- Chodung dich NaHS03 vdo honhdp con lai thi (C2H5)20 khong phan iing di/dc tach ra c6n C 2 H 5 C H O phan itng tao ket tua, Ipc lay két tua cho tdc dung vd
j iiịiii iiiai H6a hpc 11 HOu co D f• '^' 'ting
44x + 26y = 2,02 b) T a c o h ^ p h U d n g t r l n h , 2 , . i 0 8 - . 2 4 0 y = 11,04
x = 0,04 (mol)
So mol C 2 H 2 ban dau : nc^Hg = 0.04 + 0,01 = 0,05 (mol)
H i $ u suat phan uTng cong ni/dc vao axetilen : ,. . .
' 0 04 '
H % = X 100% = 80%. . u ! ; J ,! :
0,05
Bai 9. Dun 12g axit axetic v d i mpt luTdng d\i ancol etylic (c6 axit H2SO4 d | c lam xuc tac). Den khi dufng thi nghi^m thu difdc 12,3g estẹ , , . , . , , a) V i e t phifdng trinh hda hpc cua phan iJng^
b) Tinh phan trSm khoi liTdng cua axil da tham gia ph5n tfng este hdạ ^,
Giai . , ,,
a) Phi/dng tnnh phan i/ng :
C H 3 C O O H + C 2 H 5 O H C H 3 C O O C 2 H , + H 2 O ' ' • '
0 , 1 4 m o l tith iumqi: ụ,.(.,- 0 , 1 4 m o l 1, 1 ' H i l)
b) So mol ẹste : n.-sic = —^ = 0,14 (mol) ' ' * , v d t vl
=> mcH3COOn = 0 , 1 4 x 6 0 = 8,4 (g) ' , , n , , , 1
Thanh phan % khoi liTdng axit tham gia phan iJng este hoa : \\,
%mcH,COOH = ^ X 100% = 70%. *' " ' '
B a i 10. Them niTdc vao 10ml axit axetic bang (axil 100%; D = l,05g/cm') den the tich 1,75 lit d 25"C roi diing may do Ihi lhay p H = 2,9. , ^ , ^
a) Tinh nong do mol cua dung dich thu di/dc.
b) Tinh do d i ^ n l i a cua axit axetic d dung djch ndi tren (*nni t^nuu* c) Tinh gan dung hhng so can b^ng cua axit axetic d 25"C. j f , )
V fY Giai '.»aKW r;sx , ^Y)--\
a) K h o i liTdng dung djch CH3COOH : m = D . V = 1,05.10 = 10,5 (g) => So m o l CH3COOH : ncH^cooH = — = 0.175 (mol)
^„ ^ . 60 ; ,Ạ;.i:..rftv! y 7 ; £ 0 .
Nong do m o l cya dung dich CH3COOH : C M = - = = 0,1 ( M ) .
V 1,75 .i'l VÍ,,)) It"''
b) PhiTdng tnnh phan iJng : CH3COOH CH3COO" + l u i 1
.'•t^MiX^ 0,1M ..,^lh'*^^^'^•-•,>nv 0 , 1 M - f ! i > : H
pH = 2,9 => [H*] = 10""" = 10"^'' = 1,26.10"^ ' M r ' ^ a = ^ ' ^ ^ ' ^ ^ = 1,26.10-^ = 0,0126 hay 1,26%. nhii aW).!
0.1 266
c) CH3COOH -> CH3COO- + H"^
Ban dau 0,1 • ' • i ^ ' ; ^ \ . i , ' " C K P ' V H : i 0
Phanu-ng 0,1a 0,1a 0,1a C a n b h n g 0 , 1 - 0 , 1 a 0,1a 0,1a
K a =
[ C H 3 C O O " ] . [ H + ] 0,lạO,la 0,01.(0,0126)^
= 1,6.10 ,-5 [ C H 3 C O O H J 0 , 1 - 0 , 1 a 0,1-0,1.0,0126
ai 11. Cho 0,04 mol mpt hon hdp X gom C H 2 = C H - C O O H , CH3COOH va C H 2 = C H - C H O phan iJng vijfa du v d i dung djch chiJa 6,4 gam brom. M a t khdc, de ; trung hoa 0,04 mol X can diing vifa du 40 ml dung djch NaOH 0,75 M . K h o i liTdng ; cua C H 2 = C H - C O O H trong X la . a j ^ '
Ạ 1,44 gam B . 2,88 gam C. 0,72 gam D . 0,56 gam
(Trkh de thi tuycn sink Dai hoc khoi B) Giai
6,4 ^ , _ n n r n r ^ A r>T i U/ t V i )
Ta cd: = 0,04 m o l ; nNaOH = 0,75.0,04 - 0,03 j n o l
160 ciX, Gpi: CH2=CH-COOH: a mol, CH3COOH: b mol, CH2=CH-CH0: c mol
a + b + c = 0,04 (1)
Hon hdp X tac dung vdi dung dich Br2 t S . T !
CH2=CH-COOH + Br2-> C H z B r - C H B r - C O O H ^ fti i
CH2=CH-CH0 + 2Br2 + H2O ^ C H z B r - C H B r - C O O H + 2HBr j b r ,
=> a + 2c = 0,04 (2) ' ' ''^ ''^ • ' - r, u
Hon hdp X tac dung v d i dung dich NaOH
CH2=CH-C00H + NaOH - » CH2=CH-COONa + H2O
a a
CH3COOH + N a O H - > CHjCOONa + H2O
b ,.,•„, . h ,;5::,„..„,,„ flM t;fV«?Q:,)M UOBVI
=i> a + b = 0,03 (3) , m^O
Tir (1), (2), (3) la c6:a = 0,02; b = 0,01; c = 0,01
V$y khoi liTdnf cua CH2=CH-C00H = 0,02.144 = 2,88 (g) ' . C i
=> D a p an B
Bai 12: Chodung djch X chu-a hon hdp gom CH3COOH ỌIM va CHjCOONa 0,1M. [V Biet d 25"C K^ cua CH3COOH la 1,75.10 va bo qua sir phan l i cua ni^dc. Gia
tri p H cua dung djch X d 25" la uh'j fhiv n'-in]) a i M f i;<'i! n.^:
Ạ 1,00 B . 4 . 2 4 C. 2,88 D . 4,76
( ',,0! w u , (Tn'ch de thi tuyen sinh Dai hoc khoi B)
\ V I ' \Á rtK t ii,"WH\, n • V \
Phan djng va phtMng ph^p gi5i H6a hgc 11 HQu co - Dfl XuSn Hung Gidi C H 3 C O O N a - > CHjCOÓ+Na"^ B K v; >„,•,:, 0,1 >• • 0,1 " C H , C O O H CH.COO" + i * ' B a n d a u : 0,1 0,1^ ,0_, , ; ; H ^ i ; 0 2; ; i r H ^ P h a n l i : x x •••"k^' ' i i K K ) : ) ^ K::>i 1 ^ - , C a n bKng: 0 , 1 - x x + 0.1 x . . r 1,1:!:) • ,1! ' ^ ' V^ K c = ^ ^ ^ ^ ^ ^ i ^ = l,75.10-^ ( D i g u k i $ n : 0 < x < 0 , l ) ] ^'^^'^ - ^ 3 = - ' • v ; : , ; / - v ^ , ' ' 0 , 1 - x ^1, X ¥\'j iw/'i 'ỉ.;
=> X 1= 1,75.10^^ (nhan); Xa = - 0 , 1 (loai) , ^ ,.yi:A :io:;::^|/!*D(>3v-l!::)-~;l'i:)
=> pH = - l g l H ' ' ] = = - l o g (1,75.10"'*) = 4,76 => D d p a n D
Bai 13. Trung hoa 8,2 gam hon hdp gom axit fomic va mpt axit đn chiJc X can 100 ml dung dich N a O H 1,5M. N e u cho 8,2 gam hon hdp tren tac dung v d i mol
li/dng d\i dung dich AgNOs trong NH3 dun ndng t h i thu difdc 21,6 gam A g . Ten g o i c u a X l a : : A A , ; .V .1.;;^: , , mi
Ạ axit acrylic B . axit propanoic C. axit etanoic D . axit metacrylic
Gidi
Ta c6: n ^ , = ^ = 0,2 m o l ; n, . o H = 1,5.0,1 = 0.15
Goi C T T Q cua axit đn chiJc la RCOOH ti «
* Hon hdp X tac dung v d i dung dich A g N O i / N H , O l . ) ! ' H)
H C O O H +A g 2 0 - > C 0 2 + H2O + 2 A g '
0,1 ; 0,2 h <• f b
=> m R c o o H = 8 , 2 - 0 , 1 . 4 6 = 3,6 (g) ^ fit f H'V ft? il i-^z'
* H o n hdp X tac dung v d i dung djch NaOH . M l t? ' H
H C O O H + N a O H ^ HCOONa + H2O
0,1 0,1 ' ' ' ' " i U > n > \\i)u)fi ' t o i I í»
RCOOH + N a O H - > RCOONa + H2O ^«
0,05 0,05 „ . a n .
= > M« c o o H = - ^ = 7 2= > M R = 2 7^ R l a C 2 H 3 "
0,05 iudi V B V .
V a y axit do la: C H 2= C H- C 0 0 H : axit acrylic D a p a n A
Bai 14. H o n hdp X g6m axit H C O O H axit CH3COOH ( t i 1^ m o l 1:1). L a y 5,3 gam hon hdp X tac dung v d i 5,75 gam C2H5OH (c6 xuc tac H2SO4 dSc) thu diTdc m gam hon hdp este (hieu suát ciia cac phan iJng este hda deu b^ng 80%). Gia
t r i c u a m l a : . . , ^ i,,*^;,^,.^^^ iJiA! -A
Ạ 10,12. - . i i - B . 6,48. Cf'V C. 8,10. D . 16,20.
(Trich de thi tuyen sink Dai h<)c khdi A)
268
V I m o l ci5a 2 axit b^ng nhau:
— 46 + 60 5,3 ;M
=> M 2 a x i t = = 53 => n2axit = • 7 7 = ^ ' ^ ' " ° '
Goi cong thtfc chung cua 2 axit m R C O O H v d i = 53 - 45 = 8. Ta c6: ncjHsOH = ^ = 0.125 mor
Ta thay: n ^ ^ ^ j ^ = O . K nc^HjOH = 0,125 mol ^ '^'^ ''
. •: -)-•'•• 'I'fq dim: „;;••• •
=> sau phan ỉng C2H5OH AM, liTdng este sinh ra tinh theo axit ^ ,
R C O O H + C2H5OH RCOOC2H5 + H2O
Tir a x i t - » este k h ^ i l i W n g tang 0,1.(29 - 1) = 2,8 (g). v';
^ m^sic = (5.3 + 2,8).80% = 6.48 (g) . . .
D a p a n B . * ^
Bai 15. Cho hSn hdp X g o m ancol metylic v^ hai axit cacboxylic (no, đn chiJc, ke tiep nhau trong day d6ng d^ng) t^c dung het v d i Na, giai phdng ra 6,72 l i t khi H 2 (dktc). N 6 u dun ndng hon hdp X (c6 H 2 S O 4 dSc l^ m xuc tic) thi cic
chat trong hon hdp phan tfng vCfa du v d i nhau tao th^nh 25 gam hon hdp este (gi^ thiet phan iJng este hda dat hỉu suat 100%). Hai axit trong hon hdp X 1^
Ạ H C O O H va CH3COOH. B . CH3COOH v^ C2H5COOH ,
C. C2H5COOH vk C3H7COOH. D . C3H7COOH v ^ C4H9COOH
" Trich de thi tuyin sinh D^i hQc khoi A ndm 2010''
- Ta c6: = 0,3 m o l '; , J \' «i ^fts '':.' X mt iUs
• G o i C T T Q cua 2 axit cacboxylic no, đn chtfc, ke t i ^ p nhau trong day dong
d i n g m : R C O O H
• Phi/dngtrinhphaniJng: RCOOH + Na •RCOONa + iu^*'^
Ta c6 : nx = 2n^^ = 0.6 mol p^^^^-^
Cdc chat trong X phan iJng v d i nhau vita du => n^^„i = naxị = 0.6/2 = 0.3 m o l =i> Hesie = 0.3 m o l . C T T Q cua este : R C O O C H , , <
: Ta cd : M E = 25/0.3 = 83.3 R + 59 = 83.3 => R = 24.3
=> 2 axit m CH3COOH va C2H5COOH ,
= > D d p ^ n A .
Phan d^ng va phuong phAp giai H6a hpc 11 HOu cd - DC ung
^Oaiiq 5. >^fị cAiuf Húte phan tit, Jtdẹ ttisih e&nq. thite emi tqi» eua tuiitehil, axil ,, .
B A I T A P M A U " ' Bai 1. Cho 4,4g mpt andehit đn chiJc no thifc hi#n phan iJng trang gifdng thu dUdc
21,6g bac. Xdc dinh CTPT cua andehit goi ten. ,;
Gidi ^=.-K:V>^:
DatCTTQ andehit đnchU-c no la C„H2„+iCHO ^ , „, PhiTdng Irinh phan iJng :
C„H2„^,CHO + 2AgN03 + SNHj + H2O ^ CnHzn+iCOONR, + 2 A g l + 2 N H 4 N O 3
0,1 mol ' 'Ó!'^ +',.ilVK>00 H .t^v HO 0,2 mol So mol Ag : nAg = — = 0,2 (mol) '
Ta cd : Manjchị = ^ = 44 14n + 30 = 44 n = 1 ; ' / \
Vay CTPT andehit: C H 3 C H O : ctanal. 1 v - , - , . s ,
Bai 2. De trung hoa 150g dung dich 7,4% cua axit no mach hd, đn chifc X can dung 100ml dung djch NaOH 1,5M. Viet CTCT \h goi ten ciia chat X. j f ) - J i .
ahif rmi 'vu:<j ii^iníoii irdV'gfiÚnM'ii qUii nhti :i0(m
D a t C T T Q X : C„H2n+iC00H ^ ^i,i,> ^y^-,] j ^ : ) ii^iu 'S^tfl- ahdqi-M) k.
PhuTdng Irinh phan iJng : ,^ H(X)'.)B C„H2„+,C00H + NaOH > C„H2„+,C00Na + H 2 O ^(y^DJip
1 ' 0,15 mol 0,15 mol ,,
So mol NaOH : HNaOH = 0,1.1,5 = 0,15 (mol)
Kh5i lirging chát tan X : mx = ^^^^r^ = 11.1 (g) Mx = ^ = 74 "^a c6 : I4n + 46 = 74 => n = 2
Vay CTPT X : C 2 H 5 C O O H => CTCT : C H 3 - C H 2- C O O H : axit propanoic. Bai 3. Cho hdp ch^t hCTu cd A chiTa cdc nguydn to C, H, O trong d6 oxi chiem
55,17%. Trong A chl c6 mpt ioai nhdm chi^c. Khi cho 1 mol A t i c dung vdi
AgNOj/NHj dir thu difdc 4 mol Ag. Xic djnh CTPT A va viet CTCT, goi ten Ạ
Ttf AvietphiTdngtrlnhphantfngtheosđo : 1 ^ i> t
A > B — ^ dung dich m i u xanh. ! \ i !
:;;'(;:;', Nịt"
A + đ AgNOs/NHj => A g i •=> trong A c6 nh6m - C H O
Ta c6 : = - => A c6 the l i HCHO ho$c diandehit.
" A g 4
N6u A la HCHỌ Ta c6 : %0 = x 100% = 53,33% ^ 55,17% (Ioai)
Vay A la diandehit. Dat CTTQ A : R(CHO)2 (
% 0 = X 100% = 55,17% =>R = 0 R + 58
Vay CTPT A : (CHO)2 >--.H.:> r^'.-- I T * . ;V;lOO::^•-..t4J j H
CHO /. • - v . ,
CTCT : I andehit oxalic (glioxal) CHO
Phi/dng trinh phan iJng :
CHO C H 2- O H
CHO f t " C H 2- O H •« ^Mihô A jiv.-
CH —OH ^ ^
^ / ^ H - n H ^ - ^ " ^ ^ " ^ ^ — ^ l ^ " ' ^ ^ C u ^ ° " ? " ^ •^2H20 ^
^"2-<JH C H 2- 0 - ^ " - - O- C H 2
Bai 4. Ket qua phan tich nguyen to cho thay hdp chat A chiJa 55,81% C, 7,017o H, c6n lai la oxị A IS chat I6ng raft it tan trong niTdc, khong c6 v| chua, khong lam mat mau nifdc brom. l,72g A phan iJng viifa du vdi 20ml dung djch NaOH I M va tao thanh mpt hdp chat duy nhat B c6 cong thrfc C4H703Nạ Khi dun n6ng vdi dung dich axit v6 cd, tCf B lai tao th^nh Ạ ^ ^,
a) Xdc dinh cong thiJc phan tuT Ạ
b) Tir CTPT A va tinh chat cua A, cho biet A thupc Ioai hdp chat naọ
c) Viet CTCT cua A, B v i cdc phiTdng trinh h6a hpc da neụ ' '
Gidi .. - , .
% 0 = 100% - 55,81% - 7,01% = 37,18% D a t C T T Q A : C , H y O ,
T a c 6 t i l § : x : y : z = ^ : ^ : ^ =4,65:7,01:2,31 = 2 : 3 : 1 => CT A c6 dang (C2H30)„ • ,..}f :)...tlị> MOOÔH^"j Theo de A + NaOH > B (C4H703Na) T V v 1 r xjr%i-vx 11 -»
B + axit v6 cd -> A (khong c5 vj chua, khong l^m mat m^u ni/dc brom A la estev6ng).
S^ mol NaOH : nwaOH = 0,02.1 = 0,02 (mol) , ,, , ,,
= > H A = -7— S0,02 => n < 2 . u,;,; > 43n n = 1 CTPT A : C 2 H 3 O ( l o a i ) " : " " - ^ ^ " ' " ^ ' ' ' : ' n = 2 = > C T F r A: Q H e 0 2 - ' ' " ^ . " C T C T : I ^ ^"-C=Oestevdng.đnchỉcno ; , , ; f ' ' C H 2 - 0 ^
H(5a hpc 11 HOu CO - D5 XuanHimg Phifdng trinh.phan ỉng: ' 7 C H —CH o '''' I ^ ^ X : = 0 + N a O H C H 2 - C H 2 - C H 2 - C O O N a OH C H j - C H j - C H j - C O O N a + H * C H 2 - C H 2 - C H 2 - C O O H + N a ^ O H O H •f"5t-í> , ^ ,0 C H 2 - C H 2 ^ ^ >H0 ' • C H 2 - C H 2 - C H 2 - C O O H " S I ^ c = o . , , . OH ^
Bai 5. Cho dung dich A g6m hai axit đn chiJc no ke tid'p nhau trong day dong ding. De trung hoa 2(X)ml dung dich A can dilng 250ml dung dich NaOH 0,4M. Sau
khi CO can dung dich thu dtfdc 10,44g muai khan. X5c dinh CTPT, CTCT cua
mSi axit tinh n6ng dO mol tilfng axit trong dung dich Ạ ' ;' *
ui E ) a t C T T Q A : C- H j - ^ i C O O H \,^^t,n!,m ihski : s?;*ak«>.i)iiKl:si!fr,
PhiWng trinhphan iJng : 'Jj:0yr%tfh -ikriiiA, 4»*idar&fi ỉkm
i u , C- H j - ^ j C O O H + N a O H > C-H^-^jCOONa + H2O flniiU 0^
0.1 mol " o"lmol ^ •
S6mol NaOH : nwaOH = C M - V = 0,4.0.25 = 0.1 (mol) ^ ^^^'^^i ạfeX ic
10 44 > : i s l f g * ^ i f t T T O ^ T 0'
Khoi liidng phSn tit m u d i : = = 104.4 g g ^ . ^ . fSiV (.
1 4 n +6 8 = 104.4 => n = 2,6 ' i '
"1 = 2 < n = 2,6 < = 3 • , „ ; . : ' " jf^it)'«6
V5y CTCT hai axit C2H5COOH ••c- r ' i ' 5^>
C3H7COOH • ' fl
CTCT : C2H5COOH : CH3- C H 2 - C O O H : axit propanoic -fi C3H7COOH: C H 3- C H 2 - C H 2 - C O O H : a x i t b u t a n o i c A ' f i l i ^ ^ l ?
i-<:-,;:f}:yp.:^^^^ CH3- C H - C O O H : axit 2-metyl propanoic' •''«'^* »3
C2H5COOH + NaOH > CzHjCOONa + H2O a mol a mol a mol j j :
C3H7COOH + NaOH > C3H7COONa + H2O .p^^.,.^
b mol b mol b mol r
^ V , ; , , fa + b = 0,l ra = 0,04
Ta c6 phtfdng trinh <^ => i (mol) * ; ^ rn^^^^^
[ 9 6 a + 110b = 10,44 [ b = 0.06 •
Nong do mol tiifng a x i t : 'í-"' »"^^'
C M ( C 2 H 5 C O O H ) = = 0,2 ( M ) ; ( C 3 H 7 C O O H ) = - ~ = 0,3 ( M ) .
Bai 6. Hdp chat X no, mach hd c6 phan trSm khoi lifdng C va H Ian liTdt bang 66,679? va 11,11% con lai la Ọ T i khoi hdi cua X so vdi oxi bang 2,25.
a) T i m cong thu-c phan tijf cua X. t f C ; ; y'V[7.-L •'••>:!'• ^
b) X khong tac dung vdi dung djch AgNOs trong NH3 nhufng khi tac dung vdi hidrt sinh ra X|; X| tac dung diTdc vdi Na giai phong hidrọ Viét cong thiJc cáu tao va goi ten cua hdp chat X. ,, .
a) % 0 = 1 0 0 % - 6 6 . 6 7 % - 1 1 . 1 1 % = 22,22% . i i
D a t C T r Q X : Q H A ' ^ ;„ '
T a c 6 t i l ? : x : y : z = ^ : i M l : ^ =.5,56: 11,11: 1,4 = 4 : 8 : 1
^ 12 1 16 .
^ CT X c6 dang (C4H80)„ => M x = 2,25 x 32 = 72 ' '/^l^f s'irt .-if<t r-,
Ta C O : (4 X 12 + 8 + 16).n = 72 n = 1 '
Vay CTPT X : C4HgỌ - ,•
b) Theo de thi X i : ancol h^c 2 ,, ,, , ? i •
=> X : xeton. Vay CTCT X : C H 3- C - C H 2- C H 3 etyl'metyl xeton. '
' : , v
B a i 7. Lap CTPT, xac dinh CTCT va goi tSn axit trong cdc tnfdng hdp sau :