Tài liệu Kỹ thuật mới giải nhanh bài tập hóa học tập 2-hóa học vô cơ - cù thanh toàn part 2

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FeS + lOH* + 9NO3- -> Fe'" + + QNO^t + 5H,0 a 9a FeCOj + 4H* + NOf Fe'" + CO.t + N O j t + 2H.O b b b a) Goi a, b la s6' mol m6i mu6'i trong h6n hap. 3FeO + NO3 + lOH^ ^ - 3Fe^^ + NO + 5H2O ; 2. Cac thi du minh hoa: ffii du I: Hoa tan Au bang nu6c cucmg toan thi san ph^m khijf 1^ NO; hoa tan Ag trong dung djch HNO3 dac thi san p h ^ khu la NO2. 44b + (9a + b).46 ^ x . v , „ , = ^ ^ ^ L _ i _ = , 2 . 8 0 5 ^ b = 2,877a • Chon a = l , b = 2,877(mol) Tinh duoc; %FeS = 20,87%; %FeC03 = 79,13%. b) Phan ling dime hoa NO2: 2NO3 n^: ^ ' np„ :'-v : 2x A. 1:2. ' ' \ ' 11,877a-2x x, PTHH xay ra: -''^.v••.•;',.'::>;:;| . • •, 'C02 = b nco-, (ll,877a-2x + b + x).2 » Au + HNO3 + 3HC1-» AuClj + NO + 2H2O -o:... ^ Ag + 2 H N 0 3 - > A g N 0 3 + N02 + H20 ,!sft _ (11,877a - 2 x ) . 46 + 44b + 92x _ ^•^•^^"2- C. 1 : 1. D. 1 : 3. (Trich de thi tuyen sink DH khoi B) Hu&ng ddn giai :v7;;;v^^tii>p,.ett•.^^^^^ ' '^xi'' B. 3 : 1 . on. r N304.^,,, • 11,877a s6' mol NOj bang s6'mol NO thi ti 16 s6'mol Ag va Au tuong ling la ,; NO + 2H2O + NO3 -> 3Ag^ 3Ag + n^o • Di = nN02 n^g i, DapandiinglaC. ~, = n^u Thi du 2: Cho 3,2 gam Cu tac dung vdi 100 ml dung djch h6n hop HNO3 0,8M va H2SO4 0,2M, san phdm khiJt duy nha't la khi NO. S6' gam mu6'i khan thu Thay b = 2,877 ->• - = 3,762 ^ x = 3,762a dugc la S6'mol NO2 bj dime hoa la 2x. i ^ ,, A. 7,90 f n B. 8,84 C. 5,64 , , %N02 bi dime hoa = ^ ^ ^ ^ ^ = 63,35%. ^ • 11,877a • i , ... D. 10,08 ' ' ' ' Cfrich de thi du bi dai hoc) Hu&ng ddn gidi • - Phan iJng dime hoa didn ra khi lam lanh va khi do mau ciia h6n hop nhat din. De giai hdi nay hat hudc phdi sU dung phucfng trinh ion riit gon. Muoi trong - can bang dich ve phai khi ha nhiet d6 => Phan ung dime hoa la toa nhiet. dung dich thu duc/c gom c6 muoi sunfat vd muoi nitrat (di/a vdo sU hdo toan nguyen to nita ta xdc dinh duac so mol ion nitrat tao mud'i) Dang 5: Cdc chat tdc dung voi hon harp axit (HNO^, HCl, H^SO^) 1. L i thuyet van dung va phiromg phap giai: - 5 « Au + HNO3 + 3 HCl > AuClj + 4NO + 2H2O Pt + 4HNO3 + 12Ha > PtCl4 + 4 N 0 + 8H2O Ag + HNO3 + 3 H a > 3AgCl 4 + NO + 2H2O |:>n^^ =0,08+ 0,02.2 = 0,12 (mol) n 8H^ + 2NO3->3Cu^^+2NO + 4H20 NO3 -0,08(mol); n =0,02(mol) 3Cu + 8H" + 2NO3 - . 3 C u 2 U 2 N O + 4H20 0,045<-0,12->0,03 Phuong phap giai bai tap dang nay la sir dung phuong trinh ion riit gon. -finiM M6t s6' phuong trinh ion thu gon thiromg gap: 3Cu + = 0,05 (mol); nHN03 = 0,1.0,8 = 0,08(mol) nH2S04-0,1.0,2 = 0,02(mol) Nu6c cucyng thuy (con goi la nu6c cucmg toan) \h dung dich chiJa h6n hop HNO, va HCl theo ti Id 1 : 3 vd s6' mol. Nu6c cuomg thuy hoa tan dirge Au, Pt (khong hoa tan dugc kirn loai Ag vi tao ra kd't tiia AgCl can tro phan ling): ,, - S6 mol cac chat: nc^ = ^ Vi 0,12 0,05 8 Suy ra: n , 0,08 ->0,045 ' p ' n6n H* phan ling h6't. (tao mu6'i) = 0,08 - 0,03 = 0,02 (mol) NO3 , 181 180 :^n^^ Vay khd'i lucmg mu6'i khan thu duoc: "'Cu(N03)2 CUSO4 ^^2+ 5Q2- ^Q- I Trong X c o : Cu(4x mol); Ag(x mol). j.,,,,; => m = 0,045.64 + 0,05.62 + 0,02.96 = 7,90(g) Dap an diing la A. ^ . Vii B. 19,76 gam. ; C. 19,20 gam. D. 22,56 gam. , (Trich de tuyen sink Dai hoc khoiA) Hu&ng ddn gidi - Tac6: 4x.64+ x. 108 = 1,82 =:>364x = 1,82 => x =0,005 3Cu • Thi du 3: Cho 7,68 gam Cu vao 200 ml dung dich g6m HNO3 0,6M va HjSO, 0,5M. Sau khi cac phan utig xay ra hoan loan (san ph^m khu duy nha't la NO), c6 can c^n than toan b6 dung dich sau phan ting thi kh6'i luong mu6'i khan thu duoc la A. 20,16 gam. =0,015.2 + 0,06 = 0,09(mol); n^^_ =0,06(mol) + + 4H20 8H^ + 2NO3 ^ 33Cu^^+2NO Cu2+ Phan ling: 0,02 ^ 0,16/3->• 0,04/3 ^ Con: 0 3Ag Banddu: 0,005 0,11/3 0,14/3 + 4H+ + NOJ ^ 0,11/3 0,04/3 3Ag* 0,14/3 , + NO + 2H2O . 0,005/3 Phan ling: 0,005 0,04 0,005 0,045 \ „ ',7 • ---^ = 0,015(mol) Phirong phap giai bai nay: Tirong tir cau 2 cr trdn. S6'molcaccha't: ncu = — = 0,12(mol); 64 "HNOJ = 0,2.0,6 = 0,12(mol);nH2S04 - 0 , 2 . 0 , 5 - 0 , l ( m o l ) =>n^+ =0,12 + 0,1.2 = 0,32(mol) n _ =0,12(mol);n 3Cu + , =0,l(mol) 8H^ + 2NO3 ^ 3 C u 2 ^ + 2 N O + 4H20 Phan ling: 0,12 -> 0,32 Con: 0 0 -> 0,08 -> 0,12 0,04 Kh6'i luong mud'i thu duoc: m = m 9, + m • " ^ • Cu^+ =^ m = 7,68 + 0,04.62 + 0,1.96 = 19,76 =:0,1M =>pH = z = l 0,12 , ,+m NOJ(sau) , SO|" (gam) Dap an dung la B. Thi du 4: Cho 1,82 gam h6n hop b6t X g6m Cu va Ag (ti 16 s6' mol tuong ting 4:1) v^o 30 ml dung dich g6m H2SO4 0,5M v^ HNO3 2 M , sau khi cdc phan ling xay ra hoan toan, thu duoc a mol khi NO (san ph^m khu duy nha't cua Dap an dung la A. Thi du 5: Cho 3,2 gam b6t Cu tac dung vdi 100 ml dung dich h6n hop g6m HNO3 0,8M va H2SO4 0,2M. Sau khi cac phan ling xay ra hokn to^n, sinh ra V lit khi NO (san ph^m khu duy nha't, b dktc). Gia tri cua V la A. 0,448. B. 0,792. C. 0,672. D. 0,746. (Trich De thi tuyen sinh DH - CD khdi A) Hu&ng ddn gidi N*'). Tr6n a mol NO trdn vdi 0,1 mol O, thu duoc h6n hop khi Y. Cho toan De gidi nhanh hdi nay can thiet phdi su dung phucfng trinh ion rut gon. Chu b6 Y tac dung vdi H3O, thu duoc 150 ml dung dich c6 pH = z. Gia tri ciia z la tinh theo ion phan icng hit (dua vao ti le so mol cac ion vdi he sd'ciia cac ion do A. 1. trong phucmg trinh ion riit gpn, ion cd ti le nhd nhdt se Id ion phan vcng het B. 3. C. 2. D.4. (Trich de tuyen sink Dai hgc khoiB) Hu&ng ddn gidi S6'mol cdc chat: n^^^^^ =0,015(mol); nHNOj =0,06 (mol) thdng thudng trong nhieu hai tap Id ion H*). 32 86' mol cac chat: ncu = — = 0,05 (mol); v< • '• 183 nHNo 3 = 0,1.0,8 = 0,08 (mol); n ^^^so4 => = nHN0 3 = 0,1 0,2 = 0,02 (mol) + 2n H 2 S O 4 = 0-08 + 2.0,02 = 0,12 (mol) " N O T " " " N O 3 =0,08 (mol) 3Cu Band^u: 0,05 Phan ixng: 0,045 <- 0,03 <- 0,12 (mol) + 2NO; + 8H* BandSu 0,06 phan ling: 0,06-> 0,04 -> Suyra: 3Cu'* + 2NO + 4H2O + 8H^ - 0,08 n6n: 0,03 (mol) 0,16 = 0,04 (mol) nN0(2)- nNO(2) =2. nNO(i) =>V2 = 2V|. ,,,,„,, it) t : 1 E)ap an dung la B. , ,2 Thidu 6: Thuc hidn hai thf nghidm: ' -' Hu&ng dan giai 1) Cho 3,84 gam Cu phan ling vdi 80 ml dung djch H N O 3 I M thoat ra V, lit khi NO. 2) Cho 3,84 gam Cu phan ung vdi 80 ml dung dich chiia H N O 3 I M va 0,5M thoat ra lit khi NO. A. V 2 = 1 , 5 V , . B. V , = 2V,. C . V , = 2,5V,. D. V2 = V,. Au + 0,02 - > ''^'-'^f . > ' . i ^^ Ilu&ngdangiai giai nhanh bai nay cin thid't phai sir dung phuong trinh ion riit gon. Chii y tinh theo ion phan ung hd't (dua vao ti 16 s6' mol cac ion vdi hd s6' ciia cac ion do trong phuong trinh ion rut gon, ion c6 ti 16 nho nha't se la ion phan ting h6't, thdng thuomg trong nhilu bki tap la ion H"^). + HNO3 3Ha > AICI3 + NOt + 2H2O 0,02 (mol) 0,06 (mol) Vay HHC, = 0,06 (mol); n^o = 0,02 (mol). , Dap an diing la C . Cku 1: Hoa tan h6't 5,36 gam h6n hop FeO, Fe^Oj va Fe304 trong dung djch chira 0,03 mol HNO3 va 0,18 mol H2SO4, ket thiic phan ling thu duoc dung djch X va 0,01 mol khi NO. Cho 0,02 mol b6t Cu tac dung h6't vdi ^ dung djch X, thu duoc dung djch Y. Kh6'i luong Fe2(S04)3 chiia trong dung djch Y la (Bi6't NO la san ph^m khir duy nha't). A. 20 gam B. 10 gam ' 64 T h i n g h i d m l : nHNOj = 0,080.1 = 0,08 (mol) i :ac bai tap tir luyen: (Trich Be thi tuyen sink DH - CD khoi B) = 0,06(mol). " Nuoc cucmg toan la h6n hop axit H C l va HNO3, c6 l^ha nang hoa tan vang: H2SO4 Bid't NO la san ph^m khu duy nha't, cac thd' ti'ch khi do a cung di^u kidn. Quan he giira V, v^ V , la * 0,04 (mol) 0,16 'rhfdii 7: Khi hoa tan hoan toan 0,02 mol Au bang nu6c cucmg toan thi s6 mol HCl phan irng va s6' mol NO (san phdm khir duy nha't) tao thanh \ir\t la A. 0,06 va 0,01. B. 0,03 va 0,01. C. 0,06 va 0,02. D. 0,03 va 0,02. (Trich De thi tuyen sink DH - CD khoi B) = 0,03. 22,4 = 0,672 (lit). S6'mol kim loai Cu: nc^ = -> 3Cu-* + 2 N O + 4H2O E)6'i vdi chat khi, trong cung didu kien, ti 16 v^ s6' mol cung bang ti 16 v6 th^ tich 0,12 (mol) 0,12.2 - = 0,03(mol) => nNo = 8 Dap an diing la C . 3Cu + 2NO3 0,06.2 PTPir: 0,08 pXPlT: C. 24 gam D. 5 gam Khi giai hai nay can chii y mot sd'noi dung sau: Quy dot Fe,04 ->FeiOj.FeO PTPU": 3Cu + 8HNO3 Banddu 0,06 Phaniing: 0,03 <-0,08 Suyra: n. ' N O ( I ) 3Cu(N03)2 + 2 N 0 + 4H2O 0,08 (mol) 0,02(mol) HHNOJ • = 0,02(mol) = 0,08 mol; n Ion '' NO3 / H"^ CO tinh oxi hoa ntanh Hon Fe^* Dung dich Y hit ion NO3 (lii khi? thanh NO) nen trong Y c6 FesiSO,),, FeSO, va H2SO4. 0,08.2 — 8 Thf nghidm 2: > , Hu&ng ddn giai S6'mol cac ion: n H2SO4 = 0,08.0,5 = 0,04 (mol) =0,03 (mol); n , =0,03 + 0,18.2 = 0,39 (mol) NO3 Vi Fe304 = Fe203. FeO n6n c6 th6' coi h6n hop FeO, Fe203 va Fe304 la h6n hop " N O J =0,08 mol; n^^ = 0,08 + 0,04.2 = 0,16 (mol) 184 FeO (x mol) va Fe203 (y mol) 185 0,004 3 F e O + N O 3 + 10H+ -> SFe^^ + N O + 5 H 2 O PTPLT: X ^ FcjOj+eH^ y X ^ x/3 -> lOx/3 —> ' x/3 —> .^Jsn) 2 y i/' Dap an dung la A. 72x + 160y = 5,36 Theo bai ra, ta c6: 2Fe^++3H20 6y 0,4 ^ = 0,03; y = 0,02 0,5M (loang), thu duoc V 1ft khf N O (san ph^m k h u duy nha't a dktc). a) Tfnh V , biet r i n g phan ling xay ra hoan toan. Trong dung dich X c6: 0,07 mol Fe^*; 0,02 mol N O J va 0,27 m o l H * b) Gia si^ khi phan utig xay ra hoan toan, lucmg Cu k i m loai khong tan h6't thi ' lucfng mu6'i khan thu ducKc la bao nhi6u gam? Hu&ng ddn gidi => 1/2 dung djch X c6: 0,035 mol Fe^^ 0,01 mol N O J v^ 0,135 mol PTPU: 3Cu + 2NO3 + Ban ddu: 0,02 0,01 Phan ung: 0,015 < - 0 , 0 1 Con: 0,005 ; Banddu: 0,005 Phanirng: 0,005 -> 0,01 Con: 0 3Cu + 8 H * + 2 N O ; ->Cu2+ + 2Fe2+ 0,035 - , lan xet: Nha (trong do n^ci : nHN03 • "H2SO4 = 1 : 5 : 1 ) , sau khi phan ihig xay ra hoan toan tha'y thoat ra 22,4 m l khf bj hoa nfiu trong khdng k h i (san phim nha't, do a dktc). Gia t n p H cua dung dich X la ,; 1 j,\ C. 3,0. 3Cu + 8 H ^ + 2 N O 3 -> r--v 0-06 .22,4 = 1,344 (1ft). b) K h i Cu k i m loai kh6ng tan het (tiic a > 0,09) thi trong dung djch sau phan ting r hoi so vdi hidro bang 23,5. Tfnh kh6'i luong m6i k i m loai trong h6n hop X v^ kh6'i luong m 6 i mu6'i trong dung djch Z. Hu&ng ddn gidi MT 2x = 23,5. 2 = 47 ^ MNO = 30 < 47 < Mo - D la SO, ( M = 64). Suy ra s6' mol N O = 0,2 mol va SO, = 0,2 mol phan ling hS't ( N O 3 con du). Ta CO cac qua trinh sau: Al 8x = 0,004 . k h u duy T r u f n i g hofp 2: Cu d u hoSc viira d i i , H^ he't (a > 0,09) 8,96 1ft (dktc) h6n hop T g6m N O va khf D kh6ng mau. H6n hop T c6 ty kh6'i 3C\x^* + 2 N 0 + 4 H 2 O -> • nN O = ^ ( m o l ) ml dung djch Y chiia H 2 S O 4 C^, va H N O , 2 M dun n6ng tao ra dung djch Z v^ n ^ ^ = x + 5x + 2x = 8 x ( m o l ) ; n ^ ^ _ = 5 x ( m o l ) 186 -> Bai toan c6 2 trucfng hop xay ra: 2 C a u 4: Hoa tan hoan toan 18,2 gam h6n hop X g6m 2 k i m loai A l va Cu trong 100 Goi nHci = x ( m o l ) = > n H N 0 3 = 5 x ( m o l ) ; nH2S04 = x ( m o l ) Theo PTHH: 2x = 0,001 0,12 ->-m^„ai = 0 , 0 9 . 6 4 + 0,06.62+ 0,06.96 = 15,24(gam). K h f bi hod nau trong kh6ng khf la NO: n^o =0,001 (mol) V I 8x / 8 < 5x / 2 =^ 8 < g6m c6: s6' mol Cu^"" = 0,09; s6' mol N O J = 0 , 0 6 ; s6' mol SO4" = 0,06 D.3,4. Hu&ng ddn gidi 8x 0,24 (mol) 0,12 = 14,933a (1ft). C&u 2: Cho Cu (du) tac dung vdri 400 m l dung djch X chiia h6n harp H C l , HNO3, B. 1,4 , 3Cu^^ + 2 N O + 4 H 2 O T r u e m g h o p 1: Cu he't, H* du (tu-c la a < 0,09) =^"iFe2(so4)3 = 0,025.400 = lO(gam) PTHH: 0,24 Ban dSu:.; 0,025 (mol) A. 2,0. V Phirong trinh phan ung: =^nFe2(so4)3=0'025(mol) H2SO4 ,: => S6' mol H^ = 0,24; s6' mol N O J = 0,12; s6' mol SO^" = 0,06 \:;r;, 0,095 Cu + 2Fe^+- ^ Theo bai ra ta c6: n^^Qy^ = 0,12(mol);np,2so4 = 0 , 0 6 ( m o l ) 0,135 0 •i'fl^ a) Tfnh VNO- 8 H + ^ 3 C u 2 + + 2 N O + 4H20 • 0,04 ^ ' ' ' - 3e ^ A l ' \i s6' mol A l = x va s6' mol Cu = y .... • . • • . 187 Cu - 2e^Cu-* N O , + 4Fr + 3c - » N O + 2 H 2 O * SO^"+4H^+2e^S02+2H2O Dang 6: Bdi tap ve muoi nitrat v• *! Vay ta CO he phuong tnnh: Cac mu6'i nitrat d l bj nhiet phan huy, giai phong o x i . V i vay, o nhiet do cao cac niu6'i nitrat c6 tinh o x i hod manh. Cac mu6'i nitrat ciia k i m loai boat d6ng manh (kali, natri, . . . ) bj phan huy tao ra n ^ i i ,> ,i; • . ^ ,, IJ',.^,;;'^v, tnu6'i nitrit va O2. 2KNO3 — 2Cu(N03)2 — . Cu con lai ti^'p tuc cho vao b gam dung djch H N O 3 6 8 % thu duoc V , lit k h i . Sau hai lin phan ling khd'i luong Cu con lai la 1,28 gam, bid't V , + V , , = 1,12 l i t . Pha 68%, sau do pha loang dung djch thu duoc bang H j O thi duoc dung djch X . Cho 5,76 gam Cu vao dung djch X thu duoc V 3 l i t k h i N O . Bi6't cac phan ling xay ra 2AgN03 — * Hu&ng ddn gidi * * Theo b^i ra s5 m o l ddng la: nc„ = 1,92 : 64 = 0,03 (mol) NO3 X Taco: :',.•„w'-;'.'-';.!*'^ y ' ^''>V K n . :. ,:i 2x + y = 0,06 Phuong phap .sir dung phuong trinh ion thu gon. + Phuong phap bao toan electron. O^UM Mot sd P T H H chii y: Fe(N03)3 + A g 2Fe203 + 8NO2 + O , , n .: •,. = ; ^ 4FeCi3 + 5Fe(N03)3 + 3 N O + 6H2O i ling xay ra hoan toan, thu duoc dung djch X gdm hai mudi va chat rSn Y gom (2) hai kim loai. Hai mudi trong X va hai kim loai trong Y \in lugt la A. Fe(N03)2; Fe(N03)3 va C u ; A g B. Cu(N03)2; AgN03 va Cu; A g C. Cu(N03)2; Fe(N03)2 va C u ; Fe D . Cu(N03)2; Fe(N03)2 va Ag; Cu. pha loang dung djch thi dung djch thu duoc chiia H2SO4 loang va HNO3 loang. K h i cho 0,09 mol Cu vao dung djch X thi khi thoat ra la N O theo P T H H : 3Cu + 8 H " + 2 N 0 3 ' - > 3 C u - ' + 2 N O + 4 H 2 0 ( * ) ' ' (Trich de tuyen sink Dai hoc nam 2013 • Khd'i A) • Hu&ng ddn gidi •'• i Phuong trinh hoa hoc (theo thir tir): 2 Tiif (*) ta tha'y Cu phan ufng he't nen npjo = j "cu = 0,06 m o l . " ' "'W Thi du I: Cho bdt Fe vao dung djch gom AgN03 va C u ( N 0 3 ) 2 . Sau khi cac phan (0 K h i lay 2a gam dung djch H2SO4 9 8 % tron voi 3b gam dung dich H N O 3 6 8 % r6i V 3 = 0 , 0 6 . 2 2 . 4 = 1,344 lit. ' ' • C a c thi d u m i n h h o a : Giai U ta duoc: x = 0,01; y = 0,04. vay 188 + 4H2O • > , > + 9Fe(N03)2 + 12Ha y x + y = 0,05 3Cu^^ + 2 N 0 t FeCK + 3AgN03 (du) - > Fe(N03)3 + A g + 2AgCi + 2 H * + l e - > N 0 2 + 2H20 2y i SO2 + 2H2O ; 2Ag + 2NO2 t + b 2 t Phuong phap de giai nhanh cac bai tap dang nay: 4Fe(N03)2 0,06 2x 2CuO + 4 N O 2 T + O 2 t Trong moi truong axit, mudi nitrat c6 ti'nh oxi hoa manh: A g N 0 3 + Fe(N03)2 2e ^ 3Cu + 8 H ^ + 2NO3 hoan toan, cac khi do 6 dktc. Ti'nh V , . .:v».v-''v,.'f'^ 4x ^ ling, NO2 va O2: tr6n h6n hop g o m 2a gam dung djch H2SO4 9 8 % va 3b gam dung djch H N O , SO4- + 4H^ + 2e ' Mudi nitrat cua bac, vang, thuy ngan,... bj phan huy tao thanh k i m loai tuong C a u 5: Cho 3,2 gam Cu vao a gam dung djch H2SO4 9 8 % thu duoc V , l i t k h i , luong 0,03 2KNO2 + 0 2 ! loai tuong ufng, NO2 va O j . Luong AI2 { S O 4 )^ - ^ - 3 4 2 = 34,2gani, luong CUSO4 = 0,2. 160 = 32 gam. Cu-^ + ^ Mu6'i nitrat ciia magic, kem, sSt, chi, d6ng ... bj phan huy tao ra oxit ciia k i m V i NO3 phan ilmg tao N O bang luomg NO3 trong Y ntn dung djch Z kh6ng c6 -> i s-i * Vay khdi luong cac k i m loai trong h6n hop ddu 1^: tricu = 12,8g; rriAi = 5,4g. Cu u L i thuyet v a n d u n g v a p h w m g p h a p giai: Giai hd phirong trinh cho X = y = 0,2. N03vadod6chic6AF\Cu-\SO^". > » ' s J T6ng s6' mol e nhuomg = 3x + 2y = T6ng s6' mol e thu = 0,6 + 0,4 = 1 f27x + 64y = 18,2 < [3x + 2y = 1 ; Fe + 2 A g N 0 3 F e ( N O , ) 2 + 2 A g 4 '' '"^^ Fe + C u ( N 0 3 ) 2 - > Fe(N03)2 + C u i . . Ci. . ' ' i i : 189 Hu&ngddngidi => 2 kim loai la Ag, Cu. Hai mu6'i trong X la FeCNO,), va CuCNOj), con du. Dap an dung la D. ' " CM v: Vi AgNO, phan irng he't vod Fe ntn kh6ng c6 phan Ung: AgNOj + Fe(N03)2 ^ FeCNOjjj + Ag B. 800 C.400 Trong 50 gam A c6 kh6'i luong nguyfin t6' oxi bang: D. 120 (Trich de thi du bi Dai hoc) ^ ( I S6' mol cac chat: n^^ = Hu&ngddngidi w Yii hieu chung cac mu6'i nitrat trong h6n hcrp A la NfNOj. Thi du 2: Hoa tan 19,2 gam Cu vao 500 ml dung djch NaNOj I M , sau do thdm v^o 500 ml dung djch HCl 2M. Kfi't thuc phan umg thu duoc dung djch X va khi NO duy nha't, phai ihtm bao nhifiu ml dung djch NaOH I M vao X d^ ke't tua hfit ion Cu"* ? A. 600 , • Suy ra: " j ^ ^ - = 0,3 /3 = 0,1 (mol) (trong 1 g6c nitrat NO3 c6 3 nguydn til O). Sod6phanihig: 2MNQ3 )2MOH ^-^M.O (5 day ta tha'y: thay 2 ion NOJ bang 1 ion O^" => 0,1 mol NO3 thay bang 0,05mol O"". < • = 0,3(mol) Ma m ^ = m^NOj - m ^ ^ - '''''' " '• nNaN03 ="0,5.1 = 0,5(mol);nHci =0,5.2 = l(mol) =5> m = mwMo, ~ m S6'mol cac ion: n . = 1 (mol); n H+ =0,5fmo0 NO3 ^ •^•^"3 ^ C. A g , 0 , NO, O2 0,2^0,2(mol) =>n 0H~ D- Ag, NO,, O, (Trich detuyen sinh Cao dang khoi A) ' o n + (con du) = 1 - 0,8 = 0,2 (mol) Hu&ng ddn gidi Cu2^+20H-->Cu(OH)2 i 0,3-)-0,6(mol) - 0 , 2 + 0,6 = 0,8(mol) ' ' ' V ; ' II B. AgjO, NO,, 0 , A. A g , N O , 0 , 0,3 (mol) I. " Dap an dung la B. " Thi du 4: San ph^m cua phan ling nhiet phan hoan toan AgN03 la 3Cu + 8H^ + 2NO3 ^ 3Cu2+ + 2NO + 4H2O H^+0H-^H20 O'^ z:>m = 50-0,1.62 + 0,05.16 = 44,6(g) Phuong trinh ion riit gon cac phan ling xay ra: 0,3^0,8->0,2 _ + m -)_ NO3 PTHH: .0 2AgN03 —i—>2Ag + 2NO2 + O2 Dap an dung la D. Thi du 5: Cho 0,87 gam h6n hop g6m Fe, Cu va A l v^o binh dung 300 ml dung | | - djch H2SO4 0 , l M . Sau khi cac phan ung xay ra hoan toan, thu duoc 0,32 gam Vay th^ ti'ch dung djch NaOH I M c^n dung: chat ran va c6 448 ml khi (dktc) thoat ra. Them tiep vao binh 0,425 gam NaNOj, khi cac phan umg ke't thuc thi the' ti'ch khi NO (dktc, san ph^m khu B. hop cac mu6'i Cu(N03)2, Fe(N03)2, Fe(N03)3, Mg(N03)2. ThiDap duan3:dung A lala h6n Trong do nguydn t6' oxi chie'm 9,6% v^ khd'i luong. Cho dung djch K O H du A. 0,224 lit va 3,750 gam. vao dung djch chiia 50 gam mu6'i A. Loc kd't tua thu duoc dem nung trong C. 0,112 1ft va 3,865 gam. chan kh6ng de'n kh6'i luong kh6ng ddi thu duoc m gam oxit. Gia trj ciia m la A. 47,3 B.44,6 C. 17,6 D. 39,2 (Trich de thi du bi Dai hoc) 190 duy nha't) tao thanh va khd'i luong mu6'i trong dung djch la B. 0,112 lit va 3,750 gam. D. 0,224 lit va 3,865 gam. (Trich de tuyen sinh Dai hoc khoi A) Hu&ngddngidi S6 mol cac cha't: nH2S04 = O'^^ imo\)\j = 0-005 (mol) 191 Hu&ng ddn gidi n'H2 „ , =Ml^=o,02(mol) 22,4 Vi < nH2S04 11^2 IChd'i luong nguyen t6' nito trong 14,16 gam X: => axit dir n6n Fe, Al bi tan he't; cha't ran thu duoc la Cu (vi khong phan ung) =>Cu =>nN , \, i ' u Ta c6: -> =0,12(mol) = n ^ ^ . =>mk™i„ai= (condu) =(0,03-0,02).2 = 0,02(mol) Fe + H2SO4 (1) FeS04 + t X l,68(g) 100 0 32 — = 0,(X)5(mol) 64 S6'mol ion H"^ con du sau phan ling: n^^ 14,66.11,864 '-m.- ^ m ^ ^ _ =0,12.62 = 7,44(g) ' m x = m ^ ^ _ =14,16-7,44 = 6,72(gam) Thi du 7: Nhiet phSn m6t luong AgNOj duoc chat rSn X va h6n hop khi Y. D i n toan b6 Y vao m6t lucmg du HjO, thu duoc dung djch Z. Cho toan b6 X vao 2A1+ 3 H 2 S 0 4 ( l ) ^ A l 2 ( 5 0 4 ) 3 + 3 H 2 t Z, X chi tan mot phdn va thoat ra khi NO (san ph£m khiJ duy nha't). Biet cac y phan ling xay ra hoan toan. Ph^n tram khdi luong ciia X da phan ling la 1.5y A. 75%. 56x + 27y = 0,87-0,32 = 0,55 B.25%. Hu&ng ddn gidi Gia sijf nhiet phan 1 mol AgNOj: Fe^^/pe^^ mi + 8H+ + 2NO3 3Cu2+ + 2 N 0 1 + 4 H 2 O P/u-ng: 0,005->0,04/3->0,01/3-> 0,01/3(mol) Con: 0 • 3Fe^^ 0,02/3 i 0,005/3 3Fe-^^ + NO T +2H2O 0,02 / 4 <-0,02 / 3->0,02 / 1 2 (Cac chat phan ting vira hdt vdi nhau) 0,02/12 _ V^o = (0,01 / 3 + 0,02 /12).22,4 = 0,112(l) m(mu«i) = " I p ^ C u . A I + "1^^+ + Dap an diing la C. Thi du 6: H6n hop X gom Fe(N03)2, Cu(N03)2 va AgNOj. Thanh phdn % khoi luong cua nito trong X la 11,864%. Co the di^u che' duac toi da bao nhidu gam h6n hap ba kim loai tir 14,16 gam X? B. 6,72 gam. X: 1 mol Ag; C. 10,56 gam. D. 3,36 gam. (Trich de tuyen sink Dai hoc khoi B) > 0,5(mol) 1 (mol) Y: 1 mol NO, :^ 'nc!. . va 0,5 mol O, 1 2NO2 + - O 2 + H2O ^ 2HNO3 ->l(mol) Dung djch Z: chiia 1 mol HNO, 3Ag + 4HNO3 -)• 3AgN03 + NO + 2H2O 0,75 < - l ^^y'^''"'x(Ag)-- "1^2- = 0,87 + 0,005.23 + 0,03.96 = 3,865(g) A. 7,68 gam. > l(mol) 1^0,25 +NO3 + 4H^ l(mol) ,: „ + N02t+-02t AgNOj—^Ag n^^2+=0.005 (mol) 3Cu D. 60%. (Trich de tuyen sinh Dai hoc khoi B) n^^ =0,02 (mol); n^^_ =0,005 (mol); n^^ =0,005 (mol); Thii tu trong day didn hoa: Cu^"'/Cu; C. 70%. X =0,005; y = 0,01 Khi phan iJng vofi NaNO, c6 so mol cac ion: 192 " Pap an dung la B. X -> X x + l,5y=0,02 Vay •*-"-.f:,>«.,*=r4 (mol) 0,75.108.100% 1.108 = 75% ap an diing la A. hi du 8: Cho Cu va dung djch H 2 S O 4 loang tac dung vdi chat X (m6t loai phan bon hoa hoc), tha'y thoat ra khi kh6ng mau hoa nau trong kh6ng khi. Mat khac, khi X tac dung vdi dung djch NaOH thi c6 khi miii khai thoat ra. Chat Xla A. ure. I ; 5 B. natri nitrat. C. amoni nitrat. D. amophot. (Trich De thi tuyen sinh DH - CD khdJ A) K» thudt ntdi giii BT Hod HQC, tap 2 - C M i nann i oan nhanh Hu&ng ddn gidi fhi du 10: Nung 6,58 gam Cu(N03)2 trong binh km khdng chiia kh6ng khi, sau rndt then gian thu duoc 4,96 gam chat ran va h6n hop khi X. Ha'p thu hokn toan X vao nude de duoc 300 ml dung dich Y. Dung djch Y c6 pH bang A. 2. B.3. C.4. D. 1. X la amoni nitrat (NH4NO3): 2 N O ; + 3 C u + 8H* 2NO N H ; > 3Cu^" + 2NO + >• 2 N O 2 (mau + Oj 4H2O (Trich De thi tuyen sinh Dai hoc khoi A) nSu) ->i > NH3 + H , 0 + O H - Mui khai , ; . ; (NH^)^^ Dap an dung Ik C . Chujj, -Ure: * : ' CU(N03)2 ^rin) — . , ,, u, 112 192 S6' mol cac kim loai: np. = — = 0,02 (mol); nc„ = ^ = 0,03 (mol) 56 64 •It,. • S6'mol c a c ion: n^+ = 2.nH2S04 = 2. 0,4.0,5 = 0,4 (mol); r, i X n^^_ = n^aNOa = 0-4.0,2 = 0,08 (mol) P T P L T xay ra (dang ion Fe + 4H* + 0,02 -> 0,08 3Cu + 8H" + Cu-* 0,03 => 0,02 > ,, . + 2NaOH 0,06 (mol) N30H > Fe'* + 2NO3 + 3NaOH 0,06 (mol) Suy ra S nN^oH = 0,06 -, 0,02 ^ NaOH (0,4-0,08-0,08) Fe'* 0,02 NO 3 ^ 0,03 ->• 0,08 ^ + riit gon): + 0,06 > 3Cu-* ^ + 2H2O - m(N02.02) = - 4.96 = 1,62 = 2x. 46 + 0,5x.32 V' =>x = 0,015 (mol) Ulfi ^|Xll- <)|Ur; 1 2NO, + - O 2 + H,0 2x n„N03 0,5x 2N0 + > 2HNO3 • 2x(mol) = 2x = 2.0,015 = 0,03 (mol) [H*] = [ H N O 3 ] = ^ = 0,1M ,. pH = - I g [H*] = 1. Dap an diing la D. Thi du II: Cho m gam b6t Fe vao 800 ml dung dich h6n hop g6m Cu(N03)2 0,2M va H2SO4 0,25M. Sau khi cac phan ihig xay ra hoan tohn, thu duoc 0,6m gam h6n hop b6t kim loai va V lit khi N O (san phdm khii duy nha't, d dktc). Gia tri ciia m va V l^n luot la A. 10,8 va 2,24. B. 10,8 va 4,48. C. 17,8 vk 2,24. D. 17,8 vk 4,48. (Tnch De thi tuyen sinh DH - khoi B) 0,02 (mol) + 0,5x Huong ddn gidi 4H2O Sd'molcaccha't: ncu(N03)2 = 0.8.0,2 = 0,16 (mol); , 0,03 (mol) Na' + H2O nH2S04 X-i = 0.8.0,25 = 0,2 (mol) S6'mol cac ion: n^^_ = 0,16.2 = 0,32 (mol); n^^2+=ncu(N03)2 = 0.16 (mol) 0,24 (mol) > Fe(OH)., i + 3Na* n \ 0,2 . 2 = 0,4 (mol) H > Cu(0H)2 i + 2Na* Phuong trinh ion nit gon cac phan ling xay ra: Fe + 0,24 = 0,36 = 0,36/1 = 0,36 (lit) = 360 ( m l ) . Dap i n dung l a C . ^ ^ NO 2x Khd'i luong chat r l n giam bang kh6i luong ciia h6n hop khi NO2 va Oj: Hu&ng ddn gidi Cac ^ C u O ( ^ ) + 2 N 0 2 t + ^O^t x(mol) , : , Thi du 9: Cho h6n hop g6m 1,12 gam Fe va 1,92 gam Cu vko 400 ml dung dich chiia h6n hop g6m H 2 S O 4 0,5M va NaNO, 0,2M. Sau khi cac phan ihig xay ra hoan loan, thu duoc dung dich X va khi NO (san p h ^ khii duy nha't). Cho V ml dung dich NaOH I M vao dung dich X thi lugmg k6't tiia thu duoc la 16n nh^t. Gia tri t6'i thi^u ciJa V la A. 240. B. 120. C. 360. D. 400. (Tnch De thi tuyen sink DH • CD khd'i A) >. Hud^g ddn gidi . -Amophot: NH4H2PO4, (NH4)2HP04. :v ,. ! (mol) + N O J (d„) + 4 H * • Fe'* + N O t + 2H2O 0,1 < - 0,1 < - 0,4 0,1 -> 0,1 (mol) Sau phan ling, thu duoc h6n hop b6t kim loai (Fe va Cu) nftn xay ra cac phan ling: rt v|^'ii>1Q< Fe + 0,05 < - 0,1 (mol) Fe(d„) + 2Fe'* > Cu^* SFe"* > Fe^* +Cui 0,16 < - 0,16 3 C u + BH"^ + 2NO3 - p/ir: 0,3^0,8 0 Con: '^^^^^ •yrs.gV' ^ m ->0,2 1 (0,1+0,05 + 0,16) m o l Fe 56 Theo bai ra, ta c6: 0,16. 64 + m - (0,1 + 0,05 + 0,16). 56 = 0,6m 1 0,8 ^ 0,6 0,2 vay V = (0,2 + 0,2). 22,4 = 8,96 (1ft). 0,2 Dap an dung la C. A. Cu(N03)2 B. F e ( N 0 3 ) 2 B. 11,28 gam. C. 9,40 gam. D . 20,50 gam. (Trich De thi tuyen sink Cao dang khoi A) M { N O 3 )^ — Hu&ng ddn gidi ( M + 124) Goi X, y l^n lirot la s6' mol K N O 3 , Cu(N03)2. 29,6 , , 0,5x y 4NO2 + O2 2y 0,5y Theo bM ra, ta c6: l O l x + 188y = 34,65 0,5x + 0,5y + 2y Vay ^ -> -> M O + 2NO2 + ^ 0 2 ( M + 16) 8 M + 16 8 1 M + 124 29,6 3,7 M +124 = 3 , 7 M + 5 9 , 2 j V a y mu6'i nitrat la M g ( N 0 3 ) 2 |Dap an dung la D . (2) Chd v: 2 F e ( N 0 3 ) 3 — l % F e 2 0 3 + 6 N O 2 y = 0,05. m(Cu(N03)2 = 0.05.188 = 9.40 +|o2 Thi du 15: Nung m gam h6n hop X g6m Cu(N03)2 v^ A g N O j trong binh kfn (gam) kh6ng chiia kh6ng khf, sau phan ling hokn toan thu dugc ch^t rSn Y va Dap an dung la C. 10,64 1ft h6n hgp khf Z (dktc). Cho Y tac dung vdi dung djch H C l d u , ke't thiic phan Thi du 13: Cho 0,3 b6t Cu va 0,6 b6t Fe(N03)2 vao dung djch chiia 0,9 mol ling con lai 16,2 gam chat ran kh6ng tan. Gia trj ciia m la H2SO4 (loang). Sau khi cac phan ihig xay ra hoan toan, thu duoc V lit khf N O A . 44,30 B. 52,80 (san ph^m khuf duy nhat, d dktc). Gia trj cua V la A . 10,08 {' > 2 , 7 M - 64,8 => M = 2 4 ( M g ) (1) (0.5x.0,5y).32 + 2 y . 4 6 ^ ^ 3 ^ ^ ^ ^ 3 ^ ; ' Tir ( 1 , 2) ta giai ra dirac: x = 0,25; D. Mg(N03)2 . 8 ( M + 124) = 2 9 , 6 ( M + 16) 2KNO2 + O2 2CuO + C. P b ( N 0 3 ) 2 Gia sir mu6'i nitrat la M ( N 0 3 ) 2 luong Cu(N03)2 trong h6n hop ban dAu la 2Cu(N03)2 — ^ (mol) Hu&ng ddn gidi thu dugc h6n hop khi X (ti kh6'i cua X so vdi khf hidro bang 18,8). Kh6'i X 0,2 0,8 Thi du 12: Nhiet phan hoan toan 34,65 gam h6n hcfp g6m KNO3 va Cu(N03)2, 2KNO3 — ^ + 2H2O Thi du 14: Nhiet phan hoan toan 29,6 gam m6t mu6'i nitrat k i m loai, sau phan ihig thu dirge 8 gam oxit k i m loai. C6ng thiic ciia mu6'i nitrat la V = VNO = 0,1. 22,4 = 2,24 (lit). Phuong trinh phan ling: ->3Fe^* + N O P/ir: Con: 0 (mol) 0,2 Pap an dung la C 0,4m = 0,31.56 - 0,16.64 = 7,12 => m = 17,8 (gam) A . 8,60 gam. ->3Cu^^ + 2 N O + 4 H 2 O , 3Fe2^+4H++N03 0,16 (mol) Chat rSn sau phan ling c6: 0,16 mol Cu; ^ PTPU': B. 4,48 C. 8,96 Hu&ng ddn gidi D . 6,72 (Trich de thi tuyen sinh Dai hpckhoi B) I Khf Z g6m O 2 , N O , : n ^ = 10 64 Hu&ng ddn gidi S ^ m o l c d c i o n : n^^2+ = 0 , 6 ( m o l ) ; n ^ ^ . = l , 2 ( m o l ) ; n ^ ^ = l , 8 ( m o l ) C. 47,12 Chat ran khdng tan la A g : n = = 0,475 (mol) * f ' 16 2 — ^ = 0,15 (mol) D . 52,50 _ 2Cu(NOj )2 — ^ 2 C u O + 4NO2 t + O2 t X 2x -> -> a (24 + 64 + 56) = 14,4 => a = 0,1 (mol) * Phan umg vdri H,S04 (1): \ ?^ 0,5x Mg + H2SO4 0,1-> 2AgN03 — ^ 2 A g + 2 N O 2 + O2 ;:o|,t , • y -> y y -> 0,5y ; => Y g6m CuO, Ag: a::^^,^ ; Ag + HCl ^''^'^^ ^.i,C,r f.(v/-' la A. ''' .^••^.'•„ v^,^' ^; " 1 ; , f , , , ; " ' ' ''•''^•••/rJh ,• ^ " n c u , s = — = 0,03(mol); ^ 160 , /-^'-l''^'^'< "NaNOa = 0 . 1 2 ( m o l ) ; n H c i - 0 , 2 ( m o l ) S6 mol cac ion: n 3CU2S Banddu: 0,03 Phan ling: 0,03^0,1 =^VNO=V = 0,1.22,4 = 2,24 (lit) + IONO3 + 16H* 0,12 ' / 0,4/3^0,1/3 con du (0,4/3 + 0,8/3 = 0,4 < 1,2) nen Fe'* vk Cu phan umg h6t vay m NaN03 oj^a2 .85 = 8,50(gam) Ddp dn dung la D. r\0 ,Sai: 1. ,,(/ . • ' ' C&u 3: Nung 9,40 gam Cu(N03)2 sau phan ling thu dufoc 7,24 gam ch^t ran. Hafp thu toan b6 khi thoat ra vao nude duoc 0,5 lit dung djch c6 pH 1^ A. 2,10 B. 1,10 C. 1,00 D.0,98 PTPLT: m v('; -•'^^lM t^iiWo, Cu(N03),„, —!—>CuO„ +2NO2+i^02 => Khd'i lirong cha't ran giam bang t6ng kh6'i luong khi NOj v^ Oj. Do 66. 6Cu2* + 380^" + lONO + 8 H 2 O mo2+N02 =9,40-7,24 =2,16(gam) 0,2 ->0,16 Vi / Hu&ngddngidi H PTPir: • T S ; . _ =0,12 (mol); n ^=0,2 (mol) NO3 ' >\f^lfmm;:r. ^ 3Fe^^+ 4H^ + NO3->3Fe^++NO + 2 H 2 9 iv. 0,1 ^ , v ' v H u & n g d d n g m SeJmolcacch^: A 0,1 ^ 0 , 8 / 3 ^ 0 , 2 / 3 Cau 1: Cho 4,8 gam b6t Cu^S vao 120 ml dung djch NaNOj IM, sau d o them 200 ml dung djch HCl IM vao, ke't thuc phan ling thu duoc dung dich X va V lit khi NO (san ph^m khijrduy nhaft,dktc). Gia trj cua V 1& g , , .y,, A. 67,2 • B. 22,4 C. 2,24 ; D. 6,72 ; i:, 3Cu + 8H^ + 2NOJ^3Cu^^+2NO + 4 H 2 0 3. Cac bai t$p tu luyen: ':',„.,„. ,' 0,1 mol FeS04 s},iy ,;d h xfi:: , 0,1 molCu . •'•'^^'•••^^';|.; '. ' 1,6 - 0,1. 2 - 0,1. 2 = 1,2mol MW:Y: * Khi cho NaNOj vao h6n hop sau phan ling: Vay m=0,1.188 + 0,15.170 = 44,30(g) diing ).FeS04 + H j Sau phan ling c6: Tac6he:|y = ' ' ^ ' ' ^ x =0,1; y = 0,15 ^^^^^^^^^^^^^^^ • [2x + 0,5x + y + 0,5y = 0,15 5:iyVi{;::?. Dap an '•' ' '••'>:h-. ,(,;;3fr:;;' > Kh6ng xay ra. -'''^'^^ 0,1 Fe + H 2 S O 4 if;:]^.-y :. CuO + 2 H C l ^ C u C l 2 + H 2 0 >MgS04 + H2 t ^ . hk 0 , 1 (mol) Theo PTPLT tren ta tha'y: , '"/A ; Dap an dung la C . Cau 2: Cho 14,4 gam h6n hop Mg, Cu, Fe c6 s6' mol bang nhau vao 0,8 lit dung djch H2SO4 IM (loang). C5n phai them it nha't bao nhieu gam NaNOj vao h6n hop sau phan img thi khdng con khi NO (san ph^m khii duy nha't) thoat ra? A. 12,75 B. 5,67 C. 2,83 D. 8,50 Hu&ngddngidi Theobaira: n H 2 S 0 4 =0,8.1=0,8(mol) => n^^. = 2.0,8=l,6(mol) Goi a la s6' mol m6i kim loai trong h6n hop 198 ' ' nN02 : " 0 2 = ^ Goi x la s6' mol Oj => s6' mol NO2 la 4x. Ta c6: 32x + 46 . 4x = 2,16 =:> x = 0,01 (mol) PTPlT: 4NO2 + O2 + 2H2O -> 4HNO3 4x X 4x => n H N 0 3 =4x =0,04 (mol) 0,04 f=>m=CH,03=^=0,08M 0,5 Vay pH = -lg0,08= 1,10. Dap an dung la B. . ,, ^ ' » «' • ^ ^:;:|yO.y.v0.1HX #^^! .y " 199 Cau 4: Nung nong 34,6 gam h6n horp X g6m CuCNOj), va Cu trong binh km da'n kha'i lircmg khong d6i thu duofc chat rSn Y. Di hoa tan he't Y cSn viifa du 500 ml dung dich H2SO4 loang 0,5M. Kh6'i lugmg cua Cu(N03)2 trong X la , A. 18,8 gam B. 23,5 gam C. 28,2 gam D. 14,1 gam. Hu&ng dan gidi '^rV} Theobaira: nH2S04 =0'25(mol) Goi X, y lan lirot la s6' mol cua Cu(N03)2 va Cu trong h6n hop X. ' Taco: 188x + 64y = 34,6 (1) r PTPU": 2Cu(N03)2 — 4Cu + 2NO, — ^ X+y Taco: E)ap an diing la A. C^u 6: Nhifit phan hoan toan h6n hop 2 muoi KNO3 va Fe(N03)2 sau phan iJng thu duoc h6n hofp X g6m hai khf c6 ti kh6'i so vdi H2 bang 21,6. Phan tram kh6'i .jv,- . luong cua Fe(N03)2 la B. 40% A. 60% r. CUSO4 + H2O X Tiif (1,2) ta c6:x = 0,15; y = 0,1. > \ • Fe203 +4NO2 t +i-02 t 2Fe(N03)2 y t 0,5x (mol) -> "2 (2) -> 2y ^ 0,25y (mol) H6n hop X gdm O, va NO2 Vay mcu(N03)2 =0,15.188=28,2 (gam) Dap an diing la C. Cau 5: Nung hoan toan 13,96 gam h6n hop AgNOj va Cu(N03)2 thu duac cha't ran A. Cho A tac dung voi dung djch HNO3 du thu duoc 448 ml khi NO (dktc) duy nha't. Phan tram kh6'i luong cac chat trong h6n hop d^u la A. 73,066%; 26,934%. B. 72,245%; 27,755%. ^ ' C. 68,432%; 31,568%. D. 82,52%; 17,48%. Huongddngidi D. 34,3% C. 78,09% ; KNO3 — ^ K N O , + - 0 , 4CuO + Nj x+y A,' %mc„(N03)2 "26,934% PTHH: x + y = 0,25 = 73,066% Hu&ng ddn gidi 2CuO CuO + H2SO4 170.0,06.100% %mAgN03 = 2CuO + 4NO2 + 0 , ^ 2Cu + 0 , vay j , . ^ ^ ... 32.0,5x + 46.2y + 32.0,25y Theo bai ra: •—^ = 21,6.2 0,5x + 0,25y + 2y ^16x.92y.8y^ 0,5x + 2,25y =>16x + 100y = 21,6x + 97,2y •V^. ' ' => 2,8y = 5,6x => y = 2x Theobaira: n^o =0.02(mol) Fe(N03)2 AgN03 — ^ Ag + NO2 + - O 2 Dap an diing la C. y 3Ag + 4HNO3 -> 3AgN03 + NO + 2H2O -> ' ,,,;„ Cau 7: Nung h6n hop g6m 6,4 gam Cu va 54 gam Fe(N03)2 fong binh kin, chan kh6ng. Sau phan ling hoan toan thu dugc h6n hqyp khi X. Cho X phan ling h6't vdi nuoc, thu duoc 2 lit dung dich Y. pH ciia dung dich Y la A. 0,525 B. 1,3 C. 0,664 D. 1 ' Hu&ng ddn gidi ^'hieo bai ra: =0,1 (mol); np^^j^Q^^ =0,3 (mol) Cu (NO3 )2 — C u O + 2NO2 + ^ O2 X 180.2x + 101.X x /3 C U O + 2HNO3 ^ C U ( N 0 3 ) 2 + H 2 O ' ^ , [I70x + 188y = 13,96 „ ^ „^ Taco:^ ^ =i> x =0,06; y = 0,02 x / 3 = 0,02 ^ 200 4 F e ( N 0 3 ) 2 — ^ 2 F e 2 0 3 +8N02 + ° 2 0,3 -> 0,6 0,075(mol) 201 Dang 7: Xdc dinh chat 2Cu + O 2 — ^ 2CuO 0,1 => 1 ; 0,05(niol) lA thuyet \kn dung va phuong phap giai: giai cac bai tap dang nay c6 th^ su dung cac phuong phap giai nhanh: Phuong phap xac dinh chat dua vao cong thurc tinh phan tu khoi: Gia sit c6 chat cho electron la A. Ta c6: ' = 0.075 - 0,05 = 0,025 (mol) 4NO2 + O2 + 2H2O ^ 4HNO3 0,1 <- 0 , 0 2 5 . i f 0 , 1 (mol) 3NO2 + H 2 O ^ (0,6-0,1) ^ 2HNO3 + - NO 1/3 (mol) ; ^ ^, ^ r H + 1 = M^lil(mol)=>pH-0,664 * 3.2 6 .>. . ;,: ^ '-'^A^'' ' Dap an diing la C. C^u 8: Nhidt phan hoan toan 27,3 gam h6n hop rSn X gdm NaNOj, thu duoc h6n hop khf CO th^ tich 6,72 lit (dktc). ' CuCNO^), 1. Phuong trinh hoa hoc ciia cac phan ihig: 2NaN02 + Ojt mol (0 0,5x mol 2Cu(N03), — ^ ymol 2CuO + 4 N O 2 T ymol + OJT (2) 2y mol 0,5ymol 2. Goi x va y la s6 mol ciia NaNOj va Cu(N03)2 trong h6n hop X. ,,. Til ( l ) v a (2) ta CO he phuong trinh: •gSx + 188y - 27,3 0,5x + 2y + 0,5y zz,4 x - 0 , 1 (mol) 0 , 3 ^ y = 0,l (mol) PhSn tram khd'i luqmg cia m6i mu6i trong h6n hop X: %'nNaNO3-^100%-31,l% - ... r ./.:••; %mcu(N03)2 =100^"-31'1^'' = 68,9% m. .k m . -.k = n„ (cho) n. (cho) + Dua vao cac dix ki6n bai ra cho, xac dinh duoc cac dai luong m^, k, n^ (j^o)+ Tinh gia trj M ^ . .. + Th phan tir kh6'i xac djnh duoc chat cdn tim. Phuong phap trung binh: C6ng thtic tinh nguydn i\i kh6'i trung binh ciia hai* '^hh Hu&ng ddn gidi X ke kim loai lien tid'p trong ciing m6t nh6m A: M = "• i h h 1. Viet phuong trinh hoa hoc ciia cac phan ting xay ra. 2. Tinh thanh phSn % v l khoi luong ciia m6i mu6i trong h6n hop X. is'-l 2NaN03 — ^ A"* • m. , Vay: , -I rir y ' =>n^, = n H N O 3 = 0 , l + ^ - ^ ( m o l ) L A / i|vd ; (•;••;.' n,h„. N,,h„ = n„hj„. N„hj„ + Vi6t cac qua trinh oxi hoa, qua trinh khir (kh6ng cin vie't phuong trinh phan ling oxi hoa - khir). ;•• • • ;4,v/( r v • . ' i n v ' * / - : - • + Dat ^n, dua vao dinh luat bao toan electron lap duoc phuong trinh dai s6'. + Giai he phuong trinh, xac djnh duoc M chat cSn tim.. Phuong phap bao toan khoi luong: Trong m6t phan ij-ng hoa hoc, t6ng kh6'i luong ciia cac chat san phdm bang tong kh6'i luong ciia cac chat tham gia phan ling. i ;> i ' Ap dung: Trong m6t phan ting, c6 n ch&l Qui ca cha't phan ung vk san phim), nfiu bie't kh6'i luong ciia (n - 1) chai^t thi tinh duoc kh6'i luong ciia chat con lai. Phuong phap hoa tri trung binh: Trong nhirng trucmg hop h6n hop c6 hai gia trj hoa trj (hoac trang thai s6' oxi hoa) tr6 len ta c6 the' sir dung phuong phap hoa tri trung binh de' giai quye't cac yeu ci\x ciia bai toan. Thf du, khi Fe tac dung vdi H N O 3 tao ra mu6'i sat (II) va mu6'i sat (III) thi 2 < hoa trj trung binh < 3. Khi h6n hop Mg, A l tac dung voi axit H Q thi ta cQng c6 2 < hoa tri trung binh < 3. 202 203 2. Cac thi du minh hoa: Thi du I: Hoa tan hoan toan 1,805 gam h6n hop g6m Fe va kim loai X bang dung djch HCl, thu duoc 1,064 h't khi H,. Mat khac, hoa tan hoan toan 1,805 gam h6n hop trfin bang dung dich HNO, loang (du), thu duoc 0,896 lit khf NO (san ph^m khij duy nha't). Bie't cac th^ tich khf d^u do 6 di^u kifin tieu chu^n. Kim loai X la A.Zn B.Cr C. Al ' D. Mg •' (Trich de thi tuyen sink DH khoi A nam 2013) iu 2: C6ng thirc phan tijf cua hop cha't khf tao boi nguy6n t6' R va hidro la RH3. Trong oxit ma R c6 hoa tri cao nha't thi oxi chife'm 74,07% v6 kh6'i lirong. Nguyen t6'R la A. S. B. As. C.N. D. P. (Trich De thi tuyen sinh DH - CDkhoi B) I Hu&ng ddn gidi Hop chat khf ciia R voi hidro la RH3 => oxit cao nha't ciia R la RjO,. 16.5 '-''^ 74,07 2R + 16.5~ 100 => 8000 = 148,14.R +5925,6 => 148,14R= 2074,4 : ^ R = 1 4 ( N ) vay R la nguy6n t6 nito (N). Dap an dung la C. ' 0-»-.•«'vvt'*'5'''<' Hu&ng ddn gidi S6 mol cac chat: = 0,0475 (mol); Xheo bai ra ta c6: = 0,04 (mol) Cac qua trinh nhucmg electron: + Trong qua trinh hoa tan vao dung dich HQ (sat tao ra hop cha't hoa tri I I , X tao ra hgfp chat hoa trj n): , Fe - 2e ^ Fe^* '1,'"V X -> X 2x - ne-^X"* ny + Trong qua trinh hoa tan vao dung dich hc>p cha't hoa trj m): Fe - 3 e ^ F e ' * X 3x y X %0 = B. N j O v a A l . (sat tao horp cha't hoa tri III, X tao me-> X"* 940,8 Theo bai ra: n NxOy my '56x + X.y = 1,805 Mat khac: 22400 ^NxOy/H2 Ta c6: ] 2x + ny - 0,0475.2 = 0,095 = 0,042 (mol) = 22 M 3x +my = 0,04.3-0,12 ^ = 2 2 => M^^o^ =44(N30) * Gia sijf n = m => X = 0,025; ny = 0,045 Taco: => X. y = 1,805 - 56.0,025 = 0,405 2N^' 2.4e 0,336 n = 2;m = 3=>H6v6 nghifim. Vay loai trucmg hop nay. 204 > 2N*' 0,042.2 (mol) •In,„,h,,„, =0,336 (mol) 2 18 M a Sn, ..fit :! ''I* NT^^ : J'\Uf<>x/^f' C. N.O va Fe. D. NO, va A l . (Trich De thi tuyen sinh Dai hoc khoi A) Hu&ng ddn gidi - ;'>«W>J:. Thi du 3: Cho 3,024 gam m6t kim loai M tan hfe't tr6ng dung dich H N O 3 loang, thu diroc 940,8 ml khf N,Oy (san pha'm khu duy nha't, 6t dktc) c6 ti kh6'i d6'i vori H2 bang 22. Khf N^O^ va kim loai M la A. N O v a M g . HNO3 ^.n*;;;;. ^, - ne na (mol) > M"* = na (mol) = 0,336 (mol) a 0,336 Do do, gia tri thoa man la n = 3 =:> M - 27 (Al). E>ap an dung la B. Kt thuat m&igiuinhanh BTHoa hQC, tap 2 - C u thanh loan Thi du 4: K h i nhiet phan hoan loan timg mu6'i X , Y thi d^u tao ra s6' mol khf nho hem s6' mol mud'i tirong ung. E)6't m6t luomg nho tinh th^ Y trfin den khf kh6ng mau, tha'y ngon liia c6 mau vang. Hai mu6'i X , Y 15n luot \k A. KMn04, NaNOj. B. NaNO,, K N O 3 . C. CaCOj, N a N O j . ViJ^v^i D. CuCNOa)^, N a N O j . (Trich De thi tuyen sink Dai hoc kho'iB) . . . . O , - H u d n g d d n g i d i 'x + y = 0 , 0 6 9,1 2NaN03 2 (mol) , , , , , , , ( , y 2N i/'-v^H) N+3e- C.NO2. " ' " ^ = 0,1 (mol). So d6 cac qua trinh nhucmg - nhan electron: 2e ne = 0 , 1 6 + 0,12 = 0,28 fty n = 2, M = 65 (Zn). ' ,5 i• ' Dap an diing la A . phan ung thu duoc 0,2 mol N O ; 0,1 mol N j O va 0,02 mol N , . Bie't kh6ng c6 phan ling tao mu6'i NH4NO3 va HNO3 da la'y du 15% so vdi luong c^n thid't. K i m loai M va n6ng d6 phSn trSm ciia HNO3 ban d^u lin luot la A.Crvk21,96 B.Znva20 C.Crva20 D . Z n v a 17,39 Hu&ng ddn gidi Di giai bki nky ta sijr dung dinh luat bao toan electron vk djnh luat bao toan +(5-n) N 0,1 0,ln Theo nguyfin tSc bao toan electron ta c6: 0 , l n = 0,3 => n = 3. Vay san phdm khu la N O . nguydn t6'. Qua trinh nhucmg electron xay ra: M - ' 1 ne->M"* - > nx (mol) X Qua trinh nhan electron xay ra: Ddp an dung lik B. +5 N+3e- 3. C a c bai tap t u luyen: Hu&ng ddn gidi +2 ->.N 0,6 < - 0,2 (mol) Cau 1 : Cho 9,1 gam k i m loai M tan hd't v^o dung djch HNO3 loang, dir thu dtf Mg'* 0,3 (mol) +5 * 1' Cau 2: Hoa tan hd't 52 gam k i m loai M trong 739 gam dung dich HNO3, kd't thiic Hu&ng dan gidi S6 mol cac chat: nwg = ^ = 0,15 (mol); nx = 24 22,4 •aco: — M D.N2. (Trich De thi tuyen sinh Cao dang khoi A) N 1 +2 +5 1 (mol) B.NO. 0,15 -> 2 N i 1 10,12<-0,04 A.NA - + 2.4e ' - I f +1 0,16 ,),;', Thi du 5: Cho 3,6 gam M g tac dung he't vdfi dung dich HNO3 (du), sinh ra 2,24 lit khf X (san ph^m khiJr duy nha't, 6 dktc). K h i X la Mg 9, In -(mol) M +5 Dap an diing la A . la N O ( 3 0 ) +ne >'M'^ M >2NaNO, + 0 2 Y * Qua trinh nhucmg, nhan electron: t6' natri. Hai mudi X , Y c6 the la K M n 0 4 , NaNO,: • la N 2 0 ( 4 4 ) ; • X =0,02; y = 0,04 • 44x + 30y = 2,08 M 2KMn04 > K2Mn04 + MnOz + O j t 2 (mol) - . . • • V ' r v ; ? 1 (mol) rr>X ^ Xac djnh s6' mol N2O ( x m o l ) , N O ( y mol) D<5t Y tr6n ngon lijfa Ahn k h i , tha'y ngon lua c6 m^u vJing => trong Y c6 nguyg^ :V = 1,467 VI M X / M Y +5 2N+8e- +1 ->.N, 0,2 < - 0,02 (mol) lieo nguySn t i e bao tokn electron, thi: I nx = 0,6 + 0,8 + 0,2 = 1,6 (mol) => x = 0,16/n (mol) if.: 4' ^'^ ^^"^ "^^^ '^'"^ ^ trong dung djch chiJa h6n hop HNO3, H^SOi dac, nong. Sau khi cac phan umg xay ra hoan toan thu duoc 4,2 lit h6n hop san phim khir g6m NO, NO,, SO2 (6 dktc, s6' mol NO, NO2 bang nhau). Kim loai M la A.Fe(56). B.Mg(24). C. Ca (40). D. Cu (64). Hu&ng ddn gidi C&O 1,6 n M Ke't luan 2 65 Zn 1 32,5 Loai 3 97,5 Loai S6 mol h6n hop san p h ^ khir: n^h = 4,2 / 22,4 = 0,1875 (mol) Gia tri thoa man la n = 2, M^, = 65 (Zn) * Cac qua trinh khir (qua trinh nhan electron): 52 = — = 0,8(mol) => n Zn(N03)2 = 0,8(mol) 65 Ap dung dinh luat bao toan nguyen t6' cho nguyfin t6' nito, ta c6: +5 +2 N+ 3e • N(NO) +5 •N(N02) N+ le nN(HN03) = " N ( N O ) + " N ( N 2 0 ) + " N ( N 2 ) + "N(zn(N03)2) ^ "HNOJ +5 +3 N+ 2e > N (vi n^o = nN02 '^V 8'^ trung binh ciia chung). +6 Vi HNO, lay du 15% nen: S + 2e- its;,:; • '.^^ 2,04.(100+15) => niHNOj ,, M,0 = 0,2.1 + 0,1.2 + 0,02.2 + 0,8.2 - 2,04(mol) "HNO3 (ihucte-) - ,,|, Suy ra, t6ng s6' mol electron da nhan la: = 2,346(mol) ,/ \M}^'M • X ne(,) = 0,1875.2 = 0,375 (mol) = X n,(_) = 2,346 . 63 = 147,798 (g) Qua trinh oxi hoa kim loai M (qua trinh nhucmg electron): 147 7Q8 => C% (ddHN03) = 0 M . 100% = 20% 8,4M -> Vay M la Zn va C%(HN03) = 20% Dap an diing la B. ' Cau 3: X la m6t phi kim c6 s6' oxi hoa duong cao nha't bang ^ s6' oxi hoa am thap - nha't (ti'nh theo tri tuyet d6'i) va khoi luong phan tir oxi cao nha't ciia X ga'p 4 , 1 7 6 \in khoi luong phan tir hop chat khi ciia X voi hidro. X la A. clo. B. photpho. C. luu huynh. D. nito. Hu&ng ddn gidi X la phi kim nen X CO the d nhom VIIA, VIA, VA hoac IVA. - Goi n la so oxi hoa duong cao nha't (n = 7, 6, 5, 4) => trj tuyet d6'i so oxi hoa am thap nha't ciia X la 8 - n. Taco: — ^ = - = > n = 5 8-n 3 (XonhomVA) Suy ra oxit cao nha't X 2 O 5 ; hop chat khi cua X vdi hidro Taco: + ^-^^^4 175 Dap an diing la B. > -i. =31 (photpho) +n -> M + ne - XH3. Do do ta c6: 8,4n M 8,4n/M ,1. = 0,375 => M = 22,4n. Ta CO bang: Kim loai M N Kd't luan Fe 56 2,5 Thoa man Mg 24 1,07 Loai Ca 40 1,79 Loai Cu 64 2,86 Loai Khi M la kim loai Fe thi n = 2,5, tiic Ih sk tao ra hai s6' oxi hod Ik +2 vh +3. Dap an diing la A . 'viM'--' C A U 5: Khi cho oxit ciia m6t kim loai h6a trj n tdc dung vdi dung dich H N O 3 dir thi tao th^nh 34,0 g muC>i nitrat va 3,6 g nude (kh6ng c6 san phdm khdc). Hoi do 1^ oxit cua kim loai nao va kh6'i luong ciia oxit kim loai da phan ling la bao nhidu ? , im>^'''-"''^^'-" Hu&ng ddn gidi Phan iJng chl tao ra mu5i nitrat va nu6c, chiing to n la h6a tri duy nhit ciia kim loai trong oxit. Dat c6ng thiJc ciia oxit kim loai la M20„ va nguyfin tii kh6i ciia M la A. 209 Phirong tnnh h6a hoc: M,0„+2nHN03 ).2M(N03) + nHjO 2(A+62n) g 18ng • " 34,0 g Th^ tfch dung dich HNO3 c^n dung la V = — = -^1^ = 61,5 (ml) D 1,365 DdpdnA. ...,^lfi.R*J;:.•..^.m^.•.^ • . : o! C&u 7: H6n hop X g6m Fe va kim loai M hoa tri 3. Chia 38,6 gam X th^nh 2 phSn bang nhau. PhSn 1 cho tan hoan to^n trong dung dich HNO3 loang du thu duoc cic san ph^m khii chi c6 NO, NjO (h6n hop Y) vdi t6ng th^ tfch 6,72 1ft, ti kh6'i ciia Y so vdi la 17,8. P h ^ 2 cho vao dung djch ki^m sau m6t then gian tha'y lucmg H , thoat ra vuot qud 6,72 1ft. Bie't cac khf do 6 di^u kien tieu chudn. a) Xac dinh ten kim loai M va % khd'i luong ciia kim loai trong X. b) Tfnh kh6'i luong HNO3 da phan ling. Hu&ng ddn gidi 3,6 g T a c 6 t i l 6 : ^(A + 62n)^18n ->2.3,6.(A + 62n) = 34,0.18n => A = 23n. 34,0 3,6 Chi c6 nghidm n = 1, A = 23 la phil hop. ^. Vay kim loai M trong oxit la natri jfyyi Phan ling giiJa NajO va HNO,: NajO + 2HNO3 -/ ^ , , , , :, i p: * £ -V^ ^,i ;„ , >-2NaN03 + H j O ^ 62,0g ^ X ^^'Og < :;i5jy:j' I 3,6 g I 3,6.62,0 ,^ , ^ x = - ^ = 12,4(g) M + 4HNO3 (mau mau do) , v 8 96 S6' mol khi NO. = - — = 0,40 (mol). Theo phuong trinh hoa hoc: • 22,4 n 40 0 40 4 2 = 0,20 mol va nHN03 =^~^= 2 0,80 25M + 96HNO3 —!—>25M(N03)3 + 0.18.25 ' '' ' KheJi lirong mol nguyen tii ciia kim loai M : M = = 64,0 (g/mol). + (l) = 0,5 mol. 3H2O + OH" M(OH)J"+3/2H2 (2) | > 2.0,3/3 = 0,2 >0,3 4!". mol >i X la s6 mol ciia M=>s6 mol Fe: 0,5 - x mol >Mx + ( 0 , 5 - x ) 5 6 = 1 9 , 3 ^ M = ^ ^ ^ ^ ^ ^ vdi 0 . 2 < x < 0 , 5 8,7 56-M •X = 0,3 mol f&y =>kimloailaCu(d6ng). <• Kh6'i lircmg HNO3 tham gia phan ling = 0,80.63 = 50,40 (gam) 50,40.100 Kh6'i luong dung djch HNO3 60,0% da dung = — — - — = 84 (gam) ou,u + ^ ^ 2 ^ + 48H2O X tac dung voi ki^m c6 khf thoat ra nen M se phan ling. Phuong trinh hoa hoc ciia phin 2: (mol) 12 8 210 ' ' [ => Ti le mol NO/N2O = 3/2 j Phuong trinh hoa hoc ciia phdn 1: M(N03)2 + 2N02t + 2H2O J = ^ Kh<5i luong trung binh ciia Y: 35,6 g/mol. H6n hop Y la 0,3 mol; a la s6' mol > 30a + (0,3 - a ) 4 4 = 35,6 => a = 0,18 mol Kh6'i luong ciia oxit kim loai da phan ling la 12,4 gam Cau 6: Hoa tan 12,8 g kim loai hoa trj II trong m6t lucmg vita dii dung dich HNO3 60,0 % (D = 1,365 g/ml), thu duoc 8,96 lit (dktc) m6t khi duy nha't mau nau do. Ten ciia kim loai va the' tich dung dich HNO3 da phan dug la A. D6ng;61,5ml t v i / C. Thuy ngan; 125,6 ml « B. Chi; 65,1 ml . D. Sat; 82,3 ml Hu&ng ddn gidi O n M la kim loai hoa trj I I dn tim d6ng thoi cung la kh6'i lugmg mol nguyfin tir ciia no. Phan ling: ' > • i Do HNO3 du nen Fe se tao mu6'i Fe^^ => Coi Fe v^ M c6 c6ng thiic chung M . i=> ny = 0,3 mol > •, •0,2< %m^i = ^ 2 1 8.7 56-M < 0.5 => 12,5 < M < 38,6 => Chi c6 A l . 100% = 41,97%; %niF^ =58,03% Theo (1) nHNOj = ^ ^ f ^ = 1.92 (mol) => Khefi luong l i N O j phan ling = 63.1,92= 120,96 gam. 211 C&u 8: Chia 16,68 gam h6n hop X g6m Fe va kim loai R (hoa tri kh6ng d6i) thanh ba phdn bang nhau. PhSn 1 cho vao dung dich HCl du, sau phan ling xay ra hoan toan chi thu dugc dung djch va 3,136 lit Hj. Phdn 2 cho vao dung dich HNO3 loang, du sau khi phan ihig xay ra hoan toan thu duoc 2,688 lit khi NO (san ph£m khir duy nha't). Cac th^ tich khi do of dktc. a. Xac dinh kim loai R va tinh thanh phSn phin tram kh6'i luong m6i kim loaj trong h6n hop ban dSu. b. Cho phSn 3 vao V lit dung djch CUSO4 I M , sau khi phan ung xay ra hoan toan thu diroc 8,64 gam cha't rln. Tinh V. Hu&ng ddn gidi a. M6i phSn c6 kh6'i 5,56 gam; goi trong m6i phSn c6 chiia x mol Fe va y mol kim loai R. Fe + 2HCl->FeCl2 + H2 x (mol) -> x (l) ' '•'* R + nHCl->RCl„+|H2 , , ' mnM&& ^ . Khi A l phan ung hfi't, khd'i luong chat rSn tang ., = 64.0,06 - 27.0,04 = 2,76 gam < 3,08 gam=>Al phan ling he't. Khi Fe phan ling het, khd'i luong chat ran tang = 2,76 + 0,08(64-56) = 3,4 gam >3,08 gam ^ Fe phan img chua hd't. Vay AI phan utig hd't, Fe phan ling m6t phSn, goi s6' mol Fe phan ling la z (mol) 2A1 + 3CUSO4 ^ AI2 (SO4 \ 3Cu 0,04^0,06 + (5) 0,06(mol) CuS04->FeS04+Cu ^ z (6) ,a' K I ' .^00 4 ' ..^ ^ =>Z"cuS04 Phaniing =.0,06 + 0,04 = 0,1 (mol) (3) a. Xac djnh tan hai kim loai ki^m. b. Tinh n6ng d6 moi/1 ciia dung djch HCl da diing. Hu&ng ddn gidi •• , „ • X - 3R+4nHN03^3R(N03)^+nNO+2nH20 ^"f'', => V = 0,1 lit = 100 ml. Q u 9: Hoa tan hoan toan 1,62 gam nh6m vao 280 ml dung dich HNO3 I M duoc dung dich A va khi NO (san ph^m khij duy nha't). Mat khac, cho 7,35 gam hai kim loai kiem thudc hai chu ki lifin tie'p vao 500 ml dung djch HCl, duoc dung djch B va 2,8 lit khi H . (dktc). Khi trdn dung djch A vao dung djch B tha'y tao thanh 1,56 gam ket tiia. (l) Fe + 4HNO3 -> Fe(N03)3 + NO + 2H2O X Sau phan utig khd'i luong chat ran tang = 8,64-5,56 = 3,08 gam =^ 2,76 + (64 - 56) .z = 3,08 => z = 0,04 (mol) Phdn 2: • ' I. =19,33%. b.EhloJ: z(mol)->z (2) nH2 =0,14(mol)=c>x + | . y = 0,14 '. =:>%mFe =80,67%;%mAi Fe n , X = 0,08(mol);y = 0,04(mol) (4) *; |£ Phuong trinh hoa hoc: n N o - 0 , 1 2 ( m o l ) = ^ x + | . y = 0,12 (ll) 0 12 Giai hf phuong trinh (I), (H) ta c6: x = 0,08(mol);y = ; f n = l = > M R -9(loai) " n = 2=>MR-18(loai) n = 3 ^ M R =27=> K i m l o a i R l a A l ; n = 3. v (1) 2M + 2 H C 1 ^ 2 M C I + H2 (2) 2 M + 2 H 2 0 ^ 2 M O H + H2 (3) Ban dSu: s6 mol A l : 0,06 mol; s^ mol HNO3: 0,28 mol | ; jj^J Theo kh6'i luong h6n hop ban dSu ta c6: 56x + M R .y = 5,56 O M R ==9.n;n lahoatri. A l + 4 H N 0 3 ^ A l ( N 0 3 ) 3 + N O + 2H20 S!^ Sau phan ihig HNO3 con du: 0,04 mol Khi cho h6n hop hai kim loai ki^m vao dung dich H Q thi xay ra phan iJng (2) CO the CO phan ling (3): ^- Theo PTHH: s6' mol M = s6' mol H , = 0,25 mol => M = 29,4 ^> hai kim loai ki^m thu6c hai chu ki lidn tie'p nen Na, K thoa man (23<29,4<39) g , , , ,j i Khi tr6n hai dung dich A B c6 kfi't tua tao ra chumg to ban d^u c6 phan vtng (3), ta CO phucmg trinh hoa hoc: , ... H N O 3 + MOH ^ Xa CO bang: n (4) MNO3 + H 2 O Al(N03)2+3MOH-^-Al(OH)3+3MN03 (5) S6 mol k6't tua: A l ( 0 H ) 3 = 0,02 mol nho Hon s6 mol A l ( N O 3 ) 3 . Nfin c6 2 kha 2 R 28 56 3 84 Kl Lx)ai Fe Loai Chi CO n = 2 (m = 3) ihig vdi R = 56 la thoa man. Vay kim loai R la Fe. nang: Dang 8: Bdi tap ve photpho va hop chat cua photpho +) Tru&nghapl: A l ( N 0 3 ) 3 condirthi s6'mol MOH = 0,04 + 0,02.3 = 0,1 mol =>S6molMphan umg (2) =0,25-0,1=0,15 mol =^ S6 1 mol HQ = 0,15 mol =^ C M ( H C I ) = 0,3M. ,, +) Tru&ng hap 2: MOH con du, A1(0H)3 tan trd lai m6t phSn: ' A l (OH)3 + MOH M A I O 2 + 2H2O ^ . ^ ' (6) S6mol A1(0H)3 tan = 0,06 - 0,02 = 0,04 mol. ' - Tilf cdc phirong trinh (4, 5, 6) ta c6: '" •* ' S6' mol MOH = 0,04+ 0,06.3+ 0,04 = 0,26 mol (loai, vi 1dm hom s6' mol M ban dSu). v ; Khi ho^ tan cung m6t lirgmg kim loai R v^o dung djch H N O 3 loang va vao dung dich H2SO4 loang thi thu duoc khi NO va H2 c6 th^ tich bang nhau (do 6 cung di^u kien). Bie't kh6'i luong mu6'i nitrat thu duoc bang 159,21% kh6'i luong mu6i sunfat. Xac djnh kim loai R. ijj yiv A Hu&ng ddn gidi C a u 10: U thuyet van dung va phuofng phap giai: Photpho la phi kim tucmg d6'i boat ddng. Photpho trang boat d6ng hoa hoc manh hem photpho do. Trong cac hc>p chat, photpho c6 %6 oxi hoa -3, +3 va +5. Do do, khi tham gia phan ting hoa hoc photpho th^ hidn ti'nh oxi hoa hoac tinh khijf. Photpho the' hifin tinh oxi hoa khi tac dung vdi m6t s6' kim loai boat ddng tao ra photphua kim loai. .u.a •„,, :.n:i .iiV^: Thidu: 2P + 3Ca ^ Cajp' canxi phophua Photpho th^ hidn tinh khix khi tac dung vdi cic phi kim boat ddng nhu oxi, halogen, liru huynh,... va cac hop chat c6 tinh oxi hoa manh khac. Photpho chay duoc trong khdng khi khi d6't ndng: thie'uoxi: 4P + 3 0 , dix oxi: 4P + 5 0 . +3 -> 2P2O3 diphotpho trioxit 0 +5 -> 2P2O5 diphotpho pentaoxit Goi n la hoa trj ciia kim loai R, a la s6' mol R da dung of m6i phan ling. Hiotpho tac dung d^ dang vdi khi clo khi dd't nong: PTPlT: thie'u clo: 3R + 4nHN03 -> 3R(N03)„ + nNO + 2nH20 a a 2R + H2SO4 ^ R2(S04)„ + an/3 'v ''-.'J fi'' nH, 3 2 a.(R + 62m) = i ^ . - . 2 R + 96n ^ 100 2 Tiir (1) => m = 1,5n. Thay m vao (2): => R + 62. l,5n = 1,5921. (R + 48n) R + 93n = 1,5921. R + 76,42n => 0,592IR = 16,58n => R = 28n. (2) ^^ +3 2P + 302 2PCI3 photpho triclorua +5 0 du clo: a a/2 an/ 2 Theo bai ra: an/ 3 = an/ 2 => v6 If. Didu nay chihig to R c6 hoa trj thay d6i. Khi R tac dung vdi HNO3 la ch&t oxi hoa manh th^ hidn hoa tri cao la m: PTPlT: 3R + 4mHN03 3R(N03)„ + mNO + 2mH20 a a am/ 3 Ta c6 he: 0 Ah ->r 2PCI5 photpho pentaclorua 2P + 5C1.2 Axit photphoric la axit ba nSc, cd d6 manh trung binh, c6 ta't ca nhOng tinh cha't chung ciia axit. Khi tac dung vdi dung djch ki^m, tuy theo lucmg chSt t&c dung ma axit photphoric tao ra mu6'i axit, hoac mu6'i trung boa, hoac hop cac mu6i dd. Datk=-5^i^.ne'u: "H3PO4 k < 1: Tao ra KH2PO4, H3PO4 du. k=l: TaoraKH2P04. 1 < k < 2: Tao ra KH2PO4 va K2HPO4. k = 2: Tao ra K2HPO4. 2 < k < 3: Tao ra K2HPO4 va K3PO4. k = 3: Tao ra K3PO4. k>3: Tao ra K3PO4, K O H du. Bi^u d i l n tr&n true s6': K 2 H P O 4 + K O H ^ K 3 P O 4 + H2O KH2P04 KH,P04 K2HPO4 H3PO4 du K2HPO4 K3PO4 1 K3PO4 3 2 f K2HPO4 KH2P04 0,01 KOHdu Pap ^0,01 0,01 dung la B. 'jfiidu3: Thanh phSn chinh cua quSng photphorit la A.Ca3(P04)2. K3PO4 B.NH4H2PO4. Thanh phSn chinh cua quang photphorit la Ca3(P04)2 (canxi photphat). Pap an diing la A. fhidu 4: Cho 0,1 mol duoc CO cac chat Hu&ng ddn gidi Theo bai ra: np = 3,1/31 = 0,1 (mol); 0,1 C. K 3 P O 4 , K O H . , ilNaOH_ = 2l^ ' * Cac PTPir (CO thi CO): 0,1 (mol) P2O5 0,2 + 0,1 (mol) Vi 1 < J 1 K O H _ ^ + 2NaOH - > Na2HP04 + 2H2O ^ D. H3PO4, K H 2 P O 4 . Hu&ng ddn gidi nH3P04 0,1 ,* ' = 2 =^ Xay ra phan ling: H3PO4 B. K 2 H P O 4 , K H 2 P O 4 . (Trich De thi tuyen sinh DH - CD khoi B) P2O5 + 3H2O - > 2H3PO4 ^ vao dung djch chiia 0,35 m o l KOH. Dung djch thu n , 0,05 (mol) 0,05 P2O5 A. K 3 P O 4 , K 2 H P O 4 . = 0,2.1 = 0,2 (mol) nN,0H 4P + 5 0 2 ^ 2 P A I f e i €lM D . CaHP04. Hu&ng ddn gidi Thi du 1: O x i hoa hoan toan 3,1 gam photpho trong khi oxi du. Cho toan b6 san ph^m vao 200 m l dung dich NaOH IM de'n k h i phan ting xay ra hoan toan, thu duoc dung djch X. Kh6'i luong mu6'i trong X la A. 16,4 gam B. 14,2 gam C. 12,0 gam D . 11,1 gam (Trich de thi tuyen sink DH khoi A nam 2013) J*v: C. Ca(H2P04)2. (Trich De thi tuyen sinh DH - CD khoi B) 2. C a c thi du minh hoa: PTHH: m^P4pMV'-4nM: 0,1 (mol) nH3P04 Vay kh6'i luomg mu6'i (Na2HP04): m = 0,1. 142 = 14,2 (gam). 3H2O 2:2^ ^2 — > 2H3PO4 0,2 (mol) tao ra K H 2 P O 4 va K , H P 0 4 : 0-2 H3PO4+ K O H >KH2P04+H20 feaw H3PO4 + 2 K O H ). K 2 H P O 4 + 2H2O Dap an dung la B. <^ Thidu 5: Cho 100 m l dung djch KOH 1,5M vao 200 m l dung djch Dap an dung la B. Thidu 2: Cho 1,42 gam PjO, tac dung hoan toan vdd 50 m l dung djch KOH IM, thu duoc dung djch X . C6 can dung djch X thu duoc chat ran khan gdm ' *>«»i ."•Vj 1 H3PO4 0,5M. A. KH2PO4 va K2HPO4. B. K2HPO4 va K3PO4. thu duoc dung djch X . C6 can dung djch X , thu duoc h6n hop g6m cac chat la C. K3PO4 va K O H . D . H3PO4 va KH2PO4. A. K 3 P O 4 va K O H . (Trich de tuyen sinh Cao dang khoi A) Hu&ng ddn gidi B. K H 2 P O 4 va H 3 P O 4 . C. KH2PO4 va K2HPO4. . Hu&ng ddn gidi Theo bai ra: np^Oj =0,01(mol); n ^ o H =0,05 (mol) P205 + 3 H 2 0 ^ 2 H 3 P 0 4 0,01 Vi 0,02 = 2,5<3 2< nH3P04 0'02 H3PO4 + 2 K O H 0,02 Theo bai ra: n ^ o H = 0,1. 1,5 = 0,15 (mol); nHjpo^ = 0,2. 0,5 = 0,1 (mol). V. i<-ilKOH_ = 0J5 ^ j ^ ^ ^ 2 n e n : nH3P04 KOH + H 3 P O 4 Tao ra 2 mu6'i la K 2 H P O 4 va K 3 P O 4 K 2 H P O 4 + 2H2O 0,04 -> 0,02 D . KH2PO4 va K 3 P O 4 . (Trich De thi tuyen sinh Dai hoc khoi B) KOH + K H 2 P O 4 0,1 > KH2PO4 + H 2 O > K 2 H P O 4 + H2O V ''i .A Vay thu duoc h6n hop KH2PO4 va K2HPO4. ^ E>ap an dung la C . Jb.< 217 3. Cac bai tap t u luyen: Cku 1: Thuy phan hoan to^n hgrp chat 20,625 gam PQ, thu diroc dung dich X chiig h6n hcrp 2 axit H3PO3 va HCl. Th^ tich dung dich NaOH 2M d^ trung ho^ dung dich X la A. 450 ml B. 225 ml C. 750 ml D. 375 ml :.v^Ar'':^ f-^^ :A':,;V'I' Vi 2 mu6'i NaH2P04 Na2HP04 c6 n6ng d6 mol bang nhau {thd' tich cung blng Vi: phau) => 2 mu6'i c6 s6' mol bang nhau (x mol) Yi nguyfin t6'natri bao toan n6n: + 2x = 0,6 => 3x = 0,6 => X = 0,2 (mol) 1;; ; Vi nguydn t6'photpho bao toan ndn: ,>i X Hu&ngddngidi X +X- Theobaira: npcij = = 0,15 (mol) PCI3+3H2O PTPlT: 0,15 HCl + NaOH ' _ Axit photphorof ''''''' '':'''\}'^^ 0,15^ => "NaOH -> 0,3 (mol) ':[}Y]n/^'^4:f): ^ > Na2HP03 + 2H2O Mud'i trung hoa A.Na3PO4 0,4M C. NaH2P04 0,4M =0-45+0,3 = 0,75(mol) ^'•'^ • W H = ^ = 0.375 (1ft) = 375 (ml) Dap An dung la D. Chu V. -CrCTcuaaxitphotphoraHjPOj: . • ^ , ' ; ^, 21.875.1,28.25 > . W^^^^^^^ fi:0'',;f "NaOH . = 0,175(mol) 100.40 ' V ; > NaH2P04 + HjO ' = , -P-O-H 0,1 H3PO3 mdiaxit: . • Theo bai ra: n^^po^ = 0,1.1 = 0,1 (mol); NaOH + H3PO4 H-O • B. NaH2PO4 0 , l M vaNa2HPO4 0.3M D. Na2HP04 0.1M va Na3P04 0.3M. Hu&ng dan gidi 1.'-) ' fXivjij H3PO4 + NaOH ^ NaH2P04 + H2O * ' '' ' H3PO4 + 2NaOH Na2HP04 + 2H2O v^fp cau 3: Cho 100 ml dung djch H3PO4 IM tac dung vdi 21,875 ml dung dich NaOH 25% (d = 1,28 gam/ml) sau do dem pha loang bang nirdc ca't thu duoc 250 ml dung dich X. Hoi trong X c6 nhSng hop chat nao ciia photpho va n6ng d6 mol Ih bao nhidu (bo qua su thuy phan ciia cac mu6i)? ^NaCl + H j O H3PO3 + 2NaOH 0,15 P2O5 + 3H2O -> 2H3PO4 0,45 (mol) 0,45-> 0,45 (mol) m = 71.0.4 = 28,4 (gam) £)ap an dung la B. y: Cac phuong trinh phan ihig xay ra: >H3P03+3HC1 > 0,2 + 0,2 = ^ H 3 P O 3 + HjPO^ :," C6n: < - 0,1 0,075 0 '^ * 0,075 C6n: 0.1 (mol) 1 ^ > Na2HP04 + H2O < - 0,075 0,025 ( 0.1 (mol) NaH2P04 + NaOH H2P03?:iH^+HPO^' Do vay H3PO3 chi tac dung vdi NaOH theo ti Id toi da -> 0,075 0 ^ 0,075 (mol) "/^ ' ; Vay trong X c6 NaH2P04 va Na2HP04 Ndn mu6'i NajHPOj tuy con nguydn t6' hidro trong g6c axit nhimg 1^ mu6i trung hoa. Cau 2: Cho m gam PjO, vho 300 ml dung djch NaOH 2 M thi dung djch sau phan ling chi chiia 2 mu6'i NaH2P04, Na2HP04 c6 n6ng d6 mol bang nhau. Gia trj cua mia A. 14,2 . V i B.28,4 C.21,3 D.71,0 ^ O Theo bai ra: Hu&ngddngidi nNaOH = 0-3.2 = 0,6(mol) ,^.,,^^„, ,j , ,, C 'NaH2P04 0,025^,,,.^ ="5;^ = ° ' ' ^ Dap an dung la B. ^ CN,3„PO4 0,075 =-^=0,3M • ^1 v I/' " ' • ' ^^u 4: D6't chay hoan toan 6,2 g photpho trong oxi du. Cho san phim tao thanh tac dung vCta du vdi dung djch NaOH 32% tao ra mu6'i Na2HP04. 3. Vife't phuong trinh hoa hoc ciia cac phan umg xay ra. l ' . Tinh kh6'i luong dung djch NaOH da dung. c. Tinh ndng d6 phin tram ciia mu6i trong dung djch thu diroc sau phan itng.
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