dug trang bac.
2.25. Dap an dung Ik C.
Rieng nh6m - O H a Cj ciia mach vong ( - 0 H hemiaxetat) tac dung v6i
C H 3 O H (c6 H C l xiic tac) tao ra ete chi a vi t r i nay goi la metyl
a - glucozit.
2.26. Dap an dung la D.
2.27. D a p a n s a i l a D .
Nhat thiét phai dung m6t chát oxi h6a oxi h6a F thanh I2 .
V i chi CO I2 m d r i tao ra hop chat mau xanh v6i h6 tinh bdt.
O3 + 2 K I + H 2 O I2 + 2 K 0 H + O j t 2FeCl3 + 2 K I -> 2FeCl2 + I2 + 2KC1
Ch + 2KI ^ 2KC1 + I2
2.28. D5p an dung la C. > 2.29. Dap an dung la C. 2.29. Dap an dung la C.
- B6t CaC03 tac dung vdi dung dich HCl c6 bot khi bay rạ - B6t NaCl tan trong dung dich HCl tao dung dich trong su6't.
- Con lai bflt Si02 va b6t xenlulozọ Phan bidt hai b6t nay bang each d6't trong O2.
2.30. SađdiinglaB.
X l a C 6 H, 2 0 6 ; Y la C2H3OH ; Z la CH2 = CH - CH = CH2 .
2.31. Ceng thiic dung la C.
Xenluloza diaxetat la phSn til xenlulozo da thay 2 nh6m OH bang 2 g6'c axetat (CH3COO), c6 CTCT C6H702(OH)(CH3COO)2. Do do c6ng thiic phan tir la : C10H14O7.
2.32. Dap an dung la Ạ
Phuong trinh phan ling :
CH2(OH)(CHOH)4CHO +2[Ag(NH3)2]OH
^ CH2(OH)(CHOH)4COONH4+ 2Agi + 3 NH3 + HjO (1)
Theo (I) : n.p = 2.nG = 2 . — = 0,4 mol.
-» m^g = 0,4.108 = 43,2 gam.
2.33. Dap s6' diing la D.
Phuong trinh phan iirng :
C6H12O6 ™ > 2C2H5OH + 2 C 0 2 t (1)
4500
Theo (1) : nc^H^OH = "COj = 2. nc^n.^o, = -2 = 50 mol.
Hieu suát phan ling la 80% nen thu duoc : 166
V c , H , 0 „ = ^ . ^ =2300 . , = 2,3,,,. Vco, = 5 0 . 2 2 , 4 . ^ =896 lit. (adsbygoogle = window.adsbygoogle || []).push({});
2.34. Dap so diing la B.
Cac phuang trinh phan ling :
(C6Hio05)„ +nH20 ^ nC6H,206 (1) CgHijOg """'"^ > 2 C2H5OH + 2 CO2 t (2) Theo(l,2): nc,H,OH = 2. nc,„,^o, = 2. i ^ . A L . ^ = 0,75 kmol.
The tich rucru nguydn chat = .2iZ^d^ = 43,125 lit. 0,8
Th^ tich ruau 46° = "^^'^^^-^^ = 93,75 ift.
46 -p' 2.35. Dap an dung la C. 2.35. Dap an dung la C.
Xla (C6Hio05)„ ; Yla C6H,206 ;
Z la phiic d6ng glucozo (C6Hn06)2Cu ; T la Cu20 4 2.36. Dap s6'dung la C. 2.36. Dap s6'dung la C.
C6H12O6 ^"""^ ) 2C2H5OH + 2CO21 (1)
CO2 + 2NaOH ^ Na2C03 + H2O (2)
Theo (1, 2) : nc^n.^Ô = | ri^^.co, = ^ = ^'^^ mol.
mQH,20,= 1.50. 180 = 270 gam.
Hieu suát phan ling len men la : H = — . 100% = 75%. 360
F2.37. Dap s6'dung la B.
Theo phuang trinh phan iJng len men : ' -
2 . 3 6 0 ^
nCjHjOH = 2 . nc^HnOe = = ^ mol.
4 4 6 6 5
Hieu suát phan ling 6 5 % n&n : VC^H^OH = T T ^ T T ^ ^ ^^^'^
U,o lUU
2.38. Menh di dung la C.
2.39. Menh 66 sai la C.
2.40. Dap i n dung la C.
Dung nu6c phan biet dugrc hai nhom :