Phần 1 tài liệu Cẩm nang ôn luyện thi Đại học - Rèn luyện giải nhanh các đề thi ba miền Bắc - Trung - Nam Hóa học cung cấp cho người đọc các phương pháp giải nhanh bài tập Hóa học, một số bộ đề thi thử Đại học và hướng dẫn giải chi tiết. Mời các bạn cùng tham khảo.
CU THANH TOAN \iibm nddl duh K I - : ' '.,.>' CAM NANG ON LUYEN THI DAI HOC • • • R^n luien ^ nhanh th ba mi^ BiC - mm - NAM HOA HOC ^^M|Ml^^lilHRI H$thdngcacphuongphapgi^ nhanh bai tap hoa hoc X Wn chon va gifli thieu cac d6 thi thii nam 2013, cac dfi thi thii ciia cac truong THPT, cac truong Chujen va cac Trung tarn liij^n thi dai hoc uij tm ctia mien B^c - Trung - Nam cac detf«d& dupe gi^ctitiet,dehfduvatheo cac phuongphapgi^i nhanh X : i H u u A V I I S ' T D A M T K U R urtD T U A N U D U f f H f ! H H f MINH cty TNHH MTV DWH Khang Vi$t Phan LOI NOI DAU Cac em hoc sinh than me'n! Chiing t6i xin trSn gidi thiSu vdri cdc em tap sach: luyen thi Dai hoc mien Bdc - Trung - Nam Hoa hoc CAC PHCWUfG P H A P G I A I imAIUI B A I Xp> HQA 119c Cam nang on Tap sach la tai lieu cap nhat cho cac em cac bo de thi thir Dai hoc, Cao dang tren ba mi^n Bac - Trung - Nam de cac em tu ren luyen ky nang lam nhanh bai thi cho minh Cac de thi deu dugc chiing toi tuyen chon k i cang, noi dung bam sat chaong trinh thi Dai hoc, Cao dang va theo dung ca'u true de thi ciia B6 G D - D T M i bai tap d^u dugc giai chi tiet, de hi^u, theo nhieu each va dac biet c6 cac phuong phap giai nhanh, d6c dao Tap sach g6m hai phan: - Ph&i I Cac phuong phap giai nhanh bai tap hoa hoc - PhSn I I Cac b6 d^ thi thit Dai hoc va hudng dan giai chi tie't Chiing toi tin tucmg rang tap sach se cap nhat cho cac em day dii cac dang de thi tuyen sinh Dai hoc, Cao dang theo hudmg de thi cua Bg G D - D T , trang bi dSy du cho cac em toan bg kie'n thiic hoa hoc Trung hgc phd thong va quan trgng hon la mang lai cho cac em su tu tin cac ky thi sap tdi Dd cuon sach hoan thien hon, rat mong nhan dugc su dong gop y kien chan ciia cac ban dong nghiep va ciia cac em hgc sinh X i n tran trgng cam on! TAG GIA r u , PhUCSng phap 1: PHLfONG PHAP BAO TOAN KHOI LUONG I LITHUYET - Gia sir CO phan ling: A+ B >C + D Ta c6: + nig = ' Jii i •! • i> 1- + triy - Ap dung: Trong mot phan ling, c6 n chat (ke ca cha't phan ling va san phdm), n€u biet khdi lirong ciia (n - 1) chat thi tinh dugc khoi lirgng ciia cha't lai II VAN DUNG ^A Thi du 1: H6n hop X gom 0,15 mol vinylaxetilen va 0,6 mol H2 Nung nong h6n hop X (xiic tac Ni) mot thcfi gian, thu dugc h6n hop Y c6 ti kh6'i so voi H, bang 10 DSn h6n hgp Y qua dung dich brom du, sau phan ling xay hoan toan, so mol brom tham gia phan ling la A 0,1 mol B 0,15 mol C 0,05 mol D 0,2 mol (Trick dethi thu Dai hoc khdi A, B nam 2013} Hu&ngddngidi H6n hgp X c6: S6'lien ket 7t = 0,15.3 = 0,45 (mol) Kh6ilugng: m^ =0,15.52 + 0,6.2 = 9(gam) * Sdmol: nx =0,15 + 0,6 = 0,75(mol) Soddphanurng: X(C4H4,H2)-^Y Nhd Sdch Khang Viet xin trdn trgng gioi thieu toi Quy dgc gid va xin Idng nghe mgi y kien dong gop de cuon sdch ngdy cdng hay han, bo ich han.Thu xin gui ve: ^ ,1 Cty TNHH Mot vien Djch Vu Van Hoa Khang Viet 71- Dinh Tien Hoang, Phirfirng Dakao, Quan 1, TP HCM Tel:(08) 39115694 - 39111969 - 39111968 - 39105797 Fax:(08) 39110880 ^ Email: Khangvietbookstore my = m^ = 9(g) , = o n Y =9/(10.2) = 0,45(mol) =>n,,h,ii, =0,75-0,45 = 0,3(mol) = nH2 (phan umg) =>n„ (con lai) =0,45-0,3 = 0,15(mol) = nB,^ ^, Dap an diing la B Thi du 2: H6n hgp X g6m m6t hidrocacbon o the' va H (ti khd'i hoi ciia X so v6i H, bang 4,8) Cho X di qua Ni dun nong den phan iJng hoan toan, thu dugc h6n hgp Y (ti kh6'i hoi ciia Y so v6i CH4 bang 1) C6ng thiic phan tir cua hidrocacbon la: ||; || , (, A C H B C2H4 C C H D C2H2 |, (Trich de thi thdDai hoc khdi A, B nam 2013) Ca'm nang fln luy$n thi DH mign BSc - Trmg - Nam mOn H6a hpc - Cu Thanh To^n Hu&ng ddn gidi M y =16=> Y g6m ankan C„H2„+2 va Cty TNHH MTV DWH Khang Vigt ^ Ba(OH)2- Sau cac phan ung thu duoc 39,4 gam ke't tiia va khdi lucmg phSn du dung djch giam bdt 19,912 gam C6ng thurc phan i\i ciia X la: H2) Gia sufco mol X (g6m C„H2n+2-2k A CH4 C„H2n+2-2k +'^^2 , ' BandSu: x Phan ting: x Saupu: ^ -> ^ (l-x-kx) ' x (mol) CO2 x (mol) H20 ddBa(OH)2 fBaCOj i Ba(HC03)2 Theo djnh lu^t bao toan khd'i luong => my = m^^ = 9,6(g) Kettua la BaCO, :nB„co3 =39,4/197 = 0,2(mol) =>nY =9,6/16 = 0,6(mol) Taco: mco2 +mH20 =mkt = > ( l - x - k x ) + x = 0,6=i.kx = 0,4 (l) =>mco2 +^H20 -^dd{g\) =39,4-19,912 = 19,488(g) Mat khac: m ^ = (l4n + - 2k)x + 2(l - x) = 9,6 X + O2 -> CO2 + H2O ^14nx-2kx=7,6 Theo djnh luat bao toan khd'i luong, ta cd: mx + Tir(l,2)^ — = ^ k (2) 0,4 '"H20 ' ' / x + - O2 ->xC02+^H20 Dap an dung la A Thidu 3: Nung nong m6t h6n hop g6m CaCO^ va MgO tori khd'i luong khong ddi, ijjBflfji i r i i thi so gam cha't rSn lai chi bang 2/3 so gam hdn hop trudc nung Vay B 24,24% C 66,67% fi6b oflc) (}f>'J,b phdn tram theo khd'i luong ciia CaCO, hdn hop ban ddu la A 75,76% = mco2 => mo2 = 19,488 - 4,64 = 14,848(g) => no2 = 0,464(mol) = 21 =>n = 3;k = ( C , H ) \ D C3H4 So dd cac phan ling xay ra: (mol) kx , , (Trich de thi thADai hoc khdi A, B ndm 2013) Hu&ng ddn gidi CnH2n+2 1-x C C2H4 B C4H,o = ^ m x = 4,8.2 = 9.6(g) / 4,64 D 33,33% 12x + y (Trich de thi thuDai hpc khoi A, B nam 2013) > 0,464 Hu&ng ddn gidi 4,64 Gia six ban d^u c6 100 gam hdn hop CaC03,MgO Ta c6: •;< = 0,464 12x + y => Khd'i luong hdn hop lai sau nung la: 100.2/3 = 200/3(g) Cap nghiem thoa man: x = 3;y = = > X 1^ C3H4 C a C O , — ^ C a O + C02 t p?, , (lib) Dap an dung la D Theo djnh luat bao toan khd'i luong: mco2 = 100-200/3 = 100/3(gam)^nco2 = 100/132(mol) ^ ncaC03 = 100/132(mol) = nco2 , ^ mcaco., = 10000/132(g) vay %CaC03 = 10000/132 = 75,76% Dap an diing la A Thi du 4: Ddt chay hoan toan 4,64 gam mdt hidrocacbon X (chat d di^u kien thudng) rdi dem toan b6 san ph^m chay ha'p thu het vao binh dung dung dich | Thi du 5: Hdn hop X gdm 0,15 mol CH^C—CH=CH, va 0,6 mol hidro Nung nong hdn hop X (xiic tac Ni) mot thdi gian, thu dugc hdn hop Y cd ti khd'i so vdi hidro bang 10 DSn hdn hop Y qua dung dich brom du, sau phan ung xay hoan toan, khd'i luong brom tham gia phan ufng la A Ogam B 24 gam C gam D 16gam • (Trich de thi thi Dai hoc khoi A, B nam 2013) Hu&ng ddn gidi Xet hdn hop X c6: n^^^) =0,15.3 = 0,45(mol) ,j /, »,)» ,J,H,h nk qpO Cty TMHH M I V JVVH Khang Vigt C^m nang On luy?n thi BH mign B&c - Trung - Nam mfln H6a hpc - Cu Thanh ToAn m x = 0,15.52 +0,6.2 = ( g a m ) ; n x = , ( m o l ) Sodo: X-^Y I K>:i'IJA.;^ ri;;!o j{r»,/> Theo djnh luat bao toan khdi lucfng ta c6: m y = ^ Thi du 7: Dot chay hoan toan x gam h6n hop g6m hai axit cacboxylic hai chiic, ut.c = 9(gam) j^^"' ' = > n Y = / ( l ) = 0,45(mol) => nf,h giin, ^ n , => m = 0,75 - , = , ( m o l ) = n^^^ (phan ufng) c.-^ = , - , = 0,15(mol) = n B ^ T,,.' " mach ho va deu c6 mot lien ket d6i C=C phan tir, thu dugc V lit k h i C O (dktc) va y mol H , Biau thurc lien he giua cac gia tri x, y va V la B.V = C V = ^ ( x + 30y) D V = | ( x + 62y) (Trich de tuyen sinh Dai hoc khdi A) Huong dan gidi = 0,15.160 = 24 (gam) A x i t cacboxylic hai churc, mach ho, c6 m6t lien ket d6i C=C c6 CTPT dang: Dap an diing la B C„H2„_2 Chii y ; V I H2,Br2 trung hoa lien k^'t n nen: < n' (COOH)^ l,5y n^ ( c o n l a i ) =11^^^ (phan ling) hay C„,2H2n04 x + 30y = 4 V / 2 , 22,4 Theo bai ra: nf^^Q^ = , m o l ; n H =0,225mol ^ fynii-; Thi du 8: Cho 3,68 gam h6n hgp g6m A l va Z n tac dung vdri m6t lirgng vijfa dii gam m6t chat k h i Gia tri ciia m la 0,24 :lO = r''•;•:,)»! ):.'• Theo dinh luat bao toan khoi lugng, ta c6: Thi du 6: D u n nong m gam h6n hop X g6m cac chat c6 cung m6t loai nhom chiic RCOONa + N a O H ' Cn+2H2n04 + I ' n " ^ ( n + ) C + n H nH2 (phanu-ng) = n ^ (giam) A 40,60 g(x-62y) A.V = ^(x-30y) m„,,, + m,,,,^, m ^ d H2SO4 = + =>m,,,„,i 98 = "^ddmM m,,„„,i + + mH2 0,1.2 = , + - , = 101,48 (gam) Dap an diing la A ^c^.; ^ ! ' ^' elm nang On luy^n thi DH mign B^c - Trung - Nam mOn H(5a hpc - Cu Thanh Toan Thi du 9: Nung nong 16,8 gam h6n hop gom Au, Ag, Cu, Fe, Zn vdi m6t luong du O2, de'p cac phan ling xay hoan toan, thu duoc 23,2 gam chat ran X Th^ tich dung djch HCl 2M vCra dia de phan umg vdi chat rSn X la: A 200 ml B 400 ml C 800 ml D 600 ml (Trich detuyen sink Cao dang khoiA) Hu&ng ddn gidi Soddxayra: - ^ ( o x i t ) - ^ ^ H , m^-^-^ Theo djnh luSt bao toan khoi luong: Cty TNHH MTV DWH Khang Vi$t Hu&ng ddn gidi 88 Theo bai ra: n^aOH = 0.2 = 0,4 mol ; ncHjCOOCaH^ = ^ PTPLT: CH3COOC2H5(l) + NaOH(r) CHjCOONaCr) + C2H,OH t 0,1 Ta tha'y: " 0,1 mol nNaOH > ncH3COOC2H5 =^ NaOH du Sau phan ung c6: CH,COONa va NaOH du Theo djnh luat bao toan kh6'i luong ta c6: mo2 =23,2-16,8 = 6,4g=>no2 =0'2(mo!) =>n_2 =0,2.2 = 0,4(mol) =:>n ^ =0,4.2 =0,8(mol) = n^ci o " Dap an diing la B Thi du 10: Khi d6't chay hoan toan m gam h6n hop hai ancol no, don churc, mach thu duoc V lit CO (a dktc) va a gam H,0 Bieu thirc lien he gii?a m, a va V la: V 5,6 B m = 2a - V C m = 2a D m = 2a + 22,4 + l,5n02 l,5n(mol) > nCOj n (mol) + =>m+ = 22,4 V.44 +a=>m = a- 22,4 Thidu 12: Cho 13,8 gam axit A tac dung vdi 16,8 gam KOH, c6 can dung djch sau phan ling thu duoc 26,46 gam chat ran C6ng thiic ca'u tao thu gon ciia A la A.C,H(,COOH Sodo: Axit X + K O H ^ Chat ran (mudi, KOH du) + H20 5,6 Theo djnh luat bao toan khd'i luong, ta c6: V =^mH20 = 13,8 + 16,8-26,46 = 4,14(g) Suy ra: M = — 0,23 f' " " ,, j, + mj^QH - "'(r) =^nH20 =4,14/18 = 0.23 (mol) Ne'u axit A don chtic thi n ^ = n^jO = 0,23(mol) ^.^^j ^^^^^^ ^, = 60 ( C H X O O H ) V ; ^ , ' " Dap an diing la C Thi du 13: Xa phong hoa hoan toan 1,99 gam h6n hop hai este bang dung dich NaOH thu duoc 2,05 gam mu6'i ciia mot axit cacboxylic va 0,94 gam h6n hop hai ancol la d6ng dang ke tiep Cong thiic ciia hai este la A HCOOCH3 va HCOOC2H5 B C2H5COOCH3 va C2H5COOC2H5 C CH3COOC2H5 va CH3COOC3H7 D CH3COOCH3 va CH3COOCH5 5,6 (Trich de thi thu:Dai hoc khoi A, B) Hu&ng ddn gidi Thidu 11: Xa phong hoa 8,8 gam etyl axetat bang 200 ml dung djch NaOH 2M Sau phan irng xay hoan toan, c6 can dung dich thu duoc chat ran khan c6 kh6'i luong la • " -S"** V (n+l)H20 (n+l)(mol) D.HCOOH (Trich de thi du bi tuyen sinh Dai hpc khoi A) Hu&ng ddn gidi Dap an diing la A A.8,2g C.CH3COOH B.C2H5COOH Gia thie't axit tac dung het v6i KOH 11,2 5V Theo PTHH ta tha'y: no, = ISuro^ = -— (mol) ^ 22,4 Theo dinh luat bao toan khoi luong: m^,,^, + TTIQ^ = m^Q^ + m^^o 1,5V.32 « n-i < - il i V ( Trich de thi thiiDai hoc khoi A, B nam 2013) Hu&ng ddn gidi Ancol no, don churc, mach ho: C„H2„+20 Phuong trinh hoa hoc d6't chay: C„H2„,20 (mol) +0,1.46 Dip an dung la C =>VddHCi2M =0,8/2 = 0,4(1) = 400(ml) A m = a 8,8 + , = m m = 20,2 gam B 8,56g C 20,2 g D 10,4g (Trich de tuyen sinh Dai hoc khoi A) Sodd phan ung: Este + NaOH Theo djnh luat bao toan khd'i luong: => 1,99 + m^aOH = 2,05 => = 0,025 (mol) IN^OH + 0,94 ^ > mu6'i + ancol m„,e + mN^oH " mmu«i ,^ + ancol m^^oH = (gam) = n^„ (don chiJc) , ;1\H , Cgm nang 6n luygn thi DH mign BSc - Trung - Nam mOn H6a hpc - CD Thanh To^n Suy ra; M Cty TNHH MTV D W H Khang Vigt nH20(2) = ^ - 79,6 ; M „ „ „ = ^ = 82 ( C H , C O O N a ) 0,025 ' • 0,025 m,„„,, = , C H C O O Q H , (88) duoc sau phan l i n g la A.68,950g B 19,675g C 13,075g (Trich Huong G i a t r i Ion nhat ciia V l a de thi thu Dai hoc khoi A, B) A 22,4 B 11,2 C 5,6 ( Trich ^ Hu&ng niHci = m d d - C % = — 100 — — = 5,4/3g Hi A a/ 5,475 , => n ^ c i = = 0,15 m o l 36,5 ddn nv = 37,6 9,4.4 = l(mol) —^ 1(aiidehil phan lillg) ~ Idiiilro ph,iu I'nig) ~ —^ l(ancoldaiichifclaora) — Ni,." (mol) (mol) RCHO+H mco2 ( b a y r a ) = , 4 = , g R C H O H + N a - > R C H O N a + 0,5H2 t vay m muoi clorua "^C02 + " H 2O (>ao ra) = 14,2 + , - , - , rt'i 0,5(mol) V = , 2 , = 11,2 ( l i t ) ' ^ D a p an d u n g la B = 10,375g Chu D a p an d i i n g l a D Thidu >RCH,OH l(mol) T h e o d i n h luat bao toan kh6'i l u o n g : muoi clorua + , jj,,^ j j„.j|| r,, , Suy so m o l k h f g i a m - = ( m o l ) = 0,15 m o l " ^-'5 mol + "^HCl gidi V i khd'i l u o n g d u o c bao toan, nen: m y = m ^ = , ( g ) cua cac phan l i n g NH4HCO.,, N a H C O ^ v a K H C O , v o l H C l la: hoc khoi A, B) , Suy ra: mH20 O^o ra) = 0,15 18 = 2,7 g "^hh D 13,44 de thi thvcDqi T a c o : m x = ( , ) = 37,6 (g) > H.O + C O , t = , ( ot, R - O - R ' + H.O — 0,25, (mol): nH20 = - l a n c o i = H™ ^HjO Theo djnh luat bao toan kh6'i luong: m„„„| = = 0,25.18 ^ ^ = 2,25 (gam) Phaong phap 2: PHl/ONG PHAP BAO TOAN ELECTRON I L I T H U Y E T + mH20 - Trong phan ling oxi hoa - khu, xay ddng thdi qua trinh oxi hoa va qua trinh - Djnh luat bao toan electron: Trong phan ilng oxi hoa khir, tdng sd electron => 0,25(14n + 18) = m + 2,25 => 0,25(14.1,6 + 18) = m + 2,25 => m = 7,85 (gam) chat k h u cho phai diing bang tdng sd electron chat oxi hoa nhan: Dap an dung la C Thi du 18: Cho mot lucmg bot Zn vao dung dich X g6m FeCl va CuCl Kh6'i lugng chat rSn sau cac phan ting xay hoan toan nho hcfn khdi luong bdt Zn ban dSu la 0,5 gam C6 can pMn dung dich sau phan ling thu duoc 13,6 gam B 17,0 gam C 19,5 gam D 14,1 gam Huong ddn gidi FeCl2 + Zn- CuCl, -> Z n C l j + Da'u hieu de nhan bai tap c6 thd' su dung phuong phap bao toan electron de giai la cac bai tap cd phan ling oxi hoa - khu Thidu 1: Cho hoi nude di qua than nong thu dugc 15,68 lit (dktc) hdn hgp A ( Trich de thi thu: Dai hpc khoi A, B) Ta CO so 66 phan ling: - 11 V A N D U N G mu6'i khan T6ng khoi lugng cac mu6'i X la A 13,1 gam 2:e(cho)= Se(nhan) gdm C O , C O o va H o Cho toan b6 A tac dung het vdi hdn hgp M g O , CuO du, nung nong thu dugc hdn hgp chat rdn B Hoa tan toan bg B bang HNO3 dac, ndng, du Fe dugc 26,88 lit N O , (san pham khu nhat, dktc) Sd mol H , A la Cu A 0,4 B.0,2 m hh(FeCl2,CuCl2 "^Z" ~ ^ ZnCI ) +mz„ = m' ,h h ( F e a , c u c i ) = '3,6 - 0,5 +(mz„ Theo bai ra: n ^ = , ( m o l ) ; n N O " ' ' ^ ( m o l ) -0,5) =13,1 C 4- H2O ^ (gam) Thi du 19: D u n nong hdn hgp gdm 0,06 mol C,H2 va 0,04 mol H vdi xiic tac N i , sau mot thdi gian thu dugc hdn hgp Y DSn toan b6 h6n hgp Y Idi tCr ttr qua binh dung dung dich brom (du) thi lai 0,448 lit hdn hgp k h i Z ( d dktc) CO ti khdi so vdi O j la 0,5 K h d i lugng binh dung dung dich brom tang la C 1,04 gam Nhan xet: K h d i lugng binh dung dung dich brom tang chinh bang khdi lugng etilen, axetilen cd hdn hgp Y Mz = 32 0,5 = 16; n^ = 0,448/ 22,4 = 0,02 (mol) m c j H + n i H = my = mz + m ,y„h brom t.i„g) y C + 2H2 _> 2y (l) +1 - H2 ->2x 1,2 4x + 4y = 1,2 • Hu&ng ddn gidi Theo djnh luat bao toan khdi lugng, ta c6: +4 C - 2e ^ N + le D 1,32 gam ( Trich de thi thii Dai hoc khoi A, B) ! -^x +2 X ^ " •< ,^ ^ C + 2H2O ^ CO2 : ^ x + x + y + 2y = , ^ x + 3y =0,7 Dap an dung la A B 1,20 gam CO + H2 x Vay tdng khdi lugng cac mudi X bang 13,1 gam A 1,64 gam 2013) Hu&ng ddn gidi '^hh(Fe, Cu) 13,6 D.'o,5 (Trich dethi thu: Dai hpc khoi A, B nam Theo djnh luat bao toan khdi lugng, ta c6: r"hh(FeCl2,CuCl2) C.0,3 A = > X + y = 0,3 (2) ^ Tilf(l,2) = > x = , ; y = , l ' Vay sd mol H2 la: x + 2y = , + 2.0,1 = , (mol) Dap an dung la A , y,,v, ' , , -.M.u ltr,ubo'.,n elm nang On luy^n thi DH mjgn BJc - Trung - Nam mOn H6a hpc - Cu Thanh Toan ^ Thi du 2: Hoa tan hoan toan 2,44 gam h6n hop hot X gom Cu va Fe.Oy bang dung dich H S O dac, nong, du Sau phan urng thu duoc 0,504 h't khf SO, (san ph^m khir nhat, dktc) va dung dich chiia 6,6 gam h6n hop muoi sunfat Phan tram khoi luong Cu X la A 26,23% B 39,34% C 65,57% D 13,11% {Trich de thi thutDqi hoc khoi A, B nam 2013) Huong dan gidi Quy d6i h6n hop Cu,Fe^Oy Cu,Fe,0 C u - e ^ C u ^ * :./?•!• 2c Matkhac, taco: 64a + 56b + 16c = 2,44 2Fe -> Fe2 ( S O )3 a ^ a b -> HCl 2M, thu duoc dung dich Z Cho AgNOy B 76,70% du vao dung dich Z, thu duoc C 53,85% D 56,36% (1) •.-'](• H2O O2 0,24->0,12 " =>y = 0,06 Thi du 3: Hoa tan hoan toan 24,8 gam h6n hop X gom Fe va Fe.Oy bang dung djch H S O dac, nong, du thu duoc dung dich Y va 4,48 lit S O (san phim khir nhat, dktc) Phan tram khd'i luong nguyen to oxi X la A 20,97% B 16,84% C 25,73% D 32,56% ( Trich de thi thu Dai hoc khoi A, B nam 2013) Huong ddn gidi Theo bai ra: n^Q^ =0,2 (mol) z -> .FeS,Fe,S (X) Tatha^y: 18 (Y) >Fe^+,S0^-,H+,N03 ^ B a S ^ (Z) ng^^) f ^8*504 = 0,025(mol) , Xac dinh s6 mol N O (x mol), NoO (y mol) Y Hu&ng ddn gidi Theo bai ra: n^^^^ = 30 < 37 =^ lai c6 M > 37, vay la N.O ( M = 4 ) , Al a(mol) - > c AP+ +3e 3a Mg b(mol) -> > Mg^+ + 2e 2b /* f 19 Cty TNHH MTV DVVH Khang Vi^t u a i n nang On luy^n thi OH mign B^c - Trung - Nam mOn H6a hgc - Cu Thanh Toan Do vay, thii tir cac P T H H : + 3Fe'+ Al 2CO2 + B a ( O H ) Al-^+ +3Fe^+ 0,05^0,15-^ 2A1 + SCu^^ ^ 0,15(mol) 0,1^0,15 3Cu j +2Al-^+ 0,2 ^0,15(mol) ^ -(0,2-0,1) V r H = (0,1 + , + 0,2).22,4 = l,2(l) ^^^^^ v a y V CO gia trj la 2,24 (1) hoac 11,2 (1) 0,05 ^ 0,075 ^ 0,075(mol) Dap an diing la B CO2+OH= a ( m o l ) ; n^^QH = l,5a(mol) -'M^l < " N a O H / " A = l'5a/a = 1,5 < => Tao h6n hop mudi axit va mu6'i trung hoa: SO (CO.) + N a O H ^ X X ^ y Theo bai ra: X ^ ' - m H = ' a l = 9a(g) ,, Theo djnh luat bao toan kh6'i lirong: ^ + niNaOH = m ™«i + > ^ =J>a(27.2) + l,5a.40 = m + 9a ^• a + 60a = m + 9a * CO]- 0,1 - O,l(mol) + Ba2+ - iqob^ '«rtl;YMlB}Yftf, < lorn 1.0 BaC03i , — 0,1 — O,l(mol) K6t tua la BaCOj ^ n^^co^ = 19,7/197 = , l ( m o l ) ^ ' " B a c o , = 0,1 ( m o l ) 2A1 ( O H ) ^ i +3Na2S04 0,3 - * ^ 0,1 VT^H, = , 2 , = 2,24(1) nt-rtrr; Thf nghiem 2: n^aOH = , = , ( m o l ) Trong thi nghiem nay: N a O H , AI2 (SO4), ddu het; • qt.' A l (OH)^ bj hoa tan m6t phSn T r u t m g hgrp 1: OH" dir, chi xay phan ling: » > i O CO2 + B a ( H ) -^BaC03 [ + H O 0,1 ( m o l ) =^a = 0,l = 7,8(g) Cdu 14: Cdch 1: Theo bai ra: ne^^^o^)^ = , (mol); n^aOH = ' ( m o l ) ; N a O H + AI2 (SO4), 6x — X " f 2A1(0H)3 i +3Na2S04 — 2x(mol) Al(OH)^+NaOH-> Na[Al(OH)^ I- : m + m T r u ^ g hgrp 2: OH" phan ung hfe't, xay cac phuofng trinh hoa hoc sau: CO2 + Ba(OH)2 -> BaCOj [ +H2O 0,1 ^ 186 +H2O + OH- - 0,1 (mol) Trong thi nghiem nay: N a O H het, AI2 (SO4 )^ (du) Dap an dung la A * ^ n ^ ^ _ = , + 0,2.1 = , ( m o l ) ; n^2+ = , ( m o l ) J*!:' Dap an diing la C C o M i J : Theo bai ra: CO2 + N a O H N a H C O j 0,2-0,2 2A1 + 3Fe2+ -> 3Fe i + 2Al-''+ v a y m = m c u + m p ^ = , + , = 13,80(g) * -^Ba(HC03)2 0,1 ^ 0,1,^ Taco: 6x + y = 0,4 2x-y = 0,l 187 Cty cam riang n luyCn I h i DH niign B^c - Trung - Nam mOn H6a hgc - Cu Thanh Toan Giai duoc: x = 0,0625; y = 0,025 M a t k h a c : 12a + 2b = 4,872 Vay m = , - , (g) T i r ( l , ) =4>a = 0,336; b = 0,42 Dap an dung 1^ A D o d o : x : y = a : b = 0,336:0,84 = : Cdu 16: K f hieu chung cac mu6'i nitrat h6n hop A la M N O Trong 50 gam A c6 khoi lucmg nguyen t6' oxi bang: mn = 50.9,6 100 = 4,8(g) (2) C T D G N ciia X la ( C j H , ) , C T P T : ( C H , \ ' at x^v V I g6c ankyl c6 hoa tri I ndn n = vay X la C H n o = , / = 0,3(mol) Dap an diing la C Suyra: n ^ ^ _ = , / = 0,1 (mol) (trong g6c nitrat N O c63 nguydn tuO) So d6 phan u-ng: 2MNO3 +OH" -NO J CauJ9: K i m loai thu duoc catot la Cu: n^^ = 1,28/64 = 0,02(mol) S6 mol k h i thu duoc anot: n,, = 0,336/22,4 = 0,015(mol) ^2MOHdcatot(-): C u + + 2e b day ta tha'y: thay ion N O bang ion O ' ' 0,04 2H2O => 0,1 mol N O thay bang 0,05 mol O"" anot (+): o 2- 2Cr m = mMN03 - H2O + m^2- = > m = - , + 0,05.16 = 44,6(g) ^ 2x (mol) i s': 4y y ( m o l ) Gia su toan bo k h i a anot la C I , thi n, (nhucfng) = 0,015 = 0,03 (mol) < n (Cu Cdu 17: De sau phan ung kh6ng c6 k i m loai thi luong Z n (phan umg hfet) chi khir 2+ Z n + 2Fe ( N O \> Ban dSu: a Phan ung: a - > 2a Zn ( N O \ 2Fe ( N O b (mol) Goi X, y lin luot la s6' mol CU,©, thu duoc o anot Tac6: (mol) x + y = 0,015 2x + 4y = 0,04 G i a i r a d i r o c : x = 0,01; y = 0,005 n ^ ^ = y = 4.0,005 = 0,02(mol) Dap an diing la C Cdu 18: Dat hidrocacbon X la C^H^ = 0,02/2 = 0,01 = - M x+ ^ 02^xC02+^H20 iAl^ -2^ Vay pH = - l g [ H + ] = - l g ( l O - = 2>ap an dung la C "ico2+"1H20 = "Ik,-5,586 =^ m c o 2a thi Z n phan u-ng hat (bi tan hd't) va Fe^+ ^ Fe^^ (dd) Taco: Cu i ^ H + + - O T+2e ,\ Dap an diing la B Fe ^ + 2e ^ H2 + H " X Fe J M H H M T V DVVH Khang Vi?t (l) Cho rat tCr tir axit H Q v^o dung dich A ntn c6 th^ coi phan umg xay Ain tiSn 1^: 189 elm nang On luy^n thi DH mien la L H+ + a a Cdu 21: Theo bai ra: HCO3 COJ- ^ ^ = , / —»a(mol) V i CO CO2 thoat nen phan ling tiep theo xay la: HCO3 + nH2S04 CO2 T +H2O H+ 0,045 ^ 0,045 ^ n = 0,05 ( m o l ) ; n H N = 0.8 = 0,08(mol) = , , = , ( m o l ) => n ^ + = , + 0,02.2 = 0,12(mol) _ =0,08(mol);n 2-=0.02(mol) 0,045 ( m o l ) 3Cu H C O ^ du di phan ling vdi BaCOH).: HCO3 + B a ( O H ) 0,15 8H+ + N O ^ C u + + N O + 4H2O 0,12 0,15(mol) Goi a, b Idn luot la s6'mol Na2C03,NaHC03 c6 500ml dung djch A a + b = 0,045 + 0,15 = 0,195 + 0,045^0,12 BaCO, i +H2O + O H " ^ Ta c6: Cty TNHH IVITV DVVH Khang Vi§t nung - Nam mOn H6a hgc - Cii Ttianh To^n •a = 0,105; b = 0,09 a + 0,045 = 0,15 0,03 0,05 0,045 0,08 , , —— < < ndn H^ phan ling h&i Suy ra: n _ (tao muoi) = 0,08 - 0,03 = 0,05 (mol) NO3 Vay khoi luong mu6'i khan thu duoc: Vi Vay CNa2C03 = a/0,5 = 0,105/0,5 = 0,21M =^ m = 0,045.64 + 0,05.62 + 0,02.96 = 7,90(g) CNaHCO., = b / , = 0,09/0,5 = , I M Dap an dung la B (^rf^"> 4- Dap an dung la A X Cdu 22: K h i tac dung voi k i m loai (chi the' hiSn tinh khiJr) axit H C l dong vai tro la Cdch 2: V i nguyen t6' cacbon duoc bao toan nen ta c6: chat oxi hoa: "c(Na2C03) +"c(NaHC03) = " c ( C ) + "c(BaC03) =>a + b + Fe + 2HC1 = , + 0,15 = 0,195(mol) = » C N a C +CN,HCO3 = a / + b / , = (a + b ) / , = , / , = , M + FeClj + H P H Dap an diing la C Cdu 23: T h i i t u cac k i m loai day dien hoa: Mg^"*" / M g ; Z n ^ ' ^ / Z n ; Trong cac phuong an thi chi c6 0,21 + 0,18 = , M Dap an dung la B Fe^+/Fe;Cu^+/Cu;(Fe-^+/Fe^+)Ag+/Ag }i =^ Kha nang phan iJng ciia cac Chii y; Ne'u each tien hanh thi nghiem: Cho nhanh (va khong khua'y deu) axit HCl vao dung dich A thi ca C O " va HCO3 dong thcfi phan img vofi H*, va c6 the dong thdi du: C O f + H + a b —> a ^2CO^- + Ba(OH), c -^b -> b =0,045 c + d = 0,15 He khong giai dirorc a, b, c, d IQO !< V i Fe du ndn: + M g , Z n phan ling hat => c6 mu6'i M g ( N ) ; Z n ( N ) + Kh6ng tao mu6'i Fe (NO3 )^ ma tao mu6'i Fe(NO3 )^ Vay muoi X la M g ( N ) ; Z n ( N ) ; F e ( N ) d Dap an dung la D a + b = 0,15 Tacoha: Do k i m loai sau phan ling la: A g ; C u ; F e => khong CO cac muoi C U ( N ) , A g N dung dich sau phan ling c HCO3 + B a ( H ) - > BaC03 [ + H - + H2O d - '' + M u o i Cu (NO3 )^ ; AgN03 het BaCOj j + H " ^ CO2 + H2O HCOJ + H+ HCOJ kim loai: M g > Zn > Fe > Cu > Ag >„a";('^' ":'" kmLM- Theo bai ra: tXi^^ = 2,8/22,4 = 0,125(mol) Kh6'i luong ruou A nguyen cha't: m ^ = ^ • ^ " • ' ' ' ' = , ( g ) 100 191 :ani n\i6n R d c • iiii n i l n a n y fin i i i y c n TIIIIKI Mam mOn H6a h Q C - Cil Thanh Toan Cty TNHH MTV DWH Khanp Vi^t =^mH20 = , - , = l,8(g) => n^^o = 1,8/18 = , ( m o l ) H2O + Na NaOH + 0,1 H2 T 0,05 ^nH2 gliiij: ' Benzen, axeton, xiclobutan, axit axetic khong lam ma't m^u dung djch thu6'c ti'm *• • Ankylbenzen chi tac dung vod dung dich KMn04 dun nong (mol) QauJS: Theo bai ra: Dat ruou (ancol) A la R ( H ) ^ R ( O H ) ^ +aNa ^ * ^OSi.i+OOOOf Xac djnh CTPT c u a e s t e E ( C ^ H y O , ) Trong 1,6 gam E c6: R(ONa)^ + ^ H T ^ = 0,08 (mol);nH20 = 0.064 ( m o l ) "NaOH = = , ( m o l ) , - , = 0,075(mol) (sh,hradoA)= 0,15/a n^^^ = 3,521U + 0,08 mol C + 0,064 = 0,128 mol H 0,075(mol) + m o = 1.6-(0,08.12 + 0,128.l) = 0,512(g) =i^MA = , / ( , / a ) = 4,6a/0,15 : no = , / = 0,032(mol) Ta C O bang: a MA le le 92 A loai loai C3H5(OH)3 ^ D o d o : x : y : z = 0,08:0,128:0,032 = : : 'H "!•' i => CTPT cua E la C , H (vi este don chiic) mNaOH=0.15.40 = 6(g) Dap an diing la C E + NaOH chat ran khan 10 16 = ^ m E + m N a O H =m(^);(lO + = 16g) Cdu 25: " * * C3H7CI CO dong phan: 11 CHjCH^CH^Cl; CHjCHCCOCHj Do do, E la mot este vong: CjHgO C O dong phan: => axit tao ntn este E c6 the la: HOOC - (CH2 )3 - C H O H C H C H C H H ; C H C H ( H ) C H ; C H - O - CH2CH3 Dap an diing la C '''' C H N c6 d n g p h a n : Cdu 29: Theo bai ra: nHci = , = , ( m o l ) CH3CH2CH2NH2;CH3CH(NH2)CH3;CH3 -NH-CH2CH3;(CH3)^N * T h i i tir cac CTPT c6 s6' dong phan tang din: C3H7CI < C3H8O < C H N + Trung hoa axit beo tit c6 chat beo A (, Dap an diing la A Cdu 26: Theo bai ra: nNaOH = 0-04.1 = 0,04 ( m o l ) ROH 2,46g + - + Phan ung vdri chat beo (trieste cua glixerol va axit beo) , + Phan limg vdi H C l ,' S6 m o l N a O H dung de phan ling vdi axit beo tir do: NaOH ^ R O N a + H j O 0,04mol ji Tatha'y: 1,420 k g N a O H dung vao viec: 10 000.7 "NaOH(i) = " K O H = 0,04mol Theo djnh luat bao toan kh6'i luong ta c6: 2,46 + 0,04.40 = m(mu6'i) + 0,04 18 )QQQ = l,25(molj = n^^oj,) n,^j,Qj^j2) = "HCi - S6'mol N a O H phan ling vdri HCl: - S6 mol N a O H phan ling vdri chat beo: nNaOH(.3) = =0,5(mol) => m (mu6'i) = 3,34 (g) Dap an dung la D > Cdu 27: Cac chat c6 kha nSng lam mat mau dung dich thu6c t i m (KMnO^) \k: • Hidrocacbon kh6ng no: etilen, axetilen - Ankylben/en: toluen • Andehit: andehit fomic Dap dn diing 1^ A 192 " 1.25 - , = 33,75 (mol =^nc3H,(OH)3 = 33,75/3 = l l , ( m o l ) ^ So 66 phan iJng: A x i t beo + N a O H ^ % 1,25 X a phong + H O i , ; 1,25 (mol) 193 Tan] ii,i/n) ("ifi luyeii Mil f )l I iiiifMi H l r Chat beo + 3NaOH 33,75 -> HCl I UKK; MM • nini) H< i,i ho : ^ i Cty TMHH MTV DVVH Khang Vi^t , Xa phong + glixerol ll,25(mol) + NaOH -> NaCl + H j O 0,5(mol) Ta C : + niMaOH - "lNaOH(2) = "IXP + "1H20(1) + "Iglixerol • • - fb Dap an diing la A 0,1 , v/J ^;^ji54: Theobai ra: ng^ =20/160 = 0,125(mol) Caosu buna-S (polifbutadien - stiren]) + Gly: H - N C H C O O H ^ ; f COOH =J> Dap an dung la B m mil xich butadien =>m„„ Chu v.- Glu: H O O C - ( C H )^ - C H ( N H ) - COOH = 0.125.54 = 6,75(g) = 45,75-6,75 = 39(g) ^ n „ i , = 39/104 = 0,375(mol) Cdu 35: Cac chat ma phan tir chiia nhom chiic andehit RNHCl , 51,7g HHCI M Dap an diing la A Cdu 31: Dat ki hieu chung ciia hai amin la RN ^ ('""O vay ty Id mat xich butadien va stiren: 0,125:0,375 = 1:3 Lys: H N - ( C H )4 - CH ( N H ) - COOH =>mHci = , - , = 21,9(g) =aH\;''^ "Br2 = n mil xfch buiadien = " - C H - C H = C H - C H - ~ ^ ' ' CH3-CH(NH2)C00H n c , H , O ( t t ) = 0,l(mol) 194 • ' stiren • ! ] trang CgH^CH = CH2 + Br2 n„anto«>«iu) = , - ^ ^ ^ ^ = 0,05 (mol) • nau C^HjCHBr - CH2Br kh6ng mau 195I m nang 611 !iiy6n i h i n i l mien Ric iting - Nam mOn H6a hpc - CCi Thanh Toan C6H,CH20H + B r / H ^ ancol benzylic Dap an diing la A Cdu 37: Ban dSu c6 Cty TNHH MTV DWH Khang Vigt Kh6ng xay 3) C H ,^ 4) C H N H + 3Br2 x mol H,; y mol C2H4 = 13y =^ X g ^ : = my a a / (2x -a).4 dx/He _ " i x - ( x - a ) 3,75 m r-(2x.4) 'y/He 2x — a my — nix ndn 2x nob'A Rum X: C„_,H2(„_,)+20(Cn-lH2nO) T6ng Andehit Y: C„H2„0 (2x) Axit cacboxylic Z: C„H2n02 (d^u no, mach ho, don chiic, ciing s6' nguyfen tir H) (2x-a) So 66 d6't chay h6n hop X, Y, Z: Ta c6: dx/He = M x / = nix /(2x.4) Y/4 J Dap an diing la B x X a > a (x-a) (x-a) ' ^ Y / H e =^ C^Hj ( N H ) B r i + H B r 6) C H C H O + Br2 + H O ^ C H C O O H + H B r = y C2H4 Ban dSu: Phan ung: Con lai: - CHBr - GOGH 5) C^HjOH + 3Br2 ^ C H (OH)Br3 i + H B r Taco: ^ ^ ^ - i ^ = 3,75.4 = 15 ^ 2x + 28y = 15x + 15y x+ y =>13x CH2Br = CH - COOH + Brj C n - i H „ - ± ^ ( n - 1)C02 + nH20 x(mol) = 0,75 —> ~* C„H2„0 , nC02 x(mol) nx + nHjO nx CnH2n02 - ^ ^ " C + n H = 0,75 = > x - a = I,5x =»0,5x = a x(mol) —» nx —> nx Vay H = - 100% = — 0 % = 50% X ' "CO2 _ ( n - l ) x + nx + nx _ n x - x _ 11 Taco: 3nx 12 nx + nx - f nx n H20 X Dap an diing la B Cdu 38: Cac chat true tie'p tao C H C O O H : C H O H + CO ) CH3COOH C H O H + O2 — ^CH3COOH + H xl.t CH3CHO + - O 2 ^ ^ 3n Vay X, Y, Z la C3H80,C4H80,C4Hg02 Dap an dung la C Cdu 41: Tinh bazo ciia cac dung dich NaOH > Na2C03 > C H C O O N a (Vi tinh axit H O < H C O < C H C O O H ) ^CHjCOOH Dap an diing la D 7ft« v: Tir metanol ( C H O H ) va cacbon monooxit (CO), nhcr xiic tac thich hop la phuofng phap hidn dai san ph^m axit axetic: CH3OH + CO- xt,t" 2) C H h Br2 ^ => PH3 < pH, < pH2 (kiem cang manh thi pH c^ng Idn) Dap an diing la A cang 16n thi pH cang be ^CHjCOOH Vi C H O H va CO d^u duoc di^u che tir metan c6 san thien nhidn va dSu mo n6n phuong phap cho axetic vdfi gia nha't rd« 39: Cac chat phan utig duoc vori nu6c brom (dd Br,): 1) C H + B r = —=>33n = n - =>3n = 12 =>n = 12 CHBr2 - C H B r CH2Br - CHjBr I OH- c^ng 16n thi pH c^ng Idn £^u42: Gia thid't axit tac dunghd't vdi KOH So 66: Axit X + KOH ' Theo djnh luat bao toan khd'i luong, ta c6: m^ + m^oH — '"(r) + ' " H O =^mH20 96 Chat rJn (mu6'i, KOH du) + H.O = 13,8+16,8-26,46 = 4,14(g) Cty H\illll M V UVVH Khang Vi§ dm nang 6n luy^n thi SH mi^n BMc - Trung - Nam mOn H6a hpc - Cii Thanh To^n ::=^nH20 ChuJ.' =4,14/18 = 0,23(mol) "^^^ thuye't CO the' co: 14NaOH + 2CrCl, + 3CI2 ^ 12NaCi + Na2Cr207 + THjO Nd'u axit A dom churc thi Suy ra: M = = n^^^^ = 0,23(mol) * *~ C^'^'- R / t r ^ J ! «s R C O O K + HjO Cho dung dich H2SO4 vao cac mSu thtr: "H20 — t Trich cac mSu thu ="H20 = 0,015(mol) Cdu 45- Chon thuoc thijf la dung dich H2SO4 loang thi each nhan bid't don gian nhat: + (COOH)^ + aKOH ^ R ( C O O K ) ^ + aH20 =^"R(COOH) 0,01 m2 =0,015.71-l,065(g) rii; (f Chu v: R C O O H + KOH ^ ^ =^m, =0,07.40 - 2,8(g); = 60 ( C H C O O H ) Dap an diing la C =^"RCOOH 0,07 , : ^ 'a a Dang bai chat ran thu duoc lu6n la mu6'i va ki^m du (axit hOu co phan tir kh6'i nho thucmg o dang long) Cdu 43: Xet cac phuong an: A Sai, VI tinh oxi hoa Fe-^+ > Cu^+ 2Fe-^+ + Cu 2Fe2+ + Cu^+ B Sai, VI tinh khu Fe > Ni tao ket tua la Ba: , - '-• Ba + H2SO4 BaS04 i +H2 T " Ba + H ^ B a ( O H ) + H T (Loc tach kfi't tua duoc dung djch Ba(OH)2) + Cac mftu lai khOng tao ket tiia (duoc cac dung djch mu6'i sunfat) Mg + H2SO4 ^ MgS04 + H2 dnMv< — MSU ^.^^ ,^ Zn + H2SO4 ZnS04 + H2 Fe + H2SO4 ^ FeS04 + - ; Vl u ,^ ,^ , , Cho tir i\s dung djch Ba(OH)2 tdi du vao cac dung djch mu6'i thu duoc: + MSu tao ket tiia, sau tan, bot mot phdn => MSu ban dSu chiia Zn Fe + N i ^ + ^ N i + Fe^+ C Dung ^; Ba(OH) + Z n S ^ B a S i + Z n ( O H ) i Ba(OH)2+Zn(OH)2 ^ B a Zn(OH)^ D Sai, VI tinh oxi hoa Fe^+ > + MSu tao kd't tiia trlng xen iSn trdng xanh 2Fe^+ + 2r -> 2Fe^+ +12 Ba(0H)2 + FeS04 Mitu ban d^u chiia Fe: Fe(0H)2 i +BaS04 [ iW,r>4.rfl :;v Dap an dung la C trdng xanh Chu v: * Chiing minh tinh khix: Cu > F > Fe^+ + M5u chi tao ket tua trSng ^ mSu ban dSu chiia Mg: (Cu2+/Cu;l2/r;Fe'+/Fe2+) Ba(0H)2 +MgS04 ^ B a S i + Mg(0H)2 i 2+ +1 2r; O T - 'r.'^u „ „ : hoa: A I2 T > ^ r^ 2+ Cu +12 ^ Cu^+ Tinh oxi Cu 2Fe-^+ + 2r trang Dap an dung la B ChuJ: Co th^ dung thu6c thir la nu6c: 2Fe^+ + =^ Tinh khu: F > Fe^+; tinh oxi hoa: Fe^+ > I2 Cdu 44: PTHH: trSng -'^ trSng ' Ba + 2H2O ^ Ba(0H)2 (tan) + H2 T 1: Zn + Ba(0H)2 ^ B a Z n ( t a n ) + H2 T 16NaOH + 2CrCl3 + 3CI2 ^ 12NaCl + 2Na2Cr04 + 8H2O 0,08 ^ 0,01 -^0,015(mol) Vay: m, -HiN^oH =0,08.40 = 3,2(g) m2 = mci2 =0,015.71 = l,065(g) Dap an diing la A Mg + H ( h ) — ^ M g O (trang)+H2 3Fe + 4H20(h)—!-^Fe304(den) +4H2 Qau46: Dat c6ng thiic ancol no A la C„H2n+20a ( " ^ ^ ) ,0 + H if i t i i S j lif; 19< Ca'm nang On luy^n thi DH miin ii.lc I rung - Nam mOn l i j i liuc - Cu Thanh Join 3n + l - a Cty 100.46 joTha'tich CjHsOH nguyfin chat thu diroc: V = ——— = 46(ml) O2 - > n C + ( n + l ) H 3n + - a ^ 3,5 ^ (mol) Dodo: ir\iHII M I V UVVII Miang V:i;i mc2H50H = 46.0,8 ^ nc2H.,OH = 46.0,8/46 = 0,8(mol) C , H „ , - ^ C H , O H + 2CO2 T \ 0,8 ^ = 3,5z:»3n + l - a r z : = » n - a = Ta C O bang (dieu kien n>a;n,a€N*) ^ ^ 0,8(mol) , : .HO.O • CO2 + 2NaOH Na2C03 + H2O \ Qg 0,8(mol) t-V n , (con du) = - 0,8 = 0,2 (mol) E>ap an diing la B H H++0H >H20 Cu2++20H ^Cu(OH)2 i 0,2->0,2(mol) 0,3->0,6(mol) n^^_ = 0,2+0,6 = 0,8(1) = 800(ml) Ca luong axetilen Q H , va andehit axetit CH3CHO sinh d^u tac dung duoc vdri dung djch Ag20/NH3 tao kfet tua Ag2C2 va Ag , £^a5i:PTHH: SO2 ( k ) + N O ( k ) S O ( k ) + N o ( k ) Dap an diing la B 00 (X) • 0,06(mol) BandSu: Can bang: ^ 0,11 0,1 0,07 0,02 (mol) O(mol) , ; , ^nba^ 201 dm nang On luygn thi DH mjgn BJc - Trung - Nam mOn H6a hgc - CO Thanh Todn Cty TNHH MTV DWH Khang Vijt Luong N O , phan ling: 0,1 - 0,02 = , ( m o l ) Luong Luong Lucmg Do do, Theo bai ra: n^j, = 32,4/108 = 0,3(mol) S O , phan ting (bang luong N O , phan irng): 0,08 (mol) S O , sinh them: 0,08 (mol) N O sinh ra: 0,08 (mol) can bang c6: = , ( m o l ) > n x = 2.0,1 = 0,2(mol) SO = 0,15M; Vay hang s6' can bang: = NO 0,15.0,08 S02 N ^ RCHO + A g Ty 1^ landehii: 0,03.0,02 ^ Ty le ,f a.C% niH^o = a - m c H j C o o H = a - — 100a-a.C%, ^ = — (g) _ 100a-a.C% _ a(lOO-C%) ^(mol) 100.18 1800 nH20 = CHjCOONa + n„,dehi,: "Ag a.C% 6000 12000 (mol) - u^g = n,„^ + Mg + 2FeCl3 ^ MgCl2 + 2FeCl2 Dapandung'laB Cdu 55: Theo bai ra: n^^ = ^-ieillA ' ' *' 3^61 gimb n q o ^ = 0,15(mol) 0,7 ^ 0,15(mol) < =i>a = 0,05; b = 0,7 Dap an dung la A Na^NaOH + - H , t Cdu56: So phan ting: ^ a(lOO-C%) Fe2 (SO4 )^ + SO^" 2FeS2 C u S - ^ C u S + Cu^^ (mol) 1800 3600 ™ , Theo bai la, ta c6: ^ K2Cr207 + H C l - ^ K C l + C r C l + C l + H T a(lOO-C%) 1200 • ^"^^ Cdu 54: Cac k i m loai tac dung vdri FeCl,: 0,05 a.C% C% ^ = : • Do nfi'u n^g > 2.n ^^hn => Co HCHO (nd'u la h6n hop andehit don chiJc) Kh6'i luong H , c6 dung dich + ^"-^ A g i + R C O O H HCH0 + 2Ag20 a.C%, H2O • ' = 20 52: Khoi luong C H C O O H c6 dung dich: mcH^cooH = — - ( g ) 100 ^ _ a.C% a.C%, ^ncH3cooH ^ = — ( m o l ) CH3COOH + N a ' + Rieng andehit fomic H C H O Dap an diing la B Cdu ' 01UJ,: Andehit don chiic c6 triromg hop: NOl = 0,08/l = 0,08M S03 , ^ ^ ^ ^_ ^ p ^ p an diing la C N O ] = 0,02/1 = , M 0,07 + 0,08 Co andehit H C H O £56ng dang k6 tiep la C H C H O flotv SOJ = M1:;^.0,03M; ^ a(lOO-C%) Ha f ^ - = 1200 3600 240.2 100-c% 11 3600 a.C% 240.2 3C% + ( 0 - C % ) = 11.75 = > C % = 175 = ^ C % = 25 Dap an dung la B Chu v; Ca nudrc va axit (trong dung djch axit) d^u tac dung \di natri sinh kh hidro ;0,U 0,24 0,12(mol) x x(mol) C u + + S ^ - _.CuS04 X X = ^ x = 0,12 ( m o l ) Qua trinh nhucmg, nhan electron: +2 -1 FeSj 0,24 - > +3 +6 Fe + S +15e 3,6(mol) +1 - CU2S ^ +2 +6 2Cu + S + lOe 0,12 ^ T6ng s6'mol electron cho: 3,6 + , = , ( m o l ) ' l,2(mol) : ; 1'! ; 203 Cly iNHH MIV D W H Khang Vi^t dm nano On luy^n thi DH miSn B^c - Trung - Nam mOn H6a hpc - CCi Thanh Toan +5 C H N : C H C H C H N H ;(CH3 )2 C H N H ; C H N H C H C H ;(CH3 )3 N +2 N + e - N(NO) QH||N:CH3CH2CH2CH2NH2;(CH3)2CH-CH2NH2; 4,8->l,6(mol) " ' ' ^ ,: = » V N O = 1,6.22,4 = 35,84(0 : CH3CH2CH(NH2)CH3;(CH3)3C-NH2; [^(i/, Dap an diing la B (CH3)2NCH2CH3 Cd« 57: - Khi lam mat mau nvidc brom la S O i , H2S: =^C6 14d6ngphan Ddp an dung la B SO2 + Br2 + H O ^ H S O + H B r H S + 4Br2 + H O - H S O + 8HBr ,, C O + Ca(OH)2 ^ y , , Ca(HC03)2 fi^i.r , ryr Taco: (0,15 + x).44 = 15-5,1 =^0.15 + x = 0,225 Ca(OH)2 ^ C a S + 2H20 Phan biet duoc CO, va N,: C6H,206 ^ ^ ^ C H < i O H + C O T C O + Ca(OH)2 ^ CaCO^ j + H O 0,1125 trang |, , Kh6ng xay Dap an diing la C ,t X (mol) trSng N2 + Ca(0H)2 j| „ , • ' ' ^ + Ca(OH)2 ^ CaC03 [ + H O 0,15 ^ 15/100(mol) SO2 + Ca(OH)2 ^ CaSO^ i + H O + •Ji,JU>M>iAi I r ) {:: C^H^ONa + C H C O O H ^ CgHjOH + CHjCOONa > H :> - M ) J13 CH3COONa + H2O C^u 8: So dd san xua't chat giat rira tdng hop: RCH2CH2R ' ' RCOOH ^ R C H O H ^ RCH2OSO3H -> RCH20S03Na Dap an dung la A cau 9: Sodd: C^HioOj ^ C H , ^2C2H50H->2CH3COOH )2RCOONa > C.HioOc 2CH3COOH ^ , 'Jflfe Dap an diing la C 0,5mol Cau 5: Xet cac phuong an: A HCOO - CgHj + 2NaOH -> HCOONa + C^HjONa + H2O mol -> -> 1, -> CHjCOONa + C6H5CH2OH mol , * t ' ) j M i y,, Cau 6: CH2 = CH - CH (CH3) - CH2 - CH3 (3 - metylpent - - en) '-M 0,2 ^ 0,1 -> 0,1 ' -yr • Theobaira: 0,1.(R,+44 + 23) + 0,l.(R2 + 4 + 23) = 17,6 j , =^R, + R = (C3H6) =^Ri = C H ; R = C H (cdth^) :,iv Vay hai mudi cd th^ la CHjCOONa va C2H3COONa => Khd'i luomg mudi: 82 g Dap an diing la A BO,M > ,\ Dap an diing la D , y Cau 11: Theo bai ra: n^g =0,32(mol);nf.o2 =0,14(mol) => X la: CH2 ( O H ) - CH2 - C H ( C H ) - CH2 - CH3 Sodd: HCHO,E — metylpen tan— — o l =^ 0,2 + 4x = 0,32 => A Loai, VI C^H^OH (phenol) • B Loai, vl Cu ( , + 4x) X = 0,03 HCHO->C02 CaHbOc^aC02 0,05 0,03 ^0,05 0,05 + , a , C Loai, VI NaHS04 M r ; / „X YV ' > 4Ag| (0,05+ x) ^ Dap an diing la A Cau 7: Xet cac phuong an: -^0,03a a=3 Vay andehit nhi chiic E cd nguy6n tir C, nhdm - CHO D Thoa man, cac cha't d^u tac dung vdi CH3COOH Dap an diing la D 0'2(mol); nx = 15,8/158 = 0,l(mol) lancol Y + mudi 0,1 Khd'i luong mu6'i: 108 (g) y,,, R.COO - R - OOCR2 + 2NaOH ^ R,COONa + R2COONa + R ( H ) mol D CH3COO - CH2C6HS + NaOH n^aOH = - ^ ^ = z:> Este X CO dang R , C O O R O O C R C CH2 = C (CH3) - COO - CH3 + NaOH -> CH2 = C (CH3 )COONa + C H O H mol cau 10: Theo bai ra: X + NaOH mol => Khd'i luong mudi: 96 (g) mol ^^^^^ Dap an dung la D B C2H5 - COO - CH = CH - CH3 + NaOH ^ C2H5COONa + CHjCHjCHO -> Imol V l H = % = ^ m ( , ^ , , ^ , ^ ) ^ = ^ i ^ : i | : l ^ = 202,5(g) => Kh6'i luong muO'i: 68 + 116 = 184(g) mol " O J ^ Dap an dung la A Cau 13: Theo bai ra: nH2 =0,025(mol) * Ancol X—-j^^^-^^ olefin (anken) => X la ancol no, mach ho, don chiic ROH 2ROH + 2Na ^ 2RONa + H T 0,05 ^ 0,025 2ROH " ^ , R - - R + H 3) CH3-CH MO rirvrr, ,^ 0,05 C3H5 (C,7H35COO)(C,7H33COO)(C,7H3,COO) M A = 8 =!>nA =221/884 = 0,25(mol) A + 3H2 cha'tbeoran 0,25 ^ 0,75 (mol) =^V = 0,75.22,4 = 16,8(1) Dap an dung la B Chu v.- A + H (C,7H35COO)3C3H5 RONa + ^ H T ' 2,84 4,6g mol ROH phan iJng thi kh6'i lucmg cha't ran tang 22g f^^f,;'';: X mol ROH phan ling thi kh6\g chat rln tang l,76g HC^O'J,li JS =^x = 1,76/22 = 0,08(mol) = ^ M R O H =2,84/0,08 = 35,5 , „ =,R + 17 = , ^ R = 18,5(CH3; C2H5) ^, ' HOe^ x' vay ancol la C H O H va CjH^OH Jmrn nl X v Dap an dung la D «^ tiwfo i:: Cau 17: C , H 2 , , + H O C(,H^20f, + C^^H^^Of, 1, saccarozo glucozo fructozo i {'] ' I40"C ^ 0,025 M Y =2,55/0,025 = 102 =^ 2R +16 = 102 =J« R = ( C H - ) vay Y la (CH3 )2 CH - O - CH (CH3 )^ (diisopropyl etc) Dap an dung la B Cau 14: Cac PTHH xay scf d6: (C6H,o05)„+nH20^nC6H,206 tinh b6t glucozo (X) CgHijOg 2C2H5OH + 2CO2 glucozo (X) ancol etylic (Y) C2H5OH + CUO—^^CHjCHO + -Cu + H2O (Y) andehit axetic (Z) vay X, Y, Z iSn lirot la glucozcr, ancol etylic va andehit axetic Dap an diing la D cau 15: C6ng thiic A: ROH + Na X C,2H220,, t X ' mantozo y —> + H2O X 2C6H,206 glucoza 2y X J ^ '7" • Theo bai (x + 2v) li ra: -^^ — = - =>x + 2y = 4x=»2y = 3x =4>x:y = : • Dap an diing la B T/ , Cau 18: Theo bai ra: = 0,15(mol) CH3CH20H^^2_^CH3COOH + H20 H,; BandSu: x 0 Phan img: 0,5x -> 0,5x 0,5x Con: 0,5x 0,5x 0,5x |.| ; * , 2R0H + 2Na ^ 2R0Na + H2 0,3 0,15 n O =»0,5x +0,5x +0,5x =0,3=»x = 0,2 •• Vay khoi luong axit thu duoc: ma^j, = 0,5.0,2.60 = 6(g) Dap an diing la D QhuJ: Dat cong thiic chung ciia CH3CH20H,CH3C00H va H2O la ROH (vi chiing deu c6 nhom OH phan ling vdi Na) Cau 19: Xet cac phuong an: ^''' " A Loai, VI ancol etylic , B Loai, VI natri axetat C Loai, VI natri axetat C^m nang On luyQn thi DH mign Bcic - Trunq - Nam mOn H6a hpc - Cii Thanh Toan D Thoa man, vi cac chat d^u tac dung v6i C u C O H ) , Dap an diing la D { CMi 2C6H,20, + C u ( O H ) - ( C H , , , ) C u + 2H20 Cty TNHH MTV DWH Khang Vi$t (4) (CeH.oOs),, + n H — ^ n C ^ H , , m xenlulozo ' 2C3H5 (OH)^ + Cu(OH)^ -> [ C j H , (OH)^ o ] ^ Cu + H O (5) (C6H,o05)„ + n H — ^ n C H , glicogen C , H 2 , , + Cu(OH)2 - ( C , H , , , ) ^ C u + 2H,0 ' 2CH3COOH + C U ( O H ) ^ (CHjCOO)^ Cu + H O mantoza C a u 20: n ^ + , = (2.8 + - ) / = => Co chat Dap an diing la B X + N a - > Sanph^m X + NaOH Cau 23: Theo bai ra: n c , H 2 , , = 342/342 = ( m o l ) Kh6ng xay => X la ancol C,2H220,,+H20- Cac ddng phan ca'u tao cua X : H(0H)-CH3 _CH2-CH2-OH 0 CH2-OH CH2-OH l(mol) C6H,206 1,5 CH3 => Co ddng phan ca'u tao (mol) ^ 2.75 CH2-OH ddAgN03/NH3 -> £ ^2C,H,20, V i H = % => nC6HI2O6 'H3 JQO = I,5(mol) )2Agi H H l ^ oiO -^di i n , : 3mol Dap an diing la B Vay m ^ g = - ( g a m ) Dap an diing la C hien phan ling trang bac n6n kh6ng xet phan ting trang bac ciia luong mantoza (1) C H + C u ( O H ) - > ( C H , , ) C u + H (2) C H , + B r / H ^ Kh6ngxayra ::X>'JO;{ du Cau 24: Phuong phap phan bidt: HCOOH, (3) C H H ( C H O H ) C O C H H + H ^ ^ C H H ( C H O H ) C H H j ^ j ;•, CH3CHO, C2H5OH, 3iO CH3COOH dd NaHCOj " " ^ )C6H|207 + A g i => fructozcf phan umg duoc vori (1), (3), (4) (i.h ,>ii;!i:':'>' C a u 22: Cac chat b i thiiy phan: (1) ( C H , o O , ) „ + n H - H+ HCOOH, C H C O O H ^nC6H,206 amiloza (2) ( C H , o O , ) n + n H - CH3CHO, coi ->nC6H,206 khong c6 HCOOH ->C,H,206+C,H,206 CH3COOH C2H5OH fill f\L dd AgNOj/NHj dd AgNOj/NHj c6 i amilopectin (3) C , H 2 , , + H — f- /! kh6ng CO t coT Dap an diing la B saccarozo • Chii v; V i tach glucozcr khoi h6n hcfp thu ducrc sau phan l i n ^ thuy phan de thuc C ^ u 21: Xet cac phan ling: (4) C H , + A g OrSiA + OHD, M' (6) C , H 2 , , + H O — ! i ^ C H , CH,CHO kh6ng c6 i ... H - f d ) = ( V + 0 ,1) = 0, 01 ( V + 0 .1) ( m o l ) ^ -> 4HNO3 0, 015 ( m o l ) 0, 015 (mol) -> H O PTHH: H^ OH' Ban 6iu: 0, 01 0,1V (mol) Phan ling: 0, 01 0, 01 (mol) Con: (0,1V-0, 01) 0, 015 => 0 ,1. .. 1, TP HCM Tel:(08) 3 911 5694 - 3 911 1969 - 3 911 1968 - 3 910 5797 Fax:(08) 3 911 0880 ^ Email: Khangvietbookstore my = m^ = 9(g) , = o n Y =9/ (10 .2)... =0,32/2 = 0 ,16 (1 ) -1 60ml O a p an diing la D 4T- A) O ,-* '' (Trich de thi thii Dai hoc khoi A, B) 47 dm nang On luyjn thi BH mi^n Cty TNHH M W OWM KfaWQ Vigt B^c - Trung - Nam mOn H6a hpc - CCi Thanh