Môn Hóa - Tuyển chọn, phân loại các dạng bài tập Hóa hữu cơ - Bài tập toán (Tái bản lần thứ hai, có sửa chữa, bổ sung): Phần 1

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Tài liệu Tuyển chọn, phân loại các dạng bài tập Hóa hữu cơ - Bài tập toán gồm 6 chuyên đề và 200 bài toán chọn lọc có lời giải. Phần 1 tài liệu trình bày 3 chuyên đề: Xác định công thức cấu tạo và thành phần khối lượng của ancol, Anđehit, axit cacboxylic. Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Môn Hóa - Tuyển chọn, phân loại các dạng bài tập Hóa hữu cơ - Bài tập toán (Tái bản lần thứ hai, có sửa chữa, bổ sung): Phần 1

BAN GIAO VIEN CHUYEN HOA NANG KHIEU TRUdNG THI NGO NGOC AN TUYEN CHON, PHAN LOAI CAC DANG BAI TAP HOA HlfU Ctf BAI T A P TOAN • GOM 6 CHUYEN DE VA 200 B A I TOAN CHON LOC VA LCJl GlAl > BOI Dl/C(NG v A N A N G CAO > L U Y E N THI DAI HOC vA CAO DANG (Tdi ban Ian thti hai, c6 su'a chiXa, bo sung) NHA XUAT BAN DAI HQC SJ PHAM
LCfl N O I D A U Be giiip da cdc em hoc sinh cd them tdi lieu tham khdo khi luyen thi vdo cdc triidng Dai hoc vd Cao dang, chung tdi xin trdn trong giai thieu quyen 'Tuyen chon, phdn loai cdc dang bdi tap Hod hoc HQu ca Icfp 12 - Phdn Todn". Quyen sdch he thong nhitng bdi tap Todn da ra trong cdc ki thi tuyen sinh vdo cdc trudng Dai hoc vd Cao dang theo nhUng phdn sau : Phdn I: Xdc dinh cong thiic cdu tgo vd thdnh phdn khoi lUgng cua ancol. Phdn II : Xdc dinh cong thiic cdu tgo vd thdnh phdn khoi lugng cua andehit. Phdn I I I : Xdc dinh cong thiic cdu tgo vd thdnh phdn khoi liigng cua axit cacboxylic. Phdn TV: Xdc dinh cong thiic cdu tgo vd thdnh phdn khoi liigng cua este. Phdn V : Xdc dinh cong thiic cdu tgo vd thdnh phdn khoi liigng cua ainin - aminoaxit. Phdn VI : Xdc dinh cong thiic cdu tgo cua gluxit vd thdnh phdn khoi lUgng cua chung. Viec bien sogn dii tl mi, can than den ddu cung khong trdnh khoi nhUng sai sot ngoai y muon, tdc gid mong dugc nhdh tdt cd mgi y kien dong gop, phe blnh xdy diing tii phia cdc bgn dong nghiep vd cdc em hoc sinh de quyen sdch ngdy mot tdt han. TAG GIA
Chuyen de 1 xAc D|NH C O N G TUXSC C A U TAO VA T H A N H P H A N K H O I LI/0NG C U A A N C O L BAITAP 1. Hon hdp A gom axit hOu cd X va este Y cua mot axit hOu cd d d n chufc. Lay a gam hon hdp A chc phan Lfng du vdi dung dich NaOH : chUng tach h6n hdp san pham ta thu du'dc 9,3 gam met hon hdp chat hOu cd B va 39,4 gam hdn hdp muoi hOu cO l 2); M : k h o i lUdng m o l p h a n tuf cua B. PhUorng t r i n h p h a n ufng v d i N a : R(OH), + Na > R(ONa), + - H 2 T 2 1 mol — mol 2 ^ m o l 0,15 m o l M 3 36 So m o l H 2 d dktc : n ^ ^ = — — = 0,15 mol = IIQ 22,4 M x i i = 0,15 => M = M x = 31x M 2 0,3 5
93 „ => a + 0,3 = 0,4 theo (3), (4) => a = 0,1 mol Vi M < 93 31x < 93 X < — = 3 31 So' mol axit X = 0,1 m o l ; so' mol este Y = 0,15 mol. Chon X = 2 ^ M = 31x = 31 X 2 = 62 c) Cong thiCc cdu tao cua cac chat X, Y : => R(0H)2 = 62 => R + 34 = 62 K h o i lircrng h a i muoi : 0, IMujcooNa), + 0. SMRcooNa = 39,4 =^ R = 1 2 x + y = 28=i. x = 2;y = 4 =^ R + 67z + 3(R' + 67) = 394 Cong thijfc cua ancol B : CH2OH-CH2OH (etylen glycol). R + 3R' + 67z = 193 b) So mol cac chat X, Y trong a gam hdn hap A : V i chi t h u diTcfc m o t hidrocacbon D n e n : RH^ c h i n h Ik R ' H hay : V i este Y + N a O H cho mot ancol B la C2H4(OH)2 nen Y la este da chiJc R + z = R' + l ^ R ' + 1 + 3 R ' - z + 67z = 193 cua mot axit don chiJc R C O O H v6i C2H4(OH)2. • z = 1 -> R' = 31,5 (loai) Dat cong thiifc axit X : R ( C 0 0 H ) 2 va este Y : R ' C O O C H , * z = 2-> R' = 15=> 12x + y = 1 5 => x = l;y = 3 I R'COOCHa R' la C H 3 - Phifcrng trinh phan iJng tao este Y : = > R + 2 = 15 + 1 =>R=14 => 12x + y r z l 4 => x = l;y = 2 2 R ' C 0 0 H + HOCH2-CH2OH > R'COOCHa + 2H2O R la - C H 2 - • Cong thiJc cau tao cua axit X : HOOC-CH2-COOH R'COOCHa • Cong thiJc cau tao ciia este Y : CH3COO-CH2-CH2-OOC-CH3 1 mol 1 mol d) Cac phan ling cua X vai B theo tl le 1 : 1 : 0,15 mol ny mol -COOH HO-CH2 •COOCH2 ny = 0,15 mol CH, CH, + 2H2O COOH HO-CH2 COOCH2 D a t a la so mol axit R(COOH),. COOH 'HO-CH2'' -COOCH2 PhUcfng t r i n h p h a n ling ciia h 6 n hap A v d i N a O H : nCH, + n I nCH I + 2nH20 COOH HO-CHp -COOCH, R(COOH), +zNaOH R(COONa), + ZH2O (1) T r o n g m o t b i n h kin d u n g tfch la 11,2 lit chufa h o n h d p h d i 3 a n c o l d d n a mol a mol chOfc A , B, C v a 13,44g O2, n h i e t d o v a a p s u a t t r o n g b i n h la 1 0 9 , 2 ° C v a R'COOCH, + 2NaOH ^ 2 R ' C 0 0 N a + C2H4(OH)2 (2) 1,4 a t m . B a t tia l i f a d i e n d o t c h a y h e t a n c o l , s a u d o diJa n h i e t d o b i n h v e I R'COOCHa 136,5°C, a p s u a t t r o n g b i n h l u c n a y la P. 1 mol 2 mol C h o t a t c a k h i t r o n g b i n h s a u k h i d o t c h a y Ian l a d t di q u a b i n h m o t d U n g H2SO4 d a c v a b i n h h a i d y n g KOH d a c . S a u thf n g h i e m t h a y k h o i l U d n g 0,15 mol 2 X 0,15 mol b i n h m o t t a n g 3,78 g a m , c o n b i n h hai t a n g 6,16 gam. PhUcfng t r i n h p h a n iJng nung h a i muoi hufu ccf v6i v o i , N a O H a) T f n h a p s u a t P. R(COONa), + z N a O H > RH, + zNa2C03 (3) b) X a c d i n h c o n g thufc p h a n tCf c a c a n c o l A , B, C , b i e t r a n g B, 0 c 6 c u n g s o a mol a mol n g u y e n X(i c a c b o n v a s o m o l c u a a n c o l A b S n g 5/3 t o n g s o m o l c u a c a c R'COONa + NaOH > R'H + Na2C03 (4) ancol B va 0 . 0,3 mol 0,3 mol (Dai hgc Hang hai ptiia Nam - nam 1995) T o n g so' mol RH^ va R ' H cf dieu k i e n tieu chuan : Giai PV 1,1 X 8,96 a) Tinh dp suat P : n = = 0 , 4 (mol) RT 22^ 1,4x11,2 273x1,1 ^ (3 ancol oxi luc dau) — = 0 , 5 (mol) 273 0,082.(273 + 109,2) 7
So mol O2 = 13,44 : 32 = 0,42 So mol t r u n g b i n h cua h5n hcrp 2 ancol B, C : M(B+C) - 777;^ = 3 So mol (A + B + C) 0,5 - 0,42 = 0,08 MB < 59,3 < Mc ' So mol A = - so mol (B + C) => so mol A = 0,05 3 CH3OH + -O2 > CO2 + 2H2O So mol (B + C) = 0,03 2 H2SO4 dac h u t h o i nudc : CO2 + 2K0H > K2CO3 + H2O 0,05 mol 0,05 m o l So mol H2O = 3,78 : 18 = 0,21 CHpOH + 1^^-P^02 — . aC02 + P ^ H 2 0 So mol CO2 = 6,16 : 44 = 0,14 So mol O2 t r o n g C O 2 = so mol C O 2 = 0,14 X mol ax mol So mol O2 t r o n g H 2 O = - so mol H 2 O = 0,105 C , H , O H H- i ^ - ^ ^ O2 _ > aC02+i|^H20 V i ancol l a don chufc nen : y mol " ay mol So m o l CO2 = a(x + y) + 0,05 = 0,14 So" mol O2 t r o n g ancol = - so' mol h 8 n hop.ancol = 0,04 X + y = 0,03 2 0,03a = 0,14 - 0,05 = 0,09 => a = 0,09 : 0,03 = 3 So m o l O2 dir = 0,42 + 0,04 - 0,14 - 0,105 = 0,215 Co 3 ancol a = 3 : C3H7OH C3H5OH va C3H3OH S6' m o l h 6 n hop k h i c6 t r o n g b i n h sau k h i dot la : M = 60 M = 58 M = 56 0. 14 + 0,21 + 0,215 = 0,565 mol M(B.C) = 5 9 , 3 -> C p h a i l a C3H7OH Vohh = (0,565 X 22,4) l i t d 0°C, 1 a t m Vhh = Vbinh = 11,2 l i t d 136,5°C, p ? B CO the la C3H5OH hoac C3H3OH. 1. ( 0 , 5 6 5 x 2 2 , 4 ) ^ P x l l ^ ^ P . 1,695 a t m . CH3OH + - O 2 > C O 2 + 2H2O 273 409,5 2 b) Xdc dinh cong thUc A, B, C : 0,05 mol 0,1 mol 0,14 mol C O 2 => mc = 0,14 x 12 = l , 6 8 g C3H7OH + - O2 — > 3CO2 + 4H2O 0,21 m o l H 2 O => mn = 0,21 x 2 = 0,42g 2 mo t r o n g ancol = 0,04 x 32 = l , 2 8 g X mol 4x mol => m(A + B + c) = 3,38g C3H5OH + 4O2 > 3CO2 + 3H2O 2 gg y mol 3y mol So mol t r u n g b i n h ciia h 6 n hop 3 ancol A , B, C : M ( A B O = —— = 42,2 0,08 C3H3OH + - O2 > 3CO2 + 2H2O Qua day t a t h a y i t n h a t p h a i c6 m o t ancol c6 M < 42,2 2 Ancol no don chiJc CH3OH = 32; C2H5OH = 46 y mol 3y mol 2y mol A n c o l k h o n g no c6 1 n o i doi : C n H 2 n - i O H (n > 3 -> M > 58) Neu B l a C3H5OH t h i : Ancol k h o n g no c6 1 n o i ba : CnHzn-aOH (n > 3 ^ M > 56) So mol H 2 O = 4x + 3y = 0,21 - 0,1 = 0,11 Vay m p t t r o n g 3 ancol p h a i l a CH3OH - > A (vi B v a C cCmg so nguyen X + y = 0,03 => x = 0,02; y = 0,01 (thoa m a n ) tijf cacbon). Neu B l a C3H3OH t h i : m c H a O H = 0 , 0 5 x 32 = 1,6g So m o l H 2 O = 4x + 2y = 0,21 - 0,1 = 0,11 m ( B , c ) = 3 , 3 8 - 1,6 = l , 7 8 g X + y = 0,03 => X = 0,025; y = 0,005 (thoa m a n )
Vay, A : CH3OH, C : CIHTOH, B : C,iHr,OH hoac C3H3OH. b) Viet cac phan ufng dieu che X di tCf butan va tCr este c6 slin trong tu nhien. Cdch tinh khdc di tim ancol B vd C : (Dai hoc Kien true l-la Noi - nam 2000) D a t cong thiJc cac ancol B , C : B : C^HyOH, C ; C^HyOH Gidi Cong thiJc chung ciia h a i ancol l a : C ^ H - O H , v d i y l a so' nguyen tuf H a) Goi cong thufc p h a n tuf ciia X l a C x H y C O H ) ^ trung binh. P h a n ufng chay : Taco: M C ^ H . O H = 1 2 + y + 1 7 = 59,3 ^ y = 43,3 - 12x y +a a C,Hy(OH)a + X + O2 > XCO2 + ^^HaO 2 2 Vay: X 1 2 3 • y +a a y 30,3 18,3 6,3 1 mol X + mol X mol 4 2 Suy r a so' nguyen tuf C : x = 3 0,05 mol 0,175 mol 0,15 mol so' nguyen tuf H : y < y < y' hay y < 6,3 < y' 0,15 -> X = = 3 Ancol 3C dan chufc no t h i y' = 7 n h a n v i y' > 6,3 0,05 Ancol 3C dcfn chufc chua no c6 y < 6,3 t h i n h a n y = 5 va y 3 y +a a 0,175 X + - = 3,5 V a y B va C CO h a i cap n g h i e m : C3H7OH va C3H5OH 4 2 0,05 C3H7OH va C3H3OH V i X l a ancol no nen y + a = 2x + 2.Thay vao ta c6 : H6n hcfp A , B , C gom : 2x + 2 a a = 3 Nhom 1 : CH3OH; C3H7OH; C.iHsOH X + = 3, 5 4 2 Nhom 2 : CH3OH; C3H7OH; C3H3OH Cong thufc phan tuf ciia ancol X : C3H5(OH)3 Cho 11 gam hon hdp hai ancol no don chQc ke tiep nhau trong day dong Cong thiJc cau tao cua ancol X : C H o - C H - C H , : Glixerol dang, tac dung het vol Na thi thu dude 3,36 Ift H j (dktc). Xac djnh cong I I I OH OH OH thiJc phan tuf va cong thilc cau tao cua hai ancol. (Dai hoc Y Ha Noi - nam 1997) b)- Dieu che' X tii butan : Crackinh Gidi CH3-CH2-CH2-CH3 > CH4 + CH2=CH-CH3 D a t cong thufc chung h a i ancol l a : C - H g - ^ j O H 500" C C H 2 = C H - C H 3 + CI2 > CH2=CH-CH2C1 + H C l Cn%.iOH + Na — ^ C,H,,^,ONa + iH,T CH2=CH-CH2C1 + CI2 + H2O > CH2-CH-CH2 + HCl CI O H CI (14n + 18)g 0,5 mol C H 2 - C H - C H 2 + 2NaOH C H 2 - C H - C H 2 + 2NaCl llg 0,15 mol I ' I I ' I ' I I CI O H CI OH OH OH 0,15(14n + 18) = 1 1 x 0 , 5 - D i l u che X ti^ este c6 sSn trong tiT nhien : 1 < n = 1,3 < 2 Dun nong dau, md dong vat vdi xut : Cong thiJc cua h a i ancol l a : CH3OH va C2H5OH. CHo-OOCR CH2-OH R-COONa Dot chay hoan toan 0,05 mol ancol no X mach hd can 5,6 gam oxi tao ra I I 6,6 gam CO2. CH-OOCR' + 3NaOH >CH-OH + R'-COONa a) Hay xac djnh cong thufc cau tao cua X. CH2-OOCR" CH2-OH R"-COONa 11
Dun nong hon hop hai ancol mach hd vdi H 2 S 0 4 dac dupe h6n hdp cac M,3 ^ = 174 ete. Lay X la mot trong cac ete do dem dot chay hoan toan thi ta c6 ti le 0,05 H x : n o 2 ^HH^O = 0 . 2 5 : 1 , 3 7 5 : 1 . D a t cong thufc p h a n tuf ciia B : C^HyO^ Mat khac khi cho axit A la dong dang cua axit oxalic tac dung vdi mot 12x + y + 64 = 174 trong hai ancol tren khi c6 mat H2SO4 dac lam xuc tac dapc este B. De 12x + y==110 xa phong hoa hoan toan 8,7 gam este B can 200ml NaOH 0,5M. N g h i e m t h i c h hop : x = 8; y = 14 Tim cong thiJc cau tao cua hai ancol va axit A. Cong thufc p h a n tuf ciia B : C8H14O4. (Dai hgc Kien true Ha Noi - nam 1999) B la este no 2 I a n este. V i axit no nen ancol cung p h a i la ancol no : Gidi R-CHvOH Goi cong thiJc p h a n tijf cua ete X la C^HyOz (x, y, z nguyen difcfng) Cong thufc cau tao ciia B : CH3COO-(CH2)4-COOCH,3 P h a n ufng dot chay : Cong thufc cau tao cQa axit A : H O O C - ( C H 2 ) 4 - C O O H (axit adipic). y z Mot este mach hd c6 toi da 3 chufc este. Cho este nay tac dung vdi X + XCO2 + - H 2 O (1) dung dich KOH thu dUdc muoi va 1,24 gam hai ancol cung day dong 4 2 dang. Neu lay 1,24 gam hai ancol nay dem hoa hPi hoan toan, thi thu y z 1 mol X + mol X mol ^mol dUPc hdi CO the ti'ch bang the ti'ch cua 0,84 gam N 2 (do 6 cung dieu kien 2 4 2 nhiet dp va ap suat). Tim cong thQc phan tCr hai ancol. 0,25 mol 1,375 mol 1 mol 1 mol (Dai tioc Quoc gia TP.I-ICM - nam 1998) 1 X = x = 4; — y = X = 4 —> y = 8 Gidi 0,25 2 So mol 2 ancol = so' mol No = ^' = 0,03 mol y z X + 1 28 4 2 ^ z = 1 — 1 24 1,375 0,25 Mruou = - — = 41,33 0,03 Vay cong thufc p h a n tuf cua ete X : C^HsO => Co 1 ancol la CH3OH (x mol) X la ete dcfn chufc, p h a n tiif c6 mot n o i doi, n e n no p h a i difofc tao t h a n h => Ancol con l a i : CnH2n+iOH (y mol) tif m o t ancol no dcfn chufc va m o t ancol don chufc c6 n o i doi. => x + y = 0,03 Cong thufc p h a n tuf ciia ancol c6 n o i doi : CnH2nO (n > 3) Truang hap 1 : Cong thufc p h a n tuf cua ancol no : CmH2m+20 (m > 1) Co 2 chufc este => so' mol ancol bang nhau => x = y = 0,015 Vay cong thiic cau tao ciia X : C H 2 = C H - C H 2 - 0 - C H 3 va cua h a i ancol : C H 3 - O H va C H 2 = C H - C H 2 - O H . 32x + (14n + 18)y = 1,24 ^ n = 1,3 (loai) Cong thufc cau tao cua A : H 0 0 C - ( C H 2 ) n - C 0 0 H (n > 1 ) Triforng hap 2 : Goi ancol p h a n ufng v d i A l a R - O H ( R = C H - hoac C3H5-) Co" 3 chufc este => x = 2y hoac y = 2x (CH2)n(COOH)2 + 2 R 0 H > (CH2)n(COOR)2 + 2H2O (2) * X = 2y ^ X = 0,02; y = 0,01 Este B : (CH2)n(COOR)2 + dd N a O H : 32 X 0,02 + (14n + 18).0,01 = 1,24 (CH2)n(COOR)2 + 2 N a O H > (CH2)n(COONa)2 + 2 R 0 H (3) n = 3 : C3H7OH HNaOH = 0,2 X 0,5 = 0,1 (mol) * y = 2x ==> y = 0,02; x = 0,01 32 X 0,01 + (14n + 18).0,02 = 1,24 = ^iiNaOH = ^ = 0,05 (mol) => n = 2 : C2H5OH. 13
Cho mot hon hdp X gom 3 ancol AOH, BOH, B'OH trong do AOH va g^n t r e n day C, hay n o i m o t each khac t r o n g day C c6 1 l i e n k e t doi. BOH cung day dong dang, BOH va B'OH c6 cCing so cacbon va mach B ' O H l a 1 ancol don chiJc chufa no c6 m o t l i e n k e t doi n e n c6 cong thiJc cacbon thang. Dun nong 30,2 gam hon hdp X vdi mot lUdng thQa C n H 2 n - i O H (vi B O H va B ' O H c6 cung so nguyen tuf C). C H 3 C O O H C O xuc tac H2SO4 dac thi thu 6U0c 51,2 gam h5n hdp 3 este 40 (hieu suat 100%). Neu dot chay 6,04 gam hon hdp X thi thu diidc 13,64 nc„H2„_i0H = "Brj = ^ = 0.25 (mol) gam CO2. Neu lay 30,2 gam hon hdp X tac dung vdi dung dich brom C „ H 2 n - , 0 H + Br2 > C„H2n-,(0H);, + 2NaBr thay CO 40 gam Br2 tham gia phan u'ng. Lay san pham chQa brom dem thuy phan bang kiem thi thu dudc ancol 3 Ian ancol. Cac phuong t r i n h p h a n iJng dot chay X : a) Tinh khol lUdng phan tilf trung binh cua hon hdp X. Xac dinh cong thiJc CH3OH + - O2 > CO2 + 2H2O phan tCf cua 3 ancol, biet rang trong 3 ancol c6 1 ancol la metylic. Tfnh 2 so mol moi ancol. X mol X mol b) Goi D la ancol dong dang ke tiep vdi AOH. Cho D tan vao H2O thu dUdc 3n CnH2„+iOH + — O2 > nCOa + (n - DHjO 82 gam dung dich D. Cho Na da vao dung dich D thu dUdc 33,6 Ift H2 (dktc). Khoi lUdng rieng cua ancol D nguyen chat la 0,8 gam/ml, cua H2O 2 y mol ny mol la 1 gam/ml. Tinh do riidu - Cho biet khi D tan vao H2O the tfch dung dich xem nhif khong doi. 3n-l CnH2„.:0H + O2 > nC02 + n H 2 0 (Dai hoc Tong hap TP.HCM - khoi A, B - nam 1995) Gidi 0,25 mol 0,25n mol a) Khoi lugng phan tii trung binh cua hdn hap X (AOH, BOH, B'OH) : So' m o l CO2 tao t h a n h k h i do't chay 30,2g h o n hop X : D a t cong thiJc chung ciia 3 ancol l a R O H . 30,2x13,64 ^ ,^ neon = = 1,55 (mol) 6,04x44 PhiTcfng t r i n h p h a n l i n g este hoa h 6 n hop X : => X + n y + 0,25n = 1,55 CH3COOH + ROH > CH,,COOR + H2O T o n g so mol h a i ancol CH3OH va CpHvn-lOH : x + y = 0,5 - 0,25 = 0,25 a mol a mol a mol a mol X +y = 0, 25 1.3 - 0, 25n He phu'ong t r i n h : y = Goi a l a so mol cua h o n hop 3 ancol ( A O H , B O H , B ' O H ) X + n y + 0, 25n = 1,55 n-1 Theo d i n h luat bao toan k h o i lucfng : B i e n l u a n : 0 < y < 0,25 ^CHaCOOH + ™ROH = ^CHaCOOR + "^HaO 1,3-0,25n ^ 60a + 30,2 = 51,2 + 18a => 0,5 mol n-1 V i C n H 2 n + i O H la ancol dong dang ciia CHjOH nen n > 1 hay n - 1 > 0. K h o i luong p h a n tuf t r u n g b i n h ciia 3 ancol ; M = = ^ = 60,4 * 0 < 1,3 - 0,25n ^ 0,25n < 1,3 => n < 5,2 a 0,5 Cong thdc phan td * 1,3 - 0,25 < 0,25n - 0,25 =^ 1,55 < 0,5n 3,1 < n Theo de b a i c6 m o t ancol la ancol metylic nen A O H la CH3OH va B O H => 3,1 < n < 5,2. Vay chon n = 4 va n = 5 la dong dang ciia C H 3 O H n e n B O H l a m o t ankanol c6 cong thiJc t o n g - Vdi n = 4 : quat CnH,„„OH. 1,3-0,25x4 V i h 6 n hop X tac dung du'oc v6i dung dich b r o m nen t r o n g dung dich X 4-1 = 0,1 X = 0,25 - 0,1 = 0,15 (mol) p h a i CO ancol chu'a no. Vay B ' O H l a m o t ancol chua no. - Vdi n = 5 : dd kiem 1,3-0,25x5 B ' O H + Br2 > san p h a m cong > ancol c6 3 n h o m - O H tufc l a y = = 0,0125 => X = 0,25 - 0,0125 = 0,2375 (mol) CO 2 n h o m - O H m d i t h e m vao, san p h a m cong p h a i c6 2 nguyen tuf B r 5-1 15
V a y c6 h a i n h 6 m n g h i e m : - Lay phan 1 cho tac dung vdi dung dich AgNOa trong N H 3 da thu dU(?c 'CH3OH :0,15mol CH3OH 0, 2375 mol 64,8 gam Ag. (I) C4HgOH : 0,10mol (II) C5H11OH 0,0125 mol - Lay phan 2 dem trung hoa, can 30ml dung djch KOH 2M. Tfnh hieu suat C4H7OH : 0,25mol C5H9OH 0, 2500 m o l qua trinh oxi hoa A. b) Do ruqu khi cho D tan trong nUac : - Lay phan 3 cho tac dung vdi B, c6 xuc tac thfch hop thu dUdc chat D ch? D l a dong dang ke t i e p cua CH3OH nen D la C2H5OH. chLfa 1 loai nhom chufc. Tinh kho^ li/Ong cua B phan ufng, hieu suat la 100%. Goi z, t la so mol C2H5OH H2O t r o n g dung dich C2H5OH. (Dai hoc Kien true TP.HCM - nam 1995) PhUcfng t r i n h p h a n iJng : Gidi ) D a t cong thufc cua h a i ancol l a R O H va R'(OH)x. C2H5OH + N a -> CaHsONa + - H2 2 V i t i k h o i cua B so v RONa + -H2T (1) 2 t mol — mol MA (gam) 0,5 mol 2 0^ 0.5a 33,6 a (gam) mol T o n g so' mol H2 tao t h a n h = 1,5 M, 22,4 R'(OH)x + xNa > R'(ONa), +-H2t (2) - + - = 1,5 z + t = 3 2 2 4 , 2 5 M A (g) 0,5x mol T o n g so k h o i liTcfng dung dich D : 46z + 18t = 82 0,5xa [z + t = 3 a(g) Vay : 4, 2 5 M A |46z + 18t = 82 z = lmol; t = 2mol ^7 16-- . 0,5xa 16 0,5a niCaHgOH = 1 ^ 46 = 46 (gam); mn^o = 2 x 18 = 36 (gam) V i VH„(9I = — V i T n\n = 46 H2(2) H2(i) 4,25MA 17 MA V , ancol nguyen chat — = 57,5 (mol) 0,8 u X 16 Vadancoi = 57,5 + 36 = 93,5 (ml) hay =— => X = 4 57,5x100 4,25 17 Do ancol cua dung dich : = 61,5". Phan ling dot chay A : 93,5 13 6 Mot hon hop X gom 2 ancol no mach thang A va B, trong do A la ancol V i t i le kho'i lufong la 1 : 1 nen m A = me = —'— = 6,8 (gam) don chLfc. Khoi laong A va B bang nhau. Cho X tac dung het vdi Na thi 2 the tich H2 sinh ra tCf B bang 16/17 the tich H2 sinh ra tC/ A (do d cung D a t A , B la CnH2n+20 va C„H2™+204 t a c6 : dieu kien nhiet do va ap suat). Mat khac khi dot chay het 13,6 gann hon 14m + 66 = 4,25(14n + 18) hay m = 4,25n + 0,75 (3) C„H2n+20 + — O2 > nC02 + (n + 1)H20 hop X ttiu dapc 10,36 lit CO2 (dktc). 2 a) Tim cong thCfc phan tQ cua A, B, biet rang tl khoi hOi cua B so vdi A la 4,25. MA (g) n mol b) Oxi hoa 19,2 gam A c6 xuc tac thich hop. Chia hon hop sau phan Gng (hon hop C) lam ba phan bang nhau. 6,8 (g) mol MA 6 17
C,„H2„+20 + 5 ^ - - O2 > mCOa + ( m + D H g O Phdn ling C + KOH : 2 HCOOH + KOH > H C O O K + H^O 4 , 2 5 M A (g) m mol y i mol y i mol ^ „ . s 6,8m , y, = 0,03 2 = 0,06 (mol) (KOH) 6,8 (g) —• mol X Thay y i vao (5) t a c6 : Xi = 0,12 mol T a c6 : + MHI . J L = 0,4625 (mol CO^) H i e u suat p h a n iJng oxi hoa C H 3 O H : H% = = — = 90% MA MA 4,25 0,6 0, 2 6,8m 1 _ 0, 4 6 2 5 M A - 6,8n 3 MA 4,25 ~ MA Phdn ling C + CrJIi-jO^ : hay 1,6m = 0,4625 (14n + 18) - 6,8n C5H8(OH)4 + 4HC00H > (HCOO)4C5H8 + 4H2O 1,6m = 8,325 - 0,325n (4) M£ mol 0,06 mol G i a i he (3), (4) t a dugc : n = 1 (CH3OH) 4 m = 5 (C5H12O4) mach t h ^ n g mu = X 136 = 2,04 (gam). CH3OH : metanol 4 CH2-CH-CH-CH-CH3 : pentantetraol - 1, 2, 3, 4 0 6 m o t h d p c h a t hQu c d d d n chufc Y, khi d o t c h a y Y t a ch? t h u d U d c CO2 I M I I v a H2O v d i so m o l nh\J n h a u v a s o m o l oxi t i e u t o n g a p 4 Ian so m o l c u a Y. OH OH OH OH X a c d j n h c o n g t h i l c p h a n tuf, c o n g thiJc c a u t a o m a c h h d c u a Y b i e t r a n g CH2-CH-CH-CH2-CH2 • pentantetraol - 1, 2, 3, 5 Y l a m m a t m a u d u n g djch b r o m v a khi Y c o n g h d p hidro thi dUdc a n c o l d d n OH OH OH OH chiJc. V i e t c a c p h U d n g t r i n h p h a n i j n g x a y ra ( g h i ro d i e u k i e n p h a n Cfng). CI H 22 - CI H - C H 22- CIH - C H (Dai hoc Xay dUng - nam 1999) I 22 : pentantetraol - 1, 2, 4, 5 OH OH OH OH Giai {Chil y : k h o n g the c6 2 n h o m O H l i e n k e t v d i 1 nguyen t i i cacbon). Theo de bai, t r o n g p h a n tuf Y so nguyen tiJf H gap h a i Ian so' nguyen tuT C. Vay cong thufc p h a n tuf ciia Y la C x H 2 x O z . b) 19,2 gam CH3OH :r 0,6 mol Phuong t r i n h p h a n i l n g do't chay Y : D a t X i va y i la so mol CH3OH da b i oxi hoa de tao r a - h 5 n hcfp C 3 CxH2xO, + ^—-=- O2 > XCO2 + XH2O theo cac phifong t r i n h : 2 [01 CH3OH > HCHO So mol O2 = = 4 3x-z = 8 2 Xl Xi V i Y la hop chat don chiJc n e n z = 1 hoac 2. [0] CH3OH > HCOOH Khiz = l x = 3 (CsHgO) yi yi Khi z = 2 => X= — (loai) • Phan ling C + AgNOs : 3 H C H O + 4AgN03 + 6 N H 3 + 2H2O > (NH4)2C03 + 4Ag + 4 N H 4 N O 3 Vay cong thijfc p h a n tuT cua Y : C3H6O Xl moi 4 x i mol Cong thiJc cau tao mach h d : H C O O H + 2AgN03 + 4 N H 3 + HgO > (NH4)2C03 + 2Ag + 2NH4NO3 CH2=CH-CH2-OH (1) CH2=CH-0-CH3 (2) y i mol 2yi mol CH3-CH2-CHO (3) CH3-CO-CH3 (4) 4 x ! + 2yi = 0,6 mol (Ag) (5) Theo de b a i , chi c6 cong thilc (1) l a thoa m a n . 19
— L a m mat mau dung dich Br2 : Ap dung he qua cua dinh luat thanh phan khong d5'i : C H , = C H - C H 2 - O H + Bra > CHaBr-CHBr-CHa-OH x = 8; y = 18-4 = 14; 7= 7- 2 = 5 Cong thilc phan tut ciia (A) : C^HuOr, — Cong hop hidro : Cong thufc cau tao ciia (A) : C H 2 = C H - C H 2 - O H + H2 > CH3-CH2-CH2-OH. HO-CH2-CH2-COO-CH2-CH2-COO-CH2-CH3. I. Chat A chiJa hai loai nhom chQc, c6 kha nang tac dung vdi Na. b) HO-CH2-CH2-COO-CH2-CH2-COO-CH2-CH3 + 2H2O > Thuy phan hoan loan 0,1 mol (A) can 0,2 mol H2O thu dUdc 18 gam (B) > 2CH2-CH2-COOH +CH3-CH2-OH va 4,6 gam (D). Cong thufc nguyen cua chat (B), chat (D) Ian luot la (C3H603)n va (CaHeO)^. Ti le mol giufa (A) tham gia phan ufng va (B) sinh OH ra la 1 : 2. CuO,t° Cho (B) 6 the hOi di qua 6'ng su" dang CuO nung nong, hdi san pham HO-CH2-CH2-COOH + [OJ > H - C - C H 2 - C O O H +H2O (B') thoat ra khoi ong c6 kha nang tham gia phan ufng trang bac. 0 a) Dinh cong thilc phan tif, cong thilc cau tao cua (A), (B), (D). HOOC-CH2-CHO + 2AgN03 + 4NH3 + H2O > CH2(COONH4)2 + b) Viet cac phUdng trlnh phan ufng minh hoa. + 2NH4NO3 + 2 A g i (Dai hoc Y Duac TP.HCM - nam 2000) HO-CH2-CH2-COO-CH2-CH2-COO-CH2-CH3 + Na > Giai > N a O - C H 2 - C H 2 - C O O - C H 2 - C H 2 - C O O - C H 2 - C H 3 + i Hat a) =i => ne = 2 X HA = 2 X 0,1 = 0,2 ne 2 1. Mot hdp chat hiJu cd X chCfa 10,34% hidro theo khoi lUdng. Khi dot chay 1Q X chi thu dUdc CO2 va H2O vdi so mol nhu nhau va so mol O2 tieu ton MB = = 90 = 90n ^ n = 1 gap 4 Ian so mol cua X. Xac dinh cong thiJc phan tif, viet cong thiJc cau 0,2 tao cua X, biet rang khi X cong hdp H2 thi dUdc ancol ddn chQc, con khi Cong thiJc phan tuf ciia (B) : C3H6O3 cho X tac dung vdi dung dich thuo'c ti'm thi thu dUdc ancol da chufc. TCi X Cpng thu-c nguyen cua (D) : (CaHeO)^ hay C^JAemOu.. Viet cac phUdng trlnh phan Qng dieu che axit propenoic. Dieu kien : 6m < 2 x 2m + 2 hay m < 1 chi nhan m = 1. (Hoc vien Krthuat Quart sU va Dai lioc Da Ning - nam 1995) Cong thiifc phan tuf ciia (D) : CaHgO. Gidi Theo gia thiet, (A) phai la tap chufc trong do c6 chufc este nen cong thilc C H O , + ^""'^"^^02 > XCO2 + JH2O cau tao cua (D) la C H 3 - C H 2 - O H . 4 2 . 4 6 Ta CO : x = 0,5y y = 2x So mol (D) sinh ra : — = 0,1 2x 10,34 „ 46 = —'-— x = 3 14x + 16z 100 Vi (D) la ancol nen (B) phai la axit (it nhat trong (B) phai c6 nhom -COOH) 6x - 2z , , =4 z = 1 (B) phai CO chufc ancol bac I 4 Cong thufc phan tuf cua X la CaHgO. Cong thilc cau tao ciia (D) : H O - C H 2 - C H 2 - C O O H Tir de ra, X chi c6 th§ la : C H 2 = C H - C H 2 0 H Dat cong thuTc phan tijf tdng quat ciia (A) : CxHyO,, ta cd ti le mol : Ni,t» "A : HH^O : nB : no = 0,1 : 0, 2 : 0,2 : 0,1 = 1 : 2 : 2 : 1 Vi : CH2=CH-CH20H + H2 > CH3-CH2-CH2OH Tif ti le tren, ta viet dugc phuong trinh phan ufng thuy phan (A) : 3CH2=CH-CH20H + 2KMn04 + 4H2O > > 3CH2OH-CHOH-CH2OH + 2Mn02i + 2 K 0 H C , H y O , + 2H2O 2C3H6O3 + C2H6O3 + CaHeO 21
P h a n ufng tiT X ^ p r o p e n o i c : Gidi xt.t" D a t cong thurc p h a n tijT cua h a i ancol l a : R - O H va R - C H 2 O H . C H 2 = C H - C H 2 0 H + O2 > C H 2 = C H - C O O H + H^O PhUcfng t r i n h p h a n ufng v d i N a : 12. Mot ancol A mach hd, khong lam mat mau niJdc brom. De dot chay a lit hdi ancol A thi can 2,5a lit oxi d cung dieu kien. R-OH + Na > R-ONa + - H2T (1), a) Xac dinh cong thufc phan tuf va cong thufc cau tao cua A . 2 b) TL/ metan va cac chat v6 cd thi'ch hdp, hay dieu che A . R-CH2OH + Na R-CH2-0Na + - Hat (2) 2 , (Dai hgc Nong nghiep I - nam 1997) TU (1) va (2) t a t h a y tir 1 mol ancol chuyen t h a n h 1 mol ancolat n a t r i Gidi k h o i luctng t a n g 23 - 1 = 22 (gam). a) A l a ancol no : CnH2n+20n,. Vay so m o l ancol : n, = ^ ' ^' = 0,08 (mol) CnH2„^20„ + ^" ^ ~ O2 nC02 + (m + IM^O 22 2 a) Theo (1), ( 2 ) : V„ = ^ — x 22,4 = 0,896 (lit). 1 mol 0,5(3n + 1 - m) m o l a mol 2,5a mol — 2 84 (Vi t i le hori ancol / O2 = t i le so mol ancol / O2) b) Kho'i liTofng mol t r u n g b i n h cua h a i ancol : M r = —— = 35,5 Vay : 0,5(3n + 1 - m) = 2,5 => 3n - m = 4 (1) 0,08 Goi cong thiic p h a n tuT chung cua h a i ancol l a C - H - O H G i a i (1) : n = 2; m = 2. ' X y Ta CO : 12x + y = 35,5 - 17 = 18,5 A la : H O - C H 2 - C H 2 - O H b) Co t h e dieu che A tir CH^ theo cac p h a n iJng : Phiroing t r i n h n a y chi thoa m a n k h i 1 < x < 2 1500-1800°C Vay x i = 1 t a c6 ancol CH;,OH 2CH4 > CH^CH + 3H2 Idin lanh n h a n h X2 = 2 t a c6 ancol C2H5OH. /Br Br^ c) Tir C H 4 dieu che : - C H ^ C H + 2Br2 CH-CH > as i Br-" --Br Ancol metylic : C H 4 + CI2 > CH3CI + H C l j Br^ /Br HQ /OH ; _ ^^^^^ _ . CH3CI + N a O H > CH3OH + N a C l ] ^CH-CH + 4NaOH > CH-CH + 4NaBr Br/ ^Br HO^ | ^OH 1500°C Ancol etylic 2CH4 C H = C H + 3H2 CH=0 l^m lanh nhanh CH=0 Pd CH=CH + H2 -> CH2=CH2 CH=0 Ni,t" CHoOH I + H. > I ^ CH=0 " CH2OH CH2=CH2 + HOH CH3-CH2-OH 14. Cho V lit (d dieu kien tieu chuan) hon hdp khi gom 2 olefin lien tiep nhau 13. C h o 2,84 gam mot hon hdp hai ancol ddn chufc la dong dang lien tiep trong day dong d i n g hdp nude (c6 H 2 S O 4 loang xuc tac) thu dUdc 12,9 nhau tac dung vdi mot li/dng Na vC/a d u , tao ra 4,6 gam chat rSn va V li't khi H2 6 dktc. gam hon hdp A gom 3 ancol. Chia A thanh hai phan deu nhau. a) Tinh V. - Phan mot dem dun trong H 2 S O 4 dac 6 140°C thi thu du'dc 5,325 gam B gom 6 ete khan. Xac dinh cong thiJc cau tao cua cac olefin, cac ancol b) Xac dinh cong thilc phan \\i cua hai ancol tren. va cac ete. c) Viet sd do dieu che moi ancol tir C H 4 . - Phan hai dem oxi hoa bang O2 khong khi 6 nhiet do cao (c6 Cu xuc tac) (Dai lioc Nong nghiep - I
tac dung vdi AgNOg trong dung dich NH3 du (hoSc AggO trong dung dich Goi X , y l a so mol C H 3 - C H O va C H 3 - C H 2 - C H O NH3 du) th] nhan dUdc 17,28 gam Ag kim loai. Ti'nh % khdi lUdng cua 17 28 m6i ancol trong A va ti'nh gia trj V. Theo (6) va (7) t a c6 : n A g = 2(x + y) = —'•— = 0,16 (mol) 108 Neu cho them 0,05 mol mot ancol no ddn chufc, bac mot khac vao phan -> X + y = 0,08 = t o n g so' mol ancol bac I hai roi tien hanh phan Ong oxi hoa bang O2 khong khf, sau do phan (ing trang bac nhu tren thi se nhan dUdc bao nhieu gam Ag kim loai ? ^ n a„eoi bac 11 = ' 0,08 = 0,045 (mol) (Gia thiet hieu suat cua tat ca cac loai phan Ong noi tren deu dat 100%). T d n g so' gam ancol bac I p h a n 2 l a : (Dai hoc Xay dUng - nam 1999) 46x + 60y = - (0,045 x 60) = 3,75 (gam) Giai 2 Goi cong thiic p h a n tijf chung cua 2 olefin l a C - H 2 - Ta CO he phuorng t r i n h : x + y = 0,08 40x + 60y = 3,75 Phin 1: C„-H,, + H O H - i ^ C , H , , ^ ^ O H (1) G i a i he phiTomg t r i n h t a duoc : x = 0,075; y = 0,005 2C„-H,,^,0H -M2!c, C,H,,^^-0-C„-H,,^, . H,0 (2) Vay % k h o i luomg cac ancol t r o n g A l a : Theo (2) : so mol HgO tao r a d p h a n 1 l a : %(CH3CH20H) = 2^0'0'^^^46 ^ ^QQ^^^ ^ 53,49% 12,9 %(CH3CH2CH20H) = 2 ^ Q ' 0 0 5 x 6 0 ^ ^^^^^ ^ ^ ^^^^ "H.0 = ^ = 0.0625 (mol) 12,9 lo nancoi = 2nH20 = 2 X 0,0625 = 0,125 (mol) %(CH3CHOHCH3) = 41,86% Tinh gid tri cua V : — 12 9 - — Vay Mnrou = = 51,6 14n + 18 = 51,6 -> n = 2,4 Theo (1) so' mol olefin = so' mol ancol = 0,25 2x0,125 V = 0,25 X 22,4 = 5,6 ( l i t ) • ni =2 CO olefin : CH2=CH2 - K h i t h e m 0,05 m o l ancol no den chiJc, bac I khac vao p h l n 2 se c6 2 • n2 = 3 CO olefin : CH2=CH-CH3 t r u d n g hop xay r a : Cac ancol : CH3-CH2-OH (Ri-OH) • Neu ancol t h e m l a CH3OH tao r a andehit H C H O : C H 3 - C H 2 - C H 2 - O H (R2-OH) 2CH3OH + O2 > 2HCH0 + 2H2O (8) CH3-CH-CH3 (R3-OH) H C H O + 2Ag20 > CO2 + H v O + 4 A g i (9) OH Theo (8) va (9) : m A g = 0,05 x 4 x 108 + 17,28 = 38,88g Cacete: Ri-O-Ri; R2-O-R2; R3-O-R3 • Neu ancol t h e m l a R-CH3OH tao ra andehit khac HCHO : R1-O-R2; R1-O-R3; R2-O-R3 2R-CH2OH + O2 > 2R-CH0 + 2H2O (10) Oxi hoa p h a n 2 b^ng O2 k h o n g k h i (xiic tac Cu, t°) : R C H O + Ag20 > 2RC00H + 2Agi (11) 2CH2-CH2-OH + O2 > 2CH3-CHO + 2H2O (3) Theo (10) va (11) : m A g = 0,05 x 2 x 108 + 17,28 = 28,08g. 2CH3-CH2-CH2-OH + O2 > 2CH3-CH2-CHO + 2H2O (4) 15. Khi phan tfch chat hQu cd A chi chufa 0 , H , O thi c6 : mc + mn = 3,5mo. 2CH3-CHOH-CH3 + O2 > 2CH3-CO-CH3 + 2H2O (5) a) Tim cong thiJc ddn gian cua A. b) Lay hai ancol ddn chiJc X, Y dem dun nong vdi H2SO4 dam dac d nhiet P h a n ling t r d n g bac cua cac andehit : dp thich hdp thi thu dUdc A. Xac dinh cong thJc cau tao mach hd cua A, CH3-CHO + Ag20 CH3COOH + 2 A g i (6) X, Y biet rang A la ete. CH3-CH2-CHO + Ag20 CH3-CH2-COOH + 2Agi (7) (Dai lioc Quae gia TP.HCM - nam 2000) , 25
Gidi Gidi a) D a t cong thufc chat hflu co A : CxHyO,,. Dat m la so' mol ciia chat B : 12x + y = 3,5 X 16z Cho z ^ I -> 12x + y = 56 -> X = 4; y = 8 A : C^HgO C.HA ^ "^"!~'^o, xCOi; + ^^HzO 2 Cho z = 2 ^ 12x + y = 112 ^ X = 8; y = 16 A : CgHigO^ v . v . . . 4x + y - 2z Cong thufc don gian cua A : C ^ H B O . m mol .m mol mx mol 0,5ym mol b) AA = 1 CO m o t l i e n ket n, A la ete vay cong thiJc cau tao ciia A : C H v = C H - C H 2 - 0 - C H ; j , X va Y la 2 ancol don chiJc dun nong c6 H2SO4 0, 896 So mol O2 da p h a n iJng : l ^ i l - Z — x m = zizz:^ = 0 04 (1) dam dac cho A nen : 4 22,4 Cong thiJc cau tao cua X : C H a ^ C H - C H ^ - O H K h o i lucfng (CO^ + H2O) = 44mx + 9my = 1,9 (2) Cong thufc cau tao ciia Y : CH.^-OH. K h o i lirgng (C + H ) = 12mx + my = 0,46 (3) 5. Ancol no A c6 cong thiCc tong quat CnH2n+20m- Xac dinh cong thufc phan Tir (2), (3) : m x = 0,035 (4) tu" cua A. Biet rang dot chay 1 mol A can 2,5 mol oxi. Tir (3) ; m y = 0,040 (5) (Dai hoc Thuy san - nam 2000) mx 0,035 X 7 Gidi Ti^ (4) va (5) : -> —= — my 0,040 y s ' 3n + 1 - m ' CnH^n+'iOni + O2 > n C O . + (n + DHzO V i cong thufc phan t i i cung la cong thufc don gian, thay x = 7, y = 8 vao (1) : 4 X 7 + 8 - 2z f 3n + 1 - m X m = 0,04 1 mol mol (m < n) 2 , Tii (5) : m = 0,005 ^ z = 2. 1 mol 2,5 mol a) Cong thu-c p h a n tiir ciia B la CvHsOa ( M = 124) 3n 1 - m = 2,5 m = 3n - 4 a = 124 X 0,005 = 0,62 (gam) n 1 2 3 b) Cho a gam 0,05 mol chat B : m < 0 2 1 5 C7H8O2 + N a > H2 C2H«0, CI H o2— C, H o2 0,005 mol 0,005 mol OH OH (Vi so mol H2 b ^ n g so mol B nen t r o n g B phai c6 2 n h o m O H ) Mot hdp chat B chiJa C , H, O c6 cong thiJc phan tLf trung vdi cong thufc C7H8O2 + NaOH > CvHyOaNa + H.O ddn gian nhat. Khi phan tfch a gam B, thay tong khoi lUdng cacbon va hidro 0,005 mol 0,005 mol trong do la 0,46 gam. De dot chay hoan toan a gam nay can 0,896 lit O2 (Vi N a O H CO so mol phan ufng b ^ n g so mol B, nen trong B phai c6 chufa (d dktc). Cac san pham cua phan ufng chay dUdc hap thu hoan toan khi phenol). Vay : cho Chung di qua binh dUng dung dich NaOH du', thay khoi ladng binh Cong thufc cau tao ciia B : tang them 1,9 gam. OH OH OH a) Xac dinh gia tri a va cong thufc phan tuf cua chat B. 1-CH2OH b) Xac dinh cong thufc cau tao cua B, biet r§ng khi cho a gam chat do tac o dung het vdi natri, ta thu diiOc khf hidro bay ra; con khi cho a gam chat B "•CH.OH tac dung vua du vdi dung d|ch NaOH 0,01 M thi so mol NaOH can dung CH20H c) V H , = 22,4 X 0,005 = 0,112 ( l i t ) bang so mol hidro bay ra d tren va cOng bang so mol cua B da phan ufng. c) Tinh the tich khf hidro (6 dktc) va the tfch dung dich NaOH da dung. 0,005 V ,d d N a O H " = 0, 5 (lit). (Dai hoc Quoc gia Ha Noi - i
J. Chia hon hop (X) gom ancol etylic, andehit axetic, axit axetic thanh ba 4 6 x ( V i - V2) phan bang nhau : T r o n g X c6 : n^CaHrjOH - 3 gam (6) 11,2 - Cho phan thuf nhat tac dung vdi Na kim loai du, thu dudc V, lit khi H 2 . 44a - Phan thuf hai cho phan Cfng vdi lUOng dU AgNOa trong dung djch N H 3 , ™CH3CHO = 3 gam (7) 216 thu dUdc a gam Ag kim loai. - Phan thu: ba phan ufng vC/a du vdi 200ml dung dich Na2C03 nong do C M 6OV2 mCHgCOOH = 3 gam (8) (mol//) lam thoat ra V2 lit khf CO2. 11,2 a) Viet cac phUdng trinh phan ufng xay ra. c) Tinh CM cua dung dich Na^COs : b) Lap bieu thu'c de tfnh khoi li/dng cac chat trong hon hop (X) theo V i , a T h a y (4) vao (3) va thay V2 = 0,448 : va V2. 0,5 X V2 = CM C M = 0,1 mol/Z c) Tinh thanh phan phan tram theo khoi iJdng cCia cac chat trong h6n hdp 11,2x0,2 (X) va nong dp CM cua dung djch Na2C03 khi : Thay V , , V2, a bang so vao (6), (7), (8) ta diTOc : V, = 1,008 lit; a = 6,48 gam; Vg = 0,448 lit. 46.(1,008 - 0 , 4 4 8 ) ™C2H50H - = 2,3g 38,20% (Cac the tich khi do 6 dktc). 11,2 (Dai hoc Hue - nam 1997) 44 X 6,48 ^CHiCHO - = l,32g 21,93% Gidi 216 60x0,448 a) D a t X , y, z la so mol cua C2H5OH, CH3CHO, CH3COOH. = 2,40g 39,87%. ™CH.,C00H - 1X2 C2H5OH + N a > CaHsONa + - H a t 19. Mot hon hop X (gom ancol metylic va mot ancol D trong day dong dang X mol 0,5x mol ancol etylic) daoc chia thanh ba phan bang nhau. Phan I tac dung vdi Na d i / giai phong 0,672 lit khi d dktc. Phan II sau khi chuyen hoan toan CH3COOH + N a > CHaCOONa + - H , t thanh cac andehit tUdng ufng, tac dung vdi AgNOa dU trong dung dich 2 NHa giai phong 19,44 gam bac. San .pham dot chay cua phan III dUdc z mol 0,5zjnol trung hoa hoan toan vQa het vdi 0,5 lit dung dich NaOH 0,6M. So mol H 2 : (x + z) = ^ ' (1) Xac dinh thanh phan phan tram theo so mol mol ancol trong X va cong 11,2 thufc phan \\i cua ancol D. CH3CHO + 2AgN03 + 2NH3 + H 2 O > CH3COONH4 + 2Ag + 2NH4NO3 (Dai ligc Nong ngliiep - I y = — (2) Goi cong thufc p h a n tuf cua ancol D la CnH2n+iOH. 216 Cac phaong t r i n h p h a n ufng : So mol NazCOa = 0,2CM - Phan I : 2CH3COOH + Na2C03 > 2CH3COONa + CO2T + H 2 O z mol 0,5z mol 0,5z mol CHaOH + N a -> CHaONa + - H 2 t (1) 2 So mol Na2C03 = 0,5z = 0,2CM (3) C„H2n.iOH + N a > CnH2n.iONa + ^ H 2 t (2) So mol CO2 : z = (4) 11,2 Phan I I : b) Tir (1) va (4) : x + z = ^ x = — — ^ (5) H C H O + 4AgN03 + 6NH3 + 2H2O (NH4)2C03 + 4NH4NO3 + 11,2 11,2, + 4Agi (3) 8 29
C „ ^ , H 2 n - i C H O + 2 A g N 0 3 + 3NH3 + H , 0 —> C„_,H,„_iCOONH4 + 21.a) Lay hai axit A,, A2 c6 so C trong phan tCf bang nhau (A, la dong dang cua axit C4H6O2) cho tac dung vdi hai ancol don chiJc Ri, R2 dUdc hai + NH4NO3 + 2 A g i (4) este dong phan tiidng ufng la Ei, E2. Khi dot chay hoan toan 0,01 mol - Phan III : hon hdp hai este trong binh kin the tich 10 lit bang ludng khong khi gap CH3OH + -O2 > CO2 + 2H2O (5) doi If thuyet thay ap suat trong binh sau phan iJng 6 0°C la 1,97 atm. Tim cong thufc phan ttf este, biet so C trong phan tuf nho hdn 8. Nhan CnHan.iOH + - ^ 0 2 > nCOa + (n + DHaO (6) xet dac diem cau tao cac chat Ai, A2, Ri, R2, E,, E2. b) Khi hidro hoa a mol h6n hdp hai ancdl Ri, R2 roi khii nUdc hoan toan san CO2 + 2 N a O H > Na.COs + HvO (7) pham sinh ra dude mpt anken c6 so mol nho hOn a va khi d o hoa c6 dot Goi X , y l a so mol ancol CH3OH va ancol D t r o n g X. nong anken nay se dUdc mpt san pham duy nhat chJa mpt nguyen tCf do 1. , 0 , 6 7 trong phan tCf. Xac dinh cong thifc cau tao cac axit, ancol va este da cho. Theo (1), (2) : n H , = ^ (x + y ) = ^ — = 0,03 (mol) ^ 66 22,4 (Dai iioc Bacli kiioa TP.i-iCM - nam 1993) 1,, „ , 19,44 Theo (3), (4) : n ^ , = - (4x + 2y) = = 0,18 (mol) Gidi ^ 3 108 a) Cong thiic phan tvC cua este : Theo (5), (6), (7) : n c o , = ^ ( x + ny) = I n ^ . o H = ^ V i 2 axit Aj, A2 va 2 ancol Ri, R2 deu l a dcfn chiJc n e n 2 este E,, E2 G i a i he 3 phifong t r i n h t r e n t a t i m difcfc : x = 0,09; y r= 0,09; n =4 cung l a este d o n chufc. Ngoai r a 2 este Ei, E2 l a 2 dong p h a n n e n E,, E2 CO cung so nguyen tiJf C. Vay ^fncH,0H = 50% = Dat cong thiJc p h a n tur cua 2 este : CxHyO^ Cong thiJc p h a n tiif cua ancol D : C4H9OH. Axit picric (2,4,6-trinitrophenol) dUcfc dieu che bang cacti cho phenol tac C.H,0, x . ^ - 1 0. -> XCO2 + -H2O 2 dung v6i hon hdp gom HNO3 dam dac va HzS04 dam dac (xuc tac). Viet phaong trinh phan Lfng. Cho 141 gam phenol tac dung vdi h6n hdp gom 0,01 mol 0,01 x + ^ - 1 0 , 0 I x mol 600 gam dung dich HNO3 6 3 % va 750 gam dung dich HNO3 dam dac. 4 Tfnh khoi lUdng axit picric sinh ra va nong dp phan tram cua HNO3 con dU can dot chay 0,01 mol h d n hop 2 este : trong dung dich sau phan Lfng. Biet rang phan iJng xay ra hoan toan. (Dai hoc Quoc gia TP.HCM - dot 2 - nam 2000) no, =0,01 x + ^ - 1 Gidi Phirong t r i n h p h a n i J n g (Gia suf axit picric k e t tua ho^n toan) : n khong khi theo l i thuyet = 5 x 0,01 x . ^ - 1 1"*! 1 ^ n 600x63 ^ , ncsHsOH = = 1' 5 ; "HNO3 = ^QQ^gg = 6 (mol) Vi lifong k h o n g k h i t r o n g b i n h gap doi lUcJng k h o n g k h i l i t h u y e t n e n CeHjOH + 3 H 0 N 0 , > C6H2(N02)30Hi + 3H2O s a u p h a n iifng c6 k h o n g k h i du'. 1,5 mol 4,5 mol 1,5 mol ^HNOa = (6 - 4,5) X 6,3 = 94,5g n khong khi dix — X + ^ - 1 - 0,01 x + ^ - 1 niaxit p i c r i c = 1,5 X 229 = 343,5g (O2 tac dung) mdd .sau p h a n uTig = mHNOg + ^ H2SO,, + ^phenol " niaxit p i c r i c Sau p h a n iJng t° = 0 ° C , h d H2O hoa long, t r o n g b i n h chi con k h o n g k h i = 600 + 750 + 141 - 343,5 = 1147,5g du va CO2. C%(HN03)= ^1^-5^11^=8,24%. nsau phan ijfng = n khong khi dir + ^'^CO^ - ^'^^x + ^ - 1 + 0,01x 1147,5 31
-H2O +Cl2,t° PV Theo cong thiJc Clapeyron : - CH3-CH-CH2OH C H 3 —C =CH2 C1CH,-C=CH RT (II) CH, CHo CH3 0,09 + 0,01x = 22,4 „^'^7-^Q = 0,88 ( 1 san p h a m ) X 273 273 CH3 -H2O +ci,t" 388 - 40x CH3-C-OH -> CH3—C—CH2 -> C 1 C H 2 - C = C H 2 y = (III) CH3 CH, CH3 V i A i la dong d i n g ciia axit C4H6O2 n e n A i la axit chiTa no. Vay este E i (1 san pham) chiTa no va E2 la dOng p h a n cua E i n e n E2 cung chira no. V i E i , E2 l a Neu ancol R j la C4H9OH t h i chi c6 cau tao I , I I , I I I la t h i c h hop. Vay este chua no n e n : y < 2x - 2. axit A i la C H 2 = C H - C 0 0 H va este E i la : 388 - 40x < 2x-2 x > 7 ' C H 2 = C H - C 0 0 C H 2 - C H 2 - C H 2 - -CH, CH, =CH-COOCH, -CH-CH, E, I Theo de bai x < 8, vay p h a i chon x = 7 y = 12 CH, Cong thufc p h a n tilf cac este : C 7 H 1 2 O 2 CH, Nhan xet dac diem cdu tao : Cac este E i , E2 deu l a este chua no va c6 . C H 2 = C H - C 0 0)-i-C -C-CH3 1 l i e n ket n t r o n g go'c axit hoac t r o n g goc ancol. Vay cac este c6 the dufcfc CH3 tao t h a n h do : axit no va ancol k h o n g no hoac axit k h o n g no va ancol no CO tCr 3 nguyen t i l C t r d l e n . Neu ancol R2 l a C 4 H 7 O H t h i chi c6 cac cau tao phCi hop l a : b) Xdc dinh cong thiic cdu tao cac axit, ancol vd este : CH2=CH-CH2-CH2 0 H > (I) D a t cong thiJc ciia este dang C x H y C O O C ^ H n v d i x > 3; m > 3. C4H7OH CH3-CH=CH-CH20H > (I) • x = m = 3 t h i axit A i la C4H6O2 t r i i n g l a i A (loai). • X = 2 t h i m = 4 : axit c6 3C va ancol c6 4C (nhan). CH3-C-CH2OH > (II) Vay axit A i : C H 2 = C H - C 0 0 H ; ancol R i : C4H9OH CHg axit A2 : C H 3 - C H 2 - C O O H ; ancol R2 : C4H7OH Vay este E2 c6 cdc dong p h a n : K h i h i d r o hoa a mol h 6 n hop 2 ancol C4H9OH va C4H7OH t h i : CH3-CH2-COO-CH2-CH2-CH=CH2 Ni,t" CH3-CH2-CH2-COO-CH2-CH=CH-CH3 C 4 H 7 O H + H2 C4H9OH CH3-CH2-CH2-C00-C=CH2 Vay sau k h i h i d r o hoa a mol (C4H9OH va C4H7OH) t h i c h i difgc a m o l CH3 C4H9OH. Chia hon hop A gom ancol metylic va mot ancol no, dOn cht/a la dong So do khuf H2O cac dong p h a n ciia C4H9OH va clo hoa a n k a n c6 do't dang cua ancol metylic thanh ba phan b l n g nhau. nong : Phan 1 : Tac dung het vdi Na da daoc 336ml khi H2 d dieu kien tieu chuan. -H2O CH3—CH2—CH2—CH2OH -> CH3—CH2~CH=CH2 Phan 2 : Oxi hoa bang CuO thanh cac andehit (hieu suat 100%), sau do •(I) cho cac andehit tac dung vdi dung dich AgNOa trong N H 3 dU thu d L f d c +ci2,r > C1CH2-CH2-CH=CH2 (1 san p h a m ) 10,8 gam Ag. (puthe) Phan 3 : Cho bay hoi roi trpn vdi mot lUOng O2 dU thu dUOc 5,824 lit khi CH3-CH2-CH-OH CH3-CH=CH-CH3 +Cl2,t" 6 136,5°C va 0,75 atm. Sau khi bat tia \Cia dien de dot chay het ancol thi I ^ (2 san p h a m ) CH, —> CH3 —CH2 —CH=CH2 pir the thu dupe 5,376 1ft khi 6 136,5°C va 1 atm. 33 32
a) Tim cong thufc phan tuf cua ancol dong d i n g . Hoa hdi hoan toan 4,28 gam h6n hop hai ancol no A va B d 81,9°C va b) Xac dinh thanh phan phan tram khoi i L f d n g cua moi ancol trong h6n hop. 1,3 atm du'Oc the tfch 1,568 lit. Cho laong h6n hop ancol nay tac dung (Dai hoc Da Ning - nam 1995) vdi kali duf thu diJdc 1,232 lit H2 (do d dieu kien tieu chuan). Mat khac, dot chay hoan toan lUdng h6n h(?p ancol do thu d\J0c 7,48 gam khi COg. Gidi Xac dinh cong thiJc cau tao va khoi li/Ong moi ancol, biet rang so nhom a) Cac p h a n ijfng : chufc trong B nhieu hdn trong A mot ddn vi. CH3OH + N a > CHgONa + - HgT (1) (De thi tuyen sinh Dai hoc An ninh Ha Noi - nam 1998) 2 Gidi C„H2„,iOH + N a > C„H2n.iONa + ]- H2T (2) D a t cong thiJc p h a n tuf cua 2 ancol A va B la R(OH)h va R'(OH)h+i. o ' , n , 1,3x1,568x273 „ , So m o i 2 ancol : UR = = 0,07 moi CH3OH + CuO > H C H O + Cui + H2O (3) (273 + 81,9) X 2 2 , 4 x 1 CnH2n+lOH + CuO > C n _ i H 2 n - l C H 0 + Cui + H2O (4) Cac phiTOng t r i n h p h a n iJng cua h 6 n hop ancol v d i k a l i : AgNOa + 2NH3 > [Ag(NH3)2]N03 (5) R(OH)h +hK > R(0K)h+-H2t (1) 2 H C H O + 4[Ag(NH3)2]^ + 2H2O > C 0 ^ + 6 N H ^ + 2NH3 + 4 A g i (6) R'(OH)h+i + (h + 1)K > R'(OK)h^i + Hat (2) C n H 2 n - i C H 0 + 2[Ag(NH3)2]* + 2H2O > C n H 2 n - l C 0 0 - + SNH^ + 2 + NH3 + 2 A g i (7) 1 232 X 2 So m o i H la : ny = = 0,11 (moi) H CH3OH + 1,502 > CO2 + 2H2O (8) " 22,4 Gpi X , y l a so' moi ancol A va B. Ta c6 : CnH2„+iOH + — O2 > nC02 + ( n + DHaO (9) 2 X + y = 0,07 Theo (1), (2) : b) Goi X , y l a so moi ancol metylic va ancol dong d i n g . j x h + y ( h + 1) = 0,11 0,336 G i a i he phuong t r i n h t i m difpc cap n g h i e m duy n h a t t h i c h hop : Theo (1), (2) : x + y = 2 x n n = 2: = 0,03 22,4 h = 1; x = 0,03 va y = 0,04 10,8 Cong thufc p h a n tuf cua 2 ancol A : CnHan+iOH, B : CmH2n,(OH)2 Theo (6), (7) : 4x + 2y = nAg = = 0,1 108 +O2 P h a n iJng chay : CnHan+iOH > nC02 (3) Giai ra CO : x = 0,02; ,y = 0,01 X moi nx moi Gpi n va n ' l a t o n g so moi k h i trifdc va sau k h i do't chay : lxnx22,4 0,075x5,824 C„H2n,(OH)2 mC02 (4) n = 0,13 273 273 + 136,5 -> y moi m y moi lxn'x22,4 1x5,376 7,48 „ , —> n = 0,16 ncoj = n x + m y = = 0,17 (moi) 273 273 + 136,5 0,03n + 0,04m = 0,17 Vay so moi k h i t a n g : 0,16 - 0,13 = 0,03. V i m , n nguyen duong n e n phuong t r i n h nay c6 mot cap n g h i e m t h i c h The.o (3), so moi k h i t a n g do dot CH3OH = (3 - 2,5).0,02 = 0,01 va do hop l a n = 3; m = 2. dot ancol dong d i n g = 0,03 - 0,01 = 0,02. Cong thufc cau tao ciia A : CH3-CH2-CH2OH 3n" Theo (9) t a c6 : n + (n + 1) 1 .0,01 = 0,02 r u t r a n = 4. Cong thiJc cau tao cua B : CHg -CH2 OH OH Vay cong thufc p h a n tijf cua ancol dong d i n g la C4H9OH. m A = 0,03 X 60 = 1,8 (gam); mij = 0,04 x 62 = 2,48 (gam). 34 35
Xac dinh thanh phan cua hon hpp X va Y (gom cac ancol nao ? so mol 24. Hai chat hOu cd A, B tao ra bdi 3 nguyen to va deu c6 34,78% oxi ve cua moi ancpl ?), biet rang m6i hpn hdp X va Y cc chila hai ancpl. khoi lUdng. Nhiet dp sdi cua A la +78,3°C, cua B la - 2 3 , 6 ° C . 0) Dun npng mot trong hai hon hop ancol tren (hdn hdp c6 ancol chiJa nhieu a) Tim ccng thiJc phan tCr va c c n g thiJc cau tap cua A, B. cacbon nhat) vdi dung djch H 2 S O 4 dac 6 nhiet dp khpng cap (140°C) ta b) Hoan thanh phi/dng trinh phkn ufng sau : thu dUdc 3 ete khac nhau A D A , BOB, AOB (A, B la cac gcc ankyi cua ? ? ? ? ? • ? ^ A, > A2 > A3 > A4 > A5 > B hai ancci) thep tl le mpl bang 2 : 1 : 2. (1) (2) (3) (4) (5) (6) Hay tfnh khp'i li/dng cija tC/ng ete thu di/dc. Gia thiet phan Cfng tach nude (Dai hgc Ngoai thuang - nam 1998) t^o ete cua cac anccI trpng h6n hdp xay ra 100%. Giai (HQC vi$n Quoc te - nam 1997) a) A , B dgu l a C^HyOz : Giai 16z 34,78 a) Gia sijf t r o n g h 6 n hop X (hoac Y) c6 h a i ancol A va B; d a t cong thiJc 12x + y = 30z 12x + y 65,22 1,875 cua A l a C n H 2 n + i O H (x mol); B l a C^Ha^n+iOH (y mol) Khi z = 1 ^ 12x + y = 30 (x = 2; y = 6) : CaHgO C„H2n.iOH + Na > C„H2n.iONa + i H 2 T z = 2 -> 12x + y 60 (x = 4; y = 12) : C 4 H 1 2 O 2 2i •- Phu hop v d i thuyet cau tao, A va B p h a i la C 2 H ( ; 0 x mol 0,5x m o l - CaHgO iJng v d i 2 cong thiJc cau tao : C H 3 - C H 2 O H va C H 3 - O - C H 3 . C^H2„,.iOH + N a >C„H2..iONa + - H 2 t - C2H5OH p h a i CO n h i e t do soi cao h o n (do c6 tao ra l i e n k e t hidro). y mol 0,5y m o l - Vay: C2H5OH CH3OCH3 5,6 (A) (B) 0,5(x + y) = -> X + y = 0,5 22,4 b) Cac p h a n ufng : men giam 3n t" C 2 H 5 O H + O2 > CH3COOH + H 2 O (1) C n H 2 n + 2 0 + — O2 > xxCOi + (n + DHzO 2 (A) (Ai) X mol l,5nx mol CH3COOH + N a O H > CHsCOONa + H 2 O (2) (A2) CmH2m+20 + — - O2 -> mC02 + ( m + 1)H20 voi,t° CHgCOONa + N a O H > Na2C03 + CH4T (3) y mol l,5mx mol (r^n) (A3) 47,05 l,5(nx + my) = nx + m y = 1,4 (2) askt 22,4 CH4 + CI2 > CH3CI + H C l (4) nx + m y = 1,4 1,4 - 0,5n Tif (1), (2) : (3) (A4) y = X + y = 0,5j m - n CH3CI + N a O H > CH3OH + N a C l (5) 0 < y < 0,5' 1,4 - 0,5n (A5) > 0 Vi : m, n > 1 y > 0 m-n (n=l;2) H2SO4d,140°C 2CH3OH > CH3OCH3 + H 2 O (6) dat m > n 1,4 - 0,5n > 0 ^ n < 2,8 25.a) Cc hai hpn hdp X va Y di/dc pha trpn tC/ cac anccl np, ddn chufc cung y < 0,5 m > 2,8 day dcng dang cc sp nguyen tCf cacbcn < 4. Khi chp X, Y tac dung vdi Vay, theo de r a : m , n < 4 : 2,8 < m < 4 ( m = 3; 4). Na da ta deu thu diidc 5,6 li't H2 (dktc), c c n khi dp't chay hpan tpan X, Y V a y X va Y l a h a i t r o n g cac h 8 n hop sau : deu can 47,04 litOg (dktc). 37
CHgOH (x = 0,05 mol), C 3 H 7 O H (y = 0,045 mol) Ap dung dinh luat bao toan khoi lugrng ta c6 : C H 3 O H (x = 0,20 mol), C 4 H 9 O H (y = 0,30 mol) m g ancol + 3g axit = khoi lisang este + 0,9 gam nifdc C 2 H 5 O H (x = 0,10 mol), C 3 H 7 O H (y = 0,40 mol) Khoi iMng este vdi hieu suat 100% = m + 3 - 0 , 9 = m + 2,1. C 2 H 5 O H (X = 0,30 mol), C 4 H 9 O H (y = 0,20 mol) (m + 2, l ) h b) H6n hcfp c6 chiJa nhieu C nhat la : C 2 H 5 O H va C 4 H 9 O H Neu hieu suat h % = 100 Dat so' mol cac ete : A O A = a mol; BOB = 0,5a mol; AOB = a mol ,) Xet xem O2 c6 dif khong : 2(C2H50H) > (C2H5)20 + H 2 O Goi ancol no dan chufc c6 1 noi doi c6 cong thufc tong quat : C m H 2 m - i O H 2a mol a mol (m > 3). 2(C4H90H) > (C4H9)20 + H 2 O C n H 2 „ . 2 0 + — O2 > n C 0 2 + (n + 1 ) H 2 0 a mol 0,5a mol 2 C 2 H 5 O H + C 4 H 9 O H — ^ C2H5OC4H9 + H 2 O 3n a mol — a mol na mol (n + l ) a mol a mol a mol a mol 2 an So" mol ancol tao ra ete = 5a = 0,3 + 0,5 = 0,5 -> a = 0,1 •3m-1^ C„H2n,0 + Oo > mC02 + mH20 Vay khoi liTomg (C2H5)20 = 74 x 0,1 = 7,4 gam 2 ( C 4 H 9 ) 2 0 = 130 X 0,05 = 6,5 gam r3m-l b mol b mol mb mol mb mol C 2 H 5 O C 4 H 9 = 102 X 0,1 = 10,2 gam. Cho hon hop X gom 1 ancol ddn chCfc no va mot ancol ddn chCfc phan C O 2 + 2 K 0 H (dtf) >K2CO3 + H2O tCr C O 1 lien ket doi, c6 khoi li/dng m gam. Nap m gam hon hop v a o 1 binh kin Y dung tich 6 lit va cho X bay ha\ 136,5°C. Khi X bay hoi hoan 0,12 mol 0,12 mol toan thi ap suat trong binh la 0,28 atm. Neu cho m gam X este hoa vdi K2CO3 + CaCl2 > CaCOai + 2 K C 1 45 gam axit axetic thi hieu suat phan ufng dat h(%). 12 0,12 mol = 0,12 mol a) Tfnh tong khoi iJdng este thu diidc theo m va n. 100 b) Them vao binh Y 8 gam oxi, sau khi bat tia lilfa dien de dot chay het X 3na (3m - l)b ^ A = 0 25 va daa nhiet do sau phan iJng chay ve 136,5°C thi ap suat trong blnh luc 2 2 32 ' nay la 1,96 atm. Cho san pham hap thu het vao dung dich KOH du, sau 3(na + mb) - b = 0,5 do them CaCl2 vQa du vao thi thu dUOc 12 gam chat ket tua. Xac dinh cong thufc phan tCf, viet cong thiJc cau tao va ten goi cac ancol. Giai ra b am, nhu vay O2 du : (Dai hgc Hang hai - nam 1998) ncoj = na + mb = 0,12 Gidi nHgO = (n + l ) a + mb = 0,12 + a a) Hon hcfp X : n= = ^'^^ ^ ^ = 0,05 mol = n2ancoi no sau phan ring = cua 2 ancol + ban d^u = & + b + 0,25 X 2 RT 0,082 X (273 + 136,5) Ta CO phuang t r i n h : a + b + 0,5 = 0,24 + 0,12 + a + no du ncHgCOOH = ^ = '^''^^ noda = b + 0,14 De b ^ i cho ancol dcfn chuTc tac dung vdi C H 3 C O O H nen ta c6 : nco2 = 0,12 1 ancol dorn chiJc + 1 mol C H 3 C O O H 1 mol este + 1 mol H 2 O U s a u phan iJng - 0,35 nHaO = 0,12 + a 0,05 mol 0,05 mol 0,05 mol 0,05 mol b + 0,14 (0,05 X 60 = 3g) (0,05 x 18 = 0,9g) 2 39