Tài liệu Cẩm nang ôn luyện thi đại học 18 chuyên đề hóa học 2

ca'm nang 6n luyjn thi dgi hgc 18 chuyfin H(Sa hpc - NguySn Van H3i Cty TNHH MTV D W H Khang Vigt Lot gidi: Cach 1: Khi hoa tan vao dung dich H 2 S O 4 , Fe(N03)2 se phan li thanh cac ion. Do vay, truoc het cac em can tinh so mol cac ion nhu sau: O' • ' 0,2 •••^'J^/•^..-:> <>^:r'^ ^i^mrp ^''^''/'i CO the hoa tan toi da m gam Fe. Biet trong cac qua trinh tren, san pham khu A. 5,6. -> Mol; 0,6 ^ 0,8 ^ 0,3 0,2 0,6 Mol: rr,fj{),p« y 4 - 0,1 ^ ^, ^, , vr ^ Mol: Mol: Dap an B. Vi d^ 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4 0,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch -I X. Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ke't tua Ian nhat. . Gia trj nho nhat ciia V la A. 80. B.50. C.60. .v, t, • 5 njs^aOH = MjoriX^:' n + + 2n H y „ " WO..'-* "'"^ * mi 2+ + 4H* + N O ; > Fe3* + N O + 2H2O QkHi - > 0,05 ••idl,0» j „ r t S « . ^ ^ i T -.'ihtvU/ > 3Fe2q % lOH dfegfror* ,;>fefii JfiSv) Vi d^ 15: Iron 200ml dung djch X gom Ba(OH)2 0,1M va NaOH 0,3M voi 100ml dung dich Y gom Al2(S04)3 0,1M va H2SO4 0,1M, thu dugc a gam ke't tua. Gia tri cua a la B. 4,66. C. 5,82. D. 5,24. Trong Y: n 3+ = 0,02 mol; n ^ ^ j - = 0,04 mol; n^+ = 0,02 mol. + OH' 0,02 ^ > 0,02 AP* + 3 0 H " 'ol: 0,02 ± Cu 0,02 > Al(OH)3i 0,06 <- -> .^w oh 0,02 > AlO; 0,02 ,Ba2^ + S04~ ; H2O , Al(OH)3 + O H " / _> V = 0,08 lit = 80ml * ^ TrongX: n„ 2+= 0,02 mol; n,,_+ = 0,06 mol; n O H _ = 0,10 m o l . « ^ ' ^ - > ^^" = 0,04 + 2.0,06 = 0,16 mol — Dap an A. > , —> mpe = 0,125.56 = 7,0 gam - > Dap an B. Mol: > 3Cu^*+ 2NO + H 2 O ;^Mol: a06 ai6 a04 Cu tan het -> X C O chiia: n^ 2+ = 0,06; n + du = 0,04. "NaOH ., 0,05 4-0,4 Fe + 2Fe3^ 0,075 <- 0,15 H+ Phuong trinh ion thu gpn: De thu dugc luang ke't tua Idn nhat thi: 0,2 Cac phuong trinh phan ung khi pha trpn: ncu =0,06 mol; n^+ = 2.0,2.0,5 = 0,20 mol; n ^ ^ , =0,2.0,2 = 0,04 mol. 3Cu + 8H* + 2NO3- 0,1 Laigidi: Nhan xet: Dung dich chiia muoi NaNOs va H2SO4 loang -> can giai theo ;> ? 0,8 A. 6,22. D. 40. Ldi gidi: phuong trinh ion. , Fe mol V = 0,4.22,4 = 8,96 lit ^ -> Fe(N03)3 + 5NO + 2H2SO4 + 2H2O Fe3^ = 0,1 mol; H* = 0,4 mol; N O ; = 0,3 mol va S O ]' = 0,2 mol. Bao toan electron: 2ncu + lnp^2+ = Sn^o =0,4 ry, •; Cac phan ung hoa tan Fe: Chat oxi hoa: N*-^ + 3e - > N O z Lai gidi: Dung dich X gom cac ion: 0,2 ^ ^ _ Cach 2: Cac chat khu: Cu - 2e ->• Cu^^; Fe^^ - le -> Fe^*. 2.a3+a6 • FeS2 + 8HNO3 - > 3¥e^ + N O + 2H2O 0,2 D. 2,8 C.8,4.* phan ling hoa hoc: cO/; , -> V = (0,2 + 0,2).22,4 = 8,96 lit -> Dap an B. riNO= B.7,0. ' 'V', Cac em luu y, Fe^* cung bi oxi hoa thanh Fe^* va giai phong khi N O : 3Fe2* + 4H* + N O ; ' > 3Cu2- + ' ^ N O + 4H2O 3Cu + 8H* + 2NO^ • a3 <- 0,8 san pham thu dugc gom dung dich X va mot chat khi thoat ra. Dung dich X duy nhat cua N*^ deu la NO. Gia tri ciia m la Phuong trinh ion rut gon: Mol: 14: Hoa tan hoan toan 0,1 mol FeS2 trong 200ml dung dich HNO3 4M, 0,02 -> 0,02 > BaS04>l ^ 0,02 Vgy: a = 0,02.233 = 4,66 gam - > Dap an B. + 2H2O Cty T N H H M T V D V V H Khang Vi$t C&m nang fln l u y j n thi dgi hgc 18 chuy6n dg H6a hoc - MguySn Van H5i V i d v 16: H o a tan hoan toan 9,46 g a m h o n h g p g o m N a , K v a Ba vao nuoc, thu Al duQC d u n g d i c h X va 1,792 l i t k h i H2 (dktc). D u n g d i c h Y g o m H C l I M va AIO2 + - H 2 + O H - + H2O H2SO4 0,5M. T r u n g hoa d u n g dich X b o i d u n g dich Y, tong k h o i l u g n g cae Mol: m u o l d u o c tao ra la 1 ^ Ta c6: n H j = 2,5a = 0,05 m o l - > a = 0,02 m o l . , A . 14,72 gam. B. 16,14 gam. C. 19,98 gam. i D . 17,14 gam. '2 = 0,08 m o l . 22,4 H . ( r{ mi'n> tV K i 1 -H2 Na+ + O H " + + H2O K+ Ba + 2H2O i. K h i cho t u t u Y vao X: OH- + Iv 2 + OH- + -Hi 1^x4 f1 J 1, t Mol: 0,01 „,p.r, f t T < j\riv, = 2a m o l ; n , so|" T r u n g hoa X b o i Y: H + + O H " <- > Al3+ + 3H2O dnSrf/; 0,03 0,01 <- 0,01 Mol: > BaS04>l' -> 0,01 - > m = mA,(OH)3 + mBaS04 = 0-01-78+ a01.233 = 3,11 g a m . : *. " H + = " H C l + 2nH2S04 = 2a + 2.a = 4a m o l cr > A l ( O H ) 3 4' 0,02 Ba^+ + SO 4- ^ n - > 0,02 Al(OH)3 + 3 H + 1 M a t khac, n o n g d o H C l gap d o i H2SO4 - > t r o n g cung m o t the tich t h i ^ T r o n g Y: • ;;i • H+ - 0,02 -> 0,02 Mol: fin J^j ti ) N/jflnxet: n ^ „ . = 2nH, =0,16 m o l . , 1,5a . < AIO2 + H + + H2O r.i I > Ba^^ + 2 0 H " + H2 "M ' OH 0,02 Mol: P + H2O a T r o n g 100ml Y: n ^ + = 0,07 m o l ; n g Q 2 - = O'Ol m o l ; n^|_ = 0,05 m o l . Cac p h a n u n g hoa hoc: Na a T r o n g X: n^^^_ = 0,02 m o l ; n^^2^ = 0,02 m o l ; n^,Q_ = 0,02 m o l Lai giai: 1,792 a 2 ->DapanC. , , , , .m V , , , ' =amol. 9. P H l / O N G P H A P L I E N H E N G U Y E N T 6 - N H 6 M C H L T C • H2O a. Npi dung ' - > 4a = 0,16 - > a = 0,04 m o l . So m o l cac n g u y e n t u c6 t r o n g n h o m chiic l u o n t i 1^ thu|in v o i so' m o l n h o m K h o i l u g n g m u o i t h u d u g c = 9,46 + m^^. + mg^a- chuc. V = 9,46 + a08.35,5 + 0,04.96 = 16,14 gam. ; b. Cac t n r o n g hg(p t h u a n g gap —> D a p a n B. ROH d u n g djch X v a 1,12 l i t k h i H2 (dktc). D u n g dich Y g o m H C l 0 , 5 M va H2SO4 RCOOH 0,1M. C h o t u t u d e n het 100ml d u n g dich Y vao X, t h u d u g c m g a m ket tiia. Giatricuamla ^ - a - . , . i* ^ ' . , • • - , - a^r* • A . 0,78 gam. B. 1,56 gam. " H 2 = 0,05 m o l . Cac p h a n u n g hoa hoc: Mol: ^9 a ,.WUt, C. 3,11 gam.' Laigidi: ^.•vr., Ba + 2H2O • ^OJ) 2a a •* S 2 ^NaHCOg ^ R C O O N a + R_NH2 ., Vi 'fd Cach t i n h no=2nH2 > R O N a + ^ H2 CO2 + H2O ' D . 5,44. J - ' ^ ; .X^CMU. > Ba2* + 2 0 H " + H2 a • - i ^ ; •: : • - r'h-Hn M o i lien h | Phan l i n g hoa hpc V i d\ 17: C h o h o n h o p g o m Ba va A l ( t i le m o l 1:1) v a o n u a c (du), t h u dugc , , torn no=2ncooH n o = 2nco2 ^ ) R-NH3CI n N = riHCi Dv M A U V i d v 1 (A-09): K h i d o t chay hoan toan m g a m h o n h g p X g o m h a i ancol no, d o n chuc, mach h o t h u d u g c V l i t k h i CO2 (6 dktc) v a a g a m H2O. Bieu thu-c lien h? g i i i a m , a v a V la: dm nang On luygn thi dji hpc 18 chuy6n V A. m = a- — . 5,6 C . : m=2 a - ^ . 22,4 Cty TNHH MTV DWH Khang Vigt H6a hpc - IMguySn Van HSi f;,: v'X,- . , ' V B. m = 2a 11,2 . So n g u y e n t u C t r u n g b i n h = "'''"^ = 2,5 ' ; 'y.-} D. m = a + - ^ . '• ^^^^^ ,,,) 5,6 ^J,, • ^ - . n : ; j . | i T , < so n g u y e n t u C n h o h o n 2 , 5 a n c o l d o la C 2 H 4 ( O H ) 2 . "ancol - n H 2 0 - "CO2 a nancol= — - 18 D o X chua cac ancol d a n chiic - . . . . . ., , ,.^^5^ 22,4 no = noH = nx -> no Bao toan kho'i lucmg: mx = m c + mH + m o , a , a ->m=12. + 2 . — + 16.( 22,4 18 ^8 V -> D a p an A . , , ) - > m=a 22,4^ . oM < — • '^'Ht .r^j,/ -^"^^ ' fCf-if:)^rA V i d u 2: D o t chay hoan toan m g a m h o n h o p Y g o m ba ancol d o n chuc, thuoc c u n g d a y d o n g d i n g , t h u d u g c 37,4 g a m k h i C O 2 (dktc) v a 27 g a m H 2 O . Gia triciiamla P'''rfm A . 27,1. l iO,U B. 28,6. C.23,6. J ,3m + # ' n H = 3,0 m o l . -> nv = n H - o - n c o , = 1'5 - 0,85 = 0,65 m o l . , B. V i = V 2 - 2 2 , 4 a . ; C. V i = V 2 + 22,4a. D . V i = 2 V 2 - 11,2a. ,/ i, OH ; Lbigidi: V2 22,4 Bao toan n g u y e n to O: no (on) + 2 n o 2 = 2 nQQ^ + -> 2a - 2V2 ^ 2 V i _ 2V2 + ^=-^ = ^ - ^ + a 22,4 22,4 22,4 Vi = 2^ 2V2 22,4 VXH^Q 2V2 - 11,2a on nancoi = n^^Q " "CO2 ' V i d ^ 5 (B-12): Do't chay hoan toan m g a m h o n h g p X g o m h a i ancol, t h u dugc 13,44 l i t k h i C O 2 (dktc) va 15,3 g a m H 2 O . M a t khac, cho m g a m X tac d y n g - > D a p an C. V i d y 3: Do't chay hoan toan m o t l u g n g h o n h o p X g o m h a i ancol (no, d a chuc, m a c h h o , c i i n g so n h o m - O H ) can v u a d i i V l i t k h i O 2 , t h u dugc 5,6 l i t k h i C O 2 v a 6,3 g a m H 2 O (cac the tich k h i d o 0 dktc). Gia t r j ciia V la at B.3,92. C. 7,28. 63 ' r ' mol; n H 2 0 = 0 , 3 5 mol. • ''^''^.ymm-Mtifbw ; , . v o i N a (du), t h u d u g c 4,48 l i t k h i H 2 (dktc). Gia t r j cua m la A. 12,9. B. 15,3. .•v^ . C.12,3. Lbigidi: ^ • , • • - ( i =c ~J ui r*3Y^;^:- "^ol • , ^ ,a,Rt; q? i • D v a t r e n m o i q u a n h | n g u y e n to' - n h o m chuc t h i v o i ancol: no = noH n o = n o H = 2 n H 2 = 0,4mol. Meit khac: n c = n c o 2 = 0,6 m o l ; n H = 2nH20= 1/7 m o l . Theo b a i ra, X c h u a 2 ancol n o -> nx = n ^ j o " "CO2 - D . 16,9. ^^002 = 0/6 m o l ; n H 2 0 = 0,85 m o l ; nH2 = 0 , 2 m o l . D . 1,12. Laigidi: ^C02 A. V i = 2 V 2 + l l , 2 a . d o n g t h a i bie't ap d u n g bao toan n g u y e n t o oxi. = 23,6 gam. yi: V i , V 2 , a la Nhan xet: Bai n a y cac e m can n h o v o i ancol n o t h i : Bao t o a n k h o i l u g n g : mv = mc + mH + m o = 0,85.12 + 3,0.1 + 0,65.16 5,6 Cac k h i d e u d o 6 d i e u kien tieu chuan. Bieu t h u c l i e n he g i i i a cac gia t r i -> D a p an D . .-^ D o Y chua cac ancol d o n chuc -> n o = noH = nx -> no = 0,65 m o l . A. 5,60.' C. Do X chua 2 ancol hai chuc -> no = noH = 2nx -> no = 2a - ^ chtia 3 ancol no , = 0,325.22,4 = 7,28 l i t . V i d v 4 (CD-12): D o t chay hoan toan h o n hg-p X g o m hai ancol (no, h a i chuc, ^^^^^ Cac e m can thay rang, k h i d o t chay Y: n n j o > " c O a ' ' ' ' ^ ' ' ' ^ ' Theo bai ra, X chua 2 ancol no -> nx = n H 2 0 ' ^C02 ~ ^ ' D . 37,1.»- m - ' WtyfAiHi>HOXH-M v ' '^ ' 27 ^• ' , . ^2.0,25 + 0 , 3 5 - 0 , 1 ^Q 225 m o l ^ aoM i 37 4 " C O T ^ — ^ = 0,85 m o l - > n c = 0,85 m o l . 44 n H 2 0 = — = 1.5 m o l i '* mach h o ) can v u a du V i l i t k h i O 2 , t h u dugc V 2 l i t k h i C O 2 va a m o l H 2 O . cfi n ^^^'^ 5,6 ^ D a p an 22,4 ^ V V Q s ^ * •* "••'''A 18 no = noH = 2nx -> no = 0,2 m o l . Bao toan n g u y e n to O: no(OH) + 2 n o 2 = 2 n c o j + nH20 V • p o X chiia 2 ancol c u n g so'nhom - O H -> 2 ancol deu hai chuc: ^ Nhan xet: K h i do't chay cac ancol (no, d o n chuc, m^ch ho), ta l u o n c6: X chua m g t ancol d a chuc c6 Cty TNHH MTV DWH Khang Vijt' Ca'm nang 6n luy^n thi d K h i d o t A. 3,36. chay 1 m o l X, t h u d u g c 3 m o l H 2 O -> nx = • ^ r i H 2 0 = 0,04 m o l . „ B. 11,20. C.5,60..y.g; • . nco2 = 0'3mol ^ Lmgidi: D. 6,72. • 0'mti^Qj nc=a3mol. N h a n thay t r o n g X cac ancol deu c6 so'cacbon = s o ' n h o m c h i i c l "CllVjt * ijj ( nc = noH = 0,3 m o l . ,r p M a t khac, k h i cho X tac d u n g v o i Na: n o H = ^^H2 ~* ' ^ H 2 ~ O'^^ mol. s; - > V = 0,15.22,4 = 3,36 l i t ^ D a p an A . V i d u 7: H o n h o p X g o m h a i axit cacboxylic d o n chuc. D o t chay h o a n toan 0,1 m o l X can 0,24 m o l O 2 , t h u dugc C O 2 va 0,2 m o l H 2 O . C o n g t h u c h a i axit la A. H C O O H va C H 3 C O O H . i i o , rm • -f i C. C H 3 C O O H v a C 2 H 5 C O O H . , * 3iwb fefl loarifi ; dG N h | n thay X c h u a 2 axit cacboxylic d o n chiic -> chua 2 n g u y e n t u o x i ^ ,.;! > mx = m c + m H + m o -> mc = 3,08-0,24.1 -0,08.16 = 1,56gam. ->• n c o 2 = '^c = 0,13 m o l . ' - ' > • N h a n thay, k h i d o t chay 1 m o l m o i chat m e t y l axetat v a etyl f o m a t d e u t h u dugc n c o 2 = n H 2 0 ' " e n g v o i v i n y l axetat t h i : n^^^ - n^^o = 1D o v^y: nvinyi axetat = % nvinyi axetat r\QQ^ - n H 2 0 v /.. 0,13 - 0,12 = 0,01. = ^ ^ . 1 0 0 % = 2 5 % - ¥ D a p an D . ' , " ,bfniJ3»,,,A V i d\ 9 (A-11): H o n h g p X g o m axit axetic, axit fomic v a axit oxalic. K h i cho m g a m X tac d u n g v o i N a H C O s ( d u ) t h i t h u d u g c 15,68 l i t k h i C O 2 (dktc). M a t khac, d o t chay h o a n t o a n m g a m X can 8,96 l i t k h i O 2 (dktc), t h u d u g c 35,2 gam C O 2 va v m o l A. 0,2. H2O. Gia t r i cua v la B.0,3. C.0,6. D.0,8. no = 2ncooH = 2nx —> no = 0,2 m o l . Bao toan n g u y e n to O: ^ no (COOH) + 2 n o , = 2 n c o , + ^ ^ -> 0,2 + 2.0,24 = 2 n c o 2 + 0,2 n c o j = 0/24 m o l . + So' n g u y e n t u H t r u n g b i n h = —S2.= ^-^'^ = 4 "X "HOO ^ ,, ,;,fj fin qP;Uv.4™- • rtljiW _> Loai B v a C ( v i cac axit 0,1 So' n g u y e n t u C t r u n g b i n h = = - 5 ^ = 2,4 -> L o a i A ( v i cac axit chua 0,1 so n g u y e n t u C < 2). - > D a p an D . ^ '••m'r\fl•„,,•• V i d\ 8 (B-11): H o n h g p X g o m v i n y l axetat, m e t y l axetat v a etyl fomat. D o t chay hoan toan 3,08 g a m X, t h u d u g c 2,16 g a m H 2 O . Phan t r a m so m o l ciia v i n y l axetat t r o n g X la A. 75%. B. 72,08%. N/ion xet: K h i cho axit cacboxylic tac d u n g v o i N a H C O a ta l u o n c6: "C02=r»C00H -> - > no(X) = 2ncooH = 2nco2 " « ^ r , y ' "^ • ^ ^ no(X) = 2. —'-— = 2.0,7= 1,4 m o l . 22,4 Bao toan n g u y e n to O: no(X) + 2 no2 = 2 nco2 '^H20 , chua so'nguyen t u H > 4. + no = 2nx = 0,08 m o l . Bao toan kho'i l u g n g cac n g u y e n to' t r o n g h o n h g p X, ta c6: -> B. C H 2 = C H C O O H v a C 2 H 5 C O O H . D. C H 3 C O O H va CH2=CHCOOH. Cac chat t r o n g X d e u chua 2 n g u y e n t u O C. 27,92%. D . 25%. ^ 1 4 +2 ^ =2.^ ' '22,4 44 + nH,o ^ ''•••ft ^ 2,2 = l , 6 + y y = 0,6mol. nang 6n luygn thi d?! hgc 18 chuy6n d j H6a hpc - IMguygn Van Hai dm Cty TNHH MTV D W H Khang Vi^t L o t gidi: ivl^t khac, cac em l u u y dya q u a n h? nguyen to - nhom chuc: Nhan xet: Khi cho axit cacboxylic tac dyng voi NaHCOa ta luon c6: nc02= "COOH J ' ' j ,3 no(X)=2ncoOHno(x)=0,2mol. • n N ( X ) = 0,06 1 344 no(x) = 2ncooH = 2 ticoj i'r'f, • Bao toan nguyen to O: 0,12 + 2. no(x) = 2. no (x) + 2 HN = = 2.0,06= 0,12 mol. 22,4 mol. "HCl = "NH2 = " H C l ^ 0,03 mol . ^ ^ Bao toan khoi lugng: ncoj + riHjO =2 <•:'•: m = mx+ m H c i = 7,66 + 0,06.36,5 = 9,85 gam. 2,016 ^ 2 .4.84 + "Hjo ^ 22,4 44 "Hjo = 0/08 mol. r , ,• OrM^' ^ " " , _> Dap an C. Vi mH20= 0,08.18 = 1,44 gam 13: Dun nong m gam hSn hgp X gom cac este voi 350ml dung dich NaOH 2M, thu dugc dung dich Y chiia muol cua mgt axit cacboxylic don chuc va —> Dap an B. Vi d y 11: Cho m gam hon hg-p X gom hai ancol tac dung v6i Na (du), thu duoc 13,9 gam hon hgp ancol Z. Cho Z tac dyng voi Na du, thu dugc 4,48 lit khi H2 4,48 lit khi H 2 . Mat khac, dot chay hoan toan m gam X, thu dugc 13,44 lit khi (dktc). Co can Y, nung nong chat ran thu dugc voi CaO cho deh khi phan ling C O 2 va 16,2 gam H 2 O . Cac the tich do 6 dktc. Gia tri ciia m la xay ra hoan toan, thu dugc 4,8 gam mgt chat khi. Gia tri cua m la A. 14,5. ) A. 40,6. B. 15,4. ( nH2 = 0,2 I mol. • •' C.12,2. B.26,6. • DHJO^O'^ ->nH=l,8 • . .,1 4 Y n ' - a w N;'"^-" # ' r i ..*rr * M > „ i , v Dya tren moi quan h^ nguyen to-nhom chiic thi voi ancol: no (X) = noH -> no(X)= noH" 2nH2 = 0,4 mol. • ! ? m-nmi mod •«» C. 30,7. Nhan xet: Cac este tao thanh tu cung mgt axit cacboxylic don chuc. Khi ancol tac dung voi Na: «;i,ib fob ,xtM A,U,A m = 0,6.12 + 1,8.1 + 0,4.16 = 15,4 gam -> Dap an B. .\i .life = 0,7 mol; n H2 = 0,2 mol. - O H + Na -> - O N a + - H 2 Bao toan kho'i lugng: m = m(- + m H + mo ^ D. 34,5. Loigidi: rAv>r^ itT,'.' ".sn-.' Ldfigiai: , " ' nco2 = 0,6->nc=0,6; D. 13,8. Mol: 0,4 iMv^jndaEv fa; 0,2 <- noH=2nH2 =0,4 mol. • M Vi d y 12: Hon hgp X gom 2 amino axit no (chi c6 nhom chiic - C O O H va - M|it khac: n.coo-= " O H = 0,4 mol "RCOONa=0,4mol. NH2 trong phan tu), trong do ti 1^ mo : m N = 80:21. De tac dung vua du voi 7,66 gam X can 100 ml dung djch K O H I M . Cho 7,66 gam X tac dyng vtra -> du voi dung dich HCl, thu dugc dung dich Y. Co can Y thu dugc m gam Phan ling voi toi xut ( N a O H he't, R C O O N a con du): muoi khan. Gia tri ciia m la A. 11,31. B. 10,58. C.9,85. ^ I Mol: ..Q ne cjsGf ^- nKOH=0,lmol. Nhan xet: '» Tu ti 1§ khoi lugng: y mo niN 80 = 21 De thay: ncooH = ^KOH = O,lmol. no 80/16 = 0,3 M RH- <- 4,8 -= 16 0,3 ) R H + Na2C03 0,3 ^ - . ;_y^, y[ ^ mwaOH = mRCooNa -> m = 0,4.82 + 18,4 - 0,4.40 = 30,7 gam. — Dap an C. > 0,3 R H la CH4. Bao toan khoi lugng: m + 10 21/14 _ 3 mol. RCOONa + NaOH D. 9,12. Lai gidi: nwaOH dir = 0,7 - 0,4 = 0,3 + mz '" Ca'm nang 6n luygn thi dgi hpc 18 chuy§n dg H6a hqc - Nguygn Van Hki Cty TN TV DVVH Khang Vi§t V i dy 14: Do't chay hoan toan m gam hon hgp X gom hai ancol, thu dugc 11,2 h't khi C 0 2 (dktc) va 12,6 gam H2O. Mat khac, cho m gam X tac dung voi N a (du), thu dixgc 4,48 h't khi H2 (dktc). Gia tri cua m la A. 12,4. ; B. 15,2. C . 12,6. m 0 - D . 13,8. .^ Laigidi: V rico2=^=0,5mol;nH2O=-^=0'7mol. Ta c6: mc = 0,5.12 = 6,0 gam; m H = 0,7.2 = 1,4 gam. ' H K h i X tac dung voi N a : i'-^ Mol: - O H + Na 0,4 n"^-:- '< j ,j r t o ^rt, ^ ff, '' < • m c j "* '•onn ni'<»'P!' wl- - 0,2 .„. - > D a p an D . ., . ^ j,^ ^ CO mm HIND AXIT C L O H I D R I C : H C l a, L i t h u y e t + Tfnft dung voi kim loai, bazo, oxit baza, muoi. V i du: Fe + 2HC1 n? f Si > FeCh + H a t '^ ^ ''' ^ > C a C h + C O a t + H2O CaC03+ 2HC1 '^^-'^^'^-^ "'^ + Tin?z fc^""- Tac dung voi cac chat oxi hoa manh: Mn02, KMn04, KCIO3, M n 0 2 + 4HC1 ,.,^„„ „ ;„,,.,,„/, ^ - ^ - ^ M n C h + CI2 +2H2O , ^-.•cr^^' rfi > F e C h +H2. (c) 6 H C l + 2A1 ' .,;.^d..Au. qfi(} i - i -^iyti^'^^ B.(a),(b). ' '-Xji::; X: jfefb g n i i ^ X i V t . ' - W ! M i:i ,i/:}c:: i " iM>f § ' ?. " Cac phan ling trong do H C l the hi?n tinh oxi hoa la C . (b), (c), (d). • . ,„,f^eH > 2 K C l + 2MnCl2 + S C h + 8H2O. (d) 16HCl + 2 K M n a - .v,t , > 2AICI3 + 3H2. A.(b),(c). C O H / , > M n C h +CI2 +2H2O. (a) H C l + M n 0 2 • ' V i d u l : Cho cac phan ling sau: (b) 2 H C l + Fe ry^.^ ^^..^^^^ 2KC1 + 2CrCl3 + 3Cl2 +7H2O K2Cr207 + 14HC1 b. V i d u m a u " > 2KC1 + 2MnCl2 +5CI2 + 8H2O 2KMn04 + 16HC1 no = noH = 2nH2 -> no = 0,4 mol -> mo = 0,4.16 = 6,4 gam. Bao toan kho'i lugng: mx = m ^ + mpi + m o = 6,0 + 1,4 + 6,4 = 13,8 1 K2Cr207. V i d y : t riD,b :,:n.b : x m t >-ONa+ CAC A X I T vo . ,„ s f8 f i D.(a),(d). Laigidi: Nhan xet: Trong cac phan ung (b) va (c), so' oxi hoa cua hidro giam tu +1 (trong H C l ) xuohgO (khi H2). , , - > Dap an A. V i dv 2: Hoa tan hoan toan 7,6 gam hSn hgp hot FesOA va C u trong 200ml dung djch H C l 1,2M (loang). Sau khi cac phan ung xay ra hoan toan, thu dugc dung dich X (khong chua axit du). Co can X thu du^c m gam muo'i khan. Gia tri cua m la A. 10,39. B. 14,20. C . 5,16. D . 11,10. Lffigidi: Gpi so mol: Fe304 = a; C u = b. Theo bai: 232a + 64b = 7,6. • ?-•<.}( ' ^• • ^han xet: chi c6 Fe304 phan ung tryc tiep voi axit. •: • '^^j^ f| ,* g .j 'j X :,,;. i : Trong X khong con axit d u nen Fe304 phan ung vua dii voi H C l : Cac phan ung hoa hpc: Caim nang On luy$n Ihi dgi hgc 18 chuy6n dg H6a hgc - Nguygn VSn i * • Fe304 + -> 0,03 < - 0,24 a = 0,03 Cu Mol: > FeCh + 2FeCh + 2H2O 8HC1 Mol: Cty TNHH MTV DVVH Khang Vijt Hit -» 0,03 V i dV 5: Cho 2,13 g a m hon h g p X g o m M g , Cu va A l a d a n g bgt tac d y n g hoan toan v o i O2 t h u d u g c h o n h g p Y g o m cac oxit c6 k h o i l u g n g 3,33 gam. 0,06 "The tich d u n g d i c h H C l 2 M v u a d u de phan u n g het v o i Y la - > b = 0,01 m o l . + 2FeCl3 0,01 - > > CuCh • 0,02 ^ + 2FeCl2 0,01 J t a r H Q U i i >J > T i X . A. 150 m l . B.SOml. C. 75 m l . Lai 0,02 D. 90 m l . gidi: m = 0,01.135 + 0,04.162,5+ 0,05.127 = 14,2 gam - > D a p an B. 3: Day g o m cac chat deu tac d u n g d u g c v o i d u n g d i c h H C l loang la Vi 01 bai nay, cac em rJia't thiet phai ap d u n g bao toan k h o i l u g n g de h'nh khoi l u g n g oxi tham gia phan ung: A. KNO3, C a C 0 3 , Fe(OH)3. ^ B. FeS, BaS04, K O H . C. AgNCte, ( N H 4 ) 2 C 0 3 , CuS. , D. NaHCOs, FeS, C u O . ' Lai mkimiiHii + m o x i = moxit — ^ .^.^^^ = 3,33— 2,13 = 1,2 gam. moxi 12 n o o = — =0/0375 m o l -> n o = 0,075 mol -> n ^ , , , ^-^2 gidi: ^ 32 Loai A v i KNO3 k h o n g tac d u n g ; loai B, C v i BaS04 va CuS k h o n g tan trong K h i cho oxit bazo tac d u n g v o i axit tao ra nuoc: d u n g d i c h H C l loang. 02- ^ D a p an D. Cac p h u a n g t r i n h hoa hoc: NaHCOa + HCl > FeS + 2HC1 > FeCh CuO + 2HC1 > ^''^ ^ ' + H2O NaCl CuCh • + iOnM>i CO2 '' V i d u 4: H o a tan hoan toan 8,55 gam hon h o p g o m Na, K va Ba vao nuoc, t h u i , ^ ; ;, Suy ra: n ^ + = 0,15 m o l - > n ^ c i ^ 0,15 m o l - > VHCI = 0,075 l i t = 75ml A. 14,90. Dap an C. B. 13,05. H2SO4, t i 1$ m o l t u o n g u n g la 2:1. T r u n g hoa d u n g d j c h X b o i d u n g d i c h Y, C. 7,45. •<• , d u o c d u n g d i c h X va 1,792 l i t k h i H2 (dktc). D u n g djch Y g o m H C l va t o n g k h o i l u o n g cac muoi dugc .,fm i . . H2O. d u n g d i c h X. Co can X t h u d u g c m gam chat ran k h a n . Gia trj ciia m la H2O + > =0,075 m o l . V i d\ 6: Hoa tan hoan toan 7,8 gam K vao 500ml H C l 0,2M, t h u d u g c k h i H2 va H2S + + 2H^ (0x11) H K = 0,2 m o l ; tao ra la K D . 20,50. iiiX Lai gidi: = 0,1 m o l ; n H c i = 0 , 1 m o l . + HCl > KCl + ,, lH2t =. ;j «o.'HBOH n . : ,. foH ,(> 2 A. 13,81 gam. B. 11,39 gam. C. 15,23 gam. D. 19,07 gam. n^, ^ Na = 0,08 m o l . Cac p h a n l i n g hoa hgc: 22,4 + H2O )• Na* + H2O > n Ba + 2H2O K ' + OH" + -Hi 0,1 't^»^'- + H2O + OH- + iH2 > KOH + iH2t n^ r<^r r a n k h a n Y. K i m loai k i e m M la A.Rb. B. K. > M H2O - > 4a = 0,16 - > a = 0,04 m o l . K h o i l u g n g muoi t h u d u g c = 8,55 + m + m Cl •* D.Li. Jl/ion Ob Lai gidi: = a mol. H* + O H " .., D a p an B. C.Na. Cac p h a n u n g hoa hgc: T r u n g hoa X b o i Y: • - H C l 0,2M, t h u d u g c k h i h i d r o va d u n g d j c h X. Co c?n X t h u d u g c 14,73 chat mo\. i_ 0,1 - V i d y 7: Hoa tan hoan toan 8,97 g a m k i m loai k i e m M vao SOOnil d u n g djch " H * ° " H C l + 2nH2S04 = 2a + 2.a = 4a m o l n ^ , . = 2a m o l ; 0,1 • OH£' - > m = m K c i + m K O H = 74,5.0,1 + 56.0,1 = 13,05 g a m > Ba2- + 20H- + H2 Nhan xet: n^^. = 2nyi^ =0,Id > Z' . 2 Mol: • o T r o n g Y: 0,1 ' • : 2 8 K 0,1 Luu y: K con d u se tiep tuc p h a n u n g v o i nuoc: 1,792 r, Mol: 'h-,-. 2- = 8,55 + 0,08.35,5 + 0,04.96 so^ = 15,23 gam —> D a p an C. + HCl > MCI + -H2 2 N/ian xet: Bao toan k h o i l u g n g m^„. OH • •= , I^^/TU' »• .| < va M +H2O ^ , > M O H + -H2 2 m y = m i 4 + m ^ i _ + m^^^. = 1 4 , 7 3 - 8 , 9 7 - 0 , 1 . 3 5 , 5 = 2,21 g a m n ^ „ . = 0,13 m o l . UH 73 Ca^m nang 6n luy^n thi dgi hqc 18 chuy6n dg H6a hgc - Nguygn VSn Hai Cty TNHH MTV DVVH Khang Vift 8 97 r i M = n H c i + n Q ^ . = 0,23 -> M = - ^ = 3 9 ( K ) ^ ' do D a p an B. ^'^^ ^^"^ ^ V i dv 8: Cho m g a m h o n h o p X g o m C u , M g , Fe tac d u n g v o i axit H C l d u , thu d u g c d u n g d i c h Y, 448ml k h i (dktc) v a 0,64 g a m chat ran. C h o d u n g dich vao Y t h u d u g c 448ml k h i (dktc). K h o i l u g n g N a C l t r o n g X la ^ 0,585. B. 1,170. N a O H d u vao Y, loc ket tiia v a n u n g trong k h o n g k h i t o i k h o i l u o n g khong doi, t h u dugc 1,2 gam chat ran. Gia t r i ciia m la A. 0,80. B. 1,16. , D . 0,84. r• . = 0,01. Ggi so m o l t r o n g X: Fe = x; M g = y - > x + y = n H 2 = 0,02 m o l . ' So d o p h a n u n g : Mg ^ -±»£U > Fe(OH)2 - ^ ^ ^ l F e 2 0 3 MgCh > Mg(OH)2 — m F e 2 0 3 + m M g O = 1/2 ^ 80x + 40y = 1,2 - » Luu CI" + 2HC1 + 2HC1 C. 28,21%. mNaci C. 19,2. D. 9,0. ^-^••«:MfBrtf 6'> Cac p h u o n g t r i n h p h a n u n g : Mg Mol: Mol: Loigidi: i ' •' > M g C h + H2 K.i i a •t^|,\V, a > M g C h + CQ2 + H2O + 2HC1 b ''"^ b A p d u n g cong thuc cua p h u o n g phap d u o n g cheo, ta c6: "H2 = 100 - 1 5 , 7 6 ^ 84,24 gam. 44 - 2 3 2 - 23 =1 • 'iliiA:: , , , ^ , — = - -> a = b = 0,1 m o l b 1 m = m^g+ mMgcos " bandau. A )• M g C h + H2 + 2HC1 Theo bai: M = 11,5.2 = 23 va a + b = 0,2 m o l . ^ FeCh + H2 Nhan xet: L u g n g H2O t r o n g Y cung c h i n h la l u g n g H2O c6 t r o n g d u n g d i c h H C l " ^^'^ 8 ^ ^ ^ ^' 2. A X I T S U N F U R I C : H2SO4 L i thuyet M a t khac, d o n o n g do H C l bang 20% mH20 = 4mHci + 4 m H c i = 84,24. , Bao toan n g u y e n to'Cl: 2nFeci2 + 2nMgCl2 = " H C l "> Dung d/clz H2SO4/oflng: T i n h axit m a n h Fe + H2SO4 FeS + H2SO4 31 52 mMgCl2 D a p an B. Loigidi: ^C02 mMgCl2 >''•'' ^ -Hi = 0,02.58,5 = 1,17 gam B. 10,8. MgCOs D. 15,76%. Xet v o i 100 g a m d u n g d i c h Y: -> m^^^ij = 1^,76 gam. mMgci2 + m H 2 0 " ' mla 15,76%. N o n g d o p h a n t r a m cua M g C h t r o n g Y la , '''' > > HCl f. K h i n u n g ngoai k h o n g k h i , Fe(OH)2 chuyen t h a n h Fe(OH)3 v a b i B. 11,79%. + HClt KHSO4 + H C l t > ZnCh + Hat A. 13,2. d u n g d i c h H C l 20%, t h u d u g c d u n g d i c h Y. N o n g d o cua FeCk t r o n g Y la ^ NaHS04 dugc 4,48 l i t h o n h g p k h i X (dktc). T i k h o i cua X so v o i H2 la 11,5. Gia t r i cvia V i d u 9: H o a tan hoan toan h o n h g p X g o m Fe v a M g b a n g m o t l u g n g vira d i i Mg jCCl + H2SO4 Z n + 2HC1 ^ V i du 11: H o a tan het m g a m h o n h g p M g v a MgCOa t r o n g d u n g d i c h H C l , t h u x = y = 0,01 m o l . p h a n h i i y t h a n h Fe203. Fe — _» x = 0,02; y = 0,02 -> ^ MgO ' A. 24,24%. ' Ta c6: mx = 58,5x + 74,5y = 2,66 va n H 2 = 0,5(x + y) = 0,02. m = 0,64 + 0,01.56 + 0,01.24 = 1,44 gam. -> D a p an C. + H2SO4 Bao toan n g u y e n to: FeCh D . 2,340. Loigidi: jsjaCl Loigidi: Fe C. 1,755. Goi so m o l : N a C l = x; K C l = y. Cac p h u o n g t r i n h p h a n u n g : C. 1,44. N^flnxef: m c u = 0,64 g a m n c u NaC\a K C l v o i H i S 0 4 dac, d u . jChi thoat ra cho hoa tan vao nuoc t h u d u g c d u n g d i c h Y. Cho b g t Z n d u 2 + ~n^gClj 1 = ^ " ^ H C l =11'79 ^ C%(MgCl2) = 11,79% -> D a p an B. '1,1 , ^ > FeS04 + H 2 t ,j, > FeS04 + H 2 S t , "*" Dwn^djc^ H2SO4 (fflc: T i n h oxi hoa m a n h N g o a i t i n h axit m a n h , axit sunfuric dac con the h i ^ n t i n h o x i hoa m ^ n h , tac d i i n g d u g c v o i n h i e u k i m loai, h g p chat,...: ff;-!;! » {'"-l Cty TNHH MTV DWH Khang ViSt C^m nang 6n luy$n thi dgi hpc 18 chuySn dg H6a hpc - NguySn VSn HJi Cu + — — > CuS04 2H2S04(^flc) 2Fe + 6H2SO4 (dac) 2FeO + 4H2SO4 +SO2 + 2H2O Gia thie't dung djch ban dau chiia 1 mol H 2 S O 4 (tiic chiia 98 gam 100 :> / Fe2(S04)3 + 3SO2 + 6H2O ',' ' _> Khoi lupng dung djch > Fe2(S04)3 + SO2 + 4 H 2 O (dac) 2Fe304 + IOH2SO4 (dac) > 3Fe2(S04)3 + SO2 + IOH2O L u u y: Cac kim loai Al, Fe, Cr khong tac dung vdi Ji2S04 dac, nguQi. Dieu che So do: Quang pirit FeS2 hoac S H2S04.nS03 > (n+1) Cac phan ung: S + O2 4FeS2 + I I O 2 — ... ) SO2 — ^ •• • "ID • ' ^ , . fiB'lii^H :6i 2S03 + nSOs SO3 — .•..,.H,.-..»,I > SO2 ^ 2Fe203+ 8SO2 2SO2 + O2 H2SO4 H2SO4 ' ,-xm:o:.fii , > H2S04.nS03 H2S04.nS03+ n H 2 0 (Oleum) ^ > (n+l)H2S04 b. V i dv mau: Vi dy 1: Cho day cac chat sau: KBr, S, Si02, FeO, Cu va Fe203. So'chat trong day the bi oxi hoa boi dung djch axit H2SO4 (dac, nong) la CO A. 4. B.5. ' G.3.c|f: Lai gidi: S + 2H2SO4 — 2KHS04 + Br2 + SO2 + 2H2O ^ 2FeO + 4H2SO4 — Cu 2H2SO4 + 3SO2 + 2H2O : ' SO2 + 2H2O ^ fH',:^:: y: Khi tac dyng voi FeaOs + 3H2SO4 — H2SO4 d^c, FeaOs the hi^n tinh baza: H2SO410% (loang), thu dug-c dung dich muoi c6 nong dp bang 14,45%. Kim loai M i a ' : B. Fe. , ' " " C. Cu. ' ^ Lcn gidi: + H2SO4 (3): Fe -3e > Fe*-^ nenhuong = 3 np^ = 3a > Cu*^ (1) (2): O2 +4e > 2Ct^ (3): S** +2e > SO2 nenhiKmg=2ncu ^ nenhan = OsH » ; • =0,03 "^s: 4no2 = ^^^^^^=0,67-7a =0,06. - nenh,^n=2nso2 Bao toan electron: 3a + 0,03 = 0,67 - 7a + 0,06 ^ a = 0,07 mol -> Dap an C. CachZ: Bao toan khoi lupng: 56a + 16b = 6,32 - 0,015.64 = 5,36. a = 0,07; b = 0,09 , ,, . Dap an C. Fe304, Fe203 tac dung voi dung dich H2SO4 dac, nong. So truang hpp xay ra phan ung oxi hoa - khu la - ? Rf ' A . 6. . B.3. C.4. , DJfjfi Lai gidi: D. Zn. Nhan Phan ling hoa hpc: MO -> Vi dy 4 (B-12): Cho cac chat rieng bi^t sau: FeS04, AgNOa, Na2S03, H2S, H I , ^ Fe2(S04)3 + 3 H 2 O Vi dy 2: Cho oxit ciia kim loai M (hoa tri 2) tac dung vira dii vai dung dich A. Mg. Fe,Cu(l) > Y (2) > Fe^3Cu^M3) Xet su trao doi electron 6 cac giai doan: Bao toan electron: 3a + 2.0,015 = 2b + 2.0,03 Dap an A. Luu : Qui doi Y thanh: Fe (a mol); Cu (0,015 mol) va O (b mol). ^ Fe2(S04)3 + SO2 + 4 H 2 O CuS04 + ne'u dya theo phuong trinh phan ung se rat dai va kho giai. Cach 1:6 day, cac em can su dyng so do phan ung: . t y s j . j , Nhan xet: (2) CO tinh khu (chiia nguyen to' dang 6 muc oxi hoa tha'p). 2KBr + 3H2SO4 Bao toan khoi lupng: mo2 = my " "^x = 6,32 - 56a - 0,96 = 5,36 - 56a. Cu - 2 e Nhan xet: Axit H2SO4 dac, nong the hien tinh oxi hoa khi tac dung vai chat = 98. — = 980 gam. 980 + M + 16 ^^^^^•^ -=0,1445 - > M = 56 (Fe)-^ Dap an B. Vi dy 3: Nung hon hop X gom a mol Fe va 0,015 mol Cu trong khong khi mpt thoi gian, thu dupe 6,32 gam chat ran Y. Hoa tan hoan toan Y bang dung dich H2SO4 dac nong (du), thu dupe 0,672 lit khi SO2 (san pham k h u duy nha't 6 dktc). Gia tri ciia a la A. 0,04. B.0,05. C. 0,07. D. 0,06. Lm gtat: ^ 0 672 -bVT' ".'^ ncri^=—^ =0,03 mol. , , ,, , i. , (1) D.2. H2SO4 H2SO4) xef: Phan ung oxi hoa-khu xay ra khi H2SO4 tac dung vai chat c6 tinh khu (chiia nguyen to dang 0 muc oxi hoa tha'p). > M S O 4 + H2O * 2FeS04 + 2H2SO4 Fe2(S04)3 + SO2 + 2 H 2 O 77 im nang 6n luy^n thi dgi hoc 18 chuygn dg H6a hqc - Nguygn Van Hi\ HlS + 3H2SO4 ' Cty TNHH IVITV DVVH Khang Vigt > 4SO2 + 4H2O 8HI + H2SO4 ^'-^ 4I2 + H2S + 4H2O 2Fe304 + IOH2SO4 Lcngiai: , Gpi so mol: Fe304 = a; Cu = b. Theo bai: 232a + 64b = 5,28. f-SnO'w! ;oi*:^f ' 3Fe2(S04)3 + SO2 + IOH2O Cac phan ung hoa hpc: Luu y: ^eiOj, AgNOs xay ra phan ung trao doi, Na2SQj khong tr.c dung: Fe2(S04)3 + 3H2O 2AgN03 + H2SO4 a = 0,02 Ag2S04i + 2HN03 C. 11,80 gam. Vi D. 9,42 gam. y ^, ^ ^ Na + H2O + OH- + - H 2 2 1 > Na^ + O H " + - H 2 Ba + 2H2O • *• trong cung mot the ti'ch thi CI . , ii ion'ji H2O ^ V. ' , K 2- SO^ ' "H2 = 0,10 mol. ''^ Vi " ; ,1- , 1 dvi 6: Hoa tan hoan toan 5,28 gam hon hgp bpt Fe304 v a Cu trong 80ml dung d i c h c h u a m g a m m u o i . Gia tri c u a m l a A. 19,04. B. 20,54. ' C. 17,96. " D. 14,50. Lai gidi: - > m = 17,96 g a m -> Dap a n C. ' •'fm-E ' o Cach 2: Ta c6: n M = nH2S04 = " H J = 0,12 m o l . N e u 1 m o l k i r n loai M > MSO4 t h i k h o i l u g n g t a n g 96 g a m 0,12 dich H2SO4 I M (loang, vua dii). Sau khi cac phan ung xay r a hoan toan, thu tang0,12.96. duoc dung djch X. Co can X thu du(?c m gam muoi khan. Gia trj ciia m la Vay m = 6,44 + ai2.96 = 17,96. A. 8,64. -> B.7,68. '' ' 8: Hoa t a n h o a n toan 6,44 g a m h o n hgp X g o m A l , Fe v a Z n b a n g m o t Bao t o a n k h o i l u o n g : 6,44 + 0,12.98 = m + 0,12.2 ' = 6,08 + a06.35,5 + 0,03.96 = 11,09 gam. Dap a n B. D. 97,80 gam. Cach 1: Bao toan n g u y e n to H : nH2S04 = " H 2 " 0,12 m o l . \V a = 0,03 mol. . C. 101,48 gam. = 3,68 + 98 = 101,68 v a se c h g n n h a m d a p a n A! ^y^:-yyh Khoi luong muoi thu dug-c = 6,08 + m . + m Xi n.' l u o n g v u a d u d u n g d i c h H2SO4 loang, t h u d u g c 2,688 lit H2 (dktc) v a d u n g o_ = a m o l ; n _ . = 2amol. CI Dap an D. Luu y: Bai n a y c a c e m d i q u e n t r u k h o i l u o n g k h i H2 b a y r a , v a c h i t i n h : mx " H a = 2nH2S04 • Goi n H c i = 2a -» nH2S04 = a- 4a = 0,12 ' -> Dap a n C. Mat khac, nong do HCl gap 2 Ian H2SO4 > ' Bao t o a n k h o i l u o n g : 3,68 + 98 = mx + 0,10.2 ^ m x = 101,48 iiJ ' ''' SO4 ' -> Khoi luong dung dich H2SO4 = 0,10.98. ^ 10 = 98 gam. Nhan xet: n ^ ^ . = 2nH2 = 0,12 mol. H ,,.•..,,.',„ • ' Lcngiai: > Ba2+ + 2 0 H - + H2 , B. 88,20 gam. , , . > Trung hoa X boi Y: H * + O H " ' ' CuS04 + 2FeS04 Bao toan nguyen to hidro: nH2S04 = -> Trong Y: n + = 4 a m o l ; n ~ 7: Cho 3,68 gam hon hop gom A l , M g va Zn tac dung voi mot luong vua A. 101,68 gam. = 0,06 mol. Li + H2O " ' Khoi lugng dung dich X la Cac phan ung hoa hoc: ,, ^ b = 0,01 mol. du dung dich H2SO410%, thu dxxgc dung dich X va 2,24 lit khi H2 (dktc). 1 344 = •• ' 0,02 -> 0,01 0,02 0,01 - > 0,01 m = 0,01.160 + 0,01.400 + 0,04.152 = 11,68 gam Lbi gidi: "H2 0,02 Mol: dung dich X v a 1,344 h't khi H2 (dktc). Dung dich Y gom HCl I M v a H2SO4 B. 11,09 gam. ^ Cu + Fe2(S04)3 i d\ 5: Cho 6,08 gam hon hop gom Li, Na v a Ba vao nuoc (du), thu duoc A. 10,38 gam. > FeS04 + Fe2(S04)3 + 4H2O Fe304 + 4H2SO4 : Mol: 0,02 < - 0,08 0,5M. Trung hoa dung dich X boi dung dich Y, tong khoi lugng cac muoi dugctaorala , : b uAi Trong X khong con axit d u nen Fe304 phan ling vua du voi H2SO4: Dap a n C. Fe203 + 3H2SO4 ji jV/ian xet: chi CO Fe304 phan ling true Hep voi axit. , 1/ ; i ; C. 15,68. D. 11,68. Dap a n C. • .. ; , I- ' „ . , (^„j,,4i<);/:H:H , ).^- C^m nang 6n luygn thi dai hgc 18 chuy6n H6a hpc - Nguygn Van H&\ V i du 9: H o a tan hoan toan 2,81 gam hon hop X gom Fe203, FeO, C u O can 50ml axit H2SO4IM (loang). Khir hoan toan 2,81 gam X bang khi C O (nung nong) thu dugc m gam kim loai. Gia tri ciia m la A. 3,24. B.2,65. C. 3,06. D.2,41. aofl ' • ' L a i g i d i : •i-,'>r"i Nhan xet: K h i cho oxit kim loai tac dung voi axit/.ion 0^~ trong oxit se ke't hqp voi H* trong axit tao thanh H2O theo phuong trinh: ^ , ^ Ijjjy, 02- + Mol: 0,025 > H2O 2H^ 0,05 ^ Vay: m o = 0,025.16 = 0,4 gam -> m kim:o?i = 2,81 - 0,40 = 2,41 gam -> Dap an D. s^iBi J O A I flj 4-»f thu dugc dung dich X va 324,8ml khi SO2 (san pham khu duy nhat, 6 dktc). Co can dung dich X, thu dugc m gam muo'i sunfat khan. Gia tri ciia m la •HV B.5,40. •• ' ' hoac N H 4 N O 3 ; 2. Al, Fe, Cr khong tan trong HNO3 d$c ngupi. ' ' , " ' Tac dung vai hap chat 3FeO + 10HNO3 > 3Fe(N03)3 + NO + 5H2O '•'^^ 3Fe2* + N O ; + 4H* > 3Fe3* + N O + 2H2O " rl-y: £n,;, ... > Fe(N03)3 + NO2 + CO2 + 2H2O * > Fe(N03)3 + 2H2SO4 + I5NO2 + 7H2O .ciijj Dieu che Vi d\ 10: Hoa tan het 2,088 gam FexOy bang dung dich H2SO4 dac, nong (du), A. 5,22. 1. Ba kim loai kha manh (Mg, Al, Zn) c6 the khu HNO3 thanh N2O, N2 FeC03 + 4HN03 FeS2 + I8HNO3 S},.0-'K ia,,j^ u Litu y: C. 5,80. D. 4,84. So do: N2 N2 + 3H2 < 4NH3 + 5O2 2 N O + O2 > 2NH3 > NO2 . -4 HNO3 < ,i 'bCl 4y '""^•'^ > 4 N O + 6H2O > 2NO2 > 4HN03 4NO2 + O2 + 2H2O Laigidi: > NO > NH3 Nhan xet: Oxit sat FexOy tac dung voi H2SO4 dac, nong -> SO2 thi oxit la FeO hoac Fe304. Vi du 1: Cho 3,024 gam mpt kim loai M tan het trong dung dich HNO3 loang, Cac em luu y rang 1 mol FeO hoac Fe304 deu chua 1 mol Fe''- nen deu c6 kha nang nhuong 1 mol electron, do do: 0 3248 = npe^Oy = 2ns02 -> "FexOy = 2 " - ^ ^ = ^'029 mol , ns« -> Mpe o = = ^ = 72 (FeO). ' ' " ' ^•^ ^ '^• ^ . >0r ^ urn '^ thu dugc 940,8 ml khi NxOy (san pham khu duy nhat, a dktc) c6 ti kho'i doi voi H2 bang 22. Khi NxOy va kim loai M la A.NOvaMg. B. N C h v a A l . C.N20vaAl. Lai gidi: Theo bai: M ^ ^ Q = 22.2 = 44 NxOy la N2O Loai A va B . f ,(>« .,, .' n'f)) • Bao toan nguyen to'Fe: 2FeO > Fe2(S04)3 mpe2(so4)3= 0/0145.400 = 5,80 gam-> Dap an C. ; p' ,' n,,o=^Jo,042™ol. 22,4 '^o^h/r Cach 1: Phan ung hoa hgc: 3. AXIT NITRIC: HNO3 8M + 10nHNO3 A. L i thuyet _v + Tinh axit manh: Tac dung voi kim loai, bazo, oxit baza, muoi. Vi du: MgO + 2HNO3 CaC03 + 2HN03 > Mg(N03)2 + H2O > Ca(N03)2 + C O z t + H2O + Tinh oxi hoa manh + Tac dung voi kim loai: Axit nitric tac dung dugc voi hau het cac kim loai (tru Au, Pt), ke ca kim loai c6 tinh khu ye'u nhu Cu, Ag,... Cu +4HN03(f?flc) D. N20vaFe. > Cu(N03)2 + 2N02T +2H2O 3Cu + 8 H N O 3 ( / 0 f l M ^ ^ ) — - > 3Cu(N03)2+ 2 N O T + 4 H 2 O 8 ^ > 8M(N03)n + nN20 + 5nH20 0,336 3,024 nfo ' ^ Q,;.^^ „ n -> n = 3 va M = 27 (Al) Dap an C. '^'^ ' Cach 2: Nhan thay de tao dugc khi N2O thi kim loai phai kha m^nh (nhu Mg, Al, Zn) -> Loai D -> Dap an C. Vi dv 2: Hoa tan hoan toan hon hgp gom 0,04 mol FeS2 va a mol CU2S vao axit HlSf03 (vua dii), thu dugc dung dich X (chi chua hai muoi sunfat) va V lit khi duy nhat NO2 (dktc). Gia tri ciia V la A. 13,44. B. 17,92. C. 20,16. D. 22,40. Cty TNHH MTV DWH Khang Vi$t im nang 6n luy^n thi d?i hpc 18 chuySn dS H6a hpc - Nguyin Van HSi Lai Cac phan l i n g k h u : gidi: Nhan xet: cac e m l i m y la d u n g dich X chi chiia hai m u o i sunfat —> sau cac peS_9e phan l i n g , S n a m he't 6 dang goc sunfat. FeS2-15e Ta CO cac so do chuyen hoa: Bao toan electron: FeS2 Mol: 0,02 Mol: a ' > 2CuS04 ,;<:J3+, sOl-'';; *• ( O M b U ' • Cu2S -> ^aicHS: + 2a Bao toan n g u y e n to S, ta c6:2 npeSz + >^ Wlh > A' . Mol: / i n Mol: 0,005 1 ^'^'^Cu2S V ^ o =17,92 l i t 1' t t- tiVK'. 0,005 0,01 NO y t III II U ' I I 0,02 ^.4 > f y > *'l) -a /' (san p h a m khvr d u y nhat) va d u n g d i c h Y. The tich d u n g d i c h N a O H A. 6 0 m l . thay axit H N O 3 b a n g axit H2SO4 dac, nong, d u t h i the tich k h i SO2 (dktc) thu d u g c sau p h a n u n g la bao nhieu? B. 5,60 lit. B. 120 m l . > Fe*3 va C u - 2e hoac +2e Bao toan electron: n g = n f j 0 2 = 2nso2 0,05 m o l 4: H o a t a n hoan toan 2,08 ncu = D . 1,12 lit. 64 + = 0,03 m o l ; V ; • • •..M , > Cu^^ > SO2 D. 180ml. n^NOa = ^^'^^^^ 63 = .J- .0.$-/ ...... : 0,2 m o l . Phan u n g hoa hpc: 3Cu(N03)2 + 2 N O + 4 H 2 O 8HNO3 Y g o m : Cu(N03)2 = 0,03 m o l ; H N O 3 d u = 0,2 - 0,08 = 0,12 m o l . > H 2 O va Cu^^ + 2 0 H - ;j > Cu(OH)2', = n„+ + 2 n ^ 2+ =0/12 + 0,03.2 = 0,18 m o l OH H Cu - > VNaOH = 0,18 l i t = 180ml ^ D a p an D . mi y: Cac e m de q u e n p h a n u n g t r u n g hoa axit ( H * + O H ) va c h i t i n h : . = 2n^^2+ - 0/06 m o l —> VNaOH = 0,6 l i t = 60ml se chpn n h a m A ! rcitjirf »s" =' Vso2 = 0,05.22,4 = 1,12 l i t ^ C. 1 5 0 m l . Loi gidt: ... Cac p h a n u n g : H * + O H ' Ji* > NO2 ' gidi: nN02 = 0/1 m o l . Fe - 2e KI; 3Cu C. 2,24 lit. Lot "502 = ^ = ti I M can thiet de ket tua he't i o n Cu^* t r o n g d u n g d i c h Y la ->DapanB. d u n g d j c h H N O 3 dac n o n g t h i t h u d u g c 2,24 l i t k h i m a u n a u (dktc). N e u Chat oxi hoa: N^^ + i e 0,01 -> -Fe2C)3 + 2BaS04 2 3: H o a tan hoan toan m p t h o n h g p g o m h a i k i m lo?ii Fe v a C u bang Chatkhu: it 'I V i d v 5: C h o 1,92 g a m C u vao 60 gam d u n g d i c h H N O 3 2 1 % , t h u d u p e V m l k h i Bao toan electron: L n ^ O j = 15r>FgS2 A. 3,36 lit. + BaS04 -> m = 160.0,01 + 0,03.233 = 8,59 - > Dap an B. > 2Cu^2 + nN02 = 0 ' 8 m o l - > FeS2 ur > Fe*3 + 2S^ Cu2S - lOe * -> - F e 2 0 3 2 0,01 •> v , FeS ,(fWi > }A}W\ i r < , Cac phan l i n g k h u : /i (Ov^jr! - Bao toan n g u y e n to Fe va S: ' • * i^Cu2S = " S O 4 -> 0,04.2 + a = 0,02.3 + 2a ^ a = 0,02. FeS2-15e > Fe^3 + 25"* nN02 = ^ "FeS +15 npeS2 - > 9a + 15b = 0,24 ^ a = 0,01; b = 0,01. > -Fe2(S04)3 0,04 > Fe*3 + D a p an D . g a m h o n h o p g o m FeS va FeS2 t r o n g d u n g djch axit H N O 3 (dac, d u ) , t h u d u p e 5,376 l i t k h i NO2 (dktc) va d u n g d i c h X. V i d\ 6: C h o 19,2 gam k i m loai M (hoa t r i n) tan hoan toan t r o n g d u n g d i c h H N O 3 , t h u dupe 4,48 l i t k h i N O (san p h a m k h u d u y nhat, 6 dktc). K i m loai Mia: A. M g . D. Fe. C. C u . B. A l . Cho X tac d u n g v a i d u n g d j c h Ba(OH)2 d u , Ipc ket tiia va n u n g t r o n g k h o n g Lot khi d e n k h o i l u p n g k h o n g d o i t h u dupe m g a m chat ran. Gia t r i ciia m la nNo = 0,2 m o l . Phan u n g hoa hpc: A. 8,43. 3M gidi: B. 8,59. C. 10,19.7 Lai gidi: nN02= D- 6,19. , > , 1^=0,24 m o l . Gpi so m o l : FeS (a mo l ) va FeS2 (b mol). Ta c6: 88a + 120b = 2,08. + 4nHN03 N h a n thay: XXM = > 3M(N03)n + n N O + 2 n H 2 0 3n NO _ 0,6 mol n n = 2 va M = 64 (Cu) MM = 19 2 u,o = 32n n D a p an C. 83 a'm nang an luy^n thi d^i hoc 18 chuyen H6a hgc - Mguyen van Hi'i Cty TNHH MTV D W H Khang Vi?t / i d u 7: N u n g 2,94 gam hon hgp X gom cac kim loai A l , Zn, M g trong khi oxi, sau mot thai gian thu duoc 3,42 gam hSn hap Y. Hoa tan hoan toan Y vao Chat o x i h o a : N*'* + n e > X dung dich H N O s (du), thu duoc 2,016 lit khi NO2 (san pham k h u duy nhat, Bao toan e l e c t r o n : 2 n M g = n . n x ^ a dktc). So mol H N O 3 da phan ung la _^ „ = 8 A. 0,20. B.0,24. C.0,16. ' . no2 = 0,015 mol no= 0,03 mol Cachl:Sudungsad6: X Lm '^^^^ "^'^^^^^ n ^ . 2 ^^^.j^= 0,03 mol. A 0,12. ''^ , = 0,06 mol. 8A1 + 3OHNO3 ^ ' Bao toan electron: n ^ ^ x j ^ ^ n g + n j ^ Q ^ =0,15 mol —> n . =0,15 ~^ " va B. Loai C v i C u O tac dung voi c a 3 axit - > dung dich muoi mau xanh. c ,, _> UHNOS = 10.0,015 + 4.0,01 = 0,19 m o l - > D a p an D . V , B . 124,45. y.' qua: ni>jo=0,l m o l ; n N 2 o = 0 , l m o l . A\,Zn,Cu > CuS04 + SO2 + 2H2O Laigidi: img = 0,15 mol; n x = 0,0375 mol. D.NO. ^ - -'•> - N^^ + 3e > NO; 2N^-'* 2N*-^ + 8e + 8e - " S'KS : « o r i f:w 7 ^ ''^^^ > N2O > NH4NO3 ( a m o l ) K h i cho k i m loai + HNO3: " N O 3 1 du 9: Cho 3,6 gam M g tac dung het voi dung dich HNO3 (du), sinh ra 0,84 lit CNC)2. D . 112,05. ) M u o i n i t r a t + N O + N2O vacothexayracaquatrinh: B.N2O. C. 99,65. 6 bai nay, t r u a 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 CU + 4 H N 0 3 > Cu(N03)2 + 2NO2 + 2H2O D u n g dich H2SO4 d ^ c hoa tan C u va c6 khi khong mau, miii xoc thoat ra: A.N2. 1 t r i cua m la Chat oxi hoa: khi X (san p h a m k h u duy nha't, a dktc). K h i X la 3nAi(N03)3 + 2 n N 2 0 + ^^NO lit h o n h g p k h i X (dktc) g o m N O va N2O. T i k h o i ciia X so v a i H2 la 18,5. Gia lil Giai t h i c h : H N O s hoa tan C u v a c6 khi NO2 mau n a u bay ra: C u + 2H2S04d,ic ' Laigidi: Nhan xet: A l , Fe khong tac dung voi H2SO4 va HNO3 dac ngupi - > L o a i A „,.,'i;.j • Dua theo cac p h a n l i n g , n h a n thay: n H N 0 3 = 1 0 n N 2 O + ^^NO . 120,45. D. Cu. Laigidi: Dap an D . - - n f i ^ - t uu^.x^ 950ml d u n g d i c h HNO3 2 M , t h u d u g c d u n g d j c h chua m g a m m u o i v a 4,48 ^' CCuO. • '' ' ' ' V i d^ 1 1 : C h o 27,45 g a m h o n h g p g o m A l , Z n va C u tac d u n g vvra d u v o i trong ba lo bi ma't nhan, ta dung thuo'c thu la B.Fe. > A1(N03)3 + N O + 2H2O '^' n H N 0 3 = 0,05.3 + 0,015.2 + 0,01 = 0,19 m o l - > D a p an D . mol. n du 8: De nhan biet ba axit dac, nguoi: H C l , H2SO4, H N O s dung rieng bi$t A.Al. D . 0,19. * J t l ^ >' Bao toan n g u y e n to N : n n N O a ' * 3 ^ ^'^^ ' 3nAi = 8 n N 2 0 + 3 n N o = 8.0,015 + 3.0,01 = 0,15 m o l - > nAi = 0,05 m o l . -> Dap an B. Bao toan N : n n N O a " N O 3 IrV Cach 2: Bao toan electron: " ^'^'^ :ach 2: Q u i doi Y thanh X va O (0,03 mol). C. 0,17. > 8A1(N03)3 + 3N2O + 15H2O A l + 4HN03 " «li 1 "^^o " B. 0,15. each 1: Cac p h a n u n g : trong oxit se bi thay the bang 2 goc N O 3 de Bao toan nguyen to N : n H N O s = " N O 3 - ^'""^ - ' " Laigidi: 4—"- + Khi cho kim loai + HNO3: n j ^ ^ _ (muoi) — ngtraodoi— fij^jQ^ — 0,09 mol. =2n ^>,dn mJ " - va 0,01 m o l k h i N O . So m o l axit H N O 3 da t h a m gia p h a n u n g la ^ Muoinitrat + NO2 y: Y chua ca kim loai (con du) va oxitsgS + s O r l tao muoi nitrat, do do: n 0,0375.n d u n g d i c h X ( k h o n g chua N H 4 N O 3 ) va h o n h g p k h i g o m 0,015 m o l k h i N 2 O > Y + K h i cho oxit + HNOa: 1 goc ai5.2 = D a p an B. V i d i ? 10: H o a tan hoan toan m g a m A l bang d u n g d i c h H N O 3 loang, t h u d u g c D . 0,14. Lotgtat: Bao toan k h ^ luc^ng: mo^ = 3,42 - 2,94 = 0,48 gam -> K h i X la N2O ^ ^'^^^ ' ' " " " i " ' = 3 n f j o + 8 n N 2 0 + 8 n N H 4 N 0 3 = ( I ' l + 8a) mol. Bao toan n g u y e n to N : UHNOS ^ " N O 3 ^ "^^^ 0,95.2 = 1,1 + 8a + 0,1 + 2.0,1 + 2a Bao toan k h o i l u g n g : m = m ^ i , zn, C u ^ "N2O + 2nNH4N03 ,. ; a = 0,05 m o l . ' " N O S ^ "^NH4N03 = 27,45 + (1,1 + 8.0,05).62 + 0,05.80 = 124,45 gam. 85 Cty TNHH MTV DVVH Kliang Vi§t :im nang On luygn thi dgi hgc 18 chuyfin 66 H6a hgc - Nguygn Van HSi Theo cong t h i i c d u o n g cheo ta c6: —> Dap an B. Nhan xet: Can n h a n ra bai toan da " g i a u d i " s a n p h a m NH4NO3. Vi dyi 12: Cho m a n h C u phan u n g he't v o i d u n g dich H N O a , t h u d u g c 0,896 h't (dktc) h o n h g p k h i X ( g o m N O va NO2) c6 k h o i l u g n g b a n g 1,52 gam. Co can d u n g d j c h sau phan l i n g t h u d u g c a gam m u o i k h a n . Gia t r j cua a la A . 5,64. B. 7,52. C. 9,4. Lai , ; + |A D a p an B. d u g c 0,672 l i t N2 bay ra 6 d k t c va d u n g dich X. K h o i l u g n g m u o i k h a n t h u d u g c k h i c6 can d u n g dich X la B. 30,1 gam. ;s ' HMg = C. 55,8 gam. D . 15,04 Chat oxi hoa: 2N*'' + lOe > N2 — ^ nenh?n 2N*^ -> + 8e N H 4 N a (a m o l ) a = 0,105 m o l . ^^''^' - thanh m u o i NH4NO3! V i dv 15: H o n h g p X g o m M g , A l va Z n . Cho m gam X tac d u n g VvJ-i d u n g dich H C l d u , t h u d u g c 36,45 g a m m u o i clorua. M a t khac, hoa tan he't m gam X A . 5,8. . B.8,7. C.11,6. D . 14,5. • , V-\T-> 0,4 - > 2nMg = 0 , 7 m o l . , = 10n[vj2 = 0 , 3 m o l . ao toan k h o i l u g n g , ta c6: m + 35,5a = 36,45; m + 62a = 55." Dap an C. Lu-u y: N e n ap d u n g ngay bao toan electron: 2 n ^ g = 3 n ^ o + 8 nNH4N03 • i du 14: H o a tan hoan toan 12,42 gam A l bang d u n g dich H N O a loang (du), t h u d u g c d u n g d i c h X va 1,344 l i t (a dktc) h o n h g p k h i Y g o m N2O v a N2. T i k h o i cua Y so v o i h i d r o la 18. Co can X, t h u d u g c m g a m chat ran khan. Gia t r i cua m la A . 38,34. B. 34,08. C. 106,38. Lai gidi: - Ta c6: nAi = 0,46 m o l ; nv = 0,06 m o l ; M y = 18.2 = 36 H3PO4 + 2 N a O H H3PO4 + 3 N a O H mol 1:1 mol : :\ vnU W ' l -> NaH2P04 + H2O 1:2 -> Na2HP04 + 2H2O mol 1:3 / T i r ',-,->•") :'t[ ill, : > Na3P04 + 3H2O a=1 : NaH2P04 1< a < 2 : N a H 2 P 0 4 + N a 2 H P 0 4 Neu: nH3P04 'en 0-..,- 2 'Y tV :•'::„ i b. V I D y M A U ^ V i du 1 (B-08): Cho 0,1 m o l P2O5 vao d u n g d j c h chua 0,35 m o l K O H . Sau k h i D . 97,98. - ^ , , - > D a p an C. H3PO4 + N a O H ^'''^''l;:, Vay: m = mMg(N03)2 + mNH4N03 = 0,35.148 + 0,05.80 = 55,8 gam ' - > a = 0,7 m o l ^ m = 36,45 - 35,5.0,7 = 11,6 g a m "••'^***W;'yi'«^ .1 1 a05 '•••T ra \ a. L i thuyet '-•••"'Mol:'.-'' ; : m o t l u g n g i o n k i m loai giong nhau t h i : n^|_ = " N O 3 " ^ ("^^O • * • ^ Chat oxi hoa: NH4NO3 '' •'••''^ Lim y: Ba k i m loai kha m a n h ( M g , A l , Zn) tac d u n g v o i axit HNO3, c6 the tao 4. A X I T P H O T P H O R I C : H3PO4 > - - > D a p an C. san p h a m k h u da dugc "giau d i " , do la su tao thanh m u o i NH4NO3: + se -> N2 Nhan xet: D o goc Or va N O 3 deu c6 dien tich 1 - nen k l i i ket h g p v o i cung N h u vay so m o l electron trao doi chua bang nhau —> "chua o n " . 6 day, mot 2N*s ( + lOe 2N^5 Lai gidi: T u y nhien, giai theo p h u o n g phap bao toan electron se nhanh hem. - > nenhucmg = 1 } > N2O i * Bai toan nay cac e m c6 the giai k h i vie't p h u o n g t r i n h phan l i n g . Mg^2 + Se cua m la ''' 0,35 m o l ; txj^^ = • ^ ^ = 0,03 m o l . M g - 2e > AP trong d u n g d i c h HNO3 d u , t h u d u g c 55 gam m u o i n i t r a t k i m loai. Gia t r i gam. Lai gidi: Chatkhu: 2N*5 fI mol V $ y : m = mAi(N03)3 + mNH4N03 = a46.213 + 0,105.80 = 106,38gam. ,f du 13: H o a tan hoan toan 8,4 gam M g bSng d u n g d i c h HNO3 v u a d u , t h u A . 51,8 g a m . C h a t k h u : A l - 3e - > 3.0,46 = 8.0,03 + 10.a03 + 8a "^•1\::^.,iriyP4^-^,'..j>fy'{^ Bao toan n g u y e n to C u : ncu(N03)2 = " c u = 0,04 m o l . a = 0,04.188 = 7,52 gam n N 2 0 = nN2 = 0,03 Baotoanelectron:3nAi = 8nN2O+10nN2+8nNH4N03 2ncu = Sn^o + " N O Z = 0,02.3 + 0.02 = 0,08 m o l - > ncu = 0,04 m o l . ^ 1 va CO the x a y r a qua trinh: T r u o c he't cac e m can t i m so m o l m o i k h i : n ^ o = nN02 ^ ^'^^ ' "••'••H' 44 -36 nN2 1 Chat oxi hoa: D . 15,04. gidi: Bao toan electron: • '''V:/'- 28 - 3 6 nN20 phan u n g xay ra h o a n toan, d u n g dich t h u d u g c c6 cac chat la iorM. j A . K3PO4 va K2HPO4. B. K2HPO4, KH2PO4. IA nuu C.K3P04vaKOH. D. '•Q-; H3PO4 va KH2PO4. 87 Cim nang On luy^n thi dgi hoc 18 chuy6n dg H(5a hpc - Nguygn Van Lai Nhanxet: HSi gidi: V i dv 4= Cho 21,3 gam P2O5 vao d u n g djch c6 chua a g a m N a O H , t h u d u g c > 2H3PO4 P2O5 + 3 H 2 O Cty TNHH MTV D W H Khang Vi?t ^ , ' d u n g d i c h c6 chua 28,4 gam Na2HP04 va b g a m Na3P04. Gia t r i ciia a, b Ian nH3P04 = 2ni^05 = 0'2 m o l . -> l u g t la A . 2 8 ; 8,2. 1 75 ^ t a o 2 m u o i : K H ^ P a v a K 2 H P O 4 . J}mH_ ^ nH3P04 0,2 —> Dap an B. '' ' ~ -M - '•n4i; * ;finri:i ©up jvi v?,?: dvlt 65 Lieu y: Cac e m can chuyen P2OS thanh H3PO4 va xet t i I f m o l n h u tren. V i dv 2 (B-09): Cho 100ml d u n g djch K O H 1,5M vao 200ml d u n g d i c h H3PO4 0,5M, t h u d u g c d u n g d i c h X. Co can d u n g d i c h X t h u d u g c m g a m m u o i C. 2 8 ; 16,4. D . 20 ; 8,2. Lot gidi: 2;i 3 28 4 '"b = - ^ = 0,15 m o l ; nN32HP04 = 7 ^ = 0'2 m o l . „ ^ ,,. digji:.^ • r^^hnvl • * B. 2 0 ; 16,4. P2OS + 3H2O ^^^.^^ nH3P04 = 2np205 = 0 , 3 m o l . > 2H3PO4 Bao toan n g u y e n to P: nH3P04 = nNa2HP04 + nNa3P04 0,3 = 0,2 + — khan. Gia t r i ciia m la A . 15,5 gam. B. 18,2 gam. V,.- ^. "KOH = C. 12,8 g a m . - > b = 16,4 g a m . D . 16,4 gam. So do p h a n u n g : H3PO4 + K O H Lai gidi: . •. m o l ; nH3P04 = 0,10 . > M u o i + H2O. Bao toan k h o i l u g n g : 0,3.98 + a = 28,4 + b ^ . , a = 28 gam D a p an C. V i d u 5: Cho 0,05 m o l P2O5 vao d u n g d i c h chua 0,4 m o l N a O H . Sau k h i phan Cach 1 : NMn xet = • = 1,5 - > tao 2 m u o i : KH2P04va K2HPO4. - u n g xay ra hoan toan, d u n g d i c h t h u d u g c c6 cac chat la "H3PO4 H3PO4 Mol: a -> H3PO4 Mol: b A.Na3P04vaNa2HP04. + KOH a a + 2KOH -> f -> Ta c6: a + b = 0,1; a + 2b = 0,15 b ''^'r •!'> i,t:.,dMte-.. • U: - > a = b = 0,05 m o l . - > m = 0,05.136 + 0,05.174 = 15,5 gam + KOH > Muoi ' ^ ^ l ' "H-^(H3P04) = ^ ' ^ ^ " ^ " l ^ '^H20 = 0,15 m o l . t h u d u g c d u n g d i c h Y c6 chua 14,95 gam h o n h g p m u o i . Gia t r i cua a la B. 1,00. C.0,50. So do p h a n u n g : H3PO4 + K O H • ' LUYEN H C l 7,3% t h u d u g c d u n g d i c h Y va 0,672 l i t k h i H2 (dktc). N o n g dp % ciia H C l t r o n g Y la B.2,18%. C. 1,83%. D . 1,78%. Bai 2: H o a tan hoan toan h 6 n h g p X g o m Fe v a Z n b a n g m g t l u g n g v u a d i i d u n g d i c h H C l C%, t h u d u g c d u n g dich Y. N o n g do p h a n t r a m ciia FeCh va Z n C h t r o n g Y Ian l u g t la 8,05% va 8,63%. Gia t r i cua C la A . 5. O H ^ O = " K O H = 0,2. B.15. C.20. D . 10. Bai 3: H o a t a n h o a n toan 7,8 gam h6n h g p g o m M g va A l b a n g d u n g d i c h H C l > Muoi + H2O. Bao toan k h o i l u g n g : 0,la.98 + 0,2.56 = 14,95 + 0,2.18 a = 0 , 7 5 - > D a p an A . D . 0,80. c6 m u o i axit - > K O H het. ^ 5. B A I T A P 6 N A . 2,43%. gidi: NMn xet: V i Y chua h o n h o p m u o i "H+(H3P04) =°'3^ 0,1 Bai 1: H o a tan het 1,04 g a m hon h o p X g o m M g , Fe bang 40 g a m d u n g d i c h - > D a p an A . V i dv 3: Cho 100ml d u n g d i c h H3PO4 a mol/1 vao 100ml d u n g d i c h K O H 2 M ' ' O H - ( K O H ) =°'2; ,. Luu y: Cac e m can c h u y e n P2O5 thanh H3PO4 va xet t i I f m o l n h u tren. 1J Lai r; —> Dap an C. + H2O m = 0,1.98 + 0,15.56 - 0,15.18 = 15,5 gam A . 0,75. , i Sij, v«x ;^fi!j ut;.a(,j 'MO ul.-' >. nH3PO4 = 2np2O5 = 0,l mol. nH3P04 ~^ "^H3P04 + r " K O H = n ^ + rflH20 ^ > 2H3PO4 ' 4, Cdc em luu y: "OH-(KOH) = N h a n x e t : P2O5 + 3H2O -> ,,f,„: Lai gidi: i]±JaOH. ^ M = 4 > 3 ^ NaOH d u va tao 1 m u o i Na3HP04. D a p an A . Cach 2: A p d u n g bao toan k h o i l u g n g cho so do phan l i n g : H3PO4 D . H3PO4 va NaH2P04. , 0 , >K2HP04+2H20 2b B. Na2HP04, NaH2P04. C. Na3P04 va N a O H . > KH2PO4 + H 2 O ^ d u . Sau p h a n u n g t h u d u g c d u n g djch c6 k h o i l u g n g tang t h e m 7,0 gam so D v o i ban d a u . K h o i l u g n g ciia A l trong h o n h g p ban d a u la A . 5,40 gam. B. 2,70 gam. C. 1,35 gam. D . 4,05 gam. Ca'm nang 6n luygn thi d^i hgc 18 chuyfin d6 H6a hgc - NguySn van Hai Cty TNHH MTV DVVH Khang Vi§t Bai 4: Dot nong 2,80 gam hon hgp X gom Cu, Zn va M g trong khi oxi d u , thu dugc 4,08 gam hon hgp oxit Y. De hoa tan het Y can toi thieu V m l dung djch H2SO4IM. Gia tri ciia V la A. 80. r?S C I B . 60. C. 100. g D. 120. ^e;/Bai 5: Hoa tan hoan toan m gam hon hgp gom Mg, A l , Z n trong dung dich H2SO4 loang, d u thu dugc 0,672 lit khi H2 (dktc) va 3,92 gam hon hgp muoi sunfat. Gia tri cua m la A. 2,48. B. 1,84. C. 1,04. D. 0,98. Bai 6: Cho 6,45 gam hon hgp hai kim loai X va Y (deu c6 hoa tri 2) tac dung voi dung dich H2SO4 loang du. Sau khi phan ung ket thuc, thu dugc 1,12 lit khi (dktc) va 3,2 gam kim loai. Lugng kim loai nay phan ung vua du vai 3,55 gam khi CI2 khi dot nong. Cac kim loai X va Y Ian lugt la A. Zn va M g . B. Zn va Cu. C. M g va Cu. D. M g va Pb. Bai 7: Nung hon hgp X gom 0,28 mol Fe va a mol Cu trong khong khi mgt thoi gian, thu dugc 25,28 gam chat ran Y. Hoa tan hoan toan Y bang dung dich H N O 3 loang (du), thu dugc 1,792 lit khi N O (san pham k h u duy nhat 6 dktc). Gia tri cua a la A. 0,03. B.0,10. C.0,06. ' D. 0,08. ^ Bai 8: Cho 10,8 gam hon hgp gom Fe304 va Cu vao dung dich H2SO4 loang du. Sau khi cac phan ling xay ra hoan toan, t h u dugc d u n g dich X chiia m gam muo'i va chat ran Y. Hoa tan het Y trong dung djch H N O 3 loang, sinh ra 0,448 lit khi N O (san pham khu duy nhat, a dktc). Gia tri ciia m la A. 16,80. B. 18,32. C. 13,92. D . 18,48. Bai 9: Hoa tan hoan toan m gam oxit FexOy bang H2SO4 dac nong, thu dugc dung dich chua 4 gam mgt loai muoi sat duy nhat va 0,224 lit SO2 (dktc). Cong thuc ciia oxit sat va gia tri m Ian lugt la A. FeOval,44. B. Fe304 va 4,64. C. Fe203 va 3,20. D. Fe304 va 2,32. Bai 10: Hoa tan hoan toan 4,4 gam hon hgp X gom Fe, Cu, Ag trong dung dich H2SO4 dac, nong thu dugc 2,24 lit khi SO2 (san pham khu duy nhat, a dktc) va dung dich Y. Co can Y thu dugc m gam muoi khan. Gia tri ciia m la A. 14,2. B. 9,2. C.9,3. ^ D. 14,0. Bai 11: Nhiet phan hoan toan 6,32 gam KMn04, toan bg lugng khi O2 sinh ra cho phan ung het vai 3,6 gam hon hgp X gom Fe va Cu, thu dugc hon hgp Y. Hoa tan het Y trong dung dich H2SO4 dac (du), thu dugc 0,784 lit SO2 (dktc). Thanh phan % khoi lugng Fe trong X la A. 46,67%. B. 53,33%. C. 31,11%. D. 69,89%. Bai 12: Cho 0,015 mol mgt loai chat oleum vao nude, thu dugc 200ml dung dich X. De trung hoa X can dung 200ml dung dich NaOH 0,3M. Oleum do la A. H2SO4.2SO3. B. H2SO4.3SO3. C. H2SO4.4SO3. D. H2SO4.SO3. 4 Bai 13: Cho m gam hon hgp X gom A l , Cu vao dung dich HCl (du), sau khi ket thiic phan ung sinh ra 3,36 lit khi (a dktc). Neu cho m gam hon hgp X tren vao mot lugng d u axit nitric (dac, ngugi), sau khi ket thiic phan ling sinh ra 6,72 lit khi NO2 (san pham khu duy nhat, a dktc). Gia tri cua m la A. 11.5- 'yimn 12,3. C. 10,5. D. 15,6. Bai 14: Hoa tan hoan toan 1,23 gam hon hgp X gom Cu va A l vao dung dich HNO3 dac, nong thu dugc 1,344 lit khi NO2 (san pham k h u duy nhat, 6 dktc) va dung dich Y. Sue khi NH3 (du) vao Y, thu dugc m gam ket tiia. phan tram khoi lugng cua Cu trong hon hgp X va gia tri cua m Ian lugt la A. 21,95% va 2,25. B. 78,05% va 2,25. C. 21,95% va 0,78. D. 78,05% va 0,78. Bai 15: Hoa tan het 20,6 gam hon hgp gom Cu, Fe, Zn trong dung dich HNO3 du, thu dugc dung dich Z (khong chua NH4NO3) va hon hgp khi gom 0,15 mol N O va 0,25 mol NO2. Co can dung dich Z thu dugc khoi lugng muoi khan la " ' * ; ^ * ^ bv.X'-^ ;iosTJ'4io . A. 45,4 gam B. 64,0 gam. C. 51,6 gam. D. 57,8 gam. Bai 16: Hoa tan hoan toan hon hgp gom 9,75 gam Zn va 9,6 gam Cu vao 200ml dung dich X chua dong thoi H N O 3 2M va H2SO4IM. Sau khi phan ung xay ra hoan toan thu dugc khi N O (san pham khu duy nhat) va dung djch Y. Kho'i lugng muo'i CO trong Y la: A. 38,55 gam. B. 50,95 gam. C. 63,35 gam. D. 43,50 gam. Bai 17: Cho 3,2 gam bgt Cu tac dung voi 100ml dung dich X hon hgp gom HNO3 I M va H2SO4 0,1M. Sau khi cac phan ling xay ra hoan toan, sinh ra V lit khi N O (san pham khu duy nhat, 6 dktc). Gia tri ciia V la A. 0,784. B. 0,896. C. 0,672. D. 1,344. Bai 18: Hoa tan hoan toan 0,06 mol FezOs va 0,1 mol FeS2 trong 260ml dung dich H N O 3 4M, san pham thu dugc gom dung dich X va mot chat khi thoat ra. Dung dich X c6 the hoa tan toi da m gam Cu. Biet trong cac qua trinh tren, san pham k h u duy nhat ciia N*"* deu la NO. Gia tri ciia m la A. 9,60. B. 16,64. C. 11,52. D. 14,03 Bai 19: Cho 12,45 gam hon hgp X gom Fe, Mg, Zn vao dung dich H N O 3 du, thu dugc dung dich khong chua NH4NO3 va hon hgp khi gom 0,2 mol N O va 0,1 mol NO2. Dot chay 12,45 gam hon hgp X trong khi clo d u thu dugc bao nhieu gam muo'i clorua? A. 37,3.. B.23,1. )U!Ji C. 30,2. D o j D. 16,0. Bai 20: Cho 3,84 gam Cu phan ung voi 80ml dung dich chua H N O 3 I M va H2SO4 0,5M, sau phan ung thu dugc V m l khi NO (dktc). Gia tri ciia V la A. 896. B. 560. C. 448. D. 336. Q1 Cty TNHH MTV DVVH Khang Vigt dm nang On luygn thi dai hpc 18 chuySn dg H6a hpc - Nguygn van Hai Bai 21: Cho m gam hot Cu kim loai vao 200ml dung djch HNOa 2M, c6 khi NO thoat ra. De hoa tan het hot Cu, can them tiep 100ml dung dich HCl 0,8M, dong thoi Cling c6 khi NO thoat ra. Gia tri cua m la A. 9,60 gam. B. 3,84 gam. C. 10,24 gam. D. 11,52 gam. Bai 22:Hoa tan hoan toan 0,01 mol FeS2 trong 42 gam dung dich HNOs 37,5% theo phan ung: FeS2 + HNO3 > Fe(N03)3 + Isr02 + H 2 S O 4 + H2O Cho V ml dung dich NaOH 2M vao dung dich sau phan ung de thu dugc lugng ket tiia Ion nhat. Gia trj nho nhat cua V la A. 40ml. B. 70ml. C. 80ml. ' D. 100ml. '''^ , H ' .'.•••A 6. H U ' 6 N G D A N - L C J I G I A I :i : 40. 7,3 1 ^ = u,uo mol;; n H , I = = 0,08 m u i 11 H = 0,03 mol. Bail: nHni = Hr 36,5 "2 22,4 Ggi so'mol: Mg = x va Fe = y mol. Ta CO so do: Mg ) H2; Fe " ^ ^ ^ > H2 . nH2 = X + y = 0,03 mol -> nnci = '^^^2 ^ ^'^^ Bai 3: 2A1 Mg i hS^ ^'^^ + > H2O 2H* , 0,16 1 Vay: nH2S04 = 2 " H ^ =0,08 mol ^ ,,;,;.3««B(-^ i ;V n no V^d H2SO4 = ^ ^ , 4 r • <• « = 0,08 lit = 80ml >') n^oJ c: ' Bai 5: H2SO4 N/ifln xet: Bao toan nguyen to'H: Mol: Kim loai + H2SO4 Ta c6: 0,03 > Hi <- mkimiogi ;« , 0,03 '^muiBi':" <,>| )• Muoi + H2 + maxit = mmuoi + s < m^^^ -> m = 0,03.2 + 3,92 - 0,03.98 = 1,04 gam , ^ i^^^--A " "' Bao toan khoi lugng cho so do: Zn + 2HC1 > ZnCh + H2 Xet vai 100 gam dung dich Y: ' ' Ggi so mol: Al = x; Mg = y: ••?« ^ Ta c6: 27x + 24y = 7,8; l,5x + y = 0,4 x = 0,2 mol; y = 0,1 mol. '' ^ ' ' -> mAi = 0,2.27 = 5,4 gam -> Dap an A. ]• ' • = Bai 4: Bao toan kho'i lugng ta c6: mo2 = 4,08 - 2,80 = 1,28 gam • 1 78 Vay: no = - — = 0,08 mol -> n 2- = 0,08 mol. * > ^(S"' • • 16 Khi cho oxit kim loai tac dung voi axit, ion 0^~ trong oxit se ket hgp voi H* trong axit tao thanh H2O theo phuong trinh: :-m.ms nbiji .1 "1 > FeCh + H2 " . Dap an A. mv = 40 + 1,04 - 0,03.2 = 40,98 gam n 02 36 5 ^ C%Hci= ' ^ ^ .100% =1,78% -> Dap an D. 40,98 ^ Bai 2: Phan ung hoa hgc: gam va niz^ciz = 8,63 gam ' -> mH2 = 0'8 gam -> nH2 = 0,4 mol. 02- Ik ^ : Mol: 0,08 t mFeCl2 = ^ Khoi lugng dung dich tang them = Dugc - Mat -> 7 = 7,8 - mH2 XfVay: nHci di, = a08 - 0,06 = a02 mol. Fe + 2HC1 > 2A1C13 + 3H2 > MgCh + H 2 + 6HC1 + 2HC1 Dap an C. ' '*. ; ' ' -> mH20 (Y) = 100 - 8,05 - 8,63 = 83,32 gam. Nhan xet: Khi cho X va Y vao dung dich H2SO4 loang du tha'y con lai 3,2 gam Nhan xet: Lugng H2O trong Y ciing chinh la lugng H2O c6 trong dung dich kim loai khong tan -> c6 mgt kim loai dung sau H trong day dien hoa (gia su do la Y) -> mv = 3,2 gam; mx = 6,45 - 3,2 = 3,25 gam. HCl ban dau. Mat khac, do nong do HCl bang C% C m HCl _ mH20 100-C X + H2SO4 Y Bao toan nguyen to'CI: Dap an D. 2.8,06 2.8,63 83,32.C -1^^1^=36,5(100-0 ' 3 25 ' -> nv = nH-, = 0,05 m o l - » Mx = —— = 65 ^ XlaZn. 83,32.C FeCl2 +2nz, 2npea2 +2nznc.2 =nHCl ^ > XSO4 + H2 ^ ^= ^„ + CI2 —!—> YCI2 i , , . , ,, 32 -> nY= n a 2 = 0'05mol -> Mx= -rhrz = 64: - ^ Y l a C u . -> Dap an B. • 93 Cty TIMHH MTV DVVH Khang Vi$t dm nang 6n luygn thi d^i hpc 18 chuyen 6i H6a hgc - NguySrr Van Hai 1 792 Bao toan n g u y e n to Fe: O'O^ m o l . Bai 7: 11^0= - > X.0,01 = 0,01 = m y - m x = 25,28 - 0.28.56 - 64a = 9,6 - 64a. > So do phan u n g : -j^yr:^],,) '^'^M'''!^^ :P". > Y(2) ¥e,Cn{l) > Fe-^ Cu-^S) Xet su trao d o i eleclion o cac giai doan: U ) ^ ( 3 ) : Fo - . V Cu ; £3 > Cu*^ ^ -2e ( 1 ) ^ . ( 2 ) : O . +4e i ^ 20-^-> nenhjn = 4no2 = •^^^•^^= 1,2 - 8a ^ ; Y | V 8 ^2) ^ ( 3 ) : N * * +3e > N O n e n h j n = 3nNo = 0,24. ir-;:! i • Bao toan electron: 3.0,28 + 2.a = 2b + 0,24 - > a = 0,06; b = 0,36. . • ': ' • . a a Mol: a -> 2FeS04 + CuS04 2a + 8HNO3 ai + 2nH20 ^ 0,2 r i 0,2 <- ' . ? > i n = 14 g a m - > D a p an D . ^ r a r h 2: K h i cho k i m loai tac d u n g v o i axit F12S04 (dac, nong), so m o l go'c sunfat Tu-(l)->(3): Chatkhu: Chat oxi hoa: u^^m * a 2- = " S O T = —— = 0,1 m o l . 22,4 J, ,5,^ • K2Mn04 + M n 0 2 + O 2 .1' ^ ..•..'•^•.^•^.•'^.fr.: <- 0,03 Vay: m = 152.3.0,03 +160.0,03 = 18,48 gam 4 = "SO2 = > Vv*^ >20 2 mol. Nhan xei: O x i t sk FexOy tac d u n g v o i H 2 S O 4 dac, n o n g v^iiiiBi •? ^ > ' .''ft' Bai 12: G Q I cong t h i i c o l e u m la H 2 S 0 4 . n S 0 3 . ,.: Mol: S O 2 t h i o x i t la FeO hoac Fe304. T r o n g 1 m o l FeO hoac Fe304 d e u chua 1 m o l Fe*^ nen d e u c6 k h a nang = npe^Oy = 2 n s o 2 n p e ^ O y = 2.0,01= 0,02 m o l ^*^''' . H2S04.nS03 + n H 2 0 n h u o n g 1 m o l electron, d o do: > Cu^^ > SO2 • D a p an A D a p an D . = 0'°! Cu - 2e + 2e Mat khac: 56x + 64y = 3,6 ^ x = 0,03; y = 0,03 - > % m F e = 46,67%. 0,02 0 224 Fe - 3e O 2 + 4e > Fe^^ (3) •''^^^^ 3nFe+2ncu=4no2 + 2 n s o 2 - ^ 3x + 2y = ai5. > 3Cu(N03)2 + 2 N O + 4 H 2 O <<- 0,03 > Y (2) * ' " ' ' £• Bao toan electron: 232a + 64(a + 0,03) = 10,8 g a m ^ a = 0,03 m o l . Bai 9: n F e 2 ( S 0 4 ) 3 = nS02 i5ao toan k h o i h m n g . 4,4 + 98.0,2 = m + 0,1.64 + 18.0,2 mol: Fe(amol),Cu(bmol)(l) a Chat ran Y la C u d u . Phan u n g hoa tan C u bang H N O 3 loang: 3Cu ' 6 day, cac e m can s u d u n g so do phan u n g : Fe2(S04)3 + FeS04 + 4 H 2 O 4H2SO4 a ' giai. Fe2(S04)3 + C u Mol: , .f f Nhqn xet: Bai nay neu d u a theo p h u o n g trinh phan u n g se rat dai d o n g va kho Phan l i n g hoa hoc: Mol: '" > R2(S04)„ + + 2n}Lb04{dac) : ' "°2^i-i§^'^'°^"'°^'"S°2=^=0'035mol. • f Bai 8: Fe304 + t ^jj-h_L: G Q I cong t h i i c r h u n g cua cac k i m loai la R . Bai 11: 2 K M n 0 4 Bao toan k h o i l u o n g : 64a + 16b = 8,58 - 0,28.56 = 9,6. ,, .. * D o vay: m = 4,4 + 0,1.96 = 14 g a m - > D a p an D . Q u i d o i Y t h a n h : Fe = 0,28 m o l ; C u = a m o l va O = b m o l . -»Dap anC. \' SO4 Cach 2: ' ,, D a p an A . tao m u o i duQ-c t i n h n h u sau: n viy/i t ,i 2 24 • Bao toan electron: 0,84 + 2a = 1,2 - 8a + 0,24 - > a = 0,06 m o l . ' -> Dap an C. m = 72.0,02 = 1,44 g a m 2 24 Bai 10: n s o j = ^ = 2R nenhirong=2ncy = 2a x.0,01 x = 1 -> o x i t sat la FeO. ^ — > le'-* -> nenhirong=3npp =0,84 - > xFe2(S04)3 0,02 Cachl: Bao toan k h o i lucmg: 2FexOy 0,015 H2SO4 Mol: , + 0,015(n+l) 2NaOH » t f u it > (n+l)H2S04 0,015(n+l) ,, > Na2S04 + 2 H 2 O 0,03(n+l) -> 0,03.(n+l) = 0,2.0,3-> n = l ^ O l e u m la H 2 S 0 4 . S 0 3 ^ Bai 13: n^^=^^= , 0,15 m o l ; n^o2 = g = 0,3 m o l . D a p an D . ^^^ ^^ ^J^^ 95 naiKj on luyeii ilii liai iioc 18 chuy6n d6 H6a hpc - Nguyjn VSn H i i Cam Cty T N I I H M T V D V V H Khang Vi§t Nhan xet: K h i cho X + H C l ( d u ) , chi c6 A l phan u n g (Cu d u n g sau H ) . = nHNOa + 2nH2S04 = 0,4 + 2.a2 K h i cho X + H N O 3 (dac, n g u p i ) , chi c6 C u phan u n g ( A l b i t h u d o n g hoa, khongtan). ,. > 2A1C13 + 3H2 Cu + 4HN03 > Cu(N03)2 + n ^ i = ^n^^ = 0,1 m o l . Phuong t r i n h i o n r i i t gpn: 3 Z n + 8H^ + 2NO3 ^ Mol: ^ ^ ncu = 2NO2 + 2H2O m = 0,1.27 + 0,15.64 = 12,3 gam Dap an B. - " N O Z Mol: ^' G o i n c u = X m o l ; n A i = y m o l . Ta c6: 64x + 27y = 1,23 gam. > Cu(N03)2 > Cu(OH)2i Al + ""''^ ^ = «! -(/ufc «6wm OJ;v So do p h a n u n g : Cu > A1(N03)3 > [Cu(NH3)4](OH)2 > Al(OH)34' Bao toan electron: 2ncu + 3nAi = n N o , - > 2x + 3y = 0,06 m o l . , '-^'^ S<,?/' - > X = 0,015 m o l ; y = 0,01 m o l - > % m c u = ^ ^ ^ ^ ^ . 1 0 0 % = 78,05%. 1,23 Vay: m = mAi(OH)3 = 0,01.78 = 0,78 gam - > D a p an D . Chat oxi hoa: N^^ + 3e Bao toan electron: netraodoi ^O; Zn-2e N*^ + l e N03 Bao toan k h o l l u p n g : m = m ^ ^ pg 8H* 0,045 0,12 0,7.62 = 64 gam. ^ \^ / n2n= 0,15 m o l ; n ^ u ^ 0,15 m o l . t r o n g d u n g d j c h la d o 2 axit p h a n l i ra —> g i a i theo p h u a n g t r i n h i o n . ^.^^^ ^^^^ - ^ . , ,. ^ ^ giai theo p h u o n g t r i n h i o n . + 2NO3 -> , . ;A -> 0,03 BailS: Phan l i n g hoa hpc: '.fOU'" •/••••e' 0,1 Mol: .5.', , ; * 'tia \ + y > .. . . ^ 0,2 > 2Fe(Na)3 + 3H2O , 0,24 + + 4H2O .OI^S .a nil qftCl ' ' Q,\2 , < , > = 0,4 m o l ; N O ^ = 0,66 m o l v a SO ] ' = 0,2 m o l . Cac p h a n u n g hoa tan Fe: 3Cu ^j^^; ; •-t-r > Fe(N03)3 + 5 N O + 2H2SO4 + 2H2O 0,8 0,06 , ^ 0,03 , FeS2 + 8HNO3 Mol: - > 3Cu2- + 2NO V = a03.22,4 = 0,672 l i t D a p a n C. Fe3* = 0,22 m o l ; ' Nhan xet: K h i axit H N O s c6 mat d o n g t h o i v o i axit H2SO4 loang t h i l u g n g H * 96 , < > Fe203 + 6HNO3 - > Dap an B. Bai 16: " n' r > D u n g d i c h X g o m cac i o n : " ^ M O - ^ "^^'^ , 1 , : ^ =.««„,m * : I n m ££.0 - ,,->?i , ^ + M o l : 0,1 > NO2. N O 3 n a m trong m u o i dugc t i n h n h u sau: ' '' o) O ' ?; = 0,05 m o l . 3Cu — 3njyjQ + 1. n[sjQ2 = 0,7 m o l . (muoi) = n„ trao doi = 0,7 m o l . + 4H2O [ = nHN03 + 2nH2S04 = 0 4 + 2.0,01 = 0,12 m o l , ...^ Trong X: \ ^^J. ^ Q^Q^ ^^^^ .i.,££.Kl,0 = V « NO3 SO| ^Zn*2 D e n day, cac e m can l u u y la k h i cho k i m loai tac d u n g v o i HNOs, so m o l goc n 0,1 H* t r o n g d u n g d j c h la d o 2 axit p h a n l i r a ("f), tf) k h i nen can ap d u n g d i n h luat bao toan electron. > Fe«; -> 0,4 > 3Cu2* + 2 N O 3 2 nr^, = f; Nhan xet: D a y la bai toan c6 nhieu chat k h u (3 k i m loai) va tgo ra 2 san p h a m > Cu"2. p e - 3 e -> 0,15 + 2NO; Dap anB. Mol: , Cu-2e 8H* oii^d.' ai Phuang trinh ion riit ggn: t Cdc em can lieu y: Cu(OH)2 tan t r o n g d u n g d i c h N H s d u . Chatkhu: + -> N/jflM xet: K h i axit HNO3 c6 mat d o n g t h a i v o i axit H2SO4 loang t h i lug-ng '^^ ' ^^• ^ •''•^ BailS: 0,4 _> K h o i l u p n g m u o i t r o n g Y = 9,75 + 9,6 + 0,2.96 + 0,2.62 = 50,95 gam. 3nAi = 2 nH2 ^ n ^ i = 0,1 m o l ; 2ncu = nN02 ^ ncu = 0,15 m o l . ' D a p an B. ^ ,p ^ - 3 i:;n< nin^'' ' , "yi.H~'. /rai J I C L . '..i/iih ^, . + 4H2O, >2>Zn^* + 2 N O H"^ t h a m gia p h a n u n g het. ' ' " m = 0,1.27 + 0,15.64 = 12,3 g a m 0,15 3Cu = 0,15 m o l . m-Uoi + Cach 2: A p d u n g bao toan electron: + _ = 0,4 m o l ; n^ o_ = 0,2 m o l . NO3 SO4 ••'..'Hi Cach 1 : Giai theo p h u o n g t r i n h 2A1 + 6HC1 Trong X: j = 0,S m o l ^, 8H* + 2 N O ; 0,15 <- Cu + 2Fe3* IQ <, ; )• 3Cu2^ + 2 N O + 4H2O i V j / ' V 0,4 > Cu^* + 2Fe2* M o l : 0,11 <-0,22 m c u = 0,26.64 = 16,64 g a m - > DapanB.^^^ ^,.rid , ao:u.,.j, i:;^! ::M;;t: ' ^ .•''i^*''.^'^MM'^^' - '.''^'^ o-iH'W]JM-!<>.:/ 97 Cty TNHH M T V DVVH Khang Vijt Cim nang On luyQn thi dai hgc 18 chuySn dg H6a hgc - Mydygn van Hi\ Bai 19: N/ifln A:e^: K h i X t a c d u n g v o i m o i c h a t o x i h o a HNO3 v a C h , cac k i m l o a i d e u T i X H CHAT CUA CAC HIttROXIT duQic d u a l e n m u c o x i h o a c a o n h a t —> So m o l e l e c t r o n t r a o d o i t r o n g hai ^ HIDROXIT KIM LOAI KIEM, KIEM THO t r u o n g h g p la b a n g n h a u . Chat o x i h o a : N^s + 3e Chat o x i h o a : C h + 2e N*^ + i e • — - > NOz. ^ NO; + • - > D a p an A . '^^^^^ • Bai20:ncu = a06mol. CO2 ^ , d u n g d i c h . K h i do, l u g n g '^^ Neu: trong ^ >• 3Cu2* + 2 N O + 4 H 2 O + 8H^ + 2 N O ; -> - > V = 0,04.22,4.1000 = 896ml ^ + . - ,v •> ^ i - o u = ' T a c o : n „ + = 0 , 2 . 2 + 0,1.0,8 = 0,48 m o l ; n , , ^ _ = 0,2.2 = 0,4 m o l . H NO3 a ••• T ( : 3 n > 3Cu2^ + 2 N 0 + 4 H 2 O , A U ^ + 2NaOH 8 ji^f^ 2NaCl CO2 + 2H2O + H2O > NaA102 , ^ + 2H2O .JBnocfofe) ;o: / . X ) :(«:>* r i m i i ^ ' > 2 N a O H + C h t + HzT ^^ >' tn r..... .x^ry •\r,m<'f-.0 - > CaCOsi + Ca(OH)2 Neu: j^Y/bib^-,: ,I'-VT; '•('..!} '• * = 2nca(OH)2 = a : Ca(HC03)2 ^>-n^ jiSt-S) t 1< a < 2 : CaCOs + Ca(HC03)2 a >2 y.; +H2O > Ca(HC03)2 nco2 ,()r/f?Py 4. ^r\>=,'4 .^^^j, ^OD" > 2NaA102 a < 1 8 D a p an D . , * ' M ' ! - ! ^ = 0,18 m o l - > m = 0,18.64 = 11,52 g a m :Na2C03 D i e u che T i n h so m o l : n 3 0 48 =— >2 > Fe(OH)3 i + 3 N a N 0 3 Hidroxit kim loai kiem tho - n e n t i n h so m o l C u t h e o H ^ . : rt-'. : NaHC03 Tac d u n g v o i h g p chat l u a n g t i n h 2 C O 2 + Ca(OH)2 Ta c6: ncu j rififi^J rmt fcrb rr!t;;< 1< a < 2 : N a H C O s + NazCOg Al(OH)3 + N a O H • ;X gncn N h | n xet: Theo b a i , ta coi C u t a n v u a d u t r o n g h o n h g p HNO3 v a H C l . * < ;v Tac d u n g v o i d u n g d i c h m u o i : AI2O3 1 0,04 D a p an A . + + Nhan thay ^ ^ Fe(N03)3 + 3 N a O H Bai 21: 3 C u + 8H* + 2N03- = a nco2 la d o 2 a x i t p h a n l i ra nen c a n g i a i t h e o p h u o n g 0,06 < - 0 , 1 6 ^ 0,04 ; , >K o»U.; Mt^O ^ i O . ^ ' . u o g ;< r r . > NaHCOs • a < 1 Ta c6: n ^ + = nHNOg + 2nH2S04 = 0,U m o l ; n^^_ = n^NO^ = 0,08 m o l . Mol: >Na2C03 + H 2 O + NaOH > udl uimrDi trinh ion. 3Cu ^ ''•'.yt ' . ' N^flM xef: Svr c6 m a t a x i t H2SO4 l o a n g se " d o n g g o p " t h e m l u ^ n g i. •'••^'•^g.^jjj Tac d u n g v o l o x i t axit C02 + 2 N a O H = mx + m Q j = 12,45 + 0,35.71 = 37,3 g a m . ' ' ' ; j;,,^, Hidroxit kim loai kiem > 2C\- mmu6i ^^^^ ^^^ j,.Uthuyet Bao t o a n e l e c t r o n : 3 n N o + n N 0 2 = 2 n c i 2 ^ 2 n c i 2 = 0,2.3 + 0,1 = 0,7 m o l - ) • nci2 = 0,35 m o l - > ^ ^ M£t,0 ^ ^ K ' , :CaC03 Bai 22: b . V i d v mau „ , 42.0,375 „ , Ta co: nHNOg = — — — = 0,25 m o l . •'« V i dvi 1: Cho cac chat: N a H C 0 3 , CO, A l ( O H ) 3 , Fe(OH)3, H F , CI2, N H 4 C I . So chat DO FeS2 + I 8 H N O 3 Mol: > Fe(N03)3 + 15N02 + 2H2SO4 + 7 H 2 O 0 , 0 1 ^ 0,18 0,01 D u n g d i c h sau p h a n u n g gom: Fe(N03)3 = 0,01 m o l ; HNa 7 ' ' ' 08 , , B.5. ' ^'-^^ '1.0 C.3. Lbi dm. , : D.6. giai: Nhan xet: Day la cau hoi tong hqip, lien quan den k i e n thuc l o p 10 (HF, CI2), l o p 11 (CO, N H 4 C I ) va l o p 12 (NaHC03, Al(OH)3, Fe(OH)3). " N a O H ^ 3"Fe(N03)3+r>HN03 + 2 n H 2 S 0 4 = 0,14 m o l = 0,07 l i t = 70ml ' = 0,07 m o l ; H2SO4 = 0,02 m o l . tac dung dugc voi dung dich N a O H loang 6 dieu kif n thuong Ik A . 4. 0,02 De t h u d u g c k e t t i i a I o n nha't: VNaOH ^ D a p an B. Cac chalt xay ra phan ung: NaHC03 + NaOH •, > Na2C03 + H 2 O .^^j^j^ ^ i , . v ( ...^