Cẩm nang hướng dẫn ôn luyện thi đại học 18 chuyên đề Hóa học: Phần 1

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Phần 1 tài liệu Cẩm nang ôn luyện thi đại học 18 chuyên đề Hóa học giới thiệu tới người đọc các nội dung: Các phương pháp giải nhanh hóa học, các axit vô cơ điển hình, tính chất của các hiđroxit, tính chất của các muối vô cơ, tính chất của các oxit, các nguyên tố phi kim điển hình, đại cương về kim loại. Mời các bạn cùng tham khảo.

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TS. NGUYEN VAN HAI (Chu bien) NGUYEN NAM TRUNG - TRAN THE NGA - NGUYEN THj THU HA Cam naligFluyen thi dai hoc 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 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 . ^. , huyen di 1. C a c phucmg phap giai nhanh 3 a. N g i d u n g « ^. Tong khoi lu
Cty TNHH MTV DWH Khanq Vigt dm nangflnluy$n thi d H2. m k i m M + moxi =moxit -> m o x i = 30,2 -17,4 = 12,8 gam. nj^+ = H H C I + 2nH2S04 = 0,5.1 + 2.0,5.0,28 = 0,78 mol. , , , -> = ^ ^ = 0'4 mol V02 = 8,96 lit ^ Dap an B. Mat k h a c : n u , = 0,39 m o l -> n^+ het -> X c h i c h i i a c a c m u o i . , ' V i dv 4 (A-08): Cho 2,13 gam hon hgp X gom Mg, Cu va A l 6 dang bot tac Khoi lugng muoi = m^|3+ + ^^^^2+ + ^^i- + ^^^2- ' dyng hoan toan voi O2 thu dugc hon hop Y gom cac oxit c6 khoi luong 3,33 = 7,74 + 0,5.35,5 + 0,14.96 = 38,93 g a m . , ; j v ^ ^ ^ y . , _ gam. The tich dung dich HCl 2M vira dii de phan ling het voi Y la -¥ Dap a n A. ' > - A. 60ml. B. 45ml. C. 75ml. D. 80ml. 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. Lai gtat: 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 Nhan xet: bai nay cac em nhat thiet phai ap dung bao toan khoi lugng de 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 tinh kho'i lugng oxi tham gia phan ling: A. 7,84. B.4,48. C. 3,36. D. 10,08. mkimioai + m o x i = m o x i t —> m o x i = 3,33 — 2,13 = 1,2 gam. Lai giai: 12 Nhan xet: Bai n a y 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. ^ = TIT = 0,0375 mol no = 0,075 mol Lim 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 ' ^ V - ( o x i t ) = 0-075 mol. ; - ^^^^ t o a n k h o i l u g n g , cia t h e : Khi cho oxit bazo tac dung vai axit tao ra nuoc: mc^03 + i"Al=mx m A i = 23,3 -15,2 = 8,1 g a m -> n ^ r 0,3 m o l . 02- + 2H^ > H2O. .0 Suy ra: n^+ = 0,15 mol -> nnci = 0,15 mol -> VHCI = 0,075 lit = 75ml. Phuong trinh h o a h g c : Cr203 + 2A1 > 2Cr + Al2C)3. (j^p.^v,-.^ Mol: ai 0,2 0,2 0,1 ' ' . -> Dap an C. X g o m : A l d u = 0,1 m o l ; Cr = 0,2 m o l . ^^-^ V i dv 5 (A-12): Hoa tan hoan toan 2,43 gam hon hgp gom M g va Z n vao mpt 3 , lugng vira d i i dung djch H2SO4 loang, sau phan ling thu dugc 1,12 lit H2 Khi X tac d \ i n g v o i axit: A l ^ - H 2 v a C r - > H2 (dktc) va dung dich X chiia m gam muoi. Gia tri ciia m la ^ -> n H 2 = 0,35 mol VH2= 7,84 lit - » Dap a n A. A. 4,83 gam. B. 5,83 gam. C. 7,33 gam. D. 7,23 gam. V i d\ 8 (B-09): Cho 100ml dung djch K O H 1,5M vao 200ml d u n g d j c h H3PO4 Lai giai: 0,5M, 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 O bai nay, cac em c6 the giai chi tiet dua theo phan ling hoa hgc. A. 15,5 gam. B. 18,2 gam. C. 12,8 gam. D. 16,4 gam. Nhan xet: nH2S04 = r'H2 = 0,05 mol. Lai giai: So do phan ling: K i m l o a i + H2SO4 > Muoisunfat + H2 " K O H = 0,15 m o l ; nH3P04 = 0,10 . Bao toan kho'i lugng cho so do tren: ^ 6I>''• Nhan xet: Cac em c6 the xet ti le m o l K O H va H3PO4, s a u do v i e t 2 phuong ->• m = 2,43 + 0,05.98 - 0,05.2 = 7,23 gam ' ii£0tOB>; 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 . ->DapanD. , , ^'^ ^^-^^^ ' ^
Cty TKi, 1TV DVVH Khang Vi$t Ca'm nang 6n luyfn thi d^i hpc 18 ctiuySn dg H6a hpc - Nguygn Van HSi 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: nong hon hgp X (xiic tac Ni) mot thoi gian, thu dugc hon hgp Y c6 t i khoi so H3PO4 + KOH > Muoi + H2O voi H 2 bang 10. D i n hon hgp Y qua dung dich brom du, sau khi phan ung Taco: mH3P04 + rnKOH= 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. mx = 0,1.98+ 0,15.56-0,15.18 = 15,5 gam Lai gidi: —> Dap an A. Bao toan khoi lugng: mY = mx = 0,15.52+ 0,6.2 = 9 gam. 5 , 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 Mat khac: M Y = 10.2 = 2 0 "2= = 0/45 mol " '^ • tac N i , sau mot thoi gian thu dugc hon hgp khi Y. Dan toan bg Y Igi t u t u 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. • ,,^.J^:n. V - A . 1,04 gam. B. 1,32 g a m . C. 1,64 gam. D. 1,20 gam. Gia su so mol H2 tham gia phan umg = a mol So mol lien ke't n phan I' • Laigiai: ;;(;'; ling = a mol. Nhan xet: Bao toan khoi lugng: mY = mx = 0,06.26 + 0,04.2 = 1,64 gam. Mat khac, so mol khi giam = so mol H2 phan ung = n x - n Y nkfusi'^'^ 0 448 - > a = 0,75 - 0,45 = 0,3 mol. or,, Mat khac: n z = = 0,02 mol va M z = 0,5.32 = 16 , - > So mol lien ke't 7i d u = 0,45 - 0,3 = 0,15 mol = So'mol Br2 phan ung. 22,4 Khoi lugng brom phan ung = 0,15.160 = 24 gam. , mz = 0,02.16 = 0,32 gam. —> Dap an B. Luu y: Khoi lugng binh brom tang = khoi lugng cac hidrocacbon bi hap thy. V i d u 12: Hon hgp X gom 0,1 mol etilen, 0,2 mol axetilen va 0,5 mol H 2 . Nung Bao toan khoi lugng, ta c6: nong hon hgp X (xiic tac Ni) mgt thoi gian, thu dugc hon hgp Y c6 t i khoi Kho'i lugng binh brom tang = mv - mz = 1,64 - 0,32 = 1,32 gam. ,.j .j-^,^, •^i^^.j^.^ so v o i H 2 bang 10. Dan hon hgp Y qua dung dich brom du, sau khi phan , Dap an B. ling xay ra hoan toan, tha'y c6 m gam brom tham gia phan ung. Gia trj cua 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 A. 16. B.32. C.8. D.24. vao binh dung djch brom (du), sau khi ke't thiic cac phan ung, khoi lugng Laigiai: " ' , ' ; ,,s*^->r:: ':^:s,rf'^~ binh tang m gam va c6 280ml hon hgp khi Z (dktc) thoat ra. Ti kho'i aia Z so Bao toan khoi lugng: mY = mx = 0,1.28 + 0,2.26 + 0,5.2 = 9 gam. voi H2 la ia08. Gia tri ciia m la • ,iaw;. : , Mat khac: M Y =10.2 = 20 n z = = 0,45 mol " ' % A.a585. B. 0,620. C. 0,205. D. 0,328. Lbi gidi: Nhan xet: etilen c6 chiia 1 lien ke't 71, axetilen c6 chiia 2 lien ke't 71 - > So Nhan xet: Bao toan khoi lugng: mv = mx = 0,02.26 + 0,03.2 = 0,58 gam. mol lien ke't 7t ban dau = 1.0,1 + 2.0,2 = 0,5 mol. , Mat khac: n z = ^ = 0,0125 mol va M z = 10,08.2 = 20,16 Gia su so mol H2 tham gia phan ung = a mol - > So' mol lien ke't n phan 22,4 ? ling = a mol. mz = 0,0125.20,16 = 0,252 gam. Nhanthay: 'r^^.^ i i,*!.'.a " ; .m»s;c,ci ;,/•> • Mat khac, so mol khi giam = so mol H2 phan ung = nx - nY ^ ^ - > a = = 0,8 - 0,45 = 0,35 mol. 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. - > So mol lien ke't 7t d u = 0,5 - 0,35 = 0,15 mol = So mol Br2 phan ung. —>^ Dap a n D . Khoi iugng brom phan ung = 0,15.160 = 24 gam. —> Dap a n D . ' 7
dm nang On luygn thi dgi hpc 18 chuySn dg H6a hpc - Nguygn van HJi Cty TNHH MTV DWH Khang Vijt V i dv 13: Cho 2,1 gam h6n hgp X gom hai amin (no, don chiic, dong dang ke Tuy nhien, bai nay dugc giai nhanh chong khi ap dung bao toan khoi tiep) phan ling het voi dung dich HCl (du), thu du(?c 3,925 gam hon hgp lugng: maxlt + mbazo= mmuol + mnuoc muoi. Cong thuc cua hai amin trong X la _^ mH20= 3,6+ 0,06.56+ 0,06.40-8,28 = l,08gam^> nH20= 0,06 mol. A. CH3NH2 va C2H5NH2. B. C2H5NH2 va C3H7NH2. Y^ic 36 Mat khac: n H , o = Max.t = = 60 Axit la C H 3 C O O H C. C3H7NH2 va C4H9NH2. i '> • D. C H 3 N H 2 va (CH3)3N. S 0 >. ^ 0,06 • . .V Lot gidi: -> 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 ' i n = ^^^22. = =3 2 este trong X c6 cong thuc phan tu la C3H6O2. Heste 0,1 A. 8,40 gam. B. 14,80 gam. C. 10,92 gam. D. 12,90 gam. -> Cong thuc cau tao 2 este la: HCOOC2H5 va CH3COOCH3. Lai gidi: Phan ling hoa hgc: Nhan xet: n^jo = 1/1 > ^C02 = 0,6 ^ X gom cac ancol no, mach ho va: HCOOC2H5 + NaOH — ^ HCOONa + C2H5OH "X =nH20 -rtC02 =1,1-0,6 = 0,5mol -> n(0)x =nx =0,5 C H 3 C O O C H 3 + NaOH —!—> CHsCOONa + C H 3 O H .' Bao toan kho'i lugng: De tha'y: nNaOH = neste = nz = 0,2 mol. ^ mx = mc + niH + mo = 0,6.12 + 2,2.1 + 0,5.16 = 17,4 gam. Bao toan kho'i lugng: meste+mNaOH = mv + mz Trong phan ung tao ete thi: nx = Zn^jO i^H20 = 0,25 mol. -> 0,1.74 + 0,1.40 =7,85 + mz ^ mz =3,55 gam. Ht:k{OCC;i Bao toan kho'i lugng: Nhan xet: Trong phan ung ete hoa ta c6: "^x = r^ete + nHjO mete = 17,4 - 0,25.18 = 12,90 gam -> Dap an D. 1 „ , n ,mw>'^''iOfU nAoi • i H2O = T n z n H20 = 0,05 mol. V i d u 15 (B-08): Cho 3,6 gam axit cacboxylic X (no, don chiic) tac dung hoan 2 ; y 7.- . Bao toan khoi lugng: mz = mete + m H20 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 -> m = mete = 3,55 - 0,05.18 = 2,65 gam -> Dap an A. ,s :sv A.C2H5COOH. B.CH3COOH. C.HCOOH. D.C3H7COOH. V i d^ 17: Xa phong hoa hoan toan 0,1 mol este X (dan chiic, mach ho) bang Lai gidi: 100 g^m dung dich M O H 11,2% (M la kirn loai kiem). Co can dung dich sau Nhan xet: Theo bai, axit phan ung hoan toar\i baza c6 the con du, do phan ling, thu dugc 15,4 gam chat ran khan, dong thai ngung tu phan hoi vay neu giai dua vao phuong trinh phan ung se rat kho khan. 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 TIMHH MTV DVVH Khang Vigt A. CH3COOCH3. B. CH3COOC3H7.4 Lmgidi: C. CH3COOC2H5. D . C2H5COOC2H5. Ggi cong t h u c cua chat beo la (RCOO)3C3H5. , < ' i - b tfW 1,. i, Lot gidi: K h i cho chat beo tac d u n g v o i N a O H : Nhan xet: (RCOO)3C3Hs + 3 N a O H > 3RCOONa + C3Hs(OH)3 K h o i l u g n g M O H bang 11,2 g a m " " ' '' • Mol: 0,06 0,02 - > kho'i l u a n g d u n g m o i (nuac) = 100 - 11,2 = 88,8 g a m . a r i f / v i •« A p d u n g bao toan kho'i l u g n g : mch.nt beo + m N . n O H = m x a phong (muoi) + mgiixeroi gfi; Phan u n g hoa hoc: iv;y£;f{;-^,';f0 m x . i phong = 17,24 + 0,06.40 - 0,02.92 = 17,8 g a m D a p an C. ' * "• ' rfsBi Oi 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 S^'- M o l : 0,1 0,1 0,1 0,1 ^ 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. 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: Gia t r i cua m la A . 9,94. B. 9,38. C. 10,50. ' " D . 8,82. :k'^' mancoi = 93,4 - mnuoc = 92 - 88,8 = 3,2 gam. Laigiai: VinV ^MROH = 32^>R' = 15(R'lanh6mCH3-)".''*^' ^ .„ ngiixe((ji = 0,01 m o l •>• m. :,3f' ! aijfi Bao toan k h o i l u g n g : meste = 15,4 + 3,2 - 11,2 = 7,4 g a m (RCOO)3C3H5 + 3 K O H > 3RCOOK + C3H5(OH)3 ^ Meste = 7 4 R C O O C H s = 74 ^ R la C H 3 - Mol: 0,03 m = 8,82 g a m . 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 -> D a p an D . 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 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 K O H d a t h a m g i a p h a n u n g la 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 A . 140 g a m . B. 175 gam. C. 280 g a m . D . 350 g a m . Laigiai: A . CH3CH2CH(NH2)COOH. B. H2NCH2CH2COOH. C h i so axit cua chat beo = ^"^^^ ^ 8== ^"^g^ C. C H 3 C H ( N H 2 ) C O O H . D . H2NCH2COOH. rn^hatbeoCg) 700 Laigiai: m ^ o H ^ 5600 m g = 5,6 g a m nmn = 0,1 m o l . Nhan xet: Cac d a p an d e u cho aminoaxit chua m g t n h o m - C O O H Phan l i n g hoa hoc: ' ggi cong t h u c cua X la H2N-R-CC)OH. - j||||vi.^^^ ^g,; > ' RCOOH + KOH > RCOOK + H2O K h i cho X tac d u n g v o i axit H C l : f'^^, Mol: 0,1 0,1 H2N-R-COOH + HCl CIH3N-R-COOH .CJ^jinV Bao toan kho'i l u g n g : mammoaxit + mHci = mmuoi * (RCOO)3C3H.s + 3 K O H > 3RCOOK + C3H5(OH)3 Mol: 3x -> 3x -> mHci= 13,95 - 10,3 = 3,65 g a m - > UHCI = 0/1 = Bao toan kho'i l u g n g : mchat beo + m K O H = mmuoi + mgUxeroi + m ^ Mx = — = 103 ^ - H 2 N - R - C O O H = 103. - '^-'^ 0,1 • ••••vv,;::fv;„ ' - > 700 + (5,6 + 56.3x) = 764,6 + 92.x + 0,1.18 ^ R = 42 —>• Goc hidrocacbon la -C3H6- D a p an A . - > x = 0,8 m o l mKOH = (0,1 + 0,8.3).56 = 140 g a m . 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 D a p an A . 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 Vi 19 (B-08): Xa p h o n g hoa hoan toan 17,24 g a m chat beo can vvra d u 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 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 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 o n g . Gi a t r i ciia m la -'"^^^ * p h a n t u . Gia t r i a i a m la A . 16,68. B. 18,24. C. 17,80. D . 18,38. A. 51,72. B. 54,30. C. 66,00. , D . 44,48. • • • • • 11
dm nang On luy?n thi dgi hpc 18 chuySn 6i H6a hpc - Nguygn Van HJi 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 Theo bai, a mol tetrapeptit mach h a X c6 a.3 = 3a mol lien ket peptit. Trong cac phan ung hoa hpc, cac nguyen to luon dupe bao toan. 2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit b, H ? qua Tong so'mol nhom c a c b o x y l - C O O H bang a + 2a =3a. Tong so mol nguyen tu ciia mpt nguyen to c6 trong cac chat truoc phan ung Cac phan ling nit ggn cua lien ke't peptit va nhom - C O O H la: ' va sau phan ung luon bang nhau. o , ^ , . . r ' 'yinit -CO-NH- + NaOH > -COONa + H2N- , LKU y: C a n xac dinh diing va day dii cac chat c6 chiia nguyen to dang xet 6 • nnMQh u 7a 7a truoc va sau phan ling. , , VIDVMAU -COOH + NaOH > -COONa + H2O , ^ 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 Mol: 3a 3a 3a M'-.A dich K O H (du), thu dupe a mol hon hpp khi v a dung dich X. Sue khi C O 2 Theo bai: nwaOH = 7a + 3a = 10a = 0,6 —> a = 0,06 mol. (du) vao X, lupng ket tua thu dupe la 46,8 gam. G i a tri cua a la Bao toan khoi lugng: m + mNaOH = mmuai + m ' i(..>.:-i(jin A. 0,55. B.0,60. C.0,40. D . 0,45. -> m = 72,48 + 0,06.3.18 - 0,6.40 = 51,72 gam Lai gidi: D a p an A. . ,, C a c h 1: 6 bai nay, cac em c6 the giai dua vao cac phuong trinh phan ung: V i d\ 23: D u n nong m gam hon hop gom a mol dipeptit mach ho X v a 2a mol tripeptit mach ho Y voi 400ml dung djch H C l I M (vira du). Sau khi cac Al + K O H + H2O > KAIO2 + •|H2 phan ung ket thuc, c6 can dung dich thu dugc 48,1 gam muoi khan cua cac AI4C3+ 4 K O H + 4H2O > 4KA102 + 3CH4 r H - f f - + cllV^-: amino axit deu c6 mpt nhom - C O O H va mpt nhom - N H 2 trong phan ttr. KAIO2 + CO2 + 2H2O > Al(OH)3 + KHCO3 , . , G i a tri ciia m la C a c h 2: So do phan ung:^ . ^ ^ j ^ , . A . 33,5. B. 29,0. C . 30,8. D . 28,1. Al, A I 4 C 3 ) KAIO2 ^^02^^20 ^ Al(OH)3 j w , . . ,0-!: ' ' , Lai gidi: Bao toan nguyen to A l : ^, Nhan xet: K h i cho cac peptit tac dung voi H C l thi c6 cac lien ket peptit va " A l + 4 n A i 4 C 3 = I^Al(OH)3 + ^^A\4C3= nhom H 2 N - (ciia amino axit dau N ) tham gia phan ung. Theo bai, a mol dipeptit mach ho X c6 a.l = a mol lien ket peptit. Theo bai: n ^ i +nAi4C3 = 0'3 nAi =0,2; nAi4C3 = 0'l- ' 2a mol tripeptit mach ho Y c6 2a.2 = 4a mol lien ket peptit > Tong so mol nhom amino H 2 N - bang a + 2a = 3a. ' ff V a y : a = | n A i + 3 n A i 4 C 3 = ^,6 D a p an B. Cac phan ung riit gpn ciia lien ket peptit va nhom H 2 N - la: 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 , -CO-NH- + HCl + H2O )--COOH + CIH3N- H N O 3 (vua dii), thu dupe dung djch X (chi chiia hai muoi sunfat) v a khi Mol: 5a 5a 5a duy nhat N O . G i a tri cua a la H2N- + HCl > CIH3N- A. 0,075. B.0,12. C.0,06. " D.0,04. ' Mol: 3a 3a Lai gidi: 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 Bao toan khoi lupng: m + m H c i + m = mmuoi phan ung, S nam het a dang goc sunfat. T a CO cac so do chuyen hoa: ; i m = 48,1 - 0,4.36,5 - 0,25.18 = 29 gam. —> D a p an B. ^ FeS2 > iFe2(S04)3 i'ouU-,,r!>! • . • Mol: 0,12 0,06 'ih.^1bm^l^^., : ^; ].,>f:v., r v - ; v : - ' - '-vi'.- ./\ 10
C^m nang 6n luygn Ihi dgi hpc 18 chuyfin ii Hoa hpc - Nguyin van H i ! Cty TNHH MTV D W H Khang Vi$t Cu2S > 2CuS04 Bao toan nguyen to P theo so do: Mol- a 2a , ^^^^ , . j,|,|,j3i);, ii^, Ca(H2P04)2 < > P2O5 ^ i, n 0,12.2 + a = 0,06.3 + 2a -> a = 0,06 -> Dap an C. % khoi lupng: 69,62% x% V i d u 3 (A-12): Cho 18,4 gam hon hop X gom Cu2S, CuS, FeS2 va FeS tac dung 69,62.142 v r ^ '' —>x=— = 42,25 - > Dap an C. ,. het voi HNO3 (dac nong, du) thu duoc V h't khi chi c6 NO2 (6 dktc, san Vi dvi 5 (A-12): Mot loai phan kali c6 thanh phan chinh la K C l (con lai la cac pham khu duy nha't) va dung dich Y. Cho toan bo Y vao mot lugng du tap chat khong chua kali) dupe san xua't tu quang xinvinit c6 dp dinh dung dich BaCl2, thu dugc 46,6 gam ket tiia; con khi cho toan bp Y tac duong 55%. Phan tram khoi lupng ciia KCl trong loai phan kali do la dung voi dung dich NH3 du thu dugc 10,7 gam ket tua. Gia tri cua V la ,^ A. 95,51%. B. 65,75%. C. 87,18%. D. 88,52%. A. 38,08. B. 24,64. C. 16,8. D. 11,2. Loigidi: < *= - 6 day, cac em can nho la dp dinh duong cua phan kali dupe tinh theo % NMn xet: Dung dich Y chiia cac ion: F e ^ Cu^^ SO4 , va NO3. khoi lupng cua K2O. „,, . rt:. Khi cho dung dich Y + dung dich BaCh: Ap dung bao toan nguyen to'K, ta CO so do: Ba2* + SO^" — ^ BaS04i /'1 v > KCl -!-K20 v;.. V . - . Bao toan nguyen to'S: ng (X) = nBaS04 ^ mol. , Kho'i lupng mol: 74,5 gam 47 gam > Khi cho Y + dung djch NHa du: % khoi lupng: x% Cu(OH)2i 47 ^ Luu y: Cu(OH)2 tan trong NH3 du tao thanh phuc chat: 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 " tac dung voi dung dich FiNOs loang (du), thu dupe khi NO (san pham khu 10 7 ' duy nha't) va dung djch chua m gam muo'i. Gia trj cua m la Bao toan nguyen to Fe: npe (X)= nFe(OH)3 = 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: Fe = 0,1 mol; Cu = 0,1 mol va S = 0,2 mol. A iKt/tij: Fe > X (Fe, FeO, Fe203, Fe304) > Fe(N03)3 -» ne (X) = 2ncu + Snpe + 6ns = 1,7 mol n^o^ = (x) = 1,7 mol va ap dung dinh luat bao toan nguyen to'Fe: ; j ,0 mm -> V = 1,7.22,4 = 38,08 lit ^ Dap an A. Fe > Fe(N03)3 „ , Vi dv 4 (B-10): Mot loai phan supephotphat kep c6 chua 69,62% Ca(H2P04)2, con .0 f:5 lai gom cac chat khong chiia photpho. Dp dinh duong ciia loai phan Ian nay la Mol: 0,1 0,1 A. 48,52%. B. 39,76%. C. 42,25%. D. 45,75%. m =0,1.242 = 24,2 gam ^ Dap an D. Loigidi: V i du 7: Cho 31,2 gam hon hpp gom Al, Cu va Ag tac dung vua dii voi 900ml 6 day, cac em can nho la dp dinh duong cua phan supephotphat dupe tinh dung dich HNO3 1,5M, thu dupe dung djch chua m gam muoi va 4,48 lit I theo % khoi lupng cua P2O5. hon hpp khi X (dktc) gom NO va N2O. Ti khoi ciia X so voi H2 la 16,75. Gia •;• NMn xet: 1 mol Ca(H2P04)2 hay 1 mol P2O5 deu chua 2 mol P . , » , u MM. 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 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 qua: nNo= m o l ; n ^ j o " ^ O'OS i ^ o l - d y n g v a c h u y e n t h a n h NaA102. Al,Cu,Ag ) M u o i n i t r a t + N O + N2O „ /, / ' 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 . Chatoxihoa: N^^ + 3e > NO; 2N*5 + ge > N2O _^ Ban d a u : nAi = 2.0,3 = 0,6 m o l "€1203= ^^^-r^^^= 0,2 m o l . . v a CO the xay ra ca qua t r i n h : ZN*' + 8e > NH4NO3 (a m o l ) K h i cho k i m loai + H N O s : Cr203 + 2A1 — ^ AI2O3 + 2Cr , , n ^ ^ Q - ( m u o i ) = n g trao J6i = 3 nfyio + 8 nivj20 + 8 nfvjH^NOs ~ 0'^^ Mol: 0,2 ^ a4:r,^ ,R>0,2;;.. OA : ^ , 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 . Bao toan n g u y e n to N : nHNOa = + " N O + 2 nN20 + 2nNH4N03 -> HHCI =3nAi +2ncr+6nAi203 = l ' 3 m o l . ^DapanB. 0,9.1,5 = 0,85 + 8a + 0,15 + 2.0,05 + 2a ^ a = 0,025 mol. ' iTs-> K!) 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 Bao toan k h o i l u g n g : m = m ^ i , cu, Ag + m ^ ^ ^ + mNH4N03 ' 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. = 31,2 + (0,85 + 8.0,025).62 + 0,025.80 = 98,3 gam. 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 -> 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. B. 11,06. C. 11,86. D . 20,66. 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 Laigiduii ^' ' |>,s£' V - ' ( 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 = ^ = 0,18 m o l . vM^b n^B . 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 22,4 , 10,2 g a m chat ran. Gia t r i cua V la 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. A . 2,24. B. 3,36. C. 5,60. D . 4,48. ' Cac p h a n l i n g khvr: Fe304 - l e > 3Fe^3 *~ - - ,0'' ^ i'; ' . Ldigidi: G
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 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 Laigiai: 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 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 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 n g u y e n t u cacbon. ciia V la i 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 . A . 14,56. B. 15,68. C . 11,20. D . 4,48. M a t khac: M x = 21,2.2 = 42,2 - > mx = 42,4.0,1 = 4,24 g a m . ' " ' '"*V Laigiai: ' Bao toan kho'i l u g n g , ta c6: mx = mc + mn. '>rn 6,(J " f' > . - ^ 1 /•-(.'" M^' «f^ ' 11'2 , 12,6 , - > 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 . "CO2=^=0'5"^ol;nH2O=^=0,7mol. ^ 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 . 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) 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. .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 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 :gfi A . 70,0 l i t . ir sv B B. 78,4 l i t . C. 84,0 l i t . D . 56,0 l i t . Ij ^ no = 0,4 m o l . Lmgidi: > 100 •JIA «;:.«,) jfW Bao toan n g u y e n to' O, ta c6: no (OH) + 2 no^ = 2 TXQQ.^ + nyi^^Q 2.0,5 + 0 , 7 - 0 , 4 , ^^i^rW ' nc02 = - ^ = 0 3 5 m o l ; n H 2 O = ^ = 0 ' 5 5 m o l . no2=— ^ ^ = 0,65 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 ^ V02 = 0,65.22,4 = 14,56 l i t ^ D a p an A . t h i chuyen he't v a o C O 2 v a H 2 O . 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^ + n^jjo • 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 = ^'^^-^^^'^^ = 0,625 m o l V Q J = 0,625.22,4 = 14,0 l i t . axit la • timA • A. H C O O H va C 2 H 5 C O O H . Vay: Vkhongkhi = 5 V o 2 = 70,0 l i t D a p an A . s at w ;.::;;;»•* B. C H 2 = C H C O O H v a C H 2 = C ( C H 3 ) C O O H . ' ' 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, C. C H s C O O H v a C2H5COOH. thupc c u n g d a y d o n g dang, t h u dugc 3,808 l i t k h i C O 2 (dktc) v a 5,4 g a m D. C H 3 C O O H va C H 2 = C H C O O H . H 2 O . Gia t r i ciia m la A . 5,42. B.5,72. C. 4,72., D . 7,42. Laigiai: H!"\,0-,. Laigiai: ' ^ Bao toan n g u y e n t o O, ta c6: 2nx + 2 n o , = 2 nco2 "H2O ri one KA . .it • • ;loM • 2.0,1 + 2 . 0 , 2 4 - 0 , 2 , nc02 = ^ = 0^17 m o l ; n H 2 0 = ^ = 0,30 m o l . ^C02=— Y ^ = 0'24mol. //,4 10 .„ 4 •• Cac e m can thay rang, k h i d o t chay X: n^^^Q > UQQ.^ - > X chiia 3 ancol no 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 - > nx = nj^^Q - n^Q^ = 0,30 - 0,17 = 0,13 m o l . Logii A v a C. D o X chua cac ancol d a n chiic ^ no = noH = nx no = 0,13 m o l . -» 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 5' j Bao toan k h o i l u g n g t r o n g X, ta c6: 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 - > ' D a p an C . M / » t j ; ; , t'-vs'jji'ii ....lii^^ 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 V i dv 18: Hon hgp X gom hai amino axit no (chi c6 nhom chuc - C O O H va - N H 2 " h6 X va Y la dong dang ke tiep (Mx < MY). Do't chay hoan toan mot lugng trong phan tu), trong do ti 1§ mo : mN = 80:21. De tac d\ing vvra du voi 3,83 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 gam hon hgp X can 30ml dung dich HCl I M . Mat khac, do't chay hoan toan Yla 3,83 gam hon hgp X can 3,192 lit O 2 (dktc). Dan toan bp san pham chay ( C O 2 , A. Etylamin. ' ' [ '2 : •'XlS:',U,'0 B. Propylamin. - ,(•) .A H 2 O va N 2 ) vao nude voi trong d u thi kho'i lugng ket tiia thu dugc la C. Butylamin. . , D. Etylmetylamin. A. 13 gam. " B. 20 gam. C. 15 gam. D. 10 gam. '! • .'' . . , Laigidi: n .i . ., . ii..t» Laigidi: 4536 224 n ;fom t ; , i . - - - H H C I = 0,03 mol; n o 2 = ai425 mol. • C>D < - - 1 n o , = ^ ^ = 0,2025 m o l ; n r o , = — = 0,1 m o l . . • • ^. r n o ^ S O n p _ 80/16 10 ' ^ 22,4 ^ 2 22,4 ......£Dtjd3X,P'!J6do'?riT man xet: Ixx ti 1? khoi lugng: , Bao toan nguyen to'O, ta c6: 21 ^ n^ 21/14 ~ 3 • 6: 2 no2 = 2 n c o j + " H J O ^ H J O = 0/205 mol. M|it khac: n N = nNH2 = " H C I (x)= 0,03 mol -> no(X)= 0,1 mol. ^ Ggiso'mol: n ^ (x) = a i^ioJ va n H ( x ) = b m o l . Nh^n thay, khi dot chay 1 mol anken thu du(?c XXQQ^ = n H 2 0 ' vol hai Bao toan khoi lugng: mx = m ^ + mj^ + mo +T^-H amin no, don chijfc thi n H j O " ' ^ C 0 2 1'^ ph"^°''^8 ^""^^'^'^"'^S- -» m c + n i H = 3,83 - 0,1.16 - 0,03.14 = 1,81 12a+ b = 1,81. CxH2x.3N + O2 — ^ X C O 2 + ( x + l , 5 ) H 2 0 + 0,5N2 ^ ' Bao toan nguyen to O: n Q (x) + 2no2 = 2 n c o 2 + ^HjO 0,205-0,1 sot os8 'fl iiAA US' Dovay: n H 2 0 " ' ^ C 0 2 = l / 5 n a m i n - > namin=—^ '— =0,07mol. ^ 0,1 + 2.0,1425 = 2a + 0,5b -> 4a + b = 0,77 ^ a = 0,13. ,. ,. . Bao toan nguyen to C: n c ( X ) = n c o 2 = '^CaC03 = 0 , 1 3 ^ m c a C O a =13 gam. -> So'nguyen t u C trung binh trong M = — ^ ^ < — ^ 2 2 _ = ~ 1^43. Y "M I^aniin 0,07 -> 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 , ffl>| { 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) Dap an A. Khi tham gia phan ling hoa hgc, nguyen t u (nhom nguyen tu) cua chat ban V i d\ 17: H o n hgp X gom mpt anken va hai ancol (no, don chiic, mach ho). dau dugc thay the (ho^c cgng hgp) bang nguyen t u (nhom nguyen tit) mai Dot chay hoan toan mpt lugng X can vua du V lit khi O 2 , thu duQC 15,68 lit de tao thanh san pham. ^ '' khi C O 2 va 18 gam H 2 O . Cac the tich do 6 dktc. Gia t r i cua V la • *^' • Do do, khoi lugng cua chat tao thanh c6 the tang len hay giam di do chenh A. 22,40. B. 17,92. C. 16,24. D. 23,52. l|ch khoi lugng mol cua cac nguyen t u (nhom nguyen tu). Dua vao sy tang hay giam nay c6 the xac djnh so mol cac chat trong n c o 2 = 0,7mol; n H 2 0 = l m o l . mt?iV6J 1 phuang trinh hoa hgc, t u do c6 the giai nhanh nhieu bai toan. NMn 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 S\f thay the (cpng hgfp) Bien doi k h o i lugmg (tinh cho 1 mol) "ancol = " H 2 O ~ ^C02 • Tang 71-60 = 11 gam C03 > CI2 Do v ^ y : nancol = " H Z O - ^coz = 1 - 0,7 = 0,3 m o l . , „,. Tang 96-16 = 80 gam 0-2 (oxit) ""''^"^ > S O ^ 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 . CO — ^ CO2; H2 H2O Tang 16 gam 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 Tang 23-1 = 22 gam -OH > -ONa 2.0,7 + 1-0,3 , Tang 23-1=22 gam -> HQ^ = = 1 , 0 5 mol _COOH > -COONa 2 -NH2 > -NHsCl Tang 36,5 gam V 0 2 = 1,05.22,4 = 23,52 lit ^ Dap an D. 21
dm nang fln luygn thi dgi hgc 18 chuySn de H6a hpc - Nguyjn Van Hi\ Cty TNHH MTV DVVH Khang Vi§t VI DU MAU 1 mol 0 - 2 > 1 mol SO l~ Khoi lugng tang 96 - 16 = 80 gam. ff V i d\ 1 (A-08): Cho V l i t hon hop khi (dktc) gom CO va H2 phan ung vol mot a mol tang 52 - 22 = 30 gam. f. • lugng d u hon hop ran gom CuO va Fe304 nung nong. Sau khi cac phan ung _^ a = — = 0,375mol -> no (X) = 0,375 mol. '" ' '' •' ' xay ra hoan toan, khoi lugng hon hgp ran giam 0,32 gam. Gia tri ciia V la 80 A. 0,448. B. 0,112. C. 0,224. D. 0,560. Khi cho 22 gam X tac dung voi CO thi: no (X) = n^oj = riCaCOa = 0375 mol. Lot giai: - > m =0,375.100 = 37,5 gam. 1 mol CO —*° > CO2 Kho'i lugng chat ran giam 16 gam. ^^^m —> Dap an B ' '''t*''-t> i?>'; f vcr-f vi KT * •'•H'>I.. 1 mol H 2 — * ° > H2O —> Kho'i lugng chat ran giam 16 gam. V i dv 4 (A-10): Dot chay hoan toan mgt lugng hidrocacbon X. Hap thu het san -> 1 mol (CO va H 2 ) ^ ° > ( C O 2 va H 2 O ) giam 16 gam. pham chay vao dung dich Ba(OH)2 du, tao ra 29,55 gam ket tiia, dung dich Theobai: 0,02 mol V = 0,02.22,4 = 0,448 lit. phan t u cua X la —> Dap an A. A.C3H8. B.C2H6. C.C3H4. D.C3H6. V i d\ 2: Hon hgp Y gom FeO, Fe203 va CuO. Hoa tan hoan toan 6,8 gam Y 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 : Ba(OH)2 + C O 2 > BaCOs + H 2 O . A. 5,2. B.5,6. C.6,0. D. 4,8. ^ Mol: 0,15 Khoi lugng tang 71 - 1 6 = 55 gam. Vay hidrocacbon X c6: So C : so'H = nc : nn = 0,15 : 0,4 = 3 :>8. a mol no (Y) = 0^1 mol. ^ 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 Khi cho Y tac dung voi CO thi cac oxit deu bi khu thanh kim loai. ^ tac dung voi dung dich HCl du, thu dugc dung djch chua m gam muoi. Bao toan khoi lugng: my = m + mo - > m = 6,8 - 0,1.16 = 5,2 gam. Gia trj cua m la - > Dap an A. ' ' " "* A. 44,65. B. 50,65. C. 22,30. ^ D. 33,50. V i dvi 3: Hon hgp X gom CuO va Fe203. Hoa tan hoan toan 22 gam X bang Ldigiai: dung dich H2SO4, thu dugc 52 gam muoi. Mat khac, neu khu het 22 gam X 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. bang CO (du), dan hon hgp khi thu dugc vao dung dich Ca(OH)2 (du), tao Khi cho X tac dung voi dung dich K O H : • •' • thanh m gam ket tua. Gia trj cua m la :> 1 mol - C O O H ^ 1 mo! -COOK -> khoi lugng tang 38 gam. A. 45,5. B.37,5. C. 40,5. D. 50,0. Theobai: 0,3 mol a + b = 0,3 mol. Nhan xet: Khi cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va mgt ion Mat khac: 75a + 60b = 21 a = 0,2 mol; b = 0,1 mol. 0-2 trong oxit dugc thay the bang mgt ion SO 4 " . 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
elm nang fln luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hi\ Cty TNHH MTV DVVH Khang Vigt H2N-CH2-COOK + 2HC1 > ClH3N-CH2-CC)OH + K C l V i dv 8: Hon hgp X gom hai amino axit (mach ho, moi amino axit deu chiia ' CH3COOK + HCl > CHCOOH + KCl •,l.>m^V mgt nhom chiic - N H 2 va mgt nhom chuc -COOH). Cho 16,4 gam X tac Vay: m = 111,5.0,2 + 74,5.0,3 = 44,65 gam - » Dap an A . dyng vua d u vai dung dich K O H , thu dugc 24 gam muo'i. M | t khac, 16,4 Luu y: Trong bai nay cac em de quen KCl khi tinh kho'i lugng muo'i va chpn gam X tac dung vua du V lit dung djch HCl 2M. Gia t r i ciia V la nham phirang an C! A . 0,1. B.0,5. C.0,2. D.0,3. f. Vi 6 (A-12): Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chat khi a dieu Loigidi: 8""?' " " U : i'..^ , , ki^n thuong) roi dem toan bp san pham chay hap thu het vao binh dvmg Khi cho X tac dung vai dung dich KOH: {i > y,i . . i ,/ 1 mol -COOK khoi lugng tang 38 gam. ^j. lugng phan dung dich giam bot 19,912 gam. Cong thiic phan t u ciia X la Theobai: 0,2 mol Dap an A. 25
Ca'm nang 6n luygn thi dgi hgc 18 chuySn dg H6a hoc - Nguyin VSn Hii Cty TNHH MTV DVVH Khang Vi$t 4. PHL/ONG PHAP BAO T O A N E L E C T R O N 7 Lai gidi: a. Npi dung 'Sh^n 4 v si-lk-l ymi t ^'bdn vym xet: Cac em luu y la dung dich X chi chua hai muoi sunfat sau cac *• ,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. ; f-Aji, Ta C O cac so do chuyen hoa: : • ' b. Cach ap di^ng FeS2 > -Fe2(S04)3 Cu2S > 2CuS04 Cac em can xac djnh dung va day du cac chat khii va chat oxi hoa ciing nhu 2 .. \ su bien doi trang thai oxi hoa cua chiing. Mol: 0,03 0,015 Mol: a -> 2a Viet cac qua trinh oxi hoa (nhuong electron) va qua trinh khu (nhan 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 0,03.2 + a = 0,015.3 + 2a ^ a = 0,015. • ^^^ ' electron. , ; ... Cac phan ung khu: VI D U M A U FeS2-15e > Fe*^ + 25*^ Cu2S-10e > 2Cu^2 + 5-6 V i du 1 (CD-11): Hoa tan hoan toan 13 gam Zn trong dung dich HNOs loang, Bao toan electron: du thu dup-c dung dich X chua m gam muoi va 0,448 lit khi Ni (dktc). Gia 3n^o = 15npesj +10ncujS " N O = 0,2 mol V^o = 4,48 lit. c tricuamla -^DapanD. .oorfv '^^WM- A. 18,90 gam. B. 37,80 gam. C. 28,35 gam. D. 39,80 gam. Vi dv 3 (B-08): Cho 2,16 gam Mg tac dung voi dung dich HNO3 (du). Sau khi Lot gidi: phan ung xay ra hoan toan thu dugc 0,896 lit khi NO (6 dktc) va dung dich 13 0 448 nzn=—= 0,2 mol; nN2 = i : 7 T = "lol. Y chua m gam muoi. Gia tri cua m la A. 8,88. B. 13,92. C. 6,52. D. 13,32. Bai toan nay cac em c6 the giai khi viet phuong trinh phan ung. Lot gidi: 6 day cac em se dugc huong dan giai theo phuong phap bao toan electron. Chatkhu': Zn - 2e > Zn*^ —> ne nhuong = 2nzn = 0,4 mol. " M e = - ^ ^ =0,09 mol; nNo= ^^^=0,04 mol. „ ^ 24 22,4 md'hu't- Chat oxi hoa: 2N^5 + lOe > N2 ^ nenh,in = lOn^^ = 0,2 mol. Chatkhu: Mg - 2e > Mg^2 f " Nhu v$y so mol electron trao doi chua bang nhau—> "chua on". O day, Chat oxi hoa: N^^ + 3e ^ NO; 2N^5 + se > NH4NO3 ' mot san pham khu da dugc "gia'u di", do la su tao thanh muoi NH4NO3: -i Bao toan electron: 2 n^g = 3 n^o + 8 nNH4N03 " Chat oxi hoa: 2N^5 + se > NH4NO3 ' 2.0,09-3.0,04 , Mol: 0,2 -> 0,025 nNH4N03 = ^=0,0075mol. ^ Vay: m = mzn(N03)2 + mNH4N03 = 0,2.189 + 0,025.80 = 39,8 gam. -> m= mMg(N03)2 + mNH4N03 = 0,09.142 + 0,0075.80 = 13,92 gam. —> Dap an D. ->DapanB. ^ ^ ^ : , Luu y: 1- Bai toan nay c6 the giai nhanh hon bang each ap dung ngay Vi d\ 4: Hoa tan hoan toan hon hgp gom 9,75 gam Zn va 2,7 gam Al vao 200 ml phuong trinh bao toan electron: 2 n^g = 3 n-^Q + 8 n[,jH4N03 • dung dich X chua dong thoi H N O 3 I M va H2SO4 1,5M. Sau khi phan ung 2- Ba kim loai kha manh (Mg, Al, Zn) tac dung voi axit HNO3, c6 the t^io xay ra hoan toan thu dugc khi NO (san pham khu duy nha't) va dung dich thanh muoi NH4NO3! ^ . Y (chi gom cac muoi). Khoi lugng muoi c6 trong Y la: V i d y 2: Hoa tan hoan toan h6n hop gom 0,03 mol FeS2 va a mol CU2S vao axit A. 41,25 gam. B. 53,65 gam. C. 44,05 gam. D. 49,65 gam. HNO3 (vira dii), thu dugc dung djch X (chi chua hai muoi sunfat) va V lit Lot gidi: khi duy nha't NO (dktc). Gia trj ciia V la 9 75 ' n2n=-^—= 0,15 mol; nAl = 0,1 mol. : • • A. 1,12. B.5,60. C.2,24. D. 4,48.
Ca'm nang On luy$n thi dgi hgc 18 c h u y § n de Hoa hqc - Nguyin Van H&i Cty TNHH MTV DVVH Khang Vigt 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. 3,808 = '^HNOg + 2nH2S04 = 0,2 + 2.0,3 = 0,8 mol , ,^, ^ .- Trong X: 1 Ggisomol:Fe304(amol)vaFeS2(bmol). . Ta c6: 232a + 120b = 5,84. .U. ' Phuong trinh ion rut gon: Cac phan ung k h u : ^^ 3Zn + 8H* + 2NO; > 3Zn2+ + 2 N O + 4H2O Fe304 - l e > 3Fe*3 "f ^ J-j^n;}».in - ' , ; Mol: ai5 0,4-^ 0,1 s?, FeS2 - 15e > Fe- + 2S- ^, ,^ ^ jo...','- ifr^f-8 Al + 4H* + NO~ > AP + NO + 2H2O Bao toan electron: Mol: 0,1^ 0,4^ ai • :Mig(m ufifki-^iiO nN02= " F e 3 0 4 + 15nFeS2 a + 15b = 0,17 ^ a = 0,02; b = 0,01.^ ' —> H + v a N O j tham gia phan ling het. "+g': 4 - Fe(N03)3 + 2H2SO4 ) Fe(OH)3 + 2BaS04 A . 2,52. B.2,22. C . 2,62. D . 2,32. 0,5Fe2O3+2BaSO4 Laigidi: • Mol: 0,01 -> 0,005 0,02 n N O = — = 0 , 0 2 5 mol. .88;;-',A - > m = 160.0,035 + 0,02.233 = 10,24 - > Dap an D . 22,4 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) , Fe^3 (3) A. 31,05. B. 43,45. C . 55,85. D . 62,05. Bao toan electron: 3.npg = 4nQ^ + 3nj^Q Lcn gtat: 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 ^ 3.E. = 1:21 + 0,075 -> m = 2,52 gam : .^v pham khi nen can ap dung dinh luat bao toan electron. —> D a p an A . Chat k h u : Fe - 3e > Fe-; Mg - 2e > Mg^^; Z n - 2e >Zn^^. C a c h 2: Q u i doi X thanh Fe (a mol) va O (b mol). Ta c6: 56a + 16b = 3. Chat oxi hoa: N- + 3e > NO; N- + le > NO2. Qua trinh oxi hoa: Fe - 3e ^ Fe*^ Bao toan electron: netraod6i = 3njsjo + l-nN02 = ^''^"^°^- . Qua trinh k h u : O + 2e - > O-^; N*^ + 3e - > NO 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 Bao toan electron: 3a = 2b + 0,075 —> a = 0,045; b = 0,03. -> m = 0,045.56 = 2,52 gam. 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 n, (muoi) = netraod6i = 0,7mol. D a p an C . (v .•••L 7S 29
C^m nang 6n luy^n thi d D a p an D . = 8,9 + (0,36+B.0,005).62 + 0,005.80 = 34,1 gam Cach 2: Qui doi Y thanh X va O (0,03 mol). f Dap anB. Bao toan electron: ng(x)=2nQ+3nr^o =0,15 mol - > n ^ ^ . =0,15 mol. Nhan xet: C a n l u u y M g tac dung vai HNO3 c6 the tao ra muoi NH4NO3. 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, Bao toan N : nnNOs = ^^Q'^ " ^'^^ ~^ ^' '' sau mpt thoi gian thu dug-c cha't ran X va khi Y . C h o Y hap thy hoan toan 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 vao dung dich Ba(OH)2 d u , thu dugc 29,55 gam ket hia. Chat ran X phan dung dich H N O 3 1 , 5 M , thu dugc dung dich chua m gam muoi v a 5,6 lit khi ung voi dung dich H N O 3 d u thu dugc V lit khi N O (san pham k h u duy 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 ,s,i nhat, a dktc). G i a tri cua V la A. 98,20. B. 97,20. C . 98,75. D . 91,00. A. 2,24. B.4,48. C . 6,72. D . 3,36. Lai gidi: /J < Lffigiai: 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 qua: nNo= 0,20 mol; n^^o= 0,05 mol. 6 day, cac em can s u dyng so do phan ung: Al,Cu,Ag ) Muoi nitrat + N O + N2O C u O , Fe203 > X ) Muoi nitrat + N O k>r> n Chat oxi hoa: N^=^ + 3e > NO; 2N*^ + 8e > N2O CO2 + Ba(OH)2 > BaCOs + H2O va CO the xay ra qua trinh: 2N*5 + Se > NH4NO3 (a mol) , Mol: 0,15 0,15 ' K h i cho kim loai + HNO3: Nhan xet: K h i cho hSn hgp oxit tac dung vdi khi C O thi: "NO3 ^ ""^'"° ^ ^ ^ ^ " N 2 O ^ 8 riNH4N03 = (1 + 8a) mol. ,, ^ j 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 ^ T^NO =0,1 mol. Bao toan nguyen to N : nHN03 ^^oi ^^'^ ^ "N2O 2nNH4N03 ' V N O =2,24 lit Dap an A . -> 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 Cty TNHH MTV DWH Khang Vijt Bao toan khoi luqmg: m = m^i, c u , A g + "1,^0^ + mNH4N03 Xet su trao doi electron a cac giai doan: = 29 + (1+8.0,0125).62 + 0,0125.80 = 98,2 gam. (3): Fe - 3 e > Fe*^ - > nenhirang = 3 npg = 3a s ^ ; ; n u, Dap an A. *' Cu - 2 e >• Cu*^ ~^ nenhir6ng = 2ncu =0,06 '"• " ' '•' Nhan xet: C a n nhan ra bai toan da "giau d i " san pham NH4NO3. ^ , _^ . 10,72 - 56a , „ 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 (1) _^ (2): O2 +4e > IQ-^ nenh^n=4no2 = = 1,34-7a gom 10,8 gam A l va 2,4 gam Mg, thu dugc 40,9 gam hon h(?p chat ran Y . (2) -* (3): N*"' +3e > NO nenhSn=3nNo =0,12. Phan tram the tich cua khi CI2 trong X la Bao toan electron: 3a + 0,0"6 = 1,34 - 7a + 0,12 -> a = 0,14 mol. . ;, A. 80%. B.40%. C.50%. , . D . 60%. Dap an D . '.f. , Laigidi: - •', Cach 2: Q u i doi Y thanh: Fe (a mol); C u (0,03 mol) va O (b mol). ,„ lit,; n^,= M = a 4 m o l ; n M , = ^ = 0 , l m o l . X gnyb .ig :f d . i 5 Bao toan khoi lugng: 56a + 16b = 12,64-0,03.64 = 10,72. ysb n 27 ^ 24 M&ii) Igol^M m mifl:4 y i[ mil Nhan xet: Day la bai toan hai chat k h u (2 kim loai) va hai chat oxi hoa (2 phi Bao toan electron: 3a + 2.a03 = 2b + ai2 ^ a = ai4; b = ai8 \ ]. (w^ - , kim) nen can ap dung dinh luat bao toan electron. -^DapanD. .;„,, i i O ftoi v M l ' o ^ ' i ' : Chat k h u : Al - 3e > Al*^ M g - 2e > Mg*^ 5 PHl/ONGPHAPTRUNGHOADIEN ;gnu nerfq 6 u ' > a . Mol: 0,4-^1,2 M o l : 0 , 1 - > 0,2 * f'^^'^ a. N p i d u n g Dung dich cac chat dien li luon luon trung hoa ve di^n. Q si*: q&CJ < Chat oxi hoa: O2 + 4e > ICf^ CI2 + 2e > lOr b. B i e u t h i i c -'••^h ^.-.-fi .(OMI) f Mol: X 4x M o l : y —> 2y Tong so'mol di^n tich duong = Tong so mol di^n tich am: 5^! ( ;v. Bao toan electron: 4x + 2y = 1,2 + 0,2 = 1,4 mol. Bao toan kho'i luong: mx + Tn^i + m j ^ g = my Lieu y: De tinh so'mol di?n tich, cac em la'y so mol ion x di^n tich ion do. y t -> 32x + 71y + 10,8 + 2,4 = 40,9 -> 32x + 71y = 27,7. -> X= 0,2 mol; y = 0,3 mol VIDVMAU ImLrx. • Q3 iff,: ^ %Vn = ' .100% = 60% -> Dap an D . 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 m/ CI2 0,2 + 0,3 ^ S04~ (y m o l ) . C o c^n dung djch X thu dugc 46,9 gam chat ran khan Y. Gia 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 tri cua X va y Ian lugt la v • nr,^ipf\n\~n f:, thoi gian, thu dugc 12,64 gam chat ran Y . Hoa tan hoan toan Y bang dung A. 0,2 v a 0,3 . B. 0,1 v a 0,2. C . 0,4 v a 0,4. D . 0,3 v a 0,3. > dich HNO3 loang (du), thu dugc 0,896 lit khi N O (san pham k h u duy nhat 6 Laigidi: 3 orbs 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 A. 0,08. B.0,10. C.0,12. D . 0,14. r mol, do vay can lap dugc 2 phuong trinh dai so de tim so mol cua chung. Laigidi: + A p dyng dinh luat trung hoa di?n: nNO= ^ ; r — = 0,04 mol. Tong so mol di^n tich duong = Tong so mol di^n tich am '^^ 22,4 ; -> 0,1.2 + 0,2.3 = x.l + y.2 ^ X + 2y = 0,8 mol. n €0,0 H Bao toan kho'i lugng: TTIQ^ = m y - mx = 12,64 - 56a -1,92 = 10,72 - 56a. + A p dung bao toan khoi lugng: my = m p g 2 + + ^^3+ "^ci" ""^ "^sol" ' Nhan xet: Neu d y a theo phuong trinh phan ung se rat dai va kho giai. Cach 1: 56.0,1 + 27.0,2 + 35,5x + 96y = 46,9 35,5x + 96y = 35,9. 6 day, cac em can s u diing so do phan ung: ^ X = 0,2 mol; y = 0,3 mol. . ^, , ; , Dap an A. -'"JOb + > Fe, C u (1) — Y (2) """^^^ ) Fe^^ Cu^^ (3) 33
dm nang On luygn thi dgi hgc 18 chuy6n 66 H6a hgc - Nguygn Van Hai Cty TNHH MTV DVVH Khang Vigt 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 Tirdo—> b = 0,04 m o l . , >,r l u ^ t t r u n g hoa d i e n tich.:t.'e'-f ,ji*i„trt,»sj:fr '^4f^- :>''J*!;/i .J.K ..•:i.«jSJi':'iie|' 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 : , 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 — ^ CO^" + CO2 + H2O. (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 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 ••• A. O H - va 0,03. B. Ch va 0,01. C. COs^- va 0,03. D . N O ^ v a 0,03. m o l ) ; C O 3 ' (0,015 m o l ) ; C I - = 0,05 m o l . *^ • ' ' ' Lai gidi: 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. Goi dien tich ion X la - n . "• < f J>0 + eC rrfor; ' ., ,(' 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 + A p d y n g d i n h luat t r u n g hoa di?n: ' ' ' ' 't;'^ 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 1.0,01 + 2.0,02 = 1.0,02 + n.a n.a = 0,03. fJom «> irtmrU Y 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 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 ion N O 3? 2a+ b a+b ^ , , _ a + 2b ^ ^ _ i ± b ' A V = ^ ^ . B.V^ — . C.V=^ ^ . D . V= X 2x + 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 Lin gidi: • "mtlJl, daudocophanung: , ..i-.,-' A6i-U)MU*I-. + A p d u n g d i n h l u a t t r u n g hoa dien: 2a + 2b = c. '~ ' OH- + HCO; > C 0 ^ - + H20. '-IdU'dfi + 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: —> D a p an D . jt^n ife graiQ Ca(OH)2 + Ca(HC03)2 > 2CaC03 + 2 H 2 O U,,J,,W... 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 , v a C I " (0,1 m o l ) . Mol: a MgCOs +CaC03 + 2 H 2 0 ^ ' " •' 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 , Mol: b CO,^' + H 2 O . 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 Mol: Ca2> + C O ^ - 0,03 CaCOs „ >^;TOMb rbil - > c = 0,03 mol. 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. f, K h i N a O H d u vao 1/2 d u n g d i c h X: OH- + HCO 3" > CO 3" + H2O = 39.0,01 + 56.0,02 + 62.0,04 + 96.0,015 = 5,43 gam. .,.^j, ^ Mol: 0,03 -> 0,03 - » D a p an A . Ca2* + C O 3 " > CaCa 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 m o l cua goc sunfat. " t,Oi:
Cty TNHH MTV DWH Khang Vi$t Ca'm nang On luyjn thi dgi hgc 18 chuySn dS H6a hpc - Nguygn Van Hi'i 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 SO 4 " . Cho dung dich Ba(OH)2 du vao dung dich X thu dugc ket tiia Y. -> n^j_ = 0,01 mol. ,'^f^•v:iJiJr^;,'^^ii:... ^ ^':^^Otti--> Nung Y a nhi^t dg cao ngoai khong khi den khol lugng khong doi thu dugc + Bao toan khoi lugng: - ^, m gam chat tin Z (coi BaS04 khong hi nhi?t phan). Gia tri cua m la Khoi lugng cac chat tan trong Y = ^^^^2+ + ' " N H ^ "^cr "^sol" ' •'• ' A. 19,91. B. 18,11. C. 24,31. D. 20,31. = 2.(0,01.24 + 0,03.18 + 0,01.35,5 + 0,02.96) = 6,11 gam Lai nidi: * _> Dap an B ^ Ap dung djnh luat trung hoa dien: 2^ n^, - 2^ nj, 2. 0,05 +1.0,07 = 0,03.1 + 2n ^ n , . = 0,07mol. Lim y: Bai nay cac em de chgn nham dap an A (3,055 gam) do chi tinh khoi lugng chat tan trong mgt nua dung dich Y. , , , , Cac so do phan ung: Vi 8: Dung dich Y c6 chua dong thoi cac ion: Ba^*; Ca^*; CI" (0,01 mol) va SO 4 " """'^ ) BaS04 > BaS04 y NO J (0,03 mol). Cho V ml dung djch Na2C03 I M vao Y de thu dugc ket tiia Mol: . 0,07 , ° •' 0,07 0,07 ^ j,, Ion nhat. Gia trj nho nhat ciia V la A. 30. B.20. C.40. D. 70. Fe^- )Fe(OH)2 > F e ( O H ) 3 1 Fe203 LaigidU iftAl Mol: 0,025 •• • 0,0125 --^ Ggi so'mol ciia cac ion : Ba^* = X ; Ca^* = y. -M.J'jj \t Vay m = 233.0,07 + 160.0,025 = 20,31 gam. Ap dyng djnh luat trung hoa di?n: 2x + 2y = 0,01.1+ 0,03.1 = 0,04 mol. —>• Dap an D. • • Cac phuong trinh phan ung : . jii 0 Vi 7: Chia dung dich Y chua cac ion: Mg2% N H J , SO 4", CI" thanh hai phan Ba2* + CO3" > BaCOsi H ^ ^ - , . . . BV p : ^ ; - : bang nhau. Ca- + COr CaC03l ^' + Phan 1 cho tac dung voi dung dich NaOH du, dun nong, thu dugc 0,58 gam : Xn^^^2- = X + y = 0,02 mol. , ket hia va 0,672 lit khi (dktc). .y 3 + Phan 2 tac dyng voi dung djch BaCl2 du, thu dugc 4,66 gam ket tiia. nNa2C03 = 0,02 mol V= = 0,02 lit = 20 ml - > Dap an B . , Tong kho'i lugng cac chat tan trong Y la Vi dv 9 (A-10): Dung dich X c6 chua: Na^ (0,07 mol); SO^" (0,02 mol) va O H " (x A. 3,055 gam. B. 6,11 gam. C. 5,35 gam. D. 7,05 gam. ' Lot gtat: mol). Dung djch Y c6 chua CIO 4, NO 3 va H"" (y mol); tong so mol CIO4 NMn xet: Bai nay cac em giai dua theo cac phuong trinh ion. va NO3 la 0,04. Trgn X va Y dugc 100ml dung dich Z. Gia tri pH ciia Z la o Phan 1 + dung djch NaOH du: jj A. 1. B. 12. C. 13. D.2. Mg2^ + 2 0 H " > Mg(OH)2i o'ijhqil Votigidi: ., , NH^ +OH- >NH3t+H20 ^ + Ap dung dinh luat trung hoa dif n vai dung djch X: , - > % 2 . = nMg(OH),= ^ = a 0 1 m o l . 1.0,07 = 2.0,02+l.n _ -» n _ = 0,03 mol. " + 3.,' • OH OH i ? -tit v a n ^ . = n N H 3 = - r r - = 0,03 mol. -A •*• Ap dung djnh luat trung hoa dien vai dung dich Y: ,^ ( i n i l ) ; 4 '^"^ 22,4 Phan 2 + dung djch BaCh du: ^•"ciOi -^l-^NOS =^-V V =0'04mol. • ^ Trgn X vai Y: H* + O H - > H2O Ba^^ + sol" , BaS04; f ' -Aq«Q.^-' -> H^du = 0,01 mol -» IH^]= ^ = 0,1 = 10"^ ->.pH = l ^ 'h '"^nt^h 'qsi^^^IKi! -yin ••{in itff ->;; : ~^ "Ba2+ = "BaS04 = ^ = 0/02 mol. jf;|„yi j ^ : . , ,j - > Dap an A. rf.^ o: - 37
CtyTI\ih.i 1TV DWH Khang Vi^t Ca'm nang 6n luygn thi dgi hgc 18 chuy§n dg H6a hgc - Nguygn van H&\ 6 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 V6i hSrhgp bat ki g6m hai chat X va Y , khi biet dugc gia tri khoi lugng mol tri cua x va y Ian lugt la trung b m h ( M ) t a s e c 6 t i l e : , A. 0,02 va 0,05. B. 0,05 va 0,01. C . 0,01 va 0,03. D . 0,03 v a 0,02. M - M Y AMN i »! ^ > I ) LOT gidi: M-Mx .., ,, ,>.'i, • • ny • O bai nay, dung dich Z chiia 4 loai ion, trong do 2 loai ion chua biet so'mol, do vay can lap dugc 2 phuang trinh dai so de tim so' mol cua chung. b. So do hoa X: n> lAMv + A p dung dinh luat trung hoa dien: ^t. 1=0 AM, 0,02.2 + 0,03.1 = x . l + y.2 X + 2y = 0,07 mol. " ''^ n, W B a o t o a n k h o i lugng: m z = m ^ 2 . + m . + m^,. + m ' N ^ a n u Q :8 i#b >; Y: M, B i l l ;H^i,^>vvi. ^" ^ ^' - K i l o f n ' e O i O ) 'rOVi . ^ 64.0,02 + 39.0,03 + 35,5x + 96y = 5,435 S,, v i o y MAU 35,5x + 96y = 2,985 ^ x = 0,03 mol; y = 0,02 mol. , 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 —> D a p an D . 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 V i dvi 11 (CD-08): Chia dung d k h X chiia cac ion: Fe^^ S O | ~ , N H 4 , N O 3 thanh hai phan bang nhau. "t* , ' f • ,-,4 v ' c ! * . 1 ; ^ ,r|/ tong so nguyen t u ciia dong vj 2 9 C u la A. 27%. B. 50%. C . 54%. ' D . 73%. n : - 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; , ,. ;„)-..i^^:;; • :';©aji-4):;::.,3-plIiJ's Lot gidi: " - 'o - Phan 2 tac dung voi dung djch B a C h du, thu dugc 4,66 gam ket tua. ...^ 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:' Co can X thu dugc m gam muo'i khan. Gia tri cua m la . A . 3,52 gam. B. 7,04 gam. C . 8,52 gam. D . 4,26 gam. 65 - 63,54 1,46 73 » % § C u = 73%. a Lai gidi: I 63-63,54 0,54 27 Nhan xet: bai nay cac em giai dua theo cac phuong trinh ion. 1*' Dap an D . 63 + 65 dong v i 29 C u chiem u u the hon . Phan 1: + Bung dich N a O H d u : , .'rin»*Wi t. Cach 2: N h a n thay A c u = 63,54 < Fe3* + 3 0 H ' > Fe(OH)3l , t ' C ^ . • 6':> V fti\b nnw - > % 29 C u > 50% D a p an D (Cac dap an khac deu < 50%). ' N H ; +OH* >NH3T+H20 1'07 , 0,672 , 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 -> "Fe3^ = " M O H ) 3 = = ^'^^ "^"l- " N H ^ = ''NHg= ^ = 0'03 mol. Phan 2: + dung dich B a C h du: Ba^* + S O 4 " > BaSOU i.: hgp khi X c6 ti khoi so voi hidro bang 20. Cong thiic ciia oxit sat v a phan jA • 4 66 tram the tich ciia khi C O 2 trong X la A.FeO;75%. B. FezOs; 75%.' C . FeaOs; 65%. D.Fe304;75%. gidi: Lai Ap dung dinh luat trung hoa di?n: 0,01.3 + 0,03.1 = 0,02.2 + n^^^. .1 ^ 4 48 Theo bai: M = 20.2 = 40 v a nx = n c o = T T T = 0'2 m o l . . . , , , -> n _ = 0,02 mol. *' ' A p dving cong thiic ciia phuong phap duang cheo, ta c6: ' '' + Bao toan khoi lugng: m = m^ 3+ + + "^j>jo~ ""^ "^so^' 44 -40 1 0,05 chgn B hoac D . = 2.(0,01.56 + 0,03.18 + 0,02.62 + 0,02.96) = 8,52 gam "CO = %Vco2 =75% 28 - 4 0 3 0,15 nC02 Dap an C . , J:^.Asxu'^'^^^'^^^: i^^^y • 39