cẩm nang ôn luyện thi đại học giải nhanh đề thi miến bắc trung nam môn hóa học

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 1MHH MIV L.'VVH Khang Vigt ca'm nang 6n luyjn thi DH mign B&c - Trung - Nam mOn H6a hpc - CCi Thanh ToSn V i cac nguyen t6' (Fe, Cu va S) dugc bao toan ntn ta c6 so do: Sa 66 p h a n i r n g : - C O O H + N a H C - > - C O O N a + CO2 t + H O 0,06 *- ;0 0,06 2FeS2 2x H.ii!;OJ>rff => "-cooH(x) = " C O = , ( m o l ) ^n^^^^ -0,06.2 = 0,12(mol) CU2S =^ n H A 18,0 B.22,4 Sod6: D C H (Trich de thi tuyen sinh DH khoi A) Huong dan gidi D 24,2 ^ Fe^^Fe203,Fe304,FeO,Fe-^^^^^!^Fe(N03)3 0,1 T a c o : n g ^ ^ o , ! =0>2mol C 15,6 ,j Hu&ng ddn gidi A C H > Bao toan nguydn t6' s i t : n^^^p^^ = 0,1 f Of ^.^^.^^^^^^^^ ^""H^o,)^ = "^^ = v a y m =0,1.242 = , ( g ) = 39,4 - , = 19,488(g) "'"^ „ Dap an diing la D (l) Thi du 10: D6't chay hoan toan 0,2 mol h6n hgp g6m mdt ancol va m6t axit don V I nguyen t6' cacbon, hidro dugc bao toan nen: m^^ = nij^^^Q^j + chiic C O cung so nguydn t6' cacbon cin diing 0,45 mol O , , thu dugc 0,4 mol CO, ^^^^^^Q^ va 0,5 m o l H^O Phdn tram khoi lugng ciia axit h6n hgp trdn la: (2) A 46,5% B 32% C 50% D.49,18% (Trich dethi thvcDai hoc khoi A, B) Tir ( , 2) x = 0,348; y = 0,232 Hu&ng ddn gidi Dat X l a Q H b ta c6 t i le: a: b = x : 2y = 0,348:0,464 = 1,5:2 = : Taco: n c = n c o / n h h =0,4/0,2 = Vay X la C H Dap an dung la D Thi du 8: Hoa tan het m gam h6n hop FeS^ va C U S dung dich H N O , sau cac phan ling hoan toan thu dugc dung dich X chi c6 chat tan, \6i tong khoi lugng cac chat tan la 72 gam Gia tri ciia m la B.40 => axit don chuc phai la C H C O O H ; ancol c6 dang C^Hfi, C 20 (Trich de thi thitDai hoc khoi A, f7-a^ +^ D 80 B) V i chl thu dugc hai cha't tan dung dich sau phan ling => Hai ch^t tan la (a = hoac a = 2) C H C O O H + 2O2 - > C O + H O C2H,0, + Hu&ng ddn gidi C U S O va Fe2(S04)3: 0:1 (Trich dethi thUDai hoc khoi A, B) Theo bai ra: np^ =0,1 ( m o l ) I jg m vnios n i o i / ' nha't) va dung djch chiia m gam mu6'i Gia trj cua m la dung dich giam bot 19,912 gam C6ng thiic phan tijf cua X la A 60 => x = 0,1 ^rB.V> iM d j c h B a ( O H ) Sau cac phan umg thu dugc 39,4 gam ket tua va khoi lirgng ph&n C C H ;' b6 X tac dung vdi dung djch H N O loang (du), thu dugc NO (san ph^m khir thuofng) roi dem toan b6 san phdm chay ha'p thu h6't vao binh dung dung =>4,64 = 12x + 2y i Thidu 9: D6't 5,6 gam Fe khong k h i , thu dugc h6n hgp cha't ran X Cho toan =0-08 ^ a =0,08.18 = l , 4 ( g ) ^ Thi du 7: Dot chay hoan toan 4,64 g a m mot hidrocacbon X (cha't of di6u kifin ^ 4 x + 18y = 19,488 2x ( m o l ) DapandiinglaB "o(H20) Dap an dung la A m c o + niH20 -> X v a y m = 2.0,1.120 + 0,1.160 = ( g ) V i nguydn t6' o x i dugc bao toan n&n ta c6: nQ^^) + " ( ) = " ( 0 ) B C H , o 2CUSO4 Theo bai ra: 400.x + 160.2x = 72 => 720x = 72 C O + H O =>0,12 + 0,09.2 = 0,11.2 + n H o l + ; H20 ~ n c o V i nguyen tO'oxi dugc bao toan nen ta c6: 0,1.2+ 0,1.a+ 0,45.2 = , + 0,5.1 =^a = ( C H , ) 31 Cty \N\iH MIV UVVH Khang Vi^t dm nang On luy^n thi DH mign B&c - Trung - Nam mOn H6a hqc - CCi Thanh Toan 0,1.60 + 0,1.62 Dap an diing la D , , , „ y'Thidu 11: H6n horp X g6m axit axetic, axit fomic va axit oxalic Khi cho m gam X tac dung vdri N a H C O , (du) thi thu dugc 15,68 lit C O , (dktc) Mat khac, d6't chay hoan toan m gam X cdn 8,96 lit khf O, (dktc), thu duoc 35,2 gam C O , va y mol H i O Gia tri ciia y la A 0,3 B 0,8 C.0,2 D 0,6 ffli du 13: Dot chay hoan toan mot luong h6n hop X g6m m6t ankan va mot anken cin diing vira dii 0,7 mol O,, thu duoc 0,4 mol CO, C6ng thiJc ciia ankan la AQH,o B.QH, C.CH, D CH4 (Trich de thi thi Dai hoc khoi A, B) Huong ddn gidi Sod6: C,Hy + O ^ C + H O Vi nguyen to oxi dugc bao toan ndn: no(02) = " o ( c o ) + " o ( H ) ^ " H O = 0-6 (mol) 1, MO (Trich de thi thUDqi hoc khoi A, B) Suyra: n,,,,,^^ = n H - " 0 = , - , = 0,2 (mol) Huong ddn gidi ' Av* Theobai ra: nco2 ( l ) = 15,68/22,4 = 0,7(mol) ; =>nx>nankan^nx>0,2(mol) no2 = 0,4(mol); nco2 (2) = 35,2/44 = 0,8(mol) Do do: nc(x) < 0,4/0,2 = => Trong X c6 CH4 (ankan) CH3COOH a + HCOOH + HOOC - COOH ^ ^ ! ^ y ^ ^ C b c —• (a + b + 2c)(mol) Suy ra, s6' mol nguyfen tir nguyen to oxi X : "o(x) = 2a + 2b + 4c = 2(a + b + 2c) = 2.nco2 = 2.0,7 = l,4(mol) X + O2 - > C O + H O VI nguyen to oxi duoc bao toan nen: "o(X) + "0(02) ^ "0(C02) Dap an dung la D Thidu 14: H n hop X g6m CiH, va H , c6 cung s6' mol La'y m6t lugng h6n hop X cho qua chat xuc tac, nung nong dugc h6n hgp Y g6m C,H4, CiH^ va C,H2, H , du DSn Y qua nuorc brom tha'y binh nu6c brom tang 10,8 gam va thoat 4,48 lit h6n hgp (do b dktc), c6 ti khd'i so vdi H , la The ti'ch O, (khi a dktc) vira dii de dot chay hoan toan h6n hgp Y la A 33,61ft B 22,4 lit C 26,88 lit D 44,8 lit (Trich dethi thi Dai hoc khoi A, B) Huong ddn gidi Saddphanung: -p>bv, "0(H20) X (C,H,, H,) => 1,4+ 0,4.2 = 0,8.2+ y.l : ^ y = 0,6 - Dap an diing la D 2FeS, CU3S + 0,12 (mol) =^a= ^ a (mol) ^ = 0,06(mol) > C,H„ H,"'^* Goi X la so mol C,H, (cung nhu H,) h6n hgp X > Fe,(S04)3 + CuS04 Theo dinh luat bao toan kh6'i lugng, ta c6: mx = my - Mat khac, theo bai ra: my = m ,bi„h b^om a„g) + " i (khf 48 my = , + ^ — = 14 (gam) 22,4 - • X i (1) - hoi (2) n) (3) / ,oMi iish' , Tir (1, 2, 3) suy ra: x = 0,5 (mol) ' Theo dinh luat bao toan nguyen t6': V i nguy6n t6' C, H dugc bao to^n ntn thi tich khf O, d^ dot chay hoan toan h6n hgp Y cung bang the' tfch O, de d6't chay hoan toan h6n hgp X : C2H, + 5/20, > 1,25 (mol) 0,5 H Dap an diing la D ) Y (C,H4, C,H„ C,H,, H,) mx = (26 + 2) x = 28x (gam) Thi du 12: Hoa tan hoan toan h6n hop g6m 0,12 mol FeS, va a mol CujS vao axit HNO, (viJra dii), thu duoc dung djch X (chi chiia hai mu6'i sunfat) va nha't NO Gia tri ciia a la A 0,04 B 0,075 C.0,12 D 0,06 (Trich de thi tuyen sinh Dai hoc khoi A) Huong ddn gidi Vi cac nguyen to (Cu, Fe va S) duoc bao toan ntn ta c6 so 66: ' 0,5 + 1/20, > 2C0, + H,0 ,?;/HtX)3H / :D-,1I- y > H,0 > 0,25 (mol) 33 Ctv iMMii M i v iJVVH Khang V i § t Ca'm nang On luygn thi B H mfo.f ^ a c - Trung - Nam mOn H6a hi?c - Cu Thanh To^n => X"o2 =0,25+1,25 = 1,5 (mol) =i> Vo^ ,dk.c) = Huong ddn giai 1,5 22,4 = 33,6 (lit) Vi axit don chiic => phan tir X c6 nguySn til oxi Dap an diing la A Nguyen to oxi dupe bao toan nen: 0,1.2 + 0,24.2 = 0,2.1 + n^^^^ Chu y: Sir dung dinh luat bao toan nguyen to de giai nhanh cac bai toan hoa hoc phiic tap: C2H2 ^ ^„^^ nco2 = 0,24 (mol) ,ii;);-sr Nhan xet: C + C2H4, C2H, ^ > CO, ^ 0.50, ^ Q nco > " H , o c6 axit khOng no (loai A, C) nc =0,24/0,1 = 2,4 (=> loai B) >»A H 61 nSv,,, • Dap an diing la C Thi du 15: H6n hop M g6m andehit X, xeton Y (X, Y c6 ciing s6' nguy6n tir cacbon) va anken Z Dot chay hoan toan m gam M cdn diing 8,848 lit O2 (dktc) sinh 6,496 lit CO, (dktc) va 5,22 gam HjO Cong thuc ciJa andehit X la A CH3CHO B.C3H7CHO C C4H,CHO D C2H5CHO (Trich de thi thi Dai hoc khoi A, B) Huong dan giai Theo bai ra: no2 = 0,395 (mol); ncoj = 0,29mol; ni^^Q = 0,29mol Vi T\QQ^ = n^^o => X, Y deu no, her, dcrti chiic : Thi du 17: Cho h6n hpp X gom HCHO va H , di qua 6'ng sii dung b6t Ni nung nong Sau phan ling xay hoan toan, thu dupe h6n hpp Y g6m hai chat h&u CO Dot chay het Y thi thu dupe 11,7 gam H2O va 7,84 lit CO, (o dktc) PhSn tram theo the tich ciia H , X la „:., A 65,00% B 46,15% C 35,00% D 53,85% (Trich dethi thu: Dai hoc khoi A, B) Hu&ng ddn giai 84 117 Theo bai ra: nco2 = ^ = 0,35 (mol); nH20 " " f ^ = ("'"'^ ' '' VI X, Y CO Cling s6' nguyen tir cacbon n6n diu c6 ciing CTPT la C„H2nO(x mol) Ta CO so cac phan ling xay ra: HCHO > HCHO D a t Z l a C ^ H ^ (ymol) H, So dd phan ling: X, Y,Z + O - > C O + H O (X) V i nguyen t6' oxi dupe bao toan nen: n ^ ^ y ) + "0(02) = "0(002) ^ "O(H20) HCHO + H2 => 1.x+ 2.0,395 = 2.0,29+ 1.0,29 x = 0,08(mol) Vi nguyen t6' cacbon diroc bao toan nen ta c6: n.0,08 + m.y = 0,29 > 1, > • > CO, (0,35 mol) CH3OH H , (0,65 mol) (V) J I > CH3OH u > CO, + H , HCHO + O2 > 2CO, + H , 2CH3OH + , Ap dung djnh luat bao toan nguyen t6', cho nguyen to cacbon va hidro, ta c6: Vi m.y > = > n < 3,625 "QHCHO) Mat khac, Y la xeton ntn n > Vay n = Do andehit X la C3H6O rr;i "H(H2) = nc ( C O 2) = 0,35 (mol) = " H (HCHO) - (C2H5CHO) => n H ( H ^ + Dap an diing la D Thi du 16: H6n hop X g6m hai axit cacboxylic dcm chiic D6't chay hoan toan 0,1 vay B CH,=CHCOOH va CH2=C(CH3)COOH C CH3COOH va C2H5COOH D CH3COOH va CH2=CHC00H H (H O) 0,35 = 2.0,65 nH(H,) = 0,6 (mol) mol X cin 0,24 mol O,, thu dirac CO2 va 0,2 mol H2O C6ng thiic hai axit la A HCOOH va C H C O O H nHCHO % V H , , = "2/x M ^ nH2 = ^ =46,15% = 0,3 (mol) * , 0,3 + 0,35 Dap an diing la B (Trich de thi thicDqi hoc khoi A, B) Thi du 18: Dot chay hoan toan m gam h6n hpp g6m n hidrocacbon khac nhau, thu dupe 11 gam CO, va gam H2O Gia trj ciia m la , CtyTN f III MTV UVVH Khang Vi^l d m nang On luygn thi DH mi^n BJlc - Trung - Nam mOn H6a hqc - Cu Thanh ToSn A 4,0 B 6,2 C 8,0 Theo nguyfin tSc bao toan electron, ta c6: D 13,6 (Trich de thi thuDai hoc khoi A, B) Huong ddn gidi PTPl/:C,H, ) xCO,+ ^H.O * r(X>^ '^^ Tir(l,2)tac6: 1,5 = 3a + 8b AKNO,)., + N O t + N p t + H j O Theo djnh luat bao toan kh6'i luong: m^i + + m^dHNOj =mddsau +0.1-30 +0,15.44 ^.12 ' ^ , , , Dap an dung la B ^ m = ^ mddHN03 = " ^ d d s a u + ' " N O + ' " N =^in„ds.„- niddHNOj =13,5 - - , =3,9 (gam) n i c t C x H y ) + mH(CxHy) = n i c x H y = ITl0(C02) + " H ( H ) (2) 0,15 So 66 phan ling: A l + HNO3 ^ => 13,5 V i nguydn t6' C, H duoc bao toan, nen: a = 0,1, b = • ' Thi du 2: Hoa tan he't 52 gam k i m loai M 739 gam dung djch H N O , ke't thiic + ^ = ^ , (gam) phan iJng thu duoc 0,2 mol N O ; 0,1 mol N2O va 0,02 mol N , Bie't kh6ng c6 Dap an diing la A phan ling tao mu6'i NH4NO3 va HNO3 da la'y d u 15% so vdri luong c^n thie't K i m loai M va n6ng d6 phSn tram ciia HNO3 ban dSu Mn luot la A.Crva21,96 B.Znva20 D.Znv^ 17,39 Sa 66 cho - nhan electron LITHUYET M-ne^M"" toan (bao toan khoi lirong, bao toan electron, bao toan dien tich va bao toan X^ nguydn i6) N+3e nx +5 (mol) 0,6 < - 0,2 +1 2N+8e II V A N D U N G +2 >N +5 Rat nhi^u bai toan phiic tap c6 the' giai nhanh bang each t6 hofp cac djnh luat bao (mol) +5 N + lOe >N2(N20) 0,8 N2 0,2 < - 0,02 (mol) cac phan u-ng hoan toan thu duoc 5,6 lit h6n hop khf N O va N O (dktc) Trong Theo nguyen tac bao toan electron, t h i : nx = 0,6 + 0,8 + 0,2 = 1,6 (mol) dung djch kh6ng c6 N H N O , Kh6'i luong dung djch sau phan ling thay d6i so =>x = , / n (mol) vdi kh6'i lircfng dung djch H N O , ban ddu la 52 52n ^ Suyra:M^,= — = = 32,5n X ^1 ^ A khong thay doi B tang 3,9 gam C tang 13,5 gam D giam 9,6 gam (Trich de thi thvcDai hoc khoi A, B nam Huong ddn gidi Theo bai ra: n^, = 13,5 / 27 = 0,5 (mol) nNO.N20 = , / 2 , =0,25 (mol) 2013) Gia trj thoa man la n = 2, MM = 65 (Zn) K V A p dung djnh luat bao toan nguyfen t6' cho nguydn t6 nito, ta c6: j ''^ ^ ^N(HN03) = " N { N O ) + "N(N20) + "N(N2) Taco: =f a + b =0,25 (1) Cac ban phan ting oxi hoa - khu: +3 Al - 3e 0,5 ->1,5 -» A l ' N +2 + 3e -> 3a +5 2N +3 N(NO) a Sifi'' - V 52 "zn = ^ = 0,8(mol)=>n2„(NO3)2 = , ( m o l ) 65 Goi a, b l l n luot la s6' mol N O , N , 2013) Huong ddn gidi PHL/ONG PHAP TO HOP CAC 0!NH LUAT BAO TOAN I C.Crva20 (Trich de thi thic Dai hoc khoi A, B nam P h a o n g p h a p 5: HHNOJ "N(zn(N03)2) = 0,2.1 + 0,1.2 + 0,02.2 + 0,8.2 = 2,04(mol) V I HNO3 la-y d u 15% nen: mH,03 C % ( d d H N O ) = ^ ^ ^ ^ 0 % = 20% >u' (liA s , v x i o 31 Cty iNHH r\yii ; L V V H MicHig Ca'm nang On luy$n thi DH mign Bjc - Trung - Nam mOn H6a hpc - CCi Thanh Toan Vay M la Zn va C % ( H N O , ) = 20% T a c : 24x + 65y = 8,9 Dap an diing la B V I nguyfen t6' nito duoc bao toan nen ta c6: 2x + 2y + 2z + 0,045.2 = 0,5.1 Thi du 3: Cho 18,4 gam h6n horp X g6m Cu2S,CuS,FeS2 va FeS tac dung hd't vdi ^x (1) + y + z = 0'205 (2) ^ HNO3 (dac nong, du) thu duoc V lit chi c6 NO2 (a dktc, san ph^m khu Theo nguyen tSc bao toan electron, ta c6: 2x + 2y = 0,045.8 + 8z nha't) va dung dich Y Cho toan bo Y vao mot luomg d u dung djch BaCi2 , thu 2x + y - z = 0,36 duoc 46,6 gam ket tua; cho toan bo Y tac dung vori dung djch NH3 du thu duoc 10,7 gam ket tua Gia trj cua V la A 11,2 B 38,08 f S ' = y;: x 'il ahx nh::i' m = ni^g(N03)2 + '"zn(N03)2 + "INH4NO3 "q ^_ D 24,64 (Trich de thi tuyen sinh DH ,., * x =0,1.148 + 0,1.189 + 0,005.80 = 34,10(g) khoiA) Hu&ng ddn gidi , xn • Dap an dung la A Theobai ra: n^^^^^ - , / 3 = 0,2mol Thi du 5: De phan ling h6't a mol kirn loai M cSn 1,25a m o l H S O va sinh X (san ph^m khir nha't) Hoa tan hd't 19,2 gam k i m loai M vao dung djch np^(OH)^^= 10,7/107 = , l m o l H.S04tao 4,48 lit X (san ph^m khir nha't, dktc) K i m loai M la A.'cu Quy d6\ hop X x mol Fe; y mol Cu va z mol S =>56x + 64y + 32z = 18,4 ; ( , 2, 3) => X = 0, l ; y = 0,l;z = 0,005 Vay C 16,8 (3) B M g C.Al D.Fe (Trich de tuyen sinh dqi hoc khoi A) (l) Hu&ng ddn gidi Mat khac: z = = ng^^Q^ = , (bao toan nguydn t6' liru huynh) Ta CO cac qua trinh: +n M-ne->M Ta lai c6: x = np^ = npg^Qp^^^ = 0,1 (bao toan nguydn to sSt) 64 a-> na-> a +3 +2 T6ng s6' m o l electron Fe - > Fe; Cu - > Cu \ - , o»«n+ n S O f + 2M"^ an/2 S - > S la ne (cho) = , 3+0,1 2+0,2 = 1,7 mol +.'> ^^.^^f, Theo bai ra: nx = 4,48/ 22,4 = 0,2(mol) (^.^ff +6 +(6-m) S(H2S04) + m e ^ — m M2 (SO4 )„ M""" + ne ^ M b (3) (4) (5) (6) O, , •81 ^- V ' - ' Cac qua trinh oxi hoa khiJr xay ra: ma Theo dinh luat bao toan electron: nx = ma Theo dinh luat bao toan difen tich: nx = 2b Theo djnh luat bao toan nguydn t6': a + b = y Tir(3,4)=>ma = 2b + V a = b Thay vac (5): 2b = y =t> nx = 2b = y (thoa man di^u kidn (2)) Vay s6' mol electron Fe cho la y Dap in dung la C + Vdri m = (san phdm khir la S): Tir (6) ^ 6a = 2b 3a = b Thay vao (5): 4a = y => 6a = 1,5y 71x + 32y = 15,8 Giai ta diroc: x = 0,2; y = 0,05 +(6-m) +) S(H2S04) + me MCU M + Ci, -, x—> nx —> x ' V i n (min) = 2; n(max) - => 2x < nx < 3x +6 Theobaira: n^h = 5,6/22,4 = 0,25(mol) CI2 0,2 ^ + 2e " 0": -> 2cr 0,4 M + 4e > 20.2- 0,05 ^ , Theo djnh luat bao toan electron, ta c6: 7,2n M = 0,4 + 0,2=0,6 =>7,2n=0,6M = > M = 12n Suyra: n = ^ M = 24(Mg) vay kim loai M la Mg Dap an dung la A e Thidu 8: Cho 29 gam h6n horp g6m A l , Cu va Ag tac dung viJra dii vdi 950 ml dung djch 1,5M, thu duoc dung djch chiia m gam mu6'i va 5,6 lit h6n hap X (dktc) g6m NO va NjO TI kh6'i cua X so vdi H2 la 16,4 Gia trj ciia m la ' => nx = 6a = 1,5y = 1,5 2,5x = 3,75x (=> loai kh6ng thoa man di6u kifen (2)) + Vdi m = (san phdm khiJr la HjS): Tuf (6) => 8a = 2b =i> 4a = b A 98,20 Thay vao (5): 5a = y => 8a = l,6y => nx = 8a = 1,6y = 1,6 2,5x = 4x (=> loai) Thi du 7: D6't chay hoan toan 7,2 gam kim loai M (c6 hoa trj hai khong d6i hgrp chat) h6n hop v^ O, Sau phan iJng thu duoc 23,0 gam ch&L ran va th^ tich h6n hop da phan ihig la 5,6 lit (6 dktc) Kim loai M la: A M g B Cu CBe D Ca (Tnch de tuyen sinh Cao dang khoi A) HNO3 B 97,20 C 98,75 D 91,00 ' ' (Trich de tuyen sinh Dqi hoc nam 2011 - Khoi A) Hu&ng ddn gidi Theo bai ra: n^No^ - l'425moI; n^ =0,25mol * Xacdjnhs6'mol N0,N20 •* " N = 0.05; n^io = 0,2 (sij dung phuang phap duomg cheo hoac giai he) * Goi s6' mol N O tao mu6'i la x; s6' mol N H mu6'i la y - '^ Nguydn to nito duoc bao toan ntn: 41 dm nang On luyjn thi DH miSn B&c - Trung - Nam mOn H6a hpc - CD Thanh Toan "KHNO.-,) = "N(N20) + "N(NO) + "N(NOJ) Cty TNHH MTV CVVH Khang ViSt Theo nguyen tac bao toan electron: "N(NH+) rr> l e (cho) = 0,35 (mol) (do cac kirn loai nhucmg) 1,425 = 0,05.2 + 0,2.1 + x l + y l =^x + y - , - ,^ ^ Theo djnh luat bao toan dien tich, ta c6: (1) (cho) = So d6: K L + H N O , - > mu6'i ( K L ) "HNO., - " ^ " N H ! nH20 = + N H N O + NO + N j O + H O X^n^^_ •••• / ^.^mif A & jifurb n = 0,35 (mol) Xheo djnh luat bao toan khoi lucmg ta c6: Of:, ri 425-4V = \ ^ "iH^o = 9.(1,425 - y ) = (12,825 - 36y) = Theo djnh luSt bao toan khoi luong, ta c6: 5,75 + 0,35.62 = 27,45 (gam) Dap an dung la A 29 + 1,425.63 = 29 + 62x + 18y + 0,25.16,4.2 + 12,825 - 36y = i > x - y = 68,75 (2) Phuong phap 6: T i r ( l , ) = > x = l,1125;y =0,0125 V a y m = mK, + m +m NO3 ^ = 29 + 1,1125.62+ 0,0125.18 = , ( g ) NH^ I LITHUYET - Quy d6i la phuong phap bien ddi nham dua h6n hop nhifiu chat phiic tap m6t hay hai chat don gian, qua lam don gian hoa bai toan ca ve mat hoa hoc Ddp an dung la A iSn toan hoc Thi du 9: Cho 5,75 gam h6n hop M g , A l va Cu tac dung vdri dung dich H N O , loang, du, thu diroc 1,12 1ft h6n hop X g6m N O va N j O {b dktc, t i khd'i cua X so vori H , la 20,6; san ph^m khir khong c6 NH4NO3) Kh6'i lirong mu6'i nitrat khan thu diroc c6 can dung djch la A.27,45g B 13,13g D 17,45g Hu&ng dan gidi ' * (1) -A 30x + 44v ^ux + 44y ^ , = 41,2 x +y ' N t ; ' Al'^ +2 0,03 > < N 0,01 (mol) +5 N > Cu-^ 3e H2SO4 dac, nong, du Sau phan ling thu dugc 0,504 l i t SOj (san ph§im A 26,23% B 39,34% C 65,57% D 13,11% (Trich dethi thu Dai hoc khoi A, B nam 2013) Hu&ng ddn gidi Cu - 2e - > Cu^* Fe - 3e - > Fe"^* a ->2a b->3b Mg(NO.,)2 Mg-^ + djch Quy d6i h6n hop Cu,FexOy - > Cu,Fe,0 T i r ( l , ) = ^ x = 0,01; y = 0,04 +5 Thi du 1: Hoa tan hoan toan 2,44 gam h6n hop bot X gom Cu va Fe^Oy bang dung ' (2) Ta C O sad6: Cu toan so oxi hoa khoi luong Cu X la Tac6:x + y = — =0,05 22,4 > K h i ap dung phuong phap quy d6i c^n tuan thu sir bao toan nguydn t6' va bao khir nha't, dktc) va dung djch chiia 6,6 gam h6n hop muoi sunfat Phdn tram Goi x, y \in luot la s6' mol N O , N^O Mg - II VAN DUNG C.58,91g (Trich de thi thi Dai hoc khoi A, B) Al PHaONG PHAP QUY DOI AKNOj), + 2e-^02- CuCNOj), c->2c +6 +4 S+2e ^ S 0,045 mx = 10,8 + ^ ^ ^ = 14(g) 22,4 20,97% ^I'l =>26a + a = 14 => a = 0,5 Quy h6n hop X cacbon va hidro nha't) va dung djch Y Cho toan b6 Y vao mot luong d u dung djch BaClj , thu duoc 46,6 gam ket tua; cho toan b6 Y tac dung v i dung dich N H , du thu duoc 10,7 gam ket tiia Gia t n cua V la B 38,08 C 16,8 D 24,64 (Trich de thi thu: Dai hoc khoi A, B) Hu&ng ddn gidi = 46,6/233 = 0,2mol = 10,7/107 = , I m o l Quy d6i h6n hop X x mol Fe; y mol Cu va z mol S =>56x + 64y + 32z = 18,4 la The' tich k h i O , (khi o du de dot chay hoan toan h6n hop Y la (2) 3: Cho 18,4 gam h6n hop X gom Cu2S,CuS,FeS2 va FeS tac dung het v6i "Fe(OH),4 vvtdi A 33,6 l i t (1) HNO3 (dac nong, dir) thu duoc V lit chi c6 N O (or dktc, san ph^m khir 44 = 0.2 (bao toan nguyen to liru huynh) Thi du 4: H6n hop X gom C^H, va +4 S + 2e ^ S Theo bai ra: n^^^^ n^^st-u Dap an diing la B +6 A 11,2 = Ta lai c6: X = n p, = n ^ ^ ^ ^ ^ = 0,1 (bao toan nguyen to' sat) nN02 Quy d6i h6n hop Fe,Fe^Oy => Fe va O Thidu j^at khac: z = (1) => He = 2a = I m o l ; n^^^ = a + a = (mol) V i C va H duoc bao toan nan nc,x) = ' HCY) = (mol); nH2(X) = " H ( Y ) C + O^2, ^^^2 ^ CO, 1H1 ,+ O, ^ ^ ^ i iH2 7vO l->l(mol) 0,5(mol) or vay: Vo^ = ( I + 0,5).22,4 = 33,6 (/) Dap an diing la D = 'mol ^^^T - ' ' Thi du 5: Cho mot luong CO di qua 6'ng sir dung m gam Fe^O, nung nong Sau mot thai gian thu duoc 10,44 gam chat rdn X gom Fe, FeO, F c O , , va F e A - Hoa tan het X dung dich H N O , d i e , nong thu duoc 4,368 lit N O , (san ph^m khir nha't a d i l u kien chudn) Gia tri ciia m la A 12 B 24 C 10,8 D 16 (Trich de thi thu: Dai hoc khoi A, B) Cty TNUli M!V ijyvH Khang Vi^t Ca'm nang fln iuyen tin Ud [iniin BSC - Trung - Nam mOn H6a hgc - Cu Thanh Toan Huong ddn gidi T h e o b a i ra: nj,02 = , / 2 , = 0,195(mol) Hu&ng ddn gidi ' T a quy h6n hop X chSit: F e , F e O , Fe + vj" * ^ Theo bai ra: n ^^^^ = ' V i Fe304 = FeO.Fe203, ntn c6 the coi h6n hop g6m FeO (x mol) va FejOj (y mol) F e ( N ) + 3NO2 + H O 6HNO3 x/3 Theo bai ra, ta c6: X Fe203 + 6HNO3 ^ 72 x + 160 y = 9,12 PTPU-; 2Fe(N03)3 + 3H2O FeO + 2HC1 y Ta = X nFeCi2 ••if9 !o(n vay 2Fe + C O x/3 (mol) = 0.06 Fe203 + C O ^ Dap an diing la A s6' mol FeO bang so mol FcjO,), cin dung vita dii V lit dung djch HCl I M Gia tri ciia V la B.0,16 C.0,18 D 0,23 (Trich de thi tuyen sink Dai hoc khoi A) np,304 dung djch HCl M , thu dugc dung djch Y c6 ti \t s6 mol Fe^"" va Fe"'* la 1: Chia Y hai phSn bang C can phSn m6t thu dugc m, gam muoi khan Sue k h i clo (du) vao phSn hai, c6 can dung djch sau phan itng thu dugc m2 gam mud'i khan Biet m2 - m, = 0,71 The ti'ch dung djch HCl da diing la A 320 m l B 80 m l C 240 m l '^^ ^^6' coi h6n hop chi c6 Fe304 ( F e O Fe203): =1^-0,0 l ( m o l ) Fe304 0,01 Suyra: V,,HC, + 8HC1 Hu&ng ddn gidi Fe^O^ = FeO.Fe203 => coi h6n hop X chi gom FeO (x mol) va Fe203 (y mol) FeO + 2HC1 ^ > FeCU + 2FeCl3 + H , -> 2x FeCl2 + H O -> Fe203 + 6HC1 ^ x y -> 6y -> 2FeCl3 + H O 2y V i t i la s6 mol Fe^^ va Fe^+ la 1: 0,08 (mol) = ^ = 0,08 (lit) = ^ n F e C i • " F e c i = 1:2 Dap an dung la A • Cha v.- * Fe304 (oxit sat tit) coi la h6n hop FeO FcjO^ (s6' mol FeO bang s6' mol FeA)- => X : 2y = 1:2 '' =:>x = y + F e C l ^ 2FeCl3 x/4 Ag* + NO3 + H P Ta c6: s6' mol H^ = x + 5x + 2x = 8x (mol) -> A g " n =>[H1 Huong ddn gidi Na,K,Ba + H 8x 5x V i Y = 4x < -|- => H ' phan ting het (NO3" du) + 0,002 mol NO, + H.O Goi x la so mol H2SO4 Y => nHci = 4x m o l 0,001 mol =>n = , 0 / , = , (mol) z:>pH PTPU hoa: =2 Biet NO la san pham khij nha't, cac th6 tich k h i ci ciing dieu kidn Quan giCfa V , va V^la B V , = 2V, C V , = 2,5V, D V , = V, (Trich de thi thucDqi hoc khoi A, B) + 0,06 mol nHNOj ni.„6i 0,08 m o l I 8,94 + 4x 35,5 + x 96 = 8,94 + 0,04 35,5 + 0,04 96 = 18,46 (gam) Dap an diing la C Thi du 5: Trong dung djch X c6: 0,02 m o l Ca""; 0,05 m o l M g " " ; 0,02 m o l H C ; C r Trong dung djch Y c6: 0,12 mol O H " ; 0,04 m o l C F ; Cho X vao B 4,2 gam D 6,2 gam ,XM ^im a»iti< • ^ Hu&ng ddn gidi P T H H (ion nit gon): 0,02 Mg2+ > 3Cu-" + N O + H , C 4,9 gam (Trich de thi thii Dai hoc khoi A, B) HCO3 Thi nghiem 2: ncu = 0,06mol; n^+ = 0,16 m o l ; n ^ ^ _ = 0,08mol 0,08mol so = 0,02 V (lit) 0,16mol + m^2- = • £,0 => Cu d u , HNO3 het 3Cu + H " + 2NO3" = m,,,K.Ba + A 2,0 gam = 0,08 (mol) > 3Cu(N03)3 + 2NO + 4H2O 8HNO3 < - 0,24 mol Y , sau cac phan ling hoan toan khoi Itrcmg ke't tiia thu duoc Idn nha't la: Hu&ng dan gidi nc„ = 0,06 (mol); > H2O v a y khoi luomg mud'i thu dugc: thoat V , lit N O Thinghieml: OH" => 6x = 0,24 => X = 0,04 (mol) Cho 3,84 gam Cu phan urng vdi 80 m l dung djch HNO3 I M va H2SO4 0,5 M A V , = 1,5 V , H^ + 0,24 Cho 3,84 gam Cu phan ling vdi 80 m l dung djch HNO3 I M thoat V , lit NO 0,06mol = x + 4x = x ( m o l ) , = , 0 (mol) Thidu 3: Thirc hien thi nhgiem sau: V, > N a O H , K O H , B a ( O H ) + H2 t Ta tha'y n ^ ^ _ = n H =2.0,12 = 0,24(mo!) Dap an diing la C 3Cu D 12,78 gam Theo bai r a : n H ^ = 2,688/22,4 = 0,12(mol) S6' mol N O = 5x (mol) Ag + 2H* + NO3- C 18,46 gam (Trich de thi thi Dai hoc khoi A, B) Dat s6' mol H C l A la x thi so mol HNO3 la 5x, s6' mol H2SO4 la x 50 Khang Vijt Cu hat cho tac dung vdi A g (du), r6i dun nong tha'y thodt t6'i da 22,4 m l k h i NO^ A 1,795 IWVM + OH" - > 0,02 -> coj- + H2O > 0,02 ( m o l ) Ca^* + C O ^ 0,02 ^ 0,02 - • CaCOji 0,02 (mol) +2 H ' ->Mg(OH)2^ 0,05^0,1 - 0,05 ( m o l ) mi 51 Ciy !NHH u i v DVVH Khang Vi?t dm nang On luygn thi BH miin BJc - Trung - Nam mOn H6a hpc - Cu Thanh Todn m ^ = mcaC03 +mMg(oH)2 = + H ^ +NO3 SFe^^ Khoi lirgng kd't tiia thu duoc Idn nha't la: 0,02/4v:i tX^mh -m^^, jr « , „rD >, =0,87.0,005.23.0,03.96 = 3,865(g) +m^2- Cu,Fe^^ (FeS04) deu bj oxi h o a boi NaNO, t r o n g m6i trucmg a x i t Cu vao 200 m l dung gam phan ling toan bo xay dung dich sau p h a n umg B 19,76 g a m g6m H N O 0,6M va H2SO4 0,5M dich (san hoan toan thi H2SO4 ph^m khir khoi luong nha't la NO), c6 can mu6'i k h a n thu C 19,20 g a m duoc la D 22,56 g a m (Trich de tuyen sinh Dai hoc khoi A) (Trich de tuyen sinh Dai hoc khoi A) Huong ddn gidi Theobaira: nc^ = , / - , ( m o l ) Hu&ng ddn gidi 0.03mol; HNaNOj =0,005mol =0,2.0,6-0,12 "HNO, =>n nH2 =0,448/22,4-0,02(mol) ( m o l ) ; nH2S04 ^ =0,12 + 0,1.2 = , ( m o l ) ; =>nc„ =0,32/64 = 0,005(mol) (condu) =(0,03-0,02).2-0,02(mol) Xr.>; : i =0,2.0,5 = 0,1 n H+ (mol) _ = , ( m o l ) ; n - = - (mol) N V i n^^^ < n^^^so^ => Fe,Al bj tan hS't; chat ran thu duoc la Cu n^^ het Dap an diing la C Thi du 6: Cho 0,87 gam h6n hop gom Fe, Cu va A! vao binh dung 300 ml dung nH2S04 = vira 0,02/12 Vay V N O = ( ' / + 0,02/12).22,4-0,112(1) Chu v: KA tiia Mg(OH) c6 d6 tan be hon nhieu so vdi MgCO, Theobaira: ->.3Fe'^+NOT+2H2O O S O 3Cu + 8H^ + 2NO3 ^ C u * + N O + 4H20 Phan ling: 0,12 Con: ^ 0,32 -> 0,08 0,04 Khd'i luong mudi thu duoc: m = ^0,12 0,12 01^,^2+ + ' " N Q - C C I ) ^ ""so^" Fe + H S ( l ) ^ F e S + H T X ^ X -> 2A1 + 3H2SO4 ( l ) - » A I =:>m = 7,68 + 0,04.62 + 0,1.96 = 19,76(gam) X (504)3 Dap + ^"2 Thi y l,5y ^ du 8: Cho - x = , 0 ; y = 0,01 B 3,2 g a m bot va H S O 0,8M , [56x + 27y = 0,87-0,32 = 0,55 ^^^°4x ,5y = 0,02 an d u n g la NO (san p h d m A 0,448 0,2M Cu Sau t a c d u n g vdi N O k h i c a c p h a n l i n g xay + 8H^ + 2N0J P/utig: 0,005-).0,04/3->.0,01/3 C6n: 0,02/3 hop g d m HNO3 h o a n t o a n , s i n h V lit k h i nha't, d d k t c ) Gia trj c i i a V la B 1,792 C 0,672 D 0,746 -> C u + + N O t + H 0,005/3 ^ Huong ddn gidi =0,005mol Fe Thijf tu day difin hoa: Cu^"" /Cu;Fe^'' /Fe^"" 3Cu m l d u n g d j c h hdn (Trich de thi thu: Dai hoc khoi A, B) =0,02mol; n^,^_ =0,005mol;nr„ =0,005mol; n n 100 khir d u y Khi phan irng vdi NaNOj c6: n ^ , 0,01/3(mol) nc„ = 0,05 (mol); n = 0,08 (mol) = 0,8 0,1 + 0,2 0,1 = 0,12 PTPIT: 3Cu + 8H" + 2NO3" 0,05 0,12 => H * phan umg het 0,08 (mol) ? j f) } , 1, j s (mol) > 3Cu-" + N O + H O uv))^ h r'r A il go&b ah qW'f 53 Cty TNHH MTV DWH Khang Vi^t dm nang 6n luyQn thi DH miin BJc - Trung - Nam mOn H6a hqc - Cu Thanh Toan Do do: H N O = , / Thi du 10: Dung dich X c6 p H = chira H C l , M va = 0,03 (mol) = > V N O = 0,03 22,4 = 0,672 (lit) ml^ A 0,122; 1,006 Thi du 9: Cho 1,82 gam h6n hop b6t X g6m Cu va A g (ti Id s6' m o l tuong ling 4:1) H2SO4 , M va H N O , M , sau k h i cac phan iJug xay B 0,14; 0,932 hoan toan, thu duoc a mol k h i N O (san ph£m khir nha't ciia N*'') Tr6n a Trong 0,1 l i t dung djch X : v6i H2O, thu duoc 150 m l dung dich c6 pH = z Gia t n ciia z la B C OH = 0,1.0,02 = 0,002mol =M i z ^ Hu&ng dan gidi , = , + 0,06 = , ( m o l ) ; n •; ^ =0,1.0,1 = , I m o l H "HCl (Trich de tuyen sink Dai hoc khoi B) =:>n = 10-'M = 0,lM=>n pH = l D.4 =>nH2S04 ^ T h e o b a i r a : n^^^^ = 0,015(mol);nHNO3 =t),06mol C 0,122; 0,932 D 0,110; 0,874 (Trich de thi thi Dai hoc khoi A, B) Hu&ng ddn gidi mol N O tren vdi 0,1 mol O , thu duoc h6n hop Y Cho toan b6 Y tac dung M A Dung dich Y c6 dung djch X thu duoc dung djch c6 p H = 12 va m gam kd't ttia Gia trj ciia V va Dap an dung la C v&o 30 m l dung djch g6m H2SO4 pH = 13 chura K O H 0,025 M \h Ba(OH)2 Cho V lit dung djch Y v^o 0,100 l i t * =0,06(mol) = 0,004(mol) = n ^ - , Trong V lit dung dich Y : pH = = > p O H = - = l OH" = 10"'M=0,1M Trong X : C u ( x m o l ) ; A g ( x m o l ) n (fom)l,0 Phan ling: , 3Cu 3Ag 0,11/3 + Ban d^u: 0,005 Phan iJng: 0,005 4H+ 0,04/3 + NO; , 3Ag* + N O + + 0-, 0,015 0,1 Phan iJng: 0,015 0,0075 Con: 0,0925 0,015 ^ 0,015 -> it \0S + «tV Dung djch thu diroc: pH = 12 > 7=> m i trucmg k i ^ m => O H " dir " o 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 V - 0,01 = 0,01 ( V + 0,1) 2NO2 + NO; HNO3 = 0,0375V(mol) = n 2H2O 0,005/3 Ban ddu: + O2 + 2H2O =^"Ba(OH)2- * 0,14/3 a = , / + 0,005/3 = 0,045/3 = 0,015(mol) 4NO2 ' 0,14/3 0,11/3 N ^ ^ a i V - ^ ^ ^ - ^ C u + + N O + 4H20 0,16/3^0,04/3 0,02 i Con: 2NO3 + 8H+ + = , l V ( m o l ) ; nKOH = 0,025.V(mol) OH" T a c o : 4x.64 + x 108 = 1,82 =^ 364x = 1,82 => x = 0,005 0,015 =>0,09V ^'^'^ 0,1 V - 0,01 = O.OIV + 0,001 = 0,011 = > V = 0,122(1) PTHH: Ba^^ Ban ddu: 0,0046 Phan umg: 0,004 + SO^" • -> BaS04 i (1,; 0,004 < - 0,004 0,004 (mol) ^^^n o "T m = 0,004.233 = , ( g ) = 0,015/0,15 = 0,1M iz>pH = z = l Dap in diing la C < 'HO '^'^ " Dap an dung la A 55 dm nang On luy^n thi DH m\6n BJc - Trung - Nam mOn H6a hoc - Cii Thanh Toan Cty TNHH WTV P W H Khang Viet Thidu 11: Co V lit dung djch chiia axit HCl a ( M ) va H , S b ( M ) C^n la'y x lit dung dich chiia baza la NaOH c ( M ) va Ba(OH)2 d ( M ) de trung hoa vCra du V lit dung djch axit trdn Bie'u thiJc tinh x theo V , a, b, c, d la V(a + b) A x = B x = c+d V(a-c) 2(b-d) C.x = V(a + c) 2(b + d) D x = V(a + 2b) c + 2d (Trich de thi thuDai hoc khoi A, B) Hu&ng ddn gidi —> w + cr H2SO4 2H^ + SO4- a b 2b a n ^ Na" OH- c Ba-* + + 0,1 Q 05 0,05 Dap an dung la C OH" V(a + 2b) x(c + 2d) B 800 C.400 Hu&ng ddn gidi Theobaira: n c u = , / = 0,3(mol) "NaNO^ - , = , ( m o l ) ; n H c , =0,5.2 = l ( m o l ) H.O >n V(a + 2b) c + 2d 0,5 lit dung djch B, sau phan ling tha'y c6 m gam ket tiia Gia trj cua m la: C 11,65 D 23,30 (Trich de thi thu Dai hoc khoi A, B) Huang ddn gidi = , , =0,15 ( m o l ) ; n B , ( O H ) =0-5.0,1 = , ( m o l ) = n ^ + OH T a c o : n , , = V (0,2 + , ) = , V ( m o l ) _ =0,5(mol) NO3 ^ ' 0,3-^0,8^0,2 0,3 ( m o l ) => n ^ (c6n du) = - 0,8 = 0,2 (mol) H H^+0H ^H20 0,2->0,2(mol) Cu2++20H-^Cu(OH)2i Trong 0,5 lit dung djch B c6: Goi V (lit) la t h ^ tich dung djch A da dung: ^T 0,3 - > , ( m o l ) =>n _ = , + 0,6 = 0,8(1) = 0 ( m l ) OH E)ap an dung la B Thi du 14: Cho 4,8 gam b6t Cu.S vao 120 m l dung djch N a N O , I M , sau them 200 mi dung djch HCl I M vao, ket thuc phan ling thu duoc dung dich X va V lit Phan ling trung hoa: N O (san ph^m khir nha't, dktc) Gia tri ciia V la H* A 67,2 0,5V + OH - > H , 0,25 ' 3Cu + 8H^ + 2NO3 ^ 3Cu^* + N + 4H2O Dung djch B chura K O H 0,3M va Ba(0H)2 0,1M Cho dung djch A trung hoa v6i => n ^ _ =0,15.1+0,05.2=0,25 (mol) , =lmol;n H+ Thidu 12: Co hai dung djch A va B Dung dich A chufa H2SO4 , M va HCl 0,1 M B 5,825 D 120 (Trich de thi thu Dai hoc khoi A, B) Dap an diing la D A 46,60 kd't tiia hfi't Cu-^ ? A 600 _ = x.c + 2d.x (mol) = x(c + 2d) (mol) Suy ra: V(a + 2b) = x(c + 2d) => x = 56 ' Thi du 13: Hoa tan 19,2 gam Cu vao 500 m l dung djch N a N O , I M , sau them OH H" ^"""'^ 0,05 (mol) 2d n "KOH ""'^ = 20H- d lb.; = >BaSO 4- SOi 0,05 ion c BaCOH), Ba=^ nH2S04 = " - N O nha't, phai them bao nhieu m l dung djch N a O H I M vao X H - Suy ra: vao 500mi dung djch H C l M Kd't thiic phan lifng thu duoc dung djch X v^ khf + = V a + 2b V (mol) = V (a+2b) (mol) NaOH 0,5 V = 0,25 => V = 0,5 ( l i t ) Vay m = 0,05.233 = 11,65 (gam) Trong dung djch: HCl ^ B.22,4 C 2,24 D 6,72 (Trich de thi thu:Dai hoc khoi A, B) 57 C^m nang On luygn thi P H nnign B&c - Trung - Nam mOn H6a hpc - Cti Thanh Tojin Cty TNHH MTV D W H Huong dan gidi -^^^^^^r^^^bpB: T h e o b a i r a : ncu2S = , / = 0,03(mol) nNaN03 =0,12(mol);nHci = , ( m o l ) =>n phuong trinh, nhung v§n giai quyet dugc cau hoi ciia Ai bai bang phirong phap NO3 ghep 3CU2S + IONO3 0,12 + 16H+ ^ 6Cu^^ + 3SO4' + lONO + 8H2O Bandau: 0,03 Phanurng: , , - > 0,16 ^ 0,1 (mol) ,^ I M (loang) C^n phai them it nha't bao nhidu gam N a N O , vao h6n hop sau phan ling thi khong khf N O (san ph^m k h u nha't) thoat ra? C 2,83 D 8,50 loai M (c6 hoa t r i I I ) va Fe H2SO4 loang, du cung thu dugc V lit (dktc) Bifit Fe c6 kh6'i lucmg nhu hdn hgp va kh6'i lugng cua M bang nira t(5ng kho'i lugng ciia Na va Zn h6n hgp ban ddu K i m loai M la A Mg B N i C Ca Hu&ng din gidi Goi a la s6' m o l m6i k i m loai h6n hop * Phan ung vdti H2SO4 M g + H2SO4 (Trich de thi thii Dai hoc khoi A, B nam 2013) 0,1 -> Sau phan ling c6: Fe + H2SO4 0,1 ^ > FeS04 + -> b ^ Fe + H2SO4 ^ :m) d t , u =: a 0,5a : i fil'goASlfifeqj-C! ZnS04 + H2 " b ' FeS04 + H2 ' c 0,1 mol FeS04 c M + H2SO4 ^ d 1,6-0,1 - , = l , m o l H " -> ;/ MSO4 + H2 d K h i cho N a N O , vao h6n hop sau phan ung: T a c : 0,5a + b + c = c + d = > d = , a + b 3Cu + H ^ + N O - > C u ^ + N O + 4H2O M a t k h a c : M d =0,5.(23a + b ) = > M d = 23a + 65b (2) 0,1 0,8/3^0,2/3 Thay (1) vao (2): 2M(0,5a + b) = 23a + 65b -> SFe^* + 0,1 H ^ + NO3 3Fe-''* + N O + 2H2O 0,4/3-^0,1/3 d,C) + • g 23a + 65b •M = a + 2b _^ 23a + 23.2b Vi (1) 23a + 32,5.2b a + 2b 32,5a + 32,5.2b ^ dir (0,4/3 + 0,8/3 = 0,4 < 1,2) nfen Fe"* va Cu phan iJng he't a + 2b Vay mNaN03 = ( , l / + 0,2/3).85 = 8,50(gam) ' 0,1 0,1 mol Cu * ' + 2Na + H2SO4 - > Na2S04 + H2 Zn + H2SO4 ^ (1): 0,1 PTHH: a a = 0,1 (mol) > MgS04 + H2 t D Ba Hu&ng dan gidi T h e o b a i r a : nH2S04 = , = , ( m o l ) = > n ^ ^ = , = l , ( m o l ) a (24 + 64 + 56) = 14,4 1/ 11 V A N D U N G (Trich de thi thvcDqi hoc kho'i A, B) ^ H^S04 loang, du thu dugc V lit H2 (dktc) Hoa tan hoan toan h6n hgp g6m k i m Thidu 15: Cho 14,4 gam h6n hop M g , Cu, Fe c6 s6' mol bang vao 0,8 lit dung B.5,67 '^e.tX) Thi du : Hoa tan hoan toan h6n hgp k i m loai Na, Z n va Fe dung djch ,^ A 12,75 r r'" chinh la cau tra Icri ciia cau hoi bai tap) 0,2 Dap an dung la C H2SO4 (tiJc la khong c6 ket qua tirng an ma chi t i m dugc cum ^n, cum ^n cung Da'u hi^u nhan biet la he s6' c6 so in nhi^u hon so phuong trinh = ^ V ; ^ o = V = 0,1.22,4 = 2,24 (lit) djch PHUONG PHAP GHEP AN Nhieu ht toan hoc ciia b^i tSp hoa hoc kh6ng giai dirge v i s6' ^n n h i l u hon s6' _ =0,12mol;n , =0,2mol PTPlT: Khang Vi^t 23{a + 2b) a + 2b 32,5(a + 2b) Dap an diing la D = > < M < , = > M = 24(Mg) Dap an dung la A -'^ [...]... Sqj^.nj = SqT-"! II VAN DUNG Thi du 1: Dung dich X c6 chiia: 0,07 mol Na"; 0,02 mol S O 4 " va x mol OH" Dung djch Y c6 chiia C104,N03 va y mol t6ng s6' mol C I O 4 va NO3 la 0,04 Tr6n X va Y duoc 100 ml dung djch Z Dung djch Z c6 pH ( bo qua sir dien li cua H^O) la A 1 B.2 C 12 D 13 ' ' (Trich de thi thi Dai hoc khoi A, B nam 2013) M 21 d m nang On luy$n thi DH 3 mign Bjc - Trung - Nam mOn H6a hpc - Cii... 0,35 Dap an diing la B (Trich de thi thicDqi hoc khoi A, B) Thi du 18: Dot chay hoan toan m gam h6n hpp g6m n hidrocacbon khac nhau, thu dupe 11 gam CO, va 9 gam H2O Gia trj ciia m la , CtyTN f III MTV UVVH Khang Vi^l d m nang On luygn thi DH 3 mi^n BJlc - Trung - Nam mOn H6a hqc - Cu Thanh ToSn A 4,0 B 6,2 C 8,0 Theo nguyfin tSc bao toan electron, ta c6: D 13,6 (Trich de thi thuDai hoc khoi A, B) Huong... (Trich de thi thii Dai hoc khoi A, B) 47 dm nang On luyjn thi BH 3 mi^n Cty TNHH M W OWM KfaWQ Vigt B^c - Trung - Nam mOn H6a hpc - CCi Thanh Toan Khdi luong binh tang chinh bang kh6'i luong san p h ^ chay (C02,H20) Thi du 9: Hoa het 2,32 gam h6n hop FeO, Fe,04, Fe.Oi (trong do FeO, Fe.O, c6 s6' mol bang nhau) trong 80 ml dung djch HCl I M thu duoc dung dich X Cho X tac dung vdri dung dich AgNO, du thi thu... diing la C < 'HO '^'^ " Dap an dung la A 55 dm nang On luy^n thi DH 3 m\6n BJc - Trung - Nam mOn H6a hoc - Cii Thanh Toan Cty TNHH WTV P W H Khang Viet Thidu 11: Co V lit dung djch chiia 2 axit HCl a ( M ) va H , S 0 4 b ( M ) C^n la'y x lit dung dich chiia 2 baza la NaOH c ( M ) va Ba(OH)2 d ( M ) de trung hoa vCra du V lit dung djch 2 axit trdn Bie'u thiJc tinh x theo V , a, b, c, d la V(a + b) A... 0,45.2 = 0 , 4 2 + 0,5.1 =^a = 2 ( C 2 H , 0 2 ) 31 Cty \N\iH MIV UVVH Khang Vi^t dm nang On luy^n thi DH 3 mign B&c - Trung - Nam mOn H6a hqc - CCi Thanh Toan 0,1.60 + 0,1.62 Dap an diing la D , , , „ y'Thidu 11: H6n horp X g6m axit axetic, axit fomic va axit oxalic Khi cho m gam X tac dung vdri N a H C O , (du) thi thu dugc 15,68 lit khi C O , (dktc) Mat khac, d6't chay hoan toan m gam X cdn 8,96... 0,1 -> 0,1 (mol) 49 dm nang On luy$n thi OH 3 mign BJic - Trung - I u i ; l i o i ; t i o a i v u Cty INHH M i v C i ihanh Tocin Thi du 2: La'y 200 m l dung djch A chiia H C l , HNO3, H , S 0 4 c6 t i \t so mol 1: 5 : 1 V , = 0,04 V (lit) (duy nha't, do of dktc) Gia trj pH cua dung djch A la B 2,79 Vay V , = 2V, C 2 D 3 (TrichdethithuDaihgckhoiAyB " ndm2013) Dap an diing la B Thi du 4: Hoa tan hoan... 3Cu-" + 2 N O + 4 H 2 O uv))^ h r'r A il go&b ah qW'f 53 Cty TNHH MTV DWH Khang Vi^t dm nang 6n luyQn thi DH 3 miin BJc - Trung - Nam mOn H6a hqc - Cu Thanh Toan Do do: H N O = 0 , 1 2 2 / 8 Thi du 10: Dung dich X c6 p H = 1 chira H C l 0 , 0 2 M va = 0,03 (mol) = > V N O = 0,03 22,4 = 0,672 (lit) ml^ A 0,122; 1,006 Thi du 9: Cho 1,82 gam h6n hop b6t X g6m Cu va A g (ti Id s6' m o l tuong ling 4:1) H2SO4... + y ) (Trich de thi tuyen sinh Dai hoc khoi A) Hu&ng ddn gidi Dung djch X chiia 3 ion la Theo djnh luat bao toan didn tich ta c6: z = 2x + 2y A Na^ A g ^ N O , - B Na\^ C f C Ba-^ Ag\- D V = ^ ^ ^ ^ a D Na^ Ba-^ NO," (Trich de thi thii Dai hoc khoi A, B nam 2013) HCO3 + OH" -> CO^" + H2O I z -> z =>z = 2.V.a Huong ddn gidi (l) (2) i:.V; ... 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... sir dien li cua H^O) la A B.2 C 12 D 13 ' ' (Trich de thi thi Dai hoc khoi A, B nam 2013) M 21 d m nang On luy$n thi DH mign Bjc - Trung - Nam mOn H6a hpc - Cii Thanh Toan Cty TNHIi MIV DWH Khang... 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