Tài liệu 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* + 2 N O ; + 2Fe3* -> > 3Cu2* + 2 N O + 4 H 2 O > Cu^^ + 2Fe2- = 0,03 + 0,02 = 0,05 m o l 4. P H A N L T N G D I E N P H A N * ' >'* ; £0.0 - ""P"' > 2 N a + CI2 l^»d m i uO -y. . ... , "^""^ ) M g + CI2 MgCh 2NaCl + 2 H 2 O - ^ E ^ ' 2 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, 6 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. 2 Phan u n g d i e n p h a n : 2 A g N 0 3 + H2O - ^ E ^ 2 A g + ^ 0 2 + 2HN03 MSO4 + H2O - J P ^ M + i o 2 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) - , ™ T: A A.I.t Theo cong t h u c Faraday: m = > 1 , 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 * + H2SO4 Dinh luat Faraday ( t i n h l u p n g chat t h u d u o c 6 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<4 f - t: t h a i g i a n d i ? n p h a n (giay) - B. 4,8. C . 2,4.- y: T i r cong t h u c d i n h luat Faraday, so mol electron trao d o i t r o n g qua b. V i d y mau . .;• • CUSO4 j lii.as;? y u,*£.0 i V i d y 1: K h i d i e n p h a n d u n g djch N a C l ( v o i d i ? n c\rc t r a , c6 m a n g n g a n xo'p) t h i cac qua t r i n h xay ra a eye a m va cue d u o n g Ian l u g t la A . o x i hoa H2O va k h u i o n C h . B. k h u i o n Cl~ va o x i hoa i o n Na*. Cui +|02t a aSa +H2SO4 |!^: • - Orii* Nhan xet: Kho'i l u g n g d u n g djch g i a m = m ^ u + n i o j = ^ Day la cong thiec giiip gidi nhanh chong nhieu bdi todn dien phan. - J s ^ a 64a + 32.0,5a = 8 ^ . . Laigiai: Phan u n g d i | n p h a n : CuS04 + H2O Mol: t r i n h d i e n p h a n d u g c t i n h theo cong thuc: D . 3,2. ' F: h l n g so Faraday (F = 96.500) Luu .On-- . d u sau d i ^ n p h a n can d u n g 1,12 l i t H2S (dkte). Gia t r j cua C la: n: so' electron trao d o i m a i o n da n h a n de tao t h a n h n g u y e n t u ( p h a n t u ) I: .jUI BI nf.r[:: V i d y 3: D i f n p h a n 500 g a m d u n g djch CuS04 eo n o n g d g C%. Sau k h i d u n g CuS04 eon - n.96500 D a p an A . A . 5,6. '< _ , _ - u t uri?,, = 0,672 - 4 A = 32n - > n = 2 va A = 64 (Cu). m : kho'i l u ^ n g chat t h u duQfc 6 d i f n cvrc (gam) |;,; A.1,5.1351 d i ^ n p h a n thay kho'i l u g n g d u n g djch g i a m 8 g a m . De ke't tua he't l u g n g nF t u a n g u n g 6 d i | n cue. ' ' Mol: a = 0,1. 0,05 "-^^ ' ' ' • ' >CuSi +H2S <- +H2S04^ 0,05 ' ^nubficricrr.."' " ncuso4 = 0,1 + 0,05 = 0,15 m o l ^ C % = ^'^^-^^^.100% = 4 , 8 % . 500 D a p an B . - •'•"•"^^ - - ^ •''''''^' •" • Cty TNHH MTV DWH Khang Vi?t dm nang On luy^n thi dgl hpc 18 chuySn ai H6a hpc - NguySn van H5i V i dv 4: Hoa tan 12,5 gam tinh the muoi ket tinh CuS04.nH20 vao nuac, thu dugc 1 lit dung dich X. Dien phan X (voi di^n cue tro) den khi bat dau c6 khi thoat ra 6 catot thi dimg dien phan, thay khoi lugng catot tang 3,2 gam va dung dich sau dien phan c6 p H bang a. Gia t r i ciia n va a Ian lugt la A.5;l. B.5;2. C. 3; 3. D. 7; 1. Lod gidi: phan ling "quy doi" gia dinh: CuS04 Phan ung dien phan: CuS04 + H2O "P^" > Cu + ^ 0 2 + H2SO4 CuCh - ^ S ^ Cu + CI2 gam McuS04.nH20 = 61' ^ ncu = O C U S O A = i ^ C u S 0 4 . n H 2 0 = 0/05 [H^]=M = o,l = 10"^ pH = l Mol: ' ' ^ = 250 -> 160 + 18n = 250 ^ n = 5. Dung dich sau dien phan: n^^^ - 2nH2S04 = 2ncu = 0,1 mol. ^ mol. , B. 0,672. C. a896. D. 1,120. < Laigidi: 1 t,,;, -A .ns NMn xet: Day la phan ung dien phan hon hgp —> nen dung cong thuc h'nh ... u u XX . It 2,68.2.3600 , nhanh so mol electron: n,,= — = = 0,2 mol. " F 96500 X chua: n^^2+ = 0,12; n^^. = 0,04; n^^^^ = 0,04 va n^^z- = 0,12 (mol), Tai anot: 4H2O - Mol: 4e 0,02 > 02! 0,16 ^ a04<-0,02 r v i x . ; ^ B. k h u H2O va oxi hoa ion Cu^*. D . k h u C u va oxi hoa H2O. -' + dJ'/::^': o r / r .,, :> ^.A..,. " A n h la anot nhuong e - E m la catot nhan e thoi m a " + Cue am (catot) eo Cu^* va H2O -> H2O bi dien phan: > Cu Cu^* + 2e {Qua trinh khu) + Cvre duong (anot) eo H2O va ion SO4' 4H2O - 4e ,_tL. H2O bi di?n phan: > O 2 1 + 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 trong t gia, 100%), thu dugc chat ran X, dung dich Y va khi Z . C h o 12,6 gam Fe vao Y , sau khi 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 ••' cuong do dong dien khong doi 2,68A (hi^u suat qua trinh di^n phan la > Cht , ^ ^ Mol: 0,04 - > 0,04 - > 0,04 Van dung cau tho: r-'v-nh. 2e C u + - O 2 + H2SO4 Laigidi: di^n phan la 100%. Gia tri ciia m la: 2C1- - + H2O -^S^ C. k h u Cu2* va oxi hoa 504'. 2,68A, thu dugc V lit khi (dktc) bay ra a anot. Biet hieu suat ciia qua trinh '-'I'lh 0,04 A. oxi hoa H2O va k h u ion Cu^"^. ,,. s n i j nerf 1;,:;, 0,04 thi cae qua trinh xay ra 6 cue am va cue duong Ian lugt la cue tra, mang ngan xop) trong thoi gian 2 gio voi dong di$n c6 cuong dp la A^c; -> ' i::;/ ^ '••'^•^> V i dy 7: K h i dien phan dung dich CuS04 (vai di^n cue tro, eo mang ngan xop) V i du 5: Dien phan dung dich X chua 0,12 mol CuS04 va 0,04 mol NaCl (dien .. . ; -> Dap an D . DapanA. 1 A. 0,560. ' •J _> Khoi lugng dung dich giam = 64.0,08 + 0,04.71 + 0,02.32 = 8,6 gam. ny^Q^^-J 0,04 CuS04 Mol: bj dien phan het (bat dau den H2O bi di^n phan). m c u = 3,2 + Na2S04 T h u tu dien phan n h u sau: Theo bai, dung dien phan khi 6 catot bat dau c6 khi thoat ra —> CuS04 vua Ta c6: > CuCh + 2NaCl Mol: 0,04 , „ , A. 0,8. , B.1,2. C.1,0. D.0,3. V i d v 6: Dien phan dung dich X gom 0,08 mol NaCl va 0,12 mol CuS04 (dien eye tro, mang ngan xop) den khi thay thoat ra 6 anot 1,344 lit khi (dktc) thi Phan ung di?n phan dung dich AgNOa: / ngung dien phan. Kho'i lugng dung dich sau dien phan giam di bao nhieu 2AgN03 gam? A. 3,48 gam. B. 6,04 gam. C. 2,56 gani. D. 8,60 gam. ^ Mol; 2a + H2O ;^ T > 2Ag + i 0 2 + 2HNO3 . 2a ^. , ': • _ 2a 123 Cty TNHH MIV UVVH Khang Vi$t Ca'm nang an luy$n thi djii hgc 18 chuy6n ai 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 < — 2a Fe Mol: b . 0,5a + , Fe(N03)2 + 2 A g 2b ^ CaC03 — Theo bai: 12,6 -56(0,75a + b) + 2b.l08 = 14,5. ' 2a = 0,1 '' ' ' 'fii'''' L i thuyet * MMOZ nitrat: AgNOs — A. 40%. if• CaC03.MgC03 Mol: C84%. ^ — ^ 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 3 . M g C O 3 = ^ - 1 0 0 % = 92% ^ n.i . :*-;H- mCaC03.MgC03 = 0,2.(100 + 84) = 36,8 g a m io2 io2 A g + NO2 + P Lin giai: Phan u n g nhiet p h a n : 2KNO2 + 02 ^ _ B.50%. •"'^ " ' ' C u O + 2NO2 + , k h o i l u g n g ciia C a C 0 3 . M g C 0 3 t r o n g loai quang tren la O^Hf BOH ixo ^ n tap cha't t r o sinh ra 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 ••!'•! nfiifc a. Cu(N03)2 — i;.(f- ....>ttf^ V i dv 2 (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 ^ ^.^.^^^^.^ 5. P H A N L f N G N H I E T P H A N , D O T C H A Y 2KNO3 + O2 . > —> D a p an A . - = 0,1 m o l - > t = 3600 giay = 1 gio. -^DapanC. : to f s i i s u y i r t .-n'u ,.3,.., 2KC1 + 3O2 ^ 2KC103 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, CaO + CO2 ^ . 2b + M n 0 2 + O2 C u O +2NO2 + i o 2 2 Cu(N03)2 tOiO' • 0,5a 2AgN03 2NaN03 — 2 N a N 0 2 •--vtJuD iiTiS;. X i^**"'s'^'iq''3'?:< 01 i<: > 3Fe(N03)2 2Fe(N03)3 0,25a 2KMn04 — K 2 M n 0 4 r O , y :h > Fe(N03)3 + N O + 2H2O 4HN03 , 5 . ^ 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 do 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 - + KNO3 (2); Fe + S (3); MgCOs + M g (4). So t r u o n g h g p xay ra s u o x i hoa C a O + CO2 2FeC03 + -O2 k i m loai la: A. 1. 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 2 ,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 ra 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. 4. D . 5. '' C. 3. + O2 '" > 2 C u O 2KNO2 + O2 (2) 2 K N 0 3 — ^ 2Cu + O2 — + 2CuO '" > FeS (4)MgC03 — ^ Mg ^ CO2 '|f D . 4. ^, j |,,^ j^^,^, CuO + 2NO2+ (1) Cu(N03)2 (•^) Fe + S V i du 1: Che day cac m u o i : K M n 0 4 , N a N 0 3 , Cu(N03)2, CaCOs, A. 2. B. 2. 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 3 ^ ^ 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: 6 hon hg-p (3), Fe la chat khir — bi oxi hoa; 6 hon hgp (4), M g c6 the > chay trong khi 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*') <- " - - -,00 10,0 „ ^ , "Cu(NO3)2=-^=0,lmol.^ Nhan xet: Khoi luong chat ran giam = khoi lugng khi bay ra. Phuong trinh hoa hoc: CU(N03)2 • Mol: X m^02 ^ CUO + 2 N O 2 + - O 2 nSritt finfifli 2x 0,5x 5 , ,;t r t f i k o i t .tsrir)'qi + r " 0 2 = 18,8 - 12,32 = 6,48 gam -> 46.2x + 32.0,5x = 6,48 -> x = 0,06 mol ^ H = - ^ . 1 0 0 % = 60% Dap an B. 6. B A I TAP O N L U Y E N Bai 1: Dung djch X gom Ba^ Na" (0,01 mol) va OH" (0,05 mol). Dung dich Y gom Na^ HCO3 (0,02 mol) va CO 3' (0,01 mol). Trpn X voi Y thay tao ra m gam ket tua. Gia t i l cua m la A. 5,91. B.3,94. C. 7,88. D. 1,97. Bai 2: Hoa tan hoan toan 9,2 gam hon hop MCO3 va M'C03 vao dung dich HCl, thu duoc V lit khi (6 dktc) va dung dich chiia 10,3 gam muoi. Gia tri cua V la A. 1,12. B. 1,68. C.2,24. D. 3,36. Bai 3: Cho 5,76 gam hon hop X gom muoi cacbonat va hidrocacbonat cua kim logi kiem M tac dung het voi dung dich HCl (du), sinh ra 1,12 lit khi (0 dktc). K i m loai M la A. Na. • -v^-^' B. K. C. Rb. D. L i . Bai 4: Cho 14,9 gam hon hgp hot X gom Zn va Fe vao 600ml dung dich CuSO* 0,25M. Sau khi cac phan ling xay ra hoan toan, thu dug-c dung dich va 15,2 gam hon hgp k i m loai Y. Phan tram ve khoi lugng cua Z n trong hon hgp dau la A. 56,38%. B. 37,58%. C. 64,42%. D. 43,62%. Bai 5: Cho m gam hon hgp bgt Zn va Fe vao lugng d u dung dich CuS04. Sau khi ket thiic cac phan ung, Igc bo phan dung djch thu dugc m gam bgt ran Thanh phan phan tram theo khoi lugng cua Zn trong hon hgp bgt ban dau l3 A. 90,28%. B. 85,30%. C. 82,20%. D. 12,67%. gai 6: Cho 5,1 gam hon hgp X gom Mg va Fe vao 250ml dung dich CuS04 0,3M. Sau khi cac phan ung xay ra hoan toan, Igc thu dugc 6,9 gam chat ran Y va dung dich Z chua hai muoi. Khoi lugng cua M g trong X la: A. 0,9 gam B. 1,2 gam C. 1,6 gam. D. 2,4gam. Bai 7 (B-07): Khi cho Cu tac dung voi dung dich chiia H 2 S O 4 loang va NaNCh, vai tro cua NaN03 trong phan ling la A. chat xiic tac. B. chat oxi hoa. C. moi truong. D. chat khu. Bai 8: Di?n phan dung dich X chua 0,2 mol CuS04 va 0,12 mol HCl trong thoi gian 4 gid voi dong dien c6 cuong do la 1,34A, thu dugc m gam kim loai bam vao catot. Biet hif u suat cua qua trinh dien phan la 100%. Gia tri cua m la: A. 6,4 gam. B. 3,2 gam. C. 12,8 gam. D. 9,6 gam. Bai 9: Di^n phan dung dich X gom 0,1 mol KCl va 0,15 mol CuS04 (di^n cue tro, mang ngan xop) den khi khoi lugng dung dich giam d i 10,75 gam thi ngimg dien phan. The tich khi (dktc) thoat ra 6 anot la A. 1,12 lit. B. 1,68 lit. C. 2,24 lit. D. 2,80 lit. Bai 10 (CD-08): Nhiet phan hoan toan 34,65 gam hon hgp X gom K N O 3 va Cu(N03)2, thu dugc hon hgp khi Y. Ti khoi cua Y so v o i hidro bang 18,8. Khoi lugng ciia Cu(N03)2 trong hon hgp ban dau la A. 20,50 gam. B. 11,28 gam. C. 9,40 gam. D. 8,60 gam. Bai 11 (A-09): Nung 6,58 gam Cu(N03)2 trong binh kin khong chua khong khi, sau mgt thoi gian thu dugc 4,96 gam chat rSn va hon hg-p khi X. Hap thu hoan toan X vao nuoc de dugc 300ml dung dich Y. Dung djch Y c6 p H bang A. 4. B. 2. C. 1. D. 3. Bai 12 (B-08): Nung mgt hon hgp ran gom a mol FeC03 va b mol FeS2 trong binh kin chua khong khi (du). Sau khi cac phan ung ket thiic, dua binh ve nhi|t dg ban dau, thu dugc chat ran duy nhat la Fe203 va hon hgp khi. Bie't ap suat khi trong binh truoc va sau phan ung bang nhau, moi lien h$ giiia a vabla io;, A . a = 0,5b. B.a = b. C.a = 4b. D . a = 2b. 7. H l / O N G D A N - L 6 I G I A I Bail: " Dung dich X trung hoa di?n: 2 n 2+ + 0,01 = 0,05 -> n ^ 2+ = 0,02 mol. HCOi Mol: 0,02 + OH" > CO3" + H 2 O 0,02 ,iv0,02 ^ =r:/ n^^2-= 0,02 + aOl = a03 m o l . CO3 Ba2* Mol: 0,02 + CO 3" 0,02 > BaC03i 0,02 mBaC03 = 0,02.197 = 3,94 gam. Dap an B. 197 Ca'm nang 6n luygn thi dgi hgc 18 chuy6n dg H6a hpc - Nguygn Van Hii Cty TIMHH MTV DWH Khang Vi$t Bai 2: Nhati xet: T r o n g p h a n u n g , m u o i cacbonat bien t h a n h m u o i clorua, nen: > 2 m o l CI- i s 1 '"ol Theobai: a m o l 1,1 0,1 m o l -> a = - » khoi Iirong tang 71 - 60 = 11 gam. , , , tang 1 0 , 3 - 9 , 2 =1,1 gam. - > 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 ^ -lnr,i ^ 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 2 0 . - . t - a .roBSli,.^ : N h a n thay : nx = n c o j = 0/05 m o l . < ' i ^ b S""^' f ^ — 5 76 -> M x = ^ — = 115,2 -> M H C O 3 < 115,2 < iVbCOs. 0,05 M + 6 K 115,2 < 2 M + 60 -> 27,6 < M < 5 4 , 2 ^ M l a K - ^ i:;is::S c.;j,K, f i & o l ns g^j^, Dap anB. ;\«l,t > a mol Cu b m o l Fe Theo bai: > h mol Cu mtang 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 3 —» hoa tan C u theo p h a n u n g : + 2NO: > 3Cu^* + 2 N O t Fe = 56 da p h a n u n g m o t phan tao thanh 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: 2 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 ' r ; -.^'^ : 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. 1 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 2 = n ^ ^ 2 - = 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: 2 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 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 mwng ^^ ^ * '• '^^zn + ^^Ve{pu) ~^ « --^ 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 = 0 -> x = 8a. 128 , 0 y f - » k h o i l u g n g g i a m (65-64)x = x. I -> Cu ' ^-'^^'^^^ SO^ 4r^-~m^^<^ ai ; ^^^^ * • ^ „ j^H < < f*,0 • Phan u n g " q u y d o i " gia d i n h : ^,15 = x + a. a = 0,05; x = 0,1 ^ %my„ = ^ ^ ^ . 1 0 0 % = 43,62% . ^" 14,9 H -> m = ai.64 = 6,4 gam -> Dap an A. ^ „ ^ CuS04 ^ CI Bai 9: = mv - mx = 0,3 g a m -> 8a - x = 0,3. Bao toan electron: 2 n ^ 2+ = Cu Catot: x mol Zn Theo bai: u u I . . I t 1,34.4.3600 „ ^ f n h a n h so m o l electron: np= — = ==0,2 m o l . ^ F 96500 X chua: n ^ 2+ = 0,2 m o l ; n ^ , . = 0,12 m o l ; n^+ = 0,12 m o l v a n^^2- = 0,2 m o l . tang (64-56)a = 8a. ' Mol: + 0,05 < - 2KCI > ' > CuCh 0,10 0,05 + K2SO4 . + ,n')<,'^r 0,05 - > X g o m : C u C h = 0,05 m o l ; CuS04 = 0,10 m o l va K 2 S O 4 = 0,05 m o l . T h u t u d i ^ n p h a n n h u sau: CuCh Mol: 0,05 '^P"' > C u + C k -> 0,05 0,05 ' mgiim '-^"^ ' ' = 0,05.(64 + 71) = 6,75 gam. K h o i l u g n g d u n g d i c h con g i a m tiep 10,75 - 6,75 = 4 gam. !^ ' 129 elm nang On luyQn thi dgi hpc 18 chuySn dg H6a hoc - Nguyin VSn HSi Mol: dpdd ^ „ 1 „ > C u + - O 2 + H2SO4 + H2O CuS04 a aSa ^ a 64a + 32.0,5a = 4 ^ a = 0,05 mol. ' ^ - > Vci2 + Cty TNHH MTV DWH Khang Vi$t = (0,05 + 0,025).22,4 = 1,68 lit Bai 10: Phuang trinh hoa hoc: l i X H CHAT CIJA CAC OXIT ; Dap an B. , ., iiCL- ' , ion" Cir, r ; a* ... * phan ung trao aoi CuO + 2 N O 2 + I 0 2 AI2O3 + 6HC1 Mol: Mol: = ^'^S-^vd " i(.*m rifid o'X: ° y ^^^'^-^ = '^^"S '^""g * u c cua so do duong cheo: "'^^^'^^ ° ' n NO2 32-37,6 5,6 2 2x 2 46-37,6 8,4 3 ' 0,5(x + y ) ~ 3 ^ ^ ~ ^ ' ' ' , , ,. ' K Mat khac: I88x + l O l y = 34,65 x = 0,05 mol; y = 0,25 mol. Fe304 + IOHNO3 - > m^joj + ,>:••...'.• + O2 + 2 H 2 O ,SMV', A. 80. ^ Mol: a Dap an B. 130 it o r „. p-g,a^ ^O'^"' ~* " o 2 - . 0,1 2H^ .H20 0,2 ' *'' • ' ' '' •.MAm^Oi'A:. "' = 0,1 mol -> V^d H2SO4 = Vay: n H 2 S 0 4 = :; ' = 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 khi H2 (nung nong) thu dugc m gam kim loai. Gia tri cua m la a ^ I'^'-^^'^'P^tm&lig^^fim.d'^bmy^^ = 02Mol: 0,25a + 2,75b = a + 2b ^ D . 120. trong axit tao thanh H2O theo phuong trinh: , j i t . .*,tj '••,,,r „ j : , ;»»u*iD}: A. 4,32. 4FeS2 + 1 1 0 2 — ! ^ -> 2Fe203 + 8SO2 b 2,75b - > 2b ^ N;zfl«^e^ V i a p s u a t k h o n g t h a y d o i ' ' C. 100. B.60. Khi cho oxit kim loai tac dung v a i axit, ion O^" trong oxit se ket hg-p voi ''^ -> Fe203 + 2 C O 2 0,25a ; Way. ' . f ^ ^ ' ^ n f M l g ' Bao toan kho'i l u o n g t a c6: m o 2 = 5,82 - 4,22 = 1,6 gam I H 1 = ^ = l O - ' - > p H = l->Dapand:^ 0,3 + - 0 2 ^ 2 ' ' Lcngidi: V ^ " " N 0 3 = n N 0 2 - > n j ^ , = 0,03 m o l . 2FeC03 > Fe2(S04)3 + SO2 + 4H2O •• i V i dv 1: Dot nong 4,22 gam hon hop X gom Mg, A l va C u trong khi oxi d u , thu -> 4 H N 0 3 Bai 12: Cac phan ung hoa hoc: v. ' b. V i du mau 0,5a V^JJ-M . i j,,, H 2 S O 4 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. •• • • „ > 3Fe(N03)3 + N O 2 + 5H2O 2FeO+ 4H2S04(^flc) CuO + 2 N O 2 + -O2 2a .->gfti^O??^P'-fv'iv *n'Mvni,» 'J .4 if,, > 3Fe(N03)3 + N O + 5H2O 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 ra. Phirong trinh hoa hpc: '' '^ymii..fmf'Mttml. t « MiirM CU(N03)2 > 2Fe(N03)3 + 3H2O 3FeO+ IOHNO3 m c u ( N 0 3 ) 2 =0,05.188 = 9,4gam ^ D a p a n C . Bai 11: Mol: > CuS04 + H2O » Phan ung oxi hoa-khu ••'-'-^ , . . « r . ; n o i t o t e a « m c . ,'. } i i W i t. ' . • • 1 L :'.i.ih 'itn.ih tfrti ' • > 2 A I C I 3 + 3H2O Fe203 + 6HNO3 KNO3 — ^ KNO2 +-O2 4N02 ' t' /I i^ififel I >''"' ^ H 2 S 0 4 = 0,08.1 =0,08 m o l - ^ n ^ + = 0,16 mol. 'Nhan xet: K h i cho oxit kim loai tac dyng voi axit, ion O^- trong oxit se ket hgp voi H * trong axit t^o thanh H2O theo phuong trinh: Ccim nang On luygn thi dgi hpc 18 chuySn ai H6a hpc - Nguygn VSn Hii 02- + 2H* ai6 M o l : 0,08 > . 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 ra 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 + 0 . 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 cua m la 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 «. > • A. 15,36. , C. 17,28. . 1 B. 23,36. j i a! Laigidi: (du), t h u d u g c d u n g d i c h X v a 1,008 h't k h i SQz (san p h a m k h u d u y nhat, 6 MI, D . 13,28. 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 + M o l : 0,04 SO2 t h i oxit la T r o n g 1 m o l FeO hoac Fe304 deu chua 1 m o l Fe^* n e n d e u c6 k h a nang 0,02 ''' " x.0,01 m = 232.0,02 = 4,64 gam - > Dap an B. n-BO,'. C. 75ml. D. 90ml. = 4,46 - 3,5 = 0,96 gam Oey, OK>1 BR.': • -> . H20. ~ nH2S04 = a045 m o l ' '' * ^fTfeisb ,i;:ilb D a p an A . 2. OXIT BAZO + C H A T KHU" (NHIETLUY5N) a. L i t h u y e t : ^h- lorn Tac dung voi CO, Hz ,?t B M fj-j'iS ^ 2Fe + 3C02 Fe304 + 4 C O — ,n ' ^'"'^^ + moxi = m o x » Fe203 + 3 C O — > xFe2(S04)3 X = 3 ^ O x i t sat la Fe304. moxi - ^ VHCI = 0,045 l i t = 45ml ^ * n h u o n g 1 m o l electron, d o d o : |§. -> x.0,01 = 0,03 ^ mkimid Suy ra: n^+ = a 0 9 m o l ^ « »OPrH'i : y « ' = 2-0,01= 0,02 m o l B.60ml. o ^ + 2H^ — > FeO hoac Fe304. 2FexOy 0,04 K h i cho oxit bazo tac d u n g v o i axit tao ra nuoc: Lmgtat: 12 0 224 n F e 2 ( S 0 4 ) 3 = 7 ^ = ^'^^ m o l ; nso2 = - r r — = 0,01 m o l . BaotoannguyentoFe: 0,02 , > 0 96 = - r r - = 0,045 m o l 32 vjiAii p a^rtfJ Ywi> 'sr-i.' -> n o = a 0 9 m o l - > nQ2-(„^i,) = 0,09 m o l . g . , . u . j,^ £, D . Fe304 va 2,32. "Fe^Oy + 2FeS04 , • • 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 6 d a n g b g t tac d u n g hoan -» , , C. Fe203 va 3,20. = npe^Oy = 2nso2 ^ ' 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 22,4 0,04 > CuS04 + 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 2 S O 4 I M vira d u de p h a n u n g het v o i Y la 0 , 0 9 . | .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 2 S O 4 dac nong, t h u d u g c dung .i 0,16 M o l : 0,02 - > 0,02 1 008 " F e x O y = 2 - ^ 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 ^ 3Fe + 4C02 CuO + H2 — ^ C u + H2O lim y: K h i CO, H2 chi k h u d u g c cac oxit ciia k i m loai tir Z n ve sau. Tac dung oai Al (nhiet nhom) 2A1 + Fe203 — ^ AI2O3 + 2Fe . . ' .1/;) .J' " - 133 Cty Ti\ : '/ITV DVVH Khang Vi?t Cim nang 6n luygn thi dgi hgc 18 chuygn dg H6a hgc - Nguygn Van HSi + 3Fe304 '** 8A1 3CuO 14 2A1 + Cr203 Laigidi: + 9Fe — 4 A I 2 O 3 + Fe203 + 3 C O 2Cr CuO -!,.•., B. 8,3 g a m . .'. V' - .'• + CO + CO — ^ C u + CO2 > 2NaA102 + D a p an D . Vi i f • I'lKG t, < ' 1 f I f + fit i ! n Vay Z g o m : Fe, C u D . 4,0 g a m . Lbigiai: Cu '° > 2Fe + 3CO2 AI2O3 + 2 N a O H C. 2,0 gam. Nhan xet: K h i C O chi k h u d u g c C u O theo phan u n g : CuO n e n k h i cho k h i C O d u tac d u n g v o i 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: , , 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. A . 0,8 g a m . AI2O3 , X, xay r a 2 p h a n u n g : AI2O3 - + 3Cu V i d\ 1 (A-09): C h o l u o n g k h i C O ( d u ) d i q u a %1 g a m h o n h g p g o m C u O va AI2O3 y: K h i C O k h o n g k h u d u g c im AI2O3 -> „ H2O " •' 3: K h u h o a n toan m g t oxit sat FexOy 6 n h i f t d g cao can v i r a d u V l i t k h i 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 thiic ciia oxit sat va gia t r i ciia V la + CO2 y n o = ~ = 0 , 0 5 m o l - > n^uo = 0,05 mol. SO.O :!o A. Fe304 va 0,224. B. Fe304 v a 0,448. ^^'l ' C. FeO v a 0,224. T a n g - g i a m k h o i l u g n g : m o (trong CuO) = 9 , 1 - 8 , 3 = 0,8 g a m D . Fe203 v a 0,448. ;n ; m e g e,fl o t O * - - » n c u o =0/05.80 = 4 , O g a m - > D a p an D . ^ nung Giai theo p h u o n g t r i n h hoa hgc '! d u , t h u d u g c 7,5 g a m k e t t u a . Chat A . 2,24. B.4,48. C. 1,12. a075.4 0,075 m o l traodoi -> VNO 3 n jvjo '^NO > Fe, C u . B. Fe203,Cu. C. A l , Fe, C u . "CO2 = 0,02 - » Vco2 = 0,448 B.4,48. C. 2,24. 1 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 2 - > 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, V c o = 0,02.22,4 = 0,448 l i t - > D a p an B. iD 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 2 O 3 , 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 con A. 0,015 3 - 5 ^ = 4 :k.; A . 6,72. = 0,15 m o l . ~ 0,05 mol. = 1,12 l i t - > D a p an C. - = 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 — ay 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^ = ^ D a p an B. . 2nQQ. M a ••'••;,'.(1J:' M a t k h a c : n p e i n o = 0,015:0,02 = 3 : 4 - > 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 = ...8 ax = 0,015; ay = 0,020 ^ 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 2 a •> X ) Muoinitrat + N O Mol: 56 yC02 Cach 2: Laigidi: CO2 + Ca(OH)2 ^ X = 0,015 m o l -> O x i t sat la Fe304 D . 3,36. , ncaco3 = 0,075 m o l yCO — a 0,84 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 3 d u t h u d u g c V l i t k h i N O (san p h a m k h u d u y nhat, 0 dktc). Gia t r i ciia V la ; Mol: + xFe + ax FexOy 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 h o a n toan v a o d u n g d i c h Ca(OH)2 gidi: Cach 1 : ' ^ 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 Lai ' .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 khi C O (dktc) di tu tir qua ong su nung nong dung 10,44 3, jChi tac dung voi axit: A l -> - H 2 gam mpt oxit sat den khi phan ung xay ra hoan toan, thu duoc hon hgp khi Y CO ti khoi so v a i hidro bang 18,8. Cong thiic cua oxit sat va phan tram the tich cua khi CO2 trong 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 = 2,688 lit ^CO 44 - 37,6 2 _ 0,12 S' riC02 28 - 37,6 3 ~ 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 trong phan 2: nAi = 0,1 mol; nFe= 0,2 mol; nAi203 = 0,1 mol. n H c i =3nAi +2nFe+6nAi203 = l , 3 m o l . .Cr203 + 2A1 -> AI2O3 + 2Cr - » Dap an B. 1,5 m^i = 1,5.27. 100 90 2. = 45 gam Dap an D OXIT L I / O N G Sau khi 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 ra V lit khi H2 (dktc). G i a trj cua ...J ah qsQ' < f Cac oxit luong tinh thuong gap: Z n O , AI2O3 '< Tac dung voi axit ^Al203 + 6HC1 IznO ^ Via B. 2,688. T I N H a. L i thuyet: V i dv 7: N u n g hon hgp bpt gom 6,96 gam Fe304 va m gam A l a nhi^t d^ cao. A. 1,344. = 0,3 mol. Do vay, bao toan nguyen to'Al, ta c6: n A i = nfsjaOH = 0,3 mol. 0^ Mol: D.1,5. d y n g va chuyen thanh NaA102. Dap an D . 78 , , ; ncr= — = 1,5 mol ' C.0,5. Nhan xet: K h i cho phan 1 + dung dich N a O H , tat ca A l v a AI2O3 deu tac phap nhi^t nhom voi hi^u suat cua phan ung la 90% thi can toi thieu m gam , B. 1,3. n N a O H = 0,3.1 . V i dv 6 (CD-09): De dieu che dug-c 78 gam C r tu Cr203 (du) bang phuong bpt A l . G^a trj cua m la gom A l va Fe203 (trong dieu ki?n vira dii voi 150 ml dung dich N a O H 2M. De hoa tan het phan hai can v u a %Vco2 =72% ^ C h Q n B h o a c D . - > npe : n o = 0,135 : 0,18 = 3:4 ^ ' thu dugc sau phan ung thanh hai phan bang nhau. Phan mot phan ung =0,3mol. n c o , = 0-18 mol ^ nj^,= 10,44 ' khong CO khong khi) den khi phan ung xay ra hoan toan. C h i a hon hgp 22,4 A p dung cong thuc cua phuong phap duong cheo, ta c6: in ^-DapanB. V i dv 8: N u n g nong 48,2 gam h6n hgp , Theobai: M y =18,8.2 = 37,6 va n Y = n c o = va Fe -> H2 C . 1,792. D . 2,016. + H2SO4 -> 2A1C13 -> ZnS04 Nhan xet: Bai nay cac em se gap kho khan neu khong tinh dugc so mol A l ban 3H2O + H2O \Tdcdung vai baza lAhOs Lai gidi: + + 2NaOH —> ZnO + 2NaOH - -> 2NaA102 + 2H2O Na2Zn02 + H2O dau. b-Vid\imau L u u y rang, trong phan ling nhi|t nhom, thuong ap dyng djnh lu|t bao toan V i d^ 1: C h i diing dung dich K O H de phan bi^t dugc cac chat rieng bi?t trong khoi lugng, cu the: i"Fe203 + m A i = m x - » m^i= 9,66 - 6,96 = 2,7gam ^ Phuong trinh phan ung: 3Fe304 + 8A1 Mol: 0,03 0,08 n A i = 0,1 mol. > 9Fe + 4Ah03. 0,09 0,04 nhom nao sau day? A.Al203,AL B.K,Na. C. Zn, Al. D . Fe, Mg. Laigidi: Loai B: K , N a deu tac dung v a i H2O trong dung djch K O H va sui bgt khi. ^r^7 Cty TNHH MTV DVVH Khang Vijt Ca'm nang 6n luy^n thi d ^ i hgc 18 chuy§n dg H6a hgc - Nguyln Van Hai yi Loai C : Z n , A l deu tan trong dung dich K O H v a siii bpt khi. 4: C h o 6,63 gam hon hop gom BaO v a AI2O3 (ti 1§ mol 3:2) vao nude (du), Loai D : Fe, M g deu khong tac dung voi dung dich K O H . — • D a p ? n A. Hien tugng n h u sau: > . thu duoc dung djch X. D u n g dich Y gom H C l 0,3M va H2SO4 0,2M. C h o t u 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 2 t < A. 5,44. m.^abn'iiM -'ii + H2O > 2KAIO2 b r a S 1,0- ^ .,,5,:.,,,, , ^J®^' " • khi 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 khi CO2: NaA102 Mol: ^ > NaA102 + - H 2 t 2 0,1 ' 0,1 + CO2 + 2H2O 0,1 V i d\ 3: C h o khi 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 ra 2 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 do c6 tinh AI2O3 + 2 K O H 0,01 Mol: a02 V a y Z gom: Fe, C u —> D a p an C . " ' :' , .'HOu- > 2AIO2 + 2H2O • ' + 3H* <- ^ t i, , <, i c^,v 1 - u. < y TIXAl. > Al(OH)3>l. 0,02 ' \^i\. > A P - + 3H20'^'''^ 0,03 " ' ' ^ H*-^^''' • ' + ^ BaSOii <-0,02 a02 . • ' H ' - > 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 2 v a dung dich X trong suo't. T h e m tu t u dung dich H C l I M vao X, khi het 100ml thi bat dau xuat hien ket tua; khi 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 con lai phan khong tan Z . Phan khong tan Z gom ^ , Ba> + 2 0 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 0 3 " ' a = mAi(OH)3 = 0,1.78 = 7,8 gam; A . AI2O3, Fe, C u . B a O + H2O AI2O3 + 2 0 H - AI2O3 CO tinh luong tinh nen tan trong 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 ngay voi nuac: ^ ^ ^ , C.3,11. Cac phan ung hoa hoc: dugc 200ml dung dich Y chi chua chat tan duy nhat c6 nong do 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 2 + 2NaOH ' > 2NaA102 ' • .^^^^ + H2O .HcJi/ri/''''. "^r. , HORK;: . . ' < 139 dm Cty TNHH MTV DWH nang On luygn thi dgi hqc 18 chuy6n 66 H6a hpc - Nguyjn Van Hi\ SO2 sc + 2 N a O H Nhan xet: Dung dich X trong suo't AbOa tan het. N a O H + HCl > NaCl + H2O Mol: 0,1 <- ai NaA102 + HCl + H2O Mol: 0,1 > Al(OH)3>l' + NaCl 0,1 ' > P2O5 + 6 N a O H CI2O7 + 2NaOH .14-.c . ' > 2Na3P04 + 3H2O > 2NaC104 + H2O - > a = 0,1.78 = 7,8 gam — Loai phuang an A ua C. > Nhan thay: Khi cho 400ml dung dich HCl vao X, toan bp NaA102 se chuye'n het thanh ket tiia Al(OH)3, va sau do bi hoa tan mot phan trong HCl: NaAlCh + H C l + H2O Mol: X X Al(OH)3 + 3HC1 MZI Mol: X nAi(OH)3 " x - x = 0,15 mol. + + Br2 + 2H2O Dap > CaCOsI 2CO2 + Ca(OH)2 Phan ieng oxi hoa-khu SO2 b. V i + Br2 , > s H 2 S O 4 + 2HBr ] gnoi B. 224. n B a S 0 4 = - ^ =0,01 SO2 C. 560. D. 336. mol. + 2Br2 + 2H2O H2SO4 + BaCb > H 2 S O 4 + 4HBr > BaSOd + 2HCI • - . i , -> Dap an B. Vi dy 4: Hap thu hoan toan 336ml khi CO2 (dktc) vao 1 lit dung dich Ca(0H)2 0,01M, thu dugc dung dich X va m gam ket tiia. Gia tri cua m la A. 1,0. loang? B.1,5. li C. 6. D.5. nco2 = ^ Laigidi: Cac oxit Si02, CuO khong tac dyng voi dung djch NaOH loang. Cac phan ung hoa hpc aia cac oxit voi dung dich NaOH loang (du): > —-> D.0,5. = 0,015 mol; n ^ ^ . = 2nc,(OH)2 = 2.1.0,01 = 0,02 mol. ^ , a = - S H I = -2l£2. ^ nco, 0,015 Nhan xet: Day la cau hoi kha tong hop ve tat ca cac loai oxit. L u u y them: C.2,0. Laigidi: ,«.^.,;v..:v B. 8. Cr203 + 2NaOH gtat: ciia V la + 2HBr CuO. Co bao nhieu oxit trong day tac dyng dup-c voi dung djch NaOH + 2NaOH ' * -> nso, = ngaSOA = O'Ol mol -> V = 0,01.22,4 = 0,224 lit = 224ml. V i d^ 1: Cho day cac oxit: NO2, Cr203, SO2, CrOs, C O 2 , P2O5, CI2O7, Si02, NO2 * dich X. Them dung dich BaCh d u vao X, thu dugc 2,33 gam ket tiia. Gia tri . mlu A. 7. "''' ' Laigidi: + H2O > H2SO4 it'z-w 2 33 > Ca(HC03)2 + H2O a • ikrlt ,r an D. A. 112. > NaHC03 CO2 + Ca(OH)2 ^ Vi dy 3: Hap thu het V m l khi SO2 (dktc) vao nuoc brom d u , thu dugc dung HO > Na2C03 + H2O CO2 + NaOH * >• —> Tdc dutiQ vai baza CO2 + 2 N a O H •ri< D. Nuacbrom. SO2 ^ <-- - > CO the lam mat mau nuoc brom: X = 0,1 . f^,/>- ^tini.>>;^ C Dung dich NaOH. ^'"-^ — > , ! iO^' B. CaO. Lot T u day suy ra ban dau: Na = 0,25 mol; AI2O3 = 0,075 mol. ^ m = 0,25.23 + 0,075.102 = 13,4 gam Dap an A. ; 4. OXIT AXIT a. L i thuyet * n ^ . - w n n ^ Nhan xet: Khi C O 2 va SO2 deu c6 tinh axit, tuy nhien khi SO2 con c6 tinh khu > mol -> an A. ^ f : A. Dung djch Ba(OH)2. > AICI3 + 3H2O a3-x Dap - Na2C03+H20 " 0 3— Ta c6: —> > Na2Cr04 + H2O Vi du 2 (CD-09): De phan biet CO2 va SO2 c6 the dung thuoc t h u la > Al(OH)3^ + NaCl • Na2S03+ H2O > Cr03 + 2 N a O H Cr r 0 2 + 2NaOH rt--.-v;f;Vijt!^-• Khang Vijt 33 _^ tao 2 loai muoi: Cacbonat va hidrocacbonat. Cach 1: Giai theo phuang trinh phan ung. CO2 NaN02 + NaNOs + H2O + Ca(OH)2 — — > CaCOsi + H2O 2NaCr02 + H2O J ^ ^ ^ 141 C^m n a n g O n l u y g n t h i d j i i UQC 1 8 c h u y § n 6i 2CO2 Mol: 2y H6ahpc- + Ca(OH)2 -i: N g u y § n van H S i , Cty TNHH M T V DVVH > Ca(HC03)2 vy , . .nvw y nc02 =0,12 mol; nggcog = 0,04 mol. Phuong trinh hoa hgc: X = 0,005; y = 0,005. Ba(OH)2 —> m c a c o a = 0,005.100= 0,5 g a m - > Dap an D . C a c h 2: T i n h n h a n h : n ^ o = n ^ ^ . - n r o o = 0,005 m o l . CO-^ Mol: ..y. OH Nhan t h a y : n^^2- > \^2^- Mol: 3 -> m = 0,005.100 = 0,5 g a m :r'- •! Dap an B. V i d u 5: Hap t h u h o a n t o a n 0,672 l i t k h i CO2 ( d k t c ) vao 200ml d u n g d i c h X g o m N a O H 0,1M va K O H 0,1M t h u dugc d u n g d i c h Y . C o c a n c a n t h a n Y B.3,06. C . 2,54. ^ + = 0,02 mol; n^+ = 0,02 n^^2- = n^^^. - nco2 = 0,01 nhanh: • ,. ' , , | . ,r . ; ,v > Ba(HC03)2 0,08 " ' '. ' -> riBa(OH)2 ~ 0,08 mol -> a = 0,1 M -> Dap an D . 5, B A I T A P O N L U Y E N Bai 1: E)6't 4,2 gam Fe trong khi oxi, thu dugc hon hgp X gom FeO, Fe203, (san pham k h u duy nhat) va dung dich chua m gam muoi. G i a tri ciia m la A. 12,10. mol. mol; 0,04 . B. 18,15. C.24,2. D . 20,57. Bai 2: Hon hgp X gom C u O va Fe203. Hoa tan hoan toan 16 gam X bang dung , dich H2SO4, thu dugc 36 gam muo'i. Mat khac, neu k h u het 16 gam X bang , C O (du), dan hon hgp khi thu dugc vao dung dich Ca(OH)2 (du), tao thanh m gam ket tiia. G i a tri cua m la N h a n t h a y : ^ =^ = i-.Taora2mu6i. n(_02 0,03 3 Tinh + 2CO2 0,04 ,.', ' V' D . 2,88. n ^ ^ . = n N a O H + " K O H = 0'02 + 0,02 = 0,04 m o l n ^ ' Fe304. Cho X tac dung voi dung dich HNO3 loang (du), thu dugc khi N O Lai gidi: Trong X: 0,04 > BaC03 + H2O ... ' ' .j, t h u d u g c a g a m c h a t r a n k h a n . G i a t r i Ion n h a t c i i a a la A. 2,26. + CO2 0,04 Ba(OH)2 ( I l l ' riQ^co^ = ^^o^- = ^''005 m o l . 3 Vi$t Lad gidi: : Ta c6: X + 2 y = 0,015 ; X + y = 0,01 ^ Khang A. 10. ' n^^^^ = n c o 2 " "^^2- = 0,02 mol. B. 15. C.25. D.20. Bai 3: Hon hgp X gom MgO, Fe203 va C u O . Hoa tan het 9,6 gam X bang dung djch H2SO4 (du), thu dugc dung dich chua 22,4 gam muoi. Mat khac, neu Y gom: Na^ = 0,02 = 0,02 mol; Bao t o a n k h o i l u g n g : a = m mol; CO ++m ++ m Na K = 0,01 mol; HCO gam -> Dap an B. V i d^ 6: Cho 0,1 mol P2O5 vao dung dich , A. 16. , , ,, K O H 2M. Sau khi phan A. K3PO4 v a K2HPO4. Gia tri ciia V la D . H3PO4 va KH2PO4. -> + 3H2O P2O3 "H3PO4 = 0,2 > 2H3PO4 = 0,2 mol. Vi dv 7: H a p t h u h o a n thanh ran X phan ling voi dung dich HNO3 d u thu dugc V lit khi N O (san pham khu duy nhat, 0 dktc). Gia tri ciia V la H3PO4 va xet ti 1$ m o l n h u tren. t o a n 2,688 l i t k h i CO2 ( d k t c ) vao 800 m l d u n g d i c h Ba(OH)2 a mol/1, t h u d u g c 7,88 gam k e t hia. G i a tri ciia a la A. 0,03. 142 B.0,04. D . 4,48. hoan toan vao dung dich Ba(0H)2 du, thu dugc 29,55 gam ket tiia. Chat A. 2,24. P2O5 C . 3,36. nong, sau mot thoi gian thu dugc chat ran X va khi Y . Cho Y hap thu nH3P04 = 2np205 = 4 > 3 ^ K O H d u va t a o ra muoi K3PO4. can chuyen B.2,24. Bai 5 (B-12): D a n luong khi C O di qua hon hgp gom C u O v a Fe203 nung - > Dap an C . Luu y: Cac em D.24. HNO3 du, thu dugc 1,12 lit NO (san pham k h u duy nhat ciia N*^ a dktc). A. 1,68. Lcngidi: Nhanxet: C . 12. dugc hon hgp X gom FeO, Fe203 va Fe304. Cho X tac d y n g voi dung dich ling B. K2HPO4, KH2PO4. :\. K3PO4 va K O H . B.20. Bai 4: Cho V lit khi C O (dktc) tac dung het voi 10 gam Fe203 nung nong, thu x a y ra h o a n t o a n , d u n g d i c h t h u d u g c c6 cac c h a t la f khu hoan toan 9,6 gam X bang CO (du), cho hon hgp khi thu dugc Igi tu tu qua dung dich Ca(OH)2 (du) thi thu dugc m gam ket tua. G i a tri ciia m la : HCO3 ^ 400ml mol. 2- + rn 3 = 0,02.23 + 0,02.39 + 0,01.60 + 0,02.61 ->• m = 3,06 = 0,02 i C0,05. D . 0,10. B.4,48. C . 6,72. D . 3,36. Sai 6 (B-10): K h u hoan toan m gam oxit MxOy can vira dii 17,92 lit khi C O ( Dap an C. c O * - - ^ - • - m , . : Y S A D. 1,344. Bai 10: Hoa tan hoan toan 8,16 gam hSn hgp gom Fe304 va FeS2 trong dung Bai 4: , dich axit HNO3 (dac, du), thu dugc 4,032 lit khi NO2 (dktc) va dung dich X. • 'I r . O - S . 0 • €.0 -,,f0(1 - f r - . 5 n :finfirin j ' .:, " 22^4 " Cho X tac dung voi dung djch Ba(OH)2 du, Igc ket tua va nung trong khong +3 +3 khi den khoi lugng khong doi thu dugc m gam chat ran. Gia tri ciia m la So do phan ling: FezOg A. 16,18. Nhan xet: So oxi hoa ciia Fe 6 san pham cuoi cimg hoan toan giohg vai ban B. 15,86. C. 12,66. D. 30,26. >X dau. 6. HI/6NG DAN-L61 G I A I V' B a i l : nFe = 0,075 mol. ^^ ^ , ,, "i^foi """^"^ > Fe(N03)3 + NO ' ^ -"^^ ' ' lAc dau C O nhuong electron cho Fe203 ^ .^.^ > X, sau do X nhuong so electron vua nhan dugc cho N"^^ de tra thanh Fe(in). Nhan xet: Bai nay neu dua theo phuong trinh phan ung se ra't ciai dong va ton nhieu thai gian. 2nco = 3r»NO "'^"i.meH" Bao toan electron: > X (FeO, Fe203, Fe304) Fe Mol: 0,075 > Fe(N03)3 2g gg : 91 fl im 6'/ »OHfJi Atmh aUsH rrayuitj nK> ma 3 E 3 ;\ U . . '""^"^ > Fe(Na)3 (£l-lf) v vaap dung bao toan nguyen to Fe: «'P ff,tt^-r'iih '^.P . ^ I K ' ' 0,075 ricr -> m = 0,075.242 = 18,15 gam -> Dap an B. Bai 2: Nhan xet: Khi cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va m ? <^ nguyen tu O^" trong oxit dugc thay the bang mgt ion SO 4 . Tang giam khoi lugng: 1 mol 0 2 a mol > 1 mol SO 4" <- Dap an A. Bai 5: O day, cac em can su dung so do phan ung: Fe " c o = 0'075 mol -> V = 1,68 lit ~ * " C O = " C O 2 = " B a C 0 3 = O'^S moi " 0'^^ So do phan ling: ' Khoi lugng tang 96 - 16 = 80 gam. '^^ ' * ' * :. cOD . : • +3 +2 +3 + 2 Pe2 0 3 , C u O >X ^"^"^3 ) Fe(N03)3 + Cu(N03)2 ^ NO N/ian xet; So' oxi hoa ciia Fe va Cu a san pham cuoi ciing hoan toan giong voi ban dau. .^ f _ , . ,,,, , Luc dau C O nhuang electron cho Fe203, CuO ' tang 36 - 1 6 = 20 gam. "BaC03 " > X, sau do X nhuong so' electron vua nhan dugc cho N'^ de tra thanh Fe(III), Cu(II). Bdo toan electron: 2nco =3nNo ^ Dap an C. - > nNO r I'.i7f-t. = 0,lmol V = 2,24lit md<.>SvO- , . , ' ^ 145 Cty TNHH MTV DVVH Khang Vigt Ca'm nang On luy^n thi dgi hpc 18 chuyen 66 H6a hqc - Nguyjn Van Hi) Bai6: 1 iw. nco =0/8 mol; nso2 =0/9 mol. Gpi so mol: Fe304 (a mol) va FeSz (b mol). Ta c6:232a + 120b = 8,16. Nhan xet: Vi oxit MvOy bj khu boi khi C O -> Logi B va C vi C O khong khir dugc oxit ciia crom. O cac phuong an con lai, M deu la Fe —> M la Fe. Cac phan ung khu: Ta co: no (oxit) = nco = 0,8 mol. » Bao toan electron: 3npe = 2nso2 = 0,6 mol. -> i}F^ = M = l ^ OxitsatlaFe304 no 0,8 4 '" '"' 'v^ Fe304-le n„ Nh^n thay: Mol: a03 Na °" FeS2 = — = 1,5 > 1 -> Phan ling t?o ra 2 muoi cacbonat va a2 nc02 ^, > -Fe203 • 2 2nB3(OH)2 = 0,2 + 2.0,05 = 0,3 mol : .^^^ hidrocacbonat. !> , , 1 , an C , 1/0", u i i ; i i j i i < sfUw'EK* ,1 ..ti !i • > ) 11 ' f J'fl ' '> fj||.ii[| ' .; '4..,,,,, ' wsV;--Vvh f M i s - . P * f fi.jt.'i. IM • Hi: *- - J a045 > -Fe203 + 2BaSa Mol: 0,01 0,005 0,02 m = 160.0,05 + 0,02.233 = 12,66 -> Dap Tinh nhanh: n^^2- = "QJ^- - " 0 0 2 = 0,3 - 0,2 = 0,1 mol! j„ ':;•.;..>. . a + 15b = 0,18 a =0,03; b = 0,01. Bao toan nguyen to'Fe va S: DapanD. = 0,05 mol; n^, + = 0,2 mol. 2+ Ba ^ nN02 = "Fe304 + 15nFeS2 Fe304 Trong Z: >3Fe^3 FeS2-15e > ¥e*^ + 2S^ Bao toan lectron: Bai7: " O H - = ""NaOH + OH' •* ... , > U • • ifh i/1 „ .jjfg,^^ -^^^ ; 3 -> Nhan thay: n^^2^ < n^^2- nBaCOa = "332+ 3 = 0,05 mol. ->m = a05.197 = 9 , 8 5 g a m ^ DapanB. Bai 8: Nhan xet: > P2O5 + 3H2O 2H3PO4 -> Tao2mu6i:K2P04vaK3P04 Lim y: Cac em can chuyen ........... DapanB. P2O5 thanh H3PO4 , u ',(h va xet ti 1? mol nhir tren.' Bai 9: npaCOa = 0,06 mol. 6 day, cac em can sir dung so do phan ung: CuO,Fe2a > X C O 2 + Ba(OH)2 ) Muoinitrat + NO > B a C a + H2O % Mol: a06 ,. a06 ' ' . ,, Nhan xet: Khi cho hon hgp oxit tac dyng voi khi C O thi: ng trao doi = 2nco • Ma n^o = " 0 0 2 ~ ^'^^ ^'^^ ~^ Bao toan electron: ^VNo=a8961it ngtraod6i =3 n]sjo-> nj^o = 0,12 mol. 0,04 mol. I -» Dap an C. U7 elm nang 6n luy^n thi dgi hgc 18 chuySn dg H6a hpc - Nguyjn Van H5i Cty TNHH M T V D V V H Khang Vigt Chuyen de 6 phat b i e u diing la: CAC mmm TO P H I KIM mm umn 1, H A L O G E N a. L i thuyet * + 3Cl2 F2 + H2 * s.i^w 2FeCl3 Cu+ C k Tdc dung vai hidro > 2HF c ^ > K C l + KCIO + H2O 3Cl2 + 6 K O H * * 2 K M n 0 4 +16HC1 '! + 5.(X0.Ooi a i , .Dn!5qM< + I2 ^ - ;. Loai C vi F2 chi c6 tinh oxi hoa, khong c6 tinh khu. MnCh + C h t ^ . Loai D v i CI2, Br2,12 deu it tac dung voi nuoc. A.a24. '~rA'^ , , 4 - B.0,48. ^•.f^n^r C.0,40. Mol: -> a = V i dv mau t'. < , t^^ n . »> A. Halogen la nhung chat oxi hoa manh. B. K h a nang oxi hoa cua halogen giam tir flo den iot. •^M^ 3Cl2 + 6 K O H ' •, V i di^ 1: Phat bleu nao sau day la sai? D . 0,2C. Laigiai: +2H2O > 2KC1 + 2 M n C l 2 + S C h t + 8 H 2 O ""'^ ) 5KC1 + KCIO3 + 3H2O 0,03 <- 0,25 0,06 = 0,24 mol/1 , '^'''^^ 0,05 , Dap an A. d\ 5: C o cac thi nghi^m sau: . , . , (1) Cho dinh sat vao dung djch H2SO4 {loang, nguoi); ,^ " ' (4) Cho la A l vao dung dich H2SO4 (dac, ngupi). -C>C ^ nV,t >tsit:'. So' thi nghi?m xay ra phan ung oxi hoa-khu la LMgidi: Nhan xet: D o flo la nguyen to'co dp am di^n manh nhat -> flo chi the hi§n so , ;. (3) Cho nuoc brom vao dung dich N a l ; D. Cac halogen c6 tinh chat hoa hpc tuong t\ nhau. , H;>lt!JH UlJJ >X (2) Sue khi SO2 vao nuoc brom; : C . Cac halogen deu c6 the c6 so oxi hoa : -1, +1, +3, +5, +7. oxi hoa -1 trong cac hop chat ^ ffj (jjyC'j - sec. Sau khi phan ung hoan toan, thu dupe 3,725 gam K C l . G i a tri cua a la mn Q , , | j r ^ '! . A-2- •\:,p,4.. B.I. c3. LMgidi: Phuong an C sai. , (|/; ... Cac phan u n g hoa hpc a cac thi nghi^m: -> Dap a n C . V i d\ 2: C h o biet cac phan ung xay ra sau: 2FeBr2 + Br2 > FeBra • 2NaBr + C h .tOnlV'. f , Laigiai: -^DapanB. * ' '-IdSwiv.}, ! Vi dv 4: Cho 1,344 lit khi CI2 (dktc) d i qua 250ml dung dich K O H a mol/1 0 2NaCl + 2 H 2 O - ^ S ^ 2NaOH + C I 2 T + H 2 t b. , V|SOnM .S I >2HBr+H2S04 — iCfiq D. Tac dung manh voi nuoc. Dieu che'clo M n 0 2 + 4HC1 ^^nub wb .gnv/ui i o v g m i n wMrt d 9 IdA gm-ml t-n uh' Lo^i A v i F2, CI2 la cha't khi; Br2 la chat long va I2 la chat ran. > 2FeCl3 • , ,' ('» r i i / l : V i dv 3: C h p n nhan dinh dung. Cac halogen deu c6 tinh cha't chung la {Clorua voi) Tdc dung vai hap chat Br2 + SO2 + 2 H 2 O CaOCh + H2O > 2NaBr CI2 + 2FeCl2 •—^DapanD B. Co tinh oxi hoa. {Nu&c Gia-ven) > 2 N a C l + Br2 Br2 + 2NaI •() li / : ; \ Tinh oxi hoa CI2 > Br2; tinh k h u Br- > CI", A. Chat khi 6 dieu kien thuong. ^.^ ^ "Day" halogen ditngsau CI2 + 2NaBr t^han xet: (1) —>• Tinh oxi hoa Br2 > Fe'^; tinh k h u Fe^* > Br". (2) \is ri\ i ijof* tinh oxi hoa CI2 > Br2 > Fe^*; tinh k h u Fe^* > B r > Ch. . > 5KC1 + KCIO3 + 3 H 2 O CI2 + Ca(OH)2 {sua voi) D . CI2 c6 tinh oxi hoa hon Fe^*. 01, > 2HC1 t o : CI2 + H2 Tdc dung v&i dung dich kiem CI2 + 2 K O H * CuCh j'nrv„>^.ff)fy^-. ., , « r o... r't;» r + , B. Br2 c6 tinh oxi hoa manh hon CI2. Q B r CO tinh k h u manh han Fe^*. Laigiai: ,,,,;.|. , Tdc dung vai kim loai: 2Fe * t-g^i ^ ^ ., . ^ A. CI" CO tinh k h u manh han B r . > 2 N a C l + Br2 0 ) Fe + H2SO4 (loang, ngupi) ' 'ii • ^ (1) (2) , (2) S02 + Br2 + 2H2O (3) Br2 + 2NaI > FeS04 + H 2 t ^ > H2SO4 + 2HBr > 2NaBr + I2 ' ' • * ' ' - * ~ '' ' " ' 149 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 trong H2SO4 loang, nguoi; 2- A l , Fe khong tac di^ng voi 5 + 2H2SO4 (dac) H2S04dac, nguQi. Vi ^ : * 6: Hon hgp khi nao sau day khong ton tai 6 nhi|t dp thuong? A. C 0 v a 0 2 . B. Cl2va02. C. H2SvaN2. Nhan xet: H2 va F2 c6 the phan ling manh liet ngay ca trong bong to'i va 6 nhi^t do rat thap, c6 the gay no va toa nhieu nhi^t, tao thanh HF. 5 ^DapanD. b. Ozon: .„;,", ' 1 ri Ozon la mpt chat oxi hoa manh, manh han ca oxi. 2Ag + O3 £ »,; '° > Ag20 + O2 O3 + 2KI + H2O phan ling vai lugng d u dung dich HCl dac, toi phan ling hoan toan thi * , M f^ , „ > O2 + I2 + 2 K O H c, Hidrosunfua A.KMn04. '" B.Mn02. ' C. CaOCh. D. K2Cr207. > Mn02 + 4HC1 MnCk + C h t KMn04 + 8HC1 + 2HCI * +2H2O 2KC1 + 2CrCl3 + SCh -—> + VHiO Dap an D. * Fe + S — ^ * FeS * * 4Fe(OH)2 + O2 + 2H2O 2H2S + O2 > Na2S03 + H2O Dieiiche '° > 2SO2 + 2H2O , , , -f, A: > 2HBr + H2SO4 — ^ , ' • ,j, ^ ^ , ^ | , . > 3 S + 2H2O : l;;,M>c.ri 4FeS2 + I I O 2 > 4Fe(OH)3 ( > 2S03 Tinhoxihoa s + 02 CO2 * Tinhkhie SO2 + 2H2S Tac dung v&i hap chat > H2SO4 + 8HC1 > NaHSOs SO2 + Br2 + 2H2O SO2 f > FeCh + H 2 S t • 2SO2 + O2 < > HgS H2S > 2S + 2H2O 2NaOH + SO2 '' S + H2 — ^ CO + O2 — ^ ' 2SO2 + 2H2O NaOH + SO2 2H2O , " *' e. Luu huynh dioxit a. Oxi - luu huynh O2 + 2H2 — ^ ' Dieii che FeS + 2KC1 2. OXI - L l / U H U Y N H Hg + S + 2HN03 + H2SO4 Voi clo: H2S + 4Ch + 4H2O K2Cr207 nhan electron nhieu nhat -> tao khf Ch nhieu nhat 3Fe + 2O2 '° > Fe304 Tac dung vai phi kim 1 * ' 1 2H2S + O2 — ^ Mn02 CO the nhan vao Ian lugt la 1, 5, 6, 4. MgO .» 1 6 dieu k i f n thuong, dung dich H2S bi oxi hoa dan thanh S: CaCh + Ch + H2O Tac dung vai kim loai i- Tinhkhiemanh 2H2S + 3O2 — ^ , Cach 2: Nhm xet: So mol electron ma 1 mol moi chat: CaOCh, KMn04, K2Cr207, O2 + S — ^ > FhSi > CuSi j Voi oxi: H2S chay trong khong khi voi ngpn lua mau xanh nhat: Dap an D. M g + O2 — ^ '•:(') H2S + CuS04 > KCl +MnCl2 + - C I 2 T + 4H2O 2 K2Cr207 + 14HC1 CaOCh Tinhaxityeu H2S + Pb(N03)2 Laigidi: Cach 1: Dua theo cac phuang trinh phan ung: * K2Mn04 + Mn02 + O2 > 2KC1 + 3O2 chat tao ra lug-ng khi CI2 nhieu nhat la: * 3S02 + 2H2O 2KC103 V i d^ 7: Neu cho 1 mol moi chat: CaOCh, KMn04, K2Cr207, Mn02 Ian luot * > SO2 + NO2 + H2O Dieu cheoxi 2KMn04 — ^ D. H2vaF2. 4 ' Laigidi: '° • SO2 J.I - • 2Fe203 + 8SO2 Na2S03{rin) + 2 H 2 S 0 4 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 ^ : .2. D a p an C. Cac p h a n u n g hoa hoc: > H2S04.nS03 • (Oleum) ---H', VIDVMAU 2SO2 + 0 2 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 0 4 , KNO3 v a A g N O s . Chat tao ra 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 + 1 " H + " ^ " 0 2 + 2nH2 = 0,28 m o l . A . 3. J o m € J .0 - ' 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 3 (dac nong, d u ) t h u d u g c V l i t k h i c h i c6 N O 2 (dktc, san p h a m dich BaCl2, t h u d u g c 23,3 g a m ket tua; con 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 trong NH3 d u tao thanh 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 0 4 . 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 0 4 . 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 2 chat oxi hoa la O va H2SO4 (H*). 2 D . H2S, O2, nuoc b r o m . KK;. Led gidi: • VK. Vay: n = 2 va M = 24 ( M g ) ^ D a p an D . ^ • n = 0,28 ^ iM nrij : , o l f -.if i ' ih) $H Jif S f « , S x/ub u d t ! (d) M o o c p h i n va cocain la cac chat ma t u y . Ba2- + sol n > K2SO4 + 2MnS04 + 2H2SO4 , • > M2(S04)n - ^ d u n g d i c h N H 3 d u t h u d u g c 5,35 gam ke't tua. Gia t r i ciia V la > 2MCln + n H 2 0 ' • + nH2S04 4<^;« k h u d u y nhat) v a d u n g d i c h Y. Cho toan b g Y vao m g t l u g n g d u d u n g Nhqn xet: K h i cho Y tac d y n g v o i axit H C l giai p h o n g k h i —> M c o n d u . 4M 5SO2 + 2 K M n 0 4 + 2H2O Lai gidi- '^'''^ ^'^^^ * Mg. ^ Lai gidi: Cach 1 : Giai theo p h u o n g t r i n h hoa hoc. < > 2HBr + H2SO4 " t h u d u o c chat r a n Y. H o a tan het Y t r o n g d u n g d}ch H2SO4 loang, t h u dugc 0,896 l i t k h i H2 (dktc). K i m l o ^ i M la , (c) K h i d u g c thai ra k h i quyen, freon (chii y e u la CFCI3 v a CF2CI2) p h a h u y Vi dv 2: C h o 3,36 g a m k i m loai M (hoa t r i k h o n g doi) tac d u n g v o i 0,05 m o l O2, A.Ca. SO2 + Br2 + 2H2O tangozon. N/ifln xef: KCIO3 cho lu(?ng O2 nhieu nhat . ^ '.^ i - (b) K h i SO2 gay ra hien t u g n g m u a axit. ' * '^'^^^^'^ * Ag . NO2 + - O2 AgN03 ' (a) K h i CO2 gay ra h i ^ n t u o n g hieu u n g n h a k i n h . > 2KC1 +3O2 2KC103 < ^^ V i d v 4 : C h o cac p h a t b i e u sau: gidi: K 2 M n 0 4 + Mn02 + O2 , + ? ^ Cu(OH)2 + 4NH3 > [Cu(NH3)4](OH)2 5^35 ^''^^ -./CiSS.i Bao toan nguyen to Fe: npg (X) = ripe(OH)3 ^ ^'^^ m o l . - .-^ Bao toan k h o i l u g n g : m x = m c u "^Fe "^s ~» mcu = 9,2 - 0,05.56 - 0,1.32 = 3,2 gam ' ' , , ', dm nang On luygn thi djii hpc 18 chuyfin ii H6a hgc - Nguyin van H5i 32 ricu(x)==0'05mol Cty TNIHH MTV DWH Khang Vi$t yi >. i Qui doi X ve hon hop gom cac don chat: Fe = a05 mol; Cu = 0,05 mol vaS = ai 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 trong dung djch 8: Hon hgp khi 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 khi di ra. Cac the tich do 6 ciing dieu kien nhi^t dg, ap suat. Phan tram the tich ciia O2 trong 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 48 - 38,4 M hon hg-p X trong oxi, thu dugc m gam oxit. Gia tri cua m la tniih u& (a; no3 A. 5,84. 3 2 x = l,5y. Phuong trinh hoa hgc: B.4,80. C. 4,28. D. 4,96.,^,: .....qiiOoM ( h : Laigidi: 2 912 nH2==0,13mol. ' M-gniAu%Mlsfki'cy:. Nhan xet: Khi cho X + HCl, ion H* da nhgn electron de tro thanh khi Hz: :;,•{„.?„,,';;. 2H^ + 2e , ilr«.;'t?v,. v Goithetichcua:02 = xml;03 = yml. Mol: v.bn ^ > Hat 0,26 <- ai3 .-|'|sf',^ itp^^,)(.if^gfi.,.(i^, > Oit + 2KI + H2O Mol: y -» Thetichkhithoatra:x + y = 75 -> x =45;y = 30. gnsorW OB-> gb %Vo, = - . 1 0 0 % = 60%. ^ Dap an C. Mol: a065 <- 0,26 * V i d\ 7: Nhi^t phan hoan toan 4,03 gam hon hop X gom KCIO3 va KMn04, thu dugc 0,784 lit khi O 2 (dktc). Phan tram khoi lugng cua KMn04 trong X la €.39,2%,^.:^^ , B.19,6%. Fe voi S la A. 40%. B.60%. 2KMn04 — ^ Fe + S Mol: ) 2Ka y ' + 302 Fe + H2SO4 • FeS + H2SO4 • = 0,02; y = 0,01. s'J =--(X, ..i< 2 .r^nn^a'AM -> %mKMn04 = °^"^Q3^^-100% = 39,2%. , r, - ,,fr ,1.8 -^DapanC. Sf:! ,0.-;^. . ' '" > FeS X -> FeS04 + H2t > FeS04 + H2St - > H2 = (0,l-x)mol. ^ H2S = xmol. Ap dung cong thiic cua phuong phap duong cheo, ta c6: l,5y Theo bai: 122,5x + 158y = 4,03 va l,5x + 0,5y = 0,035 -> X X Cho Y tac dung voi HCl: K2Mn04 + MnOa + O2 0,5x 2KC103 X - > Y gom: Fe = (0,1 - x) mol; FeS = x mol; S = (0,15 - x) mol. ! X D. 80%. Cach 1: Giai theo phuong trinh hoa hgc: D. 58,8%. ; Vj Mol: C. 70%. np = M = 0,1 mol; nc= ^ = 0,15 mol; My = 10,6.2 = 21,2. 56 32 • Mol: Cac phan ung dieu che O2: ' ' r-. Laigidi: ' „ •'^'^ ............ ,,. • Bao toan khoi lugng: m = mx + moj = 3,76 + 0,065,32 = 5,84 gam. A. 78,4%. , 75 "--2 I^I'-'M hgp khi Y. Ti khoi cua Y doi voi hidro la 10,6. Hi^u suat cua phan ung giua 2(y- - > Dap an A. + 2KOH + hi CO oxi), thu dugc chat ran X. Cho X vao dung dich H C l du, thu dugc hon Mat khac, khi cho X + O 2 , 0 2 da nhan electron de tro thanh 02-: + 4e O3 6,4 Vi dv 9: Nung 5,6 gam Fe voi 4,8 gam S 6 nhiet dg cao (trong dieu kien khong = "HCl = 2nH2 = 0,26. O2 32-38,4 ^H2S _ 34-21,2 2-21,2 x = 0,06 mol 3.(0,l-x) = 2x. 19,2 3 H = 0,06 .100% = 60%. 0,1 • Dap an B. IKK Cty TNHH MTV DVVH Khang Vigt dm nang 6n luygn thi d^ii hpc 18 chuySn dg H6a hpc - Nguyin Van H^i Cach 2: NMn xet: Do Fe 11^2 + ^H2S > Hi; Fe "^Fe O'l 'Tiol. Ap dung cong thuc duong cheo, ta c6: >»• 2 0,04 2-21,2 3 0,06 n H2S + .{fei^,8- • H = 60% Dap an B. ^ 2NH4CI + ^^.^.^ .j,, _ ^ .^.^^ „ * 6Li + N2 1 > 2Li3N N2 + 3H2 * _ 2NH3 ' , •" , : r 3 M g + N2 — 4 , ^ ^ MgsNa 2P+ * „ J^|,; , 6 nhi^t do cao khoang SOOO'^C (hoac nhiet do cua 16 ho quang dien), nito \ NH4NO2 .it.! ,) > , , . J l (> .(< ^ ' i ^ , . , , ,(;,-! ,f i , ^ ^ ^, . Tinh khu . 2P + 5Cl2 .-.ti-f-i ff>|f( 5r,!/f, (-pfK^ v^no.*' i i i i i i -i. '° > 2PC15 n-ii ,ni>. .ir,u>-\ ^ H3PO4 + 5NO2 + H2O > N2 + 2H2O > N2 + N a C l + 2H2O , V i du 1: C h o can bang hoa hpc: N2 {k) + SHi (k) < . ;i Tflc dung vai axit: NHa + H C l C . Thay doi nhiet dp. ' D . Them chat xiic tac. Nhan xet: Theo bai, phan ung toa nhif t (AH < 0) - > can bang se chuyen dich khi thay doi nhi^t dp —> Loai C . > Al(OH)3 i + 3NH4C1 ''rloralx- HI'' > [Cu(NH3)4](OH)2 ' ^,. ^ > [Zn(NH3)4](OH)2 > [Ag(NH3)2]Cl„, , , Thay doi nong dp N2 ciing se lam can bang chuyen djch - > Lo^i B. . - > D a p an C . ' Luu y: Chat xiic tac chi lam anh huong deh to'c dp phan ling, khong lam ^ chuyen dich can bang. 4NH3 + 3O2 ——> 2N2 + 6H2O > N2 + 6HC1 Tac dung vai mot sooxit kim loai: 3 C u O + 2NH3 — ^ 3Cu + N 2 +3H2O ,, , ^ ^ i d\ 2: C h o cac thi nghiem sau: (a)DotkhiH2Strong02 > 4 N O + 6H2O Tac dung vai do: 2NH3 + 3CI2 can bang chiu tac dpng cua ap sua't Loai A . Tfn/i khu 4NH3 + 5O2 ; gn< * '•' V nnrf nhrl - m » ' Lot gidi: > NH4CI > 2NH3 (fc); phan ling B. Thay doi nong dp N2. Tong so mol khi 6 hai ben khac nhau •• • Zn(OH)2 + 4NH3 Tac dung vai oxi: ' thuan la phan ling toa nhiet. C a n bang hoa hoc khong bi chuyen dich khi: A. Thay doi ap sua't ciia hf. ;•' Tflc dun^ vai muo'v. A I C I 3 + 3NH3 + 3H2O A g C l + 2NH3 ,{>m ./„ ^ NH3 + H 2 0 < = ± N H ; + O H - Cu(OH)2 + 4NH3 ) 6CaSi03 + P 4 + l O C O VIDUMAU Tinh baza yen: Tao phuc chat: + IOC (J fi6 b. Amoniac + <- ' ' ^^^^^^^ 3C. ^ ^ , C a 3 P 2 2Ca3(P04)2 + 6Si02 NH4CI + N a N 0 2 + risds 2NO Dieii che * ^ N2O + 2H2O P + 5HNCh (dac) — ket hgp trvc tiep voi oxi, tao thanh khi nito monoxit: N2 + O2 CaCh + N H s t + Ca(OH)2 NH4NO3 — Tac dung vai kirn loai: > 2NH3t Phdn itng nhiet phan NH4CI — 3.NITO-PHOTPHO ^5 - Tac dung v&i dung dich kiem (NH4)2S04 + 2 N a O H •J(W 34-21,2 ^ Muoiamoni > H2S nen > FeS • du; ( c ) D a n k h i F 2 vaonuocnong; 4v - > t- (b) Nhi^t phan K C I O 3 (xiic tac M n 0 2 ) ; (d) Do't P trong O2 d u ; (e) K h i N H 3 chay trong O2; ( g ) D a n k h i C 0 2 d u vao dung djch Na2Si03. 157 Ca'm nang 6n luy^n thi dai hgc 18 chuy§n dg H6a hgc - Nguyin van Hit Cty TNHH MTV DWH Khang Vi§t So thi nghi^m tao ra cha't khi la A. 5. y i dv 4: N u n g n o n g h o n hg-p g o m 1,6 gam Ca va 0,62 g a m P t r o n g b i n h k i n B.4. C.2. D.3. k h o n g CO k h o n g k h i t a i phan l i n g hoan toan, t h u d u g c h o n h g p cha't ran Y . Laigiai: ; Cac phan ung hoa hoc: (a) 2H2S + 3O2 , • , , i . , '° > 2SO2 + 2H2O • (b) 2KC103 > 2KC1 + 3O2 > 4HF + O2 Via A . 672. ' , .,„,, ^„ UO t B. 560. nca = 0/04 m o l ; n p = 0,02 m o l . 3Ca '° > 2P2O5 502 (e) 4NH3 + 3O2 — ^ Mol: 2N2 + 6H2O (g) 2CO2 + Na2Si03 + H2O (*) > Si02 + 2 N a H C 0 3 2P 0,03 ung hoan toan, thu dug-c V lit khi (dktc). Gia trj cua V la B. 560. C. 336. D. 448. > 3 C a C l 2 + 2PH3t A . 71,3%. 0,1 !., A ; • f i t i, C.73,1%. D . 65,9%. dugc hon hop Y . Ti khoi ciia Y so voi hidro bang 4. H i | u suat cua phan ung tong h g p N H s la B. 25%. ft C. 30%. Lai D. 10%. .1 giai: 1 m o l canxi dihidrophotphat Ca(H2P04)2 hay 1 m o l P2O5 deu chua 2 m o l P. Ca(H2Pa)2 < Khoi l u g n g mol: % khoi lugng: > P2O5 234 gam x% ! 142 g a m X = - — = 65,9-> Dap an D . ,;(„ , >;fl , ^Vv;-' V i d\ 6: H o n h g p k h i X g o m N2 va H2 c6 t i k h o i so v o i h i d r o bang 3,6. N u n g Gia thiet trong X: n^j = l m o l ; nH2 =4 m o l . Ta c6: mx = m^j + m^j = 28.1 + 4.2 = 36 gam. , Bao toan n g u y e n to'P theo so do: trong binh kin 6 nhi?t dp khoang 4500C c6 bot Fe xuc tac. Sau phan ung thu n o n g X t r o n g b i n h k i n c6 bgt Fe xuc tac. Sau p h a n u n g t h u dugc h o n h g p k h i • * X CO so m o l g i a m 8% so v o i ban dau. H i § u suat ciia p h a n l i n g t o n g h g p N H 3 36 Nhan xet: Bao toan k h o i lup-ng: my = mx = 36 gam -> ny = — = 4,5 m o l . la . , A . 25%. B. 20%. g ,r Phan ling hoa hpc: 3x M Nhan xet: V i dy 3: Hon h^p X gom N2 va H2 c6 ti 1$ m o l tuong ung la 1:4. Nung nong X X r, B.69,0%. Phan supephotphat kep CO t h a n h phan c h i n h la Ca(H2P04)2. / " - > nN2 = nNH4CI = 0,1 m o l - > V = 0,1.22,4 = 2,24 l i t - > Dap an A . Mol: -i tS. , (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 nay la N a C l + N2t + 2H2O 0,1 3H2 - ' Laigiai: N H 4 C I + NaN02 — ^ N2 + ' > " P2O5 ve k h o i l u g n g . H a m l u g n g p h a n t r a m ve k h o i l u g n g cua Phan ling hoa hoc: A . 20%. " ' ' i - > V = (0,02 + 0,01).22,4 = 0,672 l i t - > D a p an A . 40% = 0 ' 0 1 r n o l ; nNaN02 = 0 ' 2 m o l . 0,1 ' , i i if , ' i "' 0,01 ' 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: nu • ' > CaCk + H 2 t Ca + 2HC1 A . 224. • ' Ca3P2 0,02 Ca3P2 + 6HC1 V i d^ 2: Dun nong 100ml dung dich chua N H 4 C I I M va NaN02 2 M d e n phan "NH4a + M Y gom: Ca3P2 = 0,01 m o l ; Ca = 0,01 m o l . -> Dap an D. ,j D . 448. (*) (d) 4 P + fj C. 336. Laigiai: (*) » ' (c) 2F2 + 2H2O Cho Y tac d u n g v o i n u o c d u , t h u d u g c V m l h o n h g p k h i (dktc). Gia t r j ciia , 1 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 trong X: n ^ j = 1; ~ ^- 2-7,2 -> n x - n Y = 2 x - ^ ( l + 4 ) - 4 , 5 = 2 x - > x = 0 , 2 5 m o l - > H = 2 5 % - ) - D a p a n A . IRQ 159