NGUYEN TUYENH A CAC DANG fillN HiNH VA PHJdNG PHAP GIAI NHANH BAI TAP TRAC NGH||M I Ji 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 H OCHI MINH CAC DANG O I E N HINH VA BAI PHl/dNG P H A P GIAI NHANH TAP TRAC NGHIEM L C l I N(&I D A U H6AHOCI2 Nguyen T u y e n H a N H A X U A T QUdc DAI H O C De h6 trg viec hoc tap va on t h i T U T A I - D A I HOC - CAO D A N G hang nam, toi viet cuon sach n^y theo tinh than giiip cac em trang bi du kien thiJc de tham diT k i t h i dat ket qua tot nhat B A N G I A TP H O C H I M I N H Khu 6, phirdng Linh Trung, quan Thu Dutc, TP.HCM So Cong tradng Quoc te, quan 3, TP ROM D T : 38 239 172, 38 239 170 F a x : 38 239 172 - E m a i l : vnuhp@vnuhcm.edu.vn Chiu track nhiem xudt bdn TS H U Y N H BA L A N T6 chiic bdn thao vd chiu track nkiem vi tac quyin DOAN V A N K H A N U Bien tap NGUYEN T H I NGOC H A N S^a bdn in T H A N T H I HONG + Phan I I : Cac phiicfng phap giai bai tap trac nghiem + Phan I I I : Gidi thieu 05 de t h i thuf Tu t a i va 06 de t h i thuf Dai hoc de cac em thijf siJc minh Cac de t h i thuf mang tinh he thong day du cau hoi l i thuyet va hki tap Sau cac em tham khao ddp an de rut kinh nghiem k i t h i sSp tdi nham dat ket qua nhuf mong doi Chiing toi h i vong cuon sdch la tai lieu tham khao hCJu ich giup cac em ren luyen k i nSng, nang cao kien thiJc cua minh va dat ket qua tot nhat k i t h i T U T A I - D A I HOC - CAO DANG s^p t d i 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 Tac gid DIEM KHANH Nguyen Tuyen Hd TK.02.H(V) 494.2013/CXB/07-25 DHQG.HCM-13 TK.H.453-13(T) 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 21 thang nam 2013 cua Nha xuat bdn DHQGTPHCM In t?i Cong ty In Song Nguy6n, n^p lau c h i l u quy I V nSm + Phan I : Cac dang bai tap dp dung cong thiJc tinh nhanh Xin tran cam cfn Trink bay bia Ve noi dung cuon sach cd phan: 2013 T CAC D4NG BAI A P D^JNG CACH TiNH NHANH C a c h tinh nhanh so dong phan ciia: - Ancol no, dctn chiiCc - Andehit dcfn chvCc, no - Axit cacboxylic (CnH2„0): dcfn chiic, no - Este no, dcfn chtic - Ete darn chiic, no (C„H2n02): (1< n < 2n-3 (2< n < 7) 2n-3 (2< n < 7) 2n-2 (C„H2„02): (C„H2n+'20): - Xeton dam chiic, no - Amin dctn chiic, no 2n-2 (C„H2n+20): (C„H2„0): (1 n < 5) (n-l)(n-2) (2< n < 6) (n-2)(n-3) (2< n < 7) (n< 5) 2n-l (CnHzn+sN): < 6) Dang 1: KIM LOAI TAG DUNG VOl AXIT A H C I , H2SO4 ( l o a n g ) Doi vdi hat loai axit tren thi chi phdn ling voi nhUng kim loai diing trade H day hoat d6ng hoa hoc K Ca Na Hg Ag Au Mg Al Zn Cr Fe Ni Sn Pb Phucfng trinh tong quat: (Vdi M la kim loai) 2M + nH2S04 (loang) -> M ( S ) n + nHzl ta luon c6 2M + 2nHCl ^ 2MC1„+ nHg n„ = n^^ J 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,^ H Cu BAI TAP AP DUNG B a i 1: Cho l,04g hon hop hai k i m loai tan hoan toan dung dich H2SO4 loang du thodt 0,672 Ht k h i H (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 kim loai + H2SO4I dan gidi hh muoi sunfat + B a i 4: Bok tan 1,19 gam hSn hop A gom A l , Zn b&ng dung dich HCl v i T a du thu difoc dung dich X vk V l i t k h i Y (dktc) Co can dung dich X di/oc 4,03 gam muoi khan Gia t r i cua V 1^ , A 0,224 lit B 0,448 lit Ap dung cong thiJc: m^ua-i ciorua H2 4,03-1,19 + 96 n„^ ^ = 1,04 + 0,03.96 = 3,92 (gam) Chon dap an A B a i 2: Hoa tan het 3,53 gam hon hop A gom ba kim loai Mg, A l va Fe dung dich HCl, c6 2,352 l i t k h i hidro thoat (dktc) va thu daoc dung dich D Co c a n dung dich D, thu difdc m gam h6n hop muoi k h a n T r i so ciia m la: B 10,985 gam C 11,195 gam Hitdng dan gidi nn^so, = 0,25.0,5 = 0,125 mol =^ n^,, = 0,125 = 0,25 mol n„ = 0,25 mol => tong so mol H* = 0,25 + 0,25 = 0,5 mol thuTc: m m u e i doma = nihSn iwp kim loai q + n„^ Chon ddp 6n B B a i 3: Hoa t a n hoan to^n 2,17 g a m h o n hap k i m loai A, B, C dung dich HCl da thu diTOc 2,24 l i t k h i H (dktc) va m g a m muoi Gia t r i cua m 1^ B 5,72 gam C 6,85 gam D 6,48 gam Hiidng dan gidi m„us-iciorua = nihSn h;>p kim lo^i + 71 n,,^ = 2,17 n„, no =1^ = 0,175 (mol) Ta c6: 2H^ ^ Ban dau: 0,5 mol PU: 0,35 X = 0,0075 (mol) = 0,03 (mol) B a i 7: Hoa tan 23,4 gam G gom A l , Fe, Cu bang mot lirong vCra du dung dich H2SO4 d5c, nong, thu di/dc 15,12 lit SO2 (dktc) va dung dich ^ H N O j (pu oxi ho4 - khut) ^NO A 153,0 gam B 95,8 gam C 88,2 gam Hii&ng ddn gidi Mat khac: HHNO^ (pu o t ^ n g oxiu = ( cho n H N o , ,pu ox, ho - uha, = 3.n^o ( Do H^O); N^^ — + 3nf^o+ n^o = i^so, ^ = 15,12 22,4 = 0,675mol NO) nimuol simfat = ^T^o( chon B mo = m , , - m„ = 2,71 - 2,23 = 0,48(g) -> n^ - ^ '^HNOj ~ 0,04 Khoi lirong muoi khan thu dtfoc lam bay hoi dung dich X la Bao toan khoi liTOng: ^ 3.0,04 8.x So mol electron dLfOc bao to^n 0.672 (l)«5|^=0,03(mol) ^ ' l a ^ N^2(N0) - HI kjn, loai '^^m + molgpk.SO nhan — 96 = 23,4 + 0,675.2 \6 = 88,2 gam Chon dap an C | D 75,8 gam 3Cu + 8H* + 2NO3" B a i 8: Kok t a n h o ^ n t o ^ n m gam h o n X gom A l , Fe, Cu vko dung dich H N O d S c nong du, t h u duoc dung dich Y chufa 39,99 gam m u o i va 7,168 Ht k h i N O (dktc) Gia t r i cua m la A 20,15 gam B 30,07 gam C 32,28 gam Hiicfng ddn Dau b a i : P h a n iJng: 0,06 ^ D 19,84 gam m =^ m 0,16 0,08 0,16 ^ gidi B a i 10: (DH khoi B-2009): raM nitrat k i m loai = m = m + molspk.so e nhan.62 k i m loai " molspk-SO muoi nitrat G nhan.62 gam 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 Au Cho 3,84 gam Cu p h a n iJng v d i 80 m l dung dich H N O I M t h o d t r a -> 0,02 M hoa t r i vifa du vao dung dich c h d a H N O va H2SO4 va dun n o n g , thu dtroc 2,94 g a m h o n hap k h i B gom NO2 va SO2 The t i c h cua h n Quan he giOfa V i va V la B V = V i C V = , V i Hiictng dan 0,08 0,08 -> l i t (dktc) K h o i l u a n g m u o i k h a n t h u diroc l a A 6,36g B 7,06g HUdng n„, = , m o l , 3Cu + H " + 2NO3- D V2 = l,5Vi hop k h i B l a 1,344 gidi = 0,06 m o l =0.08 mol 0,06 > A U C I + N O + 2H2O B a i 11: Hoa t a n gam h o n hgfp A gom k i m loai R hoa t r i va k i m l o a i 0,5 M t h o a t r a V Ht N O B i e t N O la san p h a m khijf n h a t , cac the t i c h k h i d cCing dieu k i $ n Dau b a i : 3HC1 + H N O Chon dap a n B Cho 3,84 gam Cu p h a n ling v d i 80 m l dung dich chufa H N O I M va nHN03 + 0,02 -> 0,06 Vi lit NO n c „ = ^ b4 gidi • Ni/dc CLfdng toan l a t i le : giufa H C l va H N O ThUc h i e n h a i t h i n g h i e m : TNI: K h i h o a t a n hoan toan 0,02 m o l A u b a n g ntfdc cudng toan t h i so m o l H C l p h a n i l n g va so m o l N O (san p h a m 2007) A V = V i 0,04 m o l khuf n h a t ) tao t h a n h I a n liToft l a Chon dap a n A H2SO4 -> Nhir vay V = V i Chon dap a n B = 39,99 - 0,32.1.62 = 20,15 B a i 9: (DHKB 0,04 -> Cu va H"^ p h a n iJng h e t => V ti/ong urng v d i 0,04 m o l N O H M O = ^ ^ ^ = 0,32mol """^ 22,4 A p dung cong thiJc: 0,06 >3Cu^* + 2NO^ + H O ^ 1^344 ^ 22,4 n^^_ = 0,08 m o l x + y = 0,06 p h a n iJng h e t P h a n iJng: 0,03 0,02 ^ 0,02 m o l => V i tirang ufng v d i 0,02 m o l N O TN2: ncu = 0,06 m o l ; nn^o, = 0,08 m o l ; nn^so = 0,04 m o l => T o n g n^^, = 0,16 m o l ; n ^ ^ = 0,08 m o l • dan D 12,26g gidi oGmol dSt NO2 x m o l , SO2 y m o l r46x + 64y = 2,94 > 3Cu^^ + N T + H O C 10,56g f x = 0,05 [ y = 0,01 nimuai = + (0,05 62 + , - 96) = 7,06 gam Dap a n B B a i 12: Cho 8,3 gam h n hcfp A l va Fe tac dung v d i dung dich H N O loang dtr t h i t h u dircfc 45,5 gam m u o i n i t r a t k h a n The t i c h k h i N O (dktc, san p h a m khuf n h a t ) t h o a t r a l a : ! A 4,48 l i t B 6,72 l i t C 2,24 l i t D 3,36 l i t Qua t r i n h khii: Hiicmg dan giai Ap dung cong thiJc: ^ DNO = => m musi nitrat = ni "-^-"^ ~ so e nhan.62 ,0,1 = 0,3 + molspk-so e nhan.62 N*^ + l e = 0,2 mol 3.62 0,1 Chon dap an A B a i 13: Hoa tan hoan toan 1,23 gam hon hop X gom Cu va A l vao dung dich HNO3 dSc, nong thu dixoc 1,344 l i t k h i NO2 (san pham khuf nhat, d dktc) Phan trSm ve khoi liiong cua Cu hon hop X la C 68,05% D 29,15% Hiidng ddn giai n NO, Cu^^ + 2e 2x X Al • y Ta c6: N"" 0,2 S"^ + 2e S*^ 0,2 0,1 0,1 24x + 27y = 15 fx = 0,4 mol 2x + 3y = l , [y = 0,2 mol 97 n %A1 = ^ ^ ^ 0 % = 36% 15 %Mg = 100% - 36% = 64% Bai 15: Cho 2,8 gam hon hop bot kim loai bac va dong tac dung v6i dung Thanh phan phan tram cua bac va dong h6n hop Ian liTOt la: A 73% ; 27% B 77,14% ; 22,86% N"^ + 3e -> C 50%; 50% D 44% ; 56% 0,18 Tuang tu bai 14 0,06 Al'^ + 3e 3y 64x + 27y = l,23 |2x + y - , Dang 2: OXIT KIM LOAI TAG DUNG VOI AXIT HQ, H2SO4 [bang) fx = 0,015 Phi/dng trinh tdng quat [ y = 0,01 MzOn + 2nHCl %Cu=M15:6i.78,05% 1,23 M20„ + nH2S04 Chon dap an B 2MCln+ nHzO (loang) ^ dich Y gom H N O va H2SO4 dSc thu diioc 0,1 mol moi k h i SO2, NO, NO2, N2O Phan tr5m khoi Itfong cua A l va Mg X Ian lacft la * Doi vdi axit H2SO4 (loang) Khoi li/ong muoi thu dugc la: A 63% va 37% B 36% va 64% Doi vdi a x i t H C l C 50% va 50% D 46% va 54% Khoi luong muoi thu duoc la: HU&ng dan gidi HMg = X mol; M2(S04)„ + nHzO Cach tmh nhanh cho trie nghi?m B a i 14: Hoa tan 15 gam hon hop X gom hai k i m loai Mg va A l vao dung Dat 0,8 2N*^ dich HNO3 dSc, dtr t h i thu duoc 0,896 l i t NO2 nhat (d dktc) = M M = 0,06mol (Dat Cu X mol, A l y mol) 22,4 Cu 0,1 Theo dinh luat bao toan electron: B 78,05% 2N*^ + x 4e => Tdng so mol e nhan bkng 1,4 mol = 0,2.22,4 = 4,48 l i t A 21,95% + 3e -> N*=^ DAI x 2x => Tong so mol e nhudng b i n g (2x + 3y) + n,[^so, -80 • m,„„a-i ciorua = n i h o n hop oxit kim loai + 27,5 n,,^;, BAl TAP VAN DgNG = y mol Ta c6: Qud t r i n h oxi ho^: Mg -> Mg^* + 2e m^^ai ^unfat = m h S n hap o x u k i m loai Al y A l ' " + 3e 3y B a i 1: Cho 50g hon hop hot oxit kim loai gom ZnO, FeO, Fe203, Fe304, MgO tac dung het vdi 200ml dung dich H C l M (vCra du) thu duoc dung dich X Luong muoi c6 dung dich X bSng.A 79,2g B 78,4g C 72g D 72,9g HUcfng ddn Ap dung gidi C O n g t h u f C : nimudi cloma = nihon h A 87,5 m l B.125 So hop thiJc: 2Fe ^ C.62,5 m l D.175 m l gidi - chat rin la Fe203 = Y | ^ = 0,01875 mol So hop thufc: => 0,13mol ml Hii&ng dan nFe,03 >H20 => mpe = 7,68 - 0,13.16 = 5,6 gam => ^fe = = = 12 - 0,12.16 = 10,08 gam B a i 5: Hoa tan hoan toan 2,8 gam hon hop FeO, Fe203 va Fe304 can vCra du V ml dung dich HCl I M , thu di/oc dung dich X Cho tii tii dung dich NaOH dir vao dung dich X thu dugfc ket tua Y Nung Y khong den khoi liiOng khong doi thu dirge gam chat r ^ n Tinh V? gidi > muoi clorua + H O O'- H2O ->0,12mol = 0,26.1 =0,26 mol + > Chon dap an A = 2,81 + (0,3.0,1.80) = 5,21 gam Hii&ng dan n„^ = mo,ox,t) 2Fe Fe203 0,0375 l"iol FeaOg 0,1 -> 0,05 mol => mp^^o^ = 160.0,05 = gam Chon dap an C B a i 4: Oxi hoa cham m gam Fe ngoai khong k h i 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 I M , dong thdi giai phdng 0,672 Ht (dktc) Tinh m? A.10,08 B.8,96 C.9,84 D.10,64 2H^ + 0,0875 x = 0,03 mol B 54,0 ^ Fe^* HiC&ng dan 0,015 0,03 0,3 mol Ag" gidi 0,01 A 64,8 + A 12,8g; 32g Theo bai A l diX: 2A1 + 6HC1 -> 2AICI3 + 3H2 3x 3Ag Sau cac phan ilng chat ran la Ag c6 so mol 0,3 + 0,2 + 0,05 = 0,55 mol Chon dap an A X Fe mol Ag^ + le + 0,3 Fe'" Al^^ + 3e Al'^ Dir Ag^= 0,25 - 0,2 = 0,05 mol 0,1 mol Hi^&ng dan ^ vdi Fe Khoi luong k i m loai lai la khoi ItfOng cua Cu: 0,1.64 = 6,4g Al 3Ag* gidi Sau phan ufng vdi A l , Ag* c6n 0,55 - 0,3 = 0,25 mol dung phan ufng 0,1 mol ncucondu = 0,1 + 0,1 > 2Fe^" + Zn^^ 2Fe^^ + Cu 0,2 Al t h i so mol cua Cu la 2x = 19,3 mol ; n c u = 0,2 mol) mol 2Fe'^ + Zn 0,2 Hii&ng dan gidi D 32,4 Zn + CUSO4 2,5x 2,5x gidi = '^MF.SO, ZnS04 + C u i 2,5x 2,5x + Cui Fe + CUSO4 -^FeS04 X + e 1- T a c d u n g v6i p h i k i m : T h i du: 4Na + - ^ N a 2Na + C ] ^ N a C l T a c d u n g v d i a x i t ( H C l , H2SO4 l o a n g ) : tao muoi gam T h i du: 2Na + 2HC1 ^ N a C l + H t T a c d u n g v d i n\i6c: tao dung dich k i e m va H2 T h i du: N a + 2H2O N a O H + Hgt H2 C DANG T O A N CO2 (HOAC SO2) VAO DUNG DICH KIEM NaOH (HOAC KOH) I I I Dieu che: Nguyen tSc: khuf ion kim loai kiem nguyen tur C a c pht^ofng trinh xay ra: Phifcfng phap: dien phan nong chay muoi halogen hoSc hidroxit cua chung Thi du: dieu che Na b^ng each dien phan n6ng chay NaCl PTDP: B MQT SO HOP CUA CO2 + O H - ^ KIM LOAI CO2 + OH" ^ KIEM: + Tac dung vdi axit: tao muoi va niTdc 2 CO2 + NaOH ^ NaHC03 0,1 moi -> 0,1 moi ^ii^ico, = 0.1-84 = 8,4 gam + Tac dung v d i axit: + Tac dung vdi dung dich bazd: =2 B a i 1: Sue 2,24 l i t k h i CO2 vao 100ml dung dich NaOH I M , t i n h khoi lUdng muoi thu duac T i n h litofng tinh: Thi du: • NaHCOa HUdng 2NaHC03 — ^ Na2C03 + H2O n Natri hidrocacbonat - NaHCOg T h i du: CO3'- + H2O San phani + Tac dung vdi dung dich muoi: T h i du: HCO3- ^NaOH + Tac dung vdfi oxit axit: tao mtioi va niidc T h i du: (2) Ta can lap t le I Natri hidroxit - NaOH T h i du: (1) Thiic chat ta dung hai pt sau ) 4Na + 2H2O + O2 CHAT QUAN TRQNG NaHCOa CO2 + 2NaOH ^ Na2C03 + H2O NaOH 2NaCl — ^ ^ ^ ^ ^ 2Na + CI2 4NaOH CO2 + N a O H Hii&ng dan 2NaOH+ SO2 -> 0,4 m 0,2 muoi NazSOg 0,2 = 0,2.126 = 25,2 gam Chon dap an D C.23,0g gidi D.25,2g B a i 3: DSn 10 l i t h n hap k h i gom N2 va C d or dktc sue vao l i t dung dich Ca(OH)2 0,02M t h u difgrc I g k e t t u a T i n h p h a n t r a m theo the t i c h CO2 t r o n g h n hap k h i A.2,24% va 15,68% B 2,24% C 15,68% D 2,24% va 7,84% Hitcfng ddn B a i 5: Cho 6,72 l i t k h i CO2 (dktc) vao 380 m l dd N a O H I M , t h u duac dd " A Cho 100 m l dung dich Ba(0H)2 I M vao dung dich A di/ac m gam giai k e t tua Gia t r i m bang: A 19,7g nco^ = n c , c o =Y^ = B 15,76g 0,01mol C , l g Hiidng = 0,02.2 = 0,04mol nca.oH), m^uo = 0,1.23 + 0,1.39 + 0,1 81 + 0,05.80 = 18,3 gam Chon dap a n D Triforng hofp 1: • Theo d i n h luat bao toan nguyen tuT (tdng khoi lugng ion muoi bang long khoi lugng muoi) "N^OH = M i : ^ 0 % = 2,24% = Hj^^, = n^,,^ = 0,38 dan mol gidi "^n^iom, = n^^^, = , E n ^ j ^ = 0,38 + 0,2 = 0,58 m o l nco, = -^Q CO TrvLUns hc/p 2: • - n^^co^ = 2.0,04 - 0,01 = 0,07mol n^o^ = ^ 0,07.22,4 ^ ^ g ^ CO2 + O H X ^ D.55,16g 22,4 - 0,3 mol mol HCO3- X X CO2 + H - -> C O ' - + H O y Chon dap a n A B a i 4: H a p t h u 3,36 l i t SO2 (dktc) vac 0,5 l i t h n hgp gom N a O H 0,2M Ta c6: va K O H 0,2M Co can dung dich sau p h a n ufng t h u duc/c k h o i I'Jcfng muoi k h a n l a A 9,5gam B 13,5g HU&ng dan nwaOH nKOH = n ^ , = n^j,, = 0,5.0,2 = 0,1 m o l SO2 + H - - > X HSO3 D 18,3g y x + y = 0,3 fx = 0,02 x + 2y = 0,58 [ y = 0,28 B a ' " + CO^- BaCOsi 0,1 0,1 0,28 K h o i lirang k e t t u a l a : m B ^ c O j = 0,1.197 = 19,7 gam Chon dap a n A B a i 6: D o t chay hoan toan 0,1 m o l etan r o i hap t h u t o a n bo san p h a m chay vao b i n h chufa 300 m l dd N a O H I M K h o i liiong m u o i t h u difgrc sau phan iJng? A 8,4g va 10,6g B 84g va 106g C 0,84g va l , g D 4,2g va 5,3g X Hiidng SO2 + H - - ^ S0^-+ H O y gidi = n^^ = n^„_ - 0,5.0,2 = 0,1 mol Z n ^ ^ = 0,1 + 0,1 = 0,2 m o l X C 12,6g 2y 2y y Taco: C2H6 ^ 2CO2 0,1 - > 0,2mol x + y = 0,15 fx = 0,1 " x + 2y = 0,2 |y = 0,05 CO2 + N a O H -> NaHCOa X X X ddn gidi b) Tdc dung vai axit COa + N a O H ^ NazCOg + H2O y 2y y Ta c6: < fx = 0,1 => < x + 2y = 0,3 ,.y = , l V4y -mN,HC03 H2 t\i rx + y - , = 0.1-84 = 8,4 M + 2H^ gam M^* + H2t - D o i v d i axit c6 t i n h oxi hod m a n h nhiT HNOs, H2SO4 gam -> N"'(N02), N*^(NO), N-^(NH4N03) Chon dap a n A S^*^ ^ S^'(S02), S-=^(H2S), S°(S) D DANG TOAN CO2 (HOAC SO2) VAO DUNG DjCH Ca(0H)2 h o g c Ba(0H)2 KIM LOAI PHAN NHOM CHINH NHOM II TAT MSO4 + H z t M + H2SO4 mNa,co, - 0,1.106 = 10,6 T6M dung d i c h a x i t ( H C l , H2SO4 loang) t h a n h _ D e d a n g khilf ion LITHUYET M + IOHNO3 c) Tdc dung vai T r o n g H2O, 4M(N03)2 + NH4NO3 + 3H2O H2O B e k h o n g p h a n urng, M g khuf c h a m , c a c k i m loai l a i khuf m a n h Vj tri bang thong t u a n hoan, tinh chat v^t If M + 2H2O ^ a) Vi tri M(0H)2 + Ca(0H)2 + C a + 2H2O K i m loai p h a n n h o m I I gom: B e r i (Be); Magie (Mg); Canxi (Ca); T r o n g cac chu k i cac nguyen to diing l i e n sau k h i loai k i e m b) Tinh chdt vat li - M g day c a c k i m loai hoat dong y e u hcfn r a k h o i dung d i c h muoi M g + CUSO4 ^ - C a c k i m loai l a i t a c dung vdi H2O dung dich D i e n p h a n n o n g c h a y muoi halogenua cua chiing - La k i m loai m e m (mem hon n h o m ) MX2 - K h o i li/Ong r i e n g tiidng doi nho X : halogen Jig" phan n6ng chay ^ M + X2 hoc Cac nguyen to p h a n n h o m c h i n h n h o m I I c6: - electron hoa t r i (s^) - Co ban k i n h nguyen tuf \6n - L a nhOfng chat khuf m a n h M - 2e -> M^^ T r o n g cac hap chat c^c nguyen to c6 so oxi hoa +2 a) Tdc dung vai phi kim M ( M la nguyen tut k i m loai) 2CaO C a n x i oxit: CaO C a x i oxit l a oxit b a z a - T a c dung m a n h l i e t v d i H2O tao b a z a C a O + H2O ^ Ca(0H)2 C a O + 2HC1 ^ CaCl2 + H2O - T a c dung vdi oxit a x i t tao muoi tuang iJng C a O + CO2 - V d i CI2: M + CI2 -> MCI2 M g + CI2 ^ MQT SO HOP CHAT QUAN TRONG CUA CANXI - T a c d u n g vdi n h i e u a x i t tao muoi tuang ufng - V d i oxi k h i dot nong: 2Ca + ^ MgS04 + C u i D i e u che - N h i e t n o n g chay n h i e t soi thap M + O2 H2T d) Tdc dung vai dung dich muoi S t r o n t i (Sr); B a r i (Ba) va Radi (Ra) T i n h chat hoa H2t MgCl2 CaC03 - C a n x i oxit dirge dieu che bkng phijang phap nhiet p h a n muoi cacbonat C a C — ^ C a O + CO2 Nifofc ciJtng C a n x i h i d r o x i t : C a ( O H ) Nx^oTc cufng L a c h a t r S n it t a n t r o n g H2O Nadc curng la nUdc c6 chura n h i e u ion Ca^*, Mg^* Nifdc k h o n g chiJa hoSc chura i t nhOfng ion t r e n , goi la niTdc m e m D u n g d i c h C a ( H ) c6 t i n h bazO y e u h o n N a O H - T a c dung v d i a x i t v a oxit a x i t tao m u o i tUofng ting Ca(0H)2 + 2HC1 ^ C a ( H ) + CO2 ^ P h a n l o a i nvCdc C a C l a + 2H2O Nude curng chia t h a n h loai C a C O a i + H2Q Nifde ciJng tarn t h d i : la nirdc curng c6 chura ion HCO3" C a ( H ) + 2CO2 ^ C a ( H C ) , CaCOH), N e u t i le m o l CO2 NiXdc curng v i n h cijfu: la niJdc curng c6 chufa ion CI" hoSe SO^ ^ < — tao muoi a x i t • Nifdc curng toan p h a n : L a niidc cufng c6 chiJa dong thcfi cac ion HCO3", C r hoSe SO^- N e u t i le m o l ^^^^^^2 < ^ ^.^Q jjjy5'j t r u n g t i n h CO2 N e u t i le m o l 9^^^k CO2 k h o a n g i < ^^^^^ T a c h a i c i i a nitofc ciifng < tao d6ng - Xa phong k h o n g t a n - V a i soi mau muc n a t t h d i muoi - Nau thurc a n lau c h i n , g i a m m u i v i - T a c dung v d i dung d i c h muoi - Tao chat c&n t r o n g n o i h a i l a m l a n g p h i n h i e n lieu Ca(0H)2 + NaaCOs C a ' ^ + CO^- ^ C a n x i c a c b o n a t cufng CaCOai + N a O H C a c h l a m m e m ni^cfc CaCOai Nguyen tSc: L a m g i a m n o n g cac ion Ca^^ va Mg^* t r o n g nufdc bang each chuyen nhij'ng ion t i i vao t h a n h p h a n chat k h o n g t a n CaCOa C a n x i cacbonat l a c h a t r S n m a u t r a n g k h o n g t a n H2O C a C O s l a muoi c u a a x i t y e u v a k h o n g b e n CaCOa + H C CaCl2 + H2O + C O z t CaCOa + 2CH3COOH -> Ca(CH3COO)2 + H2O + C O a t ot n h i e t t h a p CaCOa t a n d a n H2O c6 CO2 CaCOa + H2O + CO2 -> C a ( H C ) C a n x i s u n f a t : C a S C a S goi l a t h a c h cao, m ^ u t r ^ n g , i t t a n t r o n g H2O C a S H : t h a c h cao song C a S H : t h a c h cao n u n g n h o lufa C a S : t h a c h cao k h a n T h a c h cao n u n g duoc dung de due tiiOng.dung y hoc de bo bot Phirong phap: Phuong phap hoa hoc va phuong phap trao ddi ion a) Phuang phap hod hoc * D o i v d i ntrdc cufng t a m t h d i Dun nong trUdc k h i dung Ca(HC0a)2 — ^ C a C O a i + H2O + C t Loc bo chat k h o n g t a n , dirge nifdc m e m - D u n g Ca(0H)2 v t o du de t r u n g hoa Ca(HC03)2 + Ca(0H)2 C a C a i + 2H2O Loc bo chat k h o n g t a n dtrgc nude m e m * D o i v d i nude cufng v i n h cufu va ni/dc cufng t o a n p h a n Dung dung dich Na2C0a, Na3P04 C a S + Na2C03 -> C a C i + Na2S04 Ca(HC03)2 + Na2C03 ^ CaCOgi + 2NaHC03 Ca?* + CO3 ^ CaC03>l Mg'* + CO3 ^" MgCOa b) Phuang phdp trao doi ion Cho nutdfc cufng di qua chat trao doi ion (ionit) chat se hap thu cac ion Ca^^ va Mg^* the vao la ion N a \a duoc nude mem D^ng bai tap CO2 vao dung dich Ca(OH)2 hoqc Ba(OH)2 thi: Dang 1: Biet nc^o,,,,^ ncaco, "co, TrUang hap 1: rico, = ntettua TrUdng hop 2: 2nb«,„ HCO,= = 0,02 mol - ntettua = 2.0,03 - 0,02 = 0,04 mol Chon dap an A B a i : (DH A - 2008): Hap thu hoan toan 4,48 l i t k h i CO2 (d dktc) vao 500 m l dung dich hon hop gom NaOH 0,1M va Ba(0H)2 0,2M, sinh m gam ket tua Gia t r i cua m la A 19,70 g B 17,73 g C 9,85 g D 11,82 g Hiidng dan gidi ^Ca(0H)2 Dang 2: ^^^'^ T H I : n^^o^ - nc^co^ nNaoii = " N a * = " o H " " ^ ' ^ ^ T H : n^o^ - 2.nc^,o„)^ -''^c^co^ I n^„ = 0,05 + 0,2 =0,25 mol; B i e t n^o^, nc^oH)^ t i m nc^co^ CO2 + OH" ^ X o W ^^CatOlllj CO2 T H I : nco, = nc.co, TH2 : ncacos - •^^cMom, ~ "co^ ^ BAI T A P V A N B 0,02 mol va 0,05 mol C 0,01 mol va 0,03 mol D 0,03 mol va 0,04 mol HiC&ng dan gidi = 0,02 mol C O ' " + H2O y x + y = 0,2 rx = 0,15 " x + 2y = 0,25 [ y = 0,05 0,05 0,05 Chon dap an C A 0,02 mol va 0,04 mol = HCO3" Khoi laong ket tua la: ra^^co, =0,05 197 = 9,85 gam ^CMOMU B a i Hap thu toan bo x mol CO2 vho dung dich chiJa 0,03 mol Ca(0H)2 diroc gam ket tua gia t r i x? nr,ro 2y 0,1 DUNG M | = 0,2 mol Ba'" + CO3'- ^ BaCOai _ , "Ca(OIl)o ~ • Neu n^o, ^ ric^co, =^ n^o, X + 20H- ^ y Dang 3: Biet nc^co^ "00, t i m nCaC03 X "^u^iom, = ^ , = , mol Bai 3: (BH A - 2009): Cho 0,448 l i t k h i CO2 (d dktc) hap thu het vao 100 ml dung dich chuTa hon hop NaOH 0,06M va Ba(0H)2 0,12M, thu duoc m gam ket tua Gia t r i cua m la A 1,182 g B 3,940 g C 1,970 g D 2,364 g Hit&ng dan gidi nNaOH = n^^ = nQ„_ =0,006 mol; I n ^ j j =0,006 + 0,012.2 = 0,03 mol; HB^IOH), nco, = = n^^,, = 0,012mol = 0,02mol C02 + O f f ^ X X X 2y x + y = 0,02 [ y = 0,01 Ba^^ + CO^- BaCOai 0,012 0,01 0,01 A.3,136 l i t B 1,344 l i t C 1,344 l i t hoSc 3,136 l i t D 3,36 l i t hoSc 1,12 l i t Hit6ng K h o i laang k e t t u a l a : mg^co, = , 197 = 1,97 gam Ca(0H)2 + CO2 Chon dap a n C 2,5 l i t dung dich B a ( H ) nong a mol/1, t h u difdc 15,76 gam k e t gidi CaCOsi + H2O Ca(0H)2 + 2CO2 ^ tua Gia t r i cua a l a Ca(HC03)2 0,08 < - 0,04 B 0,048 M C 0,06 M D 0,04 M Ca(HC03)2 — Hiiafng d&n gidi ^ 0,04 2,688 ^ , , „ „ , _ _ ^ = 0,12mol; ng^co, = = 0,08mol nco, > neaco, dan 0,06 < - 0,06 B a i 4: (DH A - 2007;.- H a p t h u hoan t o a n 2,688 l i t k h i CO2 (or dktc) vao = = 2.0,15 - 0,1 = 0,2 m o l B a i 6: T h o i V l i t (dktc) CO2 vao 100 m l dung dich Ca(0H)2 I M , t h u diroc gam k e t tiia Loc bo k e t t u a lay dung dich t h u ducfc dun n o n g l a i c6 4gam k e t t i i a nufa Gia t r i V l a : fx = 0,01 ' x + 2y = 0,03 ^ n^f> ntettua Chon dap a n D y A 0,032 M - Vco, = 0,2.22,4 = 4,48 l i t CO2 + H - - > COa^" + H O y n^^^ = 2nb,,o TriCdng hap 2: HCO3" CaCOai + H2O + C t Vco,= 0,14 22,4 = 3,136 l i t Chon dap a n A chi xay r a t r i / d n g hap B a i 7: Hap thu toan bo 0,3 mol CO2 vao dung dich chijfa 0,25 mol Ca(0H)2 Khoi lifofng dung dich sau phan uTng tang hay giam bao nhieu gam? C„^,„„, = - ^ = 0,04M C G i a m 16,8gam D G i a m 6,8gam gidi Be tinh khoi lixang chat sau phan i2ng tang hay giam B a i 5: D a n V h t CO2 (dkc) vao 300ml dung dich Ca(0H)2 0,5 M Sau p h a n l i n g duac lOg k e t tua Gia t r i cua V bSng: B 3,36 l i t Hildng TriCang hap 1: B T a n g 20gam Hii&ng dan Chon dap a n D A 2,24 l i t A T a n g 13,2gam C 4,48 l i t d&n gidi n^o^ = nkettoa = 0,1 m o l Vco = 0,1 22,4 = 2,24 l i t a lay khoi liigng vao trie khoi lilgng khdi dung dich niu > la tang n^u < la giam Ta c6: n^o^ > n^^^Q^^^^ D Ca A , C deu dung J^COj - 2nbaza ~ nen chi xay TH2 ^kgt tua nc,co, = 2nc„oH,, - ^co, = 2.0,25 - 0,3 = 0,2mol Am = mco, - ixic^co, = 0,3.44 - 0,2.100 = - , gam Chon dap a n D B a i 8: Cho 5,6 h t h o n hdp X gom Ng va CO2 (dktc) d i cham qua K t dung dich Ca(0H)2 0,02M de p h a n ilng xay hoan toan t h u dUdc gam k e t tua T i n h t i k h o i h o i cua h n hcfp X so v d i H2 A 18,8 B 1,88 C 37,6 Hiidng = n^ Truang hap 1: dan = O.lmol n^o^ = gidi 100 =3^ ""'"^ j^>,15.44.0,1.28 ^ 0,25 = 15,6 = 18,8 Chon dap a n A B a i 9: Sue CO2 vao 200 m l h n hgfp dung dich gom K O H I M va Ba(0H)2 0,75M Sau k h i k h i h i hap t h u hoan toan t h a y tao 23,6 gam k e t t u a C 2,24 l i t HU&ng dan ^on- Nguyen hap 1: chuyen hap 2: = 0,2 m o l ; Pt: _ 23,6 -^BaCOj ~ CO2 + O H " -> HCO3" X CO2 + y X 20H- 2y ^ 0 ' - + => X (1) (2) 2YCI3 + 3H2O +3CO2T Theo phuong t r i n h (1),(2) t a t h a y •^muo'i clorua ~ ^ Pt: h h rauoi cacbonat H2O 197 = 0,12mol A C O + H2SO4 -'^cOo B2CO3 + H2SO4 ^ | y = 0,12 "co, =0,12 + 0,26 = 0,38 mol (loang) ^ (loang) ^ A S O + CO2T + H O A S O + CO2T + H 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 VAN DUNG B a i 1: Cho 115g hon hcrp gom A C O , B C O , R C O tac dung het vdi dung dich H C I thay thoat r a 0,4481 CO2 (dktc) K h o i liicfng muoi clorua tao r a dung dich la: A 115,22g B.151,22g C 116,22g HUcfng dan fx = 0,26 Vay Vco^ =0,38.22,4 = 8,512 l i t Chon dap a n D XCO3 + 2HC1 -> X C I + H2O + CO2T Y2(C03)3 + 6HC1 y y = 0,12 " x + 2y = 0,5 chdt 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 tii chat sang chdt khac Ung vdi mol lugng (71-60) M u o i cacbonat tac dung v d i dung dich H S O loang Vco.= 0,12 22,4 = 2,688 l i t Truang D 44,8 tdc cua phuang phdp Id dua vdo sU tang, gidm khoi D Ca A va B dung gidi I n ,' ,O, i, r =0,2 +' 0,15.2 = 0,5 m o l ;' Truang C 224 n c f = 2n^^,_ (ma M^, = 35,5; M^^, = 60) B 2,688 l i t - B.44,8 hoSc 224 Dang MUQI CACBONAT TAG DUNG VO! DUNG D|CH HCI, H S loang T i n h Vco^ da dung d (dktc) - D 16,745g tuf cac b a i t r e n TMng = 0,05 mol hap 2: n^^^ = 2nb^,„ - nkeuaa = 2.0,1 - 0,05 = 0,15 m o l n KOH C 9,85g G i a i tiicfng t i i b a i t r e n A 44,8 hoSc 89,6 Khac dap an l o a i A 8,512 l i t B 14,775g duac 0,2g k e t tua Gia t r i V m l l a : = 0,05mol ^ 0,05.44 + 0,2.28 = 31,2 =^ d , / , , = ^ M = 0,25 ' "'"^ Truang A 23,64g B a i 11: T h o i V m l (dktc) CO2 vao 300 m l dung dich Ca(0H)2 0,02M, t h u n CRCOO nkettua D 21 a i 10: Sue 4,48 h t (dktc) CO2 vao 100ml hon hop dung dich gom K O H I M va Ba(OH)2 0,75M Sau k h i k h i h i hap t h u hoan t o a n t h a y tao m gam k e t tua T i n h m D 161,22g gidi C a c h g i a i 1: A C O + 2HC1 - ACI2 + H^O + CO2 t B C O + 2HC1 > 2BC1 + H2O + CO2 t R C O , + 2HC1 ^ 2RC1 + H.,0 + CO, t • " c o , = nH,o = • rijici = 0,02 (mol); m r S n B = nihon hop muoi + m„,„6i cacbonat + rnnci - ^ rriniuoi clorua = nimuoi cacbonat + n i H C l ~ nimuo'; clorua + ^Hfi '^COj ^^H^O ~ ^ C O ^ = 115 + 0,04.36,5 - 0,02 (18 + 44) = 115,22 (g) Chon dap an A C a c h giai 2: Ap dung cong thiic rtlmuoi clorua = hh muoi cacbonat + I^cOj -C^-^ B a i 2: Hoa tan 5g h6n hop 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 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 KIM LOAI SAT B| 0X1 HOA THANH HON HPP (Fe, FeO, FegOa Fe304) Cdch 1: Vi Fe bi oxi hoa cac exit s^t c6 the s^t dif (Fe, FeO, FeaOs, Fe304) Tom lai ta c6 the coi hon hop chi c6 sat va oxi nen ta c6 CTPT chung la Fe^Oy Fe^Oy - (3x - 2y)e D 22,2g xFe^^ + yO^' Ta luon c6 he phiiong trinh sau gidi 56x + 16y - = nihh muoi cacbonat + n^o.^ (71 - 60) = + 0,05.11 = 5,55 gam Chon dap an B B a i 3: Cho 42 gam hon hap muoi MgCOg, CuCOs, ZnCOg tac dung vdi dung dich H2SO4 loang, thu diTcrc 0,25 mol CO2, dung dich A va chat ran B Co can dung dich A, thu diTdc 38,1 gam muoi khan Dem nung liiong chat rSn B tren cho den khoi liiong khong doi t h i thu duac 0,12 mol CO2 va 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 T A P V A N DyNG Bai 1: (De KA-2008): Cho 11,36 gam mot hon hap gom Fe, FeO, FegOs, Fe304 phan ufng het vdi dung dich HNOg loang da, thu dtfac 1,344 l i t NO (la san pham khuf nhat d dktc) va dung dich X Co can dung dich X thu dufoc m gam muoi khan Gia t r i cua m la A 49,09 gam B 34,36 gam C 35,50 gam D 38,72 gam Hii&ng dan gidi Cdch 1: Ap Dat h5n hop muoi la A CO3 0,25 - hop oxit sat rtir^^ii clorua A CO3+ H2SO4 ^ ^^muei khan ^ihO Chon dap an B 12 n^o = — — = 0,05mol CO, 22,4 Ap dung cong thufc: "^COj ~ Dang c6 rat nhieu each giai khac nhitog qua dai nen chi dUa each giai sau ~ = 115 + 0,02.11 = 115,22 (g) Chon dap an A A l l l g ~ = 42 + 0,25.98 - 0,25.44 - 0,25.18 - 38,1 = 12,9 gam = 2nco^ = 2.0,02 = 0,04 (mol) Ap dung DLBTKL: "^HjSO^ mpe A SO4+ H2O 0,25 + CO2T 0,25mol = dung cong thitc 0,7.mh6n hap oxit s2it + (mol san pha'm khii-so = 0,7.11,36 + ( i ^ , ) = 8,96 gam nhan.5,6) n,,, =0,16mol 56 Fe B a i 3: N u n g n6ng 16,8 gam bot sSt t r o n g k h o n g k h i t h u duoc m gam h n hop X gom bon chat r ^ n Hoa t a n h e t m gam X bang H S O dSc nong diT thoat r a 5,6 l i t S O (dktc) Gia t r i cua m : Fe(N03)3 0,16 -> 0,16 m o l m„u6i = 0,16.242 = 38,72 gam A.22g B.26g Chon dap a n D Hitdng 2: H n hdp chi c6 s^t va oxi nen t a c6 the coi la FexOy ap dung Cdch cong thufc V g " dung cong •56x + 16y = m,.„ y,,^,,;,,^, mpe < 3x - 2y = mol san p h a m khuf.so e n h a n 56x + 16y = 11,36 Ta c6: ' ^ | y = 0,15 Ta CO scf hop thufc: Fe m^uoi Chon dap a n D N u n g m gam hot s^t t r o n g oxi t h u dtfoc gam h o n hop chat r ^ n X Hoa t a n h e t X t r o n g dung dich H N O dU t h a y thoat r a 0,56 l i t (dktc) k h i N O n h a t Gia t r i cua m? A.2,52g B.2,62g C.2,22g Hii&ng dan Cdch 1: Van dung cong mpe = 0,7.m/,6-„ = 0,7 + ( 0,56 22,4 Cdch 2: Ap dung cong (molsdnphdm khit-so c nhdn.5,6) 5,6) = 2,52 gam thiic 56x + 16y = fx = 0,045 3x - 2y = 0,025.3 [ y = 0,03 m p e = 0,045.56 = 2,52 gam oxit sdt + (moljdn pArfm *Ai?-so nhdn.5,6) A.10,08 g va 34,02 g B 10,8 g va 34,02 g C.10,8 g va 40,32 g D 10,08 g va 40,32 g Hiidng mpe = 0,7.mA H2O ( H C l ^ H^ + C D B 0,4M hoac 1,6M C 1,6M hoac 2,8M D 0,4M hoSc 1,2M Hit&ng dan Vg.n dung cong "AMSO,,, DANG MUOI NHQM TAG DUNG VOI BAZQ Lf THUYET A l ( H ) i + 3NaCl AICI3 + 3NaOH A1(0H)3 + N a O H N a A l O j + 2H2O THI: n^„ = 3.n.,,oH, =>C =>CM = ^ ^ ^ ^ = 1,2M 0,025 T H 2: n„„ = CM 4.n,,3 = ^'"'"^ Mf^iOH Q "'"^ = 2,8M Q25 2: Cho 150 m l dung dich K O H 1,2M tac dung v d i 100 m l dung dich 2,34 gam k e t tua Gia t r i cua x la A 1,2 M B 0,8 M - CO2 day duoc goc aluminat r a k h o i muoi CO2 k h o n g hoa t a n diioc A1(0H)3 nen p h a n ling dCrng l a i d k e t tua keo t r S n g TNI: HKOH TN2: H K O H = 0,21 tan + muoi H A K O H ) = 0,03 30H3(a + 0,03) + a A1(0H)3 + 3HC1 ^ A I C I + 3H2O tac dung v d i mol; a + 0,03 Al(OH)3 A l ( H ) i + NaCl giai = 0,18 m o l ; n^^oH), = 0.06 m o l Al^" - Neu suf dung a x i t m a n h day t h i tao k e t tua keo t r ^ n g sau loai b a i t a p baza C 0,9 M Hiicfng dan NaA102 + CO2 + 2H2O ^ A l ( H ) i + NaHC03 (Al^*) AKOlUj bo k e t tua, t h e m t i e p 175 m l dung dich K O H 1,2M vao Y , t h u difoc A l O " + 2H2O y: A1(0H)3 khong tan dUdc dung dich NH3, axit cacbonic nhom n^,,oH) A I C I nong x mol/1, t h u difOc dung dich Y va 4,68 gam ke't tua L o a i Al^^ + H - ^ A1(0H)3 Cach t i n h n h a n h cho - Chon dap a n A Bai NaAlOa + H C l + H2O giai thUc 42 = ^342 = , m o l ; n ^ = 2.0,01 - 0,02mol Al'' PhiiOng t r i n h i o n : A1(0H)3 + H - tac dung v d i 25 m l dung dich N a O H tao r a A 1,2M hoSc 2,8M mx = mcuvi Fe + mo(ox.t) = 17,44 +0,3.16 = 22,24 gam Chon ddp a n A Chu Al2(S04)3 di/oc 0,78 gam k e t t u a T i n h nong m o l cua N a O H da diing 0,3mol mcu va Fe = m^.ai - m^, = 38,74 - 0,6.35,5 = 17,44 gam ^ 1: Cho 3,42gam OH" mol > A1(0H)3 a + 0,03 m o l > A l O - + 2H2O a mol a + 3(a + 0,03) = 0,21 => a = 0,03 0,06 + 0.03 + 0,03 , 04 Chon dap a n A D 1,0 M B a i 3: Trong mot coc dUng 200ml dung dich AICI3 2M Rot vao coc 200ml dung dich NaOH c6 nong a mol/lit, ta duac mot ket tua; Dem say kho va nung den khoi Itfgrng khong ddi diioc , l g chat r ^ n Tinh a A 1,2M hay 3,5M B I M hay 0,5M C 1,5M hay 7,5M B a i 5: Cho 200ml dung dich AICI3 1,5M tdc dung vdi V l i t dung dich NaOH 0,5M, Itfong ket tua thu diTcfc la 15,6 gam Tinh gia t r i Idn nhat cua V A l i t ^M^' - = 0,2mol n^ic, =0,2.1,5 = 0,3 mol chi xay trirdng hop n A i c i , > nAuoH)3 0' 2.2 ^ 0,4mol pt: A l ' " +30H- -> A1(0H)3; n^,^03 D l i t 15 = nAi(0H)3 Vqin dying cong thitc = C 1,5 l i t Hii&ng dan giai D 1,5M hay 0,5M Hiidng dan giai nA.c.3 B 2,5 l i t = 2A1(0H)3 ^ = ^' 0^"^°^ Tacd: AI2O3 0,1 _ =4.n^,3.-nA„0H)3 nxaOH = no„- 0,05 = 4.0,3 - 0,2 = mol T T ^ = lit Chon dap an A VN«OH U,5 B a i 6: Cho V l i t dung dich NaOH 0,3M vao 200 ml dung dich Chon dap C B a i 4: (DHKA- 2007) Tron dung dich chiia a mol AICI3 vdi dung dich chufa b mol NaOH De thu daoc ket tua t h i can cd t i le A a: b = 1: B a: b < 1: C a: b = 1: D a: b > 1: A 0,4 l i t hoSc 1,2 l i t B 0,2 l i t ho^c 1,2 l i t C 0,2 l i t hoac l i t D 0,4 l i t hoSc l i t HUcfng dan giai Al=*^ + 30H- n^,3 =2.0,04 = 0,08mol > AlO- + 2H2O Al'" > A10-+2H,0 a + 40H4 mol De ket tua tan hoan toan t h i pt: A P " + H - ^ 0,06 > A1(0H)3^ A1(0H)3 + OH- THI: "^^2: > ^ - > Vay de c6 ket tua t h i - < => a : b > : a Chon dap ^ n D 0,2.0,2 = 0,04mol nA.,,so,,3= Hiidng dan giai Trpn a mol AICI3 vdi b mol NaOH de thu dtfac ket tua t h i Al2(S04)3 0,2M thu dtroc mot ket tua tr^ng keo Nung ket tua den khoi lugng lUOng khong ddi t h i diigfc l,02g r ^ n Tinh the tich dung dich NaOH da dung 1,02 ^ , n , n = — — = 0,01mol '"^"^ 102 A1(0H)3; 2A1(0H)3 ^ 0,02 0,02 n„„ = 3.n^„o„, => V =^-0,2lit n„H,n«x=4-n^, nAuon,3 nNaOH V = noir = ^-^'^^ ' 0'02 = 0,3 mol -M-iKt Chon dap An C AI2O3 0,01 B a i 7: M o t dung dich h o n hop chufa a mol NaAlOg va a m o l N a O H tac dung v d i m o t dung dich chdfa b m o l H C l Dieu k i e n de t h u ducfc k e t tua sau p h a n uTng la A a = b B a = 2b C b = 5a D a < b < 5a B a i 9: Cho 3,42gam Al2(S04)3 tac dung v d i 25 m l dung d i c h N a O H tac r a dtfOc 0,78 g a m k e t tua N o n g mol cua N a O H da dung l a ? A 1,2M B 2,8M 3,42 nAi,(so,)3 = ^ Phaong t r i n h p h a n ufng: 342 + HCl > N a C l + HzO > A l ( H ) i + NaCl A1(0H)3 + 3HC1 > AICI3 + H O NaAlOa + 4HC1 > AICI3 + N a C l + H O THI: n ^ „ - _ = 3.n,„o„, TH2: a m o l -> 4a m o l Dieu k i e n de k h o n g c6 k e t t u a k h i nnci ^ = 5a 4nN^,o^ + nNaOH HNaOH < H H C I < MN.OH 5a B a i 8: K h i cho 130 m l AICI3 0,1M tac dung v d i 20 m l dung dich N a O H , t h i t h u dLfOc 0,936gam k e t tua T i n h nong mol/1 cua dung dich NaOH A 0,6M hoac 1,95M B 0,6M hoac 1,8M C 1,95M hoSc 1,8M D 1,8M hoSc M B a i 10: T r o n 10,8g hot A l v d i 34,8g hot Fe304 r o i t i e n h a n h p h a n iJng n h i e t n h o m t r o n g dieu k i e n k h o n g c6 k h o n g k h i Hoa t a n hoan t o a n h6n hop r a n sau p h a n ufng bSng dung dich H2SO4 loang (dtf) t h u di/gc 10,752 l i t k h i H2 (dktc) H i e u suat cua p h a n uTng n h i e t n h o m l a A 80% B 90% - n ^ , = 0,13.0,1 = 0,013 m o l ; H^LOH,, - 936 10 n„„ =3.nA,,o„.=>CM A1(UH», "1,1^ = 4.n„ - n „ , „ , O H max C h o n dap a n D Al'* = AUOH), 0,012 m o l 3.0,012 ' =1,8M 0,02 ^ > C, MK.OH 4.0,013-0,012 Q Q2 + 0,4 = C 70% D 60% Hiicfng dan gidi 8A1 thiic p t : Al^* + H - ^ A1(0H)3 TH2: = 1,2M - ^ = 2,8M 0,025 Hiiofng dan gidi OH m i n 3.0,01 Chon dap a n D Chon dap a n D n„„- ^ = n^„ - 4.0,02 - 0,01 = 0,07 m o l HNaOH C, n N a A i o , + nNaOH => a < b < C^_ n^„-_=4.n^,3.-n^H0H,3 Vay suy r a dieu k i e n de c6 k e t t u a : THI: mol 78 NaA102 + H C l + H O nAic, = 0,01 n ^ = 2.0,01 = 0,02 m o l a m o l -> a m o l Van dung cong D 1,2M ho&c 2,8M HU&ng dan gidi HUdng dan gidi NaOH C 1,2 M va M 3Fe304 — ^ 4AI2O3 + 9Fe 0,15 8x 3x 4x 9x (0,4-8x) (0,15-3x) 4x 9x K h i p h a n iJng v d i H2SO4 loang: A l + H^SO, -> A l ' * + H2 ; ^ n n = ^ ? r ^ = 0.48mol 22,4 Theo d i n h l u a t bao toan electron Fe + H^SO, Fe'* + H 2> [...]... 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 T A P V A N DyNG Bai 1: (De KA-2008): Cho 11,36 gam mot hon hap gom Fe, FeO, FegOs, Fe304 phan ufng het vdi dung dich HNOg loang ... bai tap dp dung cong thiJc tinh nhanh Xin tran cam cfn Trink bay bia Ve noi dung cuon sach cd phan: 2013 T CAC D4NG BAI A P D^JNG CACH TiNH NHANH C a c h tinh nhanh so dong phan ciia: - Ancol...CAC DANG O I E N HINH VA BAI PHl/dNG P H A P GIAI NHANH TAP TRAC NGHIEM L C l I N(&I D A U H6AHOCI2 Nguyen T u y e n H a N H A X U A T QUdc DAI H... D 46% va 54% Khoi luong muoi thu duoc la: HU&ng dan gidi HMg = X mol; M2(S04)„ + nHzO Cach tmh nhanh cho trie nghi?m B a i 14: Hoa tan 15 gam hon hop X gom hai k i m loai Mg va A l vao dung Dat