Môn Hóa học - Tuyển chọn bài thi trắc nghiệm theo cấu trúc đề thi: Phần 2

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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.

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Nội dung Text: Môn Hóa học - Tuyển chọn bài thi trắc nghiệm theo cấu trúc đề thi: Phần 2

5.10. Dap an diing la A. Cac phuang trinh phan ling: Cac phirang trinh phan ling : 2AI2O3 - i E ^ 4 A I + 3O2 (1) M + Cl2^MCl2 (1) O2 + C — C O 2 (2) 2M + O2 ^ 2MO (2) O2 + 2C -> 2 C O (3) Goi X va y la s5' mol CI2 va O2 tham gia phan ling. 7 94 ? Taco: 71x + 32y = 23 - 7,2 = 15,8 = 0,1 mol X tao ra — = 0.02 mol CaCO, 22,4 - 1 0 0 y + x=-^-0,25 % nco, (trong h6n hgp X) = ^ ^ . 1 0 0 % =20%. 22,4 Giai he phircmg trinh dugc : x = 0,2, y = 0,05. S6' kmol CO2 trong hdn hgp X la 3.20% = 0,6 kmol. Theo ( I ) va (2) : n,^ = x + 2y = 0,2 + 2.0,05 = 0,3 (mol). Dat X la s6' kmol O2 c6 trong 3kmol h6n hgp X. 72 s6' kmol CO = 3 - (x + 0,6) = (2,4 - x) kmol. KhO'i lugng mol cua kim loai M la : ~ = 24 (gam). ^ . ^ 32x +44.0,6 +(2,4-x)28 Vay kim loai M la magie (Mg). Ta CO : dx/H, = — = 16 5 . 1 1 . Dap an dung la C. - » X = 0,6 k m o l ; nco = 1,8 kmol. So mol cac cha't: Theo (1, 2, 3) : S6' kmol O2 thoat ra d phan ling (1) la : ncuci, = 0,1.0,5 = 0,05 (mol) ; n^^.a = 0,5.0,5 = 0,25 (mol) 0,6 + 0 , 6 + — =2,1 (kmol). Cac phuong trinh diSn phan: Theo (1) : n^, = ^ H Q , =|.2,1 = 2,8 (kmol). CUCI2 ""^^ >Cul + C l 2 t (1) 2NaCl + 2H2O -JS!^ C L t + H z t + 2NaOH (2) m.^i = 2,8 k m o l . 27 = 75,6 (kg). 5 . 1 3 . Dap an dung la B. Phan ling (1) xay ra hoan loan, sau do phan ling (2) tiep tuc xay ra. Cac hgp kim: Cu-Fe(I) ; Fe-C(III) va Sn-Fe(IV) chiia Fe la kim loai Dua vao edng thiic Faraday : boat d6ng han kim loai va phi kim con lai nSn khi tie'p xiic vdi dung A It dich chat dien l i , Fe bi an mon. m = —.— 5 . 1 4 . Dap an diing la A. n F - Khi dien phan dung dich CuClj (vdi dien cue tra), d cue am (catoi) Tfnh dugc thdi gian dien phan hoan toan 0,05 mol CuCij la 1930 giay. xay ra phan ling: " Con lai 3860 - 1930 = 1930 giay dien phan mot phan NaCl va tao ra 0,1 Cu^* + 2e->Cu (1) mol NaOH. - Phan irng an mon dien hoa xay ra cr cue am khi nhiing hgp kim Zn- • Dung dich thu dugc chiia 0,1 mol NaOH. Hoa tan A l : Cu vao dung dich HCl: , 2A1 + 2 N a O H + 6H2O 2Na[Al(OH)4] + 3 H 2 t Z n - > Z n ' * + 2e (2) 0,1 mol 0,1 mol Phan ling (1): CO su tham gia cua ion kim loai (Cu^"^). TXIM 0,1.27 = 2,7 (gam). . = Phan ling (2): c6 su tham gia cua kim loai (Zn). 5.12. Dap an dung la C. 5 . 1 5 . Dap an dung la C. Ggi X la s6' mol Zn c6 trong 19,3 gam h6n hgp. S6' kmol h6n hap khi X = = 3(kmol). S6' mol Cu la 2x mol. 22,4 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 (1) FeCl2 + KMn04 -> 0,1 0,2 0,1 mol mol mol FeS04 + KMn04-> Fe + Fe2(S04)3 -> 3FeS04 H2S + K M n 0 4 ^ (2) 0,2 0,2 HCl + K M n 0 4 - » mol mol 3 3 5.19. Dap an dung la A. Cu + Fe2(S04)3 ^ CUSO4 + 2FeS04 (3) - Zn khur ion Fe^^: Zn + Fe'^-> Zn'^ + Fe Theo (3): SO' mol Cu tham gia phan ling = s6' mol Fe2(S04)v - Fe^^ khijr ion Ag^: Fe^^ + Ag^ ^ Fe'^ + Ag. 01^0,2 6. K I M LOAI KIEM, K I M LOAI KIEM THO, NHOM, SAT .p.u--0,2- = 0,1 mol. 6.1. Dap an diing la B. Sau khi phan ling xay ra hoan loan con du 0,1 mol Cu. Cac phuong trinh phan iJng : m = mcu = 0,1.64 = 6,4 gam. 2Na + 2H20 -» 2Na+ + 20H~ + H21 (1) 5.16. Dap an diing la D. 5.17. Dap an dung la C. Ba + 2H20 ^ Ba^^ + 20H~ + H j t (2) Phuang trinh dien phan dung dich CUSO4: Theo (1) va (2) : D Q ^ - = 2. H H , = 2 . = 0,3 (mol). CUSO4 + H,0 - i H ^ Qui + -0,t + H,SO. (1) De trung hoa 0,3 mol OH" cSn 0,3 mol H^" hay 0,15 mol H2SO4. 1 mol CUSO4 -> 64 gam Cu va 16 gam O2 kh6'i lucfng giam 80 gam. 0,15 0,1 mol • X = = 0,15 (mol). 8 0,3.(2.0,1 +0,1) = 0,09 (mol). S6 mol C U S O 4 CO trong 200 ml dung dich dien phan la: S6 mol OH" d phan ling (1) la : 0,12 - 0,09 = 0,03 (mol). ncuso4 = O'l + = 0,25 (mol). Theo (1) : S6' mol K = s6' mol OH" = 0,03 mol. 0.25 - > mK = 0,03.39 = 1,17 (gam). 'M(CuS04) = a = - 0,2 -1,25M. 165 164
6.3. DdpdndungmC. Fe, M g va Zn tac dung vdi H2SO4 loang theo cung m6t kidu phan ling : M + H2SO4 / ^ MSO4 + H21 - % V r o = — . 1 0 0 % = 75%. ^2 0,2 ~ 1 144 S6' mol SO|" tao mu6'i = s6' mol H2 = = 0,06 (mol). Theo (1) : (56x + 16y) gam Fe^Oy phan ling vdri y mol CO. 22,4 8 gam Fe^Oy phan ling vdi 0,05 mol CO. -> m^^ = 0,06.96 = 5,76 - > m = 3,22 + 5,76 = 8,98 (gam). - > - - = - CTPT cua sat oxit la Fe203 . 6.4. Dap an diing la D . y 3 Cac phuong trinh phan ling : 6.7. Dap an dung la A . 2Fe + 3Cl2 ^ 2FeCl3(X) (1) Mg -^2^04'i.n^ MgS04 , Fe -^"^^"^'S Fe2(S04)3 . Fe + 2HC1 ^ FeCl2 ( Y ) + H21 (2) . Trong dung dich con du Fe nen tat ca Fe2 ( 8 0 4 ) 3 da b i khijr het thanh Fe + 2FeCl3 - > 3FeCl2 ( Y ) (3) FeS04. Do vay, chat tan trong dung dich Y la MgS04 va FeS04. 6.5. Dap an diing la B. 6.8. Dap an diing la C. Gia six hoa tan 1 mol h6n hop X (trong do c6 x mol Fe va (1 - x ) mol Cac phuong trinh phan ling : M g ) vao dung dich HCl : XCO3 X 0 + COjt (1) Fe + 2HC1 - > FeCl2 + H21 (1) CO2 + NaOH - » NaHC03 (2) M g + 2HC1 MgCl2 + H21 (2) CO2 + 2NaOH Na2C03 + H2O (3) Theo ( 1 ) va ( 2 ) : S6' mol HCl phai dung la 2 mol. 73 100 mco2 = 13,4 - 6,8 = 6,6 (gam) nco^ = ^ = ^'15 (mol). - > mHci = 36,5.2 = 73 (gam) m^juci = = 365 (gam). S6' mol NaOH = 0,075. 1 = 0,075 mol < ncoj chi xay ra phan ling (2) Sau khi hoan tan vijfa du thi khd'i lirang dung dich ]k : 365 + 56x + (1 - x)24 - 2 = 387 + 32x va CO2 vSn con du. Kh6'i luong mu6'i NaHCOs dugc ti'nh theo NaOH : N6ng d6 % cua FeClj trong dung dich Y : mNaHCO, = 0,075 . 84 = 6,3 (gam). (56 + 71)x 15,76 , •^^ — = — > X = 0,5 mol = nw„. 6.9. Dap an diing la B. 387 + 32X 100 Theo di bai, chi c6 m6t pMn Fe phan ling (vi thie'u H N O 3 ) : %mMeCl. = •^^^^^^^^^•100% = 11,79% ^^•^'2 387 + 32.0,5 Fe + 4 H N O 3 Fe(N03)3 + N O t + 2 H 2 O 6.6. Ddp an diing la B. Fe dir se phan ling vdfi Fe(N03)3 : Goi CTPT ciia sat oxit la Fe^Oy. Phirong trinh phan iJng : Fe + 2Fe(N03)3 -^•3Fe(N03)2 Fe.Oy + y C O xFe + y C O j (1) K i m loai du la Cu va c6 thi c6 Fe. Theo (1) : CO phan ting tao thanh C O 2 , do do the' tich vSn la 4,48 lit 6.10. Dap an diing la A . (dktc), tu-c la bang 0,2 mol. nFc= = 0 , 1 2 (mol). Goi a la s6' mol CO c6 trong h6n hop sau phan Hag - > s6' mol CO2 la (0,2 - a). DO 167 166
Phucfng trinh phan dng : - K h i cho hOn hop X vao dung dich NaOH, Na va A l d^u tan he't. 2Fe + 6H2SO4 -> Fe2(S04)3 + S S O j t + 6 H 2 O (1) Vi 0,1 mol 0,3 mol 0,05 mol Theo (1) : X mol Na lam thoat ra -y-lit H2. Sau phan ling (1), Fe du 0,02 mol ntn xay ra phan ling : Fe+Fe2(S04)3 -~>3FeS04 (2) Theo (2) : y mol A l lam thoat ra 3 lit H2. Nhu vay sau (2) thu dugc : 0,05 - 0,02 = 0,03 (mol) F e j ( 8 0 4 ) 3 va 0,06 mol FeS04. Taco: ^ + 3 ^ = 1,75V (II) 6.11. Dap an dung la C. 2 2 V2 6 2 y 3Fe + 202 ^ Fe304 (Fe203 .FeO) (1) ^ ^' + 3 ^ = 1 , 7 5 . 2 V , - > ^-A-i-ii . i l l m . = 1^3 = 23 ^ = 29,87%. H6n hop X gdm FeO, Fe203 va Fe du. mAi 2.27 54 23 + 54 Dat X la s6' mol Fe du, s6' mol Fe304 la y. Ta c6 : 6.13. Dap an dung la A . 56x + 232y = 3 (gam) (I) Cac phuong trinh phan ling : Trong h6n hop X chi c6 Fe va FeO tac dung vdi dung dich H N O 3 , AICI3 + 3NaOH ^ Al(OH)3 ^ + 3NaCl (1) cho ra N O : Al(OH)3 + NaOH - > N a [ A l ( O H ) J (2) Fe + 4 H N 0 3 Fe(N03)3 + 2 H 2 O + N O t (2) Theo (1) : H A K O H ) , = " A i C b = 1'5 . 0,2 = 0,3 (mol). 3FeO + I O H N O 3 3Fe(N03)3 + SlhO + NOt (3) S6' mol Al(OH)3 tham gia phan ling (2) = 0,3 - ^ = 0,1 (mol). Theo (2) va (3) : x + - y = = 0,025 (mol). (11) Theo (1) va ( 2 ) : S6' mol NaOH da phan ling = 0,9 + 0,1 = 1 (mol). Giai (I) va ( I I ) ta dugc : x = 0,0225 ; y = 0,0075. - > np, = np, + np,,,) = x + 3y = 0,045 mol. Gia tri Idn nha't cua V la : V = = 2 (lit). mp, = 0,045.56 = 2,52 (gam). 6.14. Dap an dung la D. 6.12. Dap an dung la A. Bai tap nay gidi theo phuang phdp can bang dien tich : Cac phuong trinh phan ling : K h i tan vao dung dich H N O 3 vCra dii (chi tao thanh muO'i sunfat va 2Na + 2 H 2 0 ^ 2NaOH+H2t (1) khi N O t ) . 2A1 + 2NaOH + 6H2O - > 2Na[Al(OH)4] + 3 H 2 1 (2) 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 trong m gam X. a mol CU2S -)• 2a mol Cu^"*^ va a mol SO4". - K h i cho X vao nude, Na tan he't, A l tan khOng he't (vi thie'u NaOH). -> S6' mol dien tich duong = 3.0,12 + 2.2a = (0,36 + 4a) mol. Thi tich H2 thoat ra tijr hai phan ling (1) va (2) dugc tinh theo Na : SO' mol dien tich am = 2.0,24 + 2a = (0,48 + 2a) mol. Theo (1) : X mol Na lam thoat ra ^ l i t H 2 . Trong dung dich : 0,36 + 4a = 0,48 + 2a - > a = 0,06 mol.
Gia six c6 1 mol A l C l , . Di phan iJng xay ra ho^n toan (khdng con K [ A l ( O H ) 4 ] + CO2 -> A l ( O H ) 3 ^ + KHCO3 (3) tua) t h i phai c&n 4 mol NaOH. Do vay, c6 ke't tua t h i s6' mc 46 8 NaOH phai nho hon 4 mol. NghTa la : a : b > 1 : 4. Theo (1,2,3): nK|Ai(OH),i= x + 4y = nA,(OH),= = 0,6 (mol). 6.16. Dap an dung la B. ^ , X + y = 0,3 " Cac phuong trinh phan ling : Ta CO he phuong trinh: x = 0,2; y = 0,l. X + 4y = 0,6 F e + H2SO4 FeS04 + H j t (1) -> a = l,5x + 3 y = 1,5.0,2 + 0,1.3 = 0,6 (mol) 10FeSO4+2KMnO4+8H2SO4 ^ ^,19. Dap s6' dung la C. - » 5 Fej ( S O 4 )3 + K 2 S O 4 +2 MnS04 + 8 H 2 O (2) Cac k i m loai M g , Cu, A l tac dung hoan tokn v6i O2 tao ra cac oxit Theo (1) va (2) : So mol KMn04 = - s6' mol Fe = = 0,02 mol. MgO, CuO, AI2O3. 5 56.5 Cac oxit nay tac dung vQ\: -> VddKMn04 = ^ = 0,04 l i t (40 m l ) . M g 0 + 2HC1 MgClj + H 2 O (I) f 6.17. Dap an diing la D. CuO + 2HCI -> CUCI2 + H 2 O (2) So do b i ^ d6i: AI2O3 + 6 H C 1 2AICI3 + 3 H 2 O (3) Fe >(Fe, FeO, Fe^Oj, F e j O J - i ^ ^ ^ ^ F e C N O , ) , + N 0 t (1) S6' mol oxi nguyen tir chiia trong cac oxit: X gam 11,36 gam m gam 0,06 mol 3 3 3 - 2 13 Ho = ' ' = 0,075 (mol). 16 0 3+ 2.K Theo (1,2,3): S6' mol HCl = 2 so mol O = 0,15 m o l . - Qua trinh cho electron: Fe -> Fe+ 3e => n^(cho) = 56 V j . H c i = 0 , 1 5 : 2 = 0,075 l i t hay 75 m l . o -2 +5 +2 ,; - Qua trinh nhan electron: O2 + 4e ^ 2 0 va N + 3e -> N 6.20. Dap so diing la D. 11,36-x . „ „ , , 12,8-x ' Cac phuong trinh phan ling: =:>nnhan = —^ .4 + 0,06.3 = — FeO + 2HC1 FeCl2 + H 2 O (1) ' • 32 8 3X 12 8 — X F e 2 0 3 + 6HCI -> 2FeCl3 + 3 H 2 O (2) n,(cho) = n i n h a n ) = — = — =>x = 8,96 (gam) F e 3 0 4 (FeO. F e 2 0 3 ) + 8HC1 FeClj + 2 FeCl, + 4 H 2 O (3) .ft 56 88,96 S Dat s6' mol cua FeO la x -» s6' mol F e 2 0 3 la x; Theo so d6 (1): n^^ = n = - ^ - = 0,16 (mol). 56 Dat s6' mol ciia F e 3 0 4 la y ^ so mol FeO la y va so mol F e 2 0 3 la y. m = 0,16.242 = 38,72 (gam). Ta c6: 72(x + y) + 160(x + y) = 232(x + y) = 2 , 3 2 x + y = 0,01 6.18. Dap an diing la B. Theo (1,2,3): nnci = 8(x + y) = 8.0,01 = 0,08 mol Dat X va y la s6' mol A l va AI4C3 c6 trong 0,3 mol h6n hop. | - > V j j „ c , = 0,08 : 1 =0,08 l i t . Taco: x + y = 0,3. Cdch khdc: V i s6' mol FeO = s6' mol FejO, nen c6 the' coi 2,32g h6n Cac phuong trinh phan img: • ^ hop la chi ciia F e 3 0 4 2AI + 2 K O H + 6H2O 2K[A1(0H)4] + 3 H 2 t (1)^ Fe,0, + 8HC1 ^ FeCl, + 2FeCl3 + 4 H 2 O X mol X l,5x 0,01 -> 0,08 AI4C3 + 4 K 0 H + I 2 H 2 O 4K[A1(0H)4] + 3 C H 4 t (2) =>V,,„„=M! = 0,08 (lit) ^ y mol 4y 3y 170 ^ 171
6.21. Ddp s6'dung la A . Cac phuong trinh phan ling: 6.24. Dap s6' diing la A. Phuong trinh phan img: 2Na + 2 H 2 0 2NaOH+H2t (1) 2A1 + F e A — ^ AI2O3 + 2Fe (l) 2NaOH + 2 A l + 6 H 2 0 -> 2Na[Al(OH)4] + S H j t (2) H6n hop Y tac dung vdi dung dich NaOH sinh ra khi Hj, chung to trong Dat s6' mol Na la x mol -> s.6 mol A l la 2x mol. Y CO AI2O3, Fe va Al du (Fe203 da phan ling he't). Theo (1,2): S6' mol = - + - x = 2x = — = 0,4 (mol). Cac phuong trinh phan ling cua h6n hop Y. ^ 2 2 22,4 |p' -dphan I: 2A1 + 3H2SO4 ^ Al2(S04), + 3H2T (2) Theo (2): S6' mol A l da phan ling = so mol Na = 0,2 (mol). Fe + H2SO4-> FeS04 + H j t (3) -> S6' mol A l du = 0,4 - 0,2 = 0,2 (mol). AI2O3 + 3H2SO4 ^ Al2(S04)3 + 3H2O (4) -> m = 0,2.27 = 5,4 (gam). - dphan 2: 2A1 + 6H2O + 2NaOH ^ 2Na[Al(OH)4] + 3H2t (5) 6.22. Dap so dung la B. AI2O3 + 2 N a O H + 3H2O 2Na[Al(OH)4] (6) 78 S6' mol Al(OH);, >^ = — = 0,1 (mol). 2 ? n 84 78 Theo (5): S6' mol A l du = - s6 mol Hj = - . ^ = 0,025 (mol). Cac phuong trinh phan ling: 3 ' 3 22,4 AI2 ( 8 0 4 ) 3 + 6NaOH 2 AKOHX^xl + 3 Na2S04 (1) Theo (2 va 3): S6' mol Fe = s6' mol H2 = ^'"^"^'^"^ = 0,1 (mol). 0,1 0,6 0,2 22,4 Theo (1): S6' mol Al phan ling = s6' mol Fe = 0,1 mol. A1(0H)3 + NaOH Na[Al(0H)4] (2) (0,2-0,1) 0,1 S6' mol FcjO, phan ling = ^ s6' mol Al phan ling = 0,05 mol. H2SO4 + 2NaOH ^ Na2S04 + 2 HjO (3) -> m = 2 [(0,025 + 0,1).27 + 0,05.160] = 22,75 (gam). 0,1 0,2 6.25. Dap an dung la B. c6 lugng ke't tua 7,8 gam, thi tich dung djch NaOH Idn nha't la: Cac phuong trinh phan ung: 0,6 + 0,1 + 0,2 VjdNaOH = ^ = 0,45 lit ^ 4FeC03 + 0 2 2Fe203 + 4COjt (1) 6.23. Dap so diing la C. a mol — mol a mol 4 S6 mol cac chat: Hp,. = n^,, = 0,1 m o l ; . OAgNo, = 0,55 mol. / 4FeS2 +IIO2 -> 2Fe203 + 8S02t (2) Cac phuong trinh phan ling: I 11 A l + 3AgN03 -» A1(N03)3 + 3 A g i bmol —bmol 2b mol 4 0,1 0,3 0,3 mol Ap sua't tru6c va sau phan img bang nhau n6n s6' mol khi cung bang nhau. Fe + 2AgN03 Fe(N03)2 + 2Ag>l Taco: - + — = a + 2b-> a = b. 0,1 0,2 0,2 mol 4 4 Sau phan ling (1) va (2), AgN03 con du 0,55 - (0,3 + 0,2) = 0,05 (mol) ^ ^•26. Dap s6' dung la A. do do xay ra phan ling: Cac phuong trinh phan ling: Fe(N03)2 + AgNO, -4 Fe(N03)3 + Agi FeO + 2HC1 FeCl2 + HjO (1) 0,05 0,05 mol Fe203 + 6HC1 ^ 2FeCl3 + 3H2O (2) Vay: m = 108(0,3 + 0,2 + 0,05) = 59,4 (gam). Fe304 + HCl: xay ra nhu (1) va (2). 172 17'^
Theo (1): ma-o = ^ • 72 = 4,32 (gam). W cua axit da bay khoi dung dich du6i dang H2. - S6' mol CI = 0,5.1 = 0,5 mol -> m^, = 17,75 gam. ^ mp,^o, = 9,12 - 4,32 = 4,8 (gam). S6 mol SO^' = 0,5.0,28 = 0,14 mol -> m^^- = 1,3,44 gam. -> npco. = loU ^ =0'03 mol. Kh6'i luong mu6'i khan la: Theo (2): S6' mol FeCl, = 2 s6 mol FcjO, = 0,06 (mol). m = 7,74 + 17,75 + 13,44 = 38,93 (gam). ^ mpeci, = 162,5 . 0,06 = 9,75 (gam). Dap s6' diing la A. Cac phuong trinh phan ling: 6.27. Dap an dung la A. 0 448 2A1 + FejO, — > AI2O3 + 2Fe (1) S6' mol CO, = - ^ ^ ^ = 0,02 (mol). 22,4 0,1 mol 0,1 mol Cac phuang trinh phan img: ' H6n hop ran X gom 0,1 mol AI2O3, 0,2 mol Fe va Al du. M H C O 3 + HCl ^ MCI + CO2T + H2O (1) H6n hop X tac dung v6i dung dich NaOH: M 2 C O , + 2HC1 -> MCI2 + CO2T + H2O (2) 2AI + 6H2O + 2NaOH -> 2Na[Al(OH)4] + 3H2t (2) Theo (1) va (2): S6' mol h6n hop muoi = s6' mol COj 0,1 mol 0,15 mol (M + 61 + 2M + 60).0,02 , „ AI2O3 + 2NaOH + 3H2O --^ 2Na[Al(OH)4] (3) Va = 1,9 0,1 mol 0,2 mol 2 0,1 + 0,2 = 0,3 (lit) hay 300 ml. Glai phuong trinh diroc M = 23. 1 Vay kim loai M la natri (Na). 6.33. Dap an dung la B. 6.28. Dap an dung la C. S6' mol cac cha't: n^gNo, = HCUCNO,), =0,3.0,1 = 0,03mol. Benzen khong phan irng dirgfc v6i nude brom. Phenol (do anh hudng cua nhom -OH) phan iJng d6 dang vdi nu6c Cac phuong trinh phan iJng: brom (the Br vao cac vi tri ortho va para). Al + 3AgN03->AI(NO,)3+3Ag>l (1) 6.29. Dap an dung la C. 0,01 mol 0,03mol 0,03mol Cac phuong trinh phan ung: 2A1 + 3Cu(N03)2-^2Al(N03)3+3Cu>l (2) 2A1 + 2 N a O H + 6H2O ^ 2Na[Al(OH)4] + 3H21 (1) 0,02 mol 0,03 mol 0,03 mol 8A1 + 3OHNO3 SAKNO,), + 3N2O + I5H2O (2) Dat a la so mol Al tham gia m6i phan ting: Al du tac dung \6i dung dich HC\: 3 2Al + 6HC1 — ^ 2AICI3 + 3H2 t (3) Theo (1): n^ = x = - a 2 S6' mol Al du = - s6' mol H,= = 0,01 (mol) 3 Theo (2): 0 ^ , 0 = y = - a -> x = 4y. 3 ^ 3 22,4 " 8 ^ m, = 27(0,01 + 0,02 + 0,01) = 1,08 (gam) 6.30. Dap an diing la B. 6.31. Dap .so diing la A. = 0,03.108 + 0,03.64 + a01.27 = 5,43 (gam). - 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.
6.34. Dap an dung la D. Theo (1) va (2) : = 2 X + 6 X = 8 X = 8.0,02 = 0,16 (mol) HHCI Cac phuang tnnh phan irng: ^ Vdd HCl = 2.-0,16 = 0,16 (lit) hay 160 ml. NajO + HjO >2NaOH Dap an dung la D. AI2O3 +2NaOH + 3H20 >2Na[Al(OH)4] Cac phuong trinh phan ling : Mg + 2FeCl3 MgCl2 + 2FeCl2 (1) Chat tan duy nha't la Na[Al(OH)4] c6 s6' mol la 0,5.0,2 = 0,1 (mol). Mg + FeCl2 -> MgClj + Fe (2) Th6i khi CO2Na[Al(OH)4] vao dung dich Na[Al(0H)4] : + C02 ->A1(0H)3 i + NaHCOj Theo (1) : n^g =^np,c., =^-0,12 = 0,06 (mol). 0,1 mol 0,1 mol Theo (2) : n^g = n F e = ^ = 0,06 (mol). ^ a = 0,1.78 = 7,8 (gam). m^g = (0,06 + 0,06).24 = 2,88 (gam). Theo (1) va (2) va (3): n,^,^o, = ^ n N a | A i ( O H ) , i =0-05 mol Dap an dung la B. 3 136 "Na.O = ^ n N a | A l ( O H ) J =0.05 (mol) H6n hop khf Y g6m NO va NjO ; = 0,14(mol). ^ m = 0,05.62 + 0,05.102 = 8,2 (gam). Goi X la s6 mol NO c6 trong Y. Ta c6 : 6.35. Dap an diing !a A. 30x + 44(0,14-x) = 5,18 = 0,07 mol = 0'07 mol. So mol Ba(OH)2 = 34,2 = 0,2 (mol). X DN^O Khi cac phan irng oxi hoa - khir xay ra : 171 Goi a va b la sd mol Al va Mg c6 trong 8,862 gam h6n hop. Phuong tnnh phan irng; Taco: 27a + 24b = 8,862 (I) 3+ (NH4)2C03 + Ba(0II)2 ^ BaCO,! + 2NH3t + 2H2O a A l - 3ae aAl : Cho 3a mol electron. 0,1 mol 0,1 mol bMg-2be hMg^"" : Cho 2b mol electron. ^ m = 0,1.197 = 19,7 (gam). +5 0,07 N : Nhan 0,21 mol electron. 0,07N +0.07.3e 6.36. Dap an dung la D. +5 +1 Cac phirang trinh phan irng : 0,07.2 N +0,07.2.4e -> 0,07 N : Nhan 0,56 mol electron. FeO + 2HC1-> FeClj + H2O (1) Theo dinh luat bao toan electron : Fc^O, + 6HC1 2FeCl, + SHjO (2) 3a + 2b = 0,21 +0,56 = 0,77 (II) Fe,04 CO c6ng thirc FeO.Fe203 cung phan irng nhu tren va cung thoa man: Giai he phuong trinh (I) va (II) duoc : a = 0,042 mol. so mol Fe^^ : s6' mol Fe'^ = 1 : 2 %mA, = 0,042.27 -.100% = 12,80%. 8,862 Goi X la s6' mol ciia m6i oxit trong nira h6n hop X. Ta c6 : 6.39. Dap an diing la B. Theo (1) va (2) : m, = 127x + 325x = 452x Sue khf Clj vao phan hai, xay ra phan irng : S6' mol cac chat: n„t,^„ 47,4 = — = 0,1 mol nsoj- =0,1.2 = 0,2 mol. ' 474 2FeCl2 + CI2 -> 2FeCl3 (3) nBa(OH), =0,2.1 =0,2 m o l - n^u =0,2 mol. Theo (1), (2) va (3) : m2 = 3x.l62,5 = 487,5x m2 - m, = 487,5x - 452x = 0,71 -> x = 0,02.
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. Chat ran Z g6m BaS04 va Fe203. "Baso, = 0.2 mol -> mBaso, = 0.2 . 233 = 46,6 (gam). 6.43. Dap an diing la A . 6.40. Dap an diing la D. 3,36 nu = = 0,15 (mol). 22,4 S6 mol cha't tan (hidroxit) = = 0,02 (mol). 2> Cac phuong trinh phan ling: S6' mol B, = 0,01 (mol). 8A1 + 3Fe304 — ^ 4AI2O3 + 9Fe (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 ra cac phan ling : Cac phuong trinh phan ling: AI2O3 + 2NaOH + 3H2O ^ 2Na[Al(OH)4] (2) M + 2H2OM(0H)2 +Hjt (1) 2A1 + 2NaOH + 6H2O -> 2Na[Al(OH)4] + 3H2t (3) MO + H 2 O M ( 0 H ) 2 (2) Dung dich Y la dung dich N a [ A l ( O H ) 4 ] , cha't ran Z la Fe. Theo (1) : n ^ = XIM^OH), = "H, = 0,01 mol. Sue CO2 vao dung dich Y , xay ra phan ling: Na[Al(OH)4] + CO2 -> Al(OH)34 + NaHC03 (4) Theo(2): = 0,02 - 0,01 = 0,01 mol. 39 Taco: 0,01M + 0,01(M + 16) = 2,9 Theo(4) "Nal AI(OH), 1 = " A K O H ) , " ^ " ^' ^ (mol). -> M = 137. K i m loai M la bari. 6.41. Dap an dung la C. Theo (3) nAidu= |nH, ==|-0,15 = 0 , 1 (mol). Oxit sat tac dung vdi dung dich H2SO4 dac, nong cho k h i SO2 bay ra la FeO hoac Fe304 (FeO.Fe263). Theo (2) : n^i^o, = - ( 0 , 5 - 0 , 1 ) = 0,2 (mol). Phuang trinh phan ling: 2FeO + 4H2SO4 -> Fe2(S04)3 + S02t + 4H2O (1) Theo (1) : n^, p., = 2n^i^o, = 2-0,2 = 0,4 (mol). Theod): np^o = 2nso: = 2 . ^ ^ = 0,29 (mol) 22,4 nFeA=|nA,o, =1-0,2=0,15 (mol). mp,o p., = 0,29.72 = 20,88 (gam). Vay : m = 27(0,4 + 0,1) + 232.0,15 = 48,3 (gam). -> Oxit sk la FeO. 6.44. Dap an diing la C. mpccsoj, = 0,145.400 = 58 (gam). S6' mol cac cha't: 5.42. Dap an dung la B. n^, = = 0,46 (mol); Uy = ^ = 0,06 (mol). Cac phucfng trinh phan ling : Goi X va y la s6' mol N2O va N2 c6 trong 0,06 mol h6n hop Y . 2A1 + 3 H 2 S O 4 A l 2 ( S 0 4 ) 3 + 3H2t (1) Ta c6: x + y = 0,06 (I) Fe + H2SO4 -> FeS04 + H2t (2) 44x + 28y Dung dich X g6m Al2(S04)3 va FeS04, tac dung v6i Ba(0H)2 dir: = 18 = 18 ^'"2 2.0,06 Al2(S04)3 + 3Ba(OH)2 ^ 3BaS04^ + 2 A l ( O H ) 3 i (3) - > 4 4 x + 28y = 2,16 (II) 2 A l ( O H ) 3 + Ba(OH)2 du -> Ba[Al(OH)4]2 + 4H2O (4) , Giai he phuong trinh (I) va (II) duoc: x = y = 0,03 m o l . FeS04 + Ba(OH)2 ^ BaS04^ + Fe(OH)2i (5) 78 179
Phuang trinh plian ling: Phuang trinh phan ilng nhiet nh6m: 18A1 + 6 6 H N O 3 - > I S A K N O , ) , + 3N2O + 3N2 + 33H2O (1) 8A1 + 3 F e 3 0 4 - » 4AI2O3 + 9Fe (i) 0,18 mol 0,18 mol 0,03 0,03 mol Cac phuong trinh phan img hoa tan h6n hop ran cho bay ra: Con 0,46 - 0,18 = 0,28 mol A l phan ling vdfi H N O , tao ra NH4NO3: Fe + H2SO4 (loang) FeS04 + H2 i (2) 8A1 + 30HNO., --^ SAKNOj), + 3NH4NO, + 9H2O (2) 2 A I + 3H2SO4 ^ Al2(S04)3 + 3H2t (3) Goi X la s6' mol A l tham gia phan iJug (1), ta c6: Nhu vay trong dung dich X gom c6 0,46 mol A1(N03)3 va - .0,28 3 = 0,105 mol NH4NO3. - S6' mol H2 dugc tao thanh tur A l du la - ( 0 , 4 - x ) mol. Co can dung dich X thu dugc cha't ran khan c6 kh6'i lirgng: 9 - S6' mol H2 dugc tao thanh t i i Fe la: - x = l,125x mol. m = 0,46.213 + 0,105.80 = 106,38 (gam). 6.45. Dap an diing la C. l,125x + - ( 0 , 4 - x ) = 0,48 ^ x = 0,32. 2 So mol H2 = = 0,25 (mol). Nhu vay chi 0,32 mol A l tham gia phan ting nhiet nhom (1). So d6 phan ling: Hieu sua't phan ling: h = - — . 1 0 0 % = 80% . ^ ^ 0,4 X + HC1-^XC1+iH21 (1) 6.48. Dap an diing la D. X + 2HC1 ^ XCI2 + H j t Y + 2HC1 ^ YCI2 + H 2 1 (2) - Gia su trong h6n hop chi c6 k i m loai k i ^ m X: nx = - nuci = - .0,2.1,25 = 0,125 mol. nx = 2nH, = 2.0,25 = 0,5 (mol) Dung dich Y c6 hai mu6'i clorua ciia hai kim loai c6 n6ng d6 mol bang nhau M,= g = 14,2. nen hai kim loai trong X cung c6 so mol bang nhau va bang 0,0625 mol. Khd'i lugng mol nguySn tic trung binh ciia hai k i m loai la: - Gia sit trong h6n hop chi c6 k i m loai k i ^ m th6 Y : M x = - ^ ^ = 19,6 0,125 "Y = " H , - 0'25 mol. -> K i m loai kiem thd c6 M < 19,6 la Be (Mg, = 9). My = — = 28,4. Tilf ket qua nay xac dinh dugc k i m loai thu' hai la Ca (canxi). 0,25 6.49. Dap an diing la C. - Trong h6n hop c6 ca X va Y ntn 14,2 < Mx, My < 28,4. Cac phuong trinh phan ting: -> X la Na ( M N , = 23); Y la M g ( M ^ g = 24). 2 A l + 6HC1 2AICI3 + 3H2t (1) 6.46. Dap an dung la C. AI2O3 + 6 H C 1 2 A I C I 3 + 3H2O (2) 6.47. Dap an dung la A . t AICI3 + 3NH3 + 3H2O ^ Al(OH)34^ + 3NH4CI (3) SO' mol cac cha't: 2Al(OH)3 — A I 2 O 3 + 3H2O (4) nA,= ^ = 0,4 (mol); n,,^o, = | ^ = 0,15 (mol). ' Ggi X va y la s6' mol A l va AI2O3 c6 trong 1,56 gam hOn hop. Taco: 27x + 102y = 1,56 (I) n H 2 = ^ ^ = 0,48 (mol). 1 9 04 . Theo ( 1 , 2, 3, 4): 0^,^03 = - x + y = — = 0,02mol (II) ^ 2 102 180 161
Giai he phuong trinh ( I ) , (II) dugc: x = 0,02. 7. TONG HOP N Q I D U N G c A C K l f i N T H t C HOA V 6 CO THUQC C H U d N G T R I N H PHO THONG Theo(l): = | n A , =|.0,02 = 0,03mol. 7.1. Dap an diing la D . Cap k i m loai trong do Fe bi pha hiiy trudc la : Fe - Pb, Fe - Sn, Fe - N i . _> = 0,03.22,4 = 0,672 h't. 7.2. Dap an dung la C. 6.50. Dap an diing la A . Cac cha't khi dugc lam kho bang NaOH (ran) phai la cac chat khong tac dung vdi NaOH. K h i chuyen tur muoi hidrocacbonat thanh mudi sunfat da c6 1 g6'c 7.3. Dap an diing la D . S04~ thay the 2 gdc HCO: , khoi luong 1 mol mu6'i giam: So do chuye'n hoa : m = 2.61 - 9 6 = 26 (gam). N a C I - ^ ^ N a O H — i ^ ^ N a H C O , —i^^^J^tL^Na^CO, - )NaN03 9 125-7 5 7.4. Dap an diing la B. -> s6' mol mu6'i hidrocacbonat = — — = 0,0625 (mol). Cac phuong trinh phan ling : 26 Ba(HC03 )2 + 2 HNO3 -> Ba(N03 )2 + 2 CO2 t + 2 H 2 O 9,125 ^"^-=o:56^='''- Ba(HC03)2 + Ca(0H)2 BaCOj 4 + CaC03^ +2H2O = 146 - (2.61) = 24 ^ kirn loai M g . Ba(HC03)2 + KHSO4 BaS04 + KHCO3 + CO2 + H 2 O Cong thiJc cua mu6'i hidrocacbonat: Mg(HC03)2. Ba(HC03)2 + Na2S04 BaS04 + 2 N a H C 0 3 6.51. Dap an dung la C. 7.5. Dap an diing la B. So mol cac cha't: n^g = 0,28 m o l , nM^-o = 0,02 m o l , nx = 0,04 mol. Cho h6n hgp X vao nud'c, tao thanh dung dich h6n hgp va xay ra phan ling: S6' mol mu6'i Mg(N03)2 = s6' mol M g + s6' mol M g O . Na20 + H 2 O -> 2Na0H , ^ nMg(N03)2 = 0,28 + 0,02 = 0,3 (mol). NaOH + N H 4 C I NaCl + NH3 t + H2O mMg(N03)2 = 0,3.148 = 44,4 (gam) < 46 gam. NaOH + NaHC03 -> Na2C03 + H j O => Con mud'i NH4NO3. NajCOj + BaCl2 BaC03 i + 2NaCl mNH^NO., = 4 6 - 4 4 , 4 = 1,6 (gam). Trong dung dich chi con mudi NaCl. 7.6. Dap an diing la A . %H4N03 = ^ = 0,02 (mol). Trong dung djch X vSn con ion C03~, chiing to H C l phan ling Tinh theo phuong phap can bang electron: het, Na2C03 con du. Do vfiy, dap an A phu hgp. 0,28 mol M g cho 0,28.2 = 0,56 mol e ( M g Mg^^ + 2e). 7.7. Dap an diing la B. +5 -3 C) day luu y phan ling : C2H4 + Br2 -> C2H4Br2 0,02 mol NH4NO3 nhan 0,02.8 = 0 , 1 6 m o l e ( N + 8e ^ N ) . +5 Co ten ggi la phan ling cdng hgp brom, nhung c6 ban cha't la phan Con lai 0,56 - 0,16 = 0,4 mol e de N nhan, chuyen thanh 0,04 mol ling oxi hda - khuf, v i trong phan ling nay so o x i hda cua brom va cacbon thay ddi. N2 (2 N + lOe ^ N2). K h i X la N j . 7.8. Dap an diing la D . 6.52. Dap an dung la D . Khi nung Fe(N03)2, Fe(0H)3, FeC03 0 nhiet do cao chiing deu phan . hiiy va bi oxi hda thanh oxit Fe203 . 183 182
7.9. Dap an dung la D. Jl^. Dap s6' diing la D. Ha'p thu khi CO2 vao dung dich Ba(OH)2 c6 thi xay ra cac phan ling: Cac phirong trinh phan ling: CO2 + Ba(OH)2 BaC03>l.+ H 2 0 (1) CuO + H2 (CO) — ^ Cu + H2O (CO2 t ) (1) Nd'udu CO2 : BaCOj + CO2 + H2O -> Ba(HC03)2 (2) Fe304 + 4 H2 (4C0) 3Fe + 4 HjO (4 CO, t ) (2) Goi n la s6' inol Ba(OH)2 tham gia phan ling (1) -> ni3aco,(i) = Theo (1,2): So mol nguyen tir oxi tach ra khoi oxi = so mol h6n hgp khi = 0,32 : 16 = 0,02 (mol) -> V = 0,02.22,4 = 0,448 (lit). Goi X la s6' mol BaCO^ tham gia phan irng (2). Ta c6 : 7.16. Dap an diing la D. S6 mol BaCOg con lai = n - x = = 0,08 mol (I) Cac phucrng trinh phan ting: CaCO, — ^ CaO + CO2 So mol CO2 = n + x = = 0,12 mol (II) ^ 22,4 (X) (X,) CaO + H2O -> Ca(0H)2 Giai he (I), (II) duoc : n = 0,1 a = - - = 0,04 mol//. 2,5 7.10. Dap an diing la D. Ca(OH)2 + NaHC03 CaC03 + NaOH + H2O 4 chat CO tinh luong Ifnh: (Y) (X) (Y,) Ca(HC03 )2 , Al(OH)3 , Zn(OH)2 , (NH4 )2C03 „ •Ca(OH)2 +2NaHC03 -> CaCO, + Na^CO, + 2 H 2 O 7.11. Dap an diing la A. (Y,) 7.17. Dap an dung la C. Fe^^ khdng oxi hoa dugc Cu. 2 phan irng trong do HCI the hien tinh oxi hoa la (2) va (4). 7.12. Dap an dung la C. , (2H+ +2e -> H 2 t ) X la Fe, tac dung dugc vdi dung dich H2SO4 loang. 7.18. Dap s6'dung la B. Y la Cu, tac dung dugc vdi dung dich Fe(N03)3. Phirong trinh phan irng: Cu + 2Fe(N03)3 Cu(N03)2 + 2Fe(N03)2 3Mg + 8HN0;, 3Mg(N03)2 + 2N0 + 4H2O (1) 7.13. Dap an dung la C. So mol cac chat: Cac phan iJng tao khi la: n ^ ^ ^ = ^ ^ = 0,09 m o l ; n,o = = 0,04 mol. (2) NH4NO2 — ^ N2 + 2 H 2 O Theo(l): n„^^„ = | n , o - | . 0 , 0 4 = 0,06 (mol). (4)2NH3+3Cl2 ^^N2+6HC1 Nhu vay con 0,09 - 0,06 = 0,03 (mol) Mg thuc hien phan irng sau: (6) 2 NH3 + 3CuO — ^ 3Cu + N2 + 3 H2O 4 M g + 10HNO,->4Mg(N03)2+NH4N03+3H20 (2) 7.14. Dap an dung la C. Nhung cha't phan ling dugc v6i dung dich HCl va dung dich NaOH la: Theo (2): n „ , , o , =^n^^,,, =1.0,03 = 0,0075 (mol) Al, AI2O3, Zn(OH)2, NaHS va (NH4)2C03. Do do: m = 148.0,09 + 80.0,0075 = 13,92 (gam) 185 184
7.19. Dapan diing la D. j : lot la halogen c6 khO'i luong mol nguyfin tir 16n nha't la: I = 127. Do do, khong c6 halogen Y > 178: loai trudng hop nay. Chi CO phan img: O, ~» + O khOng phai la phan iJng oxi hoa - khu. - Xet trucfng hop c6 mOt muoi bac khong ke't tiia. Do la AgF. 7.20. Dap an dung la A. Cac phan ling tao ra don chat la: H H Mu6'i AgY la AgClk nABCi = 0,06 mol = UN^CI mN,ci = 58,5.0,06 = 3,51 (gam). (1) O-, + 2KI + H2O 2KOH + O2 + I2 . mN,F = 6,03 - 3,51 = 2,52 (gam); (2) 2F2+2H2O — ^ 4HF+O2 W %mNaF = — .100% = 41,8%. ^ 6,03 (3) M n 0 2 + 4HC1 dac MnCl2 + C l 2 t + H2O 7.30. Dap an diing la A. , 7.21. Dap an dung la B. Ba thi nghiem xay ra phan irng hoa hoc la (I), (II) va (III). Trong phong thi nghidm ngirdi ta di^u chfi' khi HjS bang phan ling: 7.31. Dap an dilng la A. FeS + 2HC1 ^ FeCl2+ H 2 S t Phuong trinh ion nit gon chung la : 7.22. Dap an dung la D. Ba^^ + SO4- -> BaS044 7.23. Dap an dung la A. 7.32. Dap an diing la C. 7.24. Dap an dung la D. Jkk So mol cac chat : 7.25. Dap an dung la C. mk ^'024 0,9408 ^ , 7.26. Dap an diing la D. = - 7 7 - ;"NO =-^:rr(moi). Phuong trinh phan ling: A I C I 3 + 3NH3 + 3H2O -» Al(OH),i + 3 N H 4 C I ^ ^ H . M 22,4 7.27. Dap an dung la C. 7.28. Dap an diing la B. Cac phirong trinh phan ling : H| Theo bai : dN^o,7H, = '"^^^^^^ = 22. 2KMn04 - - ^ K2Mn04 + M n 0 2 + O2 -> 14x + 16y = 44. NaNO, — ^ N a N 0 2 + ^ 0 2 ^ B , Cap nghiem duy nha't phu hop la: x = 2, y = 1. 7.29. Dap an diing la D. H Vay CTPT cua khi N,Oy la: N2O. Cac phuong trinh phan ling: ^B,. nN,o = 0'042 mol -> nN = 0,082 mol. NaX + AgNO,, AgX + NaNO, (D Khi phan ling: N +4e->N NaY + AgNO, -> AgY + NaNO., (2) -> So mol electron N thu la : 4.0,084 = 0,336 (mol) - Xet truorng hop AgX va AgY deu la chat ke't tua. Khi do s6' mol hai muoi la: H| • M - x.e M X . 3,024 8,61-6,03^ (mol) '0,336 3 024 108-23 -> M phu ^- > -Mmuoi, 6,03 - = ^ni201. CapSonghiem mol electron duy nha't chohop: la: xx.-^ = 3, M==0,336 27. (mol). 0,03 • Vay kim loai M la Al (nhom). M KhO'i lugng mol cua X va Y nam trong khoang: 186 X < 2 0 1 - 2 3 = 178 < Y
7.33. Dap so diing la C. >jj9. Dap an dung la B. Cac phuang trinh phan iJng: 7,40. Dap an dung la C. 7 41. Dap an diing la D . CaOCl^ + 2HC1 CI, + CaCl, + H , 0 (1) TrUc/c phan ifng: Goi x va y la s6' mol va H j c6 trong X. Ta c6: 2KMn04 + 16HC1 -> 5C1, + 2 M n C l , + 2KC1 + 8 H , 0 (2) (X - a) mol N2, (y - 3a) mol H2 va 2a mol NH3. Ta c6: K2Cr207 + 14HC1 SClj + 2CrCl3 + m.O (3) 28x + 2y , _ . M n O j + 4HC1 CI2 + MnCl2 + 2H2O (4) Tir 4 phircfng trinh phan ung ta tha'y: Neu dung 1 mol cac cha't thl 1 nio] Sail phan i(ng: K2Cr207 cho lugng khi CI2 lorn nhat (3 mol CI2). » N2 + 3H2 < » 2NH3 7.34. D a p an diing la D . Fe304 = FeO.Fe203 Gia su CO a mol phan ling. Trong h6n hcfp Y c6: Cac phuang trinh phan ting: _ 2 8 ( x - a ) + 2(y-3a) + 34a _ ^ (2) 1) FeO + H N O 3 - > FeCNO,), + N,Oy + H2O 4(x + y - 2 a ) +2 +3 Fe - le -> Fe .(5x-2y) Thay y = 4x vao (2) va tfnh dugfc - = ^ 2y/ X 4 +5 2y + /x -> m e u suat la 25%. 5- .xe-^xN 7.42. Dap an diing la B. Phuong trinh phan ling can bang: Nam chat tac dung duoc vdi dung dich NaOH loang la: (5x-2y)FeO + ( 1 6 x - 6 y ) H N 6 3 - > ' NaHC03, A l ( O H ) 3 , H F , CI2 va N H 4 C I . (5x-2y)Fe(N03)3 + N.O^ + ( 8 x - y ) H 2 0 (1) 7.43. Dap an diing la B. 2) Fe203 + 6 H N O 3 -> 2Fe(N03)3 + 3H2O (2) Phuong trinh phan ling: NaHS + HCl -> H j S t + NaCl. hay (5x-2y)Fe203 + (30x-12y)HNO3 (l0x-4y)Fe(NO3)3 + (15x-6y)H2(i Phan ling nay loai bo dugc H C l , thu dugc HjS tinh khie't. Cong phan irng ( I ) va (2) thi he s6' cua HNO3 la 46x - 18y. 7.44. Dap an diing la C. 7.35. Dap an dung la A . So mol cac cha't: 7.36. Dap an diing la A . 17,92 ^ „ . 20,16 n = 0,8 (mol); = ^ d - r = ^'^ (^ol). 22,4 22,4 Co 4 h6n hop ran: 1. NajO va AI2O3: Tan hoan loan trong HjO di(. Cac phuong trinh phan ling: M,Oy + yCO ^ x M + yC02 (D 2. Cu va FeCl3: Khong tan hoan toan trong H j O du. y = 0,8 mol. 3. BaClj va CUSO4: K h i hoa tan trong mrdrc, tao ke't tua BaS04. 2M(Cr, Fe) + 6H2SO4 -> M2(S04)3 + 3SO2 + 6H2O (2) 4. Ba va NaHC03: K h i hoa tan trong nude tqo ket tua BaCO^. 7.37. Dap an dung la A. So mol M = X = - n s o , = | . 0 , 9 = 0,6 mol. 4 phan ling oxi hoa - khir xay ra d cac thi nghiSm: (I), (II), (III) va (IV). -> X : y = 0,6 : 0,8 = 3 : 4. 7.38. Dap an dung la C. CTPT ciia M,Oy la Fe304. Flo chi CO s6' oxi hoa - 1 trong cac hop cha't. 189 188
7.45. Dap an diing la B. 9 125-7 5 D6 dinh du9ng cua phan supephotphat kep (Ca(H2P04)2) tinh khie't la: -> s6' mol mu6i hidrocacbonat = — — = 0,0625 (mol). 26 k = - ^ ' ^ . 1 0 0 % = . l l ^ =60,68%. 9,125 Mca(H2P04)2 234 M„„,:: = = 146. 0,0625 Do dinh duong cua phan lan (c6 iSn tap cha't) nay la: Akin, u,ai = 146 - (2.61) = 24 ^ kim loai Mg. k ' = 6 0 , 6 8 . - ^ ^ = 42,25%. C6ng thiJc cua mu6'i hidrocacbonat: Mg(HC03)2. 100 7.53. Dap an diing la B. 7.46. Dap an diing la B. Cac phuong trinh phan ling: 7.47. Dap an dung la D. CaO + 2 H C 1 C a C l j + H2O Hai trudng hop gay ra an mon difin hoa la: Ni - dd C U S O 4 vii Ni - dd AgNO, (Ni phai la kim loai manh hon kim loai tao mu6'i). (X) 7.48. Dap an diing la C. CaClz + 2AgN03 -> Ca(N03)2 + 2AgCl4' So d6 chuyS'n hoa: (Y) P2O, > K3PO4 > KH2PO4 > K2HPO4 '• Ca(N03)2 + (NH4)2C03 ^ CaC034' + 2NH4NO3. 7.49. Dap an diing la C. (Z) S6' mol cac chat: 8. DAI Cl/dNG HOA HOC HUU CO, HIDROCACBON nBa(OH)2 = 1-0,15 = 0,15 (mol); n^oH = 1.0,1 = O,l(mol). 8.1. Dap an diing la A. Khi Y gom : Cac phuong trinh phan ling: 1. CH = CH -> AgC = CAg I : HCH^CH = ^ " ^'^^ 4 F e S 2 + 1 1 0 2 ^ 2 F e 2 0 3 + 8S02 (1) SO2 + K 0 H ^ K H S 0 3 (2) 2. C H 2 = C H 2 CH2Br-CH2Br 1 mol 1 mol 16 SO2 + Ba(OH)2-> BaS03xl + H2O (3) ncH2=CH2 = ^BT2 = 160 = 0.1 mol ^ nH2(2) = O'l mol 0,1 mol 0,1 mol 0,1 mol 3. C H 3 - C H 3 - ^ 2 C 0 2 + 3 H 2 O : 2SO2 + Ba(OH)2 ^ Ba(HS03)2 (4) 1 1 2,24 , 0,1 mol 0,05 mol ncH3-CH3 = 2 "CO2 = 2 224 " Theo (2, 3, 4): s6' mol SO2 = 0,1 + 0,1 + 0,1 = 0,3 (mol). n H 2 ( 3 ) = 0'05 . 2 =0,1 mol; n H 2 0 ( 3 ) = O'l^ mol. Theo (1): s6 mol FeSj = - s6' mol SO2 = 0,15 mol. 4. H2 kh6ng phan ling > H2O : nH20(4) = nH2(4) = O'l "^°1- -> mp,s, = 0,15.120 =18,0 (gam). S6' mol khi trong h6n hop la : 0,05+0,1+0,1+0,05+0,1+0,1 = 0,5 mol. 7.50. Dap an diing la B. -> Vx = 0,5.22,4= 11,2 (lit). 7.51. Dap an diing la B. ^•2. Dap an diing la B. 7.52. Dap an diing la A. Benzen khOng lam mat mau nirdc brom. Khi chuy^n tijr mu6'i hidrocacbonat thanh mu6'i sunfat da c6 1 g^*^ Stiren lam mat mau nude brom. S04~ thay the' 2 g6'c H C O 3 , khC'i lugng 1 mol mu6'i giam: Anilift lam mat mau nude brom va tao ra ke't tiia trang. m = 2.61 - 96 = 26 gam. Dua vao cac hien tuong tren phan biet dugc 3 lo chat. I9n 191
8.3. D a p an dung la B. Ddp an dung 1& C. Monobrom ankan c6 CTPT la CnH2n+iBr = 75,5.2 = 151 -> n = 5. Phuong trinh phan ling dO't chay hidrocacbon C^Hy : CTPT cua ankan la : C5H,2. C^Hy + (X + ^ ) 0 2 - > XCO2 + I H2O Khi the' Br2, tao ra monobrom duy nhat n6n c6 CTCT : H6n hop khi Y g6m khi CO2 , hoi H2O va O2 du. Qua H2SO4 dac, hoi C H 3 - C ( C H 3 ) 2 - C H 3 : 2,2-dimetyl propan. H2O bi giil lai, hOn hop khi Z g6m khi CO2 va O2 du. 8.4. Dap an diing la B. Gia sit CO 1 mol C^Hy chay. Ta c6 : 4 48 - S6' mol CO2 dugfc tao thanh la x mol. SO' mol hOn hop X : nx = —— = 0,2 (mol). 1,40,5 - S6' mol O 2 du la : 10 - (x + ^ ) mol = (10 - x - ^ ) mol S6' mol Br2 da phan ling : n^^^ = ' 2^ ' = 0,35 (mol). 22,4 Vi 0,2 < 0,35 < 0,2.2 -> Trong hOn hop X c6 mOt anken va rnoi 44x + 3 2 ( 1 0 - x - ^ ) ankin (hoac ankadien). d ^ = 19 8x + y = 40 X = 4, y = 8. Dat CTPT ciia 2 hidrocacbon la : CnH2n (n > 2) va CmH2m_2 (m > 2). 2(x + 1 0 - x - ^ ) 4 Phuong trinh phan ling cua X vdi dung dich Br2 : CTPT ciia hidrocacbon X la : C 4 H 8 . C„H2n + Br2 -> C„H2„Br2 (1) 8.7. Dap an diing la D. CmH2n,-2 -'-2Br2 -> C^H2m-2Br4 (2) Phuong trinh phan ling: CjHj + 2H2 C2H,. Goi X la s6' mol cua anken CnH2n thi s6 mol cua ankin Cn^H2m_2 1^ 0'2 - x H6n hop khi Z kh6ng phan ling voi dung djch Brj chi g6m CjHf, va H2 dir. Theo (1) va (2) ta CO : x + 2.(0,2 - x) = UQ,^ = 0,35 ^ x = 0,05 0,448 n, = 22,4 = 0,02 mol. Va : 14n.0,05 + (14m - 2).(0,2 - 0,05) = 6,7. Dat s6' mol C 2 H 6 la X ^ s6' mol H 2 = 0,02 - x. n, m > 2 va nguySn nen cap nghifim duy nha't thfch hop la : n = 4, m = 2. ^ . ^ 30x + 2(0,02-x) „„, CTPT cua hai hidrocacbon trong X la : C 4 H 8 , C2H2. Ta c6: d/,n = ^=0,5 -> x = 0,01. 0,02.32 8.5. Dap an dung la C. S6' mol C 2 H 2 da phan iJng la 0,01 mol, s6' mol C 2 H 2 con du s Ta bid't: Mz = Mx + 2 M C H 2 = 2 M X M X = 2 M C H 2 = 28 dvC phan drng vdi dung dich Br2 la 0,06 - 0,01 = 0,05 mol. MY - Mx + McH2 = 28 + 14 = 42 dvC. KhO'i lircmg binh Br2 tang la: 0,05.26 = 1,3 (gam). Dat CTPT cua Y la C^Hy = 42 -> 12x + y = 42 -> x = 3 , y = 6. 8.8. COng thiJc dung la C. CTPT cua Y la C 3 H 6 . Phuong trinh phan ling chay : S6' mol cac cha't: nx = — = 0,075 (mol). 22,4 C3H6 + 4 , 5 0 2 - > 3CO2 + 3 H 2 O (1) 1,12 = 0,05 (mol). Ca(0H)2 + CO2 CaC03>l + 3H20 (2) SO' mol hidrocacbon no C„H2n+2 = —— Theo (1) va (2): ncaco., = n c o 2 = 3 P C J H ^ = 0,1.3 = 0,3 (mol). 22,4 -> mcaco3 = 0,3.100 = 30 (gam). so' mol hidrocacbon kh6ng no C^H^ = 0,025 (mol). .SO' mol Br2 phan img = = 0,025 (mol). 192 193
-> C^Hy la hidrocacbon khdng no loai anken (C^H2J 12. Dap diing Ih. A. Phuang trinh phan iJng : So d6 phan ling chay: C„H2„+2 """^ > n CO2 CH2 = C H - C = C H + 3H2 CH3-CH2-CH2-CH3 (l) C,H2, — X CO2 X mol 3x mol x mol So mol CO2 = 0,05n + 0,025x = — = 0,125. Goi X la s6' mol CH2 = CH - C = C H da phan ling. 22,4 Vi X > 2 nen cap nghiem duy nha't phu hop la: n = 1, x = 3. -> So mol h6n hop Y: CTPT cua hai hidrocacbon la: C H 4 va C^H^. (0,1 - X) + (0,3 - 3x) + X = 0,4 - 3x 8.9. Dap an dung la C. ^52(0.1-x).2(0.3-3x).58x^ ^ X la hidrocacbon, phan tir c6 6 nguyen tur C. Y/kk 29(0,4-3x) S6' mol CH2 = CH - C = C H con du la : C T C T ^ua X ia; VIH, - C H - C H - C H 3 I i 0,1 -0,0667 = 0,0333. CH, CH3 Vinylaxetilen phan ling vdi Brj : Khi X tac dung vdi CI2 (theo ti le mol 1 : 1), s6' dSn xuat monoclo la 2 (CI CH2 = C H - C = C H + 3Br2 -> CHaBr - CHBr - CBr2 - CHBr2 (2) the vao C, hoac C2). 8.10. Dap so dung la D. Theo (2) : ng,^ = 3.0,0333 = 0,0999 mol. Gia sur trong h6n hgfp M c6 a mol ankan C„H2„+2 (X) va b mol ankin me,, = 0,0999.160= 16 gam. . C,,H2„...2 (Y) 8.13. Dap an diing la B. Khi d6't chay hoan toan, tao ra: Goi X, y va z la s6' mol CH4, C2H4 va C2H2 c6 trong 8,6 gam h6n hop S6' mol CO2 = na + n'b. X . T a c 6 : 16x + 28y + 26z = 8,6 (I) S6' mol H2O = na + n'b + a - b. Cac phuang trinh phan ling cua h6ri hop X v6i dung dich brom: Khi so mol CO2 = so mol H2O thi a = b. -> Thanh pMn % ve s6' mol cua X va Y la 50% va 50%. C2H4 + Br2 ^ C2H4Br2 (1) 8.11. Dap an dung la C. C2H2 + 2Br2 -> C2H2Br4 (2) D6't chay hidrocacbon X cho s6' mol HjO > s6' mol C O 2 . 48 T h e o ( l ) v a ( 2 ) : ng,,-y + 2z = — = 0,3 (II) -> X la hidrocacbon no C„H2n+2- 160 C„H,„,, > n C 0 2 + (n + 1) H2O. Phuong trinh phan ling cua h6n hap X vdi dung dich AgNO, trong NH,: n 0,11 ^ C H ^ C H + 2[Ag(NH3)2]OH ^ AgC=CAg^ + 4 N H 3 t + 2H2O (3) = >• n = 5 36 n +1 0,132 Theo (3) : HC^H, =nc,Ag, = ^ = 0'^^("^°1) CTPT ciia X la C,H,2. i ^ = 0,25 (III) C,H,2 + C I 2 > m6t san phim huu ca. x+y+z 0,6 CH3 Giai he phuang trinh (I), (II) va (III) duac: I x = 0,2, y = z = 0 , l . CTCT cua X la: C H 3 - C - CH3 : 2,2 -dimetyl propan. %Y =-^.100% = 50%. CH3 0,4 195 194
8.14. D a p an d u n g ia A . G i a SIX C O x m o l C j H j phan l i n g . D o do, sau p h a n l i n g , h 6 n hgp k h f Y D a t C T P T cua anken C„H2„. g o m (0,02 - x ) m o l C2H2, (0,03 - 2 x ) m o l va x m o l CjHfi. Qua G o i X va y la s6' m o l anken va s6' m o l c6 t r o n g 1 m o l X . d u n g d i c h Br2, C2H2 b i giG l a i , h 6 n hgp k h f Z g 6 m (0,03 - 2 x ) m o l H2 Ta c6: x + y = 1 (I) va X m o l C2H6. ^ , . 2 ( 0 , 0 3 - 2 x ) + 30x dx/H, = 9 , l - > 14nx + 2 y = 18,2 (II) Ta co: dy/„ =~ = 10,08. 2.0,0125 D u n n o n g X vdi x i i c tac N i , xay ra phan l i n g : X = 0,00738. Q H j n + H 2 - > C„H2„ + 2 Khd'i l u g n g k h f C2H2 c o n l a i sau (1) b i ha'p t h u b d i Br2 l a : X mol X mol x mol m = (0,02 - 0 , 0 0 7 3 8 ) . 2 6 = 0 , 3 2 8 ( g a m ) . Sau phan l i n g h 6 n hop k h f Y g 6 m x m o l C„H2n + 2 va (y - x ) m o l H j . 8.17. D a p an d i i n g la C. ^ ^ ( 1 4 n + 2)x + 2 ( y - x ) ^ ^ ^ X - i ^ ^ CO2 + H2O 2(x + y - x ) San phim chay dugc ha'p thu vao dung dich B a ( 0 H ) 2 (du), tao ra B a C O ^ i . 29 55 14nx - 24y = 0 (III) G i a i he p h u a n g t r i n h ( I ) , ( I I ) va ( I I I ) dugfc : x = 0,3, y = 0,7. "Baco, = - j ^ = 0,15 m o l - > nB,o = nco^ = nc = 0,15 ( m o l ) . T h a y g i a t r i cua x va y vao ( I I ) dugc : n = 4. m H , o = m B a o - 19,35 = 1 5 3 . 0 , 1 5 - 19,35 = 3,6 ( g a m ) . C T P T cua a n k e n la : C 4 H H . 3 6 A n k e n c o n g H B r cho san p h d m h i i u ccf d u y nha't. D o d o , c 6 n g thirc " H ^ O =7^ = ("^ol) O H = 0,4 ( m o l ) . ca'u tao cua a n k e n la : C H 3 - C H = C H - C H 3 . nc: OH = 0,15 : 0,4 = 3 : 8. CSng t h i i c phan txx cua X la C^Ug. 8.15. D a p an d i i n g la D . 8.18. D a p an d u n g l a C. D a t C T P T cua M la C„H2„, cua N la C„H2„_2. S6' m o l cac chat: So m o l h 6 n h g p X : nx = ^ — = 0,3 ( m o l ) . 22,4 nx = ^ = 0,2 (mol); n^o, = ^ = 0'3 (mol). G o i s6' m o l cua M va N t r o n g h 6 n hofp X la x va y . TCr s6' m o l X , s6' m o l CO2 va t i kh6'i cua X so v d i H2 b l n g 11,2 xac Ta c6: x + y = 0,3 d i n h d u g c a n k a n l a C H 4 va anken la C^Hfi. 14n(x + y ) - 2 y = 12,4 8.19. D a p an d u n g la B. - ^ 4 , 2 n - - 2 y = 12,4 So m o l cac cha't: = 4.2n-12.4 "hh = ^ = 0,3 ( m o l ) ; HCQ^ = ^ = 0,5 ( m o l ) ; ^ 2 10,8 Cap n g h i e m d u y nha't p h u h g p la: n = 3, y = 0 , 1 . "H20 —=0,6(mol).; Vay M la C 3 H 6 , C O 0,2 mol. - n H j o >nco2 ~^ P^^' ^ h i d r o c a c b o n la h i d r o c a c b o n no. Nla C 3 H 4 , C O 0,1 mol. 8.16. D a p an d u n g l a A . - D o t 0,3 m o l h i d r o c a c b o n , tao t h a n h 0,5 m o l CO2 - > so nguyfin tir P h u a n g t r i n h phan l i n g : cacbon t r u n g b i n h t r o n g phan t ^ hai h i d r o c a c b o n la 1,667. C2H2 + 2H2 C2H, (1) ^ C T P T ciia X la C H 4 . 196 197
8.20. Dap an dung la A. 9.4. Dap an dung la B. Phuang trinh phan ling: Phuong trinh phan iJng : '-n"2n-2 + " 2 >^nn2n (1) CnlWO + C u p - ^ C„H2nO + CU + H2O (1) H 6 n hop Y g6m CJi^.-,. va C„H2„. Nhu vay, sau (1) du C„H^„^^, nghia - CuO -> Cu : Khdi krong giam 0,32 gam nc„o = = 0,02 mol. 16 l a n p H >0,lhayMcH, < ^ = 31,2. 0,02.(14n +16)+ 0,02.18 Chi CO ankin CjHj (M - 26) thoa man. Theo (1) : — — —-— = 15,5 - » n = 2. (0,02 + 0,02).2 9. DAN XUAT HALOGEN, ANCOL, PHENOL CTPT ciia ancol la : C2H5OH. 9.1. Dap an dung la D. m = 0,02.46 = 0,92 (gam). San pham dot chay ancol cho n^^o > "cOj • 9.5. Dap an dung la B. -> ancol no, CTPT la CnH2n+20x . Dat CTPT trung binh ciia 2 ancol don chu"c, ke ti^'p nhau trong day dongdangla C-H-OH. CnH2n+20x + Ol ^ nCOj +(n+l)H20(l) Phuang trinh phan ling.: " = n = 3. Thay n = 3 vao he s6' cua ta c6 : C-H-OH + Na -> C-H-0Na + - H 2 t (1) n+1 4 X y ^ y 2 in-x Phan ling: n mol n mol n mol = 1,5.3 - > X = 1. CTPT cua X la C^U^O. 92 9.2. Dap an dung la B. Sau phan ling: 0 mol (— n) mol n mol 23 Goi CTPT ciia ancol bac 2, no, don chi'rc la C„H2n+20 • Theo di bai va theo (1): (12x + y + 17)n = 15,6 (I) ^ . 12n 68,12 Ta CO : =— > n= 5. (12x + y +39)n + ( ^ - n ) .23 = 24,5 (11) i4n + 2 + 16 100 Giai he phuang trinh (I), (II) duocn = 0,3 mol -> Mhh = = 52. CTPT ciia ancol la C5H,20 c6 3 d6ng phan ancol bac 2 la : CH3 - CH(OH) - CH2 - CH2 - CH3 : pentan-2-ol. 12x + y + 17 = 52 ^ X = 2,5 ; y = 5. CH3 - CH2 - CH(OH) - CH2 - CH3 : pentan-3-ol. CTPT cua 2 ancol phu hop la : C2H5OH va C3H7OH. CH3-CH(0H)-CH(CH3)-CH3 : 3-metylbutan-2-ol. 9.6. Dap an diing la C. 9.3. Dap an diing la A. Hop chat X la ancol bac 2 c6 nhom OH dinh vao C s6' 2. = ~ = 0,175 (mol); 1 1 ^ 2 = - ^ = 0,15 (mol). Ba dong phan anken la : D6't chay 0,05 mol X cho 0,15 mol CO2 -> Phan tir X c6 3 nguyfin tir C. (1) C H 3 - C H = CH-CH3 ; (2) C H 2 - C H - C H 2 - CH3. X la ancol no, mach hd nen c6 CTPT la CjH^O,. Trong do (1) CO 2 dong phan hinh hoc. Phuong trinh phan ling : 9.7. Dap s6'dung la B. Dat CTPT trung binh cua hai ancol la: C - H 2 - ^ , a i 2 0 H . C3HA + (5-a)02 3C02 + 4 H 2 O Phuong trinh phan ting: Imol (5-a)mol ) . „ _ . 0,05mol 0,175 mol j ^ a - 3. C-Hj-^iCHjOH +CuO — - > C-H2-CHO + H2O +Cu (1) vay CTPT ciia X la C^HHO, hay C^H^COH), : Glixerol. 199 198
H6n hap hai Y g6m hai andehit C-H^-CHO va hai H 2 O , v6i s6' mol 12x^y+16 bang nhau. Ta c6 : 12x + y - 2 cap nghiem duy nha't thich hgp la: x = 2, y = 6. dv;H,'°''"'""f"°-"'=-13.75 n=0.5 CTPT cua X la CjH^O (CH3 - CH2 - OH) n, = 0 andehit HCHO. 9.11. Dap an dung la D . nj = 1 --> andehit CH3CHO. Ddt chay hai ancol X , Y cho s6' mol H j O = s6' m o l CO2 - > X , Y la - Goi X va y la s6' m o l H C H O va C H 3 C H O phan iJng vdi AgNO, trong hai ancol no. Cho 0,25 mol h6n hgp M tac dung v6i Na (du), thu dugc < 0,15 mol dung dich NH3. Ta c6: H j - > X , Y la hai ancol dan chiJc. n^g = 4x + 2y = 64,8 : 108 = 0,6 mol. Vie't phuang trinh phan iJng chay cua X , Y va xac dinh dugc: - Trong h6n hop Y c6 x mol HCHO, y mol CH^CHO v^ (x + y) - Ancol X la C2H5OH. mol hoi H j O . Do do : - Ancol Y la C3H7OH. = 30x.44y.l8(x-.y)^ 9.12. Dap an dung la A . 2.2(x + y) Ancol X chay cho s6' mol HjO > s6' mol CO2 - > X 1^ ancol no da Giai ra dugc : x = y = 0,1 mol. cMc C O cong thiJc phan tur la C2H6O2. Nhu vgy trong h6n hgp X c6 0,1 m o l CH^OH vk 0,1 m o l C2H5OH. 9.13. Dap an diing la A . mx = 0,1 .(32 + 46) = 7,8 (gam), C^H^OH CH3CHO, CH3COOH, H2O va C 2 H , 0 H du (1) 9.8. Dap an dung la B. CH3COOH + NaHC03 ^ CH3COONa + C 0 2 t + H2O (2) Phuotig trinh phan ung loai nu6c: CH3 - (:H(CH3) - CH(OH) -CH3 —^ 2 C H 3 -C(CH3) = CH - f Theo (1) va (2) : nc^H,oH - n c H , c o o H = " 0 0 , "^'^^^ (3-metylbutan-2-ol) (3-iBetylbut-2-en) Kh6'i lugng etanol bi oxi hoa thanh axit la : 9.9. Dap s6'dung la B. m = 0,025.46 = 1,15 gam. Cac phuong trinh phan ling: 9.14. Dap an dung la D . So mol cac cha't: C H 3 O H + CuO — H C H O + H2O + Cu. (1) HCHO + 4[Ag(NH3)2]OH ^ C O j t + SH^O + 4 A g ^ + 8NH3 (2) I "^"^ ^ ^ ^ ^ ' ^ "H,o =1^=0,4 (mol). Theo ( 1 . 2 ) : s6' mol C H 3 O H = - so mol A g = - . = 0,3 (mol). "co "H o pMn lit ete dem d6't chay cd 1 noi doi. 4 ^ 4 108 CTPT cua ete do la : C„H2nO Khd'i luong C H 3 O H da phan ling la: 32.0,3 = 9,6 (gam). D6't (14n + 16) gam ete, tao ra 44n gam CO2. fl|i D6't 7,2 gam ete, tao ra 44.0,4 gam COj. Hieu suat phan ling la: h = ^ .100% = 80% " ^ n = 4. Dap an D phu hgp. 9.10. C6ng thu-c dung la B. 9.15. Dap an dung la B. V i Mx > My - > Y la anken (khdng th^ la ete) So mol cac chat: Dat CTPT cua ancol X la C.H^O. Ta c6: C.H^O > C,Hy.2(Y). 3,808 „ 5,4 n- c o , = 0,17(mol); H H o = " 7 : 7 = 0-3(mol). f| - - ^ ^ - ' J ' ^ ' U " " * ^ ' "H,0 jg 201 200