Tài liệu 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 do chuyen hoa giiia cac hgp chat cua crom: = Lai gidi: Ggi so'mol trong 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 Al + 3HC1 — ^ CrCb + - H 2 t 2 -^DapanD. H * . . O-M U t 1 nci .{"^ -> T i l | m o l l : 2 Ti If mol 1:3 > v< > Cr(OH)3 + 3KBr 2Cr(OH)3 + 3Br2 + lOKOH 1 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 ra cau hinh electron cua nguyen tu Cr la: [Ar] 3d*4s2. JrfD .Y sv Tuy nhien, do 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 D. K2Cr207, Cr2(S04)3. CrCh + 3 K O H 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 trong khong khi va nuoc. B. Crom la kim loai cung nhat trong tat ca cac kim loai. C. Nhom va crom deu bj thu dpng hoa boi H N O 3 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, do 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 thanh cac muoi clorua voi hoa tri khac nhau: CrCh + H2t B. K2Cr04, Cr2(S04)3. LMgidi: Cr + H2SO4 CrS04 + H2t Tac6: nHj = 0,15 - > x + y = 0,15 x = 0,1;y = 0,05. Cr + 2HC1 — ^ , ,^ C. K2Cr04, CrS04. (I Dap an B. Z. ^ . K2Cr04, K2Cr207. 2 0,03 I ^ %mcr - "^^^"^^.100% = 30,44%. ) Y CacchatY,Zlanlugtla: NaA102 + - H 2 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 khi H2 (dktc). Mgt khac, cho 1,58 gam X vao lugng du dung djch NaOH loang, nong, thu dvigc V lit khi 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 trong X: A l = x; Cr = y. Theo bai: 27x + 52y = 1,58. 4 '-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 khi so sanh tinh cha't hoa hgc cua nhom va crom? A. Nhom va crom deu khong tan trong dung djch HNO3 d$c, ngugi. B. Nhom va crom deu ben trong khong khi va trong 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 nhau. ,. 241 elm nang 6n luygn thi dgi hpc 18 chuy6n 2A1 + 6HC1 > 2A1C13 + Cr 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+ > C r C h + H2 + 2HC1 CtyTNHH MTV DWH Khang Vigt H6a hpc - Nguyin Van Hii 3H2 ' ' - * - > D a p an C. ' • ^'^ >U'\ trong d u n g d i c h la d o 2 axit phan l i ra —> Giai theo p h u o n g t r i n h i o n . j-fr h a Trong X: 3. 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 + 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 + 2 A g • C u C h + 2FeCl2 >• 3Cu2* + 2NO n ""^UnAi.OsS^ -> + 4H2O .A D a p an B. (san p h a m k h u BaCl2/ t h u d u p e 23,3 g a m ket tiia; con k h i cho toan bp Y tac d\ing v o i d u t h u dupe 5,35 g a m ket tiia. Gia t r j cua a la A . 0,80. B.0,95. C.1,05. ,^ CuO + CO '° > 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', 3Cu + N2 + 3H2O K h i cho Y + d u n g dich BaCkBa^* + S04' D . 1,20. , va N O , . ' > BaS04>l Tan trong dung dich N H s tao dung dich mau xanh tham: 23 3 Bao toan nguyen toS: ng(x)= n^aso^ =—rr = 0,1 m o l . 233 Cu(OH)2 + 4NH3 K h i cho Y + d u n g dich NH3 d u : > d. Dong (II) sunfat + Phan ung trao doi: CuSOi [Cu(NH3)4](OH)2 + > CuS>l' > [Cu(NH3)4]S04 Phan ung dien phan: CuS04 + H2O e. D o n g (II) nitrat Cu(N03)2 Cu ? + ^02 ; I'l-' C u O + 2NO2+ i 0 2 Cu(OH)2 + 4 N H 3 V i d^ 1 : 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 ra hoan toan, sinh V l i t k h i N O (san p h a m k h u d u y nhat, 6 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 '• * > [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) = 2 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. > Cu(OH)2 i Den day cac e m l u u y: Cu(OH)2 tan trong N H s d u t^o thanh phuc chat: +H2SO4 ° > Fe(OH)3 Cu2* + 2NH3 + 2H2O {mau xanh tham) - J s ^ , Fe^ + 3NH3 + 3H2O + H2SO4 Phan ung voi dung dich amoniac CuS04 + 4NH3 242 +;lT:.">(,"' 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 + Jsi ^ 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 + 2 n H 2 S 0 4 = 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. 0 Oft So thi nghi^m xay ra phan ung la A. 3. 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 khi 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 ra. 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 khi cac phan ling xay ra 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 khi 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 6 i ) = n g t r o o d 6 i = 3nivjo + 8nN20"'' 8nNH4NO3~0'55'^^^Bao toan nguyen to N : nHNOg = " ^ 0 3 + " N O + 2 n^^O + 2nNH4N03 con 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 trong binh kin khong chiia khong khi, sau A.I. mol. > Muoinitrat + NO + N2O 0,5x H = ^ ^ . 1 0 0 % = 80% 0,1 D. 84,1. ^ ' Bao toan khoi lugng: m = mzn, Cu, Ag + ^^Q-^ + mNH4N03 m N 0 2 + m o 2 = 18,8-10,16 = 8 , 6 4 4 6 . 2 x + 32.0,5x = 8,64. ^ ^ -'^iti& + 5?':. ^ 0,95.1 = 0,55 + 8a + 0,05 + 2.0,05 + 2a ^ CuO + 2NO2 + - O 2 X . Khichokimloai + HNOs: «,0 Nhan xet: Khoi lu^ng chat ran giam = Khoi lugng khi bay ra. ' Zn,Cu,Ag va CO the xay ra qua trinh: -^.O*. C. 86,3. qua: nNo= 0,05 mol; nN20= Chat oxi hoa: D . 50%. •/nhuiliisl > inp'hr 0 bai nay, trudc het cac em can t i m so mol moi khi trong 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<7p khi X (dktc) gom N O va N2O. Ti khoi ciia X so voi H2 la 18,5. Gia tri ciia ^'j- "^-^^ + Fe(N03)3 pH = 1 dung dich HNO3 I M , thu dugc dung dich chua m gam muoi va 2,24 lit hon i > CuCb + 2¥eCh = 10-1 y i dv 6: Cho 35,6 gam hon hgp Zn, Cu va A g tac dung vira d u voi 950ml . Ta't ca cac thi nghi^m (a), (b), (c), (d) deu xay ra phan ung: Cu + 2FeCl3 fH-l = ^ 0 , 6 i ! U,; f^mtb ; a = 0,03 m d 0,15 > FeS04 + Cu 0,15 iinj fifed ,(-:':«t; Fe Mol: + 2HC1 ai • ^ > 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 3 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 ra. D u n g dich fits k h i i d u y nhat ciia N*^ deu la N O . Gia t r i ciia m la Ksl! A . 1 2 , 8 . B.6,4. H-; Laigidi: Phan u n g hoa hpc: 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 ; yioh ^ ,1 > D u n g d i c h X g o m cac i o n : = 0,4 m o l ; N O ^ = 0,3 m o l va SO 4" = 0,2 m o l . ; . Fe , Mol: > 3Cu2+ + 2 N O + 4H2O 0,15 < - 0,4 Cu Mol: ' > Cu2* + 2Fe2^ + 2Fe3* , Mol: + 8H^ + 2N03~ 0,05 < - 0,1 j • "" <- > Fe(N03)3 + N O + 2H2O 4HN03 8b " '"" «' mcu = 0,2.64 = 12,8 g a m D a p an A . -> ' ' ' 2b Mol: 3a + , <— + <- '' * > 3Fe(N03)2 2Fe(N03)3 'Mil (111; ZD 3a HI ISAf. f jj ,.n> > Fe(N03)2 + C u Cu(N03)2 -> 3a , M a t khac: nHNOa = 8a + 8b = 0,16. " , ,. ,, f, - > 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 ra hoan toan, t h u dugc iri 4 m V nuuth <. Theo bai: 56(3a + 3 b ) = 64.3a = 1,44. .,„.^,^„„ " , b Fe f , 2b '-Sftq fiBS) Xj,. 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 >I '-'f';!' 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: Fe ' + , ,1 i :K«u:> i.mi >nii*js > 3Cu(N03)2 + 2 N O + 4H2O 8a 3a nw.:,, 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 > Fe(N03)3 + 5 N O + 2H2SO4 + 2H2O 0,8 + 8HNO3 3Cu j^ol: D . 3,2 C.9,6. gidi: phan l i n g cua C u v a d u n g d i c h HNO3: 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 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 3 loang, 4. B A I T A P O N L U Y E N s i n h ra 0,448 l i t k h i N O (san p h a m k h u d u y nhat, 6 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 2 0 3 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 3 •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. ra 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 + 2 N O + 4H2O <- B. 2,016. Bai 2: Cho 6,72 g a m Fe vao 400ml d u n g d i c h H N O 3 I M , d e n k h i p h a n i m g xay a <, i Chat ran Y la C u d u . Phan l i n g hoa tan C u bang H N O 3 loang: 3Cu d u t h u d u g c V l i t k h i N O (san p h a m k h u d u y nhat, 6 d k t c ) . Gia t r i ciia V la A . 1,792. > 2FeS04 + CuS04 Fe2(S04)3 + C u Mol: sau m g t t h a i g i a n t h u d u g c chat ran X va k h i Y . Cho Y hap t h u h o a n toan D u n g d i c h X c6 the hoa t a n to'i da m g a m C u . Gia t r i ciia m la: A . 0,64. B.3,20. C. 1,92. ? ^ D . 3,84. Bai 3: H o a tan h o a n t o a n h o n h g p g o m 0,01 m o l FeCh va 0,02 m o l N a C l vao 0,02 232a +• 64(a + 0,03) = 7,84 g a m ^ a = 0,02 m o l . mgt l u g n g n u o c ( d u ) , t h u d u g c d u n g djch X. Cho d u n g d i c h A g N O s ( d u ) Vay: m = 152.3.0,02 +160.0,02 = 12,32 gam ^ Dap an A . vao, t h u d u g c m g a m chat ran. Gia t r i ciia m la V i d^ 10: H o a tan h o a n toan m g a m C u t r o n g 160ml d u n g d i c h H N O a I M , thu A . 5,74. B.2,87. C. 1,08. , i . , i (; D . 6,82. ; > d u g c d u n g d i c h X. Cho m g t t h a n h Fe vao X, sau k h i cac p h a n u n g ket thiic ^ai 4: N u n g h o n h g p b g t g o m 7,6 g a m Cr203 va m g a m A l a n h i ^ t d g cao. Sau thay k h o i l u g n g t h a n h Fe g i a m d i 1,44 gam. Biet t r o n g cac qua t r i n h tren, k h i p h a n u n g hoan toan, t h u d u g c 13 g a m h o n h g p r a n X. C h o X p h a n u n g k h i N O la san p h a m k h u d u y nhat cua N*^. Gia t r i ciia m la v o i axit H C l ( d u ) thoat ra V l i t k h i H 2 (dktc). Gia t r j ciia V la A . 0,64 A . 8,40. B.1,92. C. 1,28. D . 2,56. B.6,72. C. 5,60. D . 4,48. 247 Cty TMHH MTV DVVH Khaiig Vigt dm nang On luyQn thi d^i hgc 18 chuySn dg H6a hgc - Nguygn Van HSi Bai 5: N u n g h o n hg-p X g o m 0,14 m o l A l va a m o l C u t r o n g k h o n g k h i mQt thoj gian, t h u dxigc 8,58 g a m cha't rSn Y . H o a tan hoan toan Y b a n g d u n g djc}^ H N O s loang ( d u ) , t h u dugc 0,896 l i t k h i N O (san p h a m k h u d u y nha't 5 dktc). Gia t r i a i a a la A . 0,03. ^ B.0,10. , C.0,05. D . 0,08. 5, H l / O i N G D A N - L C J I G I A I pail: nBaCO3 = 0 ' 0 9 m o l D u n g d i c h X c6 the hoa tan t o i da m g a m C u . Gia t r i cua m la A . 3,2. B. 6,4. C.4,8. D . 8,0. Bai 7: N u n g h o n hqip hot g o m 16 g a m FezOa v a m g a m A l 6 nhiet d p cao. Sau k h i p h a n l i n g hoan toan, t h u d u g c 24,1 g a m h o n h g p r a n X. C h o X phan l i n g v o i axit H C l ( d u ) thoat ra V l i t k h i H2 (6 dktc). Gia t r i cua V la A . 3,36. B.4,48. C. 7,84. CuO,Fe2a B.11,2. k h i N O (san pham C.7,8. D.9,0. Bai 9: H o a t a n h o a n toan m g a m oxit FexOy bang H2SO4 dac nong, t h u duoc d u n g d i c h chiia 6 g a m m g t loai m u o i sat d u y nha't v a 0,112 l i t SO2 (dktc). C o n g t h u c ciia o x i t sMt va gia trj m Ian l u g t la A . Fe304 va 4,64. C. Fe304 va 2,32. D . FeO v a 1,44. Mol: A g N O a 0,2M. Sau m g t t h o i gian lay thanh k i m loai ra, r u a sach l a m kho tha'y k h o i l u g n g thanh sat tang them 1,92 gam. K h o i l u g n g sat da p h a n u n g la C. 2,80 gam. D . 1,72 g a m . Bai 11: C h o 13,5 g a m h o n h g p A l , Cr, Fe tac d u n g v o i l u g n g d u d u n g djch H2SO4 (loang, nong) t h u d u g c d u n g d i c h X va 7,84 l i t k h i H2 (dktc). Cac thi n g h i f m d u g c tien hanh t r o n g d i e u k i f n k h o n g c6 k h o n g k h i . C o can X, thu d u g c m g a m m u o i k h a n . Gia t r i ciia m la A . 48,8. B.47,1. C . 45,5. D . 42,6. • ^ • •^^ '' ' Bai 12: D e o x i h o a hoan toan 0,02 m o l C r C b thanh K2Cr04 b a n g CI2 k h i co mat K O H d u , s o ' m o l Chto'i thieu can d u n g la A . 0,01. B. 0,05. C.0,03. D . 0,02. ' Bai 13: C h o 0,14 m o l Fe tac d u n g v a i 0,4 m o l axit HNO3 t o i p h a n u n g hoaii toan, t h u d u g c d u n g d i c h Y chua m gam muo'i va k h i N O (san p h a m kh^' d u y nha't). Gia t r j cua m la (bie't cap Fe^VFe d u n g truoc cap Fe^VFe^*) A . 24,20. B. 26,44. C . 31,28. > 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 = 2 n c o • "CO = " C O 2 = «"° ~^ = 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 0 3 ) 2 + 2Fe(N03)2 " C u = ^ n F e ( N 0 3 ) 2 = ^'^^ m o l ^ - > D a p an C. - > 3Fe(N03)2 0,02 - > 0,04 2Fe(N03)3 + C u Bai 10: N h i i n g m g t thanh Fe v a o 100ml d u n g d i c h g o m C u ( N 0 3 ) 2 0,5M va 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 . B. Fe203 va 3,20. 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 do 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 thanh V m l k h i N O (san p h a m khtr d u y nha't, 6 dktc) v a d u n g dich X O i.) sb f w o - i ' Bai3: Nhqn xet: X g o m : Na^ = 0,02 m o l ; Fe^^ = 0,01 m o l ; C h = 0,04 m o l . Cac p h a n u n g hoa hgc: Ag^ > AgCl f >0 - 0,04 Mol: Ag* Mol: + CI- ==="'' a04 + Fe2* 0,01 -> )f> . . H f«Go! ^9 ^- • > Fe3" + A g 0,01 «•' ^, ' .,^;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 * thanh A g . ^. BM4: " Bai nay 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, trong 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 6 a hpc - Nguygn V S n H 5 i ncu = 0,03 + 0,02 = 0,05 mol - > mcu = 0,05.64 = 3,2 gam. mcr203 + i ^ A i = m x -> mAi= 1 3 - 7 , 6 = 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 3 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 - 1 6 = 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 6 cac giai doan: A F _^ nenhimng > Cu*^ —> nenhu«ng = 2ncu = 2 a > 2CH - > nenhan = +3e NQ-> nenh5n Bao toan electron: 0,42 + 2a = 0,6 - 8a + 0,12 Mol: = 3nNo =0/12. > Fe3*+ N O + 2H2O + 2Fe3^ 0,05 > 3Cu2- + 2 N O + 4H2O > Cu2* + 2Fe2^ m = 0,05.242 = 12,1 gam. -n. r ( f / M . 41 miUi Bai 9: "SO2 = = = 0'005 mol. ' 1 ' ' SO2 thi oxit la FeO Trong 1 m o l FeO ho§c Fe304 deu chiia 1 mol Fe'^^ nen deu c6 kha nang : ,? nhuong 1 mol electron, do 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^ + 2 N O ~ 0,05 hoac Fe304. ,^ , 0,04 ''^ Nhan xet: Oxit sat FexOy tac dung voi H2SO4 dac, nong —> D a p an A . + 4H* + NO3- :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. n ^ + = 0 , 2 4 mol; n ^ ^ _ =0,04 mol. ) Fe(N03)3 - > Dap an A . a = 0,03 m o l - > D a p an A . * Bao toan electron: 3.0,14 + 2.a = 2b + 0,12 - > a = 0,03; b = 0,18. Bai6: < < f< W > X (Fe,FeaFe203,Fesa) 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 . 0,16 .«; va ap dung dinh luat bao toan nguyen to'Fe: i Fe Dap an C. =0'05mol. Fe 4no2 = ' ^ Cach 2: 28 i'i,li,V,- ^len nj-i^ = 0,35 mol. nhieu thai gian. O day, cac em can s u dung so do phan ung: = 0,6-8a 0,04 ~^ Nhan xet: Bai nay neu dua theo phuong trinh phan ung se rat da: dong va ton = 3nAi =0,42 > (1) - » (2): O2 + 4 e (2) - > (3): 2^ ^ 2 Bai 8: So d o phan l i n g : Cu-2e l i • n^i = 0,3 mol. 3 Khi X tac dung voi axit: A l (1) - > (3): A l - 3 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) 1. Lini y rang, trong phan ung nhiet nhom, thuong ap dung dinh luat bao toan nNO= - ^ ^ = 0 , 0 4 mol. Mol: • Bai nay 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 = 3 - > Oxit sat la Fe304. '^AgN03 = 0,02 mol; ncu(N03)2 = > "^ol251 Cty TNHH M i V nVVH Khang Vi^t Ccim nang 6 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 •— < J. C A U T A G N G U Y E N T l / > 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 do 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 do : 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 nhau. > 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 + 1 6 K O H ^ So don vi di?n tich hat nhan (Z) = so proton = so'electron. Bao toan khoi lugng cho so do 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 2 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 trong 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 6 d i e u ki§n t h u d n g . . ' ^ c h u n g h o n k e m n h a u 1 d o n v i d i ^ n tich h ^ t nhan. gidi: D a p an C. D a y la ket qua da d u g c t h o n g ke, cac e m can n h o de v ^ n dyng, TudapanC-»Z < N <1,5Z. - /. Luu y: N e u bai toan cho tong so hat co ban ciia n g u y e n t u b a n g T -> N + Z + E = T h a y N + 2Z = T . M a Z < N <1,5Z -> — < Z < . >. >; ; < = ^ D . Phan l o p ngoai c u n g cua n g u y e n t u X (6 t r a n g thai co ban) co 4 electron. t o n g so hat n o t r o n ( N ) va t o n g so hat p r o t o n (Z) c6 m o i q u a n h f la A . N / Z = 1. ' 8''^'^i k tu B. D p a m d i e n cua X I o n h o n d p am d i f n cua Y. \ V i d\ 2: Cation X^* c6 t o n g so h ^ t p r o t o n , n o t r o n va electron la 80, t r o n g do ti ' ' n V|, i -.^ Theo bai, so'proton cua n g u y e n t u Y n h i e u h o n so p r o t o n cua n g u y e n t u X nen: ZY - Z x = 1. u M a t khac: Z Y + Zx = 33 -> Zx = 16 ( L u u h u y n h ) v a Z Y = 17 (Clo). !) is=. Loai A v i 6 d i e u k i f n t h u o n g , l u u h u y n h t o n tai a the r a n . Q if L o ^ i B v i d p a m d i f n cua S n h o h o n ciia C I . sM=' Loai C v i 6 l o p n g o a i c u n g cua n g u y e n t u C I CO 7 electron. 1? g i i i a so hat electron v o i so'h^t n o t r o n bang 4/5. V i t r i (chu k i , n h o m ) cua X t r o n g bang t u a n hoan la 'A D d u n g v i p h a n l o p ngoai c u n g ciia S la 3 p ^ co 4 electron. —>Dap a n D . A . chu k i 4, n h o m I I A . C. chu k i 4, n h o m V I I I B . V i dv 5 (CD-10): D o n g co 2 d o n g v i ben la ^^Cu va ^^Cu. N g u y e n t u k h o i B. chu k i 4, n h o m I I B . D . chu k i 4, n h o m V I B . Lbi t r u n g b i n h ciia d o n g la 63,54. T h a n h p h a n p h a n t r a m cua d o n g v i *^Cu la gidi: G p i so hat p r o t o n , n o t r o n va electron t r o n g n g u y e n t u X Ian lug-t la P, N va E. T a c 6 : P = E. T h e o b a i , t r o n g i o n X2*:P + N + E - 2 = 80 v a E - 2 = 0,8N | 2P + N = 8 2 v a P = 0,8N + 2. ' P = 26 v a N = 30. • ^ ^ ->DapanC. l u g n g n g u y e n t u . N g u y e n t u k h o i t r u n g b i n h ciia n g u y e n to' cacbon la: C. 12,021. N g u y e n t u kho'i t r u n g b i n h cua cacbon la: A= l ^ ^ . H x U ^ 100 D a p an B. 254 100 ^ *'v/' N g u y e n t u kho'i t r u n g b i n h ciia d o n g la: A = V ^ y a = 0,73 = 73% + 65 x (100—a) _ ^ 3 5 4 D a p an C . d o n g la 64,00. V i dv 3 : Cacbon c6 2 d o n g v i la ^^C chiem 98,9% v ^ ^^C c h i e m 1,1% ve so B. 12,011. D . 37%. Nhan xet:. N e u t i 1? 2 d o n g v i la 50:50 t h i n g u y e n t u k h o i t r u n g b i n h ciia , .''issiof A . 12,001. C. 73%. Lbi gidi: Cach 2: C a u h i n h electron cua X: Is^ 2s22p^ Ss^Sp^Sd* 4s2. X thuoc chu k i 4, n h o m V I I I B . B. 27%. Cach 1: G p i p h a n t r a m ciia d o n g v j 29 C u la a%. Nhan xet: T r o n g i o n X^*, so'hat electron la E - 2. ^ A . 30%. D . 12,031. .^^ Theo bai, n g u y e n t u k h o i t r u n g b i n h ciia d o n g la 63,54 < 64,00 -> d o n g v i nh? h o n , tuc 29 C u chiem tren 50% -> chi dap an C thoa m a n . V i dy 6: Phan t r a m k h o i l u p n g ciia n g u y e n to'R t r o n g h p p chat k h i v o i h i d r o (R CO so o x i hoa thap nhat) va t r o n g o x i t cao nhat t u o n g l i n g la a% v a b % , v 6 i a : b = l l :4. P h a t b i e u n a o s a u d a y l a d i i n g ? A . Phan t i i o x i t cao nhat ciia R k h o n g co c\rc. , , , , .. , , B. O x i t cao nhat ciia R 6 d i e u ki#n t h u o n g la chat ran. C. T r o n g bang t u a n hoan cac n g u y e n to hoa hpc, R thupc chu k i 3. D . N g u y e n t i i R (6 t r ^ n g thai co ban) co 6 electron s. 255 nang On luygn thi dji hgc 18 chuy6n dg H6a hQC - NguySn Van H i i dm Lai j, TNHii Miv nvvH Khang Vijt gidi: Lai gidi: Nhan xet: T o n g hoa t r j cua n g u y e n to' R t r o n g hqp cha't v o i k h i h i d r o vg A d u n g v i X v a Z d e u c6 ciing so k h o i bang 13. t r o n g o x i t bang 8. B sai v i X v a Z c6 s6'hi?u n g u y e n t u khac n h a u —> k h o n g la d o n g v i . G p i hoa tri cua R t r o n g hgip chat v o i h i d r o I a n C sai v i X v a Y c6 so' h i ^ u n g u y e n t u khac n h a u —> thupc hai n g u y e n to -> H o a t r i cua R t r o n g o x i t la 8 - n . r C o n g t h u c h g p chat k h i a i a R v o i h i d r o : R H n , ^ y . , rj,./,, C o n g t h i i c hgjp chat ciia R v o i oxi: R 2 0 8 . n , „ , D sai v i X c6 13 n a t r o n , Y c6 29 n a t r o n . ^ I ,•«,• ' Theo bai: -g-. =H ^ 7R = 256 - 43n. R + n 2R+16(8-n) 4 VI p h a n t u CO2 k h o n g p h a n cue. v^>r>£U'j A- xS - yS n - X^K •< - SB ^ > , 1, L o a i D v i n g u y e n t u cacbon a trang thai co b a n c6 4 electron s. —> D a p an A . V i dv 7: N g u y e n t u cua m g t n g u y e n to X c6 t o n g so h a t p r o t o n , notron, electron 46, t r o n g d o so hat m a n g d i ^ n n h i e u h a n so' h?t k h o n g m a n g di^n la 14. Cau h i n h electron n g u y e n t u X la A . [Ne]3s23p3. B. {Ne]3s'. . U m, q s i J • C. [Ne]3s23p*. Lai D . [Ne]3s23pi. gidi: l o n g so'hat m a n g d i f n v a k h o n g m a n g di?n bang 46 -> so'hat k h o n g mang d i ? n = (46 -14)/2 = 1 6 N = 16. V a y s o h a t m a n g d i ? n = 4 6 - 1 6 = 3 0 - > Z = E = 15 D a p an A . ^^ ^ , . _ ^ ' . V i dv 8: N g u y e n t u ciia n g u y e n t o X c6 t o n g so hat electron t r o n g cac phan l o p p la 7. N g u y e n to X la A . A 1 ( Z = 13). B . M g ( Z = 12). Lai C.Na(Z = ll). D . C a ( Z = 20). gidi: N h a n thay cau h i n h electron cua X c6 7 electron p ^ l o p v 6 n g u y e n tvr ciia X chua c&c p h a n l o p 2p* v a 3p'. Cac n g u y e n to' hoa hpc dupe sap xep theo chieu t a n g d a n d i $ n ti'ch hat nhan n g u y e n t u . Cac n g u y e n to' hoa hoc c6 ciing so l o p electron d u o c sap xep thanh ciing m p t hang (chu k i ) . Cac n g u y e n to hoa hpc c6 ciing so electron hoa t r i t r o n g n g u y e n tvr dupe sap xep t h a n h m o t cpt (nhom). 2. Cau tao cua b a n g h | t h o n g t u a n h o a n Dang b a n g t u a n hoan dupe sir d u n g t r o n g sach giao khoa hoa hpc p h o thong hien nay la bang dang dai. Cac khai n i e m co b a n cua bang t u a n hoan bao gom: 6 nguyen to: L a v i t r i cua m p t n g u y e n to, so t h i i t u cua 6 bang so h i ^ u nguyen tu' v a bang so d o n v j di?n tich hat n h a n n g u y e n tir. Chu ki: Co 7 chu ky, so' t h i i hf cua chu k i bang so l a p electron cua n g u y e n tvr gom: + Chu k i n h o la cac c h u k i 1, 2, 3 chi g o m cac n g u y e n to s v a cac n g u y e n to p. M o i c h u k y n h o g o m 8 n g u y e n t6'(tru' chu k y 1 chi co h a i n g u y e n to). + Chu k i I o n la cac c h u k i 4, 5, 6 ,7 g o m cac n g u y e n t o s, p, d v a f. C day, K quan t r p n g la c h u k y 4 v a chu ky 5, m o i c h u k y co 18 n g u y e n to. Nhom: Co 8 n h o m , so t h u t y ciia n h o m bang so electron hoa t r j g o m : N h o m A : So' t h i i tir cua n h o m bang so electron l o p ngoai c i m g ( g o m cac N h o m B: So t h u t y ciia n h o m B bang so electron hoa t r i (gom can n g u y e n to -> D a p a n A . d v a f). N h o m B con d u p e gpi la cac n g u y e n to n h o m p h y . Vidv9(A-10): A . X v a Z CO c i m g so'khoi. BANG TUAN H O A N n g u y e n t o s v a p). N h o m A con dupe gpi la cac n g u y e n to n h o m chinh. T u d o -> cau h i n h cua X: Is^ Is'^lp^ 3s23p' ^ Z = 13 N h a n d j n h nao sau day d u n g k h i n o i ve 3 n g u y e n t u : ^3 X, H Y, ^- N h i m g t i n h c h i t b i e n d o i t u a n hoan Z? * ^ Cac d a i l u p n g bien d o i tang dan: - B. X , Z l a 2 d 6 n g v i . C. X, Y thuQC c i i n g m p t n g u y e n to'hoa hpc. D. X v a Y CO c i m g so'notron. 25,6 UAM 1, N g u y e n tac sap xep iioloiqaa 4^*1 r Lo?i B v i a d i e u k i ^ n t h u o n g CO2 la chat k h i . Loai C v i n g u y e n to cacbon 6 c h u k i 2. —> D a p an A . , Thay n = 1 den 4, n h a n thay n = 4 v a R = 12 (Cacbon) thoa m a n . = A diing khac n h a u . ,j Trong mot chu ki: d p am d i f n, nang l u p n g i o n hoa, t i n h p h i k i m , t i n h axit ciia cac o x i t va hidro).it. - Trong mot nhom: ban k i n h nguyen tvr, t i n h k i m l o ^ i , t i n h baza ciia cac oxit va hidroxit. 257 Cly Ca'm nang On luySn thi dji hpc 18 chuyfin 66 H6a hQC - NguySn Van H5i + Lai Cac dai lugng bie'n doi giam dan: Trong mot chu kh ban kinh nguyen ttr, tinh kim loai, tinh b a z a ciia cac - oxit va hidroxit. Si- Trongmot , •' nhom: do am dien, nang luong ion hoa, tinh phi kim, tinh axit cua cac oxit va hidroxit. ' '""^^'i" ' V i di^ 1: Mot ion M^* c6 tong so' hat proton, notron v a electron la 79, trong (JQ so' hat mang di#n nhieu hon so hat khong mang dien la 19. Ca'u hinh electron cua nguyen tu M la B. [Ar]3d54si. C . [Ar]3d64si. D . [Ar]3d^4s2. Nguyen tu M co: Ins ffl c'» nhieu han so'hat khong mang dien la 19 + 3 = 22. - > So hat khong mang dien = (82-22)72 = 30 ^ N = 30. ^ * khi voi hidro la b. Moi quan h$ giiia a va b la C . a = b. Lot . TO* oe , i , D. Tinh kim loai tang dan, ban kinh nguyen tu giam dan Latgtat: B. Nita. C . Oxi. Lot B sai vi tinh kim loai tang dan, do am dien giam dan giai: R c6 hoa tr! hgip chat cua R voi hidro c6 dang R H 3 . Ta c6: 3p. Nguyen t u cua nguyen to' Y cung c6 electron 6 m u c nang lu^ng 3p va CO 1 electron 6 lop ngoai ciing. Nguyen tu X va Y c6 so' electron h a n kem B. Phi kim va kim loai. wa^;, D . Phi kim v a phi kim. . ' 'i^^m-'''' V i dv 4 (B-08): Cong thuc phan t u cua hgp chat khi 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. Lot ' giai: cimg -> L a p ngoai cung ciia Y khong the la lop thii 3 (vi c6 phan Idp 3p thi da dien day Ss^), ma la lop thu 4 -> 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 7 (A-09): Nguyen tu ciia nguyen to X c6 ca'u hinh electron lop ngoai cung la ns^np*. Trong hgp chat khi 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 2 electron so voi Y ) . cao nhat bang 5 -> H o a tri trong hop chat voi hidro bang 3 - » cong thtH R 82,35 ^ —=— 3 17,65 ' C sai vi do am dien giam dan, tinh phi kim 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 82,35 = — 17,65 ' Nhan tha'y, nguyen tu Y c6 muc nang lugng 3p va c6 1 electron 6 lop ngoai y: Khong ap dung cong thiic nay cho cac nguyen to chu ki 1 va 2. A. Photpho. ,^ C. K i m loai va khi hie'm. hop cha't voi hidro, R chie'm 82,35% ve khoi lu^ng. Nguyen to'R la %H C. D 9 am dien giam dan, tinh phi kim tang dan. ' " '^''^". ^ ' " • A. K i m logi va kim loai. giai: ~, ^ nhau la 2. Nguyen to X, Y Ian luot la D. a Dap an A. —> Dap an D . Luu i),: - D sai vi tinh kim loai tang dan, ban kinh nguyen t u tang dan. Vay s o h a i m a n g d i ^ n = 8 2 - 3 0 = 52 - > Z = E = 26 Dap c6: ^ = =— ^ R = 14 (Nito) -> Dap an C . %0 25,93 5.16 ^ A dung: Tinh phi kim giam dan, ban kinh nguyen t u tang dan , Tong so' hgt proton, notron va electron la 79 + 3 = 82; so' hat mang dien A. a - b = 8. R c6 hoa tri cao nha't Cong thuc oxit cao nha't ciia nguyen to R voi oxi c6 dang R2O5. V i di^ 5 (B-07): Trong mot nhom A (tru nhom V I I I A ) , theo chieu tang cua di^n VIDVMAU A. [ Ar]3d34s2. ^ M T V D V V H Khang Vi$t giai: Cong thuc hap chat cua R vol hidro c6 dang RH3 bang 5 IMHH , D . Photpho. 94,12% khoi lugng. Phan tram khoi lugng ciia X trong oxit cao nha't la A. 50%. >fu-,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 trong oxit ciia X bang 6 -> Hoa tri trong hgp chat voi hidro bang 8 - 6 = 2 - » Cong thuc hgp chat ciia R voi hidro c6 d^ng R H 2 . • , ^•• . „ • ^,,, 259 Cty TMHH MTV DWH Khang Viet dm nang On luy^n thi Jgi hpc 18 chuySn 66 H6a hpc - Nguyin Van Hil ^ , %X 94,12 X V A . 5,0.10^ mol/(l.s). ^^^ % H = W = T " = ^"^"^ ^"^"^'^ ' ' ' -.100% = 4 0 % - > D a p a n D . df.?. D . 2,5.10-^ mol/(l.s). Lai giai -> O x i t cao nha't ciia n g u y e n to'S v a i o x i la SOa - > % kho'i l u g n g S la: 32 B. 5,0.10-5 mol/(l.s). C. 1,0.10-3 mol/(l.s). u < a o /T „, , . , P h u o n g t r i n h hoa hoc phan h u y H2O2: < 2H2O2 Mol: > 2H2O ( V 1^5.10-3 3.10-3 + 02t ' r , , N 6 n g d p H 2 0 2 d a p h a n u n g : AC = 3.10-3/0,1 =3.10-2mol/1. a. K h a i n i ? m Toe d o p h a n u n g la d p bieh thien n o n g d p cua m p t t r o n g cac cha't p h a n ling Toe d p t r u n g b i n h ciia p h a n u n g ( t i n h theo H2O2) t r o n g 60 giay tren la hoac san p h a m t r o n g m p t d a n v i t h a i gian. V = (3.10-2 mol/l)/60s = 5,0.10^ mol/(I.s). b. Toe dp trung b i n h ciia phan ling Xet p h a n u n g : A , , . , —> D a p an A . > B . - - 1 ' • . / t2-tl = At ' * ••' ^* ^j, • "i/i^- > 2HBr + CO2 N o n g d p ban d a u ciia Br2 la a mol/1, sau 50 giay n o n g d p Br2 con l a i la 0,0^ mol/1. Toe d p t r u n g b i n h ciia p h a n u n g tren t i n h theo Br2 la 4.10-* mol/(l.s) Toe d p ciia p h a n u n g t i n h theo chat A t r o n g k h o a n g t h a i g i a n t u t i den t2 d u p e xae d i n h n h u sau: v = . _ ;i< . - V/J.'B V i dv 3: C h o p h a n u n g : Br2 + H C O O H O t h o i d i e m t i , n o n g d p chat A (chat p h a n u n g ) la C i mol/1. O t h o i d i e m t2, n o n g dp cha't A la C2 mol/1. .j i T r o n g do: v d u p e g p i la toe do trung binh cua p h a n l i n g t r o n g k h o a n g thoi Gia t r i ciia a la ,„ , A.a018. fiOff B. 0,016. ^ i ' i k r Jinfed n'i'y d>i; C. 0,012. D . 0,014. Lai giai i, P h u a n g t r i n h hoa hpe: Br2 + H C O O H > 2HBr + CO2 g i a n t u t i den ti. N o n g d p Br2 da p h a n u n g : AC = 4.10-^.50 = 2.10-3 mol/1. Y nghia: A C c h i n h la n o n g dp chat A da p h a n u n g t r o n g k h o a n g t h o i gian N o n g d p b a n d a u ciia Br2 la: 0,01 + 0,002 = 0,012 (mol/1) t u t i den t2." '• f nftv)f>>l/^ .c . c. C a c yeu to anh Huong den toe dp phan ung + Nong do: K h i tang n o n g dp chat p h a n u n g , toe d p p h a n u n g tang. + Ap suat: V a i p h a n u n g eo chat khi, k h i tang ap sua't, toe d p p h a n u n g tang. + Nhiet do: K h i tang nhiet dp, toe d p phan u n g tang. + Dien tich be mat: K h i tang d i ^ n tich be m a t cua chat p h a n l i n g (tiic l a m kich ^DapaiiC. V i du 4 (A-12): Xet p h a n l i n g p h a n h i i y N 2 O 5 t r o n g d u n g m o i C C I 4 a 4 5 ° C : N204 + - 0 2 2 Bern d a u n o n g d p ciia N 2 O 5 la 2,33M, sau 184 giay n o n g d p ciia N2O5 la 2,08M. Toe d p t r u n g b i n h ciia p h a n u n g t i n h theo N 2 O 5 la t h u d e hat n h o d i ) , toe d p phan u n g tang. + N205 — > Chat xuc tac: Cha't xiie tae la chat l a m tang toe d p p h a n l i n g . , B. 1, 36.10"^ mol/(l.s). C. 6,80.10"^ v i D V MAU A . 2, 72.10"^ mol/(l.s). D . 6,80.10~^ mol/(l.s). mol/(l.s). V i dv 1: C h o cac y e u to: (1) n o n g dp; (2) nhi#t dp; (3) ap suat; (4) d i f n tich tiep Lai giai xiie; (5) xiie tac. So' y e u to' c6 anh h u a n g den toe d p p h a n u n g la N o n g d p N2O5 da p h a n l i n g : A C = 2,33 - 2,08 = 0,25 mol/1. A . 2. To'c d p t r u n g b i n h ciia p h a n l i n g t i n h theo N2O5: B. 3. C. 4. Loi D . 5. giai: Nhqn xet: Cac yeu to anh h u o n g de'n toe d p p h a n u n g g o m : n o n g d p chat p h a n u n g , n h i ^ t dp, ap suat, d i # n tich tiep xiie va chat xiie tae. -> D a p an D . V i d v 2: C h o chat xiic tac M n 0 2 vao 100ml d u n g d i e h H2O2, sau 60 giay thu d u p e 33,6 m l k h i O2 (dkte).Toc d p t r u n g b i n h ciia p h a n u n g ( t i n h theo H2O2) ^ ^ ^ \ = 1, 36.10-3 ^ ^ 1 / ( 1 3 ) _^ D a p an B. 184s -LU VidvS: Xet p h a n l i n g p h a n h i i y Na2S203 t r o n g d u n g d i e h axit H2SO4 loang a 25°C: Na2S203+ H2SO4 > Na2S04 + S + SO2 + H2O , Ban d a u n o n g d p ciia Na2S203 la 0,2M, sau 100 giay n o n g d p ciia Na2S203 t r o n g 60 giay tren la 26 Cty TIMh, C^m nang On luy^n thi dgi hpc 18 chuySn 6i H6a hgc - Nguygn Van HSi la 0,02M. Toe do trung binh cua phan ling tinh theo N2O5 la A. 2,0.10-3 mol/(l.s). C. 1,8.10-3 mol/(l.s). B. 1,0.10-3 mol/(l.s). ' D. 1,8.10-4 mol/(l.s). Laigidi * Nong do N2O5 da phan ung: AC = 0,2 - 0,02 = 0,18 mol/1. Toe do trung binh ciia phan ung ti'nh theo Na2S203: 0,18 mol/1 -3 . ^ ^ = 1 , 8 . 1 0 3 mol/(l.s) ';; ' -» Dap an C. Lai giai: Loai A v i khi tang nhiet dp cua he, can bang se chuyen dich thao chieu lam giam nhiet dp ciia he - » tue chieu thu nhiet -> chieu nghieh. Loai B v i khi giam nong dp O2, can bang se chuyen dich theo chieu lam tang nong dp O2 -> tue chieu tao ra 0 2 ^ chieu nghieh. ' ( ' ' S'^' Loai C v i khi giam ap suat chung ciia h$, can bang se chuyen dich theo chieu lam tang ap suat -> hie chieu lam tang so'mol khi -> chieu nghieh Phuong an D: Khi giam nong dp SO3, can bang se chuyen dich theo chieu lam tang nong dp SO3 -> tiic chieu tao ra SO3 -> chieu thuan. ' >' -> Dap an D. 4. C A N B A N G H O A HQC G Can bang hoa hoe la trang thai eiia phan ung thugn nghieh khi toe dp phan ^ (1) N2 (k) + 3H2 (k) : i = > 2NH3 (k) »i. d ^ '.V J , , (2) H2 (k) + l2(k) i = ± 2HI (k) ' ' ung thuan bang toe do phan ung nghieh. b. Stf chuyen dich can bing hoa hgc ngoai do. + Khi tang nong do ciia mpt chat, can bang se chuyen dich theo chieu lam giam nong do chat do —> tue chieu chat do tham gia phan ung. + Khi tang ap suat chung cua h§, can bang se chuyen dich theo chieu lam giam ap suat ciia he —> tiie chieu lam giam so'mol khi. + Khi tang nhi^t dp cua h§, can bang se chuyen dich theo chieu lam giam nhiet tue chieu thu nhiet. Khi thay doi ap suat, nhiSng can bang hoa hpc bi chuyen dich la A. (2), (3), (4). B. (1), (3), (4). V i du 1: Cho can bang hoa hoc: N2 (k) + 3H2 (k) < = ± 2NH3 (k); phan ung thuan la phan ung toa nhiet. Can bang hoa hpc khong bi chuyen dich khi: A. Thay doi ap suat cua he. B. Thay doi nong dp N2. C. Thay doi nhiet dp. D. Them chat xiic tac Fe. Lbi giai: Nhan xet: Them chat xiie tac chi lam thay doi toe dp phan ung, khong lam chuyen dich can bang hoa hpc. D. (1), (2), (3). Nhan xet: K h i thay doi ap suat, nhung can bang c6 su khac nhau ve tong so mol khi cac chat 6 2 ve se bi chuyen dich. , -> Cac can bang b i chuyen dich khi thay doi ap suat: (1), (3), (4) -> Dap an B. Vi du 4: San xuat amoniac trong cong nghiep dua tren phuang trinh hoa hpc: AH < 0. Can bang hoa hpc se chuyen dich ve phia tao ra amoniac nhieu hon neu A. Giam ap suat chung ciia he. B. Giam nong dp khi nito. C. Tang nhi^t dp ciia h§. D. Tang ap suat chung ciia h$. Lai giai: Loai A v i khi giam ap suat chung ciia he, can bSng se chuyen dich theo chieu lam tang ap suat -> tiic chieu lam tang so mol khi -> chieu nghieh. Loai B v i khi giam nong dp N2, can bang se chuyen dich theo chieu lam tang nong dp N2 —> Dap an D. C. (1), (2), (4). Lot giai 2N2(k) + 3H2(k) 4=z± 2NH3(k) VI D U M A U ' (4) 2NO2 (k) <==± N2O4 (k) bang khi ehiu mot tac dong t u ben ngoai (nhu thay doi nong do, ap suat, nhiet do) thi can bang se chuyen dich theo chieu cho'ng Iqi tac dong ben " (3) 2SO2 (k) + O2 (k) ^ = i ± 2S03 (k) Nguyen li La Sa-ta-li-e: Mot phan ung thuan nghieh dang 6 trang thai can i; ' Vi du 3: Cho cac can bang hoa hpc: a. Khai ni^m dp cua he TV DVVH Khang Vigt tiic chieu tao ra N2 -> chieu nghieh. Loai C v i khi tang nhiet dp ciia he, can bang se chuyen dich theo chieu lam V i du 2: Cho can bang hoa hpc: 2SO2 + O2 < > 2SO3; phan ung thuan la phan ung toa nhiet. Can bang chuyen dich theo chieu thuan khi A. Tang nhi^t dp. B. Giam nong dp O2. C. Giam ap suat he phan ung. D. Giam nong dp SO3. giam nhi?t dp ciia he - » tue chieu thu nhi^t chieu nghieh. Phuong in D: Khi tang ap suat chung ciia h$, can bang se chuyen dich theo ifjl^: chieu lam giam ap suat -)• Dap an D. chieu lam giam so mol khi -> chieu thuan. Cty TIMHH MTV DVVH Khang Vi$t Ca'm nanQ On luyjn t h i d ? i hpc 18 chuy6n d g H6a tiQC - Nguygn van HSi V i d^ 5: Phan u n g : N 2 + 3 H 2 2NH3 dugc xiic tac bc« b p t Fe. Tac dpn ciia b o t ¥2 den can bang la :, A . C h u y e n d i c h can b a n g sang trai. B. C h u y e n d i c h can b a n g sang phj^j C. Tang toe d p p h a n l i n g . D . Tang h a n g so can bang. Loi gidi: Nhan xet: Bot Fe d o n g v a i t r o la chat xiic tac - > t h e m b p t Fe k h o n g 1^^^ c h u y e n d i c h can bang hoa hoc, k h o n g l a m tang hang so can bang, chi l a ^ tang toe d p p h a n l i n g . ->DapanC. ' ' ' ^ • V i d y 6: C h o can b a n g hoa hpc: H 2 (k) + h (k) < rrr,hr<:i > 2 H I (fc); A H > 0. Can b a n g k h o n g b i chuyen d i c h k h i i,'!' • A . G i a m ap suat c h u n g cua h ^ . B. Tang n o n g d p H 2 . C. Tang nhiet d p cua he. D. Giam nong dp H I . Lai ),.} gidi P^i 4 (A-10):.Cac n g u y e n to t u L i den F, theo chieu tang cua d i ^ n t i c h hat n h a n thi • A . Ban k i n h n g u y e n t u tang, d p a m d i ^ n g i a m . B. Ban k i n h n g u y e n t u v a d p a m d i ^ n deu tang. S C. Ban k i n h n g u y e n t u g i a m , d p a m d i ^ n tang. uff"!* - / I : : S; -, D . Ban k i n h n g u y e n t u v a d p a m d i e n d e u g i a m . wvh'. ; r = Bai 5: Ban k i n h n g u y e n t u cua cac n g u y e n to': sLi, sO, 9F, i i N a d u p e xep theo t h i i t y t a n g d a n t u trai sang p h a i la A.F,0,Li,Na. B. F, N a , O, L i . C.F,Li,0,Na. D.Li,Na,0,F. Bai 6: H o a t a n 3,68 g a m h o n h o p h a i m u o i cacbonat ciia h a i k i m loai M , M ' (thuoc n h o m I I A v a 6 h a i c h u k i ke tiep) bang d u n g d i c h H C l , t h u d u p e 0,896 l i t C O 2 (a dktc) v a d u n g d i c h X. H a i k i m loai M v a M ' la A . Be v a M g . B. M g v a Ca. C. Ca v a Sr. D . Sr v a Ba. Bai 7: Phan h u y m o t l u p n g ozon t r o n g b i n h k i n eo d u n g t i c h 2 l i t ( k h o n g doi), sau 30s t h u d u p e 0,045 m o l o x i . Toe d p t r u n g b i n h cua p h a n l i u g ( t i n h theo Nhan xet: Can b a n g hoa hpc d a cho c6 s u bang n h a u ve t o n g so m o l k h i cac ozon) t r o n g 30s tren la chat 6 2 v e K h i thay d o i ap sua't, can bang k h o n g b i c h u y e n d i c h . A . 2,5.10^ mol/(l.s). B. 7,5.10"^ mol/(l.s). 3MA i v - > D a p an A . C. 1,0.10-3 mol/(l.s). D . 5,0.10^ mol/(l.s) N-,nuh Bai 8: C h o can b a n g sau t r o n g b i n h k i n : 2 N O 2 (fc) ^ = ± N2O4 (k) 5. B A I T A P O N L U Y E N Bai 1: X la n g u y e n to thuoc n h o m c h i n h t r o n g bang t u a n hoan. Phan t r a m k h o i l u p n g cua n g u y e n to X t r o n g oxit cao nhat va t r o n g h a p chat k h i v a i h i d r o t u a n g u n g la a% v a b % , v a i a : b = 0,425. Phat b i e u nao sau d a y la sai? A . X thuoc c h u k i 3 t r o n g bang t u a n hoan. B. N g u y e n to'X a d i e u k i f n t h u a n g la chat ran. C. Phan t u o x i t cao nhat cua X c6 cue. D . N g u y e n t u X (6 t r a n g thai ca ban) c6 6 electron s. B a i 2: X v a Y la h a i n g u y e n to thuoc c u n g m o t c h u k y , h a i n h o m A l i e n tiep T o n g so h a t p r o t o n t r o n g n g u y e n t u X va Y la 3 9 va Zx < Z Y . N h a n xet nao sau d a y v e X, Y la dung? A . O x i t ciia X c6 t i n h baza m a n h h a n oxit cua Y. B. D p a m d i ^ n cua X I o n h o n d p a m d i f n cua Y. C. T i n h k h u cua Y m a n h h o n cua X. D . Ban k i n h n g u y e n t u cua X n h o h o n cua Y. B a i 3: H o a t a n h o a n toan 0,3 g a m h o n h p p h a i k i m lo?i (thuoc n h o m l A va 0 hai c h u k i l i e n tiep) vao nude, t h u d u p e 0,224 l i t k h i h i d r o (6 d k t c ) . H a i Vv^ loai d o la A. N a v a K . B.LivaNa. Si., ^264 • C.KvaRb. D. RbvaCs. AH Cam naiig 611 iuy(.n llii ciai hpc 18 chuySn dl Hba hgc Nijiiydn Van Hii Cty TNHH MTV DWH Khang Vi$t Cong thiic hop chat khi cua X v6i hidro: XHn Cong thuc hop chat ciia X v6i oxi: XjOg.,^ Theo bai: 2X+16(8-n) X 0,425 -> Li O va F nam trong cung mpt chu ki ma trong mpt chu ki tu trai sang X+8(8-n) 0,425 ^5,75R = 272-44n. phai ban kinh nguyen tix giam dan -> ban kinh: F < O < L i . ''^'^ ' wiiy-i,' Li va N a nam trong ciing mpt phan nhom chinh ma trong mpt phan nhom thi ban kinh nguyen tii tang dan -> ban kinh: L i < N a ! _>DapanA. Thay n = 1 deh 4, nhan thay n = 2 va X = 32 ( L u u huynh) thoa man. A diing vi nguyen to luu huynh 6 chu ki 3. 4 C sai vi phan tu SOa khong phan cue. f ' ' ' sa; B dung vi 6 dieu kien thuong, luu huynh la chat ran. •••^'^^^-'h'vf-'V^r-iHy-;'--;''', ' Gia s u 2 kim loai trong 2 chu ki lien tie'p c6 ki hieu chung la M .n 1 0' < D diing vi nguyen tu cacbon 6 trang thai co ban c6 6 electron s. <>!; i Dap an C . n n r , n = o --lufesv Bai 2: =0,04 (mol). Ta CO phuong trinh: M C O 3 + 2HC1 •(X:) tif Nhan xet: X va Y ciing thuoc mot chu ki va 6 hai nhom A lien tiep ^ ' ^ ^ ^ Mol: 0,04 Chung hon kem nhau 1 don vi dien tich hat nhan. Theo bai, so proton cua nguyen tu Y nhieu hon so' proton cua nguyen tir X Vay: M = > MCI2 <- 0,04 ., *' CO2 + H2O 1 1 < + vV : 4 0 4 ^ =32 -» Mpt kim loai can tim c6 M < 32 va mpt kim loai c6 M > 32. nen: Z Y - Zx = 1. Hai kim loai can tim la M g va C a - > Dap an B. Mat khac: Z Y + Zx = 39 Zx = 19 (Kali) va Z Y = 20 (Canxi). Bai 7: A diing vi K hoat dong manh hon nen oxit se c6 tinh bazo Ion hon. Phuong trinh hoa hpc phan hiiy O3: 2O3 Loai B v i do am dien ciia K nho hon cua Ca.. > SOi Nong dp O2 tao thanh la 0,045/2 = 0,0225 mol/1. Loai C vi tinh khii ciia C a yeu hon K. ^ Nong dp O 3 da phan ling: A C = (2/3).0,0225 = 0,015 mol/1. Loai D vi ban kinh nguyen tu ciia K Ion hon C a (trong mot chu ki, ban kinh nguyen tu giam dan). Toe dp trung binh ciia phan ling (tinh theo Os) trong 30 giay tren la V = (0,015 mol/1) / 30s = 5,0.10-^ mol/(l.s)-> Dap an D . Dap an A. Bai 8: -v-i.. Bai 3: Nhan xet: K h i gidm nhiet dp ciia he, can bang se chuyen dich theo chieu lam Gia s u 2 kim loai trong 2 chu ki lien tiep c6 ki hieu chung la M . tang nhiet dp ciia he -> chieu thu nhiet. n H j =0,01 (mol) Mat khac, theo bai ra, khi giam nhi|t dp, mau nau do ciia khi NO2 nhat . .^s - Phuong trinh hoa hoc: M Mol: + H2O dan, chiing to can bang chuyen djch theo chieu thuan. + -H2 2 0,01 > MOH 0,02 < mi <- Vay chieu thu nhif t la chieu thuan (A H > 0) ^ Dap an C . :CI:: Bai 9: (1) K h i tang nhi^t dp ciia h^, can bang se chuyen dich theo chieu lam giam nhiet dp ciia he Vay: M = =15. ^ 0,02 ^ tiic chieu thu nhi^t -> chieu thuan. (4) K h i gidm ap suat chung ciia he, can bang se chuyen dich theo chieu lam -> Mpt k i m loai c6 M <15 -> D o la L i . K i m loai ketiep la N a -> D a p an B. Bai 4: Nhan xet: C a c nguyen to'tir L i den F (la cac nguyen to trong cung mpt chi-' tang ap sua't ciia he tiic chieu lam tang so mol khi -> chieu thuan. Dap an B. Bai 10: ki 2), theo chieu tang ciia dien tich hat nhan thi ban kinh nguyen tii- giani Phuong an B: K h i tang nhiet dp ciia he, can bang se chuyen dich theo chieu dp am di^n tSng. lam gidm nhi^t dp ciia h$ -> tiic chieu thu nhiet -> chieu thuan. lar Dap an B. i - » Dap an C ; , , , > •:M-t; • 267 dm nanq On luyQn thi dgi hgc 18 chuyfln dg H6a hgc - NguySn van H5i Cty TNHH MTV DWH Khang Vi^t Chuyen de 12 Xrong t r u o n g hgip nay, hang so' can bang dug-c g p i la han g so' p h a n l i axit SV mm 1. L I - AXIT, BAZO, MIJOI Qia t r i ciia h l n g so'Ka chi p h u thuQC vao ban chat ciia axit va nhi?t d p . S I / D I E N LI AXIT, B A Z O , M U O I a. C h a t d i f n i i + Dinhnghia ^,fil/l>i.}:tWi^^ ' ) .,/<, \ •,. • .-r-..^ , , • i^j.,^,^^^^.,,^ Chat dien li manh /i .jito:S/.^:tau-sMi HCl nt::;! ., Chat di^n li manh la chat khi tan trong nuoc, cac phan tu hoa t?n deu phan > + CI- . ) i . . . . . > CH3COOH + CI<=± >; ; j i h ni, < < !/i CHsCOO- + KinsvidomfeM:-:H/fi IOC, i 1 ii.,,;af::/>: A x i t n h i e u na'c: A x i t k h i tan t r o n g n u o c m a p h a n t u p h a n l i n h i e u Ian ra i o n li ra ion, v i du: HCl , ?1 , j 1 j;xi y^xit la chat k h i tan t r o n g n u o c p h a n l i ra cation H"^: Chat dien li la nhiing chat khi tan trong niroc phan li ra ion. + K h a i n i # m axit theo thuyet A-re-ni-ut ' ^ Na2S04 > 2Na* + SO^^ • H * gpi la axit n h i e u na'c. ' ; V i d u H3PO4 la axit 3 na'c: H3PO4 ; H2PO; < = t ' ' i en, ^n'' mu NaOH Su d i ? n l i ciia cac i o n k h i hoa tan N a C l t r o n g n u a c . + -> Na^ + O H - 1 " t ((,0; Chat dien li yeu Cha't d i e n l i yeu la chat k h i tan t r o n g nuoc chi c6 m o t p h a n so' p h a n t u hoa A tan p h a n l i ra i o n , p h a n con lai v a n t o n tai d u o i dang p h a n t u , v i d y : CHaCOOH < = ± CH3COO- + ^ 5, Sy p h a n l i a i a chat d i ? n l i y e u la qua t r i n h t h u a n n g h i c h . Can bang dien H la m o t can b a n g d o n g va c u n g t u a n theo n g u y e n l i c h u y e n d j c h can bang La m M i n h hoa k h a i n i e m axit-bazo ciia A - r e - n i - u t . Sa-to-li-e. D Q d i ? n l i a (anpha) cua chat d i f n l i la t i 1§ g i i i a so p h a n t u p h a n l i ra iof lidroxit l u o n g tinh ^ H i d r o x i t l u o n g t i n h la h i d r o x i t k h i tan t r o n g nuoc, v i r a c6 the p h a n l i n h u tren t o n g so' p h a n t u hoa tan.. H a n g so'phan l i axit, vira c6 the p h a n l i n h u bazo. G i o n g n h u m p i can bang hoa hpc, can bSng di§n l i d i n g c6 hang so caf V i d u , Zn(OH)2 la h i d r o x i t l u o n g t i n h : bang, g p i la h a n g so' p h a n l i . Phan l i theo k i e u bazo: Zn(OH)2 Phan l i theo k i e u axit: Zn(OH)2 Xet can bang: CH3COOH ^=± Tac6:K.= ^ '^^COO^llH*! [CH3COOHI CHaCOO" + K. Cac h i d r o x i t l u o n g t i n h t h u o n g gap Al(OH)3, Cr(OH)3. ' <=± Zn^- + 20H- ZnO^ + 2H* la: Zn(OH)2, Sn(OH;2, Pb(OH)2, Ca'm nang 6n luygn thi dgi hpc 18 chuy6n dg H6a hqc - Nguygn Van Hai d. Muoi Muo'i la hgp chat khi tan trong nuoc phan li ra 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 con c6 hidro c6 kha nang phan li ra ion H* th; muoi do 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 con c6 hidro c6 kha nang phan li ra ion thi muoi do duoc goi la muoi axit. Vi du: NaHS04, NaHCOs, Na2HP04. 3. pH CUA DUNG DICH a. Khaini^m pH la dai lugng dac trung cho nong do ion cua dung dich. Neu bieu dien nong do H* duoi dang he thuc: [H*] = 10-'' mol/1 thi a duoc goi la pH cua dung dich. Quan he giua [ H ^ va [OH-] trong dung dich: [H*].[OH-] = 10"". b. pH va moi truong dung dich ' Moi truong trung tinh c6 pH = 7. ' sB$ki ' Moi truong axit c6 pH < 7; pH cang nho, dp axit cang Ion. ' Moi truong baza c6 pH > 7; pH cang Ian, dp baza cang Ion. 0 OS axit tang CtyTNHH MTV DVVH Khang Vi§t jj. phan liiig tao thanh chat dien li yeu ^ phan ling tao thanh nuoc NaOH + HCl > NaCl + H2O OH- + H* — > H2O Mg(OH)2 + 2HC1 — M g C l 2 + 2H2O Mg(OH)2 + 2H* — ^ Mg2* + 2H2O + phan ung tao thanh axit yeu - Trung ttnh 'I 1 llf I / H" + CHsCOO> CH3COOH c. Phanrnigtao thanh chat khi + Na2C03 -^2HC1 COf + 2H^ > 2NaCl +C02t +H2O > C 0 2 t +H2O + CaCOs +2HC1 CaCOs + 2H* + ZnS +2HC1 )• CaCh +C02t +H2O > Ca2* + C02t +H2O > ZnCh +H2St * ' • ' ^">'' ZnS + 2H^ > Zn2^ + H2St * '''' 5. CAC V I D V M A U Vi d\ 1: Theo thuyet A-re-ni-ut, phan ung nao sau day khong phai la phan ling axit-bazo: A. NaOH + HCl > NaCl + H2O B. Zn(OH)2 + 2HC1 — > ZnCh + 2H2O > Na2Zn02 + 2H2O D. NH3 + HCl — ^ NH4CI Ldigidi: Do bazd tang 14 Thang pH cua dung dich. 4. PHAN l/NG TRAO DOI ION a. Phan ung tao thanh chat ket tiia + Na2S04 + BaCh Ba^^ + X>1~ + 3NaOH + FeCh Fe3* + 30H- > BaS04 4+2NaCl > BaS04i > Fe(OH)3 4' +3NaCl 11 > CH3COOH + NaCl HCl + CHsCOONa C. Zn(OH)2 +2NaOH •7 r ,5., , Dap an D. Nhan xet: Theo thuyet A-re-ni-ut, axit la chat c6 kha nang phan li ra H*, baza la nhirng chat c6 kha nang phan li ra OH- nen phan ling axit-bazo luon tao ra H2O. Theo thuyet A-re-ni-ut, HCl la axit nhung NHs khong phai la baza vi no khong CO kha nang phan li ra 0H-. d\ 2: Cho cac phan ling xay ra trong dung dich: (l)ZnS + 2HCl > (3)(NH4)2S +2HC1 ^ > ' (2) K2S + H2SO4 (loang) (4) CaS + H2SO4 (loang) > Cac phan ling c6 phuong trinh ion riit gon: S^" + 2H* ——> H2S t la A. (1), (2), (3). B. (1), (3). C. (2), (3). D. (2), (3), (4). > Fe(OH)3 4' 271 CtyTNHH MTV PWH Khang ViQt elm nang On luy$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 2 n h o m O H : S 0 2 ( O H ) 2 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 6 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 2 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 ra OH' ,, ' 0 (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) 2Ca2* + S2+ SO 4 CaS04i + H 2 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 = 0 4 + 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 + D a p an A . Vi 'W „, fifO^if '^eif:> 6: M p t d u n g d i c h chiia: 0,1 m o l AP*; 0,1 m o l Cu^*; 0,2 m o l SO 4' v a i o n A. 28,30. B. 31,85. C. 35,40. Lai 0,08 D . 19,05. giai: G p i so'mol i o n C I " la x. A p d u n g d\nh luat t r u n g hoa d i | n -> A I O 2 fo Cl~. C o can d u n g d i c h nay 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 0 2 m o l -)• [OH-] = rfx ; ! ,, Na Ba^^ + SO^- = " H C l + 2 n H 2 S 0 4 = 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 0 1 m o l ; n^^+ = a 0 2 m o l . 0,5a > Ba(OH)2 + H 2 Mol: a T r o n g X: D . 9,34. )4! NaOH + - H 2 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 •" + 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 thanh 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 6 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 ra cation H"^ la axit. c 6 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 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... D.2,0. Laigidi: Latgtat: [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 khi C O 2 (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. 7. , Trong Y : nnfj. n„ 2+= 0/01 mol; Ba^ ^ N h a n thay: — = "CO2 A.l. fffo < iiq:3v iari hio ami md •':)<": • A. 7. A. 2,33; 0,15. C.5. ,. ^^^.^ ^ ^ j ^ ,^ ^ f t f ^ p ^ y f . ; . ^ ' - Q M r h t l B. 1,2. C . 1,0. D . 12,8. ,„,fjej B. 2,33; 0,30. C . 4,66; 0,20. D . 6,99; 0,50. „; A (4) CO ket tua; (1) tac dung voi (3) c6 sui bpt. D u n g dich B a C h va Na2S04 > * A • A. Hang so phan li axit tang. B. Hang so'phan li axit giam. C. Dp di^n li tang. theo thu tu dung trong Ip dupe danh so la A. (2), (4). c6 dp di^n li a . Phuang ' 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 D . Do dien li giam. trinh lien h^ giua hang so phan li axit Ka voi C va a la ' tac d y n g voi (2) c6 ket tiia; (2) tac dyng voi (3) c6 ket tua; (2) tac dung voi Bai 3: Axit axetic la mpt chat di^n 1 ye'u. K h i pha loang dung djch axit axetic 1 Bai 4: 6 25''C, dung dich axit axetic nong do C mol/1 D . H C O 3 , Na*, Ba^* va O H " . B a ^ CI- va SO ^ . ' ' H2SO4, Na2S04. D a n h so ngau nhien (1), (2), (3), (4). Ket qua thi nghi^m: (1) D . 17,13%. dieu nhan djnh nao sau day la dung? 'S Bai 12: C o 4 Ip hoa chat chua 4 dung djch rieng biet mat nhan: BaCh, Na2C03, 'D.2. C . 14,29%. B. K*, F e ^ C h va S O C. dung dich axit tren la J D . 6. A. K ^ C u ^ CI- v a O H - . dung djch axit fomic a007M c6 [H*] = lO"' M . D p di^n li a ciia B.3,26%. C . 12. Bai 1 1 : Trong dung djch nao sau day, cac ion eo the ton tai dong thoi? C H 3 C O O H , C a ( O H ) 2 , C H 3 C O O N H 4 . So chat di|n 1 la 1 A. 28,57%. B. 13. dich CO p H = 13. G i a trj m va a Ian lupt la 1: C h o day cac chat: KA1(S04)2.12H20, C 2 H 5 O H , C12H22O11 (saccarozo), 6 250C, D.4.,t,..,,,iv,:,f. , 300ml dung djch Ba(OH)2 a mol/L, thu dupe m gam ket tiia va 500ml dung 6. B A I T A P O N L U Y E N Bai 2: C.3. ]• , ; ^ j f / djch gom N a O H 0,2M va Ba(OH)2 0,1 M, thu dupe dung dich X. p H cua 1153003 = ^^^2- = 0,01 mol. j j ^ ^ . B.4. , . ' l . : ; * ' / •••fi:,^, Bai 10: Trpn 200ml dung djch hon hpp gom H C l 0,10M va H2SO4 0,05M voi -> m = 0,01.197 = 1,970 gam - » Dap an C . A. 3. -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 2 - " " O H - • " C O 2 = 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 3 " : 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 > 1 -> Phan ung tao ra 2 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 . 1. 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 4 dung djch: Na2C03, NaHS04, Na3P04, N a O H . So dung djch c6 Lcngiai: "oH-""NaOH+2nBa(OH)2 ; 0,1M va B a C h 0,2M, thu dupe m gam ket tua. G i a trj ciia m la 5 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 khi C O 2 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*^'^ trong coc X sau phan ung la: A . 10. 274 B. 11. , C . 12. D . 13. bay ra 6 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 4 chat bpt mau trSng: C a C h , CaS04.2H20, N a N 0 3 ; C a C 0 3 ? 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 , D a p a n C . 0/007 M mvjC: Bai 3. Loai d a p an A , B v i hang so dien l i chi p h u thupe vao nhiet d p , k h o n g p h u A.CUSO4. B.Fe(N03)3. C.AlCls. D . Ca(HC03)2. Bai 19: Cho V m l d u n g d i c h K O H I M vao coe d y n g 150ml d u n g djeh A l C b I M ta t h u d u p e 7,8 g a m ke't tua. Gia t r i nho nhat va I o n nhat eua V Ian l u p t la A . 300 va 500. , • B. 300 va 600. C. 200 va 400. D . 200 v a 500. B a i 20: Cho V m l d u n g d i c h N a O H I M vao 100ml d u n g d i c h chua d o n g thoi H C l 0,2M v a Z n C h 0,2M. Gia t r i I o n nhat ciia V de t h u d u p e 0,99 g a m ket A . 60. B.90. C.70. D . 80. Bai 2 1 : Cho V l i t d u n g d i c h N a O H 2 M vao d u n g djeh ehiia 0,1 m o l A h ( S 0 4 ) 3 va 0,1 m o l H2SO4 d e n k h i phan u n g hoan toan, t h u d u p e 7,8 g a m ket tua. Gia t r i I o n nhat ciia V de t h u dupe l u p n g ket tua tren la A . 0,45. B. 0,35. C. 0,25. thupc vao n o n g d p axit. K h i pha loang, n o n g d p axit g i a m Bai4. k h i cho 70ml d u n g d j c h K O H 2 M vao 100ml d u n g d i c h A , d e u t h u duoc Phuong trinh d i f n li C0,5. 7. H U ' 6 N G D A N - L 6 1 G I A I •.^.(•f.;- .• aC aC (l-a)C aC Can bang: K i h .Sjmi! ^ KA1(S04)2.12H20 > CH3COOH CH3COO" > C(l-a) ^ , 1-a ,^3 ' ^ '-hihi j^r'n^rirlr^'<•••) P h u o n g t r i n h phan u n g : + H2SO4 0,05 <- ' > Na2S04 + H2O 0,025 ,«;(«t>l n ... Nhan xet: so m o l N a d u = 0,10-0,05 = 0,05 m o l nen xay ra tiep p h a n u n g : Na + H2O > NaOH Na 0,05 -> p H = 1 4 - 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 = 1 _ . Bai 6. + AP* + 2SO^- + I2H2O Ca^* + 2 0 H - [CH3COOH] • 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: ^^^^ aC = [ C H 3 C O O - ] . [ H - ] ^ C o u C a ^ C o l _^ mol: D . 0,07. - ^Qf;,, :,. aC tfinOOrttC)t * . : Bail. H* C bang n h a u . Gia t r j cua x la C.0,02. + Bandau: 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 < = > CH3COO" 2Na D . 0,4. 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. 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 B.0,2. D a p an C. Bai 5. D . 0,05. Bai 22: A la d u n g d i c h Z n ( N 0 3 ) 2 x m o l / I . K h i cho 30ml d u n g d i c h K O H 2 M va A . 0,3. d p d i ^ n h tang ; Di^nli: tua la Ca(OH)2 ^ :U,0 <•' Bai 18: N h o t u t u d u n g d i c h N a O H den d u vao d u n g d i c h X. Sau k h i cac phan d u n g djeh X la ; rl . , . i o u , < . , . ; HCOOH D . 19,78. u n g xay ra h o a n toan chi t h u dupe d u n g dfch t r o n g suo't. Chat tan trorig , Phuong trinh di?n l i : Bandau: Difnli: Khang Vigt MTV D W H +N H ; — > 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 1 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 nay 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. = 2 nH2S04 + " H C I = 2.0,2.0,05 + 0,2.ai = 2 nH2S04 + " H C l = 2.0,4.0,0375 + 0,4.0,0125 = 0,035 mol. Tacophuangtrinh: OHMol: 0,03 + ^ ' ^ — > = 2 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; Na3P04 _» n Ta ^ i r»V«J> 9m,. ( d u ) = 0,6a - 0,04. > H2O -ii 0,03 ^ .,jfj., = 0,01'^' pH = 12 ^ Dap an B. " : 0,6a - 4 > H2O ' A XI 0 . [OH"] = , . i , . i 0,02 " ' " ^ = 0,01 M + 0, 0,1 0,1 p O H = - l g [ 0 , l ] = l -> p H = 1 4 - p O H = 1 3 - > D a p a n A . 278 = a05 -> a O .s, / ' ^ = ai5 ^ Dap an A., , ^^.^^^ ^'^'^ ' gj^j > H2O + C O 3 " > 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 thanh ket tua -> (2) la dung dich B a C b . Mat khac, (1) tac dung voi (3) c6 sui bgt khi ^ (1) v a (3) la cap chat Na2C03 Na2C03 + H2SO4 > N a 2 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 °" 1 Dap an A. " H + = 2 nH2S04 + " H C i = 2.0,1.0,05 + 0,1.0,01 = 0,02 m o l . + H^ ^ 0,02 IJl' va H2SO4: Bai 9. OHMol: 0,02 .^^'"^^orii," Vay a dap an D vira tao thanh chat di|n li yeu, v u a tao chat ket tiia. f Vay n ^ ^ . ( d u ) = 0,032-0,03 = 0,002-> [OH"] = ^ ->pOH =2 ^ '' [H^] = IQ-i^ -> [OH-] = 10"' = 0,1 Ba2- + C O ^ - Ta CO p h u o n g t r i n h : -» ~ ' Lo^iDvi: H C O ; + O H " " 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: * O H " d u , H * het Lo^i C v i Ba^"^ ket hgp duQfC voi S O ] ' tao ra BaS04 ket tiia. ' H* " " Loai A v i Cu^* ket hgip d u g c voi O H tao ra Cu(OH)2 ket tua. • + ^ , mpt de tao thanh chat ket tua, bay hoi hay cha't di?n li yeu. > Na* + OH~ OH- < ^.^ Bai 11. Dap an B. C a c ion trong day nay khong ket hgp du^c voi nhau tung doi + OH- P O ^ - + H2O < = > H P O ^ - + O H - Bai8. CO >H20 ^ ^ " )HJJ/. i 0,04 Theo bai ra : p H = 13 ^ > 3Na- + P O ^ NaOH ' ' * H^ <- vay : n ^ „ _ Phuong trinh di?n li tuong ung: C O 3-" + H2O < — ^ H C O 3 + Mol: 0,04 ^ 'XIA. > 2Na* + C O 3 " OH- pH =2 Cac dung dich do la: Na2C03, Na3P04, N a O H . Na2C03 ^'^^ Xa CO sau phan ung p H = 13 (moi truong baza) ,:•)«':. „ 0,03 Bai7. ' "Ba2+ = a04 mol. ^ n^^j-f = O'Ol m o l ; n^^^^ = 0,02 m o l . nBaCl2 = 0'2-ai=0,02mol -> ng^2+ = 0'02mol , ^ ' ~> ]^ng^2+=0/03mol | 0 . 0 « ^0,0 K h i trQn 2 dung dich nay voi nhau se xay ra cac phuong trinh sau: HCO; Mol: 0,02 + OH- > H2O + C O 3 V 0,02 >>"'V *- -J 1 ?V! li 279