kỹ thuật mới giải nhanh bài tập hóa học 12 vô cơ

546.076 K600T CtJ THANH TOAN NGUYEN NGOC OANH KYTHUATMdfl GIAINHANHBAITAP HOA HOC B I I 114 4 i TAP HOA H O C VO C O f i ' OCR] i wo'c lOnil , ^ ^ - Tuyen chon cac ky thuat giai nhanh ve chuyen de hoa v6 ca - Danh cho hoc sinh \6p 10, 11, 12 va luyen thi D H - CD NHA XUAT BAN DAl HOC i^UOC GIA HA NOT " X U THANH TOAN NGUYEN NGOC OANH KYTHUATMOfl GIAINHANHBAITAP HdA HOC TAP HOA H O C V C O - Tuyen chpn cac ky thuat giai nhanh ve chuyen de hoa v6 ca - Danh cho hoc sinh Idp 10,11,12 va luyen thi D H - CD NHA XUAT B A NDAI H O C Q U O C GIA H AN O I C KI THUAl L6INOIDAU Cac ban ddng nghidp va cac em hoc sinh than m6'n! Trdn CO so phan tfch kl luong cac n6i dung kien thiic v^ ki nang nam khung chuong trinh thi, ca'u true, ma tran d6 thi va cac dang bai tap thucmg gap da thi tuyen sinh dai hoc, cao dang ciia B6 GD - DT, chung toi da biSn soan tap sach: "Kithudt moi gidi nhanh bai tap Hod hoc" Cu6'n sach mof m6t co h6i cho giao vian va cac em hoc sinh nhin nhan lai mot each sau sac va toan dien cac da thi tuyan sinh DH - CD ki tii ap dung hinh thiJc thi trac nghiem, tilr giao vian c6 dinh hirdng dung dan cho cac em hoc sinh va miic kien thiic va cac dang bai tap; cac em hoc sinh se vihig vang, tu tin hom de di tran duomg den c6ng truong Dai hoc ma minh mo udfc Tac gia cung khdng quan gili gam vao eu6'n sach cac phuong phap giai hay, giai nhanh, nham giiip cac em ren luyan ki nang giai nhanh cac dang bai tap di thi tuyen sinh DH - CD Cuon sach se la nguori ban d6ng hanh than thia't vdri cac em hoc sinh qua trinh chinh phuc nhiing dinh cao vinh quang ciia tri thiic; cuon sach se la tai lieu qui cho cac ban d6ng nghiep qua trinh giang day Da' cuon sach hoan thian hon, ra't mong nhan duoc su dong gop y kia'n chan ciia cac ban d6ng nghiep va ciia cac em hoc sinh Chuc cac em dat dirge nhiau tich cao cac ky thi sap t6i Xin tran cam cm ! '* T A C G I A Nhd Sach Kkang Viet xin tran giai thieu toi Quy doc gid vd xin long nghe moi y kien ddng gop de cuon sach ngdy cdng hay hon, bo ich hon Thu xin gtH ve: Cty TNHH Mot vien - Dich Vu Van Hoa Khang Viet 71- Dinh Tien Hoang, Fhuimg Dakao, Quan 1, TP HCM Tel: (08) 1 - 1 9 - 1 - 9 - Fax: (08) 1 8 Email: khangvletbookstore@yahoo.com.vn GSAl NHANH CHUYEN ©11: cAc D^m CHUVeN HOA VO CO KT T H U A T ©SAI N H A N H BAI TJSip y g H A L O G E N A L I THUYI^T DAC D I £ M CIJA CAC NGUVfiN T6 HALOGEN Flo Qo Brom (F) (CI) (Br) ca'u hmh electron 16p 4s'4p' 3s^3p' 2s^2p' ngoai Cling Ban kinh nguyan tif D6 am dien Ai lire electron Nang lugmg ion hoa thif nhatl, (eV) Nang luong lien ket (kJ/mol) Nhiat d6 nong chay ("C)" Nhiat d6 s6i ("C) Tiang thai (dkt) Mau sac Mui Ham iKong (% s6' nguydn tir) lot (I) 5s^-5p' 0,64A° 0,99A° 1,14A° 1,33A° 3,98 3,58 eV 3,16 3,81 eV 2,96 3,56 eV 2,66 3,29 eV 17,42 12,97 11,34 10,45 159 242 192 150 -219,6 -101,0 -7.3 113,6 -188,1 luc nhat rat d6c -34,1 vang luc Xoc 59,2 long nau hdi 0,02 0,02 3.10-* 185,5 ran den tim kh6ng miJi 4.10-' • II TINH CHAT HOA HOC CUA DON CHAT III DI^U cut Br, F2 Vdikim loai - Tac dung vod tat ca cac kim loai (ca A u , Pt, ) - Phan ling toa nhiet manh nha't - Tac dung v 2EX„ Phan ling xay a nhiet d6 cao, kh6ng n6 X2 + -> 4HF + O2 Phan ling chi xay a nhiet d6 cao, thuan nghich -1 2F,+ CI, + K H dich ki^m 2NaOH„„,„3, ->KC1 + ->2NaF + OF.+ao 3X, + 6KOH Trong phong thi nghiem Kh6ng dilu che' Trong c6ng nghiep Dien phan nong > nong chay: mu6'i: F2 + CI, +2KBr 2NaCl,„^„g (.^ij,, Chi the' hien tfnh khijr Nhan xet 2NaF + O2 Kh6ng CO 3H,0 X , + MnXj + 2H2O Dien phan dung dich NaQ cd m^ng ngan: Sau phoi nudrc bien d^ la'y NaCl, ph^n lai chiia nhidu NaBr: 2Naa + 2H,0 -> Cij + H2 + 2NaOH Rong bi^n kh6 - ^ t r o —^^^'^ > dung djch c6 Nal 2NaI + CI, NaBr +CI2 -> I2 + 2NaCl 2Naa + Br2 * M n X , + X2 + H , • 16HX(,, + 2KMn04 > K X + 2MnX2 + X , + SH^O -> K C l + 3X, + 3H2O 2NaX + MnO, + 2H,S04(d) - i - ^ X2 + MnS04 + Na2S04 + 2H2O (X la CI, Br, I) Trong cong nghiep: Khu duoc r Kh6ng dung dich: Phan ling 2NaBr +1, ^2Naa + Br, Br, +5a, + 6H,0 ->2HBr03 + lOHQ > Bo > h Tinh oxi hoa gi5m dan 4HX + MnO, 6HX C I (lOfC) h6n hop 4HXF2 +H, 3a, + 6K0H rnu6'i I2 Cho dung djch H X dSc tac dung vdi chat oxi hoa (MnO,, PbOj, K C I O , KMn04, ): K F + 2HF: 5KC1 halogenua Br, CI, F2 1, Trong phong thi nghiem: 5KX + KXO3 + K C I O + H2O Khi F , kh6 khur Khir duoc Br duoc Cr, B r , I , I > X2 + 2e 2X" +1 + KCIO3 Vdri mu6'i Oxi hoa ion halogenua nguyen tir: HX + HXO H2O Phan ling kho dSn tiir CI, d6'n I, Vdi dung C6NG NGHIEF Nguyen tac chung: H , + I, ^ 2HI 2HX (X: F, Q, Br) 2F2 + H O - Tac dung vdri kim loai cf nhiet d6 cao hoac cin xiic tac TRONG P H N G THI NGHlfeM VA S A N XUAT TRONG i, + 5a,+ H , ^ 2HIO3 + lOHCl J t N a a + 2H2O m&ngngin > C l , t + H^t + 2NaOH rudrc day: 4HC1 + , 2NaBr + 400"C ^ 2Ha + Na2S04 H2(k)+ CI•'2(k) ->2HC1 (k) RH + CI2 - > RCl + HCl - D6u tan (trif CaFj, ) Mu6'i clorua (CD Mu6'i bromua (Br) - Da s6' tan (trir AgCl, PbCl,, ) - Da s6' tan (li ij AgBr, PbBr2, ) 41, + H,S + 4ao Mu6'i iotua (D - Da s6' tan (triif A g l , Pbl,, Mnlj) Tan, mau trSng Kh6ng tan, mau trang Kh6ng tan, mau vang nhat Kh6ng tan, mau vang Phan ling vori halogen Kh6ng xay F2 + 2C1 CI2 + 2Br-,,,, CI, + 2r„„ -> Phan iJng vod H,S04 dac CaF2 + HiSOiij) Phan ling vdd -> („6ngchdy) + 2F ^ Br2 + 20" Phan iJng voi I2 + ^3HX Trang thai tif nhiSn 2a- Br2 + 2r„d) - > I , + 2Br 250^ C ^ 2NaBr„y + NaCV,+ H2S04 178 => loai truong hofp nay) hop 2: X la flo, Y la clo: NaF + A g N O j y y (mol) y = 0,06 TCr ( , 2) ta c6: x = 0,06; (2) y = 0,06 6,03 1: Hoa tan hoan toan 1,1 gam h6n hop gom m6t k i m loai kiim X va m6t k i m loai ki6m thd Y (Mx < My) dung djch H C l du, thu duoc 1,12 lit H , (dktc) K i m loai X la B Rb C.Li (Trich Hudfng din 1! :::>ili A g C U + NaNOj Ta c6: 143,5y = 8,61 Cac t h i d u m i n h hoa: Theobaira: - (1) ' > khdng xay NaCl + A g N O j + NaNOj Nal + AgNO, A g l l + NaNO, V i vay cac bai tap h6n hop mu6'i halogenua tac dung vdri dung dich A g N O j tao ke't tiia cSn chia hai trucmg hop d^ giai A K i Theo bai ra, ta c6: 42x + 58,5y = 6,03 NaBr + A g N O j ^ A g B r i + NaNO, Thidu X = 178 (khdng c6 halogen nao c6 Goi X , y \in luot la sd mol ciia NaF, NaCl Fe + 2HC1 -> FeQ, + H , - Ta c6: (23 + X ) 8,61 = 6,03 (108 + X ) Tru&ng 2FeCl3 ' nn^ =0,05 (mol) Thi du 3: Hoa tan hoan toan 6,645 gam hdn hop mudi clorua ciia hai k i m loai ki^m thudc hai chu ky ke' tiep vao nude duoc dung dich X Cho toan bo dung djch X tac dung hoan toan vdi dung dich A g N O j (du), thu duoc 18,655 D.Na de tuyen sink Cao dang khoi Dap an diing la D gam ke't tiia Hai k i m loai ki6m tren la A) A R b v a C s B N a v a K gidi => 0,05 Mx < 56,67 Dap an diing la PTHH: Thi du 5: Cho 1,9 gam h6n hop mu6'i cacbonat va hidrocacbonat cua k i m loai k i ^ m M tac dung he't v6i dung dich H C l (du), sinh 0,448 1ft k h i ( dktc) K i m loai M la B K CRb D.Li (Trich De thi tuyen sinh Dai hoc khoi B) Hu&ng ddn gidi Theobaira: H, 0,03 mol Trong cac k i m loai hoa t r i I I chi c6 k i m loai M g (R = 24) thoa man Vay k i m A.Na D Ca (Trich De thi tuyen sinh Cao dang khoi A) ^' ' 0,4 (mol) nhom chinh nhom I I (hay nhom I I A ) Cho va Zn tac dung vdi luong d u dung dich H C l , Mat khac, cho 1,9 gam X tac dung v6i thi thi tich hidro sinh chua d€n 1,12 B Sr RCI2 + ViMx=32nen j ^ ^ ^ ^ [R + > > Dap an diing la A M g = 1,7/0,03 = 56,67 X + Ux Suyra: H , S (loang) (1) > XSO4 n^^^< 0,05 (mol) < 0,05 (mol) M x > 1,9/0,05 = 38 Tilr(l,2)tac6: + H^ (2) 38 < Mx < 56,67 Trong cac k i m loai nhom I I A chi c6 k i m loai canxi (M = 40) thoa man vay X la Ca Dap an diing la D nco2 = 0,448/22,4 = 0,02 (mol) Thi du 8: H n hop X g6m M3CO3, MHCO3 va M Q ( M la k i m loai k i ^ m ) Cho T a c o M h h = 1,9/0,02 = 95 32,65 gam X tac dung vijra dii vofi dung dich HCl thu duoc dung djch Y va c6 Suy ra: M + 60 > 95 > M + 61 => 17,5 < M < 34 17,6 gam C O , thoat Dung djch Y tac dung vdi dung dich A g N O j d u duoc 100,45 gam ket tiia K i m loai M la Vay M = 23 (Na) A.Rb Dap an dung la A B Na A Fe B.Al CMg D Zn (Trich De thi tuyen sinh Cao dang khoi A) Hu&ng ddn gidi Trong cac k i m loai da cho, M c6 th^ la Fe That vay: 2Fe + 3Cl2 y 2FeCl3 Fe + 2Ha > FeCl2 + H2 S6 m o l CO,: nco^ = dich mu6'i X ta cung duoc mu6'i Y K i m loai M c6 t h ^ la D K Hu&ng ddn gidi Thi du 6: Cho k i m loai M tac dung v6i C I , duoc mu6'i X ; cho k i m loai M tac dung vdfi dung dich H C l duoc mu6'i Y Ndu cho k i m loai M tac dung vdi dung C.Li I M = o,4 44 (mol) S6' mol ke't tua A g C l : n^gci = 100,45 /143,5 = , ( m o l ) Goi X , y , z Idn luot la s6' mol M^COj, MHCO3, M C I 32,65 gam X Bao toan cacbon ta c6: x + y = 0,4 Bao toan clo ta c6: 2x + y + z = 0,7 A p dung djnh luat bao toan khd'i luong ta c6: 13 n i x + niHci = mMCi(Y) + "^002 + " I H => 32,65 + 36,5(2x + y ) = ( M + 35,5).(2x + y + z) + 17,6 + 0,4.18 rr> 36,5(x + y ) + 36,5x = ( M + 35,5)(2x + y + z) = 7,85 Thay x + y = 0,4;2x + y + z = 0,7 vao, tadiroc: , , ( M + , ) - , = 22,45 = > , M - , x - - , 2,4 + , M • " Thi du 10: Cho 19 gam h6n hop b6t g6m k i m loai M (hoa tri khdng d6i) va Zn (ti IS m o l tirong umg 1,25 : 1) vac binh dirng 4,48 lit C I , (dktc), sau cac phan ufng hoan toaii thu duoc h6n hop chat rSn X Cho X tan h6't dung dich H Q (dir) tha'y c6 5,6 lit H , thoat (dktc) K i m loai M la A Na B Ca C M g D A l Huatng ddn gidi ',, => X = S6'mol k h i CI, va H , Mn lirot la: n ^ j = , ( m o l ) ; n^^^ = , ( m o l ) Dat X la s6' m o l Z n => s6' mol k i m loai M la 1,25x 36,5 Taco: V i < x < 0,4 => M < 17,4 => M = ( < 17,4) Vay k i m loai k i ^ m la L i Dap an diing la C TO ( , ) ta c6: N a X + A g N -> A g X i + N a N O j 0,1 -> 0,1 l,25n,x + 2x 0,9 65+l,25M 19 l,25n + 0,9 M le 24 le Ket luan Loai Mg Loai Gia t r i phii hop la n = => M = 24 (Mg) Dap an diing la C •> Cac bai tap t u l u y e n : Cku i : K i m loai nao sau day tac dung voi dung djch HCl loang va tac dung vdi khf ^ M ^ = - - - 0,1 CI3 cho cung loai mu6'i clorua k i m loai? 108 + X = = > X = 42 CI ( M = 35,5) Br ( M = 80) Hai halogen la Q (35,5) va Br (80) vay hai mu6'i X la NaCl va NaBr Dap an dung la D A.Fe B.Zn N a F - A g N O j -> kh6ng xay NaCl + A g N O j AgCl i + NaNOj -> 0,1 mol => "^Agca =0.1.143,5=14,35(gam) ^1.5(gam) loai tnrcmg hop C Cu D A g Hu&ng ddn gidi K i m loai la Z n : Zn + 2HC1 Zn + Cl2 Trucmg hop 2: Hai mu6'i la NaF va NaCl 0,1 19 N Trirorng hop 1: Hai halogen d^u tao diroc kfi't tiia v6i A g N O j PTPU': 65x + M.l,25x =>l,125M = 23,75n-20,5 S6 mol A g N O j : n^gNOj = - ' ("loO • NaX si (2) => 25,75n + 38 = 58,5 + 1,I25M Hu&ng dan gidi Dat c6ng thufc chung ciia mu6'i la V',' Theo nguydn t i c bao toan electron ta c6: l,25n.x + 2x = 0,4 + 0,5 = 0,9 Thi dti 9: H6n horp X chira dong ihbi hai mu6'i natri cua hai halogen lien tife'p bang tuSn hoan La'y mot luorig X cho tac dung vira dii \6'\0 m l dung dich A g N O , I M thi thu duoc 15 gam ket tiia C6ng thiJc phan tijf ciia hai muoi X la (cho F = 19; Q = 35,5; Br = 80; I = 127; Na = 23) A NaF va NaCl B NaBr va N a l C NaCl va N a l D NaCl va NaBr ' x + M l , x = 19 ( l ) >ZnCl2 + H2 >ZnCl2 Dap an diing la B '^ Cha v.- Phan ling ciia cac k i m loai v6i CI, va axit H C l : FeClj AgXi < ^^Suyra: K Y + AgNO, ^ AgY + K N O vay: M h h - - ^ ^ = , ( g / m o l ) = ^ M x , y - , ( g / m o l ) 0,0264 n^aci = - ^ ^ = 0,0522 mol 58,5 + NaNOj (mol) NaCl + AgNO, vay mudi A 1& natri bromua (NaBr) Dap an diing la C A g a i + NaNO, ** ' ' (1) * 22875 Theo (1) -> n^^ci = "Aga 0,01 (mol) 03 M^^x = ^ = 103 =>23 + M x - =^ ' " Dap an diing la A Cau 5: M6t h6n hcyp X g6m muoi halogenua ciia kim loai natri nang 6,23g hoa tan hoan toan nuoc dugc dung dich A Sue clo du vao dung dich A roi c6 can hoan toan dung dich sau phan ung dugc 3,0525g mu6'i khan B La'y m6t nua lucmg mu6'i hoa tan vao nude r6i cho phan ihig voi dung djch AgNO, du thi thu dugc 3,22875g ket tiia Tim c6ng thiic ciia cac mu^i va tinh % theo khd'i lugng m6i mu6'i X Hu&ng ddn gidi Gia sii lugng mu6'i khan B thu dugc sau cho clo du vao dung djch A chi c6 Ag + -X2 AgX 0,01 (2) •- ' S %m(KF) = — x l O O % =34,17% ^ ^ 6,79 mei2 =71.a = 14,2(gam) 0,01 AgX + K N O vay: mKci = 0,06 74,5 = 4,47 (gam) => m K P = 2,32 (gam) (24 + 2.35,5) Sd'molAg: K X + AgNOj ^ Tir (2) ^ s6' mol KCl = s6' mol AgQ = 8,61 : 143,5 = 0,06 (mol) a=Vay :, VV (1) S6 mol h6n hop KX, K Y = ^'^^ ~ ^'^^ = 0,0264(mol) 108-39 ^ ^ Theobaira: a.(24 + 2X) = 19 ^ ( X ) = ,7,8 ^ ^ C&u 4: Cho dung djch chura 6,79 gam h6n hop g6m mu6i K X , K Y (X, Y m nguydn t FeCU + HjO X X Fe^Oj + H a > 2FeCl3 + 3H2O ^ y 2y n Feci2 = X = 0-06 (mol) (2) TO (1, 2) ta c6: X = 0,06; y = 0,03 (mol) (Trich Delhi tuyen sink DH - khd'i B) vay m = m p^ci, = 2.y 162,5 = 2.0,03.162,5 = 9,75 (gam) Hudng ddn gidi S6' mol Cr^O,: Dap an diing la A 15,2 nc^o3 = T T T = 0, '(mol) 152 So 6.6 phan ling: A l + CfjOj -> H6n hop sau phan ihig •"0203 + '"AI = nihhsp => niA, = mhh,p - m C ^ Q J = 23,3 - 15,2 = 8,1 (g) => UA, PTPLT: Saupir: = — (Trich De thi tuyen sinh Dai hoc - khoi B) Hu&ng ddn gidi = 0,3 (mol) Cr,03 0,1 0,1 (mol) ; f ; A + 6Ha 0,1 Cr + H a Theobaira: n^^ = ^ =0,15(mol); n^Q= ^ 22,4 ^ > 2Cr + AI2O3 + 2A1 0,2 * X + HCl: Cu + HCl 0,15 (mol) 0,2 (mol) =0,35 22,4 = 7,84 (lit) Dap an dung la C l,5x > khdng phan ung X + HNO3 * Zn H2 = 0,15 + 0,2 = 0,35 (mol) => > 2AICI3 + 3H2t = 1,5 x = 0,15 =>x = 0,1 (mol) > C r a , + HjT 0,2 2A1 + 6HC1 X > 2k\C\^ + 3H2T - ^ = 0,3(mol) 22,4 0,1 Sau phan ling nhiet nh6m c6: 0,1 mol A l (du); 0,2 mol Cr; 0,1 mol AI2O3 ft, Cau 6: Cho m gam h6n hop X gdm A l , Cu vao dung dich HCl (du), sau V.€x thiic phan ling sinh 3,36 lit (o dktc) Neu cho m gam h6n hop X trSn vao mot luong du axit nitric (dac, nguoi), sau ka't thuc phan umg sinh 6,72 lit NO, (san phdm khir nha't, dktc) Gia tri ciia m 1^ A 11,5 B 10,5 C 12,3 D 15,6 Al + HNOj^ij, „g„oi) Cu + 4HN03, bj thu d6ng hoa > Cu(N03)2 + 2NO2 + 2H2O , n NO2 = 2y = 0,3 => y = 0,15 (mol) Vay m = niA, + mc„ = 0,1.27 + 0,15.64 = 12,3 (g) Dap an dung la C Chu v: * A l , Fe, Cr bi thu d6ng hoa HNO3 dSc, ngu6i Cau 7: Dung dich X chiia h6n hop g6m NazCOj 1,5M K H C O I M ^fh6 tir tv? timg giot cho de'n hfi't 200 ml dung djch H Q I M v^o 100 ml dung djch X, sinh V 1ft (6 dktc) Gia trj cua V la • ,vi' A 4,48 B 1,12 C.2,24 D 3.36 ^ ^'l ' * ^ (Trich Be thi tuyen sinh Dai hoc khd'i A) Hu&ngddngim Theo b^i ra: n^a = 0,2.1 = 0,2 (mol); n NajCOa = 0,1.1,5 = 0,15 (mol) ' n K H c o = 0.11 ' =0,1 (mol) ' Khi nho tir tCr dung djch H Q (H") vao X thi: % Na^COj + 0,15 HCl > + H* 0.05 < > 0.05 — > CO:t 0.15 (mol) + H2O c a u 8: Khi cho 100 ml dung djch KOH I M vao 100 ml dung djch HCl thu duot dung djch c6 chiia 6.525 gam chat tan Ndng d6 mol (hoac mol/1) ciia H Q dung djch da dung 1^ Theo bai ra: UKOH nH2S04 = 0.5 0.28 = 0,14 (mol); Ta thafy = Sord6: Kim loai+ Axit D 38,93 gam "82804 + nH^ = 8.736/ 22.4 = 0,39 (mol) ^na^ => Axit phan ling v&a he't voi kim loai > mud'i + Hj ''^ ' , Cau 10: Cho m gam b6t A l vao c6c chura V lit dung djch NaOH 2M, sau phan ling hoan tohn cho tiep dung djch HCl vao c6c 66 de'n cha't ran tan he't thafy cdn diing 800 ml dung djch H Q I M va c6 3.36 lit thoat (cj dktc) Gia tri cua m va V lin luot la C I M 2Al + N a O H + 6H2O ^ 2NaAI (OH)^ + 3H2 t D 0.25M X Suy ra, KOH phan irng chira he't (HQ he't chat tan c6 K Q v^ KOH) Goi X , y Idn luot 1^ s6' mol KOH phan ling voi HCl va c6n du Theo b i i ta c6: x + y = 0.1 (1) i 74.5 X + 56 y = 6,525 (2) Tir (1 2) ta giai duoc x = 0.05 -> x -> NaAl(OH)^ + 4HC1 X - > x y ->3y -) (mol) NaQ + AICI3 + 4H2O 4x 2A1 + 6HC1-).2A1C13+3H2 > K Q + HjO Gia thid't KOH phan img he't => n^a = I I K O H = 0.1 (mol) => mKc, = 0,1 74,5 = 7,45 (g) > 6.525 (g) HHCI B 6.075 va 0,2500 D 3,375 v^ 0,2500 Hu&ng ddn gidi - 0.8.1 = 0.8 (mol) Khi thoat la H^: n^^^= 3.36 / 22.4 - 0.15 (mol) (Trich Dim tuyen sinh Cao dang khoi A) Huong ddngidi = 0,1.1 = 0.1 (mol) Phuong trinh phan ung: KOH + H Q C 103,85 gam (Trich Di thi tuyen sinh Cao ddng khd'i A) Hudng ddn gidi Theo b^i ra: n^a = 0.5 = 0.5 (mol); TTieo bM ra: Dap an dung la B B.0.5M B 25.95 gam A 6.075 va 0.0625 C 7,425 va 0.0625 0,05 (mol) v a y V = Vco2 = 0,05.22,4 =1.12 (lit) A.0.75M A 77,86 gam =>m = 7,74 + 0,5.36.5 + 0.14.98 - 0,39.2 = 38.93 (gam) Dap An diing 1^ D Sau phan ling nay, luong HCl lai la: n^am = 0,2 -0,15 = 0.05 (mol) Do d6 xay qua trinh: HCO3 c a u 9: Hoa tan he't 7,74 gam h6n hop b6t Mg, A l b ^ g 500 ml dung djch h6n hop H Q I M va H2SO4 0,28M thu dugc dung djch X va 8,736 lit H2 (Jb dktc) C6 can dung djch X thu duoc luong mu6'i khan la - > NaCl + NaHCO, > 0,15 vay CM(„a, = x/0,1 = 0.05/0,1 = 0.5M Dap &n dung la B t l,5y v , r4x + 3y = 0.8 Theobairataco:-^ ^ ix> x = 0.125; y = 0.1 [1.5y=0.15 Tu Vay: m = (x + y).27 = (0,125 + 0,l).27 = 6.075(g) V = x/2 = 0,125/2 = 0,0625(1) J>5pdndunglaA J Cty TNHHMTV mvu C&u 11: H6n h[...]... 16NaOH + 2Cra3 + 3CI2 ^ 12NaCl + 2Na2Ci04 + 8H2O 0,08 Vay: CI2 X : 2y = 1:2 + 2FeCl2 2FeCl2 x/4