NGUYEN TUYEN HA CAC DANG fillN HiNH VA PHJdNG PHAP GIAI NHANH BAI TAP TRAC NGH||M I J i 1 BIEN SOAN THEO CHUDNG TRINH MCJI DANH CHO HQC SINH BAN CO BAN VA NANG CAO ON LUYEN THI TU TAI, DAI HQC VA CAO DANG rH(i V!EN TINHBiNH THU-V^ NHA XUAT BAN DAI HQC QUOC GIA TP. HO CHI MINH CAC DANG OIEN HINH VA PHl/dNG PHAP GIAI NHANH BAI TAP TRAC NGHIEM H6AHOCI2 Nguyen Tuyen Ha NHA XUAT BAN DAI HOC QUdc GIA TP HO CHI MINH Khu pho 6, phirdng Linh Trung, quan Thu Dutc, TP.HCM So 3 Cong tradng Quoc te, quan 3, TP ROM DT: 38 239 172, 38 239 170 Fax: 38 239 172 - Email: vnuhp@vnuhcm.edu.vn Chiu track nhiem xudt bdn TS HUYNH BA LAN T6 chiic bdn thao vd chiu track nkiem vi tac quyin DOAN VAN KHANU Bien tap NGUYEN THI NGOC HAN S^a bdn in THAN THI HONG Trink bay bia . DIEM KHANH TK.02.H(V) 494.2013/CXB/07-25 TK.H.453-13(T) DHQG.HCM-13 In 1.000 cuon, khd 16 x 24cm. S6' dang ky ke hogch xuat bin: 494-2013/CXB/07-25/DHQGTPHCM. Quyet dinh xuSt bhn so: 333/QD-DHQGTPHCM ngay 21 thang 6 nam 2013 cua Nha xuat bdn DHQGTPHCM. In t?i Cong ty In Song Nguy6n, n^p lau chilu quy IV nSm 2013. LClI N(&I DAU De h6 trg viec hoc tap va on thi TU TAI - DAI HOC - CAO DANG hang nam, toi viet cuon sach n^y theo tinh than giiip cac em trang bi du kien thiJc de tham diT ki thi dat ket qua tot nhat. Ve noi dung cuon sach cd 3 phan: + Phan I: Cac dang bai tap dp dung cong thiJc tinh nhanh + Phan II: Cac phiicfng phap giai bai tap trac nghiem + Phan III: Gidi thieu 05 de thi thuf Tu tai va 06 de thi thuf Dai hoc de cac em thijf siJc minh. Cac de thi thuf mang tinh he thong day du cau hoi li thuyet va hki tap. Sau do cac em tham khao ddp an de rut kinh nghiem trong ki thi sSp tdi nham dat ket qua nhuf mong doi. Chiing toi hi vong cuon sdch nay la tai lieu tham khao hCJu ich giup cac em ren luyen ki nSng, nang cao kien thiJc cua minh va dat ket qua tot nhat trong ki thi TU TAI - DAI HOC - CAO DANG s^p tdi. Mac du rat co g^ng de bign soan nhuhg vdi thcfi gian c6 han chac ch^n van cbn nhufng khiem khuyet. Tac gia rat mong nhan dacfc sir gop y xay diTng cua dong nghi?p, doc gid gan xa de nhufng Ian sau tai ban se duac hokn thien hdn. Xin tran trong cam cfn. Tac gid Nguyen Tuyen Hd T CAC D4NG BAI AP D^JNG CACH TiNH NHANH Cach tinh nhanh so dong phan ciia: - Ancol no, dctn chiiCc (C„H2n+20): 2n-2 (1< n < 6) - Andehit dcfn chvCc, no (CnH2„0): 2n-3 (2< n < 7) - Axit cacboxylic dcfn chiic, no (C„H2n02): 2n-3 (2< n < 7) - Este no, dcfn chtic (C„H2„02): 2n-2 (1 < n < 5) - Ete darn chiic, no (C„H2n+'20): (n-l)(n-2) 2 (2< n < 6) - Xeton dam chiic, no (C„H2„0): (n-2)(n-3) 2 (2< n < 7) - Amin dctn chiic, no (CnHzn+sN): 2n-l (n< 5) Dang 1: KIM LOAI TAG DUNG VOl AXIT A. HCI, H2SO4 (loang) Doi vdi hat loai axit tren thi chi phdn ling voi nhUng kim loai diing trade H trong day hoat d6ng hoa hoc. K Ca Na Mg Al Zn Cr Fe Ni Sn Pb H Cu Hg Ag Au Phucfng trinh tong quat: (Vdi M la kim loai) 2M + nH2S04 (loang) -> M2(S04)n + nHzl ta luon c6 2M + 2nHCl ^ 2MC1„+ nHg J n„. = 2. n^^ De tinh khoi luong muoi thu diTcfc thi •k Dung dich H2SO4: m^^^i sunfat = nihSn h(7p kim loai + 96 n,i^ * Dung dich HCI: nimudi dorua = nihSn hgp kim loai + 71 nj,^ BAI TAP AP DUNG Bai 1: Cho l,04g hon hop hai kim loai tan hoan toan trong dung dich H2SO4 loang du thodt ra 0,672 Ht khi H2 (dktc). Khoi luong hon hop muoi sunfat khan thu di/oc la: A. 3,92g B. l,96g C.3,52g D.5,88g HUdng dan gidi 2 kim loai + H2SO4I hh muoi sunfat + H2 n„ ||f = 0,03(mol) Ap dung COng thufC: mmutfi sunfat = mhSn hap kim lo?i + 96 n„^ = 1,04 + 0,03.96 = 3,92 (gam) Chon dap an A. Bai 2: Hoa tan het 3,53 gam hon hop A gom ba kim loai Mg, Al va Fe trong dung dich HCl, c6 2,352 lit khi hidro thoat ra (dktc) va thu daoc dung dich D. Co can dung dich D, thu difdc m gam h6n hop muoi khan. Tri so ciia m la: A. 12,405 gam B. 10,985 gam C. 11,195 gam D. 7,2575 gam HU&ng d&n gidi nH |^ = 0,105(mol) Ap dung cong thuTc: mmuei doma = nihSn iwp kim loai +71 n„^ = 3,53 + 0,105 . 71= 10,985 gam Chon ddp 6n B. Bai 3: Hoa tan hoan to^n 2,17 gam hon hap 3 kim loai A, B, C trong dung dich HCl da thu diTOc 2,24 lit khi H2 (dktc) va m gam muoi. Gia tri cua m 1^. A. 9,27 gam. B. 5,72 gam. C. 6,85 gam. D. 6,48 gam. Hiidng dan gidi n„ =?^ = 0,l (mol) 22,4 m„us-iciorua = nihSn h;>p kim lo^i + 71 n,,^ = 2,17 + 0,1 . 71 = 9,27 gam Chon dap an A Bai 4: Bok tan 1,19 gam hSn hop A gom Al, Zn b&ng dung dich HCl viTa du thu difoc dung dich X vk V lit khi Y (dktc). Co can dung dich X di/oc 4,03 gam muoi khan. Gia tri cua V 1^. , A. 0,224 lit. B. 0,448 lit. C. 0,896 lit. D. 1,792 lit. Hii&ng dan gidi Ap dung cong thiJc: m^ua-i ciorua = nihSn hap kim loai + 71 n^^ 4,03-1,19 r^r^A ^ ^^ nji^ = ^^^p—= 0,04 (mol) ^ V„^ = 0,04 . 22,4 = 0,896 lit Chon dap an C. Bai 5: Cho 5,35 gam h6n hop X gom Mg, Fe, Al vao 250ml dung dich Y gom H2SO4 0,5M va HCl IM thu duoc 3,921it khi H2 (dktc) va dung dich A. Co can dung dich A trong dieu kien khong c6 khong khi, thu diTOc m gam chat rSn khan. Gia tri cua m la A. 20,900 gam. B. 26,225 gam. C. 26,375 gam. D. 28,600 gam. Hitdng dan gidi nn^so, = 0,25.0,5 = 0,125 mol =^ n^,, = 0,125 . 2 = 0,25 mol HHCI = 0,25.1 = 0,25 mol n„. = 0,25 mol => tong so mol H* = 0,25 + 0,25 = 0,5 mol q no n„, =1^ = 0,175 (mol) Ta c6: 2H^ ^ H2 Ban dau: 0,5 mol PU: 0,35 <- 0,175 mol Sau pif: 0,15 mol Vi axit dir nen kho'i \\iang rin b^ng tong kho'i lifcfng kim loai va cdc ion CO trong dung dich. mr^nkhan = 5,35 + (0,125.96) + (0,25.35,5) = 26,225 gam Chon dap an B Bai 6: (DH khoi A.2010): Cho m gam hon hap bot X gom ba kim loai Zn, Cr, Sn c6 so mol b^ng nhau tac dung het vdi lirotng diT dung dich HCl loang, nong thu dugc dung dich Y khi Ha- Co can dung dich Y thu dtfdc 8,98 gam muoi khan. Neu cho m gam hon hop X tac dung hoan toan vdi O2 (du) de tao hon hop 3 oxit thi the tich khi (dktc) phan ijrng la A. 2,016 lit. B. 0,672 lit. C. 1,344 Ht. D. 1,008 lit Hii&ng dan gidi 3 kim loai tren khi phan ilng vdi HCl loang nong deu hi oxi hoa th^nh so oxi hod +2. (X + HCl thu di/Oc muoi: ZnClz, CrClz, SnClg.) - Con khi tac dung O2, Zn tao +2, Cr tao +3, Sn tao +4. (X + O2 thu daoc cac oxit: ZnO, CraOa, Sn02) - Goi so mol moi kim loai la x (mol) thi: Vi cac kim loai c6 so mol bSng nhau nen cAc muoi c6 so mol b^ng nhau va bang x 136x + 123x + 190x = 8,98 =>x = 0,02; 1 3 "oj = 2''^4^^'''^ 2,25x = 0,045 mol => Vo^ = 0,045.22,4 = 1,008 lit Chon dap an D Bai 7: Cho 11 gam hon hop Al va Fe tac dung het v6i dung dich HCl thu dirge 8,96 lit H2 (dktc). Phan tram khoi lugng cua Fe trong hon hop la. A. 49,09%. B. 50,91%. C. 40,91%. D. 59,09%. Hiiofng dan gidi Liiu y: Kim loai tac dung vdi mot chat nao do ma sinh ra thi ti le mol giufa kim loai va H2 luon b^ng hod tri kim log,i chia 2 So do Al -> - H2 2 X mol —> — X mol 2 Fe -4 H2 y mol y mol "H, =|g = 0,4(mol) Ta lap va giai he phuong trinh 0,1.56 27x + 56y =11 3 -X + y = 0,4 X = 0,2 y = 0,1 Fe = 11 X 100% = 50,91% Chon dap an B Bai 8: (De KA.2007): Cho m gam hon hop Mg, Al vao 250ml dung dich chufa hon hop HCl IM va H2SO4 0,5M thu dugc 5,32 lit khi H2 (dktc) v^ dung dich Y. Coi the tich dung dich khong doi. Dung dich Y c6 pH 1^ A. 1. B. 6. C. 2. D. 7. Hiidng dan gidi n„^o, = 0.25 . 0,5 = 0,125 mol => n,j. = 0,125 . 2 = 0,25 mol nHci = 0,25 . 1 = 0,25 mol => n„. = 0,25 mol => tdng so mol H^ = 0,25 + 0,25 = 0,5 mol 5,32 ^ nor,,-/ — .V „ . = 0,2375(mol) 22,4 Ta c6: 2H* Ban dau: 0,5 mol Pu: 0,475 <- Sau pir: 0,025mol Sau phan iJng axit diT H* =1 H2 0,2375 mol I 0,025 0,25 = 0,1M pH = -log [H^^ Chon dap an A Bai 9: (De KB.2007): Cho 1,67 gam hon hop 2 kim loai d 2 chu ky ke tiep nhau thuoc nhom IIA tac dung het vdi dung dich HCl dii, thoat ra 0,672 lit khi H2 (dktc). Hai kim loai do la (Be = 9,Ca = 40,Mg = 24, ^ Sr = 87,Ba = 137) A. Mg va Ca. B. Ca va Sr. C. Sr va Ba. D. Be va Mg. Hiictng dan gidi Dat 2 kim loai 1^ A Ti le mol gifla kim loai va hidro b^ng ho^ tri kim loai chia 2 n6n ti le mol la 1: 1 = MZ2^ 0,03 (mol) 22,4 A Ha 0,03mol 0,03mol MA =^ = 55,67 ^ Ca va Sr. Chon dap an B Bai 10: Cho 3,87gam h6n hop X g6m Mg va Al v^o 250ml dung dich X gom HCl IM va H2SO4 0,5M thu dMc dung dich B va 4,368 Ht H2(dktc). Phan trSm khoi laong Mg va Al trong X tucfng ufng la A. 37,21% Mg va 62,79% Al. B. 62,79% Mg va 37,21% Al. C. 45,24% Mg va 54,76% Al. D. 54,76% Mg va 45,24% Al. B^i nay lam tifcfng tif bai 5 Bai 11: Hoa tan m(g) h6n hop Zn va Fe can vCra du 11 dung dich HCl 3,65M (d = l,19g/ml) thu dugc mot chat khi va 1250g dung dich D. Vay m c6 gia tri: A. 65,63 (g) B. 61,63 (g) C. 63,65 (g) D. 63,61 (g) HUdng dan gidi m,,„c, = 1000.1,19 = 1190 (g);n„c, = 3,65.1 = 3,65 (mol) Zn + 2HC1 ^ ZnCl2 + H2 Fe + 2HC1 ^ FeCl2 + H2 Ap dung DLBTKL, ta c6: nihh.zn.Fe) + "^ddiici = "^udo + "^n, = 1250 + 2(^) - 1190 ^ 63,65(g) Chon dap an C. B. KIM LOAI TAG DyNG VC3l HNO3, H2SO4 (d|c) Axit H2SO4 (dSc) + kim loai = muoi sunfat + san pham khuf + H2O San phdm khii c6 thi Id SO2 (nhan 2e), H2S (nhan 8e),S (nhan 6e) * mmuoi sunfat = "1 kin, loai + molgpk.so 6 nhan. i .96 axit HNO3 + kim loai = muoi nitrat + san pham khijf + H2O San phdm khd c6 the la NO2 (nhan le), NO (nhan 3e), N2O (nhan 8e), N2 (nhan lOe), NH4NO3 (nhan 8e). * m muoi nitrat = HI kim loai + molgpk.so e nhan.62 Luu y: Al, Fe, Cr khong phan ting vcd HNO3 dSc nguoi va H2SO4 dac nguoi. BAI TAP VAN DUNG Bai 1: Cho 1,86 gam hon hdp Al va Mg tac dung vdi dung dich HNO3 loang dtf thi thu di/cJc 560 ml lit khi N2O (dktc, san pham khuf duy nhat) bay ra. Khoi liTOng muoi nitrat tao ra trong dung dich la: A. 40,5 gam. B. 14,62 gam. C. 24,16 gam. D. 14,26 gam. HtCclng dan gidi n^o =^^-0,025mol N,o 22,4 Ap dung cong thuTc: m musl nitrat = m kim loai + molspk-so e nhan.62 = 1,86 + 0,025.8.62 = 14,26 gam Chon dap an D Bai 2: Hoa tan hoan toan 9,94 gam hon hop X gom Al, Fe, Cu trong lugng du dung dich HNO3 thu dugc 3,584 lit khi NO duy nhat (dktc). Tong khoi lugng muoi khan tao thanh la: A. 39,7gam B. 29,7gam C. 39,3gam D. 27,7gam Hiidng dan gidi n„„=||M.o,16n.o. Ap dung cong thUc: mmuoi nitrat = m kim loai + molspk.so e nhan.62 = 9,94 + 0,16.3.62 = 39,7 gam Chon dap an A Bai 3: Hoa tan 4,97 gam hon hap Al, Cu, Fe trong dung dich HNOr, loang dii thu dUcfc 1,792 Ut khi NO(dktc). Tong khoi luong muoi khan tao thanh: A. 19,85 gam B. 26,5 gam C. 39,7 gam D. 40,2 gam Bai nay gidi tuang tii bai 2. Bai 4: Hoa tan hoan toan 13,68 gam hon hdp X gom Al, Cu, Fe bSng dung dich HNO3 loang, diX thu diroc 1,568 lit khi N2O (dktc) va dung dich chuTa m gam muoi. Gia tri cua m la. A. 48,40 gam. B. 31,04 gam. C. 57,08 gam. D. 62,70 gam. Bai nay gidi tuang tu bai 2. Bai 5: (DH khoi A.2010): Nung 2,23 gam hon hop X gom cac kim loai Fe, Al, Zn, Mg trong oxi, sau mot thdi gian thu difdc 2,71 gam hon hop Y. Hoa tan hoan toan Y vao dung dich HNO3 (da), thu dtfOc 0,672 Ht khi NO (san pham khuf duy nhat, d dktc). So mol HNO3 da phan ufng la A. 0,12. B. 0,14. C. 0,16. D. 0,18. HiCdng ddn gidi Tom t^t: Fe, Al, Zn, Mg + O2 -> hh ; hh + HNO3 -> muoi + NO ' ' ^'la^ 0.672 (l)«5|^=0,03(mol) Bao toan khoi liTOng: mo = m,, - m„ = 2,71 - 2,23 = 0,48(g) -> n^ - ^ = 0,03 (mol) Ta CO bao toan nguyen to N: ^ '^HNOj ~ ^UNOj (pu O trong oxit) ^HNOj (pu oxi ho4 - khut) ^NO Mat khac: HHNO^ (pu o t^ng oxiu = ( cho H^O); nHNo, ,pu ox, ho. - uha, = 3.n^o ( Do N^^ —^ NO) ^^^iiNo,= 2no,oxit) + 3nf^o+ n^o = ^T^o(<«iM+ '^^m = 2.0,03 + 4.0,03 = 0,18 mol Chon dap an D Bai 6: (BH khoi B.2008): Cho 2,16 gam Mg tAc dung vdi dung dich HNO3 (du). Sau khi cac phan iJng xay ra hoan toan, thu duoc 0,896 lit NO (d dktc) va dung dich X. Khoi luong muoi khan thu diTOc khi lam bay hoi dung dich X la A. 8,88 gam B.13,92 gam C. 6,52 gam D.13,32 gam Hii&ng dan gidi n^g = 246 : 24 = 0,09 (moDin^o - 0,896 : 22,4 = 0,04 (mol) Cdch 1: 3Mg + 8HNO3 ^ 3Mg(N03)2 + 2N0 + 4H2O < 0,06 < 0,04 0,06 < 4Mg + IOHNO3 0,03 > > 4Mg(N03)2 + NH4NO3 + 3H2O 0,03 > 0,0075 Khoi liTcfng muoi khan thu diioc khi lam bay hoi dung dich X \k m = 0,09. 148 + 0,0075.80 = 13,92 (gam) ^ chon B. Cdch 2: Diia vao dinh luat bao toan electron Mg 0,09 2e - 2.0,09 Mg^2 0,09 N +5 N +5 3e - 3.0,04 8e - 8.x N^2(N0) 0,04 N-' NH4NO3) X So mol electron dLfOc bao to^n 2.0,09 = 3.0,04 + 8x -> X = 0,0075 (mol) Khoi lirong muoi khan thu dtfoc khi lam bay hoi dung dich X la m = 0,09. 148 + 0,0075. 80 = 13,92 (gam) -> chon B Bai 7: Hoa tan 23,4 gam G gom Al, Fe, Cu bang mot lirong vCra du dung dich H2SO4 d5c, nong, thu di/dc 15,12 lit khi SO2 (dktc) va dung dich chuTa m gam muoi. Gia tri cua m la. A. 153,0 gam. B. 95,8 gam. C. 88,2 gam. D. 75,8 gam. Hii&ng ddn gidi 15,12 i^so, = 22,4 = 0,675mol nimuol simfat = HI kjn, loai + molgpk.SO 6 nhan. — .96 2 = 23,4 + 0,675.2. \6 = 88,2 gam Chon dap an C. | Bai 8: Kok tan ho^n to^n m gam hon X gom Al, Fe, Cu vko dung dich HNO3 dSc nong du, thu duoc dung dich Y chufa 39,99 gam muoi va 7,168 Ht khi NO2 (dktc). Gia tri cua m la. A. 20,15 gam. B. 30,07 gam. C. 32,28 gam. D. 19,84 gam. Hiicfng ddn gidi HMO =^^^ = 0,32mol """^ 22,4 Ap dung cong thiJc: m raM nitrat = m kim loai + molspk.so e nhan.62 =^ m kim loai = m muoi nitrat " molspk-SO G nhan.62 = 39,99 - 0,32.1.62 = 20,15 gam Chon dap an A Bai 9: (DHKB. 2007) ThUc hien hai thi nghiem: 1. Cho 3,84 gam Cu phan iJng vdi 80 ml dung dich HNO3 IM thodt ra Vi lit NO. 2. Cho 3,84 gam Cu phan ling vdi 80 ml dung dich chufa HNO3 IM va H2SO4 0,5 M thoat ra V2 Ht NO.Biet NO la san pham khijf duy nhat, cac the tich khi do d cCing dieu ki$n. Quan he giOfa Vi va V2 la A. V2 = Vi. B. V2 = 2Vi. C. V2 = 2,5Vi. D. V2 = l,5Vi. Hiictng dan gidi TNI: nc„=^ = 0,06 mol b4 n„, =0,08 mol n^^_ = 0,08 mol nHN03 =0.08 mol , 3Cu + 8H" + 2NO3- > 3Cu^^ + 2N0T + 4H2O Dau bai: 0,06 0,08 0,08 -> phan iJng het Phan iJng: 0,03 <- 0,08 -> 0,02 ^ 0,02 mol => Vi tirang ufng vdi 0,02 mol NO. TN2: ncu = 0,06 mol; nn^o, = 0,08 mol; nn^so. = 0,04 mol. => Tong n^^, = 0,16 mol; n^^. = 0,08 mol. 3Cu + 8H* + 2NO3" >3Cu^* + 2NO^ + 4H2O Dau bai: 0,06 0,16 0,08 -> Cu va H"^ phan iJng het Phan iJng: 0,06 ^ 0,16 ^ 0,04 -> 0,04 mol => V2 ti/ong urng vdi 0,04 mol NO. Nhir vay V2 = 2Vi. Chon dap an B. Bai 10: (DH khoi B-2009): Khi hoa tan hoan toan 0,02 mol Au bang ntfdc cudng toan thi so mol HCl phan ilng va so mol NO (san pham khuf duy nhat) tao thanh Ian liToft la A. 0,03 va 0,01 B. 0,06 va 0,02 C. 0,03 va 0,02 D. 0,06 va 0,01 Hiicfng ddn gidi • Ni/dc CLfdng toan la ti le 3 : 1 giufa HCl va HNO3 Au + 3HC1 + HNO3 > AUCI3 + NO + 2H2O 0,02 -> 0,06 -> 0,02 Chon dap an B Bai 11: Hoa tan 3 gam hon hgfp A gom kim loai R hoa tri 1 va kim loai M hoa tri 2 vifa du vao dung dich chda HNO3 va H2SO4 va dun nong, thu dtroc 2,94 gam hon hap khi B gom NO2 va SO2. The tich cua h5n hop khi B la 1,344 lit (dktc). Khoi luang muoi khan thu diroc la. A. 6,36g. B. 7,06g. C. 10,56g. D. 12,26g. HUdng dan gidi ^ 1^344 ^ oGmol dSt NO2 x mol, SO2 ymol 22,4 r46x + 64y = 2,94 fx = 0,05 x + y = 0,06 [y = 0,01 nimuai = 3 + (0,05 .62 + 0,01 2 .96) = 7,06 gam 2 • Dap an B. Bai 12: Cho 8,3 gam h5n hcfp Al va Fe tac dung vdi dung dich HNO3 loang dtr thi thu dircfc 45,5 gam muoi nitrat khan. The tich khi NO (dktc, san pham khuf duy nhat) thoat ra la: ! A. 4,48 lit. B. 6,72 lit. C. 2,24 lit. D. 3,36 lit. Hiicmg dan giai Ap dung cong thiJc: m musi nitrat = ni ,0,1 + molspk-so e nhan.62 ^ DNO = "-^-"^ ~ = = 0,2 mol so e nhan.62 3.62 => = 0,2.22,4 = 4,48 lit Chon dap an A Bai 13: Hoa tan hoan toan 1,23 gam hon hop X gom Cu va Al vao dung dich HNO3 dSc, nong thu dixoc 1,344 lit khi NO2 (san pham khuf duy nhat, d dktc). Phan trSm ve khoi liiong cua Cu trong hon hop X la A. 21,95%. B. 78,05%. C. 68,05%. D. 29,15%. Hiidng ddn giai n NO, = MM = 0,06mol (Dat Cu X mol, Al y mol) 22,4 Cu X Al • y . Cu^^ + 2e 2x Al'^ + 3e 3y N"^ + 3e -> 0,18 0,06 Ta c6: 64x + 27y = l,23 fx = 0,015 |2x + 3y-0,06 [y = 0,01 %Cu=M15:6i.78,05% 1,23 Chon dap an B Bai 14: Hoa tan 15 gam hon hop X gom hai kim loai Mg va Al vao dung dich Y gom HNO3 va H2SO4 dSc thu diioc 0,1 mol moi khi SO2, NO, NO2, N2O. Phan tr5m khoi Itfong cua Al va Mg trong X Ian lacft la A. 63% va 37%. B. 36% va 64%. C. 50% va 50%. D. 46% va 54%. HU&ng dan gidi Dat HMg = X mol; DAI = y mol. Ta c6: Qud trinh oxi ho^: Mg -> Mg^* + 2e Al x 2x y => Tong so mol e nhudng bing (2x + 3y). Al'" + 3e 3y Qua trinh khii: + 3e -> N*=^ 2N*^ + 2 x 4e 2N*^ 0,3 0,1 0,8 0,2 N*^ + le N"" S"^ + 2e S*^ 0,1 0,1 0,2 0,1 => Tdng so mol e nhan bkng 1,4 mol. 24x + 27y = 15 fx = 0,4 mol Theo dinh luat bao toan electron: 97 n 9 .%A1 = ^^^.100% = 36%. 2x + 3y = l,4 [y = 0,2 mol 15 %Mg = 100% - 36% = 64%. Bai 15: Cho 2,8 gam hon hop bot kim loai bac va dong tac dung v6i dung dich HNO3 dSc, dtr thi thu duoc 0,896 lit khi NO2 duy nhat (d dktc). Thanh phan phan tram cua bac va dong trong h6n hop Ian liTOt la: A. 73% ; 27%. B. 77,14% ; 22,86% C. 50%; 50%. D. 44% ; 56% Tuang tu bai 14. Dang 2: OXIT KIM LOAI TAG DUNG VOI AXIT HQ, H2SO4 [bang) Phi/dng trinh tdng quat MzOn + 2nHCl 2MCln+ nHzO M20„ + nH2S04 (loang) ^ M2(S04)„ + nHzO Cach tmh nhanh cho trie nghi?m * Doi vdi axit H2SO4 (loang) Khoi li/ong muoi thu dugc la: m^^ai ^unfat = mhSn hap oxukim loai + n,[^so, -80 Doi vdi axitHCl • Khoi luong muoi thu duoc la: m,„„a-i ciorua = nihon hop oxit kim loai + 27,5 n,,^;, BAl TAP VAN DgNG Bai 1: Cho 50g hon hop hot oxit kim loai gom ZnO, FeO, Fe203, Fe304, MgO tac dung het vdi 200ml dung dich HCl 4M (vCra du) thu duoc dung dich X. Luong muoi c6 trong dung dich X bSng A. 79,2g B. 78,4g C. 72g D. 72,9g HUcfng ddn gidi Ap dung COng thufC: nimudi cloma = nihon h<7p oxit kim loai + 27,5 n^jp, = 50 + (0,2.4.27,5) = 72 gam Chpn dap an C. Bai 2: Hoa tan hoan toan 2,81 gam hon hop A gom FezOg, MgO, ZnO bkng 300ml dung dich H2SO4 0,1M (vCra du). Co can can than dung dich thu diioc sau phan ufng thi thu difoc lirong muoi sunfat khan la: A. 3,81 gam B. 4,81 gam C. 5,21 gam D. 4,8 gam HU&ng dan gidi Ap dung COng thufC: mmum sunfat = nihSn hop oxit kim loai + J^HaSOi -^^ = 2,81 + (0,3.0,1.80) = 5,21 gam Chon dap an C. Bai 3: De tac dung vCra du vdi 7,68g hon hop gom FeO, Fe304, FegOs can dung 260 ml dung dich HCl IM. Dung dich thu diTOc cho tac dyng vdi NaOH du, ket tua thu Auac mang nung trong khong khi den khoi iLfOng khong doi duoc m gam chat r^n. Gia tri cua m la: A.6g B. 7g C.8g D.9g Hii&ng dan gidi oxit kim loai + HCl > muoi clorua + H2O nnc. = = 0,26.1 =0,26 mol 2H^ + O'- >H20 0,26 > 0,13mol 5 6 => mpe = 7,68 - 0,13.16 = 5,6 gam => ^fe == 0>l"iol So do hop thiJc: 2Fe ^ FeaOg 0,1 -> 0,05 mol => mp^^o^ = 160.0,05 = 8 gam Chon dap an C. Bai 4: Oxi hoa cham m gam Fe ngoai khong khi sau mot thcJi gian thu duoc 12 gam hon hop X g6m(Fe, FeO, FezOg, Fe304). De hoa tan het X, can vCra du 300 ml dung dich HCl IM, dong thdi giai phdng 0,672 Ht khi (dktc). Tinh m? A.10,08 B.8,96 C.9,84 D.10,64 n„^ = ' = 0,03mol .nHci = 0,3.1 = 0,3 mol Taco: 2H^ > Hg 0,06 <-0,03 2H^+ O'- > H2O 0,24 ->0,12mol m = mx- mo(oxit) = 12 - 0,12.16 = 10,08 gam Chon dap an A. Bai 5: Hoa tan hoan toan 2,8 gam hon hop FeO, Fe203 va Fe304 can vCra du V ml dung dich HCl IM, thu di/oc dung dich X. Cho tii tii dung dich NaOH dir vao dung dich X thu dugfc ket tua Y. Nung Y trong khong khi den khoi liiOng khong doi thu dirge 3 gam chat r^n. Tinh V? A. 87,5 ml B.125 ml C.62,5 ml D.175 ml Hii&ng dan gidi -4 chat rin la Fe203 nFe,03 =Y|^= 0,01875 mol So do hop thufc: 2Fe Fe203 0,0375 <- 0,01875 => mo,ox,t) = 2,8 - (0,0375.56) = 0,7 gam no,oxit)=^-0,04375mol 2H^ + O'- > H2O 0,0875 <- 0,04375 => V = 87,5 ml Chon dap an A. Bai 6: Hoa tan hoan toan 5,4 gam mot oxit sSt vao dung dich HNO3 dii thu dagc 1,456 lit hon hop NO va NO2 (dktc va khong con san pham khuf nao khac). Sau phan ufng khoi lUOng dung dich tang len 2,49 gam so vdi ban dau. Cong thufc cua oxit s^t va so mol HNO3 phan ufng la: A. FeO va 0,74 mol B. Fe304 va 0,29 mol C. FeO va 0,29 mol D. Fe304 va 0,75 mol [...]... HUcfng dan fx = 0,26 Vay Vco^ =0,38.22,4 = 8, 512 l i t h h rauoi cacbonat A C O 3 + H2SO4 y y = 0 ,12 " x + 2y = 0,5 XCO3 + 2HC1 -> X C I 2 + H2O + CO2T Cach t i n h n h a n h dung cho tr&c n g h i e m ri^^mn, = ^ B a ^ * = ^' 1^"^°^ n^o^ = nkettta = 0 ,12 m o l X Chon dap a n D lugng (71-60) M u o i cacbonat tac dung v d i dung dich H 2 S O 4 loang Vco.= 0 ,12 22,4 = 2,688 l i t hap 2: tii chat nay sang... 22,4 = 0,75 mol; So C ^ n C O , So H ~ 2.n H 2 O n„ ^ = 18 = 1 ,125 mol 0,75 ^ 1 2.1 ,125 3 Vay cong thiJc phan tuT cua X la C 3 H 9 N Chon dap an D Bai 2: Dot chay hoan toan h5n hgp 2 hidrocacbon lien tiep trong day dong dang thu dirge 22,4 l i t C02(dktc) va 25,2g H2O Hal hidrocacbon do la: A C2H6 va C3H8 B C3H8 va C4H10 C C4H10 va C5H12 D C5H12 va C S H H D 22,9% va 77,1% Hii&ng ddn gidi ^11,0 = 25,2... dung dich Y Chon dap an B Bai 5: H5n hop X gom N2 va H2 c6 Mx = 12, 4 DSn X di qua binh diTng bot Fe roi nung nong biet r^ng hieu suat tong hap N H 3 dat 40% t h i thu dixac h6n hap Y My c6 gia t r i la A 15 ,12 B 18,23 C 14,76 HiCdng dan Xet 1 mol h6n hap X D 13,48 N2 • gidi mx = 12, 4 gam gom a mol N2 vk (1 - a) mol H2 , 28a + 2(1 - a) = 12, 4 ^ a = 0,4 mol ->n„^ =0,6 mol = 40 1+a b mol I Dung dich HNO3... -^BaCOj ~ CO2 + O H " -> HCO3" X CO2 + y X 20H- 2y ^ 0 0 3 ' - + => X (1) (2) 2YCI3 + 3H2O +3CO2T Theo 2 phuong t r i n h (1),(2) t a t h a y •^muo'i clorua ~ ^ Pt: H2O 197 = 0,12mol -'^cOo B2CO3 + H2SO4 ^ | y = 0 ,12 "co, =0 ,12 + 0,26 = 0,38 mol (loang) ^ (loang) ^ A S O 4 + CO2T + H 2 O A 2 S O 4 + CO2T + H 2 O Cach t i n h n h a n h cho m u o i sunfatv nirauo'i sunfat — rnmuo'i cacbonat ^CO^ BAl TAP... -2009): Cho m gam M g vao dung dich chufa 0 ,12 m o l FeCls Sau k h i p h a n ufng xay r a hoan toan t h u dtfoc 3,36 gam chat rkn Gia t r i cua m la A 5,04 B 4,32 C 2,88 HiCdng dan npe = ^ BAI T A P A P 56 B a i 1 : Cho O,lmol Fe vao 500 ml dung dich AgNOs I M thi dung dich thu difOc chufa: gidi = 0,06 mol M g + 2FeCl3 DUNG D 2,16 MgCl2 + 2 F e C l 2 0,06 -> 0 ,12 ^ 0 ,12 M g + FeClz ^ MgCl2 + Fe A AgNOg B... I^cOj -C^-^ B a i 2: Hoa tan 5g h6n hop 2 muoi cacbonat kim loai hoa t r i I va hoa t r i I I bSng dung dich HCl thu diigfc dung dich M va 1 ,121 khi CO2 (dktc) Khi CO can dung dich M thu dircKc khoi lufdng muoi khan bang: B 5,55g C 16,5g HiC&ng dan mrinB- = 12, 9 (0 ,12 44) = 7,62 gam Dang 7 KIM LOAI SAT B| 0X1 HOA THANH HON HPP (Fe, FeO, FegOa Fe304) Cdch 1: Vi Fe bi oxi hoa thanh cac exit s^t c6 the con... Dem nung liiong chat rSn B tren cho den khoi liiong khong doi t h i thu duac 0 ,12 mol CO2 va con lai cac chat rSn B' Khoi lUcfng cua B va B' 1^: A 10,36 gam; 5,08 gam B 12, 90 gam; 7,62 gam C 15, 63 gam; 10,35 gam D 16,50 gam; 11,22 gam Hitdng dan gidi m,5„ 3x - 2y = mol san pham khuf.so e nhan Cdch 2: Van dung cong thitc tinh nhanh nipe = 0,7.mh6n hop oxit sit + (molsan phim k M - S O e nhan.5,6) BAI... uTng n h i e t n h o m l a A 80% B 90% 8A1 thiic - n ^ , = 0,13.0,1 = 0,013 m o l ; H^LOH,, - 0 936 0, 012 m o l OH m i n n„„ =3.nA,,o„.=>CM A1(UH», C h o n dap a n D Al'* = "1,1^ = 4.n„ - n „ , „ , O H max D 60% 3Fe304 — ^ 4AI2O3 + 9Fe 0,15 AUOH), 3.0, 012 ' =1,8M 0,02 ^ > C, MK.OH 4.0,013-0, 012 Q Q2 8x 3x 4x 9x (0,4-8x) 10 p t : Al^* + 3 0 H - ^ A1(0H)3 n„„- + 0,4 = C 70% Hiicfng dan gidi Hiiofng... Cho 1 ,12 gam bot Fe vh 0,24 gam bot M g tac dung v d i 250 m l dung dich CUSO4 x M , khuay nhe cho den k h i dung dich m a t mau x a n h n h a n t h a y k h o i liicfng k i m loai sau p h a n ufng la 1,88 gam Gia t r i cua x la A 0,04M B 0,06M C O I M Hii&ng ddn n i k i m io,i sau phan ling = 1,88 - 1 ,12 => Am =0,01.(64-24) = 0,4 gam n i c b n i a i = 0,52 L i (Z=3) ls'2s* hay [He]2si - 0,4 = 0 ,12. .. COzCdktc) va 25,2g H2O H a i hidrocacbon nankan = do l a : A C2H6 va C3H8 C C4H10 va C5H12 nankan B C3H8 va C4H10 D C5H12 va CgHu Hii&ng dan = 0,23 - 0,14 = 0,09; B a i 11: M o t h o n hop k h i gom C t r o n g phan tuf va c6 cung m a t mau vCfa du 80g dung chay hoan t o a n m gam h o n anken do c6 cong thuTc p h a n gidi D C5H12, CsHip = n = 2,5 z^CaHe va C3H8 C„H2n + ^ p h a m chay I a n \mi d i qua b i . mol Taco: 2H^ > Hg 0,06 <-0,03 2H^+ O'- > H2O 0,24 ->0,12mol m = mx- mo(oxit) = 12 - 0 ,12. 16 = 10,08 gam Chon dap an A. Bai 5: Hoa tan hoan toan 2,8 gam hon hop. 1,88 - 1 ,12 - 0,24 = 0,52 gam => Am =0,01.(64-24) = 0,4 gam nicbniai = 0,52 - 0,4 = 0 ,12 Mg + Cu^^ ^ Mg2" + Cu 0,01 0,01 Fe + CuS04 ^FeS04 + Cui X X Ta c6: 64x - 56x = 0 ,12 X . Dung dich Y c6 pH 1^ A. 1. B. 6. C. 2. D. 7. Hiidng dan gidi n„^o, = 0.25 . 0,5 = 0 ,125 mol => n,j. = 0 ,125 . 2 = 0,25 mol nHci = 0,25 . 1 = 0,25 mol => n„. = 0,25 mol => tdng so