andehit va H2Ọ >
Theo phiTdng trinh phan iJng ta c6: nandchu = A H J O 't * W »
M Y = (I4n +16) + 18 = 13,75.2 = 27,5
o 14n +34 = 55 n = 1,5 => 2 andehit la HCHO vi CH3CHỌ
;; Goi'1! X, y Ian liTdt la so mol cua HCHO va CH3CHỌ
X + 2y :> :> X + y CH3OH - X C2H50H - > 4x + 2x = = 1,5 => x = y HCHO 4Ag X 4x CH3CHO -> 2Ag X 2x 64,8 108 = 0,6 => x = 0,l mol
m.= mcHjOH + mcjHsOH = 0,1.32 + 0,1.46 = 7,8 (g) => Dap^nẠ ' Bai 13. Khi thi/c hi^n phan iJng este hda 1 mol CH3COOH va 1 mol C2H5OH, luTdng Bai 13. Khi thi/c hi^n phan iJng este hda 1 mol CH3COOH va 1 mol C2H5OH, luTdng
este Idn nhat thu diTdc Ih 2/3 mol. De dat hỉu suat ctfc dai 1^ 90% (tinh theo
axit) khi tien h^nh este hda 1 mol CH3COOH cln s6' mol C2H5OH Ik (hi^t cdc phan uTng este hda thifc hi^n d cUng nhi^t dp). phan uTng este hda thifc hi^n d cUng nhi^t dp).
Ạ 0,342. B. 2,925. C. 2,412. D. 0,456.
(Trich de thi tuyen sinh Dai hoc khoi A)
H * , l "
' CH3COOH + C2H5OH
Ban dau: 1 mol 1 mol Phanifng: 2/3 mol ' '•'•2/3 mol Phanifng: 2/3 mol ' '•'•2/3 mol Canb^ng: 1/3 mol 1/3 mol
CH3COOC2H5 + H2O
2/3 mol 2/3 mol 2/3 mol ^ 2/3 mol 2/3 mol ^ 2/3 mol
, Tacd: [CH3COOC,H3].[H,0] ^ 2 ^ ^ ^
^ [CHjCOOHLLCzHgOH] 1/3.1/3 1 - .
Hi$u suS't cijfc dai 90% (tinh theo axit) => n„uphinung = 1.90% = 0,9 mol.
C H 3 C 0 0 H + C2H5OH ^^£=± CH3C00C2H5 + H20
Ban dau: 1 mol a mol Phanu-ng: 0,9 mol 0,9 mol Phanu-ng: 0,9 mol 0,9 mol
can b^ng: 0,1 mol (a - 0,9)mol 0 9 0 9 0 9 0 9
Kr = ' ' = 4 ^ a = 2,925 Dap an B.
0,9 mol
0,9 mol 0,9 mol 0,9 mol
ai 14. Cho 3,6 gam axit cacboxylic no, đn chtfc X tic dung hohn toan vdi 500 ml
dung dich gom KOH 0,12M va NaOH 0,12M. Co can dung dich thu difdc 8,28 gam hon hdp chat r^n khan. Cong thuTc phan tuf cua X1^: gam hon hdp chat r^n khan. Cong thuTc phan tuf cua X1^:
A.C2H5COOH B.CH3COOH C.HCOOH D.C3H7COOH
1.4. (Trich de thi tuyen sinh dgi hpc khdi B - ndm 2008) ' Gidi , I , ' Gidi , I ,
Ta cd: nNaOH = HKOH = 0,5.0,12 = 0,06 mol. , ;
Phúdng trinh phan iJng: . ( 1,
RCOOH + NaOH -» RCOONa + H2O . f
RCOOH + KOH RCOOK + H2O
Ap dung dinh luat bao toan khoi li/dng ta cd:
niaxii + m N a O H + niKOH = nimuoi + "^^20
t ] J
= 3,6 + 0,06.40 + 0,06.56 - 8,28 = 1,08 (g) if 1
1,08
18 = 0,06 mol
Theo phiTdng trinh phan iJng: HRCOOH = n^^^Q = 0,06 mol
M R C O O H - 3,6 0,06 0,06 Vay axit do la CH3COOH
= 60 => M R= 15 (CH3-)
Dap an B.
r^liS ; r u! - 'A P I H A S K : ph p giSi H5a Rpc 11 H[7u ed - P5TuanWng
B a i 15: Dot chay hoan loan 7,6 gam hon hdp gom mot axit cacboxylic no, đn chtJc, mach hd mot ancol đn chiJc (c6 só nguyen tuf cacbon trong phan tuT khdc nhau) thu duTdc 0,3 mol CO2 va 0,4 mol H2Ọ Thifc hi§n phan u"ng este h6a 7,6 gam hSn hdp tren v d i hipu suat 80% thu diTdc m gam estẹ G i i tri cua m Ih
A . 4 , 0 8 . B.6,12. ^ C.8,16. ; D . 2,04. • ? ,;
,, , "Trkh de thi tuyen sink Dai hoc khoi A ndm 2012" Gidi
Do do't axit no, đn chtfc cho H2O = CO2 nen ancol can tlm \k ancol no, đn chiJc. So m o l ancol = 0,4 - 0,3 = 0,1 mol
So mol CO2 do ancol tao ra se < 0,3 mol. , , V § y ancol A c6 mOt hoac hai nguyen tuT C
* T H I ; Ancol c6 1 nguyen tuf C vay ancol la CH3OH *' ' '
So m o l CO2 do axit tao ra = 0,3 - 0,1 = 0,2 mol ' ' ' " ' K h o i liTdng axit = 7,6 - 0,1.32 = 4,4 gam ' " '
CT a x i t : CnHjn+iCOOH coTo mol la x mol ' ' ' "
v a y : ( n + l ) . x = 0,2 va (14n + 46)x = 4,4 ,f .
T i m diTdc: x = 0,05 v^ n = 3 => axit la C3H7COOH '
1 ;;PhaniJng:C3H7COOH + CH3OH - > C 3 H 7 C O O C H 3 + H2O
m 0,05 0,05 •• .'W;
V i H = 80% nen k h o i liTdng este = 0,05.102.80% = 4,08 gam ^ '
* T H 2 ; Ancol c6 2 nguyen tilr C vay ancol la C2H5OH ''
il'j So m o l CO2 do axit tao ra = 0,3 - 0,2 = 0,1 m o l K h o i liTdng axit = 7 , 6 - 0,1.46 = 3 gam-'^ CT a x i t : CnH2„+iCOOH c6 so mol la x m o l
V 0 y : ( n + l ) . x = O , l v a ( 1 4 n + 46)x = 3 u u. i ', , - i i * ' I T i m difdc: x = 0,05 v^ n = 1 => axit \k CH3COOH (Loai v i cung s6' C v d i ancol)
* C a c h kh^c;
K h i dot chdy axit no ta difdc so mpl nufdc va CO2 bkng nhau, sif chenh l^ch so mol d6 chinh la do ruTdu gay nen => nn^^iu = 0,4 - 0,3 = 0,1 m o l .
nc02 = 0,4 => mc = 0,4.12 = 4,8 gam, nH20 = 0,3 =:> mw = 0,3.2 = 0,6 gam. < •; => mo/7,6 gam hon hdp = 7,6 - 4,8 - 0,6 = 3,2 gam => no = 0,2 m o l
M a ta c6 so m o l n/du = 0,1 => so mol O/rufdu = 0,1
=> so m o l O/axit = 0,2 - 0,1 = 0,1 r:i> so mol axit = 0,1/2 = 0,05. Gpi so C trong nTdu la n, so C trong axit 1^ m, ta c6 :
0 , l n + 0,05m = 0,3 => n + 0,5m = 3 v l n khac m = > n = l v a m = 4 V § y nTdu la CH3OH, axit la C3H7COOH
este la C3H7COOCH3
290
Phan iJng: C3H7COOH + C H 3 O H -> C3H7COOCH3 + H2O
0,05 0,05 V I H = 80% nen k h o i liTdng este = 0,05.102.80% = 4,08 gam => D a p a n A .
p^i 16. Hoa hdi 15,52 gam hon hdp gom mpt axit no đn chiJc X va mpt axit no da chiJc Y (so mol X Idn hdn so mol Y ) , thu difdc mpt the tich hdi bang the tich cua te, 5,6 gam N2 (do ciing trong dieu k i f n nhiet dp, ap suat). Neu dot chay toan bp hon
1 hdp hai axit tren thi thu dirdc 10,752 lit CO2 (dktc). CTCT cua X , Y Ian lifdt la: i Ạ C H 3 - C H 2- C O O H va H O O C - C O O H . i Ạ C H 3 - C H 2- C O O H va H O O C - C O O H .
| B . CH3- C O O H va H O O C-CH2-CH2- C O O H ;.,
|c. H - C O O H va H O O C - C O O H ' ^••'('•'\ 5>> \ D. CH3-COOH va H O O C-CH2- C O O H , ^ ,
''Trich de thi tuyen sink Dai hoc khoi A ndm 2011'' :'ns/.:;,:,:,„.vr-^-'\r.. . Giai
ra co: nhhaxii = "^^2 ^ ^ = 0,2 mol; n ^ o j = 0,48 mol.
so nguyen tu" C trung binh = -S£2^ - = 2,4 => loai Dap ^ n C.
'hh
it no đn X c6 so nguyen tuf C la n, axit no da chiJc Y la m v d i so m ian lifdi
.a X, y , .•. I
A x i t X (x m o l ) : n m - 2 , 4 2,4
A x i t Y ( y m o l ) : m | n - 2 , 4 Theo de bai ta c6 : x > y => (m - 2,4) > (n - 2,4) Gia suT đp an B dung ta cd hp:
m - 2 , 4 n - 2 , 4 m > n =^ loai Ddp A . 60x + 118y = 15,52 x - 1 , 3 6 y = - 0 , 5 6 I = > f ' = > L o a i B 2x + 4 y = : 0 , 4 8 = > D a p a n D . ^ ' • '
17. Trung hoa 3,88 gam hh X gom hai axit cacboxylic no, đn chiJc. mach hd b^ng đ NaOH, c6 can toan bp đ sau phan iJng thu difdc 5,2 gam muoi khan. b^ng đ NaOH, c6 can toan bp đ sau phan iJng thu difdc 5,2 gam muoi khan. Neu dot chdy hoan toan 3,88 gam X thi the tich oxi (dktc) can dilng la
A . 4,48 lit. B . 3,36 lit. C. 2,24 lit. D. 1.12 l i t .
' ' "Trich de thi tuyen sink Dai hgc khd'i A ndm 2011"
Gidi
I Cong thtfc Chung ci5a cic axit \h RCOOH . .M , ,
R C O O H + NaOH RCOONa + H2O i , 1 , ' '
Phan dgng va phuang ph&p giSi H6a hpe 11 HOu c o - DO Xtian Himg
DiTa v^o phUdng trinh phan tfng ta lháy: « 1 mol R C O O H 1 mol RCOONa khoi Itfdng tSng: 23 - 1 = 22(g)
V a y : x mol R C O O H x mol RCOONa khoi li/dng tSng 5,2 - 3,88 = 1,32 (g) 1,32 , 3,88 194
X = -— = 0 , 0 6 m o l M a x i , = =
22 0,06 3 >n n & i ' j< \ i . . : - ^
G p i cong thtfc chung cda 2 axit no đn chuTc la C- H 2 - O 2 : 0,06 m o l Ta c6: 14n + 32 = =^ n = - . „ , , ^ ,^ ., , ^ ^ , CiiH2n<^2 + ^~f=- O2 > n C 0 2 + n H 2 0 3 n - 2 0,06 ^ ^ ^ ^ i — ^ .0,06 2 túV ^ - _ 3 . ^ - 2 ••••
Vay: no2 = ^^^Y^ .0,06 = — . 0 , 0 6 = 0,15 m o l > =0,15.22,4 = 3,361ft D a p a n B . '
B a i 18. Hon hdp X gom axit axetic, axit fomic va axit oxalic. K h i cho m gam X tac
dung v d i N a H C O j (diT) thi thu diTdc 15,68 l i t khi CO2 (dktc). M a t kh^c, dot chay hôn toan m gam X can 8,96 l i t khi O2 (dktc), thu dUdc 35,2 gam CO2 va y mol H2Ọ Gid t r i c O a y m . • , P'',.-. • .-.j „ , ... ....wa ,M
:: A . 0,3. B . 0.8. : ] C. 0,2. D . 0,6.
r , . i V "Trkh de thi tuyen sink Dai hoc khoi A nam 20W
T a c 6 : iXm = 0 , 7 m o l ' . u
, Cong thrfc chung cua c^c axit la R(COOH)x foj, R ( C O O H ) , + x N a H C O j - > R(COONa), + x C O j + xHzO
0,7/x 0,7 ,
=^ "Oarongoxit) = ' *
A
M a t k h a c : r i r o = 0 , 8 m o l , n^, = 0 , 4 m o l
,, Sd do phan drng: A x i t + O2 - » CO2 + H2O âpd iSfrr
^ . . . i ^ . , . ^ i , , ' 1.4 0,4 0,8 y ^^v '•"••.mU-'.
BSotoan nguyen 16" O : M + 0,4.2 = 0,8.2 + y y = 0,6 mol => D d p &n D .
B a i 19. Dot chay hoan tôn x gam hon hdp gom hai axit cacboxylic hai chiJc, m?'-'^ hd va deu c6 mot lien ket doi C=C trong phan tiJf, thu difdc V lit. k h i CO2 (dk'^' V v^ y m o l H2Ọ B i e u thiJc lien h? giffa c i c gia tri x, y va V la
292 • • T O T O A . V = | i ( x - 3 0 y ) . C. V = — ( x + 3 0 y ) . 55 9 0 B. V = — ( x - 6 2 y ) 95 ^ , D . V = ^ ( x + 6 2 y ) .
" Trich de thi tuyen xinh Dai hoc khoi A nam 2011'' Gidi
Hai axit c6 3 lien ket 71, 4 0 ^ Cong thỉc chung cua cac axit tren la: C„H2n-404 ^: ; CnH2„-404 ^ n C 0 2 + (n - 2)H20
W
— ^ ' V - y
TO phifdng trinh ta thay: naxu = — ^ ^ ^ ^ na,i, =
2 2 2 122,4 T a c d : no(ọongaxit)= 4 . - V V ,22,4 ^, —Z ,22,4 M a t khac: m H C H c = x = mQ+ mQ+ m ^ x = ^^-'^^"^ ^^-^ V 28 V = — ( X + 30y) => D a p an C. 22,4 - y + 2y t
| i 20. Hon hdp X gom hai axit cacboxylic no, mach hd Y va Z (phan tuT khoi cua Y ' nho hdn cua Z ) . Dot chay hoan toan a mol X , sau phan iJng thu diTdc a mol H2Ọ
M a t khac, neu a m o l X tac dung v d i liTdng d\i dung dich NaHCOa, thi thu dUdc ^, 1,6a mol CO2. Thanh phan % theo khoi lifdng ctia Y trong X la
! A . 46,67% B . 40,00% C. 25,41% D . 74,59%
" Trich de thi tuyen sirih Dai hoc khd'i B ncim 2011" Gidi
2nH20 2.a ^ Ta c6: So nguy6n tuf H trung binh = = — = 2
"hhx a
=> ca 2 axit deu c6 2 H 66 la H C O O H : x m o l v^ H O O C - C O O H : y m o l Chon a = 1 ta co: <• : , ' ' ->'\rt-, ;<'V'
H C O O H + NaHC03 - > HCOONa + CO2 + H2O "^i^^>^
(COOH)2 + 2 N a H C 0 3 - > ( C 0 0 N a)2 + 2CO2 + 2H2O ^.
y 2y . ( J r H SKLU i T a c o h?: x + y = l x + 2y = l,6 x = 0,4 y = o,6 la's,' ' • mHcooH = 0,4.46 = 18,4(g); m(cooH)2 = 0.6.90 = 54 (g) , . ,„ , , % Y = — i ^ i ^ .1 0 0 % = 25,41% => D a p an C . 18,4 + 54 293
Plian dang va phuong pii^t, ^j..,. ; ị^.. .ii^^. : • : iị,. Hifng B A I T A P T R A C N G H I ^ M
C a u 1. Cho cac chat sau: propanal, propanol va gUxerol. Co the dung hoa chat nao
dirdi day de nhan bict diMc 3 c h ^ tren: , > <•! | s;
Ạ Cu(0H)2 B.AgNOj/NHj C. Hj/Ni D. NaOH
c a u 2. De tdch 6\idc andehit axetic c6 Ian ancol etyHc ngi/di ta c6 the dfing nhOng
hoa chat nao difdi day: , ., },
Ạ đ NaHSOj va HCl '"',"1, B. đ HCl va NaOH " '