Cẩm nang ôn luyện thi đại học 18 chuyên đề hóa học 5

Cty TNHH MTV D W H Khang V i j t Ca'm nang 6n luygn thi dgi hgc 18 chuy6n dg H6a hgc - Nguygn Van Hai t/' d^i 5= Cho so chuyen hoa giiia cac hgp chat cua crom: Lai gidi: Ggi so'mol X: Fe (x mol); Cr (x mol) va Al (z mol) Theo bai: 56x + 52y + 27z = 8,54 Khi cho X tac dung voi NaOH loang, nong, chi c6 Al phan ung: Al + NaOH+H20—^ C,(OH)3 U t nci {"^ Luu y: Cr khong tac dung vdi dung djch kiem -s: ^ Vi dv 3: Nhan xet nao sau day la sai? A Vat dung lam bang nhom va crom deu ben khong va nuoc B Crom la kim loai cung nhat tat ca cac kim loai C Nhom va crom deu bj thu dpng hoa boi H N O dac, ngugi D Nhom va crom deu phan ung voi axit HCl theo ciing ti If so'mol Lcn gidi: A dung vi cac vat dung lam bang nhom va crom deu c6 lop mang oxit ben viing bao ve B diing vi crom rat ciing, cung bang 8,5 (so voi kim cuong bang 10) • C dung D sai vi A l va Cr tac dyng voi axit HCl va tao cac muoi clorua voi hoa tri khac nhau: Cr + 2HC1 — ^ CrCh + H2t Al + 3HC1 — ^ CrCb + - H t -^DapanD -> T i l | m o l l : Ti If mol 1:3 > v< H:,'/l Vi dv 4: Nguyen to Cr CO Z = 24 Cauhinh electron cuaCr la A [Ar] 4s2 3d^ B [Ar] 3d^4s\ [Ar] Sd^ 4si D [Ar] Sd^ ? f Lai gidi: Le cau hinh electron cua nguyen tu Cr la: [Ar] 3d*4s2 JrfD Y sv Tuy nhien, c6 sv chuyen le tir phan lop 4s sang 3d de tao cau hinh nira bao hoa som, ben han -» Cau hinh thuc te la [ArlSd"* 4s' -> Dap an C "^i 94n B K2Cr04, Cr2(S04)3 C K2Cr04, CrS04 D K2Cr207, Cr2(S04)3 > Cr(OH)3 + 3KBr 2Cr(OH)3 + 3Br2 + lOKOH H * O-M Cr + H2SO4 CrS04 + H2t Tac6: nHj = 0,15 - > x + y = 0,15 x = 0,1;y = 0,05 , ,^ ^ K2Cr04, K2Cr207 CrCh + K O H (I Dap an B Z LMgidi: 0,03 I ^ %mcr - "^^^"^^.100% = 30,44% ) Y CacchatY,Zlanlugtla: NaA102 + - H t Mol: 0,02 V > —>• z = 0,02 mol Khi cho X tac dung voi axit H2SO4 loang, nong: Fe + H2SO4 FeS04 + H2T X f, , , - > 2K2Cr04 + 6KBr + 8H2O 2K2Cr04 + H2SO4 > K2Cr207 + K2SO4 i/iiitixi> i t n ' > Y la K2Cr04 va Z laK2Cr207 Dap an A Vi dv 6: Cho 1,58 gam hon hgp bgt X gom Al, Cr tac dving voi lugng du H2SO4 loang, nong, thu dugc 1,120 lit H2 (dktc) Mgt khac, cho 1,58 gam X vao lugng du dung djch NaOH loang, nong, thu dvigc V lit khi H2 (dktc) Gia tri cua V i a A 1,344 B 0,672 C 0,896 D 1,120 Lmgtat: Ggi so mol X: A l = x; Cr = y Theo bai: 27x + 52y = 1,58 '-M + Khi cho X tac dung voi axit H2SO4 loang, nong: 2A1 + 3H2SO4 Cr + H2SO4 nH2 = 0,05 — ^ Al2(S04)3 + 3H2t CrS04 l,5x + y = 0,05 + H2t / > ' ia«t>W,:!!I)a V = 0,03.22,4 = 0,672 lit-» Dap an B ^kixU^' Luu y: Cr khong tac dung voi dung dich kiem Vi dv 7: Phat bieu nao sai so sanh tinh cha't hoa hgc cua nhom va crom? A Nhom va crom deu khong tan dung djch HNO3 d$c, ngugi B Nhom va crom deu ben khong va nuoc C Nhom va crom deu phan ung voi dung dich axit HCl theo ciing ti 1? ve so'mol D Nhom CO tinh khvr mgnh hon crom Lm gtat: , i^han xet: Al tac d^ing voi axit HCl - > Muoi AlCb Cr tac dvng voi axit H C l - > Muoi CrCk - » Al va Cr phan utig voi HCl theo ti If so mol khac , 241 elm nang 6n luygn thi dgi hpc 18 chuy6n 2A1 + 6HC1 > 2A1C13 + Cr > C r C h + H2 + 2HC1 CtyTNHH MTV DWH Khang Vigt H6a hpc - Nguyin Van Hii f\!han xet: K h i axit HNO3 c6 mat d o n g t h o i v o i axit H2SO4 loang t h i l u p n g H+ 3H2 ' ' - * - > D a p an C ' • ^'^ >U'\ d u n g d i c h la d o axit phan l i —> Giai theo p h u o n g t r i n h i o n j-fr h a Trong X: D O N G V A H O P C H A T Tac dung v&i phi kim 2Cu +02 — ^ 2CuO l a M : ^ ^l^ix:^ 3Cu + 8HNO3(/0flM^) _ t i ^ 3Cu C^Q^ > 3Cu(N03)2 + N O + 4H2O C u + mNCh{dqc,nguoi) + M o l : 0,03 < - fi.' '-^ SO| 8H^ + 2NO3" 0,080,02 2: C h o 12,4 Tinhoxihoa: ' *' >• Cu(Na)2 + A g C u C h + 2FeCl2 CuO + CO + 4H2O n ""^UnAi.OsS^ A +;lT:.">(,"' D a p an B (san p h a m k h u d u t h u dupe 5,35 g a m ket tiia Gia t r j cua a la B.0,95 C.1,05 ,^ '° > C u + CO2 » 5,^ • , Nhan xet: D u n g d i c h Y chtra cac i o n : Fe^^ C u ^ S O4', K h i cho Y + d u n g dich BaCkBa^* + S04' D 1,20 , 23 Bao toan nguyen toS: ng(x)= n^aso^ =—rr 233 = 0,1 m o l Cu(OH)2 + 4NH3 K h i cho Y + d u n g dich NH3 d u : > [Cu(NH3)4](OH)2 + > CuS>l' + H2SO4 Phan ung voi dung dich amoniac CuS04 + 4NH3 > [Cu(NH3)4]S04 Phan ung dien phan: CuS04 + H2O e D o n g (II) nitrat Cu ? + ^02 +H2SO4 ° ; I'l-' C u O + 2NO2+ i , Fe^ + 3NH3 + 3H2O > Fe(OH)3 Cu2* + 2NH3 + 2H2O > Cu(OH)2 Cu(OH)2 + N H V i d^ : Cho 3,2 g a m b p t C u tac d y n g v o i 100ml d u n g d i c h X h o n h p p gon^ HNO3 0,6M va H2SO4 0,1M Sau k h i cac phan u n g xay hoan toan, sinh V l i t k h i N O (san p h a m k h u d u y nhat, dktc) Gia t r i cua V la B 0,448 C 1,792 Laigidi: 3/2 n(7„ = — = 0,05 m o l 64' i i '• * > [Cu(NH3)4](OH)2 D 0,672 • Vnl ' ' Bao toan k h o i l u p n g : m x = n i ^ y + mpg + mg TIQU ( X ) = — = 0,1 m o l 64 Q u i d o i X ve hon h p p cac d o n chat: Fe = 0,05 m o l ; C u = 0,1 m o l ; S = 0,1 m o l (X) = n c u + 3nFe + 6ns = 0,95 -> a = 0,95 mol -» nN02 = (X) = 0,95 mol ^DapanB 3: Thi^c h i f n cac t h i n g h i ^ m sau (6 d i e u kien t h u o n g ) : ;»s r; ,! (a) Cho d o n g k i m loai vao d u n g dich sat (III) clorua (b) Sue k h i h i d r o sunfua vao d u n g dich d o n g (II) sunfat (c) Cho d u n g d i c h b^c n i t r a t vao d u n g d j c h sat (III) clorua '"•'••v''; ^ Bao toan nguyen to Fe: n F e ( x ) = n F e ( O H ) j = ^ =a05 m o l - ^ ^ f t * mcu = 12,4 - 0,05.56 - 0,1.32 = 6,4 gam VI D V M A U A 0,746 ' Den day cac e m l u u y: Cu(OH)2 tan N H s d u t^o phuc chat: {mau xanh tham) - J s ^ va N O , > BaS04>l Tan dung dich N H s tao dung dich mau xanh tham: Cu(N03)2 Jsi ^ BaCl2/ t h u d u p e 23,3 g a m ket tiia; k h i cho toan bp Y tac d\ing v o i A 0,80 3Cu + N2 + 3H2O d Dong (II) sunfat + Phan ung trao doi: CuSOi 242 >• 3Cu2* + 2NO d u y nha't) va d u n g d j c h Y Cho toan bp Y vao m p t l u p n g d u d u n g d j c h d u n g d i c h NH3 c D o n g (II) hidroxit + , Lin giai: a C u O + 2NH3 — ^ + -> , g a m h o n h g p X g o m C u , CuS va FeS tac d\ing het v o i b D o n g (II) oxit + _ iV^^t , - > V = 0,02.22,4 = 0,448 l i t Vi mol HNO3 (d§c nong, d u ) , t h u duac a m o l k h i chi c6 NO2 > Cu(N03)2 + 2NO2 + 2H2O Tac dung v&i dung dich muoi Cu + 2AgN03 C u + 2FeCl3 — > n , , ^ _ = 0,06 m o l ; n ^ o_ = 0,01 m o l p h u o n g t r i n h i o n n i t gpn: ihrimA C u tac d y n g v o i axit HNO3 d^c va loang, axit H2SO4 dac, n o n g : + nHNOg + n H S = 0,06 + 2.0,01 = 0,08 = NO3 a D o n g + r ,,, 243 ca'm nang On luyQn thi djii hpc 18 chuy6n dg H6a hgc - Nguyin Van Hi\ Cty TNHH MTV DWH Khang Vi (d) Cho bpt l u u huynh vao thiiy ngan Oft So thi nghi^m xay phan ung la A l,>fi o ,jl B.l C.4 •;?4-^,o,nf^- ^•2- Lcrigidi: H2S + C u s a — > CuSvl 3AgN03+ FeCl3 Hg + S —>DapanC + H2sa > SAgClJ' >• HgS vJ> filfHo :^ J chat ran Hi?u sua't aia phan ung nhif t phan la B 60% C 80% Loigidi: ncu,NO3„-i^-0.1mol r,-^^^ j ^ - Phuang trinh hoa hpc: Cu(N03)2 — > Mol: 2x -> x = 0,08 ^ ' mpt thai gian thu dupfc 6,16 gam chat ran va hon hg-p X Map thy hoan toan X vao nuac de dugc 600ntU dung dich Y Dung dich Y c6 p H bang B.2 C.3 D.4 Loigidi: 'i NMn xet: K h o i l u g n g chat ran giam = Khoi lug-ng k h i bay Mol: a mN02 2a 2N*5 + ge + ge > N20'~"^^ > NH4NC)3 (amol) M|«, i a = 0,025 mol = 35,6 + (0,55 + 8.0,025).62 + 0,025.80 = 84,1 gam —>DapanD OX^^-^M: l iw:- V i d\ 7: Cho m gam bot sSt vao dung dich hon hgp gom 0,15 mol CuS04 va 0,2 mol H C l Sau cac phan ling xay hoan toan, thu dugc 0,725m gam hon hgp kim loai Gia tri cua m la A 16,0 B 18,0 C 16,8 D 11,2 Loigidi: I ' « Nhan xet: Sau cac phan ling hoan toan thu dugc hon hgp kim logi -> Fe Fe ' 46.2a + 32.0,5a = 3,24 > 4HN03 " H " ^ ^'06 mol + CuS04 Mol: 0,15 aSa "^02 = 9,48 - 6,16 = 3,24 Nhan tha'y: n^+ = nHNOa ° "NO2 244 2N*s V' Cac phan ung hoa hgc: CuO + 2NO2 + ^, ^ 4N02 + O2 + 2H2O > NO; n { m u i ) = n g t r o o d i = 3nivjo + 8nN20"'' 8nNH4NO3~0'55'^^^Bao toan nguyen to N : nHNOg = " ^ + " N O + n^^O + 2nNH4N03 d u Phuong trinh hoa hpc: Cu(N03)2 — ^ N^"^ + 3e V man xet: Bai toan da "giau d i " san pham NH4NO3! Dap an C V i d v 5: Nung 9,40 gam Cu(N03)2 binh kin khong chiia khong khi, sau A.I mol > Muoinitrat + NO + N2O 0,5x H = ^ ^ 0 % = 80% 0,1 D 84,1 ^ ' Bao toan khoi lugng: m = mzn, Cu, Ag + ^^Q-^ + mNH4N03 m N + m o = 18,8-10,16 = , 4 x + 32.0,5x = 8,64 ^ ^ -'^iti& + 5?': ^ 0,95.1 = 0,55 + 8a + 0,05 + 2.0,05 + 2a ^ CuO + 2NO2 + - O X Khichokimloai + HNOs: «,0 Nhan xet: Khoi lu^ng chat ran giam = Khoi lugng bay ' Zn,Cu,Ag va CO the xay qua trinh: -^.O* C 86,3 qua: nNo= 0,05 mol; nN20= Chat oxi hoa: D 50% •/nhuiliisl > inp'hr bai nay, trudc het cac em can t i m so mol moi X de thu dupe ket V i d v 4: Nhi?t phan 18,8 gam Cu(N03)2 mot thoi gian, thu dug-c 10,16 gam n A 40% B 82,1 Loigidi: ' ^ i ^ / : , / ^ ^ ( t^^^^ Mruh t»?l! mla A 69,6 r-* £G.O / ,r Dap an A h FeS04 + Cu 0,15 iinj fifed ,(-:':«t; Fe Mol: + 2HC1 • ^ > FeCk + H2 0,2 Tac6:m-0,25.56 + 0,15.64 = 0,725m - > m = 16gam —> Dap an A '['\ 245 dm Cty TIM HH MTV DWH Khang Vigt nang 6n luygn thi d^i hpc 18 chuy§n 6i H6a hgc - NguySn van Hai V i d v 8: H o a tan h o a n toan 0,1 m o l FeS2 t r o n g 200ml d u n g d i c h H N O 41vj san p h a m t h u dugc g o m d u n g d i c h X va m g t chat k h i thoat D u n g dich fits X CO the hoa tan to'i da m g a m C u Biet t r o n g cac qua t r i n h tren, san pham fil k h i i d u y nhat ciia N*^ deu la N O Gia t r i ciia m la Ksl! A , B.6,4 H-; r )M > FeS2 + 8HNO3 f.r, M o l : 0,1 t,i f,£8.4 ijf 0,1 , 0,2 , Fe3* = 0,1 m o l ; = 0,4 m o l ; N O ^ = 0,3 m o l va SO 4" = 0,2 m o l Mol: ; > 3Cu2+ + N O + 4H2O Cu Mol: ' > Cu2* + 2Fe2^ + 2Fe3* 0,05 < - 0,1 j • "" a = 0,01 m o l ; b = 0,01 m o l V i d u 9: C h o 7,84 g a m h o n h o p g o m Fe304 va C u vao d u n g d i c h H2SO4 loang d u Sau k h i cac p h a n u n g xay hoan toan, t h u dugc iri m V nuuth < > Fe(N03)3 + N O + 2H2O 4HN03 8b '-Sftq fiBS) Xj, Theo bai: 56(3a + b ) = 64.3a = 1,44 .,„.^,^„„ " , b Fe f , 2b Fe , Mol: 0,15 < - 0,4 3a l u g n g C u b a m vao Cac p h a n u n g k h i cho Fe vao X: , + 8H^ + 2N03~ >I '-'f';!' Can l u u y Fe tac d u n g v o i Fe(N03)3: Cac p h a n l i n g hoa tan C u : 3Cu ,1 i :K«u:> i.mi >nii*js j^han xet: K h o i l u g n g t h a n h Fe tang t h e m = K h o i l u g n g Fe p h a n u n g - K h o i yioh ^> ,1 nw.:,, > 3Cu(N03)2 + N O + 4H2O 8a 3a Fe ' + , D u n g d i c h X g o m cac i o n : + 8HNO3 3Cu > Fe(N03)3 + N O + 2H2SO4 + 2H2O 0,8 gidi: phan l i n g cua C u v a d u n g d i c h HNO3: j^ol: D 3,2 C.9,6 Laigidi: Phan u n g hoa hpc: Lai , ,\ J * < j »u -^m - > m = 0,03.64 = 1,92 g a m D a p anB d u n g dich X chiia m g a m m u o ' i v a c h a t r a n Y H o a t a n he't Y t r o n g d u n g d i c h H N O loang, B A I T A P O N L U Y E N s i n h 0,448 l i t k h i N O (san p h a m k h u d u y nhat, dktc) Gia t r j ciia m la Bai 1: D a n i u o n g k h i C O d i qua h o n h g p g o m C u O v a F e n u n g nong, A 12,32 B.9,28 C 11,04 D 9,12 • Loi gidi: vao d u n g d i c h Ba(OH)2 d u , t h u d u g c 17,73 g a m ket tua Chat r a n X p h a n ung v o i d u n g djch H N O •f Phan l i n g hoa hQc: Fe304 + 4H2SO4 aii Mol: > Fe2(S04)3 + FeS04 + 4H2O a a a a a ' • 2a M o l : 0,03 + 8HNO3 C 1,120 D 1,344 hoan toan, t h u d u g c k h i N O (san p h a m k h i i d u y nhat) va d u n g d i c h X > 3Cu(N03)2 + N O + 4H2O Muoinih-at + N O > B a C a + H2O ao9 ao9 ' ' - N^flM xet: K h i cho h o n h g p oxit tac d u n g v o i k h i C O t h i : ne t»ao doi = n c o • "CO = " C O = «"° ~^ = 0,18 m o l Bao toan electron: ngiraod6i = 3ni^Q - > n^Q =0,06 m o l _> V N O =1/344 l i t ^ D a p an D v; ;v j ; ^ r.e:^ ' • , nFe= 0,12 m o l ; nHNOs = 0,4.1 = 0,4 m o l Phan u n g hoa hgc: Fe Mol: 0,1 < Fe Mol: , + 4HN03 0,4 + , ,OHH • > Fe(N03)3 -> 2Fe(N03)3 „ + 0,1 NO + 2H2O 0,1 0,06 : "^"^ ^ ^ , Q : ( £ ) >• D 21,60 > C u ( N ) + 2Fe(N03)2 " C u = ^ n F e ( N ) = ^'^^ m o l ^ - > D a p an C - > 3Fe(N03)2 0,02 - > 0,04 2Fe(N03)3 + C u A g N O a 0,2M Sau m g t t h o i gian lay k i m loai ra, r u a sach l a m kho B 2,16 gam > M ' |«J X g o m : Fe(N03)3 = 0,06 m o l ; Fe(N03)2 = 0,06 m o l Bai 10: N h i i n g m g t Fe v a o 100ml d u n g d i c h g o m C u ( N ) 0,5M va A 1,40 g a m > X CO2 + Ba(OH)2 k h u d u y nha't) v a d u n g d i c h chiia m gam m u o i Gia t r i cua m la A 12,1 , D 10,08 Bai 8: D o t 2,8 g a m Fe t r o n g k h o n g k h i , t h u d u g c h o n h g p chat ran X Cho X tac d u n g v o i d u n g d i c h HNO3 loang (du), t h u dugc , d day, cac e m can s u d u n g so phan u n g : Bai 6: Cho 2,24 g a m Fe vao 200ml d u n g d i c h g o m N a N O s 0,3M v a H2SO4 0,6Iy[ tao V m l k h i N O (san p h a m khtr d u y nha't, dktc) v a d u n g dich X O i.) H f«Go! ^9 ^- • «•' ^, ' ,^;te^:0 m = m A g c i + m A g = 0,04.143,5+ 0,01.108 = 6,82 gam , , —>• D a p an D Lim y: Fe^^ c6 the k h u A g * A g ^ BM4: " Bai cac e m se gap k h o khan neu khong t i n h dugc so m o l A l ban dau L u u y rang, phan u n g nhi?t nhom, t h u o n g ap d u n g d i n h luat bao toan khoi lugng, cy the: 248 249 Ca'm nang a n luygn thi Cty T N H H M T V D V V H K h a n g V i j t h p c 18 chuy6n tSJ H a hpc - Nguygn V S n H i ncu = 0,03 + 0,02 = 0,05 mol - > mcu = 0,05.64 = 3,2 gam mcr203 + i ^ A i = m x -> mAi= - , = 5,4 g a m - » 11^1= 0,2 mol Phuong trinh phan ung: Cr203 + 2A1 0,05 Mol: 0,1 0,1 X gom: A l d u = 0,1 mol; C r = 0,1 mol 0,05 t» I Khi X tac dung voi axit: A l -> — H2 va Cr nen n ^ j = 0/25 m o l - > ^ > 2Cr + AkOa '''' M D a p an A = 5,60 lit - > Dap an C mpeaOa + ^AI 1!^ i , , m ^ i = 24,1 - = 8,1 gam Fe203 + 2A1 = 7,84 lit > 2Fe + AI2O3 0,2 0,2 0,1 • • Y (2) ) A F , Cu-2 (3) nFe= Xet su trao d o i electron cac giai doan: A F _^ nenhimng = 3nAi =0,42 > Cu*^ —> nenhu«ng = 2ncu = a > (1) - » (2): O2 + e > 2CH - > nenhan = +3e NQ-> nenh5n Bao toan electron: 0,42 + 2a = 0,6 - 8a + 0,12 Mol: = 3nNo =0/12 + 2Fe3^ 0,05 > 3Cu2- + N O + 4H2O > Cu2* + 2Fe2^ m = 0,05.242 = 12,1 gam -n r ( f / M 41 miUi Bai 9: "SO2 = = = 0'005 mol ' ' ' SO2 thi oxit la FeO Trong m o l FeO ho§c Fe304 deu chiia mol Fe'^^ nen deu c6 kha nang : ,? nhuong mol electron, do: = npe^Oy = 2nso2 Bao toan nguyen to Fe: 0,04 X C O chua: n^^g^ = 0,04; n^+ = 0,08; n^^^, = 0,02; Na* va SO|+ 8H^ + N O ~ 0,05 hoac Fe304 ,^ , > Fe3*+ N O + 2H2O ''^ Nhan xet: Oxit sat FexOy tac dung voi H2SO4 dac, nong —> D a p an A 0,04 :kM > Fe(NC)3)3 "Fe2(S04)3 = Bao toan k h o i l u g n g : 64a + 16b = 8,58 - 0,14.27 = 4,8 + 4H* + NO3- ) Fe(N03)3 - > Dap an A a = 0,03 m o l - > D a p an A * n ^ + = , mol; n ^ ^ _ =0,04 mol < X (Fe,FeaFe203,Fesa) Fe Bao toan electron: 3.0,14 + 2.a = 2b + 0,12 - > a = 0,03; b = 0,18 0,16 «; va ap dung dinh luat bao toan nguyen to'Fe: Q u i d o i Y t h a n h : A l = 0,14 m o l ; C u = a m o l va O = b m o l Fe Dap an C =0'05mol Fe i Bai6: 28 i'i,li,V,- ^len nj-i^ = 0,35 mol nhieu thai gian O day, cac em can s u dung so phan ung: 4no2 = ' ^ Cach 2: 0,04 ~^ Nhan xet: Bai neu dua theo phuong trinh phan ung se rat da: dong va ton = 0,6-8a (2) - > (3): 2^ ^ Bai 8: So d o phan l i n g : Cu-2e l i • n^i = 0,3 mol Khi X tac dung voi axit: A l (1) - > (3): A l - e , _» X gom: A l = 0,1 mol; Fe = 0,2 mol mo2 = m y - m x = 8,58 - 0.14.27 - 64a = 4,8 - 64a Cu ^ Mol: 0,1 Bao toan k h o i l u p n g : 3Cu = phuong tiinh phan ling: C a c h 1: ) ll S T A l , C u (1) Lini y rang, phan ung nhiet nhom, thuong ap dung dinh luat bao toan nNO= - ^ ^ = , mol Mol: • Bai cac em se gap kho khan neu khong tirJi dugc so'mol A l ban dau i H2 BaiS: v'i ni^fir:^ "FexOy = 2-0/005 = 0,01 mol 2FexOy 0,01 > xFe2(S04)3 -> , , ,^ ^ x.0,005 m = 232.0,012 = 2,32 g a m - > Dap an C Bai 10: ' \: ; • V ' : ^ ' % , i, d f^i, ^ihCi ^ -> x.0,005 = 0,015-> X = - > Oxit sat la Fe304 '^AgN03 = 0,02 mol; ncu(N03)2 = > "^ol251 Cty TNHH M i V nVVH Khang Vi^t Ccim nang n luygn thi a^i hgc 18 chuySn dg H6a hpc - Nguyin Van HSi AgNCh phan ling truac roi moi den Cu(N03)2 CAC L i T H I J Y E l €tf BAN €UA HOA HOC Phan ling hoa hpc: Fe Mol: + 2AgN03 a • Fe(N03)2 + 2Ag 0,01 < - 0,02 0,02 Fe + Cu(N03)2 Mol: -> mtsng = 2,16-0,56 = 1,6 gam > Fe(N03)2 + C u a a Theo bai: mFeting = 1,92 gam J H ? t nhan nguyen t u + E)ien tich hat nhan s-: So phan ling: K i m loai + H2SO4 , do: A = Z + N , i ,^ H,£ ? H f i i o ? i-,f'i& ''^ + So hi|u nguyen tu = so'dan vj d i f n tich hat nhan cua nguyen tu, ki hi^u > 2K2Cr04 + 12KC1 + 5H2O la Z • 1% i.>».w*i,/ tA;,l! + K i hi#u nguyen tu du^c bieu dien theo qui uac: z X Trong : X la li hieu nguyen to'hoa hpc; Z la s6'hi?u nguyen tu v a A la so' • , ,20,9 *~ ,gQ,P •! Bai 13: ' ^ 1: > ' ufen'.Cl t - khoi V i du: ^ ^ O * Dong vi + Dong v i la cac nguyen tu c6 cung so proton nhung khac rUiau ve so Cac phan ung hoa hQc: 4HNa 0,4 notron, nen CO so'kho'i A khac > Fe(N03)3 + N O + 2H2O 0,1 LwM 1/: Fe d u = 0,04 mol va tiep tvic phan ling vol Fe(N03)3: + 2Fe(N03)3 0,08 > 3Fe(N03)2 0,12 Sau phan ling: Fe(N03)3 d u = 0,02 mol; Fe(N03)2 = 0,12 mol m = 0,02.242 + 0,12.180 = 26,44 gam ->DapanB IJ m i q f - b : i n h a n b a n g S ',>{w:'f^^f^Aflt,t.y^*ri(^'-t^' 0,03 - > Dap an B Mol: 0,04 MJ'i 'A ^ - i nhan V i dy: Nguyen to'O bao gom tat ca cac nguyen tu c6 di^n tich h^t ' ' Cach 2: Bao toan electron: 3ncrCl3 = 2nQ2 - » n c i j = 0,02 mol Fe ' + Nguyen to' hoa hpc la tap hgip tat ca cac nguyen tu c6 ciing d i f n tich hat , -> Dap an B Mol: 0,1 < - 'i b Nguyen to hoa hpc - D o n g v i Cach 1: D y a theo phuong trinh hoa hpc: + I * Nguyen to'hoa hoc ^ • / , : Fe ; So khoi (A) la tong so'hat proton (P) va tong so'h^t natron (N) ciia h^t nhan , - m = 13,5 + 0,35.98 - 0,35.2 = 47,1 gam 0,02 l,l\A»i/; V/.^'^.'^ + Sokho'i > Muoi sunfat + H2 ^DapanB 2CrCl3 + 3Cl2 + K O H ^ So don vi di?n tich hat nhan (Z) = so proton = so'electron Bao toan khoi lugng cho so tren: Bail2: bang Z+ va so'don vi dien tich hat nhan bang Z 1,6 + 8a = 1,92 - » a = 0,04 mol N/ifl«xet: nH2SO4 = nH2=0'05mol :'P.rv Proton mang di^n tfch 1+, neu hat nhan c6 Z proton thi dien tich hat nhan mtsng = (64-56)a = 8a gam mFe(pir) = 56.(0,01 + a) = 56.0,05 = 2,8 gam -> Dap an C Mol: d e 11 (^hriyen Nhan xet: Tinh oxi hoa A g * > Cu^* - + " Kho'i lugng nguyen tu trung binh , , N g u y e n to clo c6 dong v i tu nhien la (chiem25%) - s „ , C I (chiem 75%) v a 17CI • ^ , , ,> T Nguyen tu khoi trung binh cua d o la: A = 35 x 75 lUU + 37 x 25 lUU _ = 35,5 ^ C a u h i n h electron Thie tu cac miec nang luang Cac electron nguyen t u a trgng thai c a ban Ian lug-t chiem cac miic n5ng luQfng tu thap den cao T h u ty sap xep cac phan lap theo chieu tSng cua nSng lugmg dvtqc xac djnh bang thuc nghi^m v a li thuye't: I s 2s 2p 3s 3p 4s 3d 4p 5s Cau hinh electron nguyen tie Budc 1: Xac djnh so electron cua nguyen tijr, ^^^^ ^ 253 Cty TNHH MTV UWH Khang Vigt nang 6n luy$n thi dji hgc 18 chuy§n dg H6a hgc - Nguygn Van HSi dm v i dv 4: X v a Y la hai n g u y e n to' thupc ciing m p t chu k y , hai n h o m A lien tiep So' Bu-oc 2: D i e n cac electron vao cac p h a n m i i c n a n g l i r g n g theo t h i i t y : I s 2s 2p 3s 3p 4s 3d 4p 5s ••m%%mm*f>''W , p r o t o n cua n g u y e n t u Y nhieu h o n so' p r o t o n ciia n g u y e n t u X T o n g so' hat ''^'^-f Biroc 3: V i e t cau h i n h electron (sap xep theo trat t ^ cac l o p t i r t r o n g ngoai): I s 2s2p 3s3p3d 4s4p4d4f VIDVMAU p r o t o n t r o n g n g u y e n t u X va Y la 33 N h a n xet nao sau day ve X, Y la dung? A D o n chat X la chat k h i d i e u ki§n t h u d n g ' ^ So hat khong mang dien = (82-22)72 = 30 ^ N = 30 ^* voi hidro la b Moi quan h$ giiia a va b la C a = b Lot TO* oe giai: , ~, i , hop cha't voi hidro, R chie'm 82,35% ve khoi lu^ng Nguyen to'R la B Nita C Oxi Lot giai: Latgtat: R c6 hoa tr! hgip chat cua R voi hidro c6 dang R H Ta c6: R 82,35 ^ —=— 17,65 B sai vi tinh kim loai tang dan, am dien giam dan ' 'i^^m-'''' V i dv (B-08): Cong thuc phan t u cua hgp chat tao boi nguyen to R v^' hidro la RHs Trong oxit ma R c6 hoa trjcao nha't thi oxi chie'm 74,07% A L u u huynh B Asen ,, ; C Nita ' C sai vi am dien giam dan, tinh phi kim giam dan *^' ' ' 3p Nguyen t u cua nguyen to' Y cung c6 electron m u c nang lu^ng 3p va CO electron lop ngoai ciing Nguyen tu X va Y c6 so' electron h a n kem A K i m logi va kim loai B Phi kim va kim loai wa^;, C K i m loai va hie'm D Phi kim v a phi kim Lot ' giai: cimg -> L a p ngoai cung ciia Y khong the la lop thii (vi c6 phan Idp 3p thi da dien day Ss^), ma la lop thu -> ca'u hinh lop ngoai ciing 4s' •^m^'mr -.skr*;- X CO m u c nang lugng cao nha't la 3p ^ X c6 it electron h a n Y Theo bai -> -> X la phi kim j, D a p an B V i dy (A-09): Nguyen tu ciia nguyen to X c6 ca'u hinh electron lop ngoai cung la ns^np* Trong hgp chat ciia nguyen to X voi hidro, X chie'm „ ,XT- x r^> R = 14(Nito) Dap a n B \ V khoi luQ-ng Nguyen to R la ' ca'u hinh lap ngoai ciing cua X la 3s23p* (it hon electron so voi Y ) cao nhat bang -> H o a tri hop chat voi hidro bang - » cong thtH 82,35 = — 17,65 ,^ D Tinh kim loai tang dan, ban kinh nguyen tu giam dan Y la kim loai D L u u huynh Oxit cao nha't cua nguyen to R c6 cong thuc tong quat la R2O5 %H C D am dien giam dan, tinh phi kim tang dan ' " '^''^" ^ ' " • Nhan tha'y, nguyen tu Y c6 muc nang lugng 3p va c6 electron lop ngoai y: Khong ap dung cong thiic cho cac nguyen to chu ki va A Photpho ^ la Nguyen to X, Y Ian luot la D afu-,sy B.27% C.60% Lot D 40% giai: X CO ca'u hinh electron lap ngoai cimg la ns^np"* -> X thugc nhom V I A - » Hoa tri cao nha't oxit ciia X bang -> Hoa tri hgp chat voi hidro bang - = - » Cong thuc hgp chat ciia R voi hidro c6 d^ng R H • , ^•• „ • ^,,, 259 Ca'm nang 6n luygn thi dgi hpc 18 chuy6n dg H6a hqc - Nguygn Van Hai d Muoi Muo'i la hgp chat tan nuoc phan li cation kim loai (hoac i amoni) va anion go'c axit, vi du: NH4CI N H t + Cl~ Na^ +HCO3 Neu go'c axit cua muo'i khong c6 hidro c6 kha nang phan li ion H* th; muoi du-Q-c goi la muoi trung hoa Vi du: Na2S04, NaCl, Na2HP03 NaHCOi NaH2P02 • : • ; - ' : - r v ^ " ••;;{••.4- Neu anion go'c axit cua muo'i van c6 hidro c6 kha nang phan li ion thi muoi duoc goi la muoi axit Vi du: NaHS04, NaHCOs, Na2HP04 pH CUA DUNG DICH a Khaini^m pH la dai lugng dac trung cho nong ion cua dung dich Neu bieu dien nong H* duoi dang he thuc: [H*] = 10-'' mol/1 thi a duoc goi la pH cua dung dich Quan he giua [ H ^ va [OH-] dung dich: [H*].[OH-] = 10"" b pH va moi truong dung dich ' Moi truong trung tinh c6 pH = ' sB$ki ' Moi truong axit c6 pH < 7; pH cang nho, dp axit cang Ion ' Moi truong baza c6 pH > 7; pH cang Ian, dp baza cang Ion OS axit tang CtyTNHH MTV DVVH Khang Vi§t jj phan liiig tao chat dien li yeu ^ phan ling tao nuoc NaOH + HCl > NaCl + H2O OH- + H* — > H2O Mg(OH)2 + 2HC1 — M g C l + 2H2O Mg(OH)2 + 2H* — ^ Mg2* + 2H2O + phan ung tao axit yeu - Trung ttnh 'I llf I / H" + CHsCOO> CH3COOH c Phanrnigtao chat + Na2C03 -^2HC1 COf + 2H^ > 2NaCl +C02t +H2O > C t +H2O + CaCOs +2HC1 CaCOs + 2H* + ZnS +2HC1 )• CaCh +C02t +H2O > Ca2* + C02t +H2O > ZnCh +H2St *• ' ' ^">'' ZnS + 2H^ > Zn2^ + H2St * '''' CAC V I D V M A U Vi d\ 1: Theo thuyet A-re-ni-ut, phan ung nao sau day khong phai la phan ling axit-bazo: A NaOH + HCl > NaCl + H2O B Zn(OH)2 + 2HC1 — > ZnCh + 2H2O > Na2Zn02 + 2H2O D NH3 + HCl — ^ NH4CI Ldigidi: Do bazd tang 14 Thang pH cua dung dich PHAN l/NG TRAO DOI ION a Phan ung tao chat ket tiia + Na2S04 + BaCh Ba^^ + X>1~ + 3NaOH + FeCh Fe3* + 30H- > BaS04 4+2NaCl > BaS04i > Fe(OH)3 4' +3NaCl 11 > CH3COOH + NaCl HCl + CHsCOONa C Zn(OH)2 +2NaOH •7 r ,5., , Dap an D Nhan xet: Theo thuyet A-re-ni-ut, axit la chat c6 kha nang phan li H*, baza la nhirng chat c6 kha nang phan li OH- nen phan ling axit-bazo luon tao H2O Theo thuyet A-re-ni-ut, HCl la axit nhung NHs khong phai la baza vi no khong CO kha nang phan li 0H- d\ 2: Cho cac phan ling xay dung dich: (l)ZnS + 2HCl > (3)(NH4)2S +2HC1 ^ > ' (2) K2S + H2SO4 (loang) (4) CaS + H2SO4 (loang) > Cac phan ling c6 phuong trinh ion riit gon: S^" + 2H* ——> H2S t la A (1), (2), (3) B (1), (3) C (2), (3) D (2), (3), (4) > Fe(OH)3 4' 271 CtyTNHH MTV PWH Khang ViQt elm nang On luy$n thi djil hpc 18 chuySn dg H6a hpc - Nguygn van Hai Lai Dap B k h o n g d i i n g v i n h i e u c h a ' t c6 c h i i a O H n h u n g k h o n g p h a i l a b a z o V i d y giai: :-i'-;„ a n C a x i t H2SO4, t r o n g t h a n h p h a n c6 n h o m O H : S ( O H ) n h u n g l a a x i t P.rlrtfuh p P h u o n g t r i n h i o n t h u g o n u n g v i cac p h u o n g t r i n h (1), (4) Ian l u g t la ZnS i > Zn2* + H S t + 2H* k h o n g d i i n g v i t h e o t h u y e ' t A - r e - n i - u t , b a z o p h a i c6 k h a n a n g p h a n l i OH' ,, ' (ZnS la muo'i i t tan, nen k h o n g d i $ n l i t h a n h cac ion) Ca2* + S2+ SO 42CaS04i + H S t n e n n h a ' t t h i e t p h a i c6 n h o m O H t r o n g t h a n h p h a n p h a n t u 5: T r p n 100ml d u n g d i c h X g o m H2SO4 0,05M v a H C l 0,1M v o i 100ml dung dich Y g o m N a O H 0,2M v a Ba(OH)2 0,1M, t h u d u ( ? c d u n g d i c h Z D u n g d i c h Z c6 p H l a V i d u 3: C h o m g a m h o n h g p g o m N a va Ba (ti 1^ m o l 1:1) tac d u n g v o i nuoc A 13,0 B.1,2 (du), t h u duQC d u n g d i c h X v a 3,36 lit H2 (dktc) C h o 200ml d u n g dich Lot Al2(S04)3 0,2M v a o X, s a u k h i p h a n u n g h o a n toan, t h u d u g c b g a m ket tua B.6,24 C 24,86 ^ Lai + H2O i • I a ''5'- giai: Trong X: ^ }':;)HS + ,r:(:; a •— am a AM, " O H - = " N a O H + 2nBa(OH)2 = + 2.0,1 = 0,3 m o l n„ 2+ = O,lmol; Ijiifr:); M o l : 0,08 Al(OH)3 M o l : 0,06 , +OH- « ~ ~ - ' D H S + A I O fo v - ' i ' l - » p H = 13 -> D a p an A Lai 0,08 + 2H2O A •.nr.-/' 0,06 b = 0,02.78+ 0,1.233 = 24,86 g a m - > Dap an C V i d\f 4: Theo t h u y e t A-re-ni-ut, nhan d i n h nao sau day la diing? A M p t hgp cha't t r o n g t h a n h p h a n phan t i i c6 H la axit B MQt hpfp chat t r o n g p h a n phan t u c6 n h o m O H la bazo ^.^^ ' T o n g s o ' m o l d i ^ n t i c h d u o n g = T n g so'mol d i e n t i c h a m 0,1.3 + 0,1.2 = 0,2.2 + X.1 - > , X =0,1 •J'";' T o n g k h o i l u g n g cac muo'i = tong k h o i l u g n g cac i o n , nen: K h o i l u g n g m u o l k h a n = 0,3.27 + 0,2.64 + 0,2.96 + 0,1.35,5 = 31,85 " ' E)ap an B ^» d v 7: Can bao n h i e u g a m N a O H de pha che d u g c 500ml d u n g d j c h N a O H C M o t h o p cha't c6 kha nang p h a n l i cation H"^ la axit c p H = 13? D M p t bazo k h o n g nha't thiet p h a i c6 n h o m O H t r o n g t h a n h p h a n p h a n t u A 0,8 B.1,2 C.1,6 D.2,0 Laigidi: Latgtat: D a p an C D u n g d i c h N a O H c6 p H = 13 ^ A k h o n g d i i n g v i nhieu cha't c6 chua H n h u n g k h o n g p h a i la axit (CH-t; ( L u u y : t i c h so i o n [H*].[OH-] = lO"'") NaOH,C6H6 i^r ' Cl~ C o can d u n g d i c h se t h u d u g c bao n h i e u g a m m u o i khan? ^^'t«f>^ < 0,1 0,24 / [OH'] -> BaS04>l' + 30H- — = 10 " [H* = Phuang trinh ion: 0,1 > H2O -> n ^ ^ d u = a04 - 0,02 = a m o l -)• [OH-] = rfx ; ! ,, Na Ba^^ + SO^- = " H C l + n H S = 0,01 + 2.0,005 = a02 m o l ^^ ^ ^ j ^ ^^^^ CI SO4 Phuang trinh ion: H * + O H " n^, + = 0,1 m o l Ba AP* gtat: + • ••••'•'iiS r ,, 1,5a = 0,15 - » a = 0,1 m o l n ^ j =0,15 m o l M o l : 0,1 D.12,8 fn = ttNaOH + 2nBa(OH)2 = ^'^^ + 2.0,015 = 0,04 mol T r o n g Y: \, [ng^2+ = a m o l ; n^^+ = a m o l 0,5a > Ba(OH)2 + H Mol: a T r o n g X: D 9,34 )4! NaOH + - H 2 —^ + 2H2O Ba , i o n D o b a i t o a n c h i h o i v e p H -> c h i q u a n t a m d e n H"^ v a O H " A 29,54 Mol: a C 1,0 j^han xet: Bai n a y c h i i a n h i e u c h a ' t d i f n l i n e n g i a i d u a t h e o p h u a n g t r i n h Gia tri c u a m l a Na •" [H^] = 10-» [ O H " ] = 10-' = 0,1M ; 273 dm Vay: , Cty TNHH MTV DWH Khang Vigt nang On luy^n thi dgi hpc 18 chuy6n ei H6a hpc - NguySn Van Hil riNaOH - > iTiNaOH Vi p^i 6: Trpn 100ml dung dich (gom Ba(OH)2 0,1M va N a O H 0,1M) voi 400ml = n ^ ^ - = 0,5.0,1 = 0,05 mol dung djch (gom H2SO4 0,0375M va H C l 0,0125M), thu dupe dung djch X = 0/04.40 = 2,0 g a m - > Dap an D Gia tri p H ciia dung dich X la 8: C h o 0,448 h't C O (dktc) hap thy het vao 100 ml dung djch Y ch hon hqp N a O H 0,1M va Ba(OH)2 0,1M, thu dug-c m gam ket tua Gia tri A 1,182 * B 3,940 C 1,970 A , Trong Y : nnfj n„ 2+= 0/01 mol; Ba^ ^ N h a n thay: — = "CO2 A.l fffo m = 0,01.197 = 1,970 gam - » Dap an C A -11 > Bai 9: Trpn 100ml dung djch gom H2SO4 0,05M va H C l 0,1M voi 100ml dung dung dich X la " c o - " " O H - • " C O = 0/03 - 0,02 = 0,01 mol .r, ^ r ^, ^ ^^j;,^ B.2 A 13,0 A p dyng cong thuc tinh nhanh so'mol C O " : Bai D.6 Ba(OH)2 0,08M va K O H 0,04M Gia trj p H ciia dung djch thu dupe la sS s— = 1,5 > -> Phan ung tao muoi: Cacbonat va 0,02 Nh$n thay: n ^ ^ j - = r\^^2+ ^ , Bai 8: C h o 40ml dung djch H C l 0,75M vao 160ml dung dich chua dong thai = 0/01 + 2.0,01 = 0,03 mol hidrocacbonat as ^ C moi t r u o n g b a z o l a D 2,364 n^,+ = 0,01 mol Na /lEwp B.2 g( p^i 7: C h o dung djch: Na2C03, NaHS04, Na3P04, N a O H So dung djch c6 Lcngiai: "oH-""NaOH+2nBa(OH)2 ; ' A (2), (4) B.(2),(3) ^ C.(2),(l) D (4), (2) Bai 13: C h o dung dich K O H d u vao 100ml dung djch chua hon hpp Ba(HC03)2 0,1M va B a C h 0,2M, thu dupe m gam ket tua G i a trj ciia m la A 1,97 B.3,94 f C.5,91 , D 7,88 Bai 14: C h o 100ml dung djch X gom Na2C03 0,1M va N a H C O s 0,1M Nho tu tu tung gipt den het 25ml dung dich H C l I M vao X, thu dupe V m l C O A.K.= ^ 1-a B.K,= ^ 1-a C.K.= - ^ D K = C(l-a) " C(l-a) Bai 5: Coc X dung 500ml dung djch H2SO4 0,05M (loang) C h o 2,3 gam Na ki'^^ loai vao X (coi the tich dung djch khong thay doi) G i a tri p H ciia dung di*^'^ coc X sau phan ung la: A 10 274 B 11 , C 12 D 13 bay dktc G i a tri ciia V la A.448 B 112 a ,; ,t C.224 ;u „ D 336 Bai 15: C h i dung nuoc va dung dich nao sau day de phan biet chat bpt mau trSng: C a C h , CaS04.2H20, N a N ; C a C ? A H2SO4 loang •• B H C l loang C N a O H loang • ' : • • D Ba(OH)2 k)ang \ '; 275 Ca'm nangflnluygn thi d j i hpc 18 chuySn Cty H6a hpc - Nguyin Van HSI Bai 16: T r p n 100ml d u n g d j c h X g o m Ba(OH)2 0,1M va N a O H 0,2M vol lOGrm d u n g d i c h Y g o m MgS04 0,2M va H2SO4 0,1M, t h u d u p e a g a m ke't tiia Gj^ tricuaala i,^;, j - i i ; ' ; ' r v ' , A 3,49 ^/v;, B.2,91 ',rv'- C 5,82 i , Na2S04 + H2O ,«;(«t>l n Nhan xet: so m o l N a d u = 0,10-0,05 = 0,05 m o l nen xay tiep p h a n u n g : Na + H2O > N Naa O H 0,05 -> p H = - l + ^,,,^,ni iH2t _ 0,05 " O H - = " N a O H = 0,05 m o l A mh [OH-] = ^ = 13DapanD 0,1 -> p O H = _ Bai + AP* + 2SO^- + I2H2O Ca^* + H - C(l-a) ^^^^ nH2S04 = 0,5.0,05 = 0,025 m o l ; n^g = 0,1 m o l Mol: P h u o n g trinh d i ? n l i cua cac chat: KA1(S04)2.12H20 [CH3COOH] - ^Qf;,, :, aC = [ C H C O O - ] [ H - ] ^ C o u C a ^ C o l _^ tfinOOrttC)t * : Bail H* •.^.(•f.;- bang n h a u Gia t r j cua x la C.0,02 + < = > CH3COO" C k h i cho 140ml d u n g d i c h N a O H I M vao Y , deu t h u d u p e l u p n g ket tiia B.0,04 .fei^jfti 'ii Bandau: mol: Bai 23: Y la d u n g d j c h chua x m o l A l C b K h i cho 60ml d u n g d i c h N a O H I M va A 0,03 D a p an C y-' • tuik, b j - , Kc-i ? CH3COOH l u p n g ket tua bang n h a u Gia t r i eua X la A 0,3 d p d i ^ n h tang ; Bai D 0,05 Bai 22: A la d u n g d i c h Z n ( N ) x m o l / I K h i cho 30ml d u n g d i c h K O H M va Ca(OH)2 ^ :U,0 CH3C00" t>apanB Bai 17: Cho 2,4 g a m M g vao 200ml d u n g d i c h H2SO4 0,75M t h u d u p e d u n g dich X Cho l i t d u n g d j c h Y chua Ba(OH)2 0,06M va N a O H 0,1M vao d u n g dich TNHH Nhan xet: Bai toan cac e m can t i m so m o l O H " va so m o l H * t r o n g m o i + H* H d u n g djch 27/ Cty TNHH MTV DWH Khang Vlft Ca'm nang On luygn i h i dai hpc 18 chuyeii de H6a hpc - IMguyin Van Hi'i " O H - " 2nBa(OH)2 + "^NaOH = 2.0,1.0,1 + 0,1.0,1 = 0,03 mol = nH2S04 + " H C I = 2.0,2.0,05 + 0,2.ai = nH2S04 + " H C l = 2.0,4.0,0375 + 0,4.0,0125 = 0,035 mol Tacophuangtrinh: OHMol: 0,03 + ^'^ — > = nBa(OH)2 " ^'^^ H20 Ta CO phuong trinh: Vay n + (du) = 0,035-0,03 = 0,005 -> [H^] = ° ' ° " ^ =0,01 " 0,1 + 0,4 Dap a n B .iMtiv,'Ki'jA::f;^'vvv; Mol: 0,04 ^ vay : n ^ „ _ 'XIA C O 3-" + H2O < — ^ H C O Na3P04 _» n Ta ^ i r»V«J> 9m, CO ( d u ) = 0,6a - 0,04 f > H2O -ii 0,03 ^ ,jfj., Vay n ^ ^ ( d u ) = 0,032-0,03 = 0,002-> [OH"] = ^ ->pOH =2 ^ = 0,01'^' pH = 12 ^ Dap an B " : 0,6a - > H2O ' A XI [OH"] = , i , i 0,02 " ' " ^ = 0,01 M 0,1 + 0, 0,1 p O H = - l g [ , l ] = l -> p H = - p O H = - > D a p a n A 278 = a05 -> a ' ^ = ai5 ^ Dap an A., , ^^.^^^ ^'^'^ ' gj^j > H2O + C O " > Q , ^^Qy., , JiOilrfBsliorf BaCOsi i.l%iiliM U'h Bai 12 Truac het tim dung djch so' (2) v i dugic nh^c den nhieu nha't: (2) tac dung voi (1), (3), (4) deu tao ket tua -> (2) la dung dich B a C b Mat khac, (1) tac dung voi (3) c6 sui bgt ^ (1) v a (3) la cap chat Na2C03 Na2C03 + H2SO4 > N a S a + H2O + C O a t , ,n ; - -> (4) la dung djch Na2S04 , " ^ , « • Y J: I ^nc f; , , """^ nBa(HCO3)2 = 0'l-0'l=0'0^niol Ta CO p h u a n g t r i n h : (du) = 0,04-0,02 = 0,02 0,04 Bail3 " O H - " ^"Ba(OH)2 + "NaOH = 2.0,1.0,1 + 0,1.0,2 = 0,04 m o l I ^ Vay n °" Dap an A " H + = nH2S04 + " H C i = 2.0,1.0,05 + 0,1.0,01 = 0,02 m o l + H^ ^ 0,02 ' O s, / va H2SO4: Bai OHMol: 0,02 IJl' Vay a dap an D vira tao chat di|n li yeu, v u a tao chat ket tiia Ta CO p h u o n g t r i n h : -» ^^'"^^orii," [H^] = IQ-i^ -> [OH-] = 10"' = 0,1 Ba2- + C O ^ - " O H - " ^ "Ba(OH)2 + " N a O H = 2.0,16.0,08 + 0,16.0,04 = 0,032 m o l 0,03 ~ '' = 0,5.0,1 = 0,05 m o l " H C l = 0'4.0,75 = 0,03 m o l Mol: ' )HJJ/ i < Lo^iDvi: H C O ; + O H " ' H* 0,04 * O H " d u , H * het ' * >H20 " " Lo^i C v i Ba^"^ ket hgp duQfC voi S O ] ' tao BaS04 ket tiia • + BaS04>l -> ,' ^ 0,06 > Mg(OH)2>l' 0,06 - w / ,r.,j tti Vay: khoi lugng ket tua = 0,06.233 + 0,06.58 = 17,46 gam ' Dap an C A u i - r f i f f if, Dau tien tao ket tua theo phuong trinh phan ung: > Al(OH)3i i, CuS04 + 2NaOH ' ;lom £0,0' >• Cu(OH)2l rfHC* > + Na2S04 r' iom£0,l>-.j ^^f;'? *• > Fe(OH)34 + N a C l i ^ ,t > CaC03i + NazCOs + Bail9: C a c h l : nAi(OH)3= 0,10 mol; nAici3 = 0,15 mol 5;aO Loai D vi tao ket tua trang, khong tan: Ca(HC03)2 + N a O H ts + 3NaCl Al(OH)3 + NaOH > NaA102 + 2H2O Loai A vi tao ke't tua xanh, khong tan: FeCls + N a O H > H O ^^£^i;-,j!„ • a02 ^ a02 'i Loai B vi tao ket tua nau do, khong tan: Cac phuong trinh phan xmg pha trpn: + OH" " , ng^2- = 0,15 mol; n^+ = 0,10 mol Sau ket tiia tan, thu dugc dung dich suot: Bai 16 H^ ^^''H?' ' Trong dung dich Y: AICI3 + N a O H CaS04i + H O + C02t Da '• r^Hr^'Mt -> Dap an C Cho Ian lugt chat ran CaS04.2H20 va CaCOs vao ong nghiem dung H2SO4 loang, ong nghif m nao thay c6 thoat thi chat ran cho vao la CaCOs theo phan ung: CaCOa + H2SO4 ,^ _ H2 0,01 Ba2+ + S O " Dung nuoc thu tinh tan cua chat hot tren thay c6 chat tan (CaCh va NaNOs) va chat khong tan (CaS04.2H20 va CaCOs) ' j.^^ MgS04 = 0,10 mol va H2SO4 du = 0,05 mol Mol: 0,10 -> 0,10 Dap an A ,„.,,,,f^j£ „^ ^ MgS04+ ^ X gom: nj^^2+ = 0,10 mol; '"l > NaCl + C O + H O Mg , Vay, dung dich Xgom: ;I-{t;b) HCl du = 0,015 mol Xay phan ungketiep: ; ;r , p nMg= 0,10 mol; nH2S04 = 0/15 mol 5-^-* ; hm s d • Mol: a02 0>02 0,02 -> m = aQ2.197 = 3,94 gam -» Dap an B Bail4 , , 2H2O SO.i * - - v •i'-M^^' • ' ' = •' :fifWr? s - Trmng hap 1: Tim gia tri nho nhat cua V: Cho K O H vao dung dich AlCb den thu dugc lugng ket tiia mong muon thi dung lai Ta c6 phuong trinh: 281 H6a hpc - Nguygn VSn Hii CSm nang On luy^n thi dgi hpc 18 chuySn 3KOH Mol: _^ 0,3 + > Al(OH)34^ AICI3 0,10 Mi- e a c h 2: Ap d u n g cong t h u c t i n h n h a n h ta c6 : • Tru'ong hap 2: T i m gia t r j Ian nha't ciia V: Cho K O H d u n g djch A l C b den k h i t h u d u g c l u g n g ket t i i a I o n nha't Ta c6 _> y n N a O H = 0,06+ 0,02 = 0,08 m o l - ^ V = 80ml Mol: 0,45 > Al(OH)34 AlCb ' ' 3Na2S04 + 0,20 vi 0,10 " = > NaA102 + H O NaOH " *' ' ' 0,10 ^ ^ " " "NaOH 0,45 l i t D a p an A Cach 2: A p d u n g cong t h u c t i n h n h a n h ta c6: )l "NaOH max= 4n^,3+ - nAl(OH)3 = 4.0,20 - 0,10 = 0,70 m o l n z n c i = 0,02 m o l ; Z " N a O H = 0,70 + 0,20 = 0,90 m o l ^ 'it/ 0,99 n z n ( O H ) = - ^ = , mol + -> Z"oHV = 5- nKOHmax=4n^„2+ - n z n ( O H ) Bai 20 V = ^ = 0,45 l i t Bai 22 Cachl: K h i cho l u g n g K O H vao d u n g dich A t h u d u g c c i i n g m g t kho'i l u g n g ket HCl + NaOH M o l : 0,02 ZnCh M o l : 0,02 > H2O + 2NaOH Zn(OH)2i ' + 2NaCl day la d a n g bai tap c6 n g h i f m u n g v o i l u g n g K O H m i n va Theo bai: n R o n = 0,03.2 = 0,06 m o l ; nROH max= 0,07.2 = 0,14 m o l theo l^KOH =2nzn(OH)2 = 0,06 + 2NaOH Zn(OH)2 -> > Na2Zn02 KOH max iJng voi K O H min: phuong trinh: 0,01 + 6NaOH -> Al(OH)3 " nhanh n h u s a u : n H d =0,02 m o l ; ^^^^^^^^ , ^ 0,10 trinh: LM-M 1/; D o i v o l h i d r o x i t l u o n g t i n h Zn(OH)2, ta c i i n g c6 cac cong t h u c t i n h riKOHmin=2nzn(OH)2 u^.') De t h u d u g c 0,10 m o l ket tiia t h i can hoa tan 0,10 m o l A l ( O H ) theo p h u o n g V K Q H = ^ -> ^^^^^.^ + H2SO4 Al2(S04)3 = 0,5 l i t = 500ml ^ D a p an A Cach 2: A p d u n g cong t h u c t i n h n h a n h -> • , .,'.:„,:.«: , , 2NaOH i 0,05 VKOH= „ , - 0,15 De t h u d u g c 0,10 m o l ket tua t h i can hoa tan 0,05 m o l A l ( O H ) theo J 'M- , n N a O H m a x = n „ - n z „ ( O H ) = 4.0,02-2.0,01 =0,06 m o l - i i ' • phuong tfinh: 3KOH + ^( • • + 2H2O nzn(OH)2 = tTng v o i K O H max: 0,02 0,14+2.0,03 " K O H max = 411^^2+-2nz„(OH)2 X " o H - " ° ' ° ^ ^ ° ' ° ^ " ^ ° ' ° ^ " ° ' ° ^ ' " ° ^ ° "NaOH = 0,03 m o l n^„2+= _ , = 0,05 m o l 983 dm nang On luy^n thi dgi hgc 18 chuygn dg H6a hpc - Nguygn Van Hi\ i^Zn(N03)2 = n z „ + = 0,05 m o l - > X= ^ Cty TNHH MTV DWH Khang Vi^t = 0,5M „ W D a p an C Bai f^Kuyen de 13 - ' ANCOL - PHEIVOL ,- 23 Khi A N C O L C o n g t h i i c ancol cho l u g n g N a O H vao d u n g d i c h Y t h u duq>c cung m p t k h o i l u g n g , tiia d a y la d a n g bai tap c6 n g h i e m l i n g v o i N a O H m i n v a N a O H max Theo bai: n ^ a O H = 0,06.1 = 0,06 m o l ; n N a O H max = 0,14.1 = 0,14 m o l Lfng v o i N a O H m i n : • '^NaOHmin-'^nAi(OH)3 , = 0,U6 ' nAl(OH)3 = ^ i-;,'^ nAlci3 = n ^ , , = a m o l Al' ' ^^^'""^ 0,14+0,02 i =0,04 m o l X (n^l) A n c o l n o , h a i chvic: CnH2n(OH)2 (n>2) A n c o l no, m chuc : CnH2n + 2-m(OH)m (n > m ) * ' 01 ^t.-.^,.; HOf,i/a) * fi —* ^v,'•i ; UV,M , r, i , , >i ) w'-dm, ; • l ^ O H : propan-l-ol; CHa-CH(OH)- CHa: j;{HO)IA I CnH2n.lOH + Na > CnH2n.lONa C3H5(OH)3 + 3K > C3H5(OK)3 + CI ,0 ^ ' ( "xfirnHOc + f , - « f o ^ n i j i i i iSrl.! j ^ i u t ^ ( o n e o^J.ufi^ _ £ / ' CH3COOC2H5 + H2O Phdn itng tdch nuac tqo anken CH3-CH2-OH Ht - B ^ ' ' • ^IUJ.: + -Hz U.,' ,>i', rfo'd M f k ] i f X : I fii i l i ' i ' wM-i Hiav,-, H(l>tl*lif3S + propan-2-ol 'Mi-ia.j/ I Tdc dung voi kimloqimqnh ^ V * T - tern {)