Tài liệu Cẩm nang ôn luyện thi đại học 18 chuyên đề hóa học 1

TS. NGUYEN VAN HAI (Chu bien) NGUYEN NAM TRUNG - TRAN THE NGA - NGUYEN THj THU HA Cam naligFluyen thi dai h c o 18 CHUYEN OE ^ HOA HOC ^ He thons cac phUdng phap siai nhanh bai tap hoa hoc. ^ Duns cho on tap va thi tot nshiep THPT.,. ^ Luyen thi vao Dai hoc va Cao dans. rW VJEWTIWHSINHTHUAN D m MHA yi lAT RAN RAI HOC QUOC GIA HA NOl Cty TNHH MTV DWH Khang Vi^t r h u y e n de 1 , C A C PHlrtl^NC P H A P G I A I N H A N H MUC LUC huyen di 1. C a c phucmg phap giai nhanh ? 1. P H l / O N G P H A P B A O T O A N K H O I L l / Q N G 3 huyen de 2. C a c axit v6 ca d i l n hinh 71 huydn de 3. T i n h chat c u a cac hidroxit 99 . ^. , a. N g i d u n g « ^. Tong khoi lu M C I + -Hi 2 M +H2O > M O H + ^Hi 2 - . Nhan xet: Bao toan khoi lu^ng -> m y = mM+ m J . + m^^^. 370 e thi thir dai hoc H C l 0,1M. C o can dung dich sau phan ving thu du(?c 3,86 chat ran khan Y . 378 • 352 huyen di 18. Polime v a vat lieu polime ling - (khoi luong chat ket tua + chat bay hoi) 341 huyen d6 17. A m i n - A m i n o axit - Peptit Khoi luong dung dich sau phan ung = Khoi \ugng dung dich truoc phan ^ m ^ ^ . =3,86-1,794-0,04.35,5 = 0,646 gam ^ n^^^. = 0,038 mol. 1 794 - n M . n H a - n „ „ . = 0,078 ^ M . ^ . 2 3 (Na).^ ^a:,-., —> D a p an A. Vi 2: Cho 100ml dung dich H3PO4 a molA vao 100ml dung dich K O H 2 M thu dugc dung dich Y c6 chua 15,44 gam hon hop muoi. G i a tri cua a la A. 0,75. B.1,00. C.0,50. Lai D . 0,80. gidi: Nhan xet: V i Y chua hSn hop muoi - > chac chan c6 chiia muoi axit - > K O H phan ung het. S a d o p h a n u n g : H3PO4 + K O H > Muoi + H2O. Bao toan khoi luong: 0,la.98 + 0,2.56 = 15,44 + 0,2.18 • * - > Dap a n D . , i a = 0,80.^•' ' ^^ | . . Cty TNHH MTV DWH Khanq Vigt dm nangflnluy$n thi d + B. 8,96 lit. moxi =moxit -> = ^ ^ = 0'4 mol -17,4 = 12,8 gam. = 8,96 lit ^ Dap an B. ^ ' ^ m o x i = m o x i t —> m o x i = 3,33 12 = TIT = 0,0375 mol — 2,13 = 1,2 gam. ; + 2H^ > Truoc het, c a n x e m axit c6 p h a n l i n g het h a y - ^^ ^^ = 0,5.1 + 2.0,5.0,28 = 0,78 Khoi lugng muoi = m^|3+ + se gap kho k h a n b a n g e a c h s o s a n h so > H2. mol. , , , n^+ het -> X c h i c h i i a c a c m u o i . ^^i- ^^^^2+ + , ' + ^^^2- ' = 7,74 + 0,5.35,5 + 0,14.96 = 38,93 g a m . , ; j v ^ ^ ^ y . , _ -¥ Dap a n A. ' > V i dv 7: Nung h o n h g p bgt g o m 15,2 g a m Cr203 v a m g a m A l a nhi?t d g cao. Sau k h i p h a n u n g h o a n toan, t h u d u g c 23,3 g a m h o n h g p r a n X. Cho X p h a n l i n g v o i axit H C l ( d u ) thoat r a V lit k h i H2 (6 d k t c ) . Gia tri c i i a V l a A. 7,84. B.4,48. C. 3,36. D. 10,08. Lim t r u o c he't cac e m c a n t i n h d u g c so m o l Al ban dau. y rang, trong p h a n u n g n h i f t n h o m , t h u o n g a p d u n g d i n h luat bao t o a n k h o i l u g n g , cia t h e : mc^03 + i"Al=mx m A i = 23,3 -15,2 = 8,1 g a m -> n ^ r 0,3 m o l . .0 VHCI = 0,075 lit = 75ml. -> Dap an C. V i dv 5 (A-12): Hoa tan hoan toan 2,43 gam hon hgp gom M g va Z n vao mpt , lugng vira d i i dung djch khong m o l H " b a n d a u v a s o m o l k h i H2 b a y r a theo ti 1$: 2H* H2O. Suy ra: n^+ = 0,15 mol -> nnci = 0,15 mol -> H2SO4 loang, sau phan ling thu dugc 1,12 lit (dktc) va dung dich X chiia m gam muoi. Gia tri ciia m la A. 4,83 gam. n e u cac e m g i a i theo p h u o n g t r i n h p h a n l i n g v i can viet d e n 4 p h u o n g trinh. Nhan xet: Bai n a y no = 0,075 mol V - ( o x i t ) = 0-075 mol. Nhan xet: Bai n a y D. 77,86 g a m . Lai giai: Khi cho oxit bazo tac dung vai axit tao ra nuoc: 02- B. 103,85 g a m . C. 25,95 g a m . Lai giai: Mat k h a c : n u , = 0,39 m o l -> Lai gtat: Nhan xet: bai nay cac em nhat thiet phai ap dung bao toan khoi lugng de tinh kho'i lugng oxi tham gia phan ling: + A. 38,93 g a m . nj^+ = H H C I + 2nH2S04 V i dv 4 (A-08): Cho 2,13 gam hon hgp X gom Mg, Cu va A l 6 dang bot tac dyng hoan toan voi O2 thu dugc hon hop Y gom cac oxit c6 khoi luong 3,33 gam. The tich dung dich HCl 2M vira dii de phan ling het voi Y la A. 60ml. B. 45ml. C. 75ml. D. 80ml. mkimioai H2 ( d k t c ) . Co c a n d u n g d j c h X t h u d u g c l u g n g m u o i k h a n l a C. 17,92 lit. m o x i = 30,2 V02 V i dv 6 (CD-08): Hoa t a n het 7,74 g a m h o n h g p bgt Mg, A l b a n g 500 m l d u n g d i c h h o n h g p HCl I M v a H2SO4 0,28M t h u d u g c d u n g d i c h X v a 8,736 lit k h i B. 5,83 gam. C. 7,33 gam. ^ D. 7,23 gam. Lai giai: O bai nay, cac em c6 the giai chi tiet dua theo phan ling hoa hgc. Nhan xet: nH2S04 = r'H2 = 0,05 mol. So do phan ling: K i m l o a i + H2SO4 > Muoisunfat + H2 Bao toan kho'i lugng cho so do tren: ^ 6I>''• ->• m = 2,43 + 0,05.98 - 0,05.2 = 7,23 gam ' ii£0tOB>; ->DapanD. , , ^'^ ^^-^^^ ' ^ H2 Phuong trinh h o a h g c : Cr203 + 2A1 Mol: ai 0,2 X g o m : A l d u = 0,1 m o l ; Cr = 0,2 m o l . 3 Khi X tac d \ i n g v o i axit: A l ^ -> n H 2 = 0,35 mol VH2= V i d\ 8 (B-09): Cho 100ml 0,5M, lit - » dung djch 2Cr + Al2C)3. 0,2 0,1 (j^p.^v,-.^ ' ' . ^^-^ v a C r - > H2 - H 2 7,84 > Dap a n A. K O H 1,5M vao 200ml d u n g d j c h H3PO4 t h u d u g c d u n g d j c h X. Co c^n X t h u d u g c k h o i l u g n g c h a t r a n k h a n l a A. 15,5 gam. B. 18,2 gam. C. 12,8 gam. D. 16,4 gam. Lai giai: " K O H = 0,15 m o l ; nH3P04 = 0,10 . Nhan xet: Cac em c6 the xet ti le m o l t r i n h p h a n u n g va d a t so m o l de g i a i . K O H va H3PO4, s a u do v i e t 2 phuong V i d u 11 (B-12): Hon hgp X gom 0,15 mol vinylaxetilen va 0,6 mol H 2 . Nung Tuy nhien, ne'u ap dung bao toan kho'i lugng cho s o do: H3PO4 + KOH : nong hon hgp X (xiic tac Ni) mot thoi gian, thu dugc hon hgp Y c6 t i khoi so > Muoi + H2O Taco: mH3P04 + rnKOH= 1TV DVVH Khang Vi$t Cty TKi, Ca'm nang 6n luyfn thi d^i hpc 18 ctiuySn dg H6a hpc - Nguygn Van HSi voi H 2 bang 10. D i n hon hgp Y qua dung dich brom du, sau khi phan ung niHjO xay ra hoan toan, khoi lugng brom tham gia phan ung la J^y^^yr^^ , "H2O=0,15. A. 0 gam. B. 24 gam. •) . C. 8 gam. ; , D. 16 gam. Lai gidi: mx = 0,1.98+ 0,15.56-0,15.18 = 15,5 gam Bao toan khoi lugng: mY = mx = 0,15.52+ 0,6.2 = 9 gam. —> Dap an A. V i du 9: D u n nong hon hgp khi X gom 0,06 m o l C2H2 va 0,04 mol H2 v o l xiic tac N i , sau mot thoi gian thu dugc hon hgp khi Y. Dan toan bg Y Igi t u t u 5 Mat khac: M Y = 10.2 = 2 0 " '^ "2= = 0/45 mol , • qua binh dung dung dich brom (du) thi con lai 0,448 lit h6n hgp khi Z (6 Nhan xet: vinylaxetilen c6 chua 3 lien ke't 71 —> So' mol lien ke't K ban dau = dktc) CO t i kho'i so voi O2 la 0,5. Khoi lugng binh dung dich brom tang la 3.0,15 = 0,45 mol. V - A . 1,04 gam. I' B. 1,32 g a m . • C. 1,64 gam. Gia su so mol H2 tham gia phan umg = a mol D. 1,20 gam. Laigiai: ;;(;'; Nhan xet: Bao toan khoi lugng: mY = mx = 0,06.26 + 0,04.2 = 1,64 gam. 0 448 or,, Mat khac: n z = = 0,02 mol va M z = 0,5.32 = 16 ling = a mol. - > a = 0,75 - 0,45 = 0,3 mol. , - > So mol lien ke't 7i d u = 0,45 - 0,3 = 0,15 mol = So'mol Br2 phan ung. Khoi lugng brom phan ung = 0,15.160 = 24 gam. mz = 0,02.16 = 0,32 gam. Luu y: Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thy. Bao toan khoi lugng, ta c6: Kho'i lugng binh brom tang = mv - mz = 1,64 - 0,32 = 1,32 gam. ,.j .j-^,^, •^i^^.j^.^ Dap an B. V i d u 10 (A-10): Dun nong hon hgp khi X gom 0,02 mol C2H2 va 0,03 mol H2 trong mot binh kin (xuc tac Ni), thu dugc hon hgp k h i Y. Cho Y Igi t u t u binh tang m gam va c6 280ml hon hgp khi Z (dktc) thoat ra. Ti kho'i aia Z so % ia08. Gia tri ciia m la A.a585. B. 0,620. • C. 0,205. ,iaw;. : , D. 0,328. Lbi gidi: , Nhan xet: Bao toan khoi lugng: mv = mx = 0,02.26 + 0,03.2 = 0,58 gam. Mat khac: n z = ^ = 0,0125 mol va M z = 10,08.2 = 20,16 22,4 ? mz = 0,0125.20,16 = 0,252 gam. Nhanthay: 'r^^.^ i i,*!.'.a " ; .m»s;c,ci ;/ > • ,• Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thu. Bao toan khoi lugng: m = mv - mz = 0,58 - 0,252 = 0,328 gam. —^ Dap a n D . > , —> Dap an B. V i d u 12: Hon hgp X gom 0,1 mol etilen, 0,2 mol axetilen va 0,5 mol H 2 . Nung nong hon hgp X (xiic tac Ni) mgt thoi gian, thu dugc hon hgp Y c6 t i khoi so v o i H 2 bang 10. Dan hon hgp Y qua dung dich brom du, sau khi phan ling xay ra hoan toan, tha'y c6 m gam brom tham gia phan ung. Gia trj cua A. 16. B.32. vao binh dung djch brom (du), sau khi ke't thiic cac phan ung, khoi lugng voi H2 la So mol lien ke't n phan Mat khac, so mol khi giam = so mol H2 phan ung = n x - n Y nkfusi'^'^ 22,4 , • ,,^.J^:n. C.8. Laigiai: D.24. " ' , ' ; ,,s*^->r:: ':^:s,rf'^~ Bao toan khoi lugng: mY = mx = 0,1.28 + 0,2.26 + 0,5.2 = 9 gam. Mat khac: M Y =10.2 = 20 n z = = 0,45 mol " ' Nhan xet: etilen c6 chiia 1 lien ke't 71, axetilen c6 chiia 2 lien ke't 71 - > So mol lien ke't 7t ban dau = 1.0,1 + 2.0,2 = 0,5 mol. Gia su so mol H2 tham gia phan ung = a mol - > So' mol lien ke't n phan ling = a mol. Mat khac, so mol khi giam = so mol H2 phan ung = nx - nY ^ ^ - > a = = 0,8 - 0,45 = 0,35 mol. - > So mol lien ke't 7t d u = 0,5 - 0,35 = 0,15 mol = So mol Br2 phan ung. Khoi iugng brom phan ung = 0,15.160 = 24 gam. —> Dap a n D . ' 7 Cty TNHH MTV DWH Khang Vijt dm nang On luygn thi dgi hpc 18 chuySn dg H6a hpc - Nguygn van HJi V i dv 13: Cho 2,1 gam h6n hgp X gom hai amin (no, don chiic, dong dang ke tiep) phan ling het voi dung dich HCl (du), thu du(?c 3,925 gam hon hgp muoi. Cong thuc cua hai amin trong X la A. CH3NH2 va C2H5NH2. B. C2H5NH2 va C3H7NH2. C. C3H7NH2 va C4H9NH2. i '> • D. C H 3 N H 2 va (CH3)3N. Y^ic Tuy nhien, bai nay dugc giai nhanh chong khi ap dung bao toan khoi lugng: maxlt + mbazo= mmuol + mnuoc _^ mH20= 3,6+ 0,06.56+ 0,06.40-8,28 = l,08gam^> nH20= 0,06 mol. Mat khac: n H , o = ^ S 0 >. Lot gidi: 36 = 60 0,06 = Max.t Axit la C H 3 C O O H • . .V -> Dap an B. Nhan xet: Loai D vi hai amin khong dong dang ke tiep. Cac phuong an con V i du 16: Dot chay hoan toan hon hgp X gom hai este dong phan can diing lai deu la amin don chuc, bac I. 7,84 lit khi O2, thu dugc 6,72 lit khi CO2 va 5,4 gam H2O. Neu cho m gam Khi cho X tac dung voi axit HCl: X tac dung vua dii voi dung dich NaOH, c6 can dung dich sau phan ung R-NH2 + HCl > R-NH3CI Bao toan kho'i lugng: mamin ' i<- + mnci = mmuoi • >' . ling ete hoa hoan toan Z, thu dugc m gam ete. Cac the tich khi deu do 6 ; dieu ki?n tieu chuan. Gia tri cua m la mnci = 3,925 - 2,1 = 1,825 gam -» nnci = 0,05 mol = namm ^ A. 2,65. Mamin = 7 ^ = 42 ^ R - N H 2 = 42. 0,05 * Dap an A. V i dij 14: Dot chay hoan toan m gam hon hgp X gom 3 ancol (don chuc, thugc Cling day dong dang), thu dugc 26,4 gam khi CO2 (dktc) va 19,8 gam H2O. Neu thuc hien phan ling ete hoa m gam X (hieu suat 100%) thi tong khoi lugng ete thu dugc la B. 14,80 gam. C. 10,92 gam. D. 12,90 gam. Lai gidi: Nhan xet: n^jo = 1/1 > ^C02 = 0,6 ^ X gom cac ancol no, mach ho va: "X =nH20 -rtC02 =1,1-0,6 = 0,5mol -> n(0)x =nx =0,5 Bao toan kho'i lugng: + nHjO i^H20 = 0,25 mol. D. 3,25. este la C n H 2 n 0 2 . = 0,3 mol. ^ ^ este deu no, dan chuc -> Cong thiic cua 2 ^^ * ' • " ' ••.: Bao toan nguyen to'oxi: 2n ^^te 2n _> n = ^^^22. = Heste =3 = 2n CO2 ^ H2O ~ ^ " este 0,1 mol. 2 este trong X c6 cong thuc phan tu la C3H6O2. 0,1 -> Cong thuc cau tao 2 este la: HCOOC2H5 va CH3COOCH3. Phan ling hoa hgc: HCOOC2H5 + NaOH — ^ HCOONa + C2H5OH C H 3 C O O C H 3 + NaOH —!—> CHsCOONa + C H 3 O H nNaOH = neste .' = nz = 0,2 mol. meste+mNaOH ^ = mv + mz -> 0,1.74 + 0,1.40 =7,85 + mz ^ mz =3,55 gam. Ht:k{OCC;i Nhan xet: Trong phan ung ete hoa ta c6: mete = 17,4 - 0,25.18 = 12,90 gam -> Dap an D. V i d u 15 (B-08): Cho 3,6 gam axit cacboxylic X (no, don chiic) tac dung hoan toan voi 500 ml dung dich gom K O H 0,12M va NaOH 0,12M. Co can dung dich thu dugc 8,28 gam chat ran khan. Cong thiic cua X la A.C2H5COOH. Nhan xet: UQQ^ = nH20 Bao toan kho'i lugng: Bao toan kho'i lugng: "^x = r^ete C. 5,15. ' " " n 02 = 0,35 mol; n CO2 = 0,3 mol; n De tha'y: mx = mc + niH + mo = 0,6.12 + 2,2.1 + 0,5.16 = 17,4 gam. Trong phan ung tao ete thi: nx = Zn^jO B.3,70. ^ Lai gidi: -> R = 26 -» Hai goc hidrocabon la C H 3 - va C2H5-. A. 8,40 gam. thi thu dugc 7,85 gam chat ran khan Y va hon hgp ancol Z. Thyc hi|n phan B.CH3COOH. C.HCOOH. D.C3H7COOH. Lai gidi: Nhan xet: Theo bai, axit phan ung hoan toar\i baza c6 the con du, do vay neu giai dua vao phuong trinh phan ung se rat kho khan. 1 „ , H2O = T n z n H20 = 0,05 mol. 2 Bao toan khoi lugng: mz = mete n ,mw>'^''iOfU nAoi • i ; y 7.- . + m H20 -> m = mete = 3,55 - 0,05.18 = 2,65 gam -> Dap an A. ,s :sv V i d^ 17: Xa phong hoa hoan toan 0,1 mol este X (dan chiic, mach ho) bang 100 g^m dung dich M O H 11,2% (M la kirn loai kiem). Co can dung dich sau phan ling, thu dugc 15,4 gam chat ran khan, dong thai ngung tu phan hoi bay ra tha'y tao thanh 92 gam chat long. Cong thiic cua X la 9 Ca'm nang On luygn thi dgi hpc 18 chuyen dg H6a hpc - Nguygn VSn Hii Cty TM H MTV DVVH Khang Vigt I H A. CH3COOCH3. B. CH3COOC3H7.4 C. CH3COOC2H5. D . C2H5COOC2H5. Lmgidi: Ggi cong t h u c cua chat beo la (RCOO)3C3H5. Lot gidi: (RCOO)3C3Hs + 3 N a O H " " ' '' • - > kho'i l u a n g d u n g m o i (nuac) = 100 - 11,2 = 88,8 g a m . Phan u n g hoa hoc: gfi; RCOOR' rfsB<.m S^'- M o l : 0,1 a Mol: r i f / v i •« iv;y£;f{;-^,';f0<;j; + MOH ^ 0,1 0,1 ^MROH ^ gam. = 32^>R' = 15(R'lanh6mCH3-)".''*^' ^ Meste = 7 4 R C O O C H s d u , sau p h a n u n g , k h o i l u g n g muo'i t h u d u g c la 764,6 g a m . Kho'i l u g n g K O H d a t h a m g i a p h a n u n g la D . 350 g a m . Laigiai: ^"^g^ rn^hatbeoCg) Phan l i n g hoa hoc: ' RCOOH Mol: * 0,1 Mol: 3x -> + C3H5(OH)3 <- 0,01 -> D a p an D . V i d u 21: A m i n o a x i t X chua m g t n h o m -NH2. C h o 10,3 g a m X tac d u n g v o i axit H C l ( d u ) , t h u d u g c 13,95 g a m muo'i k h a n . Cong thuc ca'u tao t h u g g n cua X B. H2NCH2CH2COOH. C. C H 3 C H ( N H 2 ) C O O H . D . H2NCH2COOH. H2N-R-COOH > 3RCOOK + C3H5(OH)3 + HCl Bao toan kho'i l u g n g : ^ Mx = — - j||||vi.^^^ CIH3N-R-COOH mammoaxit + ^ .CJ^jinV mHci = mmuoi UHCI = 0/1 = = 103 ^ - H 2 N - R - C O O H = 103. - 0,1 mKOH = (0,1 + 0,8.3).56 = 140 g a m . ^g,; > f'^^, -> mHci= 13,95 - 10,3 = 3,65 g a m - > 3x • R = 42 —>• Goc hidrocacbon la -C3H6- '^-'^ ••••vv,;::fv;„ D a p an A . V i d u 22: D u n n o n g m g a m h o n h g p g o m a m o l tetrapeptit mach h o X va 2a 19 (B-08): Xa p h o n g hoa hoan toan 17,24 g a m chat beo can vvra d u p h o n g . Gi a t r i ciia m la A . 16,68. ! aijfi > 3RCOOK K h i cho X tac d u n g v o i axit H C l : + H2O 0,06 m o l N a O H . C o can d u n g d i c h sau p h a n u n g t h u d u g c m g a m xa ^ VinV ggi cong t h u c cua X la H2N-R-CC)OH. D a p an A . Vi 0,03 :k'^' Bao toan k h o i l u g n g : m + 0,03.56 = 9,58 + 0,92 - > m = 8,82 g a m . - > 700 + (5,6 + 56.3x) = 764,6 + 92.x + 0,1.18 - > x = 0,8 m o l D . 8,82. Laigiai: nmn = 0,1 m o l . > RCOOK * "• ' Nhan xet: Cac d a p an d e u cho aminoaxit chua m g t n h o m - C O O H Bao toan kho'i l u g n g : mchat beo + m K O H = mmuoi + mgUxeroi + m ' ' " • m. :,3f' > • 700 0,1 (RCOO)3C3H.s + 3 K O H C. 10,50. Laigiai: ' + KOH B. 9,38. A . CH3CH2CH(NH2)COOH. ^"^^^ ^ 8== m ^ o H ^ 5600 m g = 5,6 g a m D a p an C. ' 0,92 g a m g l i x e r o l va 9,58 g a m h o n h g p muo'i cua axit linoleic va axit oleic. (RCOO)3C3H5 + 3 K O H V i dv 18: C h o 700 g a m cha't beo c6 chi so axit la 8 tac d u n g v o i d u n g d i c h K O H C h i so axit cua chat beo = = 17,24 + 0,06.40 - 0,02.92 = 17,8 g a m Mol: C. 280 g a m . C3Hs(OH)3 0,02 ngiixe((ji = 0,01 m o l = 74 ^ R la C H 3 - B. 175 gam. + mch.nt beo + m N . n O H = m x a phong (muoi) + mgiixeroi .„ ^ Este la C H 3 C O O C H 3 ^ D a p an A . A . 140 g a m . m x . i phong Gia t r i cua m la A . 9,94. Bao toan k h o i l u g n g : meste = 15,4 + 3,2 - 11,2 = 7,4 g a m ^ i, 1,. V i dv 20: Xa p h o n g hoa hoan toan m g a m m o t t r i g l i x e r i t bang K O H t h u d u g c N/ian f/ifli/: Cha't l o n g sau k h i n g u n g t u g o m nuoc va ancol nen: mancoi = 93,4 - mnuoc = 92 - 88,8 = 3,2 > 3RCOONa 0,06 A p d u n g bao toan kho'i l u g n g : -> )• R C O O M + R ' O H fti ..fO !d>i Oi 0,1 < ' i - b tfW K h i cho chat beo tac d u n g v o i N a O H : Nhan xet: K h o i l u g n g M O H bang 11,2 g a m , B. 18,24. -'"^^^ * C. 17,80. D . 18,38. m o l t r i p e p t i t m a c h h o Y v o i 600ml d u n g d j c h N a O H I M ( v u a d u ) . Sau k h i cac p h a n u n g ke't thiic, c6 can d u n g d i c h t h u d u g c 72,48 g a m m u o i khan cua c^c a m i n o axit d e u c6 m g t n h o m - C O O H v a m g t n h o m - N H 2 t r o n g p h a n t u . Gia t r i a i a m la A. 51,72. B. 54,30. • • • C. 66,00. • • , D . 44,48. 11 nang On luy?n thi dgi hpc 18 chuySn 6i H6a hpc - Nguygn Van HJi dm Cty TNHH MTV D W H Khang Vi^t Lai gidi: ./.5„ A I,I Nhan xet: K h i cho cac peptit tac dung voi N a O H thi c6 cac lien ket peptit v a 2. P H L f O N G P H A P B A G T O A N N G U Y E N T O nhom - C O O H (cua amino axit dau C ) tham gia phan ung. a. N Q i d u n g Trong cac phan ung hoa hpc, cac nguyen to luon dupe bao toan. Theo bai, a mol tetrapeptit mach h a X c6 a.3 = 3a mol lien ket peptit. b, H ? qua 2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit Tong so mol nguyen tu ciia mpt nguyen to c6 trong cac chat truoc phan ung Tong so'mol nhom c a c b o x y l - C O O H bang a + 2a =3a. Cac phan ling nit ggn cua lien ke't peptit va nhom - C O O H la: -CO-NH- 'yinit 7a • nnMQh u > -COONa + H2N- truoc va sau phan ling. + NaOH 3a > -COONa + H2O 3a , 3a Theo bai: nwaOH = 7a + 3a = 10a = 0,6 ^ M'-.A ' i(..>.:-i(jin -> m = 72,48 + 0,06.3.18 - 0,6.40 = 51,72 gam D a p an A. tripeptit mach ho Y voi 400ml dung djch H C l I M (vira du). Sau khi cac phan ung ket thuc, c6 can dung dich thu dugc 48,1 gam muoi khan cua cac amino axit deu c6 mpt nhom - C O O H va mpt nhom - N H 2 trong phan ttr. G i a tri ciia m la B. 29,0. ' ' , C . 30,8. 2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit Cac phan ung riit gpn ciia lien ket peptit va nhom H 2 N - la: -CO-NH- + HCl + H2O 5a 5a + HCl )--COOH Lai gidi: C a c h 1: 6 bai nay, cac em c6 the giai dua vao cac phuong trinh phan ung: Al + K O H + H2O + •|H2 > KAIO2 AI4C3+ 4 K O H + 4H2O > 4KA102 > Al(OH)3 KAIO2 + CO2 + 2H2O . + 3CH4 r H - f f - + cllV^-: + KHCO3 , . ^^02^^20 ) KAIO2 ^ ^ j ^ ^ Al(OH)3 Bao toan nguyen to A l : , . j w , . . ,0-!: ^, + nAi n ^ i +nAi4C3 , = 0'3 ^^A\4C3= =0,2; nAi4C3 = 0'l- ' V a y : a = | n A i + 3 n A i 4 C 3 = ^,6 D a p an B. V i d v 2: H o a tan hoan toan hon hpp gom 0,12 mol FeS2 v a a mol CU2S vao axit + CIH3N- H N O 3 (vua dii), thu dupe dung djch X (chi chiia hai muoi sunfat) v a khi 5a > D . 0,45. > ' ff Tong so mol nhom amino H 2 N - bang a + 2a = 3a. duy nhat N O . G i a tri cua a la A. 0,075. CIH3N- B.0,12. Bao toan khoi lupng: m + m H c i + m C.0,06. " D.0,04. ' Lai gidi: 3a Theo bai: nHci = 5a + 3a = 8a = 0,4 —> a = 0,05 mol. --V' Nhan xet: Cac em luu y la dung djch X chi chua hai muoi sunfat - > sau cac phan ung, S nam het a dang goc sunfat. = mmuoi T a CO cac so do chuyen hoa: m = 48,1 - 0,4.36,5 - 0,25.18 = 29 gam. —> D a p an B. ^ FeS2 '-vi'.- ./\ 10 C.0,40. Theo bai: Theo bai, a mol dipeptit mach ho X c6 a.l = a mol lien ket peptit. 3a B.0,60. " A l + 4 n A i 4 C 3 = I^Al(OH)3 nhom H 2 N - (ciia amino axit dau N ) tham gia phan ung. Mol: A. 0,55. Al, A I 4 C 3 Nhan xet: K h i cho cac peptit tac dung voi H C l thi c6 cac lien ket peptit va H2N- (du) vao X, lupng ket tua thu dupe la 46,8 gam. G i a tri cua a la C a c h 2: So do phan ung:^ D . 28,1. Lai gidi: Mol: V i d u 1 (A-08): H o a tan hoan toan 0,3 mol hon hpp gom A l v a AUCs vao dung . ,, V i d\ 23: D u n nong m gam hon hop gom a mol dipeptit mach ho X v a 2a mol , , , VIDVMAU dich K O H (du), thu dupe a mol hon hpp khi v a dung dich X. Sue khi C O 2 —> a = 0,06 mol. Bao toan khoi lugng: m + mNaOH = mmuai + m A . 33,5. o , ^ , . . r ' LKU y: C a n xac dinh diing va day dii cac chat c6 chiia nguyen to dang xet 6 , 7a -COOH Mol: + NaOH va sau phan ung luon bang nhau. ' • Mol: 0,12 ; i i'ouU-,,r!>! • > iFe2(S04)3 0,06 'ih.^1bm^l^^., : . ^; ].,>f:v., r v - ; v : - ' - C^m nang 6n luygn Ihi dgi hpc 18 chuyfin ii Cu2S Mol- Cty TNHH MTV D W H Khang Vi$t Hoa hpc - Nguyin van H i ! > 2CuS04 a 2a Bao toan nguyen to P theo so do: ^^ ^^ , Bao toan nguyen to S, ta c6: 2 npeSj + "CuzS = "504 , . j,|,|,j3i);, Ca(H2P04)2 < ii^, Khoi lupng mol: -H 1 a . -> 0,12.2 + a = 0,06.3 + 2a -> a = 0,06 -> Dap an C. V i d u 3 (A-12): Cho 18,4 gam hon hop X gom Cu2S, CuS, FeS2 va FeS tac dung het voi HNO3 (dac nong, du) thu duoc V h't khi chi c6 NO2 (6 dktc, san pham khu duy nha't) va dung dich Y. Cho toan bo Y vao mot lugng du dung dich BaCl2, thu dugc 46,6 gam ket tiia; con khi cho toan bp Y tac dung voi dung dich NH3 du thu dugc 10,7 gam ket tua. Gia tri cua V la A. 38,08. B. 24,64. C. 16,8. D. 11,2. NMn xet: Dung dich Y chiia cac ion: F e ^ Cu^^ SO4 , . , , ' ,. Vi dvi 5 (A-12): Mot loai phan kali c6 thanh phan chinh la K C l (con lai la cac tap chat khong chua kali) dupe san xua't tu quang xinvinit c6 dp dinh duong 55%. Phan tram khoi lupng ciia KCl trong loai phan kali do la A. 95,51%. B. 65,75%. C. 87,18%. D. 88,52%. < *= - rt:. khoi lupng cua K2O. „,, Ap dung bao toan nguyen to'K, ta CO so do: mol. , > Fe(OH)3i > Cu(OH)2i Kho'i lupng mol: % khoi lupng: Luu y: Cu(OH)2 tan trong NH3 du tao thanh phuc chat: 10 7 74,5 gam x% <- 47 gam 55% > ..,,!' r . = 87,18 -> Dap an C. * ' ' ' 47 ^ Vi dy 6: Dot 5,6 gam Fe trong khong khi, thu dupe hon hpp chat ran X. Cho X ^ " Cu(OH)2 + 4NH3 — ^ [Cu(NH3)4](OH)2 v;.. V . - . KCl <—> -!-K20 Khi cho Y + dung djch NHa du: Bao toan nguyen to Fe: npe (X)= nFe(OH)3 % khoi lupng: 69,62% x% 69,62.142 v r ^ '' —>x=— = 42,25 - > Dap an C. v > /'1 Cu2* + 2NH3 + 2H2O 1' o. '"'!(, 6 day, cac em can nho la dp dinh duong cua phan kali dupe tinh theo % va NO3. Ba2* + SO^" — ^ BaS04i Fe3* + 3NH3 + 3H2O ^ i, n P2O5 234 gam X= tac dung voi dung dich FiNOs loang (du), thu dupe khi NO (san pham khu ' = duy nha't) va dung djch chua m gam muo'i. Gia trj cua m la A. 18,0. B.22,4. C. 15,6. D. 24,2. ; Bao toan khoi lugng: mx = m^u + mpg + mg npg=0,lmol. -» mcu = 18,4 - 0,1.56 - 0,2.32 = 6,4 gam -> ncu (X)= — = 0'^ i"ol- Nhan xet: Bai nay ne'u dua theo phuong trinh phan ung se rat dai dong va Qui doi X ve hon hgp gom cac don chat: ton nhieu thoi gian. O day, cac em can su dung so do phan ung: A iKt/tij: Fe = 0,1 mol; Cu = 0,1 mol va S = 0,2 mol. -» ne (X) = 2ncu + Snpe + 6ns = 1,7 mol n^o^ = (x) = 1,7 mol -> V = 1,7.22,4 = 38,08 lit ^ Dap an A. mm Fe va ap dung dinh luat bao toan nguyen to'Fe: Fe Vi dv 4 (B-10): Mot loai phan supephotphat kep c6 chua 69,62% Ca(H2P04)2, con lai gom cac chat khong chiia photpho. Dp dinh duong ciia loai phan Ian nay la A. 48,52%. B. 39,76%. C. 42,25%. D. 45,75%. Loigidi: I 6 day, cac em can nho la dp dinh duong cua phan supephotphat dupe tinh theo % khoi lupng cua P2O5. •;• NMn xet: 1 mol Ca(H2P04)2 hay 1 mol P2O5 deu chua 2 mol P . , » , u MM. > X (Fe, FeO, Fe203, Fe304) Mol: > Fe(N03)3 > Fe(N03)3 ; j ,0 „ , .0 f:5 0,1 0,1 m =0,1.242 = 24,2 gam ^ Dap an D. V i du 7: Cho 31,2 gam hon hpp gom Al, Cu va Ag tac dung vua dii voi 900ml dung dich HNO3 1,5M, thu dupe dung djch chua m gam muoi va 4,48 lit hon hpp khi X (dktc) gom NO va N2O. Ti khoi ciia X so voi H2 la 16,75. Gia tri ciia m la ' ' ' A. 98,3. B.97,2. C. 96,3. D. 91,0. Cty TNHH MTV DWH Khang Vi§t Ca'm nang fln luygn thi dji hgc 18 chuy6n dg H6a hpc - Nguyin Van Hai Laigidi: b a i nay, t r u o c het cac e m can t i m so' m o l m o i k h i t r o n g X de t h u d u g c ket qua: nNo= hlhan xet: K h i cho p h a n 1 tac d y n g v o i N a O H , tat ca A l va AI2O3 d e u tac m o l ; n ^ j o " ^ O'OS i ^ o l - Al,Cu,Ag d y n g v a c h u y e n t h a n h NaA102. ) M u o i n i t r a t + N O + N2O Chatoxihoa: N^^ + 3e > NO; v a CO the xay ra ca qua t r i n h : ZN*' 2N*5 + + 8e „ ge /, / D o vay, bao toan n g u y e n to A l , ta c6: n A i = njvjaOH = 0/3 n i o l . ' N2O > _^ Ban d a u : nAi = 2.0,3 = 0,6 m o l > NH4NO3 (a m o l ) K h i cho k i m loai + H N O s : Cr203 n ^ ^ Q - ( m u o i ) = n g trao J6i = 3 nfyio + 8 nivj20 + Bao toan n g u y e n to N : nHNOa = 8 nfvjH^NOs ~ + " N O + 0,9.1,5 = 0,85 + 8a + 0,15 + 2.0,05 + 2a ^ 2 mol. ' 0,2 2A1 ^ — ^ a4:r,^ AI2O3 + 0,2 m o l . . 2Cr , OA ,R>0,2;;.. , : ^ , Cac chat t r o n g p h a n 2: n ^ i = 0,1 m o l ; n c r = 0,2 m o l ; 0^1203 = 0,1 m o l . nN20 + 2nNH4N03 a = 0,025 Bao toan k h o i l u g n g : m = m ^ i , cu, Ag + m ^ ^ ^ + Mol: 0'^^ + "€1203= ^^^-r^^^= K!) iTs-> mNH4N03 -> HHCI =3nAi +2ncr+6nAi203 = l ' 3 m o l . ^DapanB. W d v 10: H o a tan hoan toan 8,16 g a m h o n h g p g o m Fe304 va FeS2 t r o n g d u n g ' d i c h axit HNO3 (dac, d u ) , t h u d u g c 4,032 l i t k h i NO2 (dktc) v a d u n g d j c h X. C h o X tac d u n g v o i d u n g d i c h Ba(OH)2 d u , Igc ket tua va n u n g t r o n g k h o n g = 31,2 + (0,85 + 8.0,025).62 + 0,025.80 = 98,3 gam. -> D a p an A . k h i d e n k h o i l u g n g k h o n g d o i t h u d u g c m g a m cha't r a n . Gia t r j cua m la NMn xet: Bai nay da " g i a u d i " san p h a m NH4NO3. A . 12,66. V i d u 8: H o a tan het 7,8 g a m h o n hgp g o m A l va AI2O3 b a n g d u n g d i c h H C l ( d u ) , t h u d u g c V l i t k h i H2 (dktc) va d u n g d j c h X. N h o d u n g d i c h N H s d u = ^ i'; ' C. 5,60. D . 4,48. G 3Fe^3 FeS2-15e ./v > Fe^^ + 2S^ *~ - li^a.W/ . . " N O z = nFe304 +15"FeSz a + 15b = ai8 - » a = a03; 3 nH2 = - H A I 2' = 0,15 m o l ^ = 0,2. Fe304 V = 3,36 l i t . D a p an B. Mol: ' ' 0,03 FeS2 V i d v 9 (B-12): N u n g n o n g 46,6 g a m h o n h g p g o m A l va Cr203 (trong d i e u Mol: I h g p t h u d u g c sau p h a n u n g t h a n h hai phan bang n h a u . Phan m o t p h a n -> m = 160.0,05 + 0,02.233 = 12,66 i u n g v u a d u v o i 300 m l d u n g d j c h N a O H I M (loang). De hoa tan het p h a n hai can v i r a d u d u n g d i c h chiia a m o l H C l . Gia t r j cua a la A . 0,9. B. 1,3. C.0,5. D . 1,5. Vi 0,01 0,005 '[ _ U m fiiJ5 ifi BfO > i > ^Fe203 2 0,045 1 > - F e 2 0 3 + 2BaS04 2 J k i f n k h o n g c6 k h o n g k h i ) den k h i phan u n g xay ra hoan toan. Chia h o n '',;t»r,' b = aOl. Bao toan n g u y e n to Fe v a S: = 0,1; y = 0,05 . - - ,0'' ^ Bao toan electron: "NH3.H2O > A i ( O H ) 3 _ ' % AI2O3 Bao toan n g u y e n to A l : x + 2y = 2nAi203 -> x + 2y = 2 . ^ ^ X n^B , Fe304 - l e ^ )AlCl3 vM^b Cac p h a n l i n g khvr: ' Ldigidi: . 0,18 m o l . D . 20,66. |>,s£' V - ' G g i so m o l : Fe304 (a m o l ) v a FeS2 (b m o l ) . Ta c6: 232a + 120b = 8,16. 10,2 g a m chat ran. Gia t r i cua V la B. 3,36. = C. 11,86. Laigiduii ^' ' 22,4 vao X, IQC ket tua va d e m n u n g den kho'i l u g n g k h o n g d o i t h u d u g c A . 2,24. B. 11,06. -8 808,£ ( \.1.0 » 0,02 -> D a p an A. 1 1 (A-08): H o n h g p X g o m p r o p a n , p r o p e n v a p r o p i n . T i k h o i cua X so v o i H2 Ja 21,2. K h i d o t chay hoan toan 0,1 m o l h o n h g p X, t o n g k h o i l u g n g c u a C C h va H2O t h u d u g c la A . 20,40 g a m . B. 18,60 . SB^JHUZ^Ji^^^j^^^ff^fpf^^ Cty TNHH MTV D W H Khang ViQt Ca'm nang On luyjn thi dgi hgc 18 chuyfin dg H6a hgc - Nguygn Van H5i Vi Laigiai: NMn xet: Cac e m can nh|in ra cac chat t r o n g h o n h o p X d e u c6 chiia 3 n g u y e n t u cacbon. ' ' Bao toan kho'i l u g n g , ta c6: mx = mc + mn. '>rn 6,(J " f' > . - no, d a chiic, m a c h h o , c6 c u n g so n h o m - O H ) can v i r a d u V l i t k h i O 2 , t h u d u p e 11,2 l i t k h i C O 2 v a 12,6 g a m H 2 O (cac the t i c h k h i d o a d k t c ) . G i a t r j ciia V la N h u vay, k h i do't chay 0,1 m o l X t h u dugc 0,3 m o l C O 2 - > nc = 0,3 m o l . M a t khac: M x = 21,2.2 = 42,2 - > mx = 42,4.0,1 = 4,24 g a m . 14 (B-10): Do't chay hoan toan m p t l u p n g h o n hgip X g o m h a i ancol ( d e u i A . 14,56. " ' '"*V ^ 1 /•-(.'" M^' «f^ ' - > m H = 4,24 - 0,3.12 = 0,64 g a m ^ nn = 0,64 m o l - > nn^o = 0/32 m o l . T o n g k h o i l u g n g C O 2 v a H 2 O bang 0,3.44 + 0,32.18 = 18,96 g a m . B. 15,68. C . 11,20. D . 4,48. Laigiai: 11'2 , 12,6 , "CO2=^=0'5"^ol;nH2O=^=0,7mol. ^ Theo b a i ra, X c h i i a 2 ancol n o - > nx = nH20 " ^COi ^ ^,2 m o l . - > D a p an C. V i d\ 12: Do't chay hoan toan m p t the tich k h i thien n h i e n g o m metan, etan, -> So'nguyen tvr C t r u n g b i n h = = 2,5 - > X chua m p t ancol d a chuc c6 p r o p a n bang o x i k h o n g k h i (trong k h o n g k h i , o x i chiem 20% the tich), t h u d u g c 7,84 l i t k h i C O 2 (a dktc) va 9,9 g a m H 2 O . The t i c h k h o n g k h i (a dktc) D o X chua h a i ancol c u n g so' n h o m chuc (hai chiic) - > n o = n o H = 2nx .X n h o nhat can d i i n g de d o t chay hoan toan l u o n g k h i t h i e n n h i e n tren la :gfi A . 70,0 l i t . ir sv B B. 78,4 l i t . C. 84,0 l i t . Lmgidi: D . 56,0 l i t . > 100 •JIA so n g u y e n t u C n h o h o n 2,5 - > ancol d o la C2H4(OH)2. ^ Ij no = 0,4 m o l . Bao toan n g u y e n to' O, ta c6: no (OH) + 2 no^ = 2 TXQQ.^ + «;:.«,) jfW 2.0,5 + 0 , 7 - 0 , 4 , no2=— ^ ^ = 0,65 m o l nc02 = - ^ = 0 3 5 m o l ; n H 2 O = ^ = 0 ' 5 5 m o l . Nhan xet: Ban d a u , n g u y e n to' o x i 0 dang O 2 t u do, c o n sau p h a n l i n g chay t h i chuyen he't v a o C O 2 v a H 2 O . = ^'^^-^^^'^^ = 0,625 m o l Vkhongkhi + n^jjo • thupc c u n g d a y d o n g dang, t h u dugc s at w timA • 3,808 l i t k h i C O 2 (dktc) v a 5,4 g a m B. C H 2 = C H C O O H v a C H 2 = C ( C H 3 ) C O O H . C. C H s C O O H v a ' ' C2H5COOH. D. C H 3 C O O H va C H 2 = C H C O O H . B.5,72. C. 4,72., //,4 10 Laigiai: D . 7,42. Laigiai: ' ^ ri one KA nc02 = ^ = 0^17 m o l ; n H 2 0 = ^ = 0,30 m o l . = .„ 4 •• 2.0,1 + 2 . 0 , 2 4 - 0 , 2 , Y ^ = 0'24mol. Nhan xet: rxyi^Q < rxQOj H o n h p p X chua i t nhat m p t axit k h o n g n o - > Logii A v a C. -» no = 0,13 m o l . 5' j Bao toan k h o i l u g n g t r o n g X, ta c6: ....lii^^ So n g u y e n t u C t r u n g b i n h = "^^^ = 2,4 - > X chua m p t axit c6 so n g u y e n t u C n h o h o n 2,4 - > L o ^ i B ( H a i axit d e u chua so'cacbon > 2,4). m x = m c + m H + m o = 0,17.12 + 0.60.1 + 0,13.16 = 4,72 g a m t'-vs'jji'ii • ^C02=— 0,30 - 0,17 = 0,13 m o l . D o X chua cac ancol d a n chiic ^ no = noH = nx H!"\,0-,. Bao toan n g u y e n t o O, ta c6: 2nx + 2 n o , = 2 nco2 "H2O • • ;loM . .it Cac e m can thay rang, k h i d o t chay X: n^^^Q > UQQ.^ - > X chiia 3 ancol no - > ' D a p an C . M / » t j ; ; , • A. H C O O H va C 2 H 5 C O O H . ;.::;;;»•* H 2 O . Gia t r i ciia m la n^Q^ V02 = 0,65.22,4 = 14,56 l i t ^ D a p an A . axit la V i d u 13 (A-10): D o t chay hoan toan m g a m h o n h g p X g o m ba ancol d o n chiic, nx = nj^^Q - ' 0,1 m o l X can 0,24 m o l O2, t h u d u p e C O 2 v a 0,2 m o l H 2 O . C o n g t h i i c h a i V Q J = 0,625.22,4 = 14,0 l i t . = 5 V o 2 = 70,0 l i t D a p an A . A . 5,42. ^ ^^i^rW V i d y 15: H o n h p p X g o m h a i axit cacboxylic d o n chuc. Do't chay h o a n toan Bao toan n g u y e n t o O, ta c6: 2 n o j = 2 TYQQ^ Vay: nyi^^Q i,-.>.,..t- —> D a p a n D . 19 Cty TNHH MTV DWH Khang Vi$t Ca'm nang 6n luy$n thi dgi hpc 18 chuyen dg H6a hgc - IMguySn VSn H5i V i dv 16 (A-12): Hon hg-p M gom mpt anken va hai amin no, don chuc, mach " h6 X va Y la dong dang ke tiep (Mx < MY). Do't chay hoan toan mot lugng r M can dung 4,536 lit O 2 (dktc) thu dugc H 2 O , N 2 va 2,24 lit C O 2 (dktc). Chat Yla A. Etylamin. ' ' C. Butylamin. [ '2 : . B. Propylamin. D. Etylmetylamin. •'XlS:',U,'0 , - ,(•) .A Laigidi: . , Laigidi: n .i . ., . ii..t» 4536 224 n ;fom t ; , i . - - n o , = ^ ^ = 0,2025 m o l ; n r o , = — = 0,1 m o l . . • • '! • ^ , .'' . 22,4 22,4 ^ 2 H H C I = 0,03 mol; n o 2 = ai425 mol. M|it khac: n N = nNH2 = " H C I ^ H J O = 0/205 mol. Nh^n thay, khi dot chay 1 mol anken thu du(?c XXQQ^ = n H 2 0 ' amin no, don chijfc thi n H j O " ' ^ C 0 2 CxH2x.3N 1'^ vol hai ph"^°''^8 ^""^^'^'^"'^S- ^ X C O 2 + ( x + l , 5 ) H 2 0 + 0,5N2 + O2 — Dovay: n H 2 0 " ' ^ C 0 2 = l / 5 n a m i n - > namin=—^ '— =0,07mol. -> So'nguyen t u C trung binh trong M = — ^ ^ < — ^ 2 2 _ = "M I^aniin 1 n p _ 80/16 10 n^ ^ Ggiso'mol: n ^ (x) = a i^ioJ va n H ( x ) = b m o l . Bao toan khoi lugng: mx = m ^ + mj^ + mo +T^-H 12a+ b = 1,81. 'fl iiAA US' ^ 0,1 + 2.0,1425 = 2a + 0,5b -> 4a + b = 0,77 ^ a = 0,13. Y 0,07 ' 21/14 ~ 3 • Bao toan nguyen to O: n Q (x) + 2no2 = 2 n c o 2 + ^HjO sot os8 ~ 1^43. C>D < - - (x)= 0,03 mol -> no(X)= 0,1 mol. -» m c + n i H = 3,83 - 0,1.16 - 0,03.14 = 1,81 ^ ' 0,205-0,1 • ^. r n o ^ S O man xet: Ixx ti 1? khoi lugng: 21 ^ ......£Dtjd3X,P'!J6do'?riT Bao toan nguyen to'O, ta c6: 6: 2 no2 = 2 n c o j + " H J O V i dv 18: Hon hgp X gom hai amino axit no (chi c6 nhom chuc - C O O H va - N H 2 trong phan tu), trong do ti 1§ mo : mN = 80:21. De tac d\ing vvra du voi 3,83 gam hon hgp X can 30ml dung dich HCl I M . Mat khac, do't chay hoan toan 3,83 gam hon hgp X can 3,192 lit O 2 (dktc). Dan toan bp san pham chay ( C O 2 , H 2 O va N 2 ) vao nude voi trong d u thi kho'i lugng ket tiia thu dugc la A. 13 gam. " B. 20 gam. C. 15 gam. D. 10 gam. ,. ,. . Bao toan nguyen to C: n c ( X ) = n c o 2 = '^CaC03 = 0 , 1 3 ^ m c a C O a =13 gam. -> Dap an A. f, -> M chua mQt chat c6 so' nguyen t u C nho hon 1,43 -> do phai la amin 3. PHl/ONG PHAP T A N G - G I A M K H O I L l / p N G ii C H 3 N H 2 (X). O day cac em can l u u y: anken chua tir 2 nguyen t u C tra len! a. N p i dung -> A m i n ke tiep la C 2 H 5 N H 2 (Y) Dot chay hoan toan mpt lugng X can vua du V lit khi O 2 , thu duQC 15,68 lit '' khi C O 2 va 18 gam H 2 O . Cac the tich do 6 dktc. Gia t r i cua V la B. 17,92. n c o 2 = 0,7mol; n H 2 0 = l m o l . NMn C. 16,24. mt?iV6J • *^' • Bao t o a n n g u y e n to O: n o (ancol) + 2 n o 2 = 2 n c o 2 + n H 2 0 -> HQ^ = 2.0,7 + 1-0,3 , = 1 , 0 5 mol 2 V 0 2 = 1,05.22,4 = 23,52 lit ^ Dap an D. dau dugc thay the (ho^c cgng hgp) bang nguyen t u (nhom nguyen tit) mai de tao thanh san pham. ^ Do do, khoi lugng cua chat tao thanh c6 the tang len hay giam di do chenh Dua vao sy tang hay giam nay c6 the xac djnh so mol cac chat trong 1 phuang trinh hoa hgc, t u do c6 the giai nhanh nhieu bai toan. "ancol = " H 2 O ~ ^C02 • Do a n c o l d o n c h u c n e n U Q (ancon ^ "ancol =0,3 m o l . { l|ch khoi lugng mol cua cac nguyen t u (nhom nguyen tu). D. 23,52. xet: K h i d o t c h a y a n k e n t h u duQC UQQ^ = XIH^Q , c o n k h i d o t c h a y a n c o l Do v ^ y : nancol = " H Z O - ^coz = 1 - 0,7 = 0,3 m o l . ffl>| Khi tham gia phan ling hoa hgc, nguyen t u (nhom nguyen tu) cua chat ban Dap an A. V i d\ 17: H o n hgp X gom mpt anken va hai ancol (no, don chiic, mach ho). A. 22,40. , C03 , „,. Bien doi k h o i lugmg (tinh cho 1 mol) S\f thay the (cpng hgfp) 0-2 (oxit) CO — -OH _COOH -NH2 Tang 71-60 = 11 gam > CI2 Tang 96-16 = 80 gam ""''^"^ > S O ^ ^ CO2; H2 H2O Tang 16 gam Tang 23-1 = 22 gam > -ONa > -COONa > -NHsCl Tang 23-1=22 gam Tang 36,5 gam 21 dm Cty TNHH MTV DVVH Khang Vi§t nang fln luygn thi dgi hgc 18 chuySn de H6a hpc - Nguyjn Van Hi\ VI DU MAU V i d\ 1 (A-08): Cho V l i t hon hop khi (dktc) gom CO va H2 phan ung vol mot lugng d u hon hop ran gom CuO va Fe304 nung nong. Sau khi cac phan ung xay ra hoan toan, khoi lugng hon hgp ran giam 0,32 gam. Gia tri ciia V la A. 0,448. B. 0,112. C. 0,224. D. 0,560. 1 mol H 2 -> —*° > CO2 — * ° > H2O 1 mol (CO va H 2 ) Theobai: Kho'i lugng chat ran giam 16 gam. ^^ ^m —> Kho'i lugng chat ran giam 16 gam. ^ ° > ( C O 2 va H 2 O ) 0,02 mol <- > 1 mol SO l~ Khoi lugng tang 96 - 16 = 80 gam. tang 52 - 22 = 30 gam. _^ a = — = 0,375mol -> no (X) = 0,375 mol. 80 Khi cho 22 gam X tac dung voi CO thi: no Lot giai: 1 mol CO 1 mol 0 - 2 a mol (X) '" ' ff f. • '' •' ' = n^oj = riCaCOa = 0375 mol. - > m =0,375.100 = 37,5 gam. —> Dap an B ' '''t*''-t> i?>'; f vcr-f vi KT * •'•H'>I.. V i dv 4 (A-10): Dot chay hoan toan mgt lugng hidrocacbon X. Hap thu het san giam 16 gam. pham chay vao dung dich Ba(OH)2 du, tao ra 29,55 gam ket tiia, dung dich giam 0,32 gam. sau phan ung c6 khoi lugng giam 19,35 gam so voi ban dau. Cong thuc - > V = 0,02.22,4 = 0,448 lit. phan t u cua X la —> Dap an A. V i d\ 2: Hon hgp Y gom FeO, Fe203 va CuO. Hoa tan hoan toan 6,8 gam Y A.C3H8. B.C2H6. C.C3H4. D.C3H6. Lot giai: bang dung djch HCI (du), thu dugc dung dich chua 12,3 gam muoi. Mat Khi hap thu hoan toan san pham chay gom C O 2 va hoi H 2 O vao dung dich khac, neu khu hoan toan 6,8 gam Y bang CO (du), thu dugc m gam kim Ba(OH)2 du: loai. Gia tri cua m la : A. 5,2. C.6,0. D. 4,8. ^ Led giai: > 2 mol CI" - > a mol ^ a= — Khoi lugng tang 71 - 1 6 = 55 gam. <— = 0,1 mol -> no (Y) = 0^1 mol. - > Dap an A. ^ ' ' " "* V i dvi 3: Hon hgp X gom CuO va Fe203. Hoa tan hoan toan 22 gam X bang thu dugc 52 gam muoi. Mat khac, neu khu het 22 gam X bang CO (du), dan hon hgp khi thu dugc vao dung dich Ca(OH)2 (du), tao thanh m gam ket tua. Gia trj cua m la A. 45,5. B.37,5. C. 40,5. D. 50,0. Lai giai: Nhan xet: Khi cho X + H2SO4 <- 0,15 thi oxit chuyen thanh muoi sunfat va mgt ion 0-2 trong oxit dugc thay the bang mgt ion SO 4 " . -f^' = Mat - Dugc = mBaCOs - -> ^ ^ Bao toan khoi lugng: my = m + mo - > m = 6,8 - 0,1.16 = 5,2 gam. H2SO4, 0,15 mH20= (r\ (mco2 + n^Hoo) = 19,35gam. 29,55 - 0,15.44 - 19,35 = 3,6 gam -> nH20 = 0,2 mol. ' *" Vay hidrocacbon X c6: So C : so'H = nc : nn = 0,15 : 0,4 = 3 :>8. tang 12,3 - 6,8 = 5,5 gam. Khi cho Y tac dung voi CO thi cac oxit deu bi khu thanh kim loai. dung dich Mol: Khoi lugng dung dich giam Nhan xet: Khi cho Y + HCl thi oxit chuyen thanh muoi clorua va mgt ion 0^~ trong oxit dugc thay the bang hai ion CI". , 1 mol O^- > BaCOs + H 2 O . Ba(OH)2 + C O 2 B.5,6. Dap an A. „.,,.n ,,a , , V i dy 5 (B-12): Cho 21 gam hon hgp X gom glyxin va axit axetic tac dung vua dii vol dung djch KOH, thu dugc dung djch Y chiia 32,4 gam muoi. Cho Y tac dung voi dung dich HCl du, thu dugc dung djch chua m gam muoi. Gia trj cua m la A. 44,65. B. 50,65. C. 22,30. ^ D. 33,50. Ldigiai: Glyxin ( H 2 N - C H 2 - C O O H ) = a mol va axit axetic ( C H 3 C O O H ) = b mol. Khi cho X tac dung voi dung dich K O H : :> Theobai: • •' • 1 mol - C O O H ^ 1 mo! -COOK -> khoi lugng tang 38 gam. 0,3 mol -> a + b = 0,3 mol. Mat khac: 75a + 60b = 21 <- tang32,4-21 = 11,4 gam a = 0,2 mol; b = 0,1 mol. Khi cho Y + HCl d u , thu dugc cac muoi: C I H 3 N - C H 2 - C O O H (0,2 mol) va KCl (0,3 mol) theo cac phan ung: 23 Cty TNHH MTV DVVH Khang Vigt elm nang fln luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hi\ H2N-CH2-COOK + 2HC1 ' + HCl CH3COOK V i dv 8: Hon hgp X gom hai amino axit (mach ho, moi amino axit deu chiia > ClH3N-CH2-CC)OH + K C l > CHCOOH + KCl •,l.>m^V Vay: m = 111,5.0,2 + 74,5.0,3 = 44,65 gam - » Dap an A . Luu y: Trong bai nay cac em de quen KCl khi tinh kho'i lugng muo'i va chpn nham phirang an C! Vi 6 (A-12): Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chat khi a dieu ki^n thuong) roi dem toan bp san pham chay hap thu het vao binh dvmg mgt nhom chiic - N H 2 va mgt nhom chuc -COOH). Cho 16,4 gam X tac dyng vua d u vai dung dich K O H , thu dugc 24 gam muo'i. M | t khac, 16,4 gam X tac dung vua du V lit dung djch HCl 2M. Gia t r i ciia V la A . 0,1. ^. A.CH4. , B. C 3 H 4 . C. C 4 H 1 0 . ^ ^ mco2 + mH20= 39,4-19,912 = 19,488 gam. ^ 12 2 M9tkhac:mx = mc + m H = — n i r o , + — ^ H - J O =4,64. '* mco2 =15,312 gam (0,348 mol); m n j o =4,176 gam (0,232 m o l ) . > 1 mol -COOK 0,2 mol <- -> Dap an C. y,i . . i ,/ <, khoi lugng tang 38 gam. VHCI=0,2 lit. ^ ,,,,, , , V i dy 9: Cho 12 gam hon hgp X gom glyxin va etylamin tac dyng vua du voi dung dich HCl, thu dugc dung dich Y chiia 19,3 gam muo'i. Cho Y tac dung voi dung dich NaOH d u , thu dugc dung dich chua m gam muo'i. Gia trj cua m la A . 9,60. B. 15,45. C. 21,40. D. 19,10. Loigidi: Dap an B. . Glyxin (H2N-CH2-COOH) = a mol va etylamin (C2H5NH2) = b mol. Lim y: Ci day bai toan khong cho dung dich Ba(OH)2 d u nen neu cac em viet phuang trinh hap thu: Ba(OH)2 + CO2 = i'..^ , , tang24-16,4 = 7,6 gam nHci=0,2mol ^ Vay hidrocacbon X c6: So C : so H = nc : nH = 0,348 : 0,464 = 3 : 4 va suy ra ncQ2 = TI^^QQ^ {i > 1 mol - C O O H nHCl=nNH2 = " x ^ = mBaCOa - (mc02 + ^ H 2 0 ) = 19,912 gam. -> X la C3H4 f. nx = ncooH = 0,2 mol. Khi cho X tac dung vai dung djch HCl: D. C 2 H 4 . Loigidi: Kho'i lugng dung dich giam = Mat - Dugc d- D.0,3. 8""?' " " U : Khi cho X tac dung vai dung dich KOH: Theobai: lugng phan dung dich giam bot 19,912 gam. Cong thiic phan t u ciia X la C.0,2. Loigidi: dung dich Ba(OH)2. Sau cac phan ling thu dugc 39,4 gam ket tua va khoi ^j. B.0,5. Khi cho X tac dung voi dung dich HCl: > BaCOa + H2O. ^ 1 mol -NH2 0,2 mol thi se bi mac sai lam ngay! V i di;i 7: Hon hgp Y gom axit axetic, axit acrylic va axit adipic. Cho 16,9 gam Y > 1 mol -NH3CI -> khoi lugng tang 36,5 gam. 0,2 mol tang 19,3-12 = 7,3 gam Theobai: 16,9 gam X tac dung vai dung dich NaHCOs d u , thu dugc V lit khi CO2 -> a + b = 0,2 mol. Mat khac: 75a + 45b = 12 ^ a = 0,1 mol; b = 0,1 mol. Khi cho Y + NaOH d u , thu dugc cac muoi: H2N-CH2-COONa (0,1 mol) va (dktc). Gia trj a i a V la NaCl (0,2 mol). tac dyng vua dti vai dung djch NaOH, thu dugc 22,4 gam muo'i. Mat khac, A. 5,60. q B. 8,40. - • - • .• p C. 4,48. n D. 7,84. Cac phan ung hoa hgc: CIH3N-CH2-COOH + 2NaOH \ Khi cho Y tac dyng voi dung djch KOH: Mol: > H2N-CH2-COONa + NaCl 0,1 0,1 0,1 1 mol - C O O H -> 1 mol -COONa - > Khoi lugng tang 22 gam. Theobai: 0,25 mol tang 22,4-16,9 = 5,5 gam ncooH =0,25 mol. Mol: Khi cho Y tac dyng vai dung djch NaHCOa: nco2 = " c o o H = 0/25 mol Luu y: Neu quen tinh khoi lugng NaCl cac em se chgn nham phuong an A ! ^ ^ C2H5NH3CI + NaOH ai > C2H5NH2 + NaCl 0,1 0,1 - Vay: m = 97.0,1 + 58,5.0,2 = 21,40 gam ^ Dap an C. Vco2 = 0/25.22,4 = 5,6 l i t -> Dap an A. 25 Cty TNHH MTV DVVH Khang Vi$t Ca'm nang 6n luygn thi dgi hgc 18 chuySn dg H6a hoc - Nguyin VSn Hii 4. PHL/ONG PHAP BAO T O A N E L E C T R O N a. Npi dung 'Sh^n 7 4 v si-lk-l Lai gidi: ymi t ^'bdn vym xet: Cac em luu y la dung dich X chi chua hai muoi sunfat * ,1 Tong so' mol electron cac chat khu nhuong = Tong so' mol electron cac cha't • phan ung, S nam he't 6 dang goc sunfat. oxi hoa nhan. b. Cach ap di^ng Ta C O cac so do chuyen hoa: ; f-Aji, Cac em can xac djnh dung va day du cac chat khii va chat oxi hoa ciing nhu FeS2 , : • ' Cu2S .. \ Mol: su bien doi trang thai oxi hoa cua chiing. Mol: Viet cac qua trinh oxi hoa (nhuong electron) va qua trinh khu (nhan electron. , ; ... V i du 1 (CD-11): Hoa tan hoan toan 13 gam Zn trong dung dich HNOs loang, du thu dup-c dung dich X chua m gam muoi va 0,448 lit khi Ni (dktc). Gia tricuamla B. 37,80 gam. C. 28,35 gam. FeS2-15e Chat oxi hoa: 2N^5 > N2 ^ • ^^ ^ ' > 2Cu^2 + 5-6 Bao toan electron: 3n^o = 15npesj +10ncujS " N O = 0,2 mol c V^o = 4,48 lit. .oorfv '^^WM- phan ung xay ra hoan toan thu dugc 0,896 lit khi NO (6 dktc) va dung dich "lol. Y chua m gam muoi. Gia tri cua m la A. 8,88. B. 13,92. nenh,in C. 6,52. D. 13,32. Lot gidi: " M e = - ^ ^ =0,09 mol; nNo= ^^^=0,04 mol. ^ 24 22,4 > Zn*^ —> ne nhuong = 2nzn = 0,4 mol. + lOe ^' Cu2S-10e > Fe*^ + 25*^ 6 day cac em se dugc huong dan giai theo phuong phap bao toan electron. Zn - 2e 2a Vi dv 3 (B-08): Cho 2,16 gam Mg tac dung voi dung dich HNO3 (du). Sau khi Bai toan nay cac em c6 the giai khi viet phuong trinh phan ung. Chatkhu': a = 0,015. -^DapanD. D. 39,80 gam. Lot gidi: 13 0 448 nzn=—= 0,2 mol; nN2 = i : 7 T = -> Cac phan ung khu: VI D U M A U A. 18,90 gam. 0,03.2 + a = 0,015.3 + 2a ^ a > 2CuS04 Bao toan nguyen to S, ta c6: 2 npeSj + ncu2S = " 5 0 4 J, electron) de xac dinh so'mol electron trao doi roi ap djmg dinh luat bao toan 0,03 > -Fe2(S04)3 2 0,015 sau cac = lOn^^ = 0,2 mol. „ md'hu'tf Chatkhu: Mg - 2e > Mg^2 " Nhu v$y so mol electron trao doi chua bang nhau—> "chua on". O day, Chat oxi hoa: N^^ + 3e ^ NO; ' mot san pham khu da dugc "gia'u di", do la su tao thanh muoi NH4NO3: Bao toan electron: 2 n^g = 3 n^o + 8 nNH4N03 Chat oxi hoa: Mol: 2N^5 + se 0,2 > NH4NO3 -> -i ' nNH4N03 = 0,025 Vay: m = mzn(N03)2 + mNH4N03 = 0,2.189 + 0,025.80 = 39,8 gam. —> Dap an D. Luu y: 1- Bai toan nay c6 the giai nhanh hon bang each ap dung ngay 2N^5 + se > NH4NO3 " 2.0,09-3.0,04 , ^=0,0075mol. ^ -> m= mMg(N03)2 + mNH4N03 = 0,09.142 + 0,0075.80 = 13,92 gam. ->DapanB. ^ ^ ^ : , Vi d\ 4: Hoa tan hoan toan hon hgp gom 9,75 gam Zn va 2,7 gam Al vao 200 ml phuong trinh bao toan electron: 2 n^g = 3 n-^Q + 8 n[,jH4N03 • dung dich X chua dong thoi H N O 3 I M va H2SO4 1,5M. Sau khi phan ung 2- Ba kim loai kha manh (Mg, Al, Zn) tac dung voi axit HNO3, c6 the t^io thanh muoi NH4NO3! ^ . xay ra hoan toan thu dugc khi NO (san pham khu duy nha't) va dung dich V i d y 2: Hoa tan hoan toan h6n hop gom 0,03 mol FeS2 va a mol CU2S vao axit HNO3 (vira dii), thu dugc dung djch X (chi chua hai muoi sunfat) va V lit khi duy nha't NO (dktc). Gia trj ciia V la A. 1,12. B.5,60. C.2,24. D. 4,48. Y (chi gom cac muoi). Khoi lugng muoi c6 trong Y la: A. 41,25 gam. B. 53,65 gam. C. 44,05 gam. D. 49,65 gam. Lot gidi: 9 75 n2n=-^—= 0,15 mol; nAl = 0,1 mol. ' : • • Ca'm nang On luy$n thi dgi hgc 18 c h u y § n de Hoa hqc - Nguyin Van H&i Cty TNHH MTV Nhan xet: K h i axit H N O 3 c6 mat dong thai voi axit H2SO4 loang thi lirgng Led giai: trong dung dich la do 2 axit phan li ra -> giai theo phuong trinh ion. = '^HNOg + 2nH2S04 = 0,2 + 2.0,3 = 0,8 mol , ,^, ^ Ggisomol:Fe304(amol)vaFeS2(bmol). + 3Zn ai5 ifr^f-8 Mol: Al 8H* + 2NO; 0,4-^ + 0,1^ 4H* 0,4^ > AP + ai NO • — H + v a N O j tham gia phan ling het. > + > 3Fe*3 FeS2 - 15e 2H2O ufifki-^iiO - Khoi luong muoi trong Y = 9,75 + 2,7 + 0,3.96 = 41,25 gam Dap an A . V i d u 5 (B-07): N u n g m gam hot sat trong oxi, thu dugc 3 gam hon hg-p chat ran X. Hoa tan het X trong dung dich H N O 3 du, thoat ra 0,56 lit N O (san > Fe- + 2S- nN02= " F e 3 0 4 + 15nFeS2 B.2,22. Fe304 Mol: ^, ,^ - ' , ; ^ jo...','- a + 15b = 0,17 ^ a = 0,02; b = 0,01.^ ' , 3Fe(N03)3 "^'^^""'^ > 3Fe(OH)3 0,03 > Fe(N03)3 + 2H2SO4 ) Fe(OH)3 + 2BaS04 0,5Fe2O3+2BaSO4 D . 2,32. Laigidi: • n N O = — = 0 , 0 2 5 mol. 22,4 1,5 Fe203 0,02 FeS2 C . 2,62. ^ J-j^n;}».in Ta CO cac so do chuyen hoa: pham k h u duy nha't 6 dktc). G i a tri cua m la: A . 2,52. "f Bao toan electron: :Mig(m "+g': 4 ' ^^ Fe304 - l e s?, + NO~ .U. Cac phan ung k h u : > 3Zn2+ + 2 N O + 4H2O 0,1 1 . Ta c6: 232a + 120b = 5,84. Phuong trinh ion rut gon: Mol: 3,808 .- Trong X: DVVH Khang Vigt .88;;-',A Mol: 0,01 -> 0,005 0,02 - > m = 160.0,035 + 0,02.233 = 10,24 - > Dap an D . V i d u 7: C h o 12,45 gam hSn hgp X gom Fe, Mg, Z n vao dung dich H N O 3 du, Bao toan khoi lugng: m.Q^ = m x - mpg= 3-m. thu dugc dung dich Y (khong chua N H 4 N O 3 ) va hSn hgp khi Z gom 0,2 mol Nhan xet: Neu dua theo phuong trinh phan ung, bai giai se rat dai va kho giai. N O va 0,1 mol N O 2 . C o c^n dung dich Y thu dugc m gam muoi khan. Gia C a c h 1: 6 day, cac em can s u dung so do phan ung: tri cua m la Fe (1) - ^ 1 ° ^ X (2) A. 31,05. , Fe^3 (3) B. 43,45. C . 55,85. Bao toan electron: 3.npg = 4nQ^ + 3nj^Q ^ 3.E. = 1:21 Lcn gtat: + 0,075 -> m = 2,52 gam : .^v —> D a p an A . C a c h 2: Q u i doi X thanh Fe (a mol) va O (b mol). Ta c6: 56a + 16b = 3. 3e ^ Qua trinh k h u : O + 2e - > Fe*^ O-^; N*^ + 3e - > dich axit H N O 3 (dac, du), thu dugc 3,808 lit khi N O 2 (dktc) va dung djch X. Cho X tac dung voi dung djch Ba(OH)2 du, Ipc ket tiia va nung trong khong khi den kho'i lugng khong doi thu dugc m gam chat ran. G i a tri cua m la 7S N- + 3e netraod6i Mg - 2e > NO; > Mg^^; Z n - 2e N- + le C . 15,86. >Zn^^. > NO2. = 3njsjo + l-nN02 = ^''^"^°^- go'c N O 3 nam trong muoi dugc tinh n h u sau: V i d\ 6: H o a tan hoan toan 5,84 gam hon hop gom Fe304 va FeS2 trong dung B.9,68. Chat oxi hoa: > Fe-; . Den day, cac em can l u u y la khi cho k i m loai tac dung voi H N O s , so mol m = 0,045.56 = 2,52 gam. A. 8,66. pham khi nen can ap dung dinh luat bao toan electron. Bao toan electron: NO Bao toan electron: 3a = 2b + 0,075 —> a = 0,045; b = 0,03. -> Nhan xet: D a y la bai toan c6 nhieu chat k h u (3 k i m loai) v a tao ra 2 san Chat k h u : Fe - 3e Qua trinh oxi hoa: Fe - D . 62,05. D . 10,24. n, (muoi) = netraod6i = 0,7mol. ^ <»i (' 1 jf'ju'/j N03 Bao toan khoi lugng: m = m p e , M g , zn+m^^oa —> D a p an C . = ^^'^^ ^ (v ^ ^^'^^ .•••L 29 Cty TNHH MTV DVVH Khang Vijt C^m nang 6n luy^n thi d Muoi nitrat + N2O 2N*^ + Be va CO the xay ra qua trinh: ' * > N2O 2N*^ + Be sau mpt thoi gian thu dug-c cha't ran X va khi Y . C h o Y hap thy hoan toan vao dung dich Ba(OH)2 d u , thu dugc 29,55 gam ket hia. Chat ran X phan ung voi dung dich H N O 3 d u thu dugc V lit khi N O (san pham k h u duy ,s,i nhat, a dktc). G i a tri cua V la 0,15 n,, trao doi = IIXQQ . M a n ^ o = ^C02 ~ 0/15 mol -> ng trao doi = 0,3 mol. Bao toan electron: ngtraodoi = 3 n N o ^ V N O =2,24 lit Dap an A . T^NO =0,1 mol. 1 goc O^^ trong oxit se bi thay the bang 2 go'c 3i' = 2n , = 0,06 mol. " ^'^^ " ^'"^^ Bao toan electron: ng(x)=2nQ+3nr^o =0,15 mol - > n ^ ^ . =0,15 mol. Bao toan N : nnNOs = ^^Q'^ " ^'^^ ~^ ^' '' V i d^ 11 (B-12): C h o 29 gam hon hgp A l , C u v a A g tac dung v u a d u v a i 950ml dung dich H N O 3 1 , 5 M , thu dugc dung dich chua m gam muoi v a 5,6 lit khi X (dktc) gom N O v a N2O. T i khoi caa X so v a i H2 la 16,4. G i a tri ciia m la B. 97,20. Al,Cu,Ag n ' Nhan xet: K h i cho hSn hgp oxit tac dung vdi khi C O thi: + HNO3: = ng trao doi = 3 n ^ o = 0,09 mol. C . 98,75. D . 91,00. qua: nNo= 0,20 mol; n^^o= 0,05 mol. 6 day, cac em can s u dyng so do phan ung: 0,15 K h i cho oxit ("""OO 6 bai nay, truoc he't cac em can tim so mol moi khi trong X de thu dugc ke't " B a C 0 3 = 0,15 mol Mol: ^ " ^ " ^ > Muoi nitrat + N O Lai gidi: Lffigiai: > BaCOs + H2O K h i cho kim loai + HNOs: T^^Q- A. 98,20. D . 3,36. ) Muoi nitrat + N O rni-ii ricirt • Cach 2: Qui doi Y thanh X va O (0,03 mol). V i dyt 9: D a n luong khi C O di qua hon hgp gom C u O v a F e 2 0 3 nung nong, C u O , Fe203 > X k>r> CO2 + Ba(OH)2 :,,0,f rrr'''.':.. —> D a p an D . Nhan xet: C a n l u u y M g tac dung vai HNO3 c6 the tao ra muoi NH4NO3. /J < v nQ.2 ^^^.jj= 0,03 mol. Bao toan nguyen to N : nnwos = " N O " ,^ ^ Dap anB. C . 6,72. + N O o de tao muoi nitrat, do do: n = 8,9 + (0,36+B.0,005).62 + 0,005.80 = 34,1 gam B.4,48. 1,;, D- 0/18. ;,; { ' W > Y + a = 0,005 mol. ' ' Bao toan khoi luong: m = mMg, Zn+ m^^^^ + mNH4N03 . j-' Ltm y: Y chua ca kim loai (con du) va oxit. > NH4NO3 (a mol) Bao toan nguyen to N : nHNOs = " N O 3 ^ ^ " ^ ^ 2 0 + 2nNH4N03 A. 2,24. C . 0,16. e a c h 1: S u dung so do: X " i^jsjQ-(muoi) = ng trao doi = 8 njs420 + 8 nNH4N03 ~ (0/36 + 8a) mol. f B. 0,14. -> n o 2 = 0/015 mol -> n o = 0,03 mol ^ '>• r -> 0,5 = 0,36 + 8a + 2.0,045 + 2a pham k h u duy nhat, a dktc). So mol HNO3 da phan ling la Bao toan khoi lugng: m o 2 = 2,71 - 2,23 = 0,48 gam , K h i cho k i m loai + HNO3: * hoan toan Y vao dung dich HNO3 (du), thu dugc 0,672 lit khi N O (san Lodgidi: nN20= ^ ^ = 0 , 0 4 5 m o l ; n H N 0 3 = 0/5 mol. Chat oxi hoa: trong khi oxi, sau mot thoi gian thu dugc 2,71 gam hon hgp Y . H o a tan A. 0,12. Loigidi: Mg,Zn V i dV 10 (B-10): N u n g 2,23 gam hon hgp X gom cac kim loai Fe, A l , Z n , Mg ) Muoi nitrat + N O + N2O Chat oxi hoa: N = + 3e ^^ va CO the xay ra qua trinh: > NO; 2N*5 2N*^ + 8e + Se > N2O > NH4NO3 (a mol) , K h i cho kim loai + HNO3: "NO3 ^ ""^'"° ^ ^ ^ ^ " N 2 O ^ 8 riNH4N03 = (1 + 8a) mol. ,, ^ j Bao toan nguyen to N : nHN03 ^^oi ^^'^ ^ "N2O 2nNH4N03 ' -> 0,95.1,5 = 1 + 8a + 0,2 + 2.0,05 + 2a -> a = 0,0125 mol. 31 elm nang On luyjn thi dgi hgc 18 chuySn dg H6a hpc - Nguyjn Van HSl Bao toan khoi luqmg: m = m^i, c u , A g + "1,^0^ + Cty TNHH MTV DWH Khang Vijt mNH4N03 Xet su trao doi electron a cac giai doan: (3): Fe - 3 e = 29 + (1+8.0,0125).62 + 0,0125.80 = 98,2 gam. Dap an A. *' Cu - 2 e > Cu*^ ~^ • Nhan xet: C a n nhan ra bai toan da "giau d i " san pham NH4NO3. (1) _^ (2): O2 +4e Phan tram the tich cua khi CI2 trong X la C.50%. , Laigidi: n^,= M = a 4 m o l ; n M , = ^ = 0 , l m o l . 27 ^ 24 , . Mol: gnyb .ig :f d . i 5 Al - 3e M g - 2e > Al*^ Chat oxi hoa: Mol: CI2 + 2e > ICf^ 4x X > Mg*^ M o l : 0 , 1 - > 0,2 * f'^^'^ O2 + 4e > lOr 5 \ ]. (w^ - , ;gnu nerfq 6 u ' > a PHl/ONGPHAPTRUNGHOADIEN . Q si*: q&CJ < .(OMI) f -'••^h ^.-.-fi 5^! ( ;v. V i dv 1: D u n g dich X chua cac ion: Fe^* (0,1 m o l ) , A P " (0,2 mol), C h (x mol) va S04~ (y m o l ) . C o c^n dung djch X thu dugc 46,9 gam chat ran khan Y. Gia D . 0,14. Laigidi: nNO= ^ ; r — = 0,04 mol. 22,4 ; Bao toan kho'i lugng: TTIQ^ = m y - mx = 12,64 - 56a -1,92 = 10,72 - 56a. Nhan xet: Neu d y a theo phuong trinh phan ung se rat dai va kho giai. v tri cua X va y Ian lugt la A. 0,2 v a 0,3 . B. 0,1 v a 0,2. • nr,^ipf\n\~n C . 0,4 v a 0,4. f:, D . 0,3 v a 0,3. > 3 orbs Laigidi: dktc). G i a tri ciia a la Nhan xet: D u n g dich da cho chua 4 loai ion, trong do 2 loai ion chvra bie't so r mol, do vay can lap dugc 2 phuong trinh dai so de tim so mol cua chung. + A p dyng dinh luat trung hoa di?n: Tong so mol di^n tich duong = Tong so mol di^n tich am -> 0,1.2 + + 0,2.3 = x.l + y.2 ^ X + 2y = 0,8 mol. A p dung bao toan khoi lugng: my = 56.0,1 + 27.0,2 + 35,5x + 96y = 46,9 Cach 1: """^^^ ) Fe^^ Cu^^ (3) lit,; VIDVMAU dich HNO3 loang (du), thu dugc 0,896 lit khi N O (san pham k h u duy nhat 6 (2) ,„ ysb n Dung dich cac chat dien li luon luon trung hoa ve di^n. thoi gian, thu dugc 12,64 gam chat ran Y . Hoa tan hoan toan Y bang dung Y ;, i i O ftoi v M l ' o ^ ' i ' : .;„,, b. B i e u t h i i c V i dv 13: N u n g hon hop X gom a mol Fe va 0,03 mol C u trong khong khi mot 6 day, cac em can s u diing so do phan ung: . Lieu y: De tinh so'mol di?n tich, cac em la'y so mol ion x di^n tich ion do. = 0,2 mol; y = 0,3 mol • Q3 %Vn = ' .100% = 60% -> Dap an D . CI2 0,2 + 0,3 ^ Fe, C u (1) — nenhSn=3nNo =0,12. Tong so'mol di^n tich duong = Tong so mol di^n tich am: X C.0,12. „ a. N p i d u n g M o l : y —> 2y B.0,10. , '.f. -^DapanD. y t -> 32x + 71y + 10,8 + 2,4 = 40,9 -> 32x + 71y = 27,7. A. 0,08. ' '•' = 1,34-7a Bao toan electron: 3a + 2.a03 = 2b + ai2 ^ a = ai4; b = ai8 Bao toan kho'i luong: mx + Tn^i + m j ^ g = my -> 10,72 - 56a nenh^n=4no2 = Bao toan khoi lugng: 56a + 16b = 12,64-0,03.64 = 10,72. Bao toan electron: 4x + 2y = 1,2 + 0,2 = 1,4 mol. ImLrx. iff,: ^ m/ . " Cach 2: Q u i doi Y thanh: Fe (a mol); C u (0,03 mol) va O (b mol). M&ii) Igol^M m mifl:4 y i[ mil 0,4-^1,2 = 2ncu =0,06 '"• Dap an D . Nhan xet: Day la bai toan hai chat k h u (2 kim loai) va hai chat oxi hoa (2 phi kim) nen can ap dung dinh luat bao toan electron. Chat k h u : > NO s ^ ; ; n u, Bao toan electron: 3a + 0,0"6 = 1,34 - 7a + 0,12 -> a = 0,14 mol. D . 60%. - •', X > IQ-^ (2) -* (3): N*"' +3e gom 10,8 gam A l va 2,4 gam Mg, thu dugc 40,9 gam hon h(?p chat ran Y . B.40%. nenhir6ng ^ , _^ V i dv 12: C h o hon hop khi X gom CI2 va O2 tac dung v u a du vol hon hgp bpt A. 80%. nenhirang = 3 npg = 3a > Fe*^ - > ^ X = 0,2 mol; y = 0,3 mol. Dap an A. m p g 2 + + ^^3+ '^^ n "^ci" "^ " €0,0 H "^sol" ' 35,5x + 96y = 35,9. . ^, , ; , -'"JOb + > 33 Cty TNHH MTV DVVH Khang Vigt dm nang On luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hai Tirdo—> Luu y: D i e m m a u cho't d bai nay l a cac e m d u a ra p h u a n g t r i n h d^ii so' cua d j n h l u ^ t t r u n g hoa d i e n tich.:t.'e'-f ,ji*i„trt,»sj:fr '^4f^- V i d y 2 (B-12):, M p t d u n g d i c h g o m : Na* (0,01 m o l ) ; Ca^* (0,02 m o l ) ; H C O , 2HCOi (0,02 m o l ) v a i o n X (a m o l ) . I o n X va gia t n cua a la ••• A. O H - va 0,03. B. Ch va 0,01. D . N O ^ v a 0,03. " < f J>0 + eC rrfor; • A p d y n g d i n h luat t r u n g hoa di?n: 1.0,01 + 2.0,02 = 1.0,02 + n.a ' n.a = 0,03. ' ' ' ' fJom «> ^ CO^" + CO2 + H2O. m o l ) ; C O 3 ' (0,015 m o l ) ; C I - = 0,05 m o l . Lai gidi: + — , D o v%y, 1/2 cha't r a n k h a n t h u d u p e chua cac i o n : Ca^* (0,02 m o l ) ; N a * (0,04 C. COs^- va 0,03. Goi dien tich ion X la - n . , >,r lim y: K h i d u n soi den can d u n g d i c h X, goc H C O 3 b i p h a n h u y : .J.K ..•:i.«jSJi':'iie|' :>''J*!;/i b = 0,04 m o l . irtmrU Y ., ,(' 't;'^ * ^ • ' ' ' m = 2(0,02.40 + 0,04.23 + 0,015.60 + 35,5.0,05) = 8,79 g a m D a p an B. V i d y 4: M p t coc n u o c chua: a m o l Ca^^ b m o l Mg^^ v a c m o l H C O 3 . Cho t o i thieu V l i t d u n g d i c h Ca(OH)2 x mol/1 vao coc de l a m g i a m t o n g n o n g d p i o n rSffr D e n day c6 2 p h u a n g an thoa m a n la A v a D . V a y c h p n i o n X la i o n O H " hay ion N O + k i m loai t r o n g coc x u o n g m u c n h o nhat. Bieu t h u c t i n h V theo a, b , x la 2a+ b A V = ^ ^ . 3? OH- + HCO; , A6i-U)MU*I-. ..i-.,-' > C 0 ^ - + H20. —> D a p an D . Lin gidi: Cac p h a n l i n g hoa hpc k h i cho V l i t d u n g d i c h Ca(OH)2 x mol/1 vao coc: Ca(OH)2 + Ca(HC03)2 v a C I " (0,1 m o l ) . Mol: Cho 1/2 d u n g d i c h X phan l i n g v o i d u n g d i c h N a O H d u , t h u d u g c 2 g a m Mol: t h u d u o c 3 g a m ket tua. M a t khac, neu d u n soi den c^n d u n g d i c h X t h i t h u fjv \ Lai gidi: kUA^gi Itfbl' b A . 3,98. + 2 H 2 0 ^ ' " a+b •' Uiidnj-ir., "'"niih. ri>\ ;iM'«vJ>'>i>t oifo \—> D a p an B. , , ,, ,,,,„,„,, -- i • B.6,87. C. 5,43. D . 4,78. O b a i nay, d u n g d i c h da cho chua 4 lo?ii i o n , t r o n g d o i o n SO 4 " chua biet so m o l , d o v a y can lap d u p e 1 p h u o n g t r i n h d a i so de t i m so m o l nay. * vao ket tua: OH- + HCO: > CO,^' + H 2 O . Ca2> + C O ^ - > CaCOs K h i N a O H d u vao 1/2 d u n g d i c h X: OH- + HCO > CO 3" 0,03 -> Ca2* + C O 3 " 0,02 3" + H2O 0,02 A p d y n g d i n h l u a t t r u n g hoa d i ^ n : * i ^ T o n g s o ' m o l d i ^ n tich d u o n g = T o n g so'mol dien tich a m 0,01.1 + 0,02.3 = 0,04.1 + x.2^^, x = 0,015 m o l . + m'^'^'^'^^^'^' Bao toan k h o i l u p n g : m = m^^^ + m^^^ + m^^^. + m^^2. = 39.0,01 + 56.0,02 + 62.0,04 + 96.0,015 = 5,43 gam. f, .,.^j, ^ - » D a p an A . 0,03 Lieu y: tai nay cac e m can ap d u n g d j n h luat t r u n g hoa di?n de t i m dupe so > CaCa <- + „ >^;TOMb rbil - > c = 0,03 mol. 0,03 MgCOs +CaC03 Lai gidi: Nhqn xet: K h i cho Ca(OH)2 d u vao 1/2 d u n g d i c h X, toan b p goc H C O 3 se d i Mol: ••ttVqed b V = A p d y n g d i n h luat t r i i n g hoa d i ^ n : 2a + b = c + 0,05. Mol: U,,J,,W... v a SO 4~ (x m o l ) . C o can Y t h u d u p e m g a m m u o i k h a n . Gia t r j cua m l a G p i so m o l t r o n g 1/2 d u n g d j c h X: Ca^* (a m o l ) ; Na* (b m o l ) , H C O 3 (c mol) Mol: 2CaC03 + 2 H 2 O a <' > '~ ' V i d v 5: D u n g d i c h Y c6 chvia: K ^ (0,01 m o l ) , Fe^^ (0,02 m o l ) , N O 3 (0,04 m o l ) va CI-(0,05 mol). + <- —>Vx = a + b—> d u g c m g a m chat ran k h a n . Gia t r j cua m la D . 8,79. a Ca(OH)2 + M g ( H C 0 3 ) 2 ket tua. C h o 1/2 d u n g d i c h X c o n lai p h a n l i n g v o i d u n g d i c h Ca(OH)2 d u , C . 9,26. "mtlJl, A p d u n g d i n h l u a t t r u n g hoa dien: 2a + 2b = c. ife graiQ B.7,47. • ' + jt^n ;V ^ ^ _ i ± b D . V= 2x + '-IdU'dfi V i d^ 3 (B-IG): D u n g d i c h X chira cac i o n : C a ^ Na*, H C O , A . 9,21. ^ , , _ a + 2b C.V=^ ^ . X N h a n thay i o n O H " k h o n g the t o n tai cung i o n H C O 3 t r o n g d u n g d i c h ban daudocophanung: ' a+b B.V^ — . - > a = 0,02 mol. m o l cua goc sunfat. " t,Oi:<; r ' i , „ i • . 35 Cty TNHH MTV DWH Khang Vi$t C ' nang On luyjn thi dgi hgc 18 chuySn dS H6a hpc - Nguygn Van Hi'i a m Ap dung djnh luat trung hoa di^n: 0,01.2 + 0,03.1 = 0,02.2 + n^^. .1 Vi dv 6: Dung djch X c6 chua: Fe^* (0,05 mol), Na* (0,07 mol), CI" (0,03 mol) va -> n^j_ = 0,01 mol. SO 4 " . Cho dung dich Ba(OH)2 du vao dung dich X thu dugc ket tiia Y. Nung Y a nhi^t dg cao ngoai khong khi den khol lugng khong doi thu dugc + Bao toan khoi lugng: Khoi lugng cac chat tan trong Y = ^^^^2+ + ' " N H ^ ^, m gam chat tin Z (coi BaS04 khong hi nhi?t phan). Gia tri cua m la A. 19,91. B. 18,11. C. 24,31. D. 20,31. Lai nidi: * ^ Ap dung djnh luat trung hoa dien: 2 n^, ^ 2. 0,05 +1.0,07 = 0,03.1 + 2n Mol: lugng chat tan trong mgt nua dung dich Y. Vi ° • 0,07 ' 0,07 Fe^- y ^ Mol: 0,025 Vay m = 233.0,07 + 160.0,025 = 20,31 gam. •• • j,, Ba2* + CO3" Ca- ket hia va 0,672 lit khi (dktc). + COr C. 5,35 gam. CaC03l A. 1. o'ijhqil >NH3t+H20 = 0,02 lit = 20 ml - > Dap an B . , B. 12. C. 13. D.2. + Ap dung dinh luat trung hoa dif n vai dung djch X: , 1.0,07 = 2.0,02+l.n _ -» n _ = 0,03 mol. OH -A OH • Ap dung djnh luat trung hoa dien vai dung dich Y: ,^ * • Phan 2 + dung djch BaCh du: = 0/02 mol. V= Votigidi: ^ - > % 2 . = nMg(OH),= ^ = a 0 1 m o l . ~^ "Ba2+ = "BaS04 = ^ , va NO3 la 0,04. Trgn X va Y dugc 100ml dung dich Z. Gia tri pH ciia Z la jj > Mg(OH)2i BaS04; ^' mol). Dung djch Y c6 chua CIO 4, NO 3 va H"" (y mol); tong so mol CIO4 o Phan 1 + dung djch NaOH du: ^ jii 0 H ^ ^ - , . . . BV p : ^ ; - : Vi dv 9 (A-10): Dung dich X c6 chua: Na^ (0,07 mol); SO^" (0,02 mol) va O H " (x D. 7,05 gam. Lot gtat: , . > BaCOsi nNa2C03 = 0,02 mol NMn xet: Bai nay cac em giai dua theo cac phuong trinh ion. Ba^^ + sol" -M.J'jj \t 3 Tong kho'i lugng cac chat tan trong Y la v a n ^ . = n N H 3 = - r r - = 0,03 mol. 4 '^"^ 22,4 iftAl : Xn^^^2- = X + y = 0,02 mol. .y + Phan 2 tac dyng voi dung djch BaCl2 du, thu dugc 4,66 gam ket tiia. NH^ +OH- D. 70. Cac phuong trinh phan ung : + Phan 1 cho tac dung voi dung dich NaOH du, dun nong, thu dugc 0,58 gam Mg2^ + 2 0 H " , Ap dyng djnh luat trung hoa di?n: 2x + 2y = 0,01.1+ 0,03.1 = 0,04 mol. bang nhau. ' C.40. Ggi so'mol ciia cac ion : Ba^* = X ; Ca^* = y. 7: Chia dung dich Y chua cac ion: Mg2% N H J , SO 4", CI" thanh hai phan B. 6,11 gam. , LaigidU • • A. 3,055 gam. , 8: Dung dich Y c6 chua dong thoi cac ion: Ba^*; Ca^*; CI" (0,01 mol) va Ion nhat. Gia trj nho nhat ciia V la A. 30. B.20. 0,0125 --^ — • Dap an D. > Vi , NO J (0,03 mol). Cho V ml dung djch Na2C03 I M vao Y de thu dugc ket tiia > F e ( O H ) 3 1 Fe203 )Fe(OH)2 "^cr "^sol" ' • ' ' • Lim y: Bai nay cac em de chgn nham dap an A (3,055 gam) do chi tinh khoi ^ n , . = 0,07mol. > BaS04 . 0,07 , ^':^^Otti--> _> Dap an B - 2 nj, ^ """'^ ) BaS04 ^ = 2.(0,01.24 + 0,03.18 + 0,01.35,5 + 0,02.96) = 6,11 gam Cac so do phan ung: SO 4 " ,'^f^•v:iJiJr^;,'^^ii:... f ' -Aq«Q.^-' 'h '"^nt^h 'qsi^^^IKi! -yin ••{in itff ->;; jf;|„yi j ^ : . , ,j : ^•"ciOi -^l-^NOS =^-V V =0'04mol. • Trgn X vai Y: H* + O H > H2O -> H^du = 0,01 mol -» IH^]= ^ = 0,1 = 10"^ ->.pH = l - > Dap an A. ., , " + 3.,' i ? -tit • ( i n i l ) ; ^ rf.^ o: 37 CtyTI\ih.i Ca'm nang 6n luygn thi dgi hgc 18 chuy§n dg H6a hgc - Nguygn van H&\ 6 1TV DWH Khang Vi^t PHI/ONGPHAPDI/6NGCHEO V i dv 10 (CD-07): D u n g dich Z chiia: Cu^* (0,02 mol), K"" (0,03 mol), C h (x mol) va S O 4 " (y mol). Tong kho'i lugng cac muo'i tan c6 trong Z la 5,435 gam. Gig tri cua x va y Ian lugt la A. 0,02 va 0,05. V6i hSrhgp bat ki g6m trung b m h ( M ) t a s e c 6 t i l e : B. 0,05 va 0,01. C . 0,01 va 0,03. D . 0,03 v a 0,02. M LOT gidi: W 0,02.2 + 0,03.1 = x . l + y.2 X + 2y = 0,07 mol. B a o t o a n k h o i lugng: m z = m ^ 2 . + m . + m^,. + m B i l l ;H^i,^>vvi. ^" " ^' ^ S,, - • ,-,4 v ' c ! * . 1 ; ^ ,r|/ Phan 1 tac dung voi dung dich N a O H du, d u n nong thu dugc 0,672 lit khi (dktc) v a 1,07 gam ket tiia; ...^ Co can X thu dugc m gam muo'i khan. Gia tri cua m la B. 7,04 gam. Lai -> "Fe3^ = " M O H ) 3 = , .'rin»*Wi , • 6':> V fti\b , = ^'^^ "^"l- " N H ^ = ''NHg= ^ > = 0'03 BaSOU Ap dung dinh luat trung hoa di?n: 0,01.3 + 0,03.1 = 0,02.2 + n^^^. .1 = 0,02 mol. Bao toan khoi lugng: m = m^ 3+ + *' mol. i: . jA • ^ ' + "^j>jo~ ""^ "^so^' Dap an C . , 1,46 73 0,54 J:^.Asxu'^'^^^'^^^: i^^^y • 'o 27 » % § C u = 73%. a an D . Cach 2: N h a n thay A c u = 63,54 < 63 + 65 dong v i 29 C u chiem u u the hon D a p an D (Cac dap an khac deu < 50%). ' V i d^;i 2 (CD-07): C h o 4,48 lit khi C O (dktc) t u tir di qua ong s i i nung nong dung 8 gam mgt oxit sat den khi phan ling xay ra hoan toan, thu dugc hon hgp khi X c6 ti khoi so voi hidro bang 20. Cong thiic ciia oxit sat v a phan tram the tich ciia khi C O 2 trong X la A.FeO;75%. B. FezOs; 75%.' C . FeaOs; 65%. D.Fe304;75%. Lai gidi: 4 48 Theo bai: M = 20.2 = 40 v a nx = n c o = T T T = 0'2 m o l . . . , , , A p dving cong thiic ciia phuong phap duang cheo, ta c6: ' '' "CO = 2.(0,01.56 + 0,03.18 + 0,02.62 + 0,02.96) = 8,52 gam " - ' D . 73%. n : Cach 1: Ggi so nguyen h i 29 C u la a v a 2^ C u la b. Ap dung cong thuc cua phuong phap duong cheo, ta c6:' - > % 29 C u > 50% , 4 66 + Lot gidi: Dap t. nnw 0,672 Phan 2: + dung dich B a C h du: Ba^* + S O 4 " -> n _ AM, tong so nguyen t u ciia dong vj 2 9 C u la A. 27%. B. 50%. C . 54%. 65 - 63,54 1*' Nhan xet: bai nay cac em giai dua theo cac phuong trinh ion. 1'07 n, 29 C u . N g u y e n t u khoi trung binh cua dong la 63,54. T h a n h phan phan tram D . 4,26 gam. gidi: > Fe(OH)3l . >NH3T+H20 lAMv M, I 63-63,54 C . 8,52 gam. . Phan 1: + Bung dich N a O H d u : Fe3* + 3 0 H ' t ' C ^ N H ; +OH* Y: • :';©aji-4):;::.,3-plIiJ's , ,. ;„)-..i^^:;; Phan 2 tac dung voi dung djch B a C h du, thu dugc 4,66 gam ket tua. . A . 3,52 gam. n> V i dxf. 1 (CD-07): Trong t u nhien, nguyen to dong c6 hai dong v i la 2 9 C u v a V i dvi 11 (CD-08): Chia dung d k h X chiia cac ion: Fe^^ S O | ~ , N H 4 , N O 3 - ,>.'i, • • v i o y MAU , —> D a p an D . "t* , ' f ,, 1=0 ''^ ' N ^ a n u Q :8 i#b > ; x = 0,03 mol; y = 0,02 mol. thanh hai phan bang nhau. .., - K i l o f n ' e O i O ) 'rOVi 64.0,02 + 39.0,03 + 35,5x + 96y = 5,435 35,5x + 96y = 2,985 ^ i »! ^ > I ) AMN X: ^t. A p dung dinh luat trung hoa dien: - M Y b. So do hoa do vay can lap dugc 2 phuang trinh dai so de tim so' mol cua chung. + , M-Mx ny • O bai nay, dung dich Z chiia 4 loai ion, trong do 2 loai ion chua biet so'mol, . ^ hai chat X va Y , khi biet dugc gia tri khoi lugng mol nC02 = 44 -40 1 0,05 28 - 4 0 3 0,15 %Vco2 =75% chgn B hoac D . 39