Cẩm nang ôn luyện thi đại học 18 chuyên đề hóa học 3

Cty TNHH MTV DWH Khang Vi$t Ca'm nang an luy^n thi d^i hgc 18 chuy6n 6i H6a hpc - Mguyjn Van Hii X CO chxia: n^^3+ = 0,04; = 0,08; n ^ ^ ^ , = 0,02; Na* va D k h u H2O v a oxi hoa i o n CI".' •* V** '"'•^ C k h u i o n Na"^ va o x i hoa i o n CI" SOl' Laigidi: 3Cu Cu + 8H* + N O ; + 2Fe3* -> > 3Cu2* + N O + H O > Cu^^ + 2Fe2- = 0,03 + 0,02 = 0,05 m o l P H A N L T N G D I E N P H A N * ' >'* ; £0.0 - ""P"' > N a + CI2 l^»d m i uO -y , "^""^ ) M g + CI2 MgCh 2NaCl + H O - ^ E ^ ' N a O H + Cht ' >: ' Cyc a m (catot) c6 H2O va Na* - > H2O b i d i e n phan: 2H2O + 2e )• H2 + 2OH- (Qudtrinhkhu) Cue d u o n g (anot) c6 Ch va H2O ;i r-,;r: , Ch b i dien phan: ^ , , , , V i d y 2: D i ^ n p h a n d u n g d i c h muo'i MSO4 ( v o i d i e n cue t r o ) v o i c u o n g d p d o n g ' + H2t ''^i*'- di?n k h o n g d o i bang 1,5A Sau 1351 giay t h i d u n g d i ^ n phan, catot ehua CO b p t k h i va kho'i l u g n g catot tang 0,672 gam C o n g thuc muo'i la T r u o n g h o p 2: C h i c6 i o n k i m loai b i d i ^ n phan: CuS04 + H2O _ i E ^ Cu + ' Nhuvay: 2C1> Ch + 2e ^j^,^^^^ {Qua trinh oxi hoa) D a p an D Dien phan dung dich T m o n g h o p 1: C h i c6 go'c axit b i dien phan: n D a p an A w.' ,n is'ufh 6-.J / : i ;, • " A n h la anot n h u o n g e - E m la catot nhan e t h o i m a " ' = 0,05.64 = 3,2 g a m VCXQ^ - ' Van d u n g cau t h o : ^ l i ;+.«fcO£ *^f^.tr1-^ a L i thuyet: m,-t.,-n * Dze« p/jflji nong chay 2NaCl s ' 't:'! fh> A.CUSO4 -O2 + H2SO4 B.ZnS04 C.FeS04 D NiS04 Phan u n g d i e n p h a n : A g N + H2O - ^ E ^ A g + ^ + 2HN03 T r u o n g h o p 3: Ca goc axit va i o n k i m loai d e u b i d i ^ n phan: CuCh * "P"" > C u + CI2T j^nu) Qua trlnh oxihoa-khie tren cdcdien cue ''^ ! i) - , + H2SO4 ™ T: A A.I.t Theo cong t h u c Faraday: m = > , gmu j^nntirf ^, Cau tho: A n h la A n o t n h u o n g e, E m la Catot n h a n e t h o i ma A n o t - qua t r i n h o x i hoa; Catot - qua t r i n h k h u Alt * MSO4 + H2O - J P ^ M + i o Dinh luat Faraday ( t i n h l u p n g chat t h u d u o c cac d i ^ n c^rc): m = f nF A : kho'i l u g n g m o l n g u y e n t u ( m o l p h a n t u ) cua chat t h u duc^c b d i e n cure, c u o n g d p d o n g d i e n (ampe) d nfi qE 02! 0,16 ^ C u + - O + H2SO4 0,04 ••' a04 O + 4H* {Qua trinh oxi hoa) -• Dap an A Vi rO': 1' -fV"': • 8: Di^n phan 1,5 lit dung dich AgNOs 0,1 M voi di^n eye tro t gia, 100%), thu dugc chat ran X, dung dich Y va Z C h o 12,6 gam Fe vao Y , sau cac phan u n g ket thiic thu dugc 14,5 gam hon hgp k i m Idai va +4H* 0,04 V = 0,04.22,4 = 0,896 lit + H2O -^S^ cuong dong dien khong doi 2,68A (hi^u suat qua trinh di^n phan la > Cht , ^ ^ Mol: 0,04 - > 0,04 - > CuS04 Laigidi: r-'v-nh 2e 0,04 C k h u Cu2* va oxi hoa 504' di^n phan la 100% Gia tri ciia m la: 2C1- - 0,04 A oxi hoa H2O va k h u ion Cu^"^ 2,68A, thu dugc V lit (dktc) bay a anot Biet hieu suat ciia qua trinh '-'I'lh -> thi cae qua trinh xay cue am va cue duong Ian lugt la ,, s n i j nerf 1;,:;, 0,04 ' i::;/ ^ '••'^•^> V i dy 7: K h i dien phan dung dich CuS04 (vai di^n cue tro, eo mang ngan xop) cue tra, mang ngan xop) thoi gian gio voi dong di$n c6 cuong dp la A^c; ; -> Dap an D V i du 5: Dien phan dung dich X chua 0,12 mol CuS04 va 0,04 mol NaCl (dien ' •J _> Khoi lugng dung dich giam = 64.0,08 + 0,04.71 + 0,02.32 = 8,6 gam ny- DapanA A 0,560 + Na2S04 T h u tu dien phan n h u sau: Theo bai, dung dien phan catot bat dau c6 thoat —> CuS04 vua Ta c6: > CuCh + 2NaCl Mol: 0,04 2Ag + i + 2HNO3 2a ^ , ': • _ 2a 123 Cty TNHH MIV UVVH Khang Vi$t Ca'm nang an luy$n thi djii hgc 18 chuy6n H6a hqc - Nguygn Van H&'\ Nhdn xet: V i cho Fe vao d u n g d i c h Y t h u d u o c h o n h o p k i m loai —> Y chiia A g N O s —> A g N O ^ chua b i dien p h a n het Mol: + 0,5a D a p an A 'fii'''' L i thuyet O^Hf BOH ixo * MMOZ nitrat: AgNOs — A 40% ^ CaC03.MgC03 P C84% Lin giai: ^ — ^ CaO ^'.ilf^y' '->•' Av.-v;ty> D 92% • + M g O + 2C02t 0,2 gnuG QA (n-ys-f ! , ' gj^^j^^-j, D a p an D " ''^'^ % n i c a C O M g C O = ^ - 0 % = 92% ^ n.i :*-;H- mCaC03.MgC03 = 0,2.(100 + 84) = 36,8 g a m io2 io2 A g + NO2 + _ Phan u n g nhiet p h a n : Mol: C u O + 2NO2 + ^ , B.50% •"'^ if• 2KNO2 + 02 Cu(N03)2 — n k h o i l u g n g ciia C a C M g C t r o n g loai quang tren la " ' ' 2KNO3 i;.(f- >ttf^ tap cha't t r o sinh 8,96 l i t k h i CO2 (6 dktc) T h a n h p h a n p h a n t r a m ve in a + O2 > V i dv (B-08): N h i e t p h a n hoan toan 40 g a m m o t loai q u a n g d o l o m i t c6 Ian ^ ^.^.^^^^.^ P H A N L f N G N H I E T P H A N , D O T C H A Y : to f s i i s u y i r t -n'u ,.3, , 2KC1 + 3O2 ^ 2KC103 ••!'•! nfiifc :f, CaO + CO2 ^ - = 0,1 m o l - > t = 3600 giay = gio + M n + O2 C u O +2NO2 + i o 2 Cu(N03)2 CaC03 — M a t k h a c : n^gNOa = 2a + 2b = 0,15 ~ ^ ^ =0,05 m o l ; b =0,025 m o l : i C i n^= , ^ f i ^ t j i , pffifu:i cac mum v6 co Cac phuong trinh phan ling: Cac p h a n u n g k h i cho Fe vao Y: i''"»ii'''V ^i'v :^.,.' ' 'f:^'.-']/!; ,, vv^'uvjr"'"' Fe Lai giai: l^han xet: D a y la dang cau hoi li thuyet kiem tra cac em ve ben nhi^t ciia J^^e V i d\ 3: N u n g n o n g cac h o n h o p bot ran t r o n g b i n h kin: C u + Cu(N03)2 (1); C u * Muoi cacbonat: CaCOs C a O + CO2 2FeC03 + -O2 * , * + KNO3 (2); Fe + S (3); MgCOs + M g (4) So t r u o n g h g p xay s u o x i hoa k i m loai la: A FeaOs +2CO2 Mudihidrocacbonaf' ,0 2NaHC03 ' ' f ' i ' ^ ^ ' V 'b.i> - m b t.t : ) H l i q ^x* ? V' ' ' -"f '(Ofl,* ( o i b '"•'^h r^*-' '}T('jr Na2C03 + CO2 + H2O , >, , Ca(HC03)2 — CaC03 + CO2 + H2O j ^ ^ , ^ 4FeS2 + I I O ,0 "• "''fidA v i j b 'Uff -^—^ 2Fe203 + 8SO2 1,1 M , , 8,(J„A b V I D V M A U KCIO3 So muoi t r o n g day k h i bj n h i ^ t phan tao so m o l k h i I o n h o n so m o l muo'i t h a m gia p h a n l i n g la: ^ B.3 ' ' C D '' C + O2 '" > C u O 2KNO2 + O2 (2) K N — ^ 2Cu + O2 — + 2CuO '" > FeS (4)MgC03 — ^ Mg ^ CO2 '|f D ^, j |,,^ j^^,^, CuO + 2NO2+ (1) Cu(N03)2 (•^) Fe + S V i du 1: Che day cac m u o i : K M n , N a N , Cu(N03)2, CaCOs, A B Cac p h u o n g t r i n h hoa hpc: 2Cu MMO? sunfua, disunfua: f , ; io2 ' HEoH f - ;M'.;e • '• - C u b } o x i hoa j , ^ , , BV / >i ^ ^ Cubioxihoa -> Febioxihoa M g O + CO2 '" > M g O + C -> M g b i o x i hoa - > D a p an D 125 Ca'm nang On luy^n thi dgi hgc 18 chuy6n dg H6a hgc - NguySn Van H5i Cty TNHH MTV DWH Khang Vigt Lieu y: hon hg-p (3), Fe la chat khir —> bi oxi hoa; hon hgp (4), M g c6 the chay CO2 V i d^ 4: Nhiet phan 18,8 gam Cu(N03)2 mot thoi gian, thu dugc 12,32 gam chat ran Hieu suat cua phan ung nhift phan la /fiv/ A 40% B.60% C.80% ^^^v^,, D.50% Ldigidi: f^OM:' -*0u*') m o l CI- i s '"ol Theobai: a m o l 1,1 0,1 m o l -> a = - » khoi Iirong tang 71 - 60 = 11 gam , , , tang , - , =1,1 gam M2CO3 + 2HC1 a mol M g {t>i> IMK imfi > 2MC1 + CO2 + H2O i'l/,, -^oifo MHCO3 + H C l > M C I + CO2 + H - t - a roBSli,.^ : N h a n thay : nx = n c o j = 0/05 m o l < ' i ^ b S""^' f ^ — 76 -> M x = ^ — = 115,2 -> M H C O < 115,2 < iVbCOs 0,05 M + K 115,2 < M + 60 -> 27,6 < M < , ^ M l a K - ^ i:;is::S c.;j,K, f i & o l ns g^j^, Dap anB ;\«l,t ^ -lnr,i ^ > a mol Cu b m o l Fe Theo bai: > h mol Cu mtang r ; -.^'^ : kho'i l u g n g tang (64-24)a = 40a -> tang (64 - 56)b = 8b = 6,9-5,1 = 1,8 gam ^ 8b + 40a = 1,8 _> a = 0,0375 m o l ; b = 0,0375 m o l - > m M g = 0,0375.24 = 0,9 (gam) Bai 7: ' '^ ^ ^ ^ ^ , D a p an A =V * — " Qua t r i n h d i ? n l i : ^ + NOJ > 2H* + H2SO4 SO I" '' D u n g d i c h di§n l i c6 m a t H * va N O —» hoa tan C u theo p h a n u n g : + 2NO: > 3Cu^* + N O t Fe = 56 da p h a n u n g m o t phan tao C u = 64 —> Y g o m Fe d u va Cu 3Cu G p i so m o l b a n d a u : Z n = x; Fe = y - > C u la chat k h u , N O J la chat o x i hoa 65x + 56y = 14,9 b + 5a = 0,225 Bao toan electron: n 2+ = 2nMg + 2npe(p.) -> 0,075 = a + b NaNOs Z n phan u n g truac r o i den Fe D o mx < m Y —> N^fln xet: T i n h k h u Z n > Fe ' - > Fe = 56 d a p h a n u n g m g t p h a n de tao t h a n h C u = 64 —> Z g o m Gpi so m o l M g ban d a u = a m o l ; npg( p.) = b N h a n thay: 22,4 V D a p an A ^ I\lhan xet: T i n h k h u M g > Fe —> M g p h a n l i n g t r u o c r o i m o i d e n Fe D o mx < MgS04vaFeS04 12 B a i 3: n(-02 = — = 0,05 m o l Cac p h i r o n g t r i n h p h a n u n g : ^ =— ^ 100% = 100% = 90,28% 65x + 56a 65.8a + 56a pai 6: jnv n c o = n ^ ^ - = 0/1 m o l -> V c o j = 2,24 l i t ->DapanC -> " o/„^ •^^^^ + 8H* + 4H2O ^ , '' , i i J g n o i< > > D a p an B Dat: nFe(p.)=a Nhan xet: Day la p h a n u n g d i ^ n p h a n h o n h g p —> nen d u n g cong t h i i c t i n h Bao toan electron: n ^ 2+ = 2n2n + 2npg(py^ -> 0,15 = x + a g M a t khac: mv = 15,2 - » 56(y - a) + 64(x + a) = 15,2 ^ % m z „ = 5:1:^.100% = 43,62% x = 0,1; y = 0,15; a = 0,05 D a p an D 2" 14,9 Cach 2: A p d u n g tang - g i a m k h o i l u g n g : ,rt: ^ x mol Zn > X mol Cu -> k h o i l u g n g g i a m (65 - 64)x = x a m o l Fe > a mol Cu ^ k h o i l u g n g tang (64 - 56)a = 8a V Theo bai: mwng ^ ' •^ ^ * '^^zn + ^^Ve{pu) ~^ Cu Catot: C u - 2e ' — ^ Mol: 0,1 < - 0,2 - > B a i 5: G o i so'mol ban d a u : Z n = x; Fe = y N h a n thay: X mol Zn > X mol Cu a m o l Fe > a mol Cu Theo bai: mtang = -> x = 8a 128 , y f - » k h o i l u g n g g i a m (65-64)x = x I -> Cu ' ^-'^^'^^^ SO^ 4r^-~m^^ C u + - O + H2SO4 + H2O CuS04 a aSa ^ a 64a + 32.0,5a = ^ a = 0,05 mol ' ^ - > Vci2 + = (0,05 + 0,025).22,4 = 1,68 lit Bai 10: Phuang trinh hoa hoc: Cu(N03)2 Cty TNHH MTV DWH Khang Vi$t l i X H CHAT CIJA CAC OXIT ; Dap an B , , iiCL- ' CuO + N O + I , ion" Cir, r ; a* * phan ung trao aoi Mol: Mol: =° ^'^S-^vd " i(.*m rifid o'X: y ^^^'^-^ = '^^"S '^""g * u c cua so duong cheo: "'^^^'^^ ° ' n NO2 32-37,6 5,6 2x 46-37,6 8,4 ' 0,5(x + y ) ~ ^ ^ ~ ^ ' ' ' , , , ' K Mat khac: I88x + l O l y = 34,65 x = 0,05 mol; y = 0,25 mol CU(N03)2 - > m^joj + ,>:•• '.• + O2 + H O ,SMV', A 80 ^ a ' ' = 02Mol: D 120 it o r „ p-g,a^ ^O'^"' ~* " o - 0,1 2H^ H20 0,2 ' *'' • ' ' '' •.MAm^Oi'A: "' = 0,1 mol -> V^d H2SO4 = Vay: n H S = :; ' = 0/1 lit = 100ml Dap a n C V i du 2: H o a tan hoan toan 5,44 gam hon hop X gom Fe203, FeO, CuO can 80 m l axit H2SO4 I M (loang) K h u hoan toan 5,44 gam X bang H2 (nung nong) thu dugc m gam kim loai Gia tri cua m la a 0,25a + 2,75b = a + 2b ^ ' ' C 100 B.60 axit tao H2O theo phuong trinh: , j i t *,tj '••,,,r „ j : , ;»»u*iD}: ^ I'^'-^^'^'P^tm&lig^^fim.d'^bmy^^ ' Khi cho oxit kim loai tac dung v a i axit, ion O^" oxit se ket hg-p voi ''^ A 4,32 4FeS2 + 1 — ! ^ -> 2Fe203 + 8SO2 b 2,75b - > 2b ^ Dap an B > Fe2(S04)3 + SO2 + 4H2O •• i Bao toan kho'i l u o n g t a c6: m o = 5,82 - 4,22 = 1,6 gam -> Fe203 + C O N;zfl«^e^ V i a p s u a t k h o n g t h a y d o i ; Way ' f ^ ^ ' ^ n f M l g 0,25a v ' Lcngidi: IH1=^ = l O - ' - > p H = l->Dapand:^ 0,3 + - ^ > 3Fe(N03)3 + N O + 5H2O i j,,, V i dv 1: Dot nong 4,22 gam hon hop X gom Mg, A l va C u oxi d u , thu V ^ " " N = n N - > n j ^ , = 0,03 m o l 2FeC03 „ b V i du mau -> H N Bai 12: Cac phan ung hoa hoc: ->gfti^O??^P'-fv'iv *n'Mvni,» 'J if,, > 3Fe(N03)3 + N O + 5H2O 2FeO+ 4H2S04(^flc) 0,5a V^JJ-M > 2Fe(N03)3 + 3H2O H S O I M Gia trj cua V la = 6,58 - 4,96 = 1,62 gam - > 46.2a + 32.0,5a = 1,62 - > a = 0,015 mol •••• 130 > CuS04 + H2O Fe304 + IOHNO3 CuO + N O + -O2 2a Mol: C u O + H2SO4 ,' } i i W i t ' • • L :'.i.ih 'itn.ih tfrti ' • dugc 5,82 gam hon hop Y De hoa tan het Y can to'i thieu V m l dung dich mart xet: K h o i lugng chat rSn giam = khol l u g n g k h i bay Phirong trinh hoa hpc: '' '^ymii fmf'Mttml t « MiirM Mol: > A I C I + 3H2O 3FeO+ IOHNO3 m c u ( N ) =0,05.188 = 9,4gam ^ D a p a n C Bai 11: 4N02 AI2O3 + 6HC1 » Phan ung oxi hoa-khu ••'-'-^ , « r ; n o i t o t e a « m c ' t' /I i^ififel I K H2O ^ ^ J • d u n g d i c h H2SO4 I M (loang, v u a du) Sau k h i cac p h a n u n g xay hoan t * * i t i, toan, t h u d u g c d u n g d i c h X C o can X t h u d u g c m g a m muo'i k h a n Gia t r j 'riXA + i f t * l i : ' ^, , „ Dap an B ^ m = 5,44 - 1,28 = 4^16 gam 5: H o a tan hoan toan 10,56 gam h o n h g p b g t Fe304 v a C u t r o n g 160 m l yi ^ Vgy: m o - =0,08.16 = 1,18 gam Vi Cty TMHH MTV DWH Khang ViQt 3: H o a tan het 20,88 gam m o t oxit sat bang d u n g d i c h H2SO4 dSc, nong cua m la « > • A 15,36 , C 17,28 B 23,36 dktc) Co can d u n g dich X, t h u dugc m gam m u o i sunfat khan Gia t r i o i a m la Gpi so m o l : Fe304 = a; C u = b Theo bai: 232a + 64b = 10,56 A 36 Nhqn xet: C h i c6 Fe304 p h a n u n g true tiep v o i axit Cac p h a n u n g hoa hgc: B.24 C 72 D 54 '^'1- Lcngidi: - i'-OlAm Nhan xet: O x i t sat FexOy tac d u n g v o i H2SO4 dac, n o n g i K, Fe304 + a = 0,04 Cac ban l u u y rang m o l FeO hoac Fe304 deu chiia m o l Fe^^ nen deu c6 Cu kha na ng n h u o n g m o l electron, d o do: = npe^Oy = 2nso2 ^ ^ F ^ x O y = ^ i"Fe2(S04)3= mm tfh • Vi = 232 (Fe304) B Fe304 va 4,64 C Fe203 va 3,20 D Fe304 va 2,32 0,02 ''' " > xFe2(S04)3 x.0,01 m = 232.0,02 = 4,64 gam - > Dap an B 0,04 n-BO,' C 75ml D 90ml ,n ' ^'"'^^ Oey, OK>1 BR.': • + moxi = m o x » = 4,46 - 3,5 = 0,96 gam -> H20 ~ nH2S04 = a045 m o l ' '' * ^fTfeisb ,i;:ilb D a p an A OXIT BAZO + C H A T KHU" (NHIETLUY5N) a L i t h u y e t : * n h u o n g m o l electron, d o d o : X = ^ O x i t sat la Fe304 moxi - ^ VHCI = 0,045 l i t = 45ml ^ « »OPrH'i : y « ' = 2-0,01= 0,02 m o l mkimid Suy ra: n^+ = a m o l ^ T r o n g m o l FeO hoac Fe304 deu chua m o l Fe^* n e n d e u c6 k h a nang |§ -> x.0,01 = 0,03 ^ 0,02 , > 96 = - r r - = 0,045 m o l 32 vjiAii p a^rtfJ Ywi> 'sr-i.' -> n o = a m o l - > nQ2-(„^i,) = 0,09 m o l g , u j,^ £, r.O Nhqn xet: O x i t sat FexOy tac d u n g v o i H2SO4 dac, n o n g - > SO2 t h i oxit la 2FexOy + 2FeS04 B.60ml o ^ + 2H^ — > FeO hoac Fe304 BaotoannguyentoFe: > CuS04 , • • K h i cho oxit bazo tac d u n g v o i axit tao nuoc: Lmgtat: 12 224 n F e ( S ) = ^ = ^'^^ m o l ; nso2 = - r r — = 0,01 m o l "Fe^Oy ' 6: C h o 3,5 g a m h o n h g p X g o m A l , Fe v a C u d a n g b g t tac d u n g hoan -» , , 22,4 0,04 Lcngidi: d i c h chiia 12 g a m m g t loai m u o i sat d u y nhat v a 0,224 l i t SO2 (dktc) Cong = npe^Oy = 2nso2 ^ + Fe2(S04)3 Bao toan k h o i l u g n g : A FeO v a 1,44 - - > b = 0,02 m o l A 45ml yjj thijrc cua oxit sat va gia t r i m Ian l u g t la + 2H2O tich d u n g d i c h H S O I M vira d u de p h a n u n g het v o i Y la , | 400 = 54 g a m - > Dap an D 400 0,04 Fe2(S04)3 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 4,46 gam The ^^.o > 3Fe2(S04)3 + - > m = 0,02.160 + 0,02.400 + 0,08.152 = 23,36 g a m - > D a p an B V i d u 4: H o a tan hoan toan m gam Fe^Oy bang H S O dac nong, t h u d u g c dung i 0,16 M o l : 0,02 - > 0,02 008 " F e x O y = - ^ J J = 0,09 m o l B a o t o a n n g u y e n t o F e : 2Fe304 > FeS04 4H2SO4 M o l : 0,04 Fe, C u B Fe203,Cu C A l , Fe, C u - = 0,015 - ^ = V c o = 0,02.22,4 = 0,448 l i t - > D a p an B iD :k.; "CO2 = 0,02 - » Vco2 = 0,448 B.4,48 C 2,24 mol CO L o a i A v a C , „, g i a m 16 gam -> lmol(COvaH2) Theobai: r (C02vaH20) G i a m 16 gam giam 4gam 0,25 m o l > V = 0,25.22,4 = 5,60 l i t D Fe, C u D 5,60 !-2-> C O - > K h o i l u g n g chat ran g i a m 16 gam lmolH2 f - - - lai p h a n k h o n g t a n Z Phan k h o n g tan Z g o m AI2O3, ax = 0,015; ay = 0,020 ^ ^ Laigidi: V i d v 2: C h o k h i C O ( d u ) tac d y n g v o i h o n h g p X g o m A I O , FezOa, C u O t h u d u g c chat r a n Y C h o Y v a o d u n g d j c h N a O H ( d u ) , k h u a y k i , tha'y A X = 0,015 m o l A 6,72 = 0,15 m o l ~ 0,05 mol = 1,12 l i t - > D a p an C ay ••'••;,'.(1J:' hoan toan, k h o i l u g n g h o n h g p r a n g i a m 4,0 gam G i a t r i ciia V !a Bao toan electron: ttg trao doi — ax .8 dv 4: C h o V l i t h o n h g p k h i (6 dktc) g o m C O va H2 tac d y n g v o i m g t l u g n g d u h o n h g p r a n g o m FeO v a Fe304 n u n g n o n g Sau k h i cac p h a n u n g xay ra" > CaCOa + H2O I\QQ = TXQQ^ = yC02 D a p an B 2nQQ M a 56 xFe + M a t k h a c : n p e i n o = 0,015:0,02 = : - > Fe304 Nhqn xet: K h i cho h o n h g p oxit tac d u n g v o i k h i C O t h i : H e traodcS = ^ a Nhdn xet: n o (oxit) = "co= _^ O day, cac e m can s u d u n g so d o phan l i n g : j ^C0 CuO, F e a •> X ) Muoinitrat + N O CO2 + Ca(OH)2 yCO — Cach 2: Laigidi: ; Mol: + -> O x i t sat la Fe304 D 3,36 , ncaco3 = 0,075 m o l gidi: Giai theo p h u o n g t r i n h hoa hgc r a n X p h a n l i n g v o i d u n g d j c h H N O d u t h u d u g c V l i t k h i N O (san p h a m B.4,48 •' 3: K h u h o a n toan m g t oxit sat FexOy n h i f t d g cao can v i r a d u V l i t k h i n o n g , sau m o t t h o i g i a n t h u d u g c chat r a n X va k h i Y C h o Y h a p t h u A 2,24 " Cach : ' ^ V i d u 3: D a n l u o n g k h i C O d i qua h o n h g p g o m C u O v a Fe203 '! H2O C O (dktc), sau p h a n u n g t h u d u g c 0,84 g a m Fe v a 0,02 m o l k h i CO2 Cong m e g e,fl o t O * - - » n c u o =0/05.80 = , O g a m - > D a p an D h o a n toan v a o d u n g d i c h Ca(OH)2 ^ C u + CO2 thiic ciia oxit sat va gia t r i ciia V la + CO2 T a n g - g i a m k h o i l u g n g : m o (trong CuO) = , - , = 0,8 g a m y t, < ' f I f + fit i ! n AI2O3 + N a O H Vi Nhan xet: K h i C O chi k h u d u g c C u O theo phan u n g : CuO D 4,0 g a m Lbigiai: n e n k h i cho k h i C O d u tac d u n g v o i '° > 2Fe + 3CO2 + CO — , K h i cho Y tac d u n g v o i N a O H d u t h i AI2O3 b i hoa tan h o a n toan: Kho'i l u g n g C u O c6 t r o n g h o n h g p ban d a u la: A 0,8 g a m AI2O3 , Chat ran Y g o m : Fe, C u , AI2O3 n u n g n o n g d e n k h i p h a n u n g hoan toan, t h u d u g c 8,3 g a m chat ran AI2O3 y: K h i C O k h o n g k h u d u g c im AI2O3 „ ' ri Dap an D 135j dm Cty TNHH MTV DVVH Khang Vi^t nang On luy$n thi djii h(?c 18 chuy6n d6 H6a hqc - Nguygn Van Hil K i m loai sau phan ung gom: A l = 0,02 mol; Fe = 0,09 mol V i dv 5: C h o 6,72 lit C O (dktc) di tu tir qua ong su nung nong dung 10,44 3, jChi tac dung voi axit: A l -> - H gam mpt oxit sat den phan ung xay hoan toan, thu duoc hon hgp Y CO ti khoi so v a i hidro bang 18,8 Cong thiic cua oxit sat va phan tram the tich cua CO2 Y la - t u.j' > A.FeO;28% B Fe203; 72% C Fe203; 28% Laigidi: D Fe304; 72% _> nH2 = 0,12 m o l - > V H = 2,688 lit ^CO 44 - 37,6 _ 0,12 S' riC02 28 - 37,6 ~ 0,18 Ta c6: no (oxit)= Fe304 dii dung dich chua a mol H C l Gia tri cua a la A 0,9 •i :E u 0,18.16 ^ ^^^^^ 56 A 81,0 ^ Ban dau: nAi = 2.0,3 = 0,6 mol "Fez03 = 48,2-0,6.27 160 , B.54,0 C.40,5 D 45,0 Mol: 0,2 AI2O3 2A1 Fe203 2Fe 0'2 0,4 • + , = 0,2 mol 1.0 0,4 Cac chat phan 2: nAi = 0,1 mol; nFe= 0,2 mol; nAi203 = 0,1 mol n H c i =3nAi +2nFe+6nAi203 = l , m o l Cr203 + 2A1 -> AI2O3 + 2Cr - » Dap an B 1,5 m^i = 1,5.27 100 90 = 45 gam Dap an D OXIT L I / O N G Sau phan ung hoan toan, thu du(?c 9,66 gam hon hgp tin X C h o X phan ling voi axit H2SO4 (loang, du) thoat V lit H2 (dktc) G i a trj cua .J ah qsQ' • D a p ? n A Hien tugng n h u sau: tu den het 100ml dung dich Y vao X, thu dugc m gam ket tiia G i a tri ciia m Jv , AhOstan: AI2O3+ 2KOH l^^^^M tan v a sui bpt k h i : , • ^»rr A l + K O H + H2O > KAIO2 + | H t < A 5,44 m.^abn'iiM -'ii + H2O > 2KAIO2 b r a S 1,0- ^ ,,5,:.,,,, , ^J®^' " • CO2 (du) vao Y thu dugc a gam ket tiia Gia tri cua m v a a Ian lugt la C 13,3 v a 3,9 • ,: Na20 + H2O i > 2NaOH A l + N a O H + H2O Mol: 0,1 Thoi CO2: NaA102 Mol: ^ > NaA102 + - H t 0,1 ' 0,1 + CO2 + 2H2O 0,1 V i d\ 3: C h o H2 (du) di vao 6'ng s u nung nong dvrng hon hgp X gom AI2O3, FezOs, C u O thu dugc chat ran Y Cho Y vao dung dich K O H (du), C Fe, C u ^ D A l , Fe, (fu Ldigiai: ' Lieu y: K h i H2 khong k h u dugc AI2O3 nen xay phan ung: Fe203 + 3H2 — ^ CuO + H2 2Fe + 3H2O C u + H2O 0,02 AIO2 Mol: ^ ' ' K h i cho y tac dung voi K O H d u thi AI2O3 bi hoa tan hoan toan c6 tinh AI2O3 + K O H 0,01 Mol: a02 V a y Z gom: Fe, C u —> D a p an C " ' :' , 'HOu- > 2AIO2 + 2H2O • ' + 3H* l 0,02 ' \^i\ > A P - + 3H20'^'''^ 0,03 " ' ' ^ H*-^^''' •+ ' ^ BaSOii m = mAi(OH)3 + mBaso4 = 0,01.78 + 0,02.233 = 5,44 gam - > D a p an A V i d\ 5: H o a tan hoan toan m gam hon hop gom N a v a AI2O3 vao nuoc, thu dugc k h i H v a dung dich X suo't T h e m tu t u dung dich H C l I M vao X, het 100ml thi bat dau xuat hien ket tua; het 200ml hoac 400ml thi deu thu dugc a gam ket tua G i a trj cua m v a a Ian lugt la A 13,40 va 3,90 C 21,05 va 3,90 B 13,4 va 7,80 ; D 21,05 va 7,80 ^^^->^'>''-^^'*' Lmgidi: v-ac phan ung hoa hgc: Na > 2KAIO2 + H2O 0,06 0,02 + H * + H2O - > Chat ran Y gom: Fe, C u , AI2O3 luong tinh: ,., ^ , pj^, > H2O • i'iii^ '^h •••• H* Ba2^ + S O)24" ' s*^"^^ ' * —> Dap an D B Fe203,Cu 0,03 Al(OH)3 « khuay ki, thay lai phan khong tan Z Phan khong tan Z gom ^ , Ba> + H - Mol: 0,02 - > 0,02 0,1 ' D 0,78 Mol: 0,02 0,04 0,04 Trong X: n ^ ^ _ = 0,02 mol; n „ 2+ = 0,03 mol; n = 0,02 mol OH Ba AIO2 - » Trong 100ml Y : n , = 0,07 mol; n^^2- = 0,02 mol; n^, = 0,03 mol H SO^ CI Mol ' va m = m N a o + mAbO-, = 0,05.62 + 0,05.102 = 8,2 gam ' Mol: 0,03 OH- + * ' '>'^ > Khi cho tu tu Y vao X: > Al(OH)3 + N a H C " ' a = mAi(OH)3 = 0,1.78 = 7,8 gam; A AI2O3, Fe, C u B a O + H2O AI2O3 + H - AI2O3 CO tinh luong tinh nen tan dung dich N a O H vtra tao ra: I • So mol moi oxit ban dau: B a O = 0,03 mol; AI2O3 = 0,02 mol D 8,2 v a 7,8 NMn xet: K h i cho X vao nuac, Na20 se tac dung voi nuac: ^ ^, ^ C.3,11 Cac phan ung hoa hoc: dugc 200ml dung dich Y chi chua chat tan nhat c6 nong 0,5M Thoi B 11,3 va 7,8 B 1,56 Laigidi: V i d\ 2: Hoa tan hoan toan m gam hon hop X gom Na20 va AI2O3 vao H2O thu A 8,3 v a 7,2 , +H2O AI2O3 > NaOH + i H + 2NaOH ' > 2NaA102 ' • ^^^^ + H2O HcJi/ri/'''' "^r , HORK;: < ' 139 Cty TNHH MTV DWH Khang Vigt Ca'm nang On luy$n thi dgi hgc 18 chuySn dg H6a hpc - Nguygn Van HSi —> Dap an C S + HNO3 (dac) Lim y: 1- Fe tan H2SO4 loang, nguoi; 2- A l , Fe khong tac di^ng voi + 2H2SO4 (dac) H2S04dac, nguQi Vi ^ : * 6: Hon hgp nao sau day khong ton tai nhi|t dp thuong? A C v a B Cl2va02 C H2SvaN2 Nhan xet: H2 va F2 c6 the phan ling manh liet ca bong to'i va nhi^t rat thap, c6 the gay no va toa nhieu nhi^t, tao HF ^DapanD 2Ag + O3 c, Hidrosunfua * ' D K2Cr207 > MnCk + C h t KMn04 + 8HC1 + 2HCI * +2H2O 2KC1 + 2CrCl3 + SCh -—> + VHiO K2Cr207 nhan electron nhieu nhat -> tao khf Ch nhieu nhat Dap an D * O2 + S — ^ * Fe + S — ^ S + H2 — ^ * FeS * * 4Fe(OH)2 + O2 + 2H2O 2H2S + O2 > NaHSOs Dieiiche '° > 2SO2 + 2H2O , , , ( -f, > 2S03 A: > 2HBr + H2SO4 — ^ , ' • ,j, ^, ^ ^ | , > S + 2H2O : l;;,M>c.ri 4FeS2 + I I O > 4Fe(OH)3 * > Na2S03 + H2O Tinhoxihoa s + 02 CO2 > H2SO4 + 8HC1 Tinhkhie SO2 + 2H2S Tac dung v&i hap chat f >• FeCh + H S t SO2 + Br2 + 2H2O H2S > 2S + 2H2O 2SO2 + O2 < > HgS SO2 CO + O2 — ^ ' 2SO2 + 2H2O 2NaOH + SO2 '' 2H2O , " *' * ' NaOH + SO2 O2 + 2H2 — ^ ' e Luu huynh dioxit a Oxi - luu huynh Hg + S Dieii che FeS + 2KC1 OXI - L l / U H U Y N H 3Fe + 2O2 '° > Fe304 Tac dung vai phi kim » + 2HN03 + H2SO4 Voi clo: H2S + 4Ch + 4H2O Mn02 CO the nhan vao Ian lugt la 1, 5, 6, MgO i- Tinhkhiemanh 2H2S + O2 — ^ Cach 2: Nhm xet: So mol electron ma mol moi chat: CaOCh, KMn04, K2Cr207, * > FhSi > CuSi j dieu k i f n thuong, dung dich H2S bi oxi hoa dan S: Dap an D M g + O2 — ^ '•:(') 2H2S + 3O2 — ^ , CaCh + Ch + H2O Tac dung vai kim loai M f^ , „ Voi oxi: H2S chay khong voi ngpn lua mau xanh nhat: > KCl +MnCl2 + - C I T + 4H2O K2Cr207 + 14HC1 CaOCh Tinhaxityeu H2S + CuS04 Cach 1: Dua theo cac phuang trinh phan ung: , > O2 + I2 + K O H H2S + Pb(N03)2 Laigidi: Mn02 + 4HC1 '° > Ag20 + O2 O3 + 2KI + H2O phan ling vai lugng d u dung dich HCl dac, toi phan ling hoan toan thi '" * > 2KC1 + 3O2 chat tao lug-ng CI2 nhieu nhat la: C CaOCh K2Mn04 + Mn02 + O2 b Ozon: „;,", ' ri Ozon la mpt chat oxi hoa manh, manh han ca oxi V i d^ 7: Neu cho mol moi chat: CaOCh, KMn04, K2Cr207, Mn02 Ian luot B.Mn02 3S02 + 2H2O 2KC103 £ »,; A.KMn04 > SO2 + NO2 + H2O Dieu cheoxi 2KMn04 — ^ D H2vaF2 ' Laigidi: '° • SO2 J.I - • 2Fe203 + 8SO2 Na2S03{rin) + H S C d j c , d u ) — ^ 2NaHS04 + SO2 + H2O 151 Cty TNHH MTV D W H Khang Vigt C^m nang On luygn thi d j i hpc 18 chuyfin 66 H6a hpe - NguySn v a n H i i f L i r a h u y n h trioxit - O l e u m SO3 + H2O Lai gidi: > H2SO4 nSOa + H2SO4 ^ : D a p an C Cac p h a n u n g hoa hoc: >• H2S04.nS03 (Oleum) -H', VIDVMAU 2SO2 + I' ' •' Vi dv 1: K h i n h i ^ t p h a n h o a n toan 100 g a m m o i chat sau: KClOa (xiic tac Mn02), K M n , KNO3 v a A g N O s Chat tao l u g n g O2 I o n nhat la A.KCIO3 C.KNO3 B.KMn04 Lai 2KMn04 — ^ D.AgNOa KNO3 — ^ KNO2 + io2 i^' rttt! ^^,,^^3 , D a p an A ' , nuh ^tvfmili B.Zn C A l tt^-$Bip' NfcOn 2M n02 — ^ + + 2nHCl ^ 2M20n -obio ' + n n nH2t " • j ^ ; - r - ^^^^^y • M n "e = 4no2 + " H + " ^ " + 2nH2 = 0,28 m o l A J o m € J - ' B.l ' C.2 Tat ca cac phat bieu d e u d u n g D a p an D ~ «> i D.4f-'" T' ' X-Offfj " JrW ;6x »^ • "^^^ V i d u 5: C h o 9,2 g a m h o n h g p X g o m Cu2S, CuS, FeS2 v a FeS tac d y n g het v o i H N O (dac nong, d u ) t h u d u g c V l i t k h i c h i c6 N O (dktc, san p h a m dich BaCl2, t h u d u g c 23,3 g a m ket tua; k h i cho toan b g Y tac d u n g v o i A 19,04 B 12,32 A D u n g d i c h BaCl2, CaO, n u o c b r o m >, ^ D 5,60 ••A rfi niXi -i- K h i cho d u n g dich Y + d u n g dich BaCh: > BaSa^^ «"soJoe;/ va N O i i ;V » • ' ^ , ^^*^^'?*^*' ^ ' " ^nol K h i cho Y + d u n g dich N H s d u : Fe3^ + 3NH3 + 3H2O > Fe(OH)3>l' > Cu(OH)2>l Lm y: Cu(OH)2 tan NH3 d u tao phuc chat: Vi dv 3: D a y chat nao sau d a y deu the hien t i n h oxi hoa k h i p h a n iVng v o i SO2? B D u n g d i c h N a O H , O2, d u n g d i c h K M n C 8,40 Nhan xet: D u n g d i c h Y chua cac i o n : F e ^ Cu^^ SO|", Cu2^ + 2NH3 + 2H2O M = 12n ^ n = 2; M = 24 ( M g ) - > D a p an D C O2, n u o c b r o m , d u n g d i c h K M n v.' i H So phat bieu d i i n g la Bao toan nguyen to S: ng (X)= r»BaS04 Cach 2: N h a n thay, M Ian lugt tac d u n g v o i chat oxi hoa la O2 va H2SO4 (H*) D H2S, O2, nuoc b r o m KK; Led gidi: • VK Vay: n = va M = 24 ( M g ) ^ D a p an D ^ • n = 0,28 ^ iM nrij : , o l f -.if i ' ih) $H Jif S f « , S x/ub u d t ! (d) M o o c p h i n va cocain la cac chat ma t u y Ba2- + sol n > K2SO4 + 2MnS04 + 2H2SO4 , • > M2(S04)n - ^ d u n g d i c h N H d u t h u d u g c 5,35 gam ke't tua Gia t r i ciia V la >• 2MCln + n H ' + nH2S04 4 M c o n d u 4M 5SO2 + K M n + 2H2O Lai gidi- '^'''^ ^'^^^ * Mg ^ Lai gidi: Cach : Giai theo p h u o n g t r i n h hoa hoc < > 2HBr + H2SO4 " t h u d u o c chat r a n Y H o a tan het Y t r o n g d u n g d}ch H2SO4 loang, t h u dugc 0,896 l i t k h i H2 (dktc) K i m l o ^ i M la , (c) K h i d u g c thai k h i quyen, freon (chii y e u la CFCI3 v a CF2CI2) p h a h u y Vi dv 2: C h o 3,36 g a m k i m loai M (hoa t r i k h o n g doi) tac d u n g v o i 0,05 m o l O2, A.Ca SO2 + Br2 + 2H2O tangozon N/ifln xef: KCIO3 cho lu(?ng O2 nhieu nhat ^ '.^ i - (b) K h i SO2 gay hien t u g n g m u a axit ' * '^'^^^^'^ * Ag NO2 + - O2 AgN03 ' (a) K h i CO2 gay h i ^ n t u o n g hieu u n g n h a k i n h > 2KC1 +3O2 2KC103 < ^^ V i d v : C h o cac p h a t b i e u sau: gidi: K M n + Mn02 + O2 , + ? ^ Cu(OH)2 + 4NH3 > [Cu(NH3)4](OH)2 5^35 ^''^^ -./CiSS.i Bao toan nguyen to Fe: npg (X) = ripe(OH)3 ^ ^'^^ m o l - -^ Bao toan k h o i l u g n g : m x = m c u "^Fe "^s ~» mcu = 9,2 - 0,05.56 - 0,1.32 = 3,2 gam ' ' , , ', dm nang On luygn thi djii hpc 18 chuyfin ii H6a hgc - Nguyin van H5i 32 ricu(x)==0'05mol Cty TNIHH MTV DWH Khang Vi$t yi > i Qui doi X ve hon hop gom cac don chat: Fe = a05 mol; Cu = 0,05 mol vaS = mol ^ n e ( x ) =2ncu +3nFe +6ns =0,85mol^^ ^ i|? Q^j.^r; nN02 " " e ( X ) =0,85 mol -> V = 0,85.22,4 = 19,04 lit ' ^.n >, ,,f).; ,0;.^ ^Dap an A V i dy 6: Hoa tan het 3,76 gam hon hop X gom Al, Mg va Zn dung djch 8: Hon hgp X gom O2 va O3, c6 ti khoi hoi so voi H2 la 19,2 Dan X syc tu tu vao dung dich KI du, thay c6 75ml di Cac the tich ciing dieu kien nhi^t dg, ap suat Phan tram the tich ciia O2 X la ^.30% B.45% C.60% D.40% Laigidi: Theo bai: M = 19,2.2 = 38,4 Ap dung cong thuc cua phuong phap duong cheo, ta c6: HCl, thu dugc 2,912 lit Ha (dktc) Mat khac, dot chay hoan toan 3,76 gam M no3 A 5,84 Phuong trinh hoa hgc: C 4,28 D 4,96.,^,: .qiiOoM ( h : Laigidi: 912 nH2==0,13mol ' M-gniAu%Mlsfki'cy: Nhan xet: Khi cho X + HCl, ion H* da nhgn electron de tro Hz: :;,•{„.?„,,';; 2H^ + 2e , 48 - 38,4 hon hg-p X oxi, thu dugc m gam oxit Gia tri cua m la tniih u& (a; B.4,80 Mol: v.bn ^ > Hat 0,26 O3 6,4 x = l,5y Oit + 2KI + H2O Mol: y -» Thetichkhithoatra:x + y = 75 -> x =45;y = 30 gnsorW OB-> gb %Vo, = - 0 % = 60% ^ Dap an C Mol: a065 CaCk + H t A 71,3% N a C l + N2t + 2H2O 0,1 N2 + " ' (PHs k h o n g tac d u n g v o i H C l ) Ca(H2P04)2 t r o n g loai p h a n b o n la - > nN2 = nNH4CI = 0,1 m o l - > V = 0,1.22,4 = 2,24 l i t - > Dap an A ,j nu • ' Laigiai: N H C I + NaN02 — ^ A 20% ' > C a C l + 2PH3t Phan ling hoa hoc: fj • ' - > V = (0,02 + 0,01).22,4 = 0,672 l i t - > D a p an A 40% = ' r n o l ; nNaN02 = ' m o l 0,1 M V i d\ 5: Phan supephotphat kep t h y c te san xuat d u g c t h u o n g chi u n g v o i Laigiai: Mol: D 448 Ca3P2 0,02 Ca + 2HC1 ung hoan toan, thu dug-c V lit (dktc) Gia trj cua V la "NH4a + Ca3P2 + 6HC1 V i d^ 2: Dun nong 100ml dung dich chua N H C I I M va NaN02 M d e n phan B 560 C 336 Y gom: Ca3P2 = 0,01 m o l ; Ca = 0,01 m o l -> Dap an D A 224 B 560 Laigiai: 3Ca 2N2 + 6H2O ,„,, ^„ UO t nca = 0/04 m o l ; n p = 0,02 m o l '° > 2P2O5 502 , C.16% D 23% ' ' , Lcngidi: Theo bai: M x =3,6.2 = 7,2 -> 2NH3 A p d u n g cong thiic a i a p h u o n g phap ducmg cheo, ta c6: 2x N h | n thay sau phan ung so m o l giam: 4x - 2x = 2x 28 - 7,2 "N2 = — - » Dat so m o l X: n ^ j = 1; ~ ^- 2-7,2 -> n x - n Y = x - ^ ( l + ) - , = x - > x = , m o l - > H = % - ) - D a p a n A IRQ 159 dm Cty nang 6n luygn thi dgi hgc 18 chuy§n dg H6a hpc - Nguyin Van HSi Khi cho Fe(OH)3 va Cu(OH)2 vao dung dich N H s du thi Cu(OH)2 tan tao Phan ling hoa hpc: N2 • Mol: X + 3MgCl2 + Mg+2Hci—>Mgci2+H2t-^ , - Cacbon • / , A: ' : „ ,; , nHi\f r r a i f ' ' pjnvwto ,^^^.^^ 2NH3 ^ '^^'H ••• M • ' '^xy u:) • Tac d u n g v o i o x i : C + O2 > CO2 CJ nhiet d p cao, cacbon lai k h u d u p e CO2 theo p h a n l i n g : > 2CO « P> •sfK 'h H -mmM;.::,' : ^ • + " ^rixfcii J ; CO2 + 4NO2 + 2H2O •iw'Bin'lYujin CO + Z n D T i n h chat hoa hpc • vff(m , rsfto • **yr4^*' i i ! * ! - " •f"'-^' J^TJS '• C + CO2 nMg = 0,5 m o l ; n^^^ = 0,1 m o l Phan u n g hoa hoc: 3Mg Ket tiia Z la Fe(OH)3 Tinh khu: vu,' 2x = 0,4 m o l -> x = 0,2 m o l -> H = 20% - » D a p an B B.4,48 phuc chat tan [Cu(NH3)4](OH)2 C A C B O N - S I L I C 100 A 3,36 MTV DVVH Khang Vi$t VIDVMAU Vi di;i : Cho C O du di vao ong su nung nong dung hon hpp A I O , Fe203, CuO Sau phan ung hpan toan, chat ran thu dupe ong su gom A A l ; F e ; Cu B A I O ; Fe203; Cu C A I O ; F e ; Cu D A l ; Fe; CuO 161 dm Cty Tr\iHM MTV n v v H Khang Vijt nang 6n luy§n thi d?i hgc 18 chuyeii Jg Hua hoc - Nguyen VSn H4i Lot gidi: NMn xet: K h i C O chi k h u dugc cac oxit ciia cac kim lo^i dung sau Z n M - 3»; H'"\ VXM ' p day di?n hoa nen AI2O3 khong bi khu Fe203 + S C O — ^ + 3CO2 2Fe •• CuO + CO — ^ Cu + • CO2 , ; ->• D a p an C -.rrod-K V , a, » Mol: SO2 + Br2 + H2O 0,02 < - a02 > H2SO4 + 2HBr >fi •- -» ms = 0,02.32 = 0,64 gam -> % m c = 95,1% - > D a p an A j ifib,,- ' each 2: •'im.:,: > SO2 Bao toan nguyen to S: S -> ns= nso2 = H B I J > H2SO4 = 0/02 mol ^ m s = 0,64 gam -> % m c = 95,1% - > D a p an A V i d v 2: D a n tu tu V h't khf C O (dktc) di qua mot ong s u dung luong d u hon V i dy 5: N u n g nong 16 ho quang di^n hon hg-p chua 11,2 gam C a O va 5,4 hgp ran gom C u O va Fe203 (6 nhi^t dp cao) Sue toan bp lugng khf thu dupe gam C (than coc) den phan ung hoan toan thu dugc chat ran X va Y () sau phan vao dung dich C a ( O H ) du, tao gam ket hia G i a tri ciia V theo phuong trinh phan ung: C a O + C 1^ A 0,224 'gniir B 0,448 C 0,896 D 1,120 LaigiAi: Baotoannguyentocacbon:CO > CO2 > CaCOs -> nco=nco2 =ncaC03 = 0/04 m o l V c o = 0,896 1ft -» - The tfch dung djch H C l 2M can thie't de tac dung het voi chat ran X la A 50ml fOVIHI* + ^ V i d v 3: C h o cac chat bot m a u trang: N a C l , Na2C03, BaCOs v a BaS04 C h i c6 A NMn - < , ^ ' B ^to^J/ C J' D B 100ml 112 54 56 12 ncaO= -TT " D a p an C H2O v a CO2 C O the nhan biet dugc toi da bao nhieu chat? >• CaC2 + C O CaO + 3C 0,15 < - 0,45 C 150ml Lot gidi: D 200ml = 0,45 mol — i - > CaC2 + C O S^Hfb Mol: •Ofb.J/i ^ X gom: CaC2 = 0,15mol; C a O = 0,2 - 0,15 = 0,05 mol xet: D u n g nude de thu tfnh tan: Hai chat tan la N a C l va Na2C03; hai CaC2 + 2HC1 CaO + 2HC1 0,15 > C a C h + C2H2t > C a C h + H2O ntyicf^fttf, , 04 chat khong tan la BaCOs va BaS04 Sue khf CO2 vao hai cha't khong tan (ngam nuac), chat nao tan la BaCOa: BaCOa + CO2 + H2O > Ba(HC03)2 Lay dung dich Ba(HC03)2 v u a tao cho vao dung dich d y n g hai chat -> VHCI = ^ =0,21ft = 0 m l ^ D a p a n D , , V i dy 6: H e n hop X gom C u O va Fe203 Hoa tan hoan toan 44 gam X bSng dung dich H C l , thu dugc 85,25 gam muoi Mat khae, neu k h u het 22 gam X tan, neu xua't hi^n ket tua la dung dich Na2C03: bang C O (du), dan hon hgp thu dugc vao dung djch Ca(OH)2 (du), tao Ba(HC03)2 + Na2C03 m gam ket tua G i a tri cua m la > BaC03l + 2NaHC03 —> N h a n dugc ea chat —> D a p an C A 45,5 B 50,0 C 40,5 Lot gidi: V i d u 4: Dot chay hoan toan 1,3 gam than (chua tap chat luu huynh) oxi du Dan hon hgp khf tao qua dung dich brom (du), thay c6 0,32 gam NMn brom da tham gia phan ung H a m lugng (%) ciia cacbon m a u than tu O oxit dugc thay the bang hai nguyen tir CI tren la mol O^- •> [h,- A 95,1% B.91,2% iJkHS + t-ifc C.93,4% Lai gidi: Cach 1: C i a i theo phuong trinh hoa hpc C + O2 — ^ S + O2 CO2 '" > SO2 Khf SO2 tac d y n g voi dung dich brom: xet: K h i cho X + H C l thi oxit chuyen muoi clorua va mot nguyen a mol D 97,8% Aljr D 37,5 a= * , > mol C h - > khoi lugng tang 71 - = 55 gam v " no(X) =0,75mol oavlj as&fU Khi cho 22 gam X tac dung vai C O thi: n o (X) = nc02 = ^C^CO^ = 0,375 mol - > m = 0,375.100 = 37,5 gam —> D a p an D 163 Ca'm nang 6n luygn thi dji hqc 18 chuyen 6i H6a hqc - Nrjiiyeii van Hai BAI TAP O N L U Y E N Cty TNIHH MTV DWH Khang Vijt n^f Bai 1: Cho hpn hg-p X gom O2 va CI2 tac dung vira du voi hon hop bgt gom 1,2 gam M g va 1,3 gam Zn, thu dugc 5,27 gam hon hgp chat ran Y Phan tram the tich cua O2 X la A 80% B.40% C.50% ' D 60% Bai 2: Di#n phan h't dung dich hon hgp gom NaCl 0,5 M va KCl 0,1M (dien cvc tro, CO mang ngan xop) den anot thu* dugc 1,12 h't (dktc) thi |i; dung di§n phan Gia tri p H cua dung dich sau di^n phan la A 12 B 13 C 10 j j^^l^, D.7 Bai 3: Hon hgp M gom hai muoi NaX va NaY (X, Y la hai nguyen to'co tu nhien, a hai chu ki lien tiep thugc nhom VIIA, Zx < ZY) Hoa tan het 6,03 gam M vao nuoc roi cho tac dung voi dung dich AgNOs (du), thu dugc 8,61 gam ke't tua Hai nguyen to X, Y Ian lugt la A F va CI B CI va Br C Br va I D CI va Bai 4: Cho nuoc CI2 (loang, du) vao dung dich chua a mol NaBr I b mol NaT den phan ling hoan toan, thu dugc dung dich chua 1,17 gam NaCl Gia trj cua (a+ b) la , A 0,01 B 0,02 C 0,03 va O2 Sau phan ung thu dugc 11,5 gam chat ran va tong so mol k h i da C.Ca D M g H^;^; Bai 6: Dot chay hoan toan 6,8 gam mgt chat v6 co X chi thu dugc 4,48 lit , SO2 (dktc) va 3,6 gam nuac The tich O2 (dktc) da dung de dot chay X la A 7,4 lit B 6,72 lit C 4,48 lit D 8,96 lit Bai 7: Hoa tan het hon hgp gom 0,02 mol FeS2 va 0,04 mol FeS vao dung djch H2SO4 (dac, nong, du), thu dugc dung dich X va V lit SO2 (dktc) Gia tri cua V la lis A 7,392 CuS04 0,2M Sau phan ung hoan toan, Igc ket tua va dem nung khong den khoi lugng khong doi, thu dugc m gam cha't ran Gia tri cua m la A 2,62 C 1,792 C 3,64 D 2,016 J f i o J p i^02 dung djch N H den d u vao dung dich Y, ban dau xuat hi?n ket tua xanh, sau ket tua tan, thu dugc dung djch mau xanh tham Chat X la A.CuO B.Cu C.Fe D FeO Bai 13: Hoa tan het 1,56 gam hon hgp gom A l va AI2O3 bang dung dich HCl (du), thu dugc V lit H2 (dktc) va dung djch X Nho dung djch N H d u vao X, Igc ket tua va dem nung den khoi lugng khong doi thu dugc 2,04 A 0,448 (2) NH3 + O2 , B 0,224 BM14: Cho cac phan ung sau: A (2); (3) t ~p C 1,344 ai- ' j > D.a672.p^^ ,8 (3)NH4N02 ) - i l ^ ;>fo o r D ; f • (4) NH3 + CuO Cac phan ung deu tao N2 la j B (2); (3); (4) — ^ * C (3); (4) sCJi- D.(l); (3); (4) Bai 15: Cho 0,448 l i t N H (dktc) d i qua ong su d\mg 16 gam CuO nung nong, thu dugc chat ran X (gia su phan ung xay hoan toan) Phan tram khoi lugng Cu X la A 85,88% B 5,600 B_ 2,04.**-''> Bai 12: Chat ran X phan ung voi dung dich HCl dugc dung djch Y Cho tir t u (1) N H + Ch tham gia phan ung la 0,125 mol Kim loai R la v , * , ^ B.C Bai 11: Cho dung djch N H a d u vao 100ml dung dich chua Al2(S04)3 0,1M va gam chat ran Gia trj ciia V i a D 0,05 Bai 5: Dot chay hoan toan 3,6 gam kim loai R (hoa trj 2) hon hgp CI2 A Be Bai 10: Co dung djch muoi rieng bi?t: CuCh, ZnCh, FeCb, A I C I Neu them dung dich NHs (du) roi them tiep K O H (du) vao bon dung dich thi so cha't ket tiia thu dugc la: (aliifo) •sf{-jri>) i ' i ' X D v h mri ,;>/b V')l J i> A.3 B.2 J A :> c.4."j-ri.s D B 14,12% C 87,63% D 12,37% Bai 16: Nen hon hgp gom lit nito va 14 lit hidro binh phan ling Bai 8: Dot chay hoan toan 18 gam FeS2 O2 d u roi hap thu toan bg lugng nhi|t dg khoang 450''C va co bgt Fe xiic tac Sau phan ung thu dugc 16,4 SO2 thu dugc vao 2,5 lit dung dich Ba(OH)2 0,1 M , tao m gam ket tiia lit hon hgp (a ciing dieu ki^n nhi^t dg va ap suat) Hi|u suat ciia phan Gia tri cua m la: ung tong hgp amoniac la A 32,55 B 21,70 C 54,25 D 43,40 Bai 9: Cho 8,45 gam mgt loai oleum vao nuoc du, thu dugc dung dich X De A 16% , B 20% "' • C 23% D 25% Bai 17: Hon hgp X gom N2 va H2 co t i khoi so voi heli bang 1,8 Dun nong trung hoa 1/10 dung dich X can 20 ml dung djch NaOH I M Oleum la X mgt thoi gian binh kin (co Fe lam xiic tac), thu dugc hon hgp Y A H2SO4.2SO3 CO t i khoi so voi heli bang H i f u suat cua phan ung tong hgp N H s la B H2SO4.3SO3 C H2SO4.4SO3 D H S O S O A 25% B.50% C.36% *' D 40% Cty TNHH MTV DWH Khang Vi?t dm nang On luygn thi d^i hgc 18 chuy6n 66 H6a hpc - Nguygn Van H5i ^ Bai 18: K h u hoan toan 10,44 mpt oxit kim loai kim loai nhi^t dp cao can x = a m o l ; y = a m o l - > %Vo^= ^ 0 % = 40%.,^ 0,18 mol C O Toan bp luong k i m loai sinh durp-c hoa tan het vao dung djch H C l d u , thu dupe 3,024 lit Hi (dktc) Cong thuc ciia oxit kim l o ^ la: A.FeO i C! B Fe203 C AI2O3 D Fe304 M Bai 19: N h o timg gipt dung dich X chiia Na2C03 0,2M v a N a H C 0,1M vao \ 100ml dung dich H C l 0,5M deh khi ngimg»thoat ra, thu dug-c V m l ^^• -'. e:,,,f, -^^P^^„ B.224 D 336 = 0,6 mol; n , , + = 0,5 mol; n , = 0,1 mol jSlhan xet: Ion C I " v a H O tham gia di^n phan Phuong trinh di^n phan dang 2C1- D 448 Bai 20: H a p thu hoan toan 448ml C O (dktc) vao 500ml dung djch gom , ,,,„„•.' c' ion riit ggn: >, C O (dktc) G i a trj cua V la A 672 • Bai 2: Mol: [trnAslroM + 2H2O > 20H- ai +CI2 ntiorouli + Hz^Q ao5 ^ ' ^ N a O H 0,02M va Ca(OH)2 0,01M, thu diegc dung dich Y C o can Y thu dugc m gam cha't ran khan G i a tri cua m la A 1,34 B 1,65 C.0,84 D 1,06 p H = 13 ^ Bai 21: C h o gam hon hop silic va than tac dung vdi lugng dung dich N a O H loang, d u , tao 1,344 lit hidro (dktc) Phan tram kho'i lugng ciia silic hon hgp ban dau la A 42,15% B 84,30% C 28,10% D 56,20% gom 2,4 gam magie va 2,7 gam nhom tao 17,35 gam cha't ran Than tram ve the tich cua clo X la - B.55% C.75% Bai 23: C h o cac phan ling sau: (a) H S + SO2 > (c) S i + M g — ^ ; i;, f (b)Na2S203 + H2SO4 (loang) ' (d) AI2O3 + B.2 C.3 D.4 a Bai 1: nMg= — = 0,05 mol; n z „ = ^ = 0,02 mol ^' '•^^ 24 65 ) (I , „ ^ ; , Nhan xet: D a y la bai toan hai chat k h u (2 kim loai) v a hai chat oxi hoa (2 phi kim) nen can ap dung djnh luat bao toan electron Mol: 2e )-Mg^2 0,05-^0,1 Chat oxi hoa: Mol: > 20-2 4x CI2 + 2e > Zn*^ - > 32x + 71y +1,2 +1,3 = 5,27 ^ 32x + 71y = 2,77 M N X = — = 198,67 -> X = 175,67 ^ n M = ^ Y =C1 Mgtnguyentoco , , • Ltm y: K h i X = F thi N a F khong tac dyng v a i A g N A g F la muoi tan " "Nacr ~ , 0/02 mol Cac phan ung hoa hgc: 2NaI + CI2 > 2NaCl + 2NaBr + CI2 > N a C l + Br2 - I2 arttq^Q i^NaBr+ " N a l = 0,02 mol -> a + b = 0,02 -> D a p an B iJd^H' Bai 5: Nhqn xet: Bai cac em c6 the vie't phuang trinh de giai T u y nhien, nen ap Chat oxi hoa: > 2C1- Mol: 2e a X £,f5 > R*^ + 4e —> - > 4x Ch + 2e ICt^ Mol: ' ,o?fi > 20r y " > 2y Bao toan khoi lugng: m o j + mciz = " ' - 3,6 = 7,9 -> 71x + 32y = 7,9 Bao toan electron: 4x+2y = 0,1+0,04 = 0,14 mol tang 8,61-6,03 = 2,58 gam • 2y Bao toan khoi lugng: m x + m j ^ g + m z n " m y khoi lugng tang 108-23 = 85 gam dving bao toan khoi lugng v a bao toan electron de giai nhanh hon 2e Mol: , ^ , O2 +4e X —> Zn - > mol A g X ^ B^4: H i r d N G D A N - L G I A I Chat khu: M g - • f,;.n •-v : -> Dap an A > > So phan ung t^o don cha't la A.I Theo bai: + VayX = F ^HMfi 'L-i'A(i'i NaOH _ nguyen tu khoi > 175,67 -> Loai v i khong c6 tu nhien D.65% • • -fsiijinji^,, !, • ^r,oan''«,g ¥ : G g i cong thuc chung cua hai muoi la N a X Nhan xet: mol N a X " Bai 22: H o n hgp X gom clo va oxi C h o X phan u n g vira het v i hon hop A.85% Bai 3: + Neu D a p a n A — Theo bai: n c i j + " = 0,125 ^ x + y = 0,125 -> x = 0,025; y = 0,1 167 Cty TNHH MTV DWH Khang ViQt elm nang On luy?n thi dgi hpc 18 chuy6n di H6a hpc - Nguygn Van HSi Bao toan electron: 2nR = 4no2 + 2nci2 -> MR= =24 R 0^5 + 2.0,1 ^ ,3^5 - » nR = GQi cong thiic oleum la H2S04.nS03 Mol: Bai6: Nhanxet-.X + O2 SO2 + H2O '» ' Bao toan khoi lugng: mx + m 02 = m -> no2 = ^ a ' i^'' ' +n^H20 - > m O2 = ^2,8 + 3,6 - 6,8 = 9,6 gam Mol: « ', v a(n+l) a(n+l) - " > Na2S04 + 2H2O H2SO4 + 2NaOH r,/ I • fj > (n+l)H2S04 H2S04.nS03 + nH20 la Mg -> Dap an D / « ,^, 2a(n+l) 2a(n+l) = 10.0,02.1 ^ a(n+l) = 0,1 ^ [o/; Theo bai: (98 + 80n)a = 8,45 - > a = 0,025; n = Cong thuc oleum la H2SO4.3SO3 = 0,3 mol -> V02 = 0,3.22,4 = 6,72 lit ^ Dap an B " ^ - > Dap an B Bai 10: Trong X: ns= nso2 = 0> mol; n H = n H = 0/4 mol -> n H : n s = :1 ^ bai nay, cac em can viet va can bang dung phucmg trinh sau: -iiK^yij,^ 0,04 -> nso2 = 0,33 mol ^ Vgoj = 0,33.22,4 = 7,392 lit ' ^ - X '(f tao ra, nen ne'u viet phuong trinh bao toan electron: ^ "^^ FeS = > Cu(OH)24 + 2KC1 ZnCh +4KOH > YULnOi ^^sOz , , ^ Mol: 0,15 '_),, ; ( , > i ; V t ' - ' - - Q, m^y^ y: A1(0H)3 va Zn(OH)2 c6 tinh luong tinh; Cu(OH)2 tan du(?c j^Wii.) ;f< '^x'y.^'^P''^ „ rnc' Ah(S04)3 ) 2Al(OH)3l — ^ 0,01 0,02 CuS04 !'>rff S9,C > Cu(OH)2i AI2O3 0,01 ICu(NH3)4](OH)2rt S « « «' m = mA,203 =0,01.102 = 1,02 gam 0,3 (Ltm y: Co the bao toan nguyen to S: ngoj = 2nFeS2 " ^'^ "^°^)^ ' " O H - 0,5 Ta co: = >^ nso2 0,3 ' dung djch N H , 2Fe203 + 8SO2 A- "' Dap an D Mol: 4FeS2 +1102 + 2KC1 + 2H2O > [Cu(NH3)4](OH)2 Bai 11: So phan ung "soa = 0'25 mol ^ Vso2 = 5,600 lit -> Dap an B - > Ket qua sai i :J?iL™ j + iHl-'U > Fe(OH)3i + 3KC1 Cu(OH)2 + 4NH3 luu Luu y: Do SO2 ca chat k h u (FeS2, FeS) va chat oxi hoa (H2SO4 dac) cung llnFeS2 €M * f H Vay ket tua lai la Fe(OH)3 i^^oy.- •/;:,->_ Dap an A *pKt rfsn mJ AICI3 + 4KOH > KAIO2 + 3KC1 + 2H2O -' Them tiep N H du, Cu(OH)2 tan hoan toan tao phuc chat: > Fe2(S04)3 + 9SO2 + I O H O 2FeS + IOH2SO4 Mol: ' C u C l + 2KOH FeCb + 3KOH > Fe2(S04)3 + I S O + I4H2O 0,15 2FeS2 + I4H2SO4 0,02 ' Khi cho tac dung v6i K O H du, thu dugc ke't tua: Hgp chat X la H2S Bai7: *' Mol: ' - > Dap an D b f i f f ? ' • ; ^ r f s ^ M ^ Tao muoi: Sunfit va hidrosunfit ' „ • • • ^ Bai 12: YS.-^*S.' m&n xet: Loai B v i Cu khong tan HCl ' it • >i i* * +*) * • ^ ' jv , Loai C, D v i ket tua Fe(OH)2 khong tan dugc NHs d u Tinh nhanh: n^^-^ = n ^ ^ - ngoj = 0,5 - 0,3 = 0,2 mol - > Dap an A Cac phan ung hoa hpc: , Nhan thay: n^ < n^^2 -> n^^so, = ^o^- = ^'^ m = 0,2.217 = 43,4 gam Dap an D ' , , of*^ t*(' > CuCh + H O CuO + 2HC1 CuCh + 2NH3 + 2H2O Cu(OH)2 + N H > Cu(OH)2i (xanh) + 2NH4CI > [Cu(NH3)4](OH)2 (xanhtham) ^ ' ' '' •M'*"'''''' dm nanq On luyjn thi dgi hpc 18 chuy6n 66 H6a hoc - NguyHn Van Hi\ Cty TNHH MTV DWH Khang Vi^t Bai 13: - , = ^ _ ^ s o ' m o l X: nN^ = 1; n^^ = 2-7,2 5/2 G Q I SO mol: A l = x; AhOs = y Ta c6: 27x + 102y = 1,56 So phan ling: ^, A l , AI2O3 AICI3 j, ^, , , "^"3^H20 ^ ^j^Qj^^^ ^j^Q^ , 04 Bao toan nguyen to'Al: x + 2y = 2nAi,o^ x + 2y = 2.—— = 0,04 » 102 ^,| - » X = 0,02; y = 0,01 ^ nH2= | n A i =0,03 mol V = a03.22,4 = 0,672 lit Bai 14: ,v." D a p an D ' ' ^| ^ ^ f , ' • • ' j^^^' r,':'- (1) 2NH3 + S C h ,^ j, (2) 4NH3 + 5O2 (3) NH4NO2 — i , ^ ^ g„y > N2 + 6HC1 "^""^•'^ ) N O + 6H2O ^ N2 + 2H2O (4) 2NH3 + C u O C))»:0 — v N2 + 3H2 iDiO dDi;*? „.^«t, ^ ' "36 ' ny = — = 4,5 mol Af(cqi.>>i 2x ^ i md dmb gnoi , 22,4 j.,,,,^ " " = 0,02 mol 0,02 * ^(HO)u:J Ji^noDfiiJlf&lYeV Jtm'gEi'iiXl^-iJi£m.()^-,;,m • Bai 18: =0,135 mol Nhan xet: V i oxit MxOy bi khii boi C O L o ? i C v i C O khong khii dupe AI2O3 ^(HO)!/ t-v^- v, r-tv;:v>,; a03 ' V , ,.,^^5 Bail6: Phan ling hoa h