Nối tiếp nội dung phần 1 tài liệu Tuyển chọn bài thi trắc nghiệm theo cấu trúc đề thi môn Hóa học, phần 2 giới thiệu các hướng dẫn chọn đáp án cho các đề thi ở phần 1. Mời các bạn cùng tham khảo nội dung chi tiết.
5.10 Dap an diing la A Cac phirang trinh phan ling : Cac phuang trinh phan ling: 2AI2O3 - i E ^ M + Cl2^MCl2 (1) 2M + O2 ^ 2MO Goi X va y la s5' mol CI2 va O2 tham gia phan ling Taco: 71x + 32y = 23 - 7,2 = 15,8 (2) 94 22,4 y + x=-^-0,25 22,4 (1) O2 + C — C O (2) O2 + 2C -> C O (3) ? = 0,1 mol X tao — = 0.02 mol CaCO, - 0 % nco, (trong h6n hgp X) = ^ ^ 0 % =20% Giai he phircmg trinh dugc : x = 0,2, y = 0,05 S6' kmol CO2 hdn hgp X la 3.20% = 0,6 kmol Dat X la s6' kmol O2 c6 3kmol h6n hgp X s6' kmol CO = - (x + 0,6) = (2,4 - x) kmol ^ ^ 32x +44.0,6 +(2,4-x)28 Ta CO : dx/H, = — = 16 Theo ( I ) va (2) : n,^ = x + 2y = 0,2 + 2.0,05 = 0,3 (mol) 72 KhO'i lugng mol cua kim loai M la : ~ = 24 (gam) Vay kim loai M la magie (Mg) - » X = 0,6 k m o l ; nco = 1,8 kmol Theo (1, 2, 3) : S6' kmol O2 thoat d phan ling (1) la : 1 Dap an dung la C So mol cac cha't: ncuci, A I + 3O2 = 0,1.0,5 = 0,05 (mol) ; n^^.a = 0,5.0,5 = 0,25 (mol) 0,6 + , + — =2,1 (kmol) Cac phuong trinh diSn phan: CUCI2 ""^^ >Cul + C l t 2NaCl + 2H2O -JS!^ (1) C L t + H z t + 2NaOH (2) Phan ling (1) xay hoan loan, sau phan ling (2) tiep tuc xay Dua vao edng thiic Faraday : A It m = —.— n F Tfnh dugc thdi gian dien phan hoan toan 0,05 mol CuCij la 1930 giay Con lai 3860 - 1930 = 1930 giay dien phan mot phan NaCl va tao 0,1 mol NaOH • Dung dich thu dugc chiia 0,1 mol NaOH Hoa tan A l : , 2A1 + N a O H + 6H2O 0,1 mol 2Na[Al(OH)4] + H t 0,1 mol 0,1.27 = 2,7 (gam) 5.12 Dap an dung la C TXIM = S6' kmol h6n hap X = = 3(kmol) 22,4 Theo (1) : n^, = ^ H Q , =|.2,1 = 2,8 (kmol) m.^i = 2,8 k m o l 27 = 75,6 (kg) Dap an dung la B Cac hgp kim: Cu-Fe(I) ; Fe-C(III) va Sn-Fe(IV) chiia Fe la kim loai boat d6ng han kim loai va phi kim lai nSn tie'p xiic vdi dung dich chat dien l i , Fe bi an mon Dap an diing la A - Khi dien phan dung dich CuClj (vdi dien cue tra), d cue am (catoi) xay phan ling: " Cu^* + 2e->Cu (1) - Phan irng an mon dien hoa xay cr cue am nhiing hgp kim ZnCu vao dung dich HCl: Z n - > Z n ' * + 2e (2) Phan ling (1): CO su tham gia cua ion kim loai (Cu^"^) Phan ling (2): c6 su tham gia cua kim loai (Zn) Dap an dung la C Ggi X la s6' mol Zn c6 19,3 gam h6n hgp S6' mol Cu la 2x mol Ta c6: 65x + 64.2X = 19,3 - > x = 0,1 163 .18 Dap an dung la C Cac phuang trinh phan ling: B6'n phan iJng oxi hoa - khic la: 3Zn + Fe2(S04)3 ^ 3ZnS04 + 2Fe 0,1 0,1 mol Fe 0,2 0,2 mol (1) FeCl2 + KMn04 -> mol + Fe2(S04)3 -> 3FeS04 FeS04 + KMn04-> H2S + K M n ^ (2) HCl + K M n - » 0,2 mol mol 5.19 Dap an dung la A ^ CUSO4 Cu + Fe2(S04)3 + 2FeS04 (3) Theo (3): SO' mol Cu tham gia phan ling = s6' mol Fe2(S04)v p.u 0,2- 01^0,2 - Zn khur ion Fe^^: Zn + Fe'^-> Zn'^ + Fe - Fe^^ khijr ion Ag^: Fe^^ + Ag^ ^ Fe'^ + Ag K I M LOAI KIEM, K I M LOAI KIEM THO, NHOM, SAT = 0,1 mol 6.1 Dap an diing la B Sau phan ling xay hoan loan du 0,1 mol Cu Cac phuong trinh phan iJng : m = mcu = 0,1.64 = 6,4 gam 5.16 Dap an diing la D 5.17 Dap an dung la C Phuang trinh dien phan dung dich CUSO4 + H,0 mol CUSO4 CUSO4: - i H ^ Qui + H,SO (1) H2SO4 ^ FeS04 + H t 0,1 mol Fe + C U S O C u s l ' + FeS04 Theo phan iJng (3): mol Fe phan ling, tao mol Cu -> khd'i lucmg tang gam X mol Cu m^^ = 0,06.96 = 5,76 - > m = 3,22 + 5,76 = 8,98 (gam) 6.4 - > - - = y Dap an diing la D 6.7 Cac phuong trinh phan ling : 2Fe + 3Cl2 ^ 2FeCl3(X) FeCl2 ( Y ) + H21 Fe + 2HC1 ^ Fe + 2FeCl3 6.5 (2) - > 3FeCl2 ( Y ) 6.8 (2) Theo ( ) va ( ) : S6' mol HCl phai dung la mol = 36,5.2 = 73 (gam) m^juci = — = — 387 + 32X , > X = 0,5 mol = nw„ 100 CO2 + 2NaOH (3) Goi CTPT ciia sat oxit la Fe^Oy Phirong trinh phan iJng : xFe + y C O j (1) Theo (1) : CO phan ting tao C O , do the' tich vSn la 4,48 lit mNaHCO, 6.9 nco^ = ^ = ^'15 (mol) chi xay phan ling (2) = 0,075 84 = 6,3 (gam) Dap an diing la B Fe + H N O Fe(N03)3 + N O t + H O Fe + 2Fe(N03)3 -^•3Fe(N03)2 K i m loai du la Cu va c6 thi c6 Fe 6.10 Dap an diing la A (dktc), tu-c la bang 0,2 mol Goi a la s6' mol CO c6 h6n hop sau phan Hag - > s6' mol CO2 la (0,2 - a) 166 Na2C03 + H2O Fe dir se phan ling vdfi Fe(N03)3 : Ddp an diing la B Fe.Oy + y C O (2) Theo di bai, chi c6 m6t pMn Fe phan ling (vi thie'u H N O ) : = •^^^^^^^^^•100% = 11,79% ^^•^'2 387 + 32.0,5 %mMeCl 6.6 CO2 + NaOH - » NaHC03 va CO2 vSn du Kh6'i luong mu6'i NaHCOs dugc ti'nh theo NaOH : N6ng d6 % cua FeClj dung dich Y : •^^ (1) S6' mol NaOH = 0,075 = 0,075 mol < ncoj 365 + 56x + (1 - x)24 - = 387 + 32x 15,76 X + COjt = 365 (gam) Sau hoan tan vijfa du thi khd'i lirang dung dich ]k : (56 + 71)x Dap an diing la C mco2 = 13,4 - 6,8 = 6,6 (gam) 73 100 - > mHci Fe2(S04)3 Trong dung dich du Fe nen tat ca Fe2 ( ) da b i khijr het XCO3 (1) MgCl2 + H21 M g + 2HC1 -^"^^"^'S Cac phuong trinh phan ling : Gia six hoa tan mol h6n hop X (trong c6 x mol Fe va (1 - x ) mol M g ) vao dung dich HCl : Fe + 2HC1 - > FeCl2 + H21 MgS04 , Fe -^2^04'i.n^ FeS04 Do vay, chat tan dung dich Y la MgS04 va FeS04 (3) Dap an diing la B Dap an dung la A Mg (1) CTPT cua sat oxit la Fe203 nFc= DO = , (mol) 167 - K h i cho hOn hop X vao dung dich NaOH, Na va A l d^u tan he't Phucfng trinh phan dng : 2Fe 0,1 mol + 6H2SO4 0,3 mol -> Fe2(S04)3 + S S O j t + H O Vi (1) 0,05 mol Theo (1) : Sau phan ling (1), Fe du 0,02 mol ntn xay phan ling : Fe+Fe2(S04)3 -~>3FeS04 (2) X mol Na lam thoat -y-lit Theo (2) : y mol A l lam thoat H2 lit H2 Nhu vay sau (2) thu dugc : 0,05 - 0,02 = 0,03 (mol) F e j ( ) va 0,06 mol FeS04 Taco: ^ + ^ 6.11 Dap an dung la C 3Fe + 202 ^ Fe304 (Fe203 FeO) 2 ^ ^' + ^ = , V , - > i l l m = 1^3 = 23 mAi 2.27 54 (1) H6n hop X gdm FeO, Fe203 va Fe du Dat X la s6' mol Fe du, s6' mol Fe304 la y Ta c6 : 56x + 232y = (gam) 6.13 (I) y ^-A-i-ii ^ = 29,87% 23 + 54 + 3NaOH ^ Al(OH)3 ^ + 3NaCl Al(OH)3 + NaOH - > N a [ A l ( O H ) J Fe + H N 3FeO + I O H N O Fe(N03)3 + H O + N O t (2) 3Fe(N03)3 + SlhO (3) Theo (2) va (3) : x + - y = + NOt = 0,025 (mol) S6' mol Al(OH)3 tham gia phan ling (2) = 0,3 - ^ (11) (2) = 0,1 (mol) Theo (1) va ( ) : S6' mol NaOH da phan ling = 0,9 + 0,1 = (mol) Gia tri Idn nha't cua V la : V = + np,,,) = x + 3y = 0,045 mol mp, = 0,045.56 = 2,52 (gam) 6.14 6.12 Dap an dung la A = (lit) Dap an dung la D Bai tap gidi theo phuang phdp can bang dien tich : Cac phuong trinh phan ling : ^ (1) Theo (1) : H A K O H ) , = " A i C b = 1'5 0,2 = 0,3 (mol) Giai (I) va ( I I ) ta dugc : x = 0,0225 ; y = 0,0075 • V2 Dap an dung la A AICI3 cho N O : 2Na + H (II) Cac phuong trinh phan ling : Trong h6n hop X chi c6 Fe va FeO tac dung vdi dung dich H N O , - > np, = np, = 1,75V 2NaOH+H2t (1) 2A1 + 2NaOH + 6H2O - > 2Na[Al(OH)4] + H (2) K h i tan vao dung dich H N O vCra dii (chi tao muO'i sunfat va N O t ) 0,12 mol FeSz ^ 0,12 mol Fe^+va 0,12.2 = 0,24 mol SO^" Goi X va y la s6' mol Na va A I c6 m gam X a mol CU2S -)• - K h i cho X vao nude, Na tan he't, A l tan khOng he't (vi thie'u NaOH) -> Thi tich H2 thoat tijr hai phan ling (1) va (2) dugc tinh theo Na : 2a mol Cu^"*^ va a mol SO4" S6' mol dien tich duong = 3.0,12 + 2.2a = (0,36 + 4a) mol SO' mol dien tich am = 2.0,24 + 2a = (0,48 + 2a) mol Theo (1) : X mol Na lam thoat ^ l i t H Theo (2) : x mol NaOH lam thoat ^ Taco: ^ + ^ 168 =2V, = V lit Trong dung dich : x ' = 8,96 (gam) 88,96 = - ^ - = 0,16 (mol) 56 S Cac phuong trinh phan img: • AI4C3 + K H + I H O y mol | x + y = 0,3 mol Cac phuong trinh phan ling: FeCl2 + H O (1) F e + 6HCI -> 2FeCl3 + H O (2) F e (FeO F e ) + 8HC1 (3) Dat s6' mol cua FeO la x -» FeClj + FeCl, + H O ft s6' mol F e la x; so mol FeO la y va so mol F e la y Ta c6: 72(x + y) + 160(x + y) = 232(x + y) = , x + y = 0,01 Dat X va y la s6' mol A l va AI4C3 c6 0,3 mol h6n hop X (3) 6.20 Dap so diing la D Dat s6' mol ciia F e la y ^ Dap an diing la B 2AI + K O H + 6H2O (2) S6' mol oxi nguyen tir chiia cac oxit: 3 - 13 Ho = ' ' = 0,075 (mol) 16 Theo (1,2,3): S6' mol HCl = so mol O = 0,15 m o l FeO + 2HC1 m = 0,16.242 = 38,72 (gam) Taco: 2AICI3 + H O f V j H c i = , : = 0,075 l i t hay 75 m l 56 o x = 0,2; y = 0,l 4y = 0,6 Dap s6' dung la C M g + 2HC1 3+ = 0,6 (mol) Cac oxit tac dung vQ\: - Qua trinh cho electron: Fe -> Fe+ 3e => n^(cho) = 170 nA,(OH),= Cac k i m loai M g , Cu, A l tac dung hoan tokn v6i O2 tao cac oxit = 0,04 l i t (40 m l ) 6.18 = Dap an diing la D Fe X gam (3) -> a = l,5x + y = 1,5.0,2 + 0,1.3 = 0,6 (mol) ^ : So mol KMn04 = - s6' mol Fe = X x + 4y X + (1) - » Fej ( S O )3 + K S O +2 MnS04 + H O 6.17 nK|Ai(OH),i= ^ , Cac phuong trinh phan ling : FeS04 + H j t A l ( O H ) ^ + KHCO3 46 Theo (1,2,3): Dap an dung la B F e + H2SO4 -> ^ 2K[A1(0H)4] + H t (1)^ 4K[A1(0H)4] + C H t 4y - > V j j „ c , = 0,08 : =0,08 l i t Cdch khdc: V i s6' mol FeO = s6' mol FejO, nen c6 the' coi 2,32g h6n hop la chi ciia F e Fe,0, + 8HC1 ^ FeCl, + 2FeCl3 + H O 0,01 -> 0,08 l,5x X Theo (1,2,3): nnci = 8(x + y) = 8.0,01 = 0,08 mol (2) 3y ^ =>V,,„„=M! = 0,08 (lit) ^ 171 6.21 Ddp s6'dung la A Cac phuong trinh phan ling: 2NaOH+H2t 2Na + H 2NaOH + A l + H (1) -> 2Na[Al(OH)4] + S H j t 6.24 Dap s6' diing la A Phuong trinh phan img: 2A1 + F e A (2) Dat s6' mol Na la x mol -> s.6 mol A l la 2x mol Theo (1,2): S6' mol = - + - x = 2x = — = 0,4 (mol) ^ 2 22,4 |p' Theo (2): S6' mol A l da phan ling = so mol Na = 0,2 (mol) Na[Al(0H)4] (2) Na2S04 + HjO (3) ^ = 0,45 lit -> m = [(0,025 + 0,1).27 + 0,05.160] = 22,75 (gam) 6.25 Dap an dung la B ^ Cac phuong trinh phan ung: 4FeC03 + a mol 6.23 Dap so diing la C n^,, = 0,1 m o l ; OAgNo, = 0,55 mol A l + 3AgN03 -» A1(N03)3 + A g i 0,3 Fe + 2AgN03 0,1 0,2 0,3 mol Fe(N03)2 + 2Ag>l 4FeS2 2Fe203 + 4COjt (1) — mol a mol +IIO2 -> 2Fe203 + 8S02t (2) 11 —bmol 2b mol Ap sua't tru6c va sau phan img bang n6n s6' mol cung bang bmol Taco: - + — = a + 2b-> a = b 4 0,2 mol Sau phan ling (1) va (2), AgN03 du 0,55 - (0,3 + 0,2) = 0,05 (mol) ^ do xay phan ling: Fe(N03)2 + AgNO, -4 Fe(N03)3 + Agi 0,05 0,05 mol Vay: m = 108(0,3 + 0,2 + 0,05) = 59,4 (gam) 172 / I Cac phuong trinh phan ling: 0,1 (6) Theo (2 va 3): S6' mol Fe = s6' mol H2 = ^'"^"^'^"^ = 0,1 (mol) 22,4 Theo (1): S6' mol Al phan ling = s6' mol Fe = 0,1 mol (1) 0,1 0,2 c6 lugng ke't tua 7,8 gam, thi tich dung djch NaOH Idn nha't la: 0,6 + 0,1 + 0,2 Hp, = (5) S6' mol FcjO, phan ling = ^ s6' mol Al phan ling = 0,05 mol H2SO4 + 2NaOH ^ S6 mol cac chat: (4) 2A1 + 6H2O + 2NaOH ^ 2Na[Al(OH)4] + 3H2t 2Na[Al(OH)4] ? n 84 Theo (5): S6' mol A l du = - s6 mol Hj = - ^ = 0,025 (mol) ' 22,4 0,1 VjdNaOH = (l) AI2O3 + N a O H + 3H2O 0,2 A1(0H)3 + NaOH (0,2-0,1) - dphan 2: AKOHX^xl + Na2S04 0,6 AI2O3 + 2Fe AI2O3 + 3H2SO4 ^ Al2(S04)3 + 3H2O -> m = 0,2.27 = 5,4 (gam) 6.22 Dap so dung la B 78 S6' mol Al(OH);, >^ = — = 0,1 (mol) 78 Cac phuong trinh phan ling: 0,1 ^ H6n hop Y tac dung vdi dung dich NaOH sinh Hj, chung to Y CO AI2O3, Fe va Al du (Fe203 da phan ling he't) Cac phuong trinh phan ling cua h6n hop Y -dphan I: 2A1 + 3H2SO4 ^ Al2(S04), + 3H2T (2) Fe + H2SO4-> FeS04 + H j t (3) -> S6' mol A l du = 0,4 - 0,2 = 0,2 (mol) AI2 ( ) + 6NaOH — ^•26 Dap s6' dung la A Cac phuong trinh phan ling: FeO + 2HC1 FeCl2 + HjO Fe203 + 6HC1 ^ 2FeCl3 + 3H2O (1) (2) Fe304 + HCl: xay nhu (1) va (2) 17'^ W cua axit da bay khoi dung dich du6i dang H2 - S6' mol CI = 0,5.1 = 0,5 mol -> m^, = 17,75 gam S6 mol SO^' = 0,5.0,28 = 0,14 mol -> m^^- = 1,3,44 gam Kh6'i luong mu6'i khan la: m = 7,74 + 17,75 + 13,44 = 38,93 (gam) Dap s6' diing la A Cac phuong trinh phan ling: 2A1 + FejO, — > AI2O3 + 2Fe 0,1 mol 0,1 mol Theo (1): ma-o = ^ • 72 = 4,32 (gam) ^ mp,^o, = 9,12 - 4,32 = 4,8 (gam) -> npco = loU ^ =0'03 mol Theo (2): S6' mol FeCl, = s6 mol FcjO, = 0,06 (mol) ^ mpeci, = 162,5 0,06 = 9,75 (gam) 6.27 Dap an dung la A 448 S6' mol CO, = - ^ ^ ^ = 0,02 (mol) 22,4 Cac phuang trinh phan img: M H C O + HCl ^ MCI + CO2T + H2O M C O , + 2HC1 -> MCI2 + CO2T + H2O ' H6n (1) (2) Theo (1) va (2): S6' mol h6n hop muoi = s6' mol COj (M + 61 + 2M + 60).0,02 , „ Va = 1,9 Glai phuong trinh diroc M = 23 Vay kim loai M la natri (Na) 6.28 Dap an dung la C Benzen khong phan irng dirgfc v6i nude brom Phenol (do anh hudng cua nhom -OH) phan iJng d6 dang vdi nu6c brom (the Br vao cac vi tri ortho va para) 6.29 Dap an dung la C Cac phuong trinh phan ung: 2A1 + N a O H + 6H2O ^ 2Na[Al(OH)4] + 3H21 (1) 8A1 + 3OHNO3 SAKNO,), + 3N2O + I5H2O (2) Dat a la so mol Al tham gia m6i phan ting: Theo (1): n^ = x = - a Theo (2): ^ , = y = - a -> x = 4y " 6.30 Dap an diing la B 6.31 Dap so diing la A - S6' mol H^ = 0,5.1 + 0,5.0,28.2 = 0,78 mol (0,39 mol H2) S6' mol H2 = 22,4 = 0,39 mol hop ran X gom 0,1 mol AI2O3, 0,2 mol Fe va Al du H6n hop X tac dung v6i dung dich NaOH: 2AI + 6H2O + 2NaOH -> 2Na[Al(OH)4] + 3H2t 0,1 mol 0,15 mol AI2O3 + 2NaOH + 3H2O ^ 2Na[Al(OH)4] 0,1 mol 0,2 mol 0,1 + 0,2 = 0,3 (lit) hay 300 ml 6.33 Dap an dung la B S6' mol cac cha't: n^gNo, = HCUCNO,), =0,3.0,1 = 0,03mol Cac phuong trinh phan iJng: Al + + (2) (3) 3AgN03->AI(NO,)3+3Ag>l (1) 3Cu(N03)2-^2Al(N03)3+3Cu>l (2) 0,01 mol 0,03mol 2A1 (1) 0,03mol 0,02 mol 0,03 mol Al du tac dung \6i dung dich HC\: 2Al + 6HC1 — 0,03 mol ^ 2AICI3 + 3H2 t S6' mol Al du = - s6' mol H,= = 0,01 (mol) ^ 22,4 ^ m, = 27(0,01 + 0,02 + 0,01) = 1,08 (gam) = 0,03.108 + 0,03.64 + a01.27 = 5,43 (gam) (3) 6.34 Dap an dung la D Cac phuang tnnh phan irng: NajO + HjO >2NaOH AI2O3 +2NaOH + 3H20 >2Na[Al(OH)4] Chat tan nha't la Na[Al(OH)4] c6 s6' mol la 0,5.0,2 = 0,1 (mol) Th6i CO2Na[Al(OH)4] vao dung dich Na[Al(0H)4] : + C02 ->A1(0H)3 i + NaHCOj 0,1 mol 0,1 mol ^ a = 0,1.78 = 7,8 (gam) Theo (1) va (2) va (3): n,^,^o, = ^ n N a | A i ( O H ) , i =0-05 mol "Na.O = ^ n N a | A l ( O H ) J =0.05 (mol) ^ m = 0,05.62 + 0,05.102 = 8,2 (gam) 6.35 Dap an diing !a A So mol Ba(OH)2 = 34,2 = 0,2 (mol) 171 Phuong tnnh phan irng; (NH4)2C03 + Ba(0II)2 ^ BaCO,! + 2NH3t + 2H2O 0,1 mol 0,1 mol ^ m = 0,1.197 = 19,7 (gam) 6.36 Dap an dung la D Cac phirang trinh phan irng : FeO + 2HC1-> FeClj + H2O (1) Fc^O, + 6HC1 2FeCl, + SHjO (2) Fe,04 CO c6ng thirc FeO.Fe203 cung phan irng nhu tren va cung thoa man: so mol Fe^^ : s6' mol Fe'^ = : Goi X la s6' mol ciia m6i oxit nira h6n hop X Ta c6 : Theo (1) va (2) : m, = 127x + 325x = 452x Sue khf Clj vao phan hai, xay phan irng : 2FeCl2 + CI2 -> 2FeCl3 (3) Theo (1), (2) va (3) : m2 = 3x.l62,5 = 487,5x m2 - m, = 487,5x - 452x = 0,71 -> x = 0,02 Theo (1) va (2) : = X + X = X = 8.0,02 = 0,16 (mol) ^ Vdd HCl = 2.-0,16 = 0,16 (lit) hay 160 ml Dap an dung la D Cac phuong trinh phan ling : Mg + 2FeCl3 MgCl2 + 2FeCl2 Mg + FeCl2 -> MgClj + Fe Theo (1) : n^g =^np,c., =^-0,12 = 0,06 (mol) HHCI Theo (2) : n^g = n F e = ^ (1) (2) = 0,06 (mol) m^g = (0,06 + 0,06).24 = 2,88 (gam) Dap an dung la B 136 H6n hop khf Y g6m NO va NjO ; = 0,14(mol) Goi X la s6 mol NO c6 Y Ta c6 : 30x + 44(0,14-x) = 5,18 = 0,07 mol = 0'07 mol Khi cac phan irng oxi hoa - khir xay : Goi a va b la sd mol Al va Mg c6 8,862 gam h6n hop Taco: 27a + 24b = 8,862 (I) 3+ a A l - 3ae aAl : Cho 3a mol electron bMg-2be hMg^"" : Cho 2b mol electron +5 0,07 N : Nhan 0,21 mol electron 0,07N +0.07.3e X DN^O +5 +1 0,07.2 N +0,07.2.4e -> 0,07 N : Nhan 0,56 mol electron Theo dinh luat bao toan electron : 3a + 2b = 0,21 +0,56 = 0,77 (II) Giai he phuong trinh (I) va (II) duoc : a = 0,042 mol %mA, = -.100% = 12,80% 0,042.27 8,862 6.39 Dap an diing la B S6' mol cac chat: n„t,^„ 47,4 = — = 0,1 mol nsoj- =0,1.2 = 0,2 mol ' 474 nBa(OH), =0,2.1 =0,2 m o l - n^u =0,2 mol Phuang trinh phan ling : Ke't tua Y g6m BaS04 va Fe(OH)2 Ba^^ + SO^"Baso, Nung ke't tua Y den khdi luang khdng doi dugc chat ran Z BaSO.^ = 0.2 mol -> mBaso, Chat ran Z g6m BaS04 va Fe203 = 0.2 233 = 46,6 (gam) 6.43 Dap an diing la A 6.40 Dap an diing la D nu S6 mol cha't tan (hidroxit) = = 0,02 (mol) 3,36 22,4 = 0,15 (mol) Cac phuong trinh phan ling: 2> S6' mol B, = = 8A1 + 3Fe304 — ^ 4AI2O3 + 9Fe 0,01 (mol) (1) H6n hop ran X g6m AI2O3, Fe va A l du -> K i m loai M hca trj I I Cho X tac dung vdri dung dich NaOH, xay cac phan ling : Cac phuong trinh phan ling: M + 2H2OM(0H)2 +Hjt (1) MO + H O M ( H ) (2) Theo (1) : n ^ = Theo(2): XIM^OH), = "H, = 0,01 mol 0,01M + 0,01(M + 16) = 2,9 -> M = 137 K i m loai M la bari Theo (3) (1) np^o = 2nso: = ^ ^ = 0,29 (mol) 22,4 Theo (2) : |nH, ==|-0,15 = , (mol) n^i^o, = - ( , - , ) = 0,2 (mol) =1-0,2=0,15 (mol) m = 27(0,4 + 0,1) + 232.0,15 = 48,3 (gam) S6' mol cac cha't: = 0,145.400 = 58 (gam) n^, = Cac phucfng trinh phan ling : 2A1 + H S O A l ( S ) + 3H2t (1) Fe + (2) -> FeS04 + H2t 39 " ^ " ^' ^ (mol) 6.44 Dap an diing la C 5.42 Dap an dung la B Dung dich X g6m Al2(S04)3 va FeS04, tac dung v6i Ba(0H)2 dir: 78 nAidu= (4) Theo (1) : n^, p., = 2n^i^o, = 2-0,2 = 0,4 (mol) Vay : -> Oxit sk la FeO H2SO4 "Nal AI(OH), = " A K O H ) , nFeA=|nA,o, mp,o p., = 0,29.72 = 20,88 (gam) mpccsoj, (3) Sue CO2 vao dung dich Y , xay phan ling: Theo(4) Oxit sat tac dung vdi dung dich H2SO4 dac, nong cho k h i SO2 bay la FeO hoac Fe304 (FeO.Fe263) Phuang trinh phan ling: Theod): 2A1 + 2NaOH + 6H2O -> 2Na[Al(OH)4] + 3H2t Na[Al(OH)4] + CO2 -> Al(OH)34 + NaHC03 6.41 Dap an dung la C 2FeO + 4H2SO4 -> Fe2(S04)3 + S02t + 4H2O (2) Dung dich Y la dung dich N a [ A l ( O H ) ] , cha't ran Z la Fe = 0,02 - 0,01 = 0,01 mol Taco: AI2O3 + 2NaOH + 3H2O ^ 2Na[Al(OH)4] Al2(S04)3 + 3Ba(OH)2 ^ 3BaS04^ + A l ( O H ) i (3) A l ( O H ) + Ba(OH)2 du -> Ba[Al(OH)4]2 + 4H2O (4) FeS04 + Ba(OH)2 ^ BaS04^ + Fe(OH)2i (5) = 0,46 (mol); Uy = ^ = 0,06 (mol) Goi X va y la s6' mol N2O va N2 c6 0,06 mol h6n hop Y Ta c6: x + y = 0,06 ^'"2 (I) 44x + 28y = 18 = 18 2.0,06 - > 4 x + 28y = 2,16 (II) , Giai he phuong trinh (I) va (II) duoc: x = y = 0,03 m o l 179 Phuang trinh plian ling: Phuang trinh phan ilng nhiet nh6m: 18A1 + 6 H N O - > I S A K N O , ) , + 3N2O + 3N2 + 33H2O 0,18 mol 0,18 mol 0,03 (1) 8A1 + F e - » 4AI2O3 + 9Fe 0,03 mol Cac phuong trinh phan img hoa tan h6n hop ran cho Con 0,46 - 0,18 = 0,28 mol A l phan ling vdfi H N O , tao NH4NO3: 8A1 + 30HNO., ^ SAKNOj), + 3NH4NO, + 9H2O m = 0,46.213 + 0,105.80 = 106,38 (gam) 6.45 Dap an diing la C l,125x + - ( , - x ) = 0,48 ^ = 0,25 (mol) x = 0,32 Hieu sua't phan ling: h = - — 0 % = 80% ^ ^ 0,4 X + HC1-^XC1+iH21 (1) Y + 2HC1 ^ (2) YCI2 + H 6.48 Dap an diing la D X + 2HC1 ^ - Gia su h6n hop chi c6 k i m loai k i ^ m X: g M,= = Dung dich Y c6 hai mu6'i clorua ciia hai kim loai c6 n6ng d6 mol bang nen hai kim loai X cung c6 so mol bang va bang 0,0625 mol Khd'i lugng mol nguySn tic trung binh ciia hai k i m loai la: 14,2 - Gia sit h6n hop chi c6 k i m loai k i ^ m th6 Y : "Y = " H , M x = - ^ ^ = 19,6 0,125 - 0'25 mol — 0,25 XCI2 + H j t nx = - nuci = - 0,2.1,25 = 0,125 mol nx = 2nH, = 2.0,25 = 0,5 (mol) -> K i m loai kiem thd c6 M < 19,6 la Be (Mg, = 9) = 28,4 Tilf ket qua xac dinh dugc k i m loai thu' hai la Ca (canxi) - Trong h6n hop c6 ca X va Y ntn 6.49 Dap an diing la C 14,2 < Mx, My < 28,4 Cac phuong trinh phan ting: -> X la Na ( M N , = 23); Y la M g ( M ^ g = 24) A l + 6HC1 6.46 Dap an dung la C 6.47 Dap an dung la A t SO' mol cac cha't: = 0,4 (mol); n H = ^ ^ = 0,48 (mol) 2AICI3 + 3H2t (1) AI2O3 + H C A I C I + 3H2O (2) AICI3 + 3NH3 + 3H2O ^ (3) Al(OH)34^ + 3NH4CI 2Al(OH)3 — A I O + 3H2O n,,^o, = | ^ = 0,15 (mol) ' (4) Ggi X va y la s6' mol A l va AI2O3 c6 1,56 gam hOn hop Taco: 27x + 102y = 1,56 (I) 04 Theo ( , 2, 3, 4): 0^,^03 = - x + y = — = ^ 180 (3) Nhu vay chi 0,32 mol A l tham gia phan ting nhiet nhom (1) So d6 phan ling: nA,= ^ (2) - S6' mol H2 dugc tao t i i Fe la: - x = l,125x mol Co can dung dich X thu dugc cha't ran khan c6 kh6'i lirgng: My = FeS04 + H2 i Goi X la s6' mol A l tham gia phan iJug (1), ta c6: - S6' mol H2 dugc tao tur A l du la - ( , - x ) mol .0,28 mol NH4NO3 So mol H2 = bay ra: A I + 3H2SO4 ^ Al2(S04)3 + 3H2t (2) Nhu vay dung dich X gom c6 0,46 mol A1(N03)3 va = 0,105 Fe + H2SO4 (loang) (i) 102 0,02mol (II) 161 19.30 Dap an diing la C Crom (VI) oxit la oxit axit 20 PHAN BIET CHAT V6 Cd, HOA HOC VA VAN DE PHAT TRIEN KINH TE, XA HOI, MOI TRUING 20.1 Dap an diing la C 20.2 Dap an diing la C 20.3 Dap an diing la C Cu kh6ng tan HCI dac ngu6i Cu tan HNO3 dac, ngu6i cho NO2 mau nSu bay Cu tan H S O dSc, ngu6i cho SO2 kh6ng mau bay can cii vao hifin tuong nSu tren phan biet duoc axit 20.4 Dap an diing la A (CH3)2CO la axeton, chi phan ling dugc \di Hj va HCN : CH3-(j:-CH3 + H2 O CH3-C-CH3 + H-CN > CH3-CH-CH3 OH CH3-C(OH)-CH3 O CN 20.5 Dap an diing la A 20.6 Dap an diing la B tac dung vori Hg tao HgS la hop cha't kh6ng bay hoi ntn lotv dugc su d6c hai cua thuy ngSn 20.7 Dap an diing la A nhirng 6'ng nghiem xay ket tiia la: (NH4)2S04, Cr(N03)3, KjCO, 20.8 Dap an dung la C 20.9 Dap an diing la C 20.10 Dap an diing la D 20.11 Dap an diing la D 20.12 Dap an diing la D 20.13 Dap an diing la A 20.14 Dap an diing la D 20.15 Dap an diing la B Cho mot bdt Cu vao dung dich N H N O va dung djch ( ^ ) , roi cho tie'p axit HCI vao ong nghiem c6 kh6ng mau, (sau N chuyen H N O sang Vi dmau da nSuxaydo)rabay phanra,ling: la dng nghiem dung dung dich It 3Cu + 8H* + N O ; 3Cu^* + N O t 2NO + ^ 2NO2 (mau nau + 4H2O do) 0,16 Dap an diing la A Cho BaCO, tac dung vdi titng dung dich: Dung dich cho ket tiia trang la dung dich NaHS04 Dung dich cho thoat la dung dich HCI Dung dich khdng c6 hien tugng xay la dung dich NaCl 21 DAN XUAT HALOGEN, ANCOL, PHENOL 21.1 Dap an diing la B Nhirng ancol da chiic c6 nhom OH lien ke't v6i cacbon li6n ki thi tac dung dugc vdi Cu(0H)2 tao dung dich mau xanh lam 21.2 Dap an diing la B X la ancol no, don chirc, c6 CTPT CnH2n+20 Ddt chay X : > n CO2 + (n + 1) H2O CnH2n+20 n 0,25 -> n = n + 0,3 CTPT cua X la CsH.iOH d6ng phan ciia X loai H2O chi cho anken nha't: CH3-(CH2)3-CH20H ; CH3 - CH(CH3) - CH2 - CH2OH ; CH3 - CH2 - CH(OH) - CH2 - CH3 ; CH3 - CH2 - CH(CH3) - CH2OH [1.3 Dap an diing la B Isopentan: C H , - C H ( C H ) - C H - C H , + Clj — ^ Clo CO the the' hidro d 4C (1,2,3,4) tao san phdm monoclo 21.4 Dap an diing la A Qua cac dir kien cua di bai, xac dinh dugc c6ng thiic ca'u tao cua este E la : H N C H C O O C H , + NaOH ^ H N C H C O O N a + QH.OH 0,25 mol 0,25 mol 0,25 mol Sau phan ling tao 0,25 mol mu6'i H N C H C O O N a va du 0,05 mol NaOH -> m = 0,25.97 + 0,05.40 = 26,25 (gam) H2NCH2COOC2H5 21.5 Dap an diing la D , Hidrocacbon X la m6t anken hoac mdt xicloankan c6 s6' nguyen tir C < Goi CTPT cua X la C„H2„ C„H2„ + Br^ C„H2„Br2 (Y) 160 74,08 ^— > n = 14n + 160 100 Hidrocacbon X la but-1-en (CH3-CH2-CH=CH2) 21.6 Dap an dung la B Dat CTPT cua ancol no X la: C„H2„+2-a(OH), Phuong trinh phan ting chay: C„H2„,2.,(OH):-H mol ^"""^^"^O^ -> nC02 -F (n + 1)H20 (1) 3n + l - a , mol 17 92 0,2 mol i^=0,8mol 22,4 ^ , n - h 0,1-0,1a = 0,8 0,7 +0,1a n= 0,3 Cap nghiSm nha't phii hop la: a = 2, n = CTPT cua ancol X la: C3H,(OH)2 X tac dung ducfc vofi Cu(OH)2 -> Hai nhom OH dinh vao hai nguyen duac tir cacbon canh -> CTPT cua X la CH3-CH(OH)-CH2(OH): propan-l,2-diol Mat khac: mol X phan ling v6i mol Cu(OH)2 0,1 mol X - - - 0,05 mol Cu(OH)2 -> mcu(OH), = 98.0,05 = 4,9 (gam) 21.7 Dap an dung Ik C 256! 21.8 di^u Hcfp CH, Dap che' =an cha't CCH3 -dung bang C H4-metyl 2la-phan COH B.H -ling pentan-2-ol C Hcdng , ; C HHj^ -tiirCH3 CjcHj ^5 Cchat H- CH3 -CH OH Csau: H- CHj - C H-OH , CH - CH3 j CH3 - CH - C H , - C - CH3 ; C H , = C - CHj - C - C H , CH3 O CH3 o CH3-C = CH-C-CH3 I II CH3 o 21.9 Dap an diing la D 21.10 Dap an dilng la A Phuang trlnh phan ling: CH3 - CH - OH + C u O — ^ C H , - C = O + Cu + HjO F CH3 CH3 22 AMIN, AMINO AXIT, PROTEIN 22.1 Dap an dung la B X la mot amino axit hay mSt este cua amino axit X CO c6ng thiic : H2N - CHj - COOC^Hy Phuong trlnh phan ting : I HjN - CH2 - COOC^Hy + NaOH ^ H2N - CH2 - COONa + C^HyOH _ u - V ' sdnguyentirC 3,36 ^ Trong phan tu X co : ^-^ = —-— = s6'nguyen tir N 0,56.2 Trong phan tir X c6 nguyen tu C (vi c6 nguyen tir N) _>x = ; y = CTCT thu gon cua X la : H2N - CH2 - COO - CH3 22.2 Dap an dting la C Tir CTPT C2HXO3N2 va tinh chat cua X ta tim dugc CTCT cua X la: [CH3 - CH2 - NH3]^ N O Phuang trlnh phan ling vdi NaOH: CH3CH2NH3NO3 + NaOH -> CH3CH2NH2 + NaNO, + H2O (X) (Y) MY = d v C 22.3 Dap an diing la A 0,02 mol X tac dung vdi 0,02 mol HCl -» X c6 nhom chiic am.l > 0,02 mol X tac dung vdi 0,04 mol NaOH -> X c6 nhom chiic xit -> CTPT cua amino axit la : H2N-C,Hy(COOH)2 Phuang trlnh phan ting cua X vdi HCl: H2N-C,Hy(COOH)2 + HCl ^ ClH3N-C,H,(COOH)2 25' 0,02 mol 0,02 mol - > M „ „ , i = ^ = 183,5 2 Dap an dung la C 2 Dap an diing la C 23 CACBOHIDRAT Dap an diing la B Phuong trinh phan ling: 12x + y+ 52,5 + 90= 183,5 cap nghiem nha't phu hop la : x = 3, y = C6ng thiic phan ttr cua X la : H2N-C3H5(COOH)2 2 Dap an dung la A 10 S6' mol QH9O2 = = 0,1 (mol) Cong thu-c ca'u tao cua X la: CH2 = CH-COO-NH3-CH3 Phirang trinh phan ling: CH2=CH-COO-NH3 CH^ + NaOH ^ CH2=CH-COONa + CH3-NH2t + HjO (1) Khi CH20H[CHOH]4CHO + H2 - ^ ^ ^ CHzOHECHOHLCHjOH (glucozo: 180) (sobitol: 182) D^ CO 1,82 gam sobitol v6i hiSu su&X 80% cAn: m„uco», = 180.100/80 = 2,25 gam Dap an diing la A X la glucozo, Y la ancol etylic va Z la axit axetic Dap an diing la D Dap an diing la D Cac phirong trinh phan ting : CfiH,206 > 2C2H,OH + 2C02t Y la CH3-NH2 Chat tan Z la CH2=CH-C00Na, c6 s6' mol la 0,1 mol Khi c6 can dung dich Z, thu duoc mu6'i khan c6 kh6'i lugng: m = 0,1.94 = 9,4 (gam) 2 Dap an diing la B S6' mol HCl = 0,175.2 = 0,35 mol Cac phuong trinh phan ling: H2NC3H.,(COOH)2 + HCl 0,15 mol ^ H3ClNC3H,(COOH)2 0,15 mol 0,15 mol NaOH +HC1 ->NaCl + H20 0,2 mol 0,2 mol H3C1NC3H,(C00H)2 + 3NaOH ^ H2NC3H.,(COONa)2 + NaCl + 2H2O J CO2 + Ca(0H)2 ^ CaCO,^ + H^O 1 4n v (1) (2) Theo (1) va (2) : n^^H.A = 2"caco, = ^ - ] ^ = 0'2 (mol) (1) (2) (3) 0,15 mol 0,45 mol Theo (2) va (3): S6' mol NaOH = 0,2 + 0,45 = 0,65 (mol) 2 Dap an diing la C Tijr d6 bai xac dinh dugc: - Amino axit no, mach hcJ, c6 nhom NH2 va nhom COOH - Amin no, don chiic, c6 nhom NHj Do do, suy y = (1 mol N2) Tdng s6' nguyen tu: C amino axit va amin Ik 6, d6 nguyen tir C a nhom chiJc axit (COOH), phSn g6'c no ciia amin va amino axit c6 nguySn lit C ling vdi nguySn tit H Nhu vSy s6' nguydn ttr H amino axit va amin la: (d g6'c) + (or chiic axit) + (6 chiic amin) = 14H Do s6' mol H2O tao x = no -> mc^H„o, =0,2.180.— = 48,0 (gam) Dap an diing la D Dap an diing la C Dap an diing la A Dap an diing la C (Ban doc tu viSt phuong trinh phan ung trang bac) I Dap an diing la D II Ta chi cAn nhd: Saccarozo tao tijf a-glucozo va fructoza la loai i dugc cac dap an A, B va C ^ Dap an diing la B , B So qua trinh di^u che' gia'm tii glucozo: C,H,20, ) 2C2H,OH ) 2CH3COOH W 180 gam 73,6 gam TCf 73,6 gam C2H5OH, theo If thuyfit dilu cM dugc m gam gia'm: t 60.73,6 i m= = 9,6 (gam) 46 I 7^^" .4 Dap an diing la B Phuong trlnh phan ling : I n u c le a i u wk«c d u g c : m' = 0,72.0,2.60 = 8,64 gam Vay hieu sua't qua trinh I6n men gia'm la: 8,64.100% h = = 90% 9,6 2H,0, ^""^ > H , + , t Theo (1) : D H ^ ^ =2no, = ^ ^ ^ = 6,003 mol 23.11 Dap an dung la B AC(H202) = 0,003.10 = 0,03 mol// = 3.10"^ mol// 23.12 Dap an dung la A - AC V =• I I THEO CHl/ONG T R I N H NANG CAO 24 T O C D O P H A N U N G , C A N B A N G H O A H O C , S l / D I E N L I 24.1 Dap an dung ia A - can bang (1) : - d can bang (2) : K, 7T = V ^ H.]'"[I.]" - can bang (3) : K, H,]'".[I = ^/M=8, Cac phuang trinh phan l i : CH3COOH ^ CH3COO- + H* CHjCOONa CH3COO" + Na^ Ta H^ CO : CH3COO- J ^ X ( , + X ) ^ 0,1 CH3COOH Tmh dugc x ^ pH = 4,76 1/2 HI t Khi can bang : [H^] = x, [CHjCOO] = (0,1 + x) -64 HI 3.10"^ -= 5,0.10"^ mol/(/.s) 60 Dap an ddng la D 24.2 Dap an dung la C [Hlf (1) ' 4-0,125 24.3 Dap an diing la B S6' mol H"^ 100 ml dung djch h6n hop hai axit: n^^ = 2.0,1.0,05 + 0,1.0,1 = 0,02 mol Dap an diing la A Phuong trinh phan ling: N j + 3H2 n 25 A N D E H I T , X E T O N , A X I T C A C B O X Y L I C +0, -> nCOj 2n mol nmol 0,5 mol 0,7 mol 25.1 Dap an dung la D Oxi hoa cumen r6i che' hoa bang axit H2SO4 thu dirge axeton : (CH3 )2 - CH - CgHs Andehit X c6 CTPT la: HCHO CH3 - CO - CH3 + C6H5OH Andehit Y c6 CTPT la: CH3CHO 25.2 Dap an dung la C Tijr hai c6ng thitc phan tut ixtn, tinh dugc phSn tram kh6'i lugng cua X , Y la ancol, Z la andehit phan tur chi c6 nguyfin tur cacbon (tijt CH^l) > loai dap an A, B va D Y la ancol CH3OH Z la andehit HCHO h6n hgp 25.8 Dap an diing la B 25.3 Dap an dung la D Cac phuong trinh phan iJng: - H6n hgp X phan ling vdi Brj: Cac phuang trmh phan ling : CH2=CH2+[Q] P'^ci^/cuci^ ^ CH3CHO CH2 = C H - C O O H + Br2 (1) CH2Br-CHBr-COOH CH2 = CH-CHO + 2Br2 + H2O 0,2 moi (1) CH2Br-CHBr-C00H + 2HBr - H6n hgp X phan ling vdi NaOH: CH3CHO + H - C N - > CH3CH(CN)0H 7,1 71 (2) = 0,1 mol Hieu sua't cua qua trinh la : h = — 100% = 50% 0,2 25.4 Dap an dung la D Este thuy phan m6i trucfng axit cho axetandehit phai la este CO gg'c ancol kh6ng no chiia nguyfin tur C C6ng thiic D phu hgp I CH2 = CH-COOH + NaOH - > CH2 = CH-COONa + H2O (3) CH3COOH + NaOH ^ CHjCOONa + H2O (4) Ggi X , y va z l^n lugt la s6' mol CH2 = CH - COOH, CH3COOH va CH2 = C H - C H O CO 0,04 mol h6n hgp X Ta co: X + y + z = 0.04 Theo (1) v^ (2): n^,^ = x + 2z = ^ Theo (3) va (4): nNaOH (I) = 0,04 (mol) = x + y = 0,04.0,75 = 0,03 (II) (III) 25.5 Dap an diing la C Giai he phuong t r i n h (I), ( I I ) va ( I I I ) dugc : x = 0,02 mol 25.6 Dap an diing la D Kh6'i lugng cua CH2=CH-COOH X la : m = 72.0,02 = 1,44 (gam) X la axit (tac dung dugc v6i Na va NaHCO,) Y la hgp chat tap chiJc, c6 m6t chiic ancol (tac dung dugc vdi Na) va m6t chiJa andehit (tac dung dugc \6i AgN03 dung dich N H ) ->• chgn dap an D 262 (2) 25.9 , Dap an diing la D Hv -• = 0,05 (mol) X la este don chiic 100 Hop cha^t huu C O X tac dung v d r i dung dich NaOH cho mud'i C.HyCOONa v6i s6' mol la 0,05 m o l 26.3 Dap an diing la B Day E" nho dSn: E"(Y - Cu) > E"(Z - Cu) > E"(Cu - X ) (12x + y + 44 + ) , = 3,4 ^ X Tinh k h i tang din theo day: X , Cu, Z , Y = 0, y = 26.4 Dap an diing la C C6ng thiic phan l\x cua mu6'i la : HCOONa = , - ( - , 4 ) = 0,78 V Chon dap an D v i : H C O O C ( C H ) = C H C H + N a O H -> HCOONa + C H , - C - C H C H 26.5 Dap an dung la D Cap oxi hoa - khit M" V < Fe V , thi kirn loai M khir ducJc ion Fe^\ 'M" /Fe O Xeton C H - C - C H C H kh6ng lam mat mau nirdfc brom 26.6 Dap an diing la A O Sua't dien dOng ciia cAc p i n : 25.10 Dap an diing la B Pin Z n - C u : E" = + , - (-0,76) = 1,1V Ta da bia't: Hidro hoa andehit cho ancol bac I Pin A l - Z n : E" = - , - ( - l , 6 ) = 0,9 V Hidro hoa xeton cho ancol bac I I CH, - C H - C - C H + H , -> Q i , - C H CH3 O CH3 (3-metyl butan-2-on) Pin Z n - P b : E" = - , - ( - , ) = 0,6 V CII-CH, OH f (3-metyl butan-2-ol) iP" 25.11 Dap an diing la D So d6 phan itng: ^"QH > c H , C H - C H — C H , - C- OH CH3 II O 25.13 Dap an diing la A Cac phirong trinh phan ling: C H = C H + O2 (Y) CH2=CH2 + H2O E^y = 0,8 - 0,46 = 0,34V E'znV = , - 1,1 = - , V /Zn O C.H.-C-CHjBr, (X) -> Pin Z n - Cu CO sua^t dien d6ng Idn nha't .7 Dap an diing la D 25.12 Dap an diing la A QH,-CH=CH2 Pin P b - C u : E" = +0,34 - (-0,13) = 0,47 V P''^'2.CuCl2 ^ C H - C H O (andehit axetic) ) CH3CH2OH (ancol etylic) 26 D A Y T H E D I E N CUC C H U A N 26.1 Dap an diing la D 26.2 Dap an diing la D Qua trinh khCr Cu^"^ d cue duong pin dien h6a Z n - Cu pS.S Dap an diing la A K h i dien phan dung dich X (g6m C U S O va NaCl), d anot c6 anion c r phong dien truofc tao k h i CI2 He't anion C I " , anion O H " se phong dien tao k h i O2: - Thdi gian cdn de' dien phan he't 0,12 mol C I " ( C r C l j + 2e) la: ^ = • ^ • = 4,26 (gam) m.n.F 4,26.1.96500 , ^ t^,_ = = = 5790 (giay) c A.I 35,5.2 -> Thdi gian dien phan anion O H " la: 9650 - 5790 = 3860 (giay) A It 16 2.3860 u n n - i1 mo = — — = — = 0,64 gam hay 0,02 m o l °2 n F 96500 ^ ^ Thi tich k h i (dktc) thoat cf anot sau 9650 giay dien phan la: V = (0,06 + 0,02).22,4 = 1,792 ( l i t ) 26.9 Ddp an dung Ih C Phuomg trinh phan iJng: Fe304 + SHI ^ 3Fel2 + I2 + 4H2O 26.10 Dap an diing la B 26.11 Dap an diing la C Sua't dien ddng chudn = E°-E°_) po _ pO _ pO E°„ =-0,76-(-2,37)= 1,61V 27 CROM, DONG, NIKEN, CHI, KEM, BAC, VANG, THifi'c 27.1 Dap an dung la D Chat ran kh6ng tan ki^m dac la Fe203 -> mp^^Q^ =16 gam Cac phuomg trinh phan iJng nhift nh6m : 2AI + FejOj AI2O3 + 2Fe (1) 2A1 + CrjOj ^ AI2O3 + 2Cr (2) Theo (1) : D^ khu duac 16 gam Fe203 cin 5,4 gam Al -> Kh6'i luong Al tham gia phan ling (2) la : 10,8 - 5,4 = 5,4 (gam) Theo (2): Kh6'i luong Cr203 la : ^ ^ ^ ^ = 15,2 (gam) 2.27 ^mcrjo, = ^41,4 -100% = 36,71% 27.2 Dap an dung la B 84 S6' mol Cu = — = 0,06 mol 64 Phucfng trinh phan ling : 3CU + N O + H + - > 3Cu2+ + N O t + H O TNI : Sfi'mol H+ = 0,08.1 = 0,08 mol D^ phan urng vdi 0,08 mol H"^ chi c&n 0,03 mol Cu -> Cu du, the' tich NO bay ti'nh theo H+: 166 V, = I n^^ 22,4 = 0,02.22,4 = 0,448 (mol) 7W2 : S6' mol H+ = 0,08 + 0,08.0,5.2 = 0,16 mol Theo phuong trinh phan irng Cu phan iJng vijra he't V2= I .22,4 = 0,04.22,4 = 0,896 mol -» Vj = 2V, 17.3 Dap an diing \h C ncr203= ^ = a (mol) Phuang trinh phan ling : 2A1 + CrjOs AI2O3 + 2Cr (1) Vi kh6ng bie't gia tri ciia m nfin phai xet hai trucfng hop : Tritdng hap I : Du Al, Cr203 phan ling he't: I H6n hop ran X g6m 0,1 mol A I O , 0,2 mol Cr va Al du mx = 0,1.102 + 0,2.52 + m^i^ = 23,3 gam -> m^i^ = 2,7 gam ->nA,a„ = 0,l mol Cr + 2HC1 -» CrClj + H t 0,2 mol 0,2 mol (2) >• (3) 2A1 + 6HC1 2AICI3 + 3H2t 0,1 mol 0,15 mol nH2 = 0,2 + 0,15 = 0,35 mol ^ VH^ = 0,35.22,4 = 7,84 Trudng hap : Du Cr203, Al phan ling hS't: H6n hgrp ran X g6m AI2O3 , Cr va Cr203 du Vi kh6ng bie't gia tri cua m ntn khong tinh dugc 27.4 Dap an dung la A phan ling tao kim loai Cu 1^ (1), (3) va (4) : 1) 2CU2O + CujS ^ C u + S t 2) CuO + CO ^ Cu + C O 3) C u O + 2NH3 3Cu+ + 3H2O f 27.5 Dap an diing la B CrO la oxit baza; Cr(0H)2 la m6t baza Chi CO Cr203 va Cr(OH)3 c6 tinh luong tinh 27.6 Dap an dung la B Cac phuang trinh phan ting : AI2O3 + H N O ^ 2A1(N03)3 + H O a mol 6a 2a CuO + H N O Cu(N03)2 + H2O b mol 2b b Ag20 + H N O ^ 2AgN03 + H2O c mol 2c 2c lit (1) (2) (3) Theo (U 2, 3) : Axit HNO3 va cac exit phan ling vCra du Trong dung dich CO cac muoi A1(N03)3 , Cu(N03)2 , AgN03 Mu6'n thu duoc Ag tinh khiet phai dung b6t Cu theo phan ling : Cu + 2AgN03 -> Cu(N03)2 +2Ag>l' 2c S6' mol Cu = ~ s6' mol AgN03 = — = c mol 2 27.7 Dap an diing la B C U C I va Z n C l chuydn phiic tan : [Cu(NH3)4]OH va [Zn(NH3)4](OH)2 A I C I chuy^n mu6'i tan K A I O Chi CO FeClj chuyin Fe(OH)3 khdng tan 27.8 Dap an dung la B 2/.12 Dap an diing la A IB I Cac phucfng trinh phan ling: m Zn + FeClj-> ZnCl2 + Fevl' (1) Fe + C U C I - > FeCl2 + Cu4 (2) Zn + C U C I - > ZnCl2 + Cui (3) • Theo (1, 2, ) : Zn da B V i khe'i luong Cu tao nho hon kh6'i lucmg Zn la 0,5 gam ntn kh6'i H luong CuClj (mat di) cung nho hon kh6'i luong ZnCl2 (tao thanh) la 0,5 gam ^ v a y khd'i luong mud'i X (FeCl2, CuClj) la: m = 13,6 - 0,5 = 13,1 (gam) the' Cu CUCI2 (FeCl2 kh6ng thay doi) 27.13 Dap an dung la B Khi X la CO2; = nco, C O + Ca(OH)2 (du) -> CaCO, + H O Phirong trinh phan ling : C U + N O +8H^ -> S C u ^ ^ + N O + H O S6' mol Cu = 3,2 : 64 = 0,05 (mol) Sd'mol =0,1.(0,8+ 0,2.2) = 0,12 (mol) Theo (1): De phan ling he't vdi 0,05 mol Cu cSn: + |.0,05 = 0,033 (mol) NO3 < 0,1.0,8 = 0,08 (mol) (1) %o, - "caco, = ^'04 mol I -> Vco, = = 0,04 22,4 = 0,896 (lit) 17.14 Dap an dung la B Hai chat CO tinh cha't luofng tinh la: Cr(OH)3 va Zn(0H)2 du N O 17.15 Dap an diing la C + - 0,05 = 0,133 (mol) > 0,12 mol -> thie'u Cac phuong trinh phan ung: 2A1 + 3H2SO4 -> Al2(S04)3 + Do vay ta tinh VNotheo H"^ 27.9 Dap an dung la C (1) (2) Fe + H2SO4-» FeS04 + H t (3) S6' mol H2 = — - = 0,35 (mol) 27.10 Dap an dung la B Phucmg trinh phan ling : 2CrCl3 + C I + 16KOH -> 2K2Cr04 + 12KC1 + Cr + H S O - > CrS04 + H t ^ Theo(l): V^o = - -0,12 22,4 = 0,672 (lit) 3H2t ' 22,4 Theo (1, 2, 3): S6' mol 804" tao mudi = s6' mol H2 = 0,35 mol 8H2O S6' mol CI2 = - s6' mol CrCl, = | 0,01 =0,015 (mol) 2 Sd'mol KOH = — sd'mol CrCl, = —.0,01 = 0,08 (mol) 2 27.11 Dap an dung la B Ba dung dich phan ung duac vdfi Cu(OH)2 la: HCl, NaOH dac va N H " - > Khd'i luong mudi khan la: m= 13,5 + 0,35.96 = 47,1 (gam) 27.16 Dap an diing la A Khdi luong oxi da phan ling la : 23,2 - 16,8 = 6,4 gam 64 Sd mol oxi nguyen tit phan ting la : no = —^ = 0,4 mol 16 268 26 Giai he phuong trinh (I) va (II) dugc : x = y = 0,15 mol Khi cac oxit phan ung vdri HCl, m6i mol nguySn tu- oxi duac thay the' bang mol CI 3.0,15" I"Cu(N03), - 0,15 + - - > nHci = 0,4.2 = 0,8 mol Vau Hci = mpeCNo,), =3.0,15.180 = 81 (gam) % = 0.4 (lit) hay 400 ml .188 = 70,5 (gam) Khdi luong mu6'i khan la : m = 70,5 + 81 = 151,5 (gam) 27.17 Dap an dung la B 27.19 Dap an dung la D S6' mol cac cha't: S6' mol cac cha't: npeci, = 0,1.1,2 = 0,12 mol -> n^,- = 0,24 mol ncu(No,), = 0,2.0,8 = 0,16 (mol); i "AgNo, = 0,2.2 = 0,4 mol ->• n^^* = 0,4 mol Cac phuang trmh phan ling : nH,so, =0'25.0,8 = , ( m o l ) ^ n H = 0,4mol Cac phuang trinh phan ling: Fe'-^ + Ag^ ^ Fe^^ + A g ^ (1) Fe Ag^ + C r - > AgCl^ (2) 0,16 mol T h e o ( l ) : n^g = np^, = 0,12 mol Theo (2): n^gci = Fe = 0,24 mol Cu(N03)2Fe(N03)2 + Cui + 4H^ + NO3 -> Fe'^ + 0,1 -> m = 0,12.108 + 0,24.143,5 = 47,4 (gam) 27.18 Dap an diing la D 0,16 mol 0,4 mol Fe + 3Fe304 + 28HNO3 -> 9Fe(N03)3 + NO + I H O (1) 3Cu + 8HNO3 ^ 3Cu(N03)2 + N + 4H2O (2) Tac6:n,o=^x + | y = ^ Ta c6: 56 = 0,15mol (I) Sau phan iJng (1) va (2), Cu dir nen tiep tuc xay phan ling: _3x_ va ncu(NO,),(3) "'^ ~ "Cu(3) Thay (1), (2) va (3): 232x + 3x^ 64 = 61,2-2,4 = 58,8 (II) y+- m 1,56 -0,31 +0,16.64= 0,6m -> m = 17,8 gam (3) Phan ling (3) xay hoan toan (vi v5n dir Cu), dung dich Y 3x (2) 0,1 mol Goi X va y la s6' mol Fe304 va Cu tham gia phan irng (1) va (2) = "FC(NO,), = + 2H2O 3Fe'* S6'mol Fe du la : —-(0,16 + 0,1 + 0,05) 56 Sd' mol Cu tao la : 0,16 mol Theo (1) va (3) : np,(No,K + NO 0,1 mol 0,1 mol 2Fe'*^ 0,05 CO hai mu6'i la Cu(N03)2 va Fe(N03)2 0,16 mol Fe du nSn xay phan ling: Cac phuong trinh phan irng: Cu + 2Fe(N03)3 ^ Cu(N03)2 + 2Fe(N03)2 (1) = 0,1.22,4 = 2,24 (lit) F27.2O Dap an dung la C 'NO 1,344 = 0,06 (mol) 22,4 Cac phucfng trinh phan iJng cua X v6i HNO3 dac, nong A l + 6HNO3 A1(N03)3 + 3NO2 + 3H2O (1) Cu + 4HNO3 Cu(N03)2 + 2NO2 + 2H2O (2) - Goi X va y la s6' mol A l va Cu c6 1,23 gam X 071 Tac6: 27x + 64y = Theo(l): 1,23 IINO, = 3x + 2y = Khi x m o l C u O 0,06 Giai he phircfng trinh duoc : x = 0,01, y = 0,015 -> X 0.015.64.100% ^" Dung djch Y g6m A1(N03)3 va Cu(N03)2 Sue k h i N H (du) vao, xay cac phan ling : Al(OH)3>l^ + N H N O Cu(N03)2 + N H + 2H2O -> [Cu(NH3)4](OH)2 + N H N O 86' mol Al(OH)3^^ = s6' mol A l = 0,01 ^ (3) (4) mol 16 - = 0,05 H N O + 3HC1 + A u -> A U C I + NO + 2H2O Phuong trinh phan ling: 2Cu + 4HC1 + O2 - > 2CUCI2 + 2H2O Dap an diing la A '""^ 239 2PbO + S t (1) m-a 239 C u O + N O + O2 N O + O2 + H O - > H N O " ^ " ^ mol PbO 239 223 + a = 0,95m 16a = 4,05m (1) a (2) Theo ( ) : mol Cu(N03)2 tao mol CuO, kh6'i luong giam 108 gam x m o l - - - - - ymol - - - 1,62 gam = 0,015 mol PbS bi dot chay, tao KhO'i luong h6n hgp ran la: mol Phuofng trinh phan ling: (mol) Theo (1) va (2): S6' mol H N O , = s6' mol CuO = 2.0,015 = 0,03 m Z n C l + H t (1) Cr + 2HC1 ^ CrCl2 + H t (2) Sn + 2HC1 ^ SnCl2 + (3) Ggi X la s6 mol m6i kim loai c6 m gam h n hgrp X Ta c6: ->pH=l Dap an dung la A (65 + 71)x + (52 + 71)x + (119 + 71)x = Phirong trinh phan ling: CuO + CO — ^ ( A I O + CO 272 X Cu + C t '° > phan ting kh6ng xay ra) (1) (2) = 0,02 = nz„ = ns„ = 8,98 ncr Cac phirong trinh phan ling cua h6n hgp X vdi 2Zn + O2 -> 2ZnO Sn + O, SnO, Of (4) (5) 273 4Cr + 3O2 2Cr20, (6) 27.32 Dap an diing la D Cac phuong trinh phan ling: Theo (4, 5, 6): S6' mol O2 da phan ling la: +0,02+ -.0,02 = 0,045 (mol) ^ Cr + 2HC1-> CrClj + H2t 27.27 Dap an diing la C 0.448 ""^ " ilA " ° ' ^'""'^; "cu = — = (1) Nhu vay cha't rdn X g6m: 0,03 mol Cu va 0,17 mol CuO du PhAn tram kh6'i lixgfng cua Cu X la: „ 64.0,03.100% %mp„ = = 12,37% ^" 64.0,03 + 80.0,17 S6'miligam KOH cin dung l a : 0,015.0,1.1000.56 = 84 (mg) Chi s6' axit cua cha't beo la : — = 6,0 28.3 Dap an diing la B CH3COOH + C2H5OH < 2A1 + 6HC1 -> 2AICI3 + 3H2 (n^i : n HCI = : 3) (nc,: ; 28.2 Dap an diing la B Cau C kh6ng diing vi hai phan ling c6 t i 16 s6 mol khac nhau: CrCl2 + H2 28.1 Dap an dung la C ' '\ Chi s6' axit cua cha't beo la s6' miligam KOH cSn thie't Ai trung hoa: axit tu CO gam cha't beo 14 27.28 Dap an dung la C HHCI + 8H^ 0,8 mol + 4H^ 0,8 mol > CH3COOC2H5 + H2O Gia sic phan ling xay thi tich lit Ta c6 : = 1:2) \ - Trudng hcfp ddu : trang thai can bang, n6ng d6 cac cha't: i [ CH3COOH ] = [ C2H5OH ] = (1 - 2/3) mol// Cac phuong trinh phan irng: 0,6 mol mol H2 = s6 mol Cr = 0,2 mol pS DAN XUAT HALOGEN, PHENOL, ANCOL Theo (1): phan ting het vdi 0,02 mol N H , c^n 0,03 mol CuO va tao 0,03 mol Cu 3Fe^^ = 0,2 (mol) = 0,2.22,4 = 4,48 (lit) I 0,2 (mol) 2NH3 + 3CuO ^ N2 + 3H2O + 3Cu Cr + 2HC1 27.29 Dap an diing la B Theo (1): B 16 Phirong trinh phan iJng: 0,3 mol (2) Theo (2): S6 mol Cr = s6' mol Cr203 = — So mol cac cha't: * (l)i 4Cr + 3O2 -> 2Cr203 = 0,045.22,4 = 1,008 (lit) 3Cu | ; + 2NO3 ^ 3Cu^^ + 2N0 + 4H2O 0,2 mol 0,2 mol + NO3 -> 3Fe'^ + NO + 2H2O 0,2 mol (1) [ CH3COOC2H5 ] = [ H2O ] = 2/3 mol// - Trudng hcfp sou : trang thai can bang, n6ng d6 cac cha't: Phan ling (1) va (2) xay hoan toan S6' mol NO la 0.,4 mol ^ V = 0,4.22,4 = 8,96 lit 27.30 Dap an dung la B Mu6'i PbClj la cha't it tan ntn Pb kh6ng dt phan ung vdri dung dich HCI 27.31 Dap an dung la D ; | [ CH3COOH ] = (1 - 0,9) mol//; [ C2H5OH ] = (x - 0,9) mol// (2) 0,2 mol \ [ CH3COOC2H5 ] = [ H2O ] = 0,9 mol// \ Hang s6' can bang KCB kh6ng thay d6i b hai truofng hop va bang : 2 ^ _ [CH3COOC2H5].[H20] _ JS.PR — [CH3COOH].[C2H50H] — 33 _ — (l-0,9).(x-0,9) ^ X = 2,925 S6' mol C2H5OH can d^ hieu sua't phan ling dat 90% la 2,925 moL 28.4 Dap an diing la A d6ng phan xeton c6 CTPT CjHioO la: CH3-CO-CH2-CH2-CH3; CH3 CH2-CO-CH2-CH3; CH3 - C O - CH(CH3) - CH3 28.5 Dap an dung la A X la ancol bac 11, don chiic Dat CTPT cua X la C^H^O QH^O (ancol X) > C,H,_20 • (xeton Y) My I2x I y i- 16 = 29.2 = 58 x = 3, y = CTPT cua ancol X la: CJHRO CTCT cua ancol X la: CH3 - CH(OH) - CH3 28.6 Dap an dung la A 28.7 Dap an diing la B So d6 chuydn hoa : ' CH3-CH2-CHOH-CH3 —i°-> CHjCH^CHCHj ^"^^ >CH3-CH2-CHBr-CH3 _ A f " = ' ^ h a n 0,3 -» 14n = 24,33 -^ n = 1,74 n, = CTPT ciia axit thil nha't la C H C O O H n2 = ^ CTPT cua axit thir hai la C H C O O H 28.12 Dap an diing la A >cH3-CH2-CH(MgBr)-CH3 28.8 Dap an dung la C 54 S6 mol Ag = — = 0,5 mol > Mn s6' mol andehit don chiic 108 -> H6n hop Y c6 andehit fomic HCHO va d6ng ding ke' tie'p la CH3CHO Goi so mol HCHO h6n hop Y la x -> s6 mol C H C H O Ik (0,2 - x) Ta CO : x mol HCHO cho 4x mol Agi va (0,2 - X) mol C H C H O cho 2(0,2 - x) mol mol ancol etylic cho mol HjO mol ancol Y cho mol HjO :=> phan tii ancol Y c6 lOH -> ancol Y la C4H,OH (CH3CH2CH2CH2OH) I 28.11 Dap an diing la A 6,72 = 0,3 (mol) S6 mol H2 = 22,4 -> S6' mol C H O H = s6' mol C-Hj-COOH = 0,3 mol C6ng thiJc phan tit chung cua hai este la: C-Hj-COOCH, (0,3 mol) Khd'i lugng mol trung binh ctia cac este la: Mcste-14n + 44 + - — - , 3 Agi -> 4x + 0,4 - 2x = 0,5 -> X = 0,05 mol Trong h6n hop X cd 0,05 mol C H O H va 0,15 mol CjHjOH m = 0,05.32 + 0,15.46 = 8,5 (gam) 28.9 Dap an diing la C 28.10 Dap an dung la B Tach nude ancol Y chi tao anken: Loai dap an A va D (tao anken) Xet dap an B va C: Gia sii d6't mol m6i ancol: S6 mol cac cha't: „ nrn = T8,96 r~7=0,4 ''°2~22,4 , (mol); n H , o = ^ = ' (mol) "2- 18 Dat CTPT chung ciia ba ancol la: C-K^-^p C;iH2-,20- +0, ->nC02f(n + l)H20 ->n = l,6 n n + l 0,65 CTPT trung blnh cua ba ancol la: C, ^H, jO 04 -> S6' mol ancol = - ~ = 0,25 mol 1,6 S6' mol H,0 bi loai tao ete la: 0,25 = 0,125 mol Khd'i luong ete td'i da thu dupe la: m = (12.1,6 + 5,2 + 16).0,25 - 18.0,125 = 7,85 (gam) 28.13 Dap an diing la A 29.4 Dap an dung la D Dat CTPT cua amino axit X la: H N - C.Hj, - COOH Phuong trinh phan ling: 29 A M I N , AMINO AXIT, PROTEIN Dap an diing 1^ C Dat CTPT cua a - amino axit la : HjN-CxHy -COOH Phuong trinh H N - C,Hy - COOH + NaOH -> H N - C.H^ - COONa + H O phan ling : H N - C x H y - C 0 H +HC1 ^ 19 4-15 QHjN-C^Hy-COOH mol X tac dung v6i HCl khO'i luong tang 36,5 gam 16 + 12x + y + 45 = 75 ^ a mol X tac dung v6i HCl khO'i lirong tang 13,95 - 10,3 = 3,65 gam CTCT cua X la: HjN - C H - COOH 12x + y + 16 + 45 = 103 x = , y = CTCT ciia X la : C H - CHj - CH(NH2) - COOH thiic diing la D Mx = 89 ^ nx = — = 0,1 mol ^89 CO CTPT la: 0,1 mol muO'i Y Con du 0,15 - 0,1 = 0,05 mol NaOH -> Hop cMt hiru co X la mu6i ciia axit cacboxylic don chiic va g6c amin don chiic I Khi tac dung v6i NaOH, tao 0,02 mol muefi C^HyCOONa I T a c o : (12x + y + 44 + 23)0,02= 1,64 |: 12x + y = 15-> II H2NCH2COOCH3 0,1 mol X + 0,1 mol NaOH I I m m B my = 11,7 - (0,05.40) = 9,7 gam • M = ^ ^0,1 = 97, ling CTPT Vay CTCT thu gon cua X la: H2NCH2COONa H2NCH2COOCH3 Dap s6' diing la C Phuong trinh phan ung: CfiHjNHj + NaN02 + 2HC1 -> C^U^ CI + NaCl + H2O S6' mol CgHjNHj = s6' mol NaNOj = s6' mol CeHsNjCl = = 0,1 (mol) ^ ^ ^ 140,5 278] X = 1, y = thiic ca'u tao ciia X la I Dap an dung la D Dap an diing la B H2NCH2CH2COOH - Este cua amino axit c6 CTPT la: = 0,02 (mol) -> C6ng % a o H = 0,1.1.5 = 0,15 mol Chat X phan img v6i NaOH nSn c6 th^ Ik: - Amino axit = 1, y = 89 nx = ^ CTPT cua X la : H j N - CjHg - COOH C6ng X Dap an dUng la B a = 0,1 mol -> Mx = — = 103 dvC 0,1 29.2 15 SO' mol benzen = CH3COO-NH3CH3 = (mol) 78 TCr mol C^Hfi di^u che' duoc 1,2 mol Tit 1,2 mol C6H.,N02 ""QHJNH, C6H,N02 dieu che' duoc 0,6 mol = 0,6.93 = 55,8 (gam) C6H5NH2 B9.8 H • I Dap an dung la C Amin X la amin bac I : Loai dap an A va D Xet dap an B va C (d^u la amin don chiic, bac I , phan tuf c6 nguySn tii C) -> Khi dO't chay V lit hoi amin X cho 0,5 V lit N2, V lit khf C O va 4,5 V lit hoi H O (= V lit) Dap dn C phii hop | Dap an diing la D B " Dat CTPT ciia amin la C,Hy(NH2), - Khi a = 1: Giai tha'y v6 l i - Khi a = 2: S6' mol amin = ^'^'^'^'^^ =0,12 (mol) 2.36,5 279 12x + y + 32 = 74 C T P T cua amin l a : H2NCH2CH2CH2NH2 10 D a p an dung Ik D B6'n d6ng phan amin therm bac m6t cua C7H9N la: CH3-C6H4-NH2 (c6 ddng phan: ortho, meta va para) va MUC LUC X = 3, y = 6: phil hap Ldi noi dau Trang Cau true de thi tuyen sinh Dai hoc - Cao ding Khai quat ve phu'ong phap trie nghiem C H , - C H - N H (nfiu oho chat l a amin thom) PHAN I TUY^N CHpN BAI THI T R A C NGHIEM THEO CAU TRUC OE THI A PHAN CHUNG CHO TAT CA CAC THf SINH Nguyen tCf, bang tuan hoan, lien ket hoa hoc 11 Phan ilmg oxi hoa - khijf, t6'c phan img va can bing hoa hoc 14 Sirdienli 17 Phi kirn (cacbon, silic, nito, photpho, oxi, luu huynh, halogen) 20 Dai cU'Ong ve kirn loai 24 Kim loai kiem, kirn loai kiem tho, nh6m, s i t 27 Tong hop noi dung cac kien thifc hoa v6 co thuoc chi/dng trinh p h d thong 35 Dai CU'Ong hoa hoc hCiru co, hidrocacbon 44 D i n xuat halogen, ancol, phenol 47 10 Andehit, xeton, axit cacboxylic 51 11 Este, lipit 55 12 Amin, amino axit, protein 58 13 Cacbohidrat 62 14 Polime va vat lieu polime 15 Tong hop noi dung cac kien thCrc hoa hOru cd thupc chirong trinh p h d thong 65 67 B PHAN RIENG I Theo churdng trinh Chuan 79 16 Tdc phan tjfng, can bang hoa hoc, sir di§n 11 79 17 Andehit, xeton, axit cacboxylic 80 18 Day the dien ciic chuan 19 Crom, dong, niken, Chi, kern, bac, v^ng, thiec 82 84 20 Ph§n bi§t chat v6 co, hoa hoc va van de phat trien kinh tg', xa hoi, moi trifdng 21 D i n xuat halogen, ancol, phenol 88 9° 22 Amin, amino axit, protein ^2 23 Cacbohidrat II Theo chirdhg trinh Nang cao 24 Toe phan Crng, can bang hoa hoc, sir dien li 25 Andehit, xeton, axit cacboxylic 26 Day the dien cifc ehuan 27 Crom, dong, niken, chi, kern, bac, vang, thiec 28 Dan xu§'t halogen, phenol, aneol 29 Amin, amino axit, protein PHAN II: Hl/dNG DAN 95 95 97 99 101 106 108 CHON DAP AN BAI THI T R A C N G H I E M T H E O C A U T R U C D E THI A M O T S P H U O N G P H A P G I A I N H A N H C A C BAI T O A N T R A C NGHI§M HOA HOC I Phirong phap bao to^n khoi Itfong II Phirong phap tang - giam khoi lifOng III Phi/ong phap bao toan electron IV Phirong phap bao to^n dien ti'ch V Phirong phap bao to^n nguyen to VI Phirong phap giai cac bai toan dien phan VII Phirong phap trung binh VIII Xac dmh cong thirc phan tCT cDa cae hop chat v6 co IX Phirong phap xac djnh cong thufc hop chS't hijfu co 111 115 117 120 123 125 129 137 140 B PHAN RI^NG I Theo chirong trinh Chuan 16 Toe dp phan iTng, can bang hoa hoc, sir dien li 17 Andehit, xeton, axit cacboxylic 18 Day the dien circ chuan 19 Crom, dong, niken, chi, kern, bae, vang, thiec 20 Phan biet chat v6 co, hoa hpc va v§'n de phat trien kinh te, xa hpi, moi trirdng 21 Dan xuat halogen, ancol, phenol 22 Amin, amino axit, protein 23 Caebohidrat II Theo chirong trinh Nang cao 24 Toe dp phan irng, can bSng hoa hpc, sirdi^n li 25 Andehit, xeton, axit cacboxylic 26 Day the dien cire chuan 27 Crom, dong, niken, chi, kern, bae, vang, thi§'e 28 Dan xuat halogen, phenol, aneol 29 Amin, amino axit, protein 241 241 242 245 248 254 255 257 259 260 260 262 264 266 275 278 B Hl/dNG DAN CHON DAP AN Nguy§n [it, bang tuan ho^n, lien ket hoa hoe Phan img oxi hoa - khCr, toe dp phan Cmg c§n bang hoa hoc Sir dien li 146 149 150 Phi kirn (cacbon, silie, nito, photpho, oxi, liru huynh, halogen) Dai cirong ve kirn loai 155 160 Kim loai kiem, kirn loai kiem thd, nhom, s^t Tong hop noi dung cac kien thirc hoa v6 co thu6c chirong trinh ph6 thdng 165 183 Dai cirong hoa hoc hCru CO, hidrocacbon Dan xuat halogen, ancol, phenol 10 Andehit, xeton, axit cacboxylic 11 Este, lipit 12 Amin, amino axit, protein 13 Caebohidrat 14 Polime vat li§u polime 15 Tong hop noi dung cac kien thiJfe hoa hOfu co thuoc chu^ong trinh thong 191 198 203 210 217 224 227 228 283 '89 ... la : CH2-A CH2-B CH2-A CH2-B CH2-A CHo-B CH - A CH - B CH - A CH - B CH -B CH - A CH2-A CH2-B CH2-B CH2-A CH2-A CH2-B I 11.5 Dap an diing la C 2, 2 M E = 16 5,5 = 88 dvC ; n,,,, = ^ = 0, 025 (mol)... NH2 (1) NH2 fiOOC - [CH2 ]2 - CH - COOH + 2NaOH -> NaOOC - [CH2 - CH - COONa + 2H2O NH2 CH3 CH - COOH + HCl CH3 - CH - COOH NH2 NH2 (3) NCIH3 HOOC - [CH2 ]2 - C H - COOH + HCl -> HOOC - [CH2 -. .. la : CnH2n (n > 2) va CmH2m _2 (m > 2) Phuong trinh phan ling cua X vdi dung dich Br2 : C„H2n + Br2 -> C„H2„Br2 CmH2n , -2 -' -2 Br2 -> C^H2m-2Br4 (1) (2) Goi X la s6' mol cua anken CnH2n thi s6 mol