- 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
4. Cho 0,86g mpt andehit đn chiJc n oX dem hidro h6a thudifdc mpt ancol Cho Y tac dung v d i natri thu drfdc mpt the tich hidro c6 kha nSng cong hdp vifa
dii v d i 112ml k h i elilen (dktc).
PhSn d?ng phoang phap giai H6a hpc 11 HQu cO - P 8 Xuan Hung
a) Xac djnh CTFT cua X . Viet cong thiJc cau tao va goi ten cac dong phan X .
b) Xac dinh cong thuTc cau tao dung cua X , Y biet k h i dun nong v d i H 2 S O 4 cjâ i70"C thi Y khong the tao thanh olefin.
Gidi ,
Dat C T T Q cua X : C„H2„+,CHO ' ,
Phifdng trinh phan iJng :
C„H2„ . , C H O + H2 ^ C„ H 2 „ . , C H 2 0 H ( Y ) 0.01 m o l 0,01 mol C„H2n+i C H 2 O H + Na > C„H2„+1 CHzONa + ^ H j t 0,01 m o l ' 0,005 m o l ' C 2 H 4 + H 2 C 2 H , ^ ^ ^ j J, , , ( , H , . , I ' 0,005 mol 0,005 mol 1, m ^ ^ A S6'molC2H4: nc2H4 = = ("^"D nnÓ 0,86 M x = ^ = 8 6 T a c 6 : 1 4 n + 30 = 86 =>n = 4 , H . ) 0 . ' f I , ! V a y C T P T X : C4Fi;CH0. ««' H ( I, ; C T C T X : C H 3 - C H 2 - C H 2 - C H 2- C H O :pentanal C H 3- C H- C H 2- C H O : 3-metyl butanal • - i , * i ' ' C H . - C H . - C H - C H O : 2-m6'lyi butanol ' I CH3 ' • C H - i - C - C H O ' : 2,2-dimetyl propanal. C H 3
b) Thco de thi C T C T dung cua X la : C H 3- C - C H O
C H 3 ':LrfHnJ.|RM, M i c i$i:H)-
C H 3
=> C T C T Y : C H 3- C- C H 2 O H ( V I Y > khong tao olefin).
Bai 5. Hai chat hffu cd A va B (dcu chiJa C. H , O) la dong đng ke tiep. Th^nh phi" % khoi liTdng o x i trong A , B Ian liTdt la 53,33% va 43,24%.
.in
a) Xdc dinh cong thỉc phan tuT c6 the c6 cua A , B .
b) Hay viet cong thiJc cau tao cua A , B , biet A , B deu phan iJng v d i Na v^ deu cho phan iJng trang gUdng. ^ ' . -
(Trích de thi CDSP TPHCM)
Gidi
a) Dat C T T Q A : C.HyO,; B : C,HbO, . ^' ' - " H' KV), H'>
Theo de : % 0 (A) = — X 100% = 53,33% M A = i^^l^ ( i ) 53,33
% 0( B ) = — X 100% = 43,24% r > M B = i^ ^ i ^ (2)
M B 43,24
V i A , B la dong dang ke tiep nen M Q = M A + 14
162.100 , < p n ) M M a t k h a c t a c d : ^ = ^P^ = i ^ ' , , (1) 16Z.100 M A , , . f • o . . . u : n ' ' ' ' ' « - Hay ^ = 1,233 => M A ± i l. i , 2 3 3 ^ | ^ - = ' ' ' ' , M A 43,24 M ^ [ M B = 7 4 \ A = 60 => 12x + y + 16z = 60 , , . ^ 1 => 12x + y = 44 y = 44 - 12x f t w 1 . j. r X 1 2 . 3 4 y 32 20 8 - 4
(loai) (loai) (nhan) (loai)
=> CTPT A : CsHxO; B : C 4 H i ( ) 0 :..,u .-,,0 Hij~H(ý •
Neu z = 2 => 12x + y = 28 ^ y = 28 - 12x ,..^.^,jtj,,j,j,ai,%t^;ql.'|,f;.^-^
1 2 3
16 4 - 8 ,, . ^ H ? ' : ;
(loai) >: (nhan) -.y^w) (loai) jv. ' j ' ' CTPT A : C 2 H 4 O 2 ; B : C3H6O2 i
N e u z = 3 12x + y = 12(loai). . i ; , ^ - ^ ,
b) V i A , B + Na => A , B phai CO H linh dong - , . , i i i / i „ ( h V
A, B cho phan u:ng trang giTdng => A , B phai CO nhom-CHO I, , r , ,
V§y CTPT A la C2H4O2. C T C T : C H 2- C H O . i
OH
CTPT B la C3H6O2. C T C T : C H 3- C H - C H O ; C H j - C H j - C H O OH O H
B^i 6. Cho 1 mol axit axetic tdc dung v d i 1 m o l ancol propylic. O tai nhỉt dO can b^ng se dat difdc khi c6 0,6 mol este tao thanh. Néu sau do cho them 1 m o l axit
luM uaiiij phuong phip Q\i\a hoc 11 HOu c a - P5 Xuan Hung
axelic thi thanh phan ve so mol cac chat trong hon hOp sau khi can b i n g mdi thanh lap Ik bao nhieu 7 Biét r i n g hhng so toe do cua phan uTng thuan gap 2,25 Ian hhng
SO toe dp cua phan i^ng nghich. • r ' . i , . » f, . j , , : \
Giai
PhiTdng trinh phan iJng :
C H 3 C O O H + C H 3 - C H 2 - C H 2- O H C H 3 C O O C H 2 C H 2 C H 3 + H 2 O
Can bang 0,4 0,4 i : „ . r^M 0,6
Theode : K,h= —!- = 2,25 - "''^
Neu them vao 1 mol axit axetic => So mol axit axetic : k
" C H j C O O H =^1 + 0,4 = 1 , 4 (mol) if A '"/ C H 3 C O O H + C 3 H 7 O H C H 3 C O O C 3 H 7 + H 2 O Bandau 1,4 0,4 0,6 0,6 sdji 1,^4 Phanu-ng x x ' x Canb^ng ( 1 , 4 - x ) ( 0 , 4 - x ) (0,6+ x) (0,6+ x) [ C H 3 C O O C 3 H 7] . [ H 2 Q ] ^ (0,6 + x).(0,6 + x) ^ ^ 25 - ^•':7 lCH3COOH].[C3H70H] (1,4 - x).(0,4 - x) * i . ^ " A * ' ' =>x = 0,18(mol) ,,,,„ ,,„,.„^, „ m^M:-O^^T
So mol C H 3 C O O H : ticHjCOOH = 1.4 - 0,18 = 1.22 (mol) ' ' ' - "'•'''^
So mol este : n c H j C O O C s H y = 0 , 6 + 0,18 =0,78 (mol)
So mol H 2 O : nH20 = 0,6+ 0,18 =0,78 (mol) |; . i t
So mol C3H7OH : n c 3 H 7 0 H = 0,4 - 0,18 = 0,22 (mol). H, j A VVlO c.::
ai 7. Cho hdp chat hiJu cd X (phan tu* chi chu-a C, H, O va mpt loai nhom chiJc). Xac dinh cong thiJc cau tao cua X, biét 5,8g X tac dung vdi dung dich AgN03 trong NH3 tao ra 43,2g Ag. Mat khdc 0,1 mol X sau khi hidro hda hoan toan phan iJng vCfa du vdi 4,6g Nạ
(Trkh TSDHQG TPHCM) Gidi
V i X + A g N 0 3 / N H 3 -> A g (43.2g) => X la andehit ' '
Dat C T T Q cua X : CxHy(CHO)„ hay R(CHO)„ ''^^^^^
PhiWng trinh phan u-ng : -'-^ •' . i - V ' ' ( - • r'' - ' 'ÓM^'' i» ^
R(CHO)„ + 2nAgN03 + SnNHj + nHj > R(C00NH4)n + 2 n A g i + 2nNH4N03 — mol ' 1 ; , r t 0,4 mol n So mol Ag : n A g = — = 0,4 (mol) ' ' " • "''"^'"-'íWl \:--KM:-^^'SS. R4 ' ' R ( C H O ) „ + n H 2 R C C H z O H ) 0,1 mol R( C H 2 0 H ) „ + nNa 0 , 1 mol -)• R ( C H 2 0 N a ) „ + - H j t 2 0 , 1 mol So mol Na : n N a = 4,6 23 0,ln mol = 0,2=:>0,ln = 0,2=>n = 2 Trong X C O 2 nhom - C H O X c6 dang R(CH0)2
Ta C O : Mx = 5,8 0,2
yf,. ,
= 29.n = 2 9 x 2 = 58 « R + 2.29 = 58 => R = 0
CHO
Vay CTPT X : (CH0)2. CTCT : I andehit oxalic. CHO
Bai 8. Z la mot axit hiJu cd đn chiJc. De dot chay 0,1 mol Z c i n 6,72 lit O2 (dktc) Xac dinh cong thiJc phan tuf, cong thiJc cáu tao va goi ten Z.
(Trich TSDHDLKTCN Gidi
Dat CTTQ Z : C.H^Oz (Z : axit đn chiJc)
PhiTdng trinh phan iJng : C x H y 0 2 + ' y ^ x + ^ - l 1? * 1 0,1 mol 0,3 mol O2 > X C 0 2 + ^ H 2 0 2 6,72 S^ mol O 2: n o , = -22,4 — = 0,3 (mol) Ta cd ti 1§ : 0.1 x + ^ - 1 o x + — = 4 =>y + 4 x = 1 6 = > y = 1 6 - 4 x X t. 1 3 4 y 12 8 4 0
(loai) (loai) (nh|n) (loai)
Vay CTPT Z:C3H402 :"'^^'V V'
CTCT : C H 2 = C H - C 0 0 H : axit acrylic. * ^ " ' , V'
Bai 9. Hdp chat hffu cd X (chi chiJa C, H, O). T i k h ^ i c i a X so vdi H2 b^ng 30. X khong tac dung vdi Na de giai phong H2, X tdc dung vdi AgNOs trong dung dicl
N H 3 giai phong ra Ag. Viet cong thtfc cau tao cua X va cac dong phan ciia X, Cho biel iinh chat hda hpc d|c tnfng cua cdc dong phan n^ỵ
(Trich TSDHBCVT} Gidi
M x = 3 0 x 2 = 60 ' J:
D a t C T T Q X : CJlyO,. T a c6 : 12x + y + 16z = 60 N e u z = 1 => 12x + V = 44 X 1 2 3 4 " " y 32 (loai) 20 (loai) 8 (n a m 4 n ) (loai) H2 X khong c6 H linh dpng. ^' " * - ^ X + AgNOj/NHs >• A g => X c6 nh6m - C H O nghia Ik c6 1 lien k^t n
V a y tnrdng hdp z = 1 (loai) - N e u z = 2 12x + y = 28
X 1 ' 2 3
y 16 4 a m
(loai) (nhan) (loai)
C T P T X : C2H4O2 , ~, , <;.n ":1ui^f'}:ih 0,! n'i;;( i i ^ f i J C I K i i ! A . « ! = ;
<^ v „ _ uaoa dn0> 3feX
O H '
D i e u ki$n bai thi C T C T X la H C O O C H 3 . j ; , ^ , , r ) J l , : ): S 0 H : »m
- N ^ z = 3 = > , 2 x - . y = 1 2 ( l o a i ) n ^ r ' ^ d a h i gn^AnH Phildng trlnh phan i J n g : ^, ? "i ' ' : ; ; V
H C O O C H 3 + 2 A g N 0 3 + 4NH3 + 2H2O > (NH4)2C03 + 2 A g i +
+ 2NH4NO3 + C H 3 O H H o c s i n h trlnh bay tinh chat h6a hoc ciia c a c dong phan X . H o c s i n h trlnh bay tinh chat h6a hoc ciia c a c dong phan X .
Bai 10. C o n g thtfc cua X c6 dang (CHjCDn va cua Y c6 dang ( C H z O ) ^ . H a y bỉn lu§n de t i m C T P T cua X , Y . B i e t :
X + N a O H U A va Y + H2 A A + C u ( 0 H ) 2 > dung dich mau xanh lam.
X:(CH2Cl)„hayCnH2„Cln
D i e u kiOn : 2n + n < 2n + 2 =:> n < 2 - N e u n = 1 => C T P T X : C H 2 C I (loai) - N e u n = 2 ^ C T P T X : C2H4CI2
1^ijf)rf'> Uo) X to yt)!'! ihh: qt>H, At . • —• Xv' "'^^ ^ '-'^^ '^"^^^'^'^ C T C T X : ' """ .^.^S/^;,#^.$'?^>';'q;j%'*.^ C H 2 - C I , jvj j;^f4 jifjil o d " ' i-r X + N a O H ^ A ] Y + H 2 > A J i j ' i => Y c 6 2nguyentijrC=i> m = 2 Z86
X va Y phai cCing so nguySn tijf C
V $ y C T P T Y : C2H4O2
Y + H2 > A => A la C2H6O2 "*,, ^ • : • • • "] - ^ • • • A + Cu(OH)2 > dung dich mau xanh Ihm A + Cu(OH)2 > dung dich mau xanh Ihm
=> A la ancol no da chiJc c 6 2 nhom O H ot 2 C ke can nhaụ C H 2 - O H C H O C T C T A la I ^ C T C T Y la I
C H 2 - O H C H 2 O H j:., Í
PhiWng trlnh phan iJng : X (1# W " ' • '
C H 2 - C 1 ,0 C H 2 - 0 H í, ,
2 + 2 N a O H ^ I ^ + 2 N a C I ''*vv^«^ ' •
C H 2 - C I C H 2 - O H .J,,,, , , , , . ,. C H O ^ C H 2 - O H .-ri;'/;;;:! JiiKK.;ii7«j.5^ ' ' C H O ^ C H 2 - O H .-ri;'/;;;:! JiiKK.;ii7«j.5^ ' '
C H 2 O H ^ ~ t ° ^ CH2-OH;>^^>:;? ,i^.>;..fi
• (A) ' ' . ' ^ u ,
C H 2 - O H ^ „ H .
^ „ + C u ( O H ) 2 > I 2 " " ^ C u 1 2 + 2H2O - . ™ 2 - O H , , ^ ^ 2 - 0 - ^ - ^ - ^ 2
dung dich mau xanh lam
Bai 11. H o n hdp X g o m 1 ancol va 2 san pham hdp niTdc cua propen. T i khoi hdi cua X so vdi hidro b^ng 23. C h o m gam X di qua ong su" difng C u O (du") nung n6ng. S a u khi c a c phan iJng xay ra hoan toan, thu diTdc h6n hdp Y gom 3 chat hCTu c d va hdi niTdc, khoi luTdng ong svt giam 3,2 gam. C h o Y t^c dung hoan toan vdi liTdng dir dung dich A g N 0 3 trong NH3, tao ra 48,6 gam A g . Phan trSm khoi lifdng cua propan-l-ol Irong X l a : , , . ;:; > > .., t r . . . ,
A . 6 5 , 2 % . B . 16,3%. C . 4 8 , 9 % . D . 83,7%.
„ ;„,{ ,,. (;, J " Trich de thi tuyen sinh Dai hoc khoi Bnam 2010''
Gidi
• Hai san pham hdp niTdc cua propen 1^ C2H5CH2OH v^ C H 3 C H P H C H 3 (M = 60) T a c6 : M X = 2.23 = 46 => Trong hh X c6 C H 3 O H
" Khói liTdng ong sỉ di/ng C u O giam 3,2 g chinh la khoi lUdng nguyen to oxi lham gia phan úng
l a CO : nx = no(CuO)= — = 0,2_mol J_Q « ^ c t S.O " G o i a la so mol c u a C3HKO -> n^njon = 0,2 - a ^ , .3 ,
T a c o : 3 2 ( 0 . 2 - a ) + 60a = 46.0,2 r:> a = 0,1 . q nb^!'
n C H 3 0 H =0.1 "^"1 . $ 1 1 ..(í'
f at khac: HAg = 0,45 m o l • m 'f '^V' * ^ '
Phan d?ng va phaang ph^p gi^i H6a hpc 11 HQu c o - D5 XuSn Hang
Sd do pu": C H 3 O H —» HCHO —> 4Ag 0,1 0,1 0,4 mol 0,1 0,1 0,4 mol
C2H5CH2OH —> C2H5CHO 2Ag
0,025 mol 0,05
=> %C2H5CH20H = Ml^J^. 100% = 16,3%
^ => = 48,6/108 = 0,45 mo]
Dap an B. n
46.0,2