N-C3H5-(COOH)2

D^ tim dap an diing cua cau nay, ta diing phaong phap loai trir la nhanh nha't: Dua vao dac diem cua phan ling giua amino axit vai HCl va giiJa amino axit vdi NaOH, ta thay khi > m, thi trong phan tijf amino axit s6 chiic axit phai Idm han s6' chiic amin Tit nhan xet nay chi dap

Dat CTPT cCia amin dan chiic la C^H^-NHj

Phuong trinh phan irng: C,Hy-NH2 + HCl C,Hy-NH3Cl

10 gam x g a m 15 gam -> So mol C J L- N H 2 = s6' mol HCl = - - ^ ^ =0,137 (mol)

" ^ ^ 36,5

'•^ Mr u MH = = 73

12x + y + 16 = 73

Cap nghiem duy nha't phii hgp la : x = 4, y = 9

C6ng thiic ca'u tao cua amin X la : C4Hc ,-NH2

S6' d6ng phan cau tao ciia X la 8 (ban doc t u viet cac d6ng phan) 12.20 Dap an diing la B

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