5450-0 NHA GIAO LfU TU PHAM SY LL/U THI/VIEN Vm Nhiem vu cua cac trUdng chuyen, 16p chon clUdc nganh Giao due giao nhiem vu la dao tao nguon nhan lUc, boi dudng nhan tai de gop ph^n day manh sU nghiep cong nghiep hoa va hien dai hoa dat nudc. De dap ilng yeu cau do, Nha sach Khang Viet xin tran trong gidi thieu cuoh sach: Bp de Hoa hoc 9 on thi vdo 10 cua Nha giao Uu tii Pham Sy Luu. Sach gom 3 noi dung chinh: - Phan 1; Cac phUdng phap giai nhanh bai tap hoa hoc. - Phan 2: Tuyen chon gi6i thieu de thi tuyen sinh ciaa mot so'trUdng chuyen. - Phan 3 : Ba mUdI de thi thii. Phan phiftfng phap : giiip cac em c6 each giai dung, hay va nhanh gon nhat, thich hdp v6i tufng dang bai tap. Moi phUdng phap dUdc minh hoa bang nhieu vi du dl hieu de cac em c6 the van dung dUdc mot each nhanh nhat. De dam bao yeu cau chuan kien thiic, ki nang nen cac vi du da dUdc can nhfic, Ixia chon chu yeu tii cac de thi DH & CD cua bo GD & DT cung vdi Idi giai dUdc trinh bay chi tiet theo each giai tU luan phu hdp vcti kieu ra de thi tuyen vao cac trUdng chuyen va 16p chon. Phan giofi thieu de thi tuyen sinh cua cac trtfofng chuyen : gom 30 de cua nhieu trUdng chuyen tren ca nUdc trong 5 nam gdn nhat. Moi de bao gom nhieu kien thiic trong tam va nhieu ddn vi kien thiic nang cao, doi hoi tinh ta duy sang tao, sU van dung Unh boat cac phUdng phap giai toan hoa hoc hien dai. Phan nay giup cac em trUcfc ki thi tuyen can nhdc, chuan bi cho minh kien thiic va ki nang lam bai hieu qua de dat thanh tich tot nhat. Phan de thi thii: gom 30 de. Moi de c6 kien thiic bao quat nhieu van de trong tam va nang cao cua chUdng trinh hoa hoc THCS c6 do kho nhat dinh, doi hoi hoc sinh phai hieu biet, suy luan mcl rong, c6 kl nang van dung linh boat cac phiidng phap giai toan hoa hoc. Nhieu van de dUdc lap lai ci cac de khac nhau nhSm giiip cac em nim dUdc kien thiic va phUdng phap giai tot nhat. Cuo'n sach nay la cong cu hiiu ich khong the thieu doi v6i hoc sinh trong qua trinh hoc tap va on luyen mon hoa chuan bi cho ki thi tuyen vao 16p 10 chuyen, sach cung giiip cho cac bac phu huynh c6 dUdc tU lieu de c6 phUdng hu6ng tU van, giiip dd cac em hoc tap dat ket qua tot trong ki thi. Mac du da rat c6 gkng va dau tU nhieu thbi gian cung v6i nhiing tich luy trong nghe ciia tac gia nhUng chac chan khong tranh khoi nhiing thieu sot, chiing toi rat mong nhan diidc nhiing dong gop cua ban doc cho noi dung ciia cuoh sach hoan thien hdn trong Ian tai ban. ; Xin chan thanh cam dn! O/ TNHHMTVDVVnKhan^ Vi6l Phan I: PHimNG PHAP GIAI TOAN HOA HOC 1. Quy tac thCr ta uu tien cua phan Crng (/) Phan ifng oxi hoa khu: - Phan irng theo chi^u: chat o.xi hoa manh licfii oxi hoa chat khit nuiuh hc/ii tao thanh chat o.\i hoa yen hem vd chat khifyeh hon. vu , - Khi xay ra phan irng ciia h6n hop chat oxi hoa vori h6n hap cha't khir: Phan I'Oi^ mi tien la chat o.xi hoa manh nlid't o.xi hoa chat klu't nianli nhat tao thdnli chat o.xi hoa yen nhat vd chat khi'f yen nhat. (2) Phan itng cua cac chat dien li trong dung dich theo thii tit uu tien sau: - Phan ii-ng trung hoa: H"^ + OH" -> HgO - Phan irng tao ket tiia hidroxit: M"^ + nOH" -> M(OH)„ - Phan irng hoa tan hidroxit lu5ng ti'nh trong k'lim dir: M(OH)„ +(4-n)0H ->[M(0H)4f Hay : M(OH)„ + (4 - n)OH -> MO^^ + 2H2O i'i Vi du 1. Cho 29,8 gam h6n hop bot gom Zn va Fe vao 600ml dung dich CUSO4 0,5M. Sau khi cac phan ling xay ra hoan toan, thu diroc dung djch X va 30,4 gam h6n hofp kirn loai. Phdn tram va khoi lucmg ciia Fe trong h6n hop ban ddula A. 56,37% B. 64,24% C. 43,62% D. 37,58% Gidi 29 8 29 8 "CUS04 = 03 moK ig- = 0,450 < (n^,, + np,) < ^ = 0,532 Zn + CUSO4 ^ Cu + ZnS04 x->x-> X mol 56 (I) Fe + CuS04 ^Cu + FeS04 CUSO4 het. (2) y^y-> y mol Nd'u chi CO Zn phan urng, Fe kh6ng phan ling thi: n^,, = n^u = 0,3 mol. Kh6'i luong kim loai sau phan urng giam so vai ban diJu: y; =>mKL =mp, +mc., - 29,8 - Am = 29,8-0,3 x (65-64)-29,5 gam < 30,4 gam . Neu Fe cung tham gia phan iing het thi kim loai chi c6 Cu . Trai d^ ra. Vay Zn het, Fe con du. Zn phan ung lam kh6'i luong kim loai sau phan ii:ng giam, Fe phan ling lam khd'i luong kim loai sau phan irng tang: He PT: ncuS04 =(x + y) = 0,3mol fx = 0,2 Jmz„ =0,2x65 = 13 gam Am = (-lx + 8y) = 0,6gam ^[y = 0,l ^[mp^ = 29,8-13 = 16,8 gam >%mp^ = —xl00% = 29,8 56,37% Chon A. //. V( Ik• . V; //// v',-;.' !0 - Phm S/Liru Vi du 2. Cho 1,68 gam Fe va 0,36 gam b6t Mg tac dung v6i 375ml dung djch CUSO4, khua'y nhe cho de'n khi dung djch mat mau xanh. Nhan tha'y khd'i lugnig kim loai thu duoc sau phan ling la 2,82 gam. N6ng d6 mol cua dung dich CUSO4 ia A.0,15M B.0,10M C.0,20M D. 0,25M Gidi =0.03mol; n^g =0,015mol Phan ling xay ra theo thir tu uu tien sau : • , ! Mg + CuS04->.MgS04 + Cu (I) • ' Fe + CUSO4 ^ FeS04 + Cu . (2) Sau phan irng dung dich mS't mau xanh chung to rang CUSO4 da phan ung het, CO the con du kim loai sau phan ling. Gia sij Fe phan iJng het khdng con du thl khoi luong Cu thu dugc se la : mc„ = (0,03 + 0,015) X 64 = 2,88 gam > 2,82 gam (theo gia thie't). Fe con dir sau phan ung. Goi X la so mol Fe phan ling. Am : do tang khoi luong kim loai. Ap dung phuong phap TGKL ta c6: Am =(64 - 24).0,015 + (64 - 56)x = 2,82 - (1,68+0,36) = 0,78gam 0,015 + 0,0225 0,375 >x = 0,0225mol => CM(CU.SO4) = -—rrri = 0,1 M Chon B Vidu 3. Cho m, gam Al vao 100ml dung djch gom Cu(N03)2 0,3M va AgNOj 0,3M. Sau khi cac phan ling xay ra hoan toan thi thu duoc m2 gam chat rdn X. Neu cho m2 gam X tac dung v6i luong du dung dich HCl thi thu duofc 0,336 lit khi (a dktc). Gia trj ciia m, va m2 Idn iuot la A.8,10va5,43. B. 1,08 va 5,43. C. 0,54 va 5,16. D. 1,08 va 5,16. Gidi ncu(N03)2 " "AgNO, = 0,1-0,3 = 0,03 mol; n^^ = 0,015 mol. Phan ling khir AgNO^ va Cu(NO,)2bori Al theo thii tu uu tien, sau do chat rdn 2 tac dung v6i HCl tao khf H2=> Al duda khijf HCl: n^nu^, =-n^^ Xet toan qua trinh : - Chat khir: Al. - Chat oxi hoa : AgNO,, Cu(N03)2, HCl. Al + 3AgNO,-^3Ag + Al(N03)3 (1) 0 0,01-^0,03^ 0,03^ 0,01 2Al + 3Cu(NO,)2 ^2Al(N03)3+3Cu (2) 0,02 -> 0,03 -> 0,02 0,03 •-y iiHiii i-ii f uy Til luiuiig Tici 6HC1 + 2A1 2AICI3 + 3H2 0,03 <- 0,01 <- 0,01 <- 0,015 mol (3) => n^i = 0,01+ 0,02 + 0,01 =0,04mol =>m, =0,04x27 = l,08gam m2 = m^g + + niAKDu) = 0,03(108 + 64) +0,01 x 27 = |5,43 gam => Chon B. Vi du 4. (DHB 2009) Cho 2,24 gam b6t sdt vao 200ml dung dich chiia h6n hop gom AgNOj 0,1M va Cu(N03)2 0,5M. Sau khi cac phan ting xay ra hoan toan, thu duoc dung djch X va m gam chat rdn Y. Tinh gia trj cua m. A. 5,08 gam B. 4,08 gam C. 3,72 gam D. 6,24 gam Gidi npe =0,04 mol; nAgwo, = 0,02 mol; ^c^^no^,)^ =^,1 mol Su khijf xay ra theo thii tu uu tien: AgNO, tac dung het sau do Cu(N03)2 bi khii. Fe + 2AgN03 ^2Ag + Fe(N03)2 0,01<-0,02-). 0,02 (1) (2) ,01/::.:; 14 Fe + Cu(N03)2 ^Cu + Fe(N03)2 0,03 0,10 0,03^ 0,03-> 0,03 , j.| 0,00 0,07 Theo cac PTHH : Cu(NO02 con du 0,07 mol, Fe phan ling hd't. Chon B. m,.„ = 0,02 X 108 + 0,03x64 = 4,08 gam Vidu 5. Hoa tan 13,8 gam Na2C0, vao nude. Vira khuay vifa them tirng giot dung djch HCl IM cho t6i dii 180ml dung dich axit, thu duoc V lit khi. Viet phuong trinh phan ufng xay ra va tmh V (dktc). ' "K2CO3 = 0,1 mo! =^ HHCKcdndung) = 0,2mol > HHCKCIC^ra) = 0,18mol PTHH theo thii tu uu tien: HCl + Na2C0, -> NaCl + NaHCO, HCl + NaHCO, NaCl + CO2 + H2O = 0,18-0,10 = 0,08 mol m nco2 = n HCl ~"Na2C03 V = 0,08x22,4= 1,792 lit Vi du 6. Them tir tir dung djch NaOH 2,5M vao 400ml dung djch X chiia HQ IM va AlCl, 0,5M. Ti'nh the tich dung djch NaOH cin dung de thu duoc ket tiia 16n nhat. Gidi Phan ling theo thii tu uu tien: ; • NaOH + HCl ^ NaCl + H2O (1) ' Dp dS/l^>a hoc y on Iht y^o IP-Phaa 3/ MT 3NaOH + AlCl, Al(OH), + 3NaCl (2) NaOH +Al(OH),->Na|Al(OH)4l (3) H* + OH ->H:0 (1) Ar'* + 30H ->Al(OH), (2) "nci = 0,4x1 =0,4mol; 11^10, =0,4x0,5 = 0,2 mol De thu dirge lircnig ket tiia nhieu nhat: khoiig xiiy ra phau iriig (3). Tir(l) va (2) ta c6: nf^yQ^ = 3nA|Q^ + =3X0,2 + 0,4 = 1 mol Vay the ti'ch dung dich NaOH can dung la : |V = 400 ml Vidu 7. Dung dich D g6m cac chat NaAlO, 0.16 mol; Na,S04 0,56 mol; NaOH 0,66 mol. Can them bao nhieu ml dung dich HCl 2M vao dung djch D de: (a) Dugc khoi lirctng ket tua Idii nhat. (b) Dugc ket tiia ma sau khi nung den khoi lugng khong d6i. thu dugc chat rdn can nang 5,1 gam. Gidi (a) •NaCl + H,0 NaOH + HCl 0,66 0.66 NaAlO, + HCl + HpNaCl + AI(0H)3 ' X X X AI(OH), + 3HC1 ^ AlCl, + 3H2O y 3y y D6 khoi lugng ket tiia thu dugc loii nhA't thl .so mol HCl can them : "HCl ="Na()H +"N;.AI()2 = 0,66 + 0,16 = 0,82 (mol) V^^ci = ^ = 0,41 lit = |410ml (I) (2) (3) (b) n. 5,1 = 0,05 (mol) 'Ai20,-,y2 2Al(OH),->Al20,+ 3H3O (4) 0,1 0,05 THI: Khong xay ra pluin irng (3) : x = 0,05 x 2 = 0,1 mol ; y = 0 mol S6 mol HCl Clin dung: n^ci = 0,66 + 0,1 = 0,76 mol ' V-i = — = 0,38 lit = 1380ml TH2: Co phiin irng (3): x= 0,16mol ;y =0,16-0,1 =0,06 mol So mol HCl can dung: = 0,66 + 0,16 + 3.0,06 = 1 mol ' ' ^HCi =-=0.5 lit = I 500 mil 2. Phiiong phap bao toan khoi luong. Dinh ludt: Trong phdn ung hod hoc long khoi lumg cdc sdn phdin bdng long khoi luang cdc chat tham gia phdn dug. A + B^C + D , m^ + mp = m^ + m^j (A, B : vira dii hoac con du) m^, la khoi lugng ciia A, B tham gia phan ung m^, m^: la khoi lugng ciia C, D tao thanh ' Apdung: - Phan ung c6 n chat ma biet dirge khd'i lugng cua (n - 1) chat => khoi lugng chat con lai. - Trong cac bai toan Xiiy ra nhieu phan I'rng, c6 the khong ciin viet day dii cac phuong trinh phan itng, chi can lap so do phan irng de thay moi quan he ti le mol giOa cac chat can xac djnh va nh&ng chat ma de cho. Sau do lip dung dinh luat de tim ket qua. - Khi CO can dung dich thi kh6'i lugng mu6'i thu dugc bang tong khoi lugng cac cation kim loai va anion goc axit. - Tfnh khoi lugng dung djch sau phan ihig: '"(dd s;ui pif) - "^(dd IriAK- pit)"'' ''"(clut l:m) '^(chiVl ki lliia) '^(chal bay hai) Vi du I. Cho mot luong khf clo du tac dung vdri 9,2 gam kim loai sinh ra 23,4 gam muoi kim loai hod trj I. Hay xac djnh kim loai hoa trj I va muoi kim loai do. Gidi Dat M la ki' hieu hoa hoc ciia kim loai hoa trj I. ' i PTHH : 2M + Clj -> 2MCI , 23,4-9,2 DLBTKL:9,2 + mci, =23,4g^ n^ =2nc,2 =2x. 9,2 71 = 0,4 mol =>M = 0,4 • = 23(Na)=^Mu6i: NaCl Vi du 2. Cho 7,8 gam h6n hgp kim loai Al va Mg tac diing vdi HCl thu dugc 8,96 li't H, (6 dktc). Hoi khi c6 can dung dich thu dugc bao nhieu gam muoi khan. Gidi =(8,96:22,4) = 0.4 mol PTHH: Mg + 2HC1 MgCl, + H:t (I) 2AI + 6HCI 2A1CI, + 3H,T (2) "HCI =2.nH2 =0.8mol So mol goc Ch n^._ =0,8mol => m^i- = 0,8X 35,5 = 28,4gam cr Vay khoi lugng mu6'i khan thu dugc la: m, ^^- = 7,8 + 28,4 = 36,2gam Vi du 3. Hoa tan hoan toan 3,22 gam h6n hop X g6m Fe, Mg va Zn bang m6t luong vCra du dung dich H2SO4 loang, thu diroc 1,344 1ft hidro (0 dktc) va dung djch ' chiia m gam mu6'i. Tinh m? Gidi nH2S04 =(1-344:22,4) = 0,06 mol PTHH: M + H2SO4MSO4 + H2 DLBTKL: m^^g, = nix + "1H2S04 -"IRZ =3,22 + 98 X 0,06 - 2 X 0,06 = |8.98 gam Vidu 4. Hoa tan 10 gam h6n hop 2 mudi cacbonat cua cac kim loai hoa trj 11 va III bang dung djch HCI du thu dugc dung djch A va 0,672 lit khf (dktc). Hoi c6 can dung dich A thu dugc bao nhieu gam muoi khan? Gidi n^„^ = (0,672:22,4) = 0,03mol Goi 2 kim loai hoa tri II va III Ian lugt la X va Y ta c6 cac PTHH: XCO, + 2HC1 -> XCl, + CO, + H,0 (1) , Y,(C03), + 6HC1 2YC1., + 3CO, + SH.O (2) ' ' Tir(l) va(2) nH20=nco2 =^'^3 mol; = 2.0,03 = 0,06 mol ^ "iHCi(ph:min,g) =0,06x36,5 = 2,19gam Ggi X la khoi lirgng mudi khan: XCK va YCl, ! Theo djnh luSt bao toan khdi lugng ta c6: , 10 + 2,19 = x + 44.0,03 + 18.0,03 x = 10,33gam Vidu 5. Cho 4,96 gam hdn hgp rdn gdm Ca va CaC, tac dung vdi nude dir thu dugc 2,24 1ft hdn hgp khf A (dktc). D3n A qua dng dung bdt Ni nung ndng mot thdi gian thu dugc hdn hgp khf B. Tiep tuc dfin hdn hgp B qua dung dich nude brom du thl khdi lugng bhih dung dung dich brom tang m gam va cd 0,896 1ft hdn hgp khf D (dktc) thoat ra, ti khdi ciia D so vdi H2 la 4,5. Tfnh m. Gidi Cac PTHH : Ca + 2H2O -> Ca(0H)2 + Hj CaC2 + 2H2O ^ Ca(0H)2 + C2H2 HC^CH + H2 ''^y"^ )H2C = CH2 H-C = C-H + 2H2 )CH3-CH3 CH = CH + 2Br2 > ^v^fH - CHBr2 Ni,,» CH2 = CH2 + Br2 ' > BrCH; -CH.Br Tacd: n^ =^ = 0,1 mol; n^ = = 0,04 mol 22,4 ° 22,4 , Goi X va y liin lugt la sd mol Ca va CaC, Theo cac PTHH ta cd : sd mol H2 va C2H2 liin lugt la x va y. >HePr: '40x + 64y = 4,96 _ [x = 0,06 x+y =0,1 => =0,06x2 + 0,04x26 = 1,16 gam Theo DLBTKL: m^ = mg = l,16gam Va: m = mB-m^ = l,16-0,04x2x4,5 = [oj8jam y = 0,04 Vi du 6. (DHA 2009) Xa phdng hda hoan toan 1,99 gam hdn hgp hai este bflng dung djch NaOH thu dugc 2,05 gam mudi ciia mot axit cacboxylic va 0,94 gam hdn hgp hai ancol la ddng dAng ke tiep nhau. Cdng thuc ciia hai este dd la A. HCOOCH, va HCOOC.H,. B. C,H.,COOCH., va C2H5COOC2H5. C. CHjCOOC^H, va CH^cboCH^. D. CH,COOCH, va CH.COOC^H,. Gidi DLBTKL : m^^oH = "imuoi + "lancoi - niesie - 2,05 + 0,94 -1,99 = 1 gam =^nN.oH= (1:40) = 0,025 mol Do tao mdt mudi va hdn hgp 2 ancol nen 2 este cd cung gdc axit vdi dang RCOOR': RCOO R' + NaOH -> RCOONa + R^OH Mnax^N. = ^ = 82 M, = 15 (-CH,) ^ROH = Trbr = 37,6 => 2 ancol la CH,OH va C^H^OH 0,94 0,025 2este: CH,COOCH, va CH.COOC^H, => Chgn D. Vidu 7. Cho 200 gam mot loai chii't beo cd chi sd axit la 7 tac dung vdi mdt lugng NaOH vua du thu dugc 207,55 gam hdn hgp mudi khan. Tfnh khdi lugng NaOH da tham gia pluin ung. Gidi Chi so axit la .\<'>'niii KOH de tiling hoa axit heo co tron}> /.i; chat heo. Sd mol KOH trung hda axit beo : l^!:!^—-^T^ _ Q Q25,,^Q| 56 PTHH: RCOOH + KOH ^ RCOOK + H.O (I) C,H,(0C0R)3 + 3NaOH ^ C3Hs(OH)3 + 3R'COONa (2) Ta cd : nM.,oH drujig hoa axil beo) = "KOH (Irung hoa axit beo) = "HitJ ^ ^'^^^ Goi X la sd mol triglixerit cua chii't beo=> nNaOH(.huy phA,> trigiixci.) = 3x "NaOH(ph;iii linj.) = iNaOHCIhiiy phiii Iriglixeril) + "Na()H(lrunp hoa axil boo) ~ (^X +0,25) DLBTKL : mc,,,, + mN,OH(phan m.g) = ^ 61 + "Iplixerol + mH20 200 + (3x + 0,025) X 40 = 207,55 + 92x + 18 x 0,025 => x = 0,25 mol niNaOH = (3 X 0,25 + 0,025) x 40 = |31gam bg de H6a kx; 9 on Ihi viio lO- Pham S/ltSu 3. Phaong phap tang, giam khoi luong Nguyen tdc: So siinh khoi lugiig cua chii't can xac dinh v6i luoiig san pham cua no ma gia thiet cho biet, de tir khoi lirong tang hay giam nay, ket hop \d\n he ti le mol giiJa 2 chat de tlm ra iiroiig cha't can xac djnh (c6 the la so nhom chuc, so mol, ). Pham vi su dung: Doi v6i cac bai toan phiin irng xay ra thu6c phan ling phfin huy, phan u'ng giua kim loai manh, khong tan trong nirdc diiy kim loai yeu ra khoi dung djch muoi phan ung, phan ling trung hoa axit cho biet krgng muoi tao thanh, Dac biet klii chira biet ro phan u'ng xay ra la hoan toan hay khong thl phirong phap nay to ra ra't hieu qua. Vi du I. Nhung mot thanh sdt naiig 8 gam vao 500ml dung djch CUSO4 2M. Sau m6t thofi gian UYy la sdt ra can hi thay nang 8,8 gam. Xem the tfch diuig djch khong thay doi thi nong do mol/li't cua CUSO4 trong dung djch sau phan ii"ng la bao nhieu? Gidi "cu,S()4 =0,5.2 = 1,0 mol Fe + CUSO4 ^ FeS04 + Cu x mol X mol PTHH : TGKL=> Am = (64 - 56)x = 8,8 - 8,0 ^ x = — = 0,1 mol 8 I 111- 0,9 • "cu.so4(ci.r) = (1,0-0,1) = 0,9mol Cc„.s(),(du, - ^ - [l.HM Vi du 2. Hoa tan 20 gam h6n hop hai muoi cacbonat kim loai hoa tri 1 va 2 bang dung djch HCl du thu duoc dung djch X va 4,48 lit khi (odktc). Ti'nh khoi luong muoi khan thu dirge khi c6 can dung djch X. Gidi Goi kim loai hoa trj 1 va 2 Ian lirgt la A va B. "a)2 =(4,48:22,4) = 0,2 mol Cac PTHH : A.CO, + 2HC1 ^ 2ACI + CO,T + H.O (I) BCO, + 2HCI BCK + CO,! + H^O (2) Theo (I) va (2) ta nhiui thay cur I mol CO, bay ra tuc la c6 1 mol muoi cacbonat chuyd'n thanh muoi clorua va khoi luong tang them 11 gam (goc CO^" la 60 gam cluiyen thanh goc 2C1 c6 khoi luong 71 gam). Vay CO 0,2 mol khi bay ra thi khoi krgng muoi tang la: 0,2.1 1 = 2,2 gam Vay tdng khoi lugng muoi clorua khan thu dirge la: m (muoi ;i khan) =20+ 2,2= 22,2 gam ci/ imnmvuwnKhang vie/. Vidu 3. Nhung mot thanh kim loai M hoa trj II vao 0,5 li't dung djch CUSO4 0,2M. Sau mot thcfi gian phan itng, khoi lugng thanh M tang len 0,40 gam trong khi nong d6 CUSO4 con lai la 0,1M. (a) Xac djnh kim loai M. (b) Lay m gam kim loai M cho vao 1 li't dung djch chira AgNO, va Cu(NO,)2, nong do m6i muoi la 0,1M. Sau phan urng ta thu dugc eha't rdn A c6 khoi lugng 15,28 gam va dung dich B. Tinh m gam? Gidi '•'}^''• (a) ncuso4(b<i) =0,5.0,2 = 0,1 mol; ncuso4(d.) =0,5.0,1 =0,05mol •> > • : ncu.s()4(|«) =0,10-0,05 = 0,05 mol PTHH: M + CUSO4 -> MSO4 + Cu Am = 0,05.(64-M) = 0,4 = 56 =>M la Fe (b) Ta chi biet so mol eija AgNO, va so mol cua Cu(NO,),. - Nhung khong biet so mol ciia Fe. > ' Day HDHH : PTHH theo tM tir iru tidn : Fe^* Cu^* Ag^^ rFe + 2AgNO, ^Fe(NO,)2+2Ag Fe Cu Ag |Fe + Cu(N03)2-^Fe(N0,)2+Cu - Neu chi c6 AgNO, phan irng: Fe thieu, Cu(N03)2 khong phan vtng, AgNO, vira du (hoac con du). => m,,.„) =mAj. =0,1.108 = 10,8g< 15,28 gam - Neu Cu(NO,),cung phan ung het, Fe vua dij hoae eon du. ' „ => m(rnn)="lAp+mCu+mFc(d,r) = 10,8 + 0,1.64 + mp,,,,, = 17,2 + mp,,,,, > 15,28 V;)y Fe phan I'rng het, CUSO4 da tham gia phan ting nhung eon dir. Goi X la so mol CUSO4 tham gia phan urng. m rAn) =^'"Ai. + 'Ticu = 10,8 + X.64 = 15,28 => X = 0,07 mol 1 "K- = "OK, r) + 2 "Ag = 0,07 + 0,05 = 0,12mol => m = 0,12.56 = |6,72 gam Vi du 4. Nhung mot thanh sdt va mot thanh kem vao cung mot coc chi'ra 500ml dung djch CUSO4. Sau mot thtJi gian lay hai thanh kim loai ra khoi coc thi m6i thanh c6 them Cu bam vao, khoi lugng dung djch trong coc bj giam mat 0,22 gam. Trong dung djch sau phan ung, nong do mol ciia ZnS04 gap 2,5 liin n6ng do mol cua FeSOj. Them dung dich NaOH du vao c6c, Igc lay ket tua roi iiung ngoai khong khi den khoi lugng khong d6i, thu dugc 14,5 gam chdt rdn. So gam Cu bam tren m6i thanh kim loai va n6ng do mol cua dung djch CUSO4 ban dau la bao nhieu? • Gidi ••••'^ PTHH: Fe + CUSO4-> FCSO4 + Cu (1) Zn + CUSO4 -> ZnS04 + Cu (2) Goi a la so mol cua FeS04 Trong cuiig mot dung djch nen tl le v6 nong do mol ciia cac chat trong dung dich cung chmh la ti le ve so mol. Theo bai ra: CM/z„s(i4 = 2.5 x CM/CUS04 => nznS04 = 2,5 x n^^^^^ . Khoi luong thanh sat tang : (64 - 56)a = 8a gam Khoi luong thanh kem giam: (65 - 64)2,5a = 2,5a gam Khoi luong ciia hai thanh kim loai tang: 8a - 2,5a = 5,5a gam Mii thirc te bai cho la : 0,22g 5,5a = 0,22 ^ a = 0,04 mol Vay khoi luong Cu bam tren thanh sdt la: 64.0,04 = 2,56 gam va khoi luong Cu bam tren thanh kem la: 64 x 2,5 x 0,04 = 6,4 gam Dung djch .sau phan ling 1 va 2 c6 : FeS04, ZnS04 va CUSO4 (neu c6) Ta CO so do phan u'ng: 2FeSO, ->2Fe(OH)2 a -> +OT,I .0 ^Fe20, amol-> a—> 0,5a TIFCJO., = 160.0,5a = 160.0,5.0,04 = 3,2gam 11-3 CUSO4 bmol ->.Cu(OH)2 b • CuO b moio =^0b = 14,5-3,2 = ll,3g=>b = 0,14125 mol IncuS(34(b;m L) = a + 2,5a + b = 0,28125 mol 0,28125 -CUSO4 0,5 0,5625M Vi du 5. Nung 6,58 gam Cu(N0,)2 trong blnh kin khong chua khong khi, sau mot thori gian thu dugc 4,96 gam chat rfui va h6n hop khf X. Hap thu hoan toan X vao nude de' duoc 300ml dung djch Y. Dung djch Y c6 pH bang A. 2. B. 3. C.4." D. 1. Gidi tv, 2Cu(NO,)2 X mol -> ->2CuO + 4N02 t+O, t X -> 2x —> 0,5x 4NO2 + O2 + 2H2O ^ 4HNO3 2x 0,5x -> 2x mol Do giam khoi luong: Am = ni^(j^l +mo,t = 2x.46+ 0,5x.32 = 6,58-4,96 = 1,62 gam 1,62 HNO, 0,1M- 108 = 0,015 mol => n^^fgo^ =2x =0,03 mol =>CHMO, = 0.03 ,' J MS'"'™ ->H^ +NO3 0,1M pH = l Chon D. Cl/ mmMiyDVVnKhang K/c, Vidu 6. (f)HA 2011) Trung hoa 3,88 gam h6n hop X g6m hai axit cacboxylic no, don chufc, mach ho bang dung dich NaOH, c6 can toan bo dung djch sau phan iJng thu dugc 5,2 gam muoi khan. Neu d6't chay hoan toan 3,88 gam X thl the tich oxi (dktc) cdn dung la A. 4,48 lit. B. 3,36 lit. C. 2,24 lit. D. 1,12 lit. Gidi +23-1=22 - \ in = RCOOH + NaOH RCOONa + H,0 a mol a mol - n Axil = a = • Tang 22 gam 5,2-3,88 -^COONa 22 = 0,06 mol Axit no, don chiic, mach ho => 2 dong dang lien tie'p: C-H2-O2 - , - 3,88 - 7 MAXU = 14n + 32 = => n = - 0,06 3 ^;.^2u^2 + ^02 nC02 + nH20 Chon li. 4. Phuong phap gia In trung binh 3.^-2 X0,06x22,4= 3,36 lit 1. Khdi niein gid tri trung binh : Co cac so duong A, B, a, b > 0. Trong do A < B. Ta CO: A.a < B.a A.b < B.b (A.a + A.b) < (A.a + B.b) < (B.a + B.b) Chia tat cii cho (a+b), ta duoc: A < i^^-^ii^l^^ <B <; • (a + b) A = -—'• 5^ (A : la gia tri trung binh ciia A va B). (a+b) 2. Cac gid tri trung binh trong bai tap hoa hoc: Nguyen tu khoi trung binh : nguyen tir khoi ciia nguyen to c6 iihiSu d6ng vj la nguyen tu khoi trung binh ciia h6n hgp cac dong vj c6 tinh deii ti le phan tram so nguyen tu ciia m6i d6ng vj. Cong thuc : -_Xi.A| +X2.A2 +- + x„.A„ 100 (vdi:X|+X2+ x„ = 100) Neu CO 2 dong vj: -_x,.A, +(100-x,).A2 /\ -~ 100 Ta CO the thay the ti le phdn tram s6' nguyen tii (X|,X2,X3 )bang s6' nguyen Bp d: Hda hoc 9 on Ihi vAo lO - Pham S/ Ltfu tir (n|,n2,n, )cua m6i dong vj : — _ n|.A| + 112^2 + n^.A^ n,+112+113. Khoi Imng mol trung binh : 'i! ' J Khoi lugiig mol trung blnh ciia h6n hop la khoi luoiig ciia 1 mol h6n horp c6 tinh den % so mol ciia m6i chat trong h6n hop. - Xet h6n hop X gom 3 chat A, B, C: Chat A B C Phan tu khoi : MA MB Mc mol a b c % so mol : XA — aMi + bMn +cMp Mx = (a + b + c) 100 Mx : khoi luong mol trung blnh ciia h6n hop X. Neu h6n hop gom i cha't ta c6 cong thiic tong quat tinh Mx : -__^IiiiMi_Zx,Mi 100 Thucnig gap h6n hop gdm 2 chat: A va B. Taco: %A + %B = 100%B = (100-%A) - aM^+bM, X^M^+(100-X^)MB (a + b) 100 - Trong cac h6n hop khi: % the tich = % so mol. Cdc gid tri trung binh khdc: S6 nguyen tu trung binh (C, H , O , N), d6 kh6ng no trung binh (k ), so nhom chiJc trung binh, phan til khoi trung blnh (M ) (a.MA+b.MB) H6n hop X A:C^HyO,N, (amol) B:CpHqO,N,(bmol) M = C = H = 0 = N = (a + b) (xa + pb) (^ + b) (y^+qb) (a + b) (za + rb) (a + b) (ta + sb) (a + b) O/ nun MTV DVVn Khang Viet Vi du 1. Trong tir nhidn, nguyen to dong c6 2 d6ng vj la 29Cu va 29Cu. nguyen tijr khoi trung binh ciia d6iig la 63,54. Thanh phan philn tram tong so nguyen tir ciia d6ng vi 29Cu la A. 27%. B. 50%. C. 54%. Gidi 73%. Goi phan tram s6' nguyen tirciia dong vj 29Cu la x Phan tram so nguyen tir ciia dong vj 29Cu la : (100% - x) Chon A. T.c6:I=''^^^ =63,54^ X = 27% Vi du 2. Trong tu nhien nguyen to brom c6 2 dong vj trong do dong vj 33 Br chiem 54,5% ve so lugng. Biet nguyen tu khoi ciia brom la 79,91. Xac djnh s6' khoi ciia dong vj con lai. Gidi Goi s6' kh6'i ciia dong vj con lai la A. Phan tram so nguyen tii ciia dong vj c6 so khoi A la: 100% - 54,5% = 45,5% „ , - 54,5X79 +45.5XA ^ ^, Ta co: A = = 79,91 => A = 81 100 Vay dong vj con lai c6 so khoi |A = 81 Vi du 3. Cho 6,2 gam h6n hop 2 kim loai kiem thuoc 2 chu ky lien tiep trong bang tudn hoan hoa hoc phan ling vdri HiO du, thu dugc 2,24 lit khi (dktc) va dung djch A. Ti'nh thanh phdn % \i khoi lugng tirng kim loai trong h6n hgp ban ddu. Gidi = (2,24:22,4) = 0,1 mol =>n,,, =2nH2 =0,2 mol =>M(A,B) = ^ = 31g/mol Gia su: A < B => A < 31 < B ^ A = 23(Na);B = 39(K) 39 23 'Na 8" 1 %K=—. 100% = 62,9% 6,2 %Na = 100-62,9 = 37,1% ."K =nNa =0,1 mol Vidu 4. Cho 28,1 gam quang d6l6mit g6m MgCOj va BaCOj trong do %MgC03 la a% vao dung djch HCl du thu dugc V (lit) CO, (o dktc). Xac djnh V (lit). Gidi ,j PTHH: 2HCl + MC03->MCl2+C02 t+ HjO ' ' 0.143 = ^snco2 =n3,eo3+nMgC03 <^ = 0,3345mol =>0,143.22,4 = 3,2 lit < Vco2 ^0,3345.22,4 = 7,491it Bp dSH6a hoc 9 on Uii v6o lO- Phm Sy Lmi Vay: |3,21it <Vco, <7,49Iit Vi du 5. Hoa tan 115,3 gam h6n hap gom MgCO, va RCO, bang 500ml dung djch H2SO4 loang ta thu duoc dung djch A, chat rdn B va 4,48 lit CO. (dktc). C6 can dung djch A thi thu diroc 12 gam muoi khan. Mat khac dem nung chat rdn B tori khdi luong kh6ng doi thi thu duac 11,2 1ft CO. (dktc) va chat rdn B,. Tinh nong do mol/lit ciia dung dich H2SO4 loang da dung, khdi luong ciia B, B, va nguydn tir khdi ciia R. Biet trong hdn hap ddu sd mol ciia RCO, ga'p 2,5 Idn sd mol cua MgCO,. Gidi Cdng thiic trung binh ciia hdn hop: MCO3 . Tdng sd mol khi COj = ^'^^^^ ' ''^ = 0,7mol 22,4 M = 11^-60 = 104,71 1.24 + 2,5.R 0,7 3,5 Mudi khan khi cd can dung djch A la MgS04 nMgSO4=(12:120) = 0,lmol Chat rdn B gom mudi cacbonat con dir va BaS04 Theo phuang phap TGKL: = 104,71 =^R = 137 =^RlaBa V'« Co 0,2 mol gdc CO3' da chuyen thanh gdc SO 2- . '4 • => Khdi luong tang: 0,2.(96-60) = 7,2 gam Do CO 12g MgS04 tach ra ntn khdi luong chat rdn B la: me = 115,3 + 7,2-12= 110,5gam 0\^X rdn B, gom cac oxit CaO, BaO va BaS04 11 2 m„. =110,5-—.44 = 'Bl 22,4 88,5 gam Vidu 6. A la hdn hap gom hai chat ddng ddng lien tia'p cua ancol etylic. Cho 7 gam hdn hap A tac dung het vdi Na thu dugc 1,12 lit khi H, (dktc). Tinh % sd mol va % khdi lugng mdi chat trong hdn hap A. Gidi "• Goi 2 chat ddng ddng lien tiep A, B => MR = M^ + 14 CrPT cua A, B la C„H2n+,0H (a mol) va C„+,H2n+30H (b mol) CT trung binh cua hdn hop C-H2-^,0H (a + b) mol ^u^2^.P^ + Na -> C-H^-^ONa + -H, 0,1 mol 1,12 00 A = 0,05 mol (14n + 18)< 14n + 18 = — = 70< 14n + 32 2,71 <n<3,71=>n = 3 ^ 0,1 CTPT cua A, B la CjH^OH va C4H,OH . Ta c6 : (a + b) = 0,10 fa = 0,029 H6PT: • |60a + 74b7,0 [b = 0,071 Thanh phdn hdn hap : oh XJ/: %Vc3H70H=29% %Vc4H90H-71% %m %m C3H7OH C4H9OH 24,86% 75,14% Vi du 7. A va B la hai chat trong day ddng dang ciia etylen. Ca hai d^u la chat kh trong dieu kien thudng. 5 gam hdn hap A, B chid'm th^ ti'ch bang thi tich ciia 4,4 gam khi CO. trong cung ditu kien ve nhidt dd va ap sua't. Xac djnh % v^ thd' tich ciia mdi chd't trong hdn hc»p. Gidi Cong tht(c: CT ciia A, B la : C„\{.„ (a mol) va C^H,, (b mol). CT trung binh ciia hdn hop C-H2- (a + b) mol Gia sir n < m va A, B deu la chat khi trong dilu kien thudng =:>2<n< X <m<4 (1) 4,4 "(hh)- "CO2 M= 14x^ 44 = 0,1 =>(a + b) = 0,l mol -=50(g/iTiol)=^ x= —=3,57 0,1 ' 14 Tir(l)=>2<n< x=3,57<m<4 (2) Tir (2) va n, m d^u la cac sd nguydn, ta c6: • n nhan 2 gia tri la n = 2 va n = 3 CTPT ciia A la C2H4 hoac CjH^. • m nhan 1 gia tri la m = 4 CTPT ciia B la C4Hg vay CTPT ciia A, B la: C3H4vaC4H8 (THI) hoac : C,H, va C4H8 (TH2) Thanh phcin hdn hep: % ihi tich = % sd mol Khh)=(a + b) = 0,l Ja = 0,021 b = 0,079 THI: %C.H^ =121% TH2:- m(hh)=28a + 56b = 5 [n(hh)=(a + b) = 0.1 [•"(hh)=42a + 56b = 5' %C^Hg =179% %C^H^ =[43% %C^Hg =|57% a = 0,043 b = 0,057 Vi du 8. Ddt chay hoan toan 2,76 gam hdn hap X gdm C^H^COOH C^H^COOCH,, CH,OH thu duoc 2,688 lit C02(dktc) va 1,8 gam H,0. Mai khac, cho 2,76 gam X phan ung vto dii vdi 30ml dung dich NaOH IM, thi dugc 0,96 gam CH,OH. Cdng thiic ciia C^H COOH la A.C,H,COOH. , D. CILCOQH. C. C^HaCOOH. D. C^H^COOH. THU VIEN TIWH IINM THU.m 17 B6 dSOda hga^dalhivio lO-Phm3ylHtu Sod6:X + Y + Z + HNO3 XCNOj )3 + YCNOj )2 + ZNO3 + NO2 + NO Bao toan nguyen to nito : "HN03(PII) ="N/HN03 " "N/NO + "N/NO2 +"N/X(N03)3 "N/Y(N03)2 "N/ZNO3 Vlit(dktc) Jons ^ V - + 3x + 4x + 3x 2x mol 3x mol 63x1,25 22,4 ( V ' y = l,25x + 10x x63 y = l,25x 122,4 , VI du 3. Khir het m gam Fe304 bang CO thu diroc h6n hop A gom FeO va Fe. H6n hop A tan viTa du trong 0,3 lit dung dich H2SO4 IM giai phong 4,48 lit khf (dktc). Tinh gia tri cua m. Gidi Sodd: Fe304- Fe + H2S04H 0,2 <-0,2 Fe0 + H2S04 0,1<-0,1 CO FeO ->A Fe FeS04 + H2 <-0,2 FeS04 + H2O 0.3 mol H2SO4 1,2 mol H2 Fe 2+ "Fe2+ ""H2SO4 =0,3 mol "Fe304 =^nFe=0,l mol m =0,1x232= 23,2 gam Vi du 4. (DHB 2012) D6't chay hoan toan 20ml hoi hop chat hihi co X (chi g6m C, H, O) cdn vira du 110ml khi O, thu duoc 160ml hdn hop Y g6m khi va hoi. Din Y qua dung djch H2SO4 dac (dir), con lai 80ml khi Z. Biet cac the tich khi va hoi do o Cling didu kiSn. C6ng thiic phan tix ciia X la A. C4HA B. C4H,oO C.CjHgO D.C4H«0 Gidi H2SO4 dac ha'p thu hoi nu6c, khi con lai la CO, I lOmI O2 X:C^HyO,(20ml)- ->C02 (80ml)+ H2O (80ml) Ap dung dinh luat Avogadro va bao toan s6' nguydn tir, ta co: Oxi: 20z + 2xll0 = 2x80 + lx80=>z = l Cacbon: 20x = 1 x80=> x = 4 Hidro: 20y = 2x80^y = 8 Vay CTPT : C4H8O => Chon D. Vidu 5. (CDAB 2010) H6n hop Z g6m hai este X va Y tao boi cung m6t ancol va hai axit cacboxylic ke tiep nhau trong day dong dang (M^ < M^). Dot chay hoan toan m gam Z cdn dimg 6,16 lit khi 02(dktc), thu dugc 5,6 lit khi CO2 (dktc) va 4,5 gam H,0. C6ng thiic este X va gia tn ciia m tuong ling la . A. (HCOO),C2H, va 6,6. B. HCOOCH, va 6,7. C. CH3COOCH, va 6,7. D. HCOOC^Hj va 9,5. Gidi 'iiKf'iu - nco2 = 0,25 mol; nH20= 0,25 mol; no2 = 0,275 mol nco2 = "H20 =^ Z gom 2 este no, don chiic: ^.H^.p, S^d6 : C,.H2„02 -»^"'-^^-^-> >{nH^o=neo2 =0,25 mol Bao toan so nguyen tijf oxi: no(esie) + "0(02 p/u) = no(C02) "OCHJO) => no(es.c) + 2no2 (p/u) = 2nco2 + nH20=> "o(estc) =(2nco2 + "H2o)-2no2(p/u) =-!^^^^= (0,25 + O,125)-O,275 = 0,lmol 2 , • • \ ^ 0,25 => C = 0,1 = 2,5=>Cx < C = 2.5 < Cy ^ tX: HCOQCHJ , m = 0,1.(14n+32) = 0,1 .(14.2,5+32) = 6,7 gam Chon B. Vidu 6. H6n hop X g6m 2 chat hiru co A, B (chi co chua C, H, O, N trong phan tir), trong do ti le m,,: m^, = 80 : 21. Thanh phdn % khoi luong cua nito trong h6n hop X la 10,966%. Dot chay hoan toan 3,83 gam h6n hop X can 3,192 lit O, (dktc), thu diroc cac san phdm chay gom khi va hoi la CO,, HjO va N,. Toan b6 san phdm chay trdn dem sue vao binh dung dung djch nuoc v6i trong du. Tinh khoi luong ket tiia thu duoc. A. 13 gam. B. 20 gam. C. 15 gam. D. 10 gam. I. i vi., Gidi m^ = 3,83X 10,97% = 0,42 gam mo = — x0,42 = 1,6 gam => mx =mc +m„ +mo +mf^ => "1^ = 3,83-(0,42 +1,6) = 1,81 gam Goi a la so mol C va b la s6' mol H => 12a + b = 1,81 gam (1) ' [CO2 (a mol) Xet phan ling chay ciia X : 3,83 gam X—"•''^^-'''""'"2 > H2O — mol Bao toan kh6'i luong oxi: 32a + 8b = 1,6 + 0,1425 x 32 = 6,16 gam (2) a = 0,13 Giaih6(l),(2): ' o m™, =0,I3x 100= 13 gam [b = 0,25 ^ 6. Phuong phap ty chon luong chdt ' Chon A. Co mot so bai toan ta cho dudi dang gia trj tong quat nhu a gam, V lit, n mol hoac cho ti le the tich hoac ti \t s6 mol cac chat Nhu vay ket qua giai bai toan khong phu thu6c vao luong chtft tong quat da [...]... Giai Cach 1: phuoiig phap duonig eheo Quy doi: CuS0 5H20=>%Cf.„c(> ^ V • So do dudng ciieo: = 160 100 % = 64% 250 'Cu.sO4.5H2O ~ ::ib vtlu i'.n k m CUSO4 8% 560 A = 79, 3 19 3 19- 79 0,3 19 % ( ^ { B r ) " 8 1 - 7 9 , 3 1 9 " 1,681 (A,-A) '3^Bt%S61ugng: %(^sBr) = 0,3 19 - x l 0 0 % = 15 ,95 % 1,681 + 0,3 19 Vi du 2 Mot h6n hgp gom O., O, b didu kien tieu chudn c6 ti kh6'i hcri vdri hidro la 18 Ti'nh thanh... '"Ai2(S04)3 = " I A I + nig^caxi = 10, 8 + 0,3x96 + 0,3x2x35,5= 60 ,9 Theo d^: m + 44,34 = 60 ,9 => |m = 16,56 gam| m -10, 8 16,56 -10, 8 ^ , „ , Bao toan khoi luong ta co: no2 = — — — = — = 0,18 mol NaOH MSu thu nao khi cho vao cac mau thir kia tao duoc 1 ket tiia la NaOH, MgCK, BaCU MSu thir nao khi cho vao cac mSu thu kia tao 1 chat khi thoat ra: NaHCO, MSu thu nao khi cho vao cac mSu thir kia tao duoc 1 chat khi... Vidu 2 Cho 0,448 1ft khf CO, (0 dktc) htfp thu het vao 100 ml dung djch chi'ra h6n hop NaOH 0,06M va Ba(OH), 0,12M, thu duoc m gam ket tiia Ti'nh gia tri ciia m A 3 ,94 0 B 1,182 C 2,364 D 1 ,97 0 Gidi 2' 0 ,10 = 0,04M=>Ch()n 1) => a = • 2,5 "ceo, CO3" => m = 0,1 X 197 = 19, 7 gam r:^ Chon A i - Thuc te trong dung djch c6: ion (3) na.(OH)2=:7(0.08 + 0,l2) = 0,10mol - = 0 , 4 - 0 , 3 = 0,1 mol Do lofp 9 chua... 0,09mol M A = 3 2 ; M g = 81oai 69 Phan I I : BA MirOfl DE THI VAO LQTP 10 0 CAC TINH SdGD&DT T/PDANANG KYTHITUY^NSINHVAOLCJPIOTRL;ONG THPT CHUVeN le QUtDON rinP NAM HOC: Di CHINH THirC 2012-2013 MON: HOA HOC Th()1 gian lam bai: 150 phut (khong keth&i gian giao de) Cau 1 (3,0 diem) (1) Viet cac phudig trinh hoa hoc ciJa phan ling xay ra trong qua trinh san xuat thiiy tinh tir cat trdng, s6da, da... T thu duoc 0, 896 lit CO, (dktc) (a) Xac djnh thanh phdn hdn hop Y, Z, T (b) Tinh ti khdi cua Y so vdi heli ^ Cau V {2,50 diem) (1) Neu hien tuong vii giai thi' ch: (a) Nhd vai giot iot vao mat mdi cdt cua cii khoai lang (b) Cho vai giot chanh vao cdc sua bd (c) Cho mot mdu cao su tu nhien vao xang (2) Len men giam 115ml dung djch rugu etylic 10 mdt thdi gian thu duoc dung djch A Neu cho toan bd dung... CO, (dktc) va 3,42 gam H,0 Mat khac, cho 3,74 gam X phan ung het vdi 40ml dung djch NaOH I M , thu duoc dung djch Y va 0,05 mol C.HyOH Cd can dung djch Y, thu duoc 2,86 gam cha't rdn khan (a) Xac djnh c6ng thiJc phan tir ciia ancol C^H^OH (b) Tinh % theo khdi luong cac chat trong X Cho C = 12, H = 1, N = 14, O = 16, Na = 23, Ba = 137, K = 39, Li = 9, A l = 27, Ag = 108 , He = 4, S = 32, CI = 35,5, Fe... Them m gam kali vao 300ml dung djch chura Ba(OH), 0,1M va NaOH 0 , i M thu duoc dung djch X Cho tir tu dung dich X vao 200ml dung dich AU(SO^), 0,1M thu duoc ket tua Y De thu duoc luofiig ket tiia Y Ion nha't thi gia tri ciia m la (ChoH= 1 ; 0 = 16;Na = 23; A l = 27;S = 32;K = 39; Ba= 137) A 1, 59 B 1,17 C 1 ,95 D 1,71 Gidi "ft,2+ +"1AI(OH)3(1) ~ ' " A 1 ( O H ) 3 ( 2 ) 233 X 0,3V+ 78(0,2V-(0,1-0,6V)) . '3^Bt<A2= 79) A = 79, 3 19 >%S61ugng: %(^sBr) = (A,-A) 0,3 19 3 19- 79 0,3 19 %(^{Br)"81- 79, 3 19& quot; 1,681 -xl00% = 15 ,95 % 1,681 + 0,3 19 Vi du 2. Mot h6n hgp gom O., O, b didu kien tieu . < 31 < B ^ A = 23(Na);B = 39( K) 39 23 'Na 8" 1 %K=—. 100 % = 62 ,9% 6,2 %Na = 100 -62 ,9 = 37,1% ."K =nNa =0,1 mol Vidu 4. Cho 28,1 gam quang d6l6mit g6m MgCOj. =0,06x36,5 = 2,19gam Ggi X la khoi lirgng mudi khan: XCK va YCl, ! Theo djnh luSt bao toan khdi lugng ta c6: , 10 + 2, 19 = x + 44.0,03 + 18.0,03 x = 10, 33gam Vidu 5. Cho 4 ,96 gam hdn