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Cach 1: Khi hoa tan vao dung dich H2SO4 , Fe(N03)2 se phan li thanh cac ion. Do vay, truoc het cac em can tinh so mol cac ion nhu sau: O • Phuong trinh ion rut gon: i •••J•..:> :r imrp Mol: a3 < 0,8 0,2 > 0,3 0,2 cO ; , Cac em luu y, Fe cung bi oxi hoa thanh Fe va giai phong khi NO : Mol; 0,6 0,8 0,2 0,6 0,2 _ > V = (0,2 + 0,2).22,4 = 8,96 lit > Dap an B. rr,fj{),p« y 4 Cach 2: Cac chat khu: Cu 2e >• Cu; Fe le > Fe. Chat oxi hoa: N + 3e > N O Bao toan electron: 2ncu + lnp2+ = Sno riNO = z =0,4 mol V = 0,4.22,4 = 8,96 lit Dap an B. Vi d 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4 0,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch I X. Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ket tua Ian nhat. . Gia trj nho nhat ciia V la A. 80. B.50. C.60. D. 40. 3Cu + 8H + 2NO • > 3Cu2 + N O + 4H2O 3Fe2 + 4H + NO ; > 3¥e + N O + 2H2O 2.a3+a6 , v r Ldi gidi:
ca'm nang 6n luyjn thi dgi hgc 18 chuyfin H(Sa hpc - NguySn Van H3i Lot gidi: Cach 1: Khi hoa tan vao dung dich H2SO4, Fe(N03)2 se phan li thanh cac ion. Do vay, truoc het cac em can tinh so mol cac ion nhu sau: O' • Phuong trinh ion rut gon: ' ^''^''/'i •••^'J^/•^ :> <>^:r'^ ^i^mrp 3Cu + 8H* + 2NO^ • > 3Cu2- + '^NO + 4H2O Mol: a3 <- 0,8 0,2 -> 0,3 0,2 cO/; , Cac em luu y, Fe^* cung bi oxi hoa thanh Fe^* va giai phong khi NO: 3Fe2* + 4H* + NO; > 3¥e^ + NO + 2H2O Mol; 0,6 ^ 0,8 ^ 0,2 0,6 0,2 ^ ^ _ -> V = (0,2 + 0,2).22,4 = 8,96 lit -> Dap an B. rr,fj{),p« y 4- Cach 2: Cac chat khu: Cu - 2e ->• Cu^^; Fe^^ - le -> Fe^*. Chat oxi hoa: N*-^ + 3e -> NO Bao toan electron: 2ncu + lnp^2+ = Sn^o 2.a3+a6 , vr ^ riNO= z =0,4 mol ^ V = 0,4.22,4 = 8,96 lit ^ Dap an B. Vi d^ 13: Cho 3,84 gam Cu vao 200ml dung djch gom NaNOs 0,2M va H2SO4 0,5M, tao thanh Vml khi NO (san pham khu duy nhat, 0 dktc) va dung djch -I X. Cho Vml dung dich NaOH 2M vao X de thu dugc lugng ke't tua Ian nhat. . Gia trj nho nhat ciia V la A. 80. B.50. C.60. D. 40. Ldi gidi: Nhan xet: Dung dich chiia muoi NaNOs va H2SO4 loang -> can giai theo phuong trinh ion. ;?> .v, t, •5 ncu =0,06 mol; n^+ = 2.0,2.0,5 = 0,20 mol; n^^, =0,2.0,2 = 0,04 mol. Phuong trinh ion thu gpn: 3Cu + 8H* + 2NO3- > 3Cu^*+ 2NO +H2O ;^Mol: a06 ai6 a04 MjoriX^:- Cu tan het -> X CO chiia: n^ 2+ = 0,06; n + du = 0,04. ' De thu dugc luang ke't tua Idn nhat thi: njs^aOH = n + + 2n 2+ / H Cu "NaOH = 0,04 + 2.0,06 = 0,16 mol y _> V = 0,08 lit = 80ml —> Dap an A. ; .^w Cty TNHH MTV DWH Khang Vigt 14: Hoa tan hoan toan 0,1 mol FeS2 trong 200ml dung dich HNO3 4M, san pham thu dugc gom dung dich X va mot chat khi thoat ra. Dung dich X CO the hoa tan toi da m gam Fe. Biet trong cac qua trinh tren, san pham khu duy nhat cua N*^ deu la NO. Gia tri ciia m la A. 5,6. ' B.7,0. C.8,4.* ' 'V', • Lai gidi: ry, •; D. 2,8 -> Fe(N03)3 + 5NO + 2H2SO4 + 2H2O 0,1 0,2 phan ling hoa hoc: FeS2 + 8HNO3 - Mol: 0,1 0,8 Dung dich X gom cac ion: Fe3^ = 0,1 mol; H* = 0,4 mol; NO ; = 0,3 mol va SO ]' = 0,2 mol. ^, ^, , ., , „ " WO '-* "'"^ * mi Cac phan ung hoa tan Fe: Fe + 4H* + NO; > Fe3* + NO + 2H2O QkHi Mol: 0,05 4-0,4 -> 0,05 ••idl,0» j„rtS«.^^iT 'ihtvU/ Fe + 2Fe3^ > 3Fe2- q% lOH dfegfror* ,;>fefii JfiSv) Mol: 0,075 <- 0,15 —> mpe = 0,125.56 = 7,0 gam -> Dap an B. Vi d^ 15: Iron 200ml dung djch X gom Ba(OH)2 0,1M va NaOH 0,3M voi 100ml dung dich Y gom Al2(S04)3 0,1M va H2SO4 0,1M, thu dugc a gam ke't tua. Gia tri cua a la A. 6,22. B. 4,66. C. 5,82. D. 5,24. Laigidi: * ^ TrongX: n„ 2+= 0,02 mol; n,,_+ = 0,06 mol; n^^_ = 0,10 mol.«^'^-> OH" Trong Y: n 3+ = 0,02 mol; n^^j- = 0,04 mol; n^+ = 0,02 mol. Cac phuong trinh phan ung khi pha trpn: H+ + OH' > H2O 0,02 ^ 0,02 , AP* + 30H" > Al(OH)3i 0,02 0,06 -> 0,02 Al(OH)3 + OH" > AlO; + 2H2O 0,02 <- 0,02 ,Ba2^ + S04~ > BaS04>l 0,02 -> 0,02 ^ 0,02 Mol: 'ol: ± oh Vgy: a = 0,02.233 = 4,66 gam -> Dap an B. C&m nang fln luyjn thi dgi hgc 18 chuy6n dg H6a hoc - MguySn Van H5i Vi dv 16: Hoa tan hoan toan 9,46 gam hon hgp gom Na, K va Ba vao nuoc, thu duQC dung dich X va 1,792 lit khi H2 (dktc). Dung dich Y gom HCl IM va 1^ H2SO4 0,5M. Trung hoa dung dich X boi dung dich Y, tong khoi lugng cae muol duoc tao ra la , i A. 14,72 gam. B. 16,14 gam. C. 19,98 gam. D. 17,14 gam. Lai giai: 1,792 = 0,08 mol. . '2 22,4 Cac phan ung hoa hoc: Na + H2O 1 K + H2O Na+ + OH" + -H2 2 K+ + OH- + -Hi H ( r{ mi'n> tV i i. Iv P J r.i I 1^x4 f 1 1, t fin J^j ti ) Ba + 2H2O > Ba^^ + 20H" + H2 "M ' 1 N/jflnxet: n^„. = 2nH, =0,16 mol. OH ^ Mat khac, nong do HCl gap doi H2SO4 -> trong cung mot the tich thi „,p.r, ft T< j\riv, *. , ^ Trong Y: • "H+ = "HCl + 2nH2S04 = 2a + 2.a = 4a mol n = 2a mol; n , =amol. cr so|" H2O Trung hoa X boi Y: H++ OH" • -> 4a = 0,16 -> a = 0,04 mol. Khoi lugng muoi thu dugc = 9,46 + m^^. + mg^a- V = 9,46 + a08.35,5 + 0,04.96 = 16,14 gam. ; —> Dap an B. Vi d\ 17: Cho hon hop gom Ba va Al (ti le mol 1:1) vao nuac (du), thu dugc dung djch X va 1,12 lit khi H2 (dktc). Dung dich Y gom HCl 0,5M va H2SO4 0,1M. Cho tu tu den het 100ml dung dich Y vao X, thu dugc m gam ket tiia. Giatricuamla ^ - a , i,*-^a'^.r,*••• , .WUt, •-i^; •: : A. 0,78 gam. B. 1,56 gam. C. 3,11 gam.' ' D. 5,44. ^.•vr., Laigidi: J-' ^ ; .X^CMU. ., "H2 = 0,05 mol. Cac phan ung hoa hoc: ^OJ) • * S , Ba + 2H2O > Ba2* + 20H" + H2 Mol: a a 2a a ^9 Cty TNHH MTV DVVH Khang Vi$t Al + OH- + H2O AIO2 + -H2 2 ;;i • Mol: a a a 1,5a Ta c6: nHj = 2,5a = 0,05 mol -> a = 0,02 mol. Trong X: n^^^_ = 0,02 mol; n^^2^ = 0,02 mol; n^,Q_ = 0,02 mol Trong 100ml Y: n^+= 0,07 mol; ngQ2-= O'Ol mol; n^|_ = 0,05 mol. Khi cho tu tu Y vao X: . < OH- + H+ - 0,02 -> 0,02 Mol: AIO2 + H+ + H2O > Al(OH)3 4' Mol: 0,02 -> 0,02 0,02 Al(OH)3 + 3H+ > Al3+ + 3H2O Mol: 0,01 <- 0,03 dnSrf/; Ba^+ + SO 4- > BaS04>l' Mol: 0,01 <- 0,01 -> 0,01 -> m= mA,(OH)3 + mBaS04 = 0-01-78+ a01.233 = 3,11 gam. : .m ->DapanC. , , , , V,,,' 9. PHl/ONG PHAP LIEN HE NGUYEN T6 - NH6M CHLTC a. Npi dung ' So mol cac nguyen tu c6 trong nhom chiic luon ti 1^ thu|in voi so' mol nhom chuc. , torn • • - r'h-Hn 'fd b. Cac tnrong hg(p thuang gap Phan ling hoa hpc Moi lien h| Cach tinh ROH > RONa + ^ H2 2 no=2nH2 RCOOH ^NaHCOg ^ RCOONa + CO2 + H2O ^ no=2ncooH no= 2nco2 R_NH2 ) R-NH3CI nN = riHCi Vi Dv MAU Vi dv 1 (A-09): Khi dot chay hoan toan m gam hon hgp X gom hai ancol no, don chuc, mach ho thu dugc V lit khi CO2 (6 dktc) va a gam H2O. Bieu thu-c lien h? giiia m, a va V la: dm nang On luygn thi dji hpc 18 chuy6n H6a hpc - IMguySn Van HSi V V A. m= a- —. . , B. m = 2a . 5,6 : . f;,: v'X,- 11,2 ' ; 'y } C m = 2a-^. ' D. m = a + -^. ' • ^^^^^ 22,4 ,< ,,,) 5,6 ^J,, • ^ n :;j.|iT Nhan xet: Khi do't chay cac ancol (no, don chuc, m^ch ho), ta luon c6: a V "ancol - nH20 - "CO2 nancol= — - 18 22,4 - ., , ,.^^5^ Do X chua cac ancol dan chiic no = noH = nx -> no 18 22,4 Bao toan kho'i lucmg: mx = mc + mH + mo , ^ V , a , a V , V oM ->m=12. +2.— + 16.( )->m = a . cfi n .r^j,/ 22,4 18 ^8 22,4^ 5,6 ^^^'^ -^"^^ -> Dap an A. Qs^* •* "••'''A <—• '^'Ht ' fCf-if:)^rA Vi du 2: Dot chay hoan toan m gam hon hop Y gom ba ancol don chuc, thuoc cung day dong ding, thu dugc 37,4 gam khi CO2 (dktc) va 27 gam H2O. Gia triciiamla iO,U liJ#' aoM A. 27,1. P'''rfm B. 28,6. C.23,6. ,3m + D. 37,1.»- m 37 4 i "COT^—^ = 0,85 mol-> nc = 0,85 mol. 44 27 '-'^ ' WtyfAiHi>HOXH-M v' nH20= — = 1.5 mol nH = 3,0 mol. ^^^^^ Cac em can thay rang, khi dot chay Y: nnjo > "cOa ^ chtia 3 ancol no -> nv = nH-o - nco, = 1'5 - 0,85 = 0,65 mol. , , ^ Do Y chua cac ancol don chuc -> no = noH = nx -> no = 0,65 mol. Bao toan khoi lugng: mv = mc + mH + mo = 0,85.12 + 3,0.1 + 0,65.16 = 23,6 gam. -> Dap an C. Vi dy 3: Do't chay hoan toan mot lugng hon hop X gom hai ancol (no, da chuc, mach ho, ciing so nhom -OH) can vua dii V lit khi O2, thu dugc 5,6 lit khi CO2 va 6,3 gam H2O (cac the tich khi do 0 dktc). Gia trj ciia V la a t A. 5,60.' B.3,92. C. 7,28. D. 1,12. Laigidi: ' r • 5,6 6 3 ' ''^''^.ymm-Mtifbw yi: ^C02 mol; nH20=0,35 mol. ;, . ^ Theo bai ra, X chua 2 ancol no -> nx = n^jo " "CO2 "^ol • Cty TNHH MTV DWH Khang Vigt So nguyen tu C trung binh = "'''"^ = 2,5 X chua mgt ancol da chuc c6 so nguyen tu C nho hon 2,5ancol do la C2H4(OH)2. • ' * po X chiia 2 ancol cung so'nhom -OH -> 2 ancol deu hai chuc: ^ • ' ^ no = noH = 2nx -> no = 0,2 mol. , . Bao toan nguyen to O: no(OH) + 2no2 = 2ncoj + nH20 ''''^'''^' ^ ^2.0,25 + 0,35-0,1 ^Q 225 mol ^ = 0,325.22,4 = 7,28 lit. Dap an C. Vi dv 4 (CD-12): Dot chay hoan toan hon hg-p X gom hai ancol (no, hai chuc, mach ho) can vua du Vi lit khi O2, thu dugc V2 lit khi CO2 va a mol H2O. Cac khi deu do 6 dieu kien tieu chuan. Bieu thuc lien he giiia cac gia tri Vi,V2, a la A. Vi =2V2 +ll,2a. B. Vi =V2 -22,4a. ; C. Vi = V2 + 22,4a. D. Vi = 2V2 - 11,2a. , / i , Lbigidi: OH ; V2 Theo bai ra, X chua 2 ancol no -> nx = nH20 ' ^C02 ~ ^ ' 2V2 Do X chua 2 ancol hai chuc -> no = noH = 2nx -> no = 2a - Bao toan nguyen to O: no (on) + 2 no2 = 2 nQQ^ + VXH^Q 2V2 ^ 2Vi _ 2V2 22,4 2^ 22,4 -> 2a - + ^=-^ = ^-^ + a Vi = 2V2 - 11,2a on 22,4 22,4 22,4 -> Dap an D. Nhan xet: Bai nay cac em can nho voi ancol no thi: nancoi = n^^Q " "CO2 ' dong thai bie't ap dung bao toan nguyen to oxi. Vi d^ 5 (B-12): Do't chay hoan toan m gam hon hgp X gom hai ancol, thu dugc 13,44 lit khi CO2 (dktc) va 15,3 gam H2O. Mat khac, cho m gam X tac dyng voi Na (du), thu dugc 4,48 lit khi H2 (dktc). Gia trj cua m la - A. 12,9. B. 15,3. C.12,3. D. 16,9. .•v^ . Lbigidi: • , • • -(i =c ~J ui r*3Y^;^:- ^^002 = 0/6 mol; nH20= 0,85 mol; nH2 = 0,2mol. , ^ ,a,Rt; q? i • Dva tren moi quan h| nguyen to' - nhom chuc thi voi ancol: no = noH no=noH=2nH2 = 0,4mol. Meit khac: nc= nco2 = 0,6 mol; nH = 2nH20= 1/7 mol. Ca'm nang 6n luy^n thi d<ii hpc 18 chuy6n 6i H6a hgc - Nguyin Van H^i Bao toan kho'i lugng: mx = mc + mn + mo = 0,6.12 + 2.0,85.1 + 0,4.16 = 15,3 gam. Dap an B. Vi du 6 (B-12): Cho hon hop X gom metanol, etylen glicol va glixerol. Do't chay hoan toan m gam X thu dugc 6,72 lit khi CO2 (dktc). CQng m gam X tren cho tac dung voi Na du thu dugc to'i da V lit khi H2 (dktc). Gia tri aia V la A. 3,36. „ B. 11,20. C.5,60 y.g; D. 6,72. • Lmgidi: • . nco2 = 0'3mol ^ nc=a3mol. 0'mti^Qj Nhan thay trong X cac ancol deu c6 so'cacbon = so'nhom chiicl "CllVjt * ijj ( nc = noH = 0,3 mol. ,r p Mat khac, khi cho X tac dung voi Na: noH= ^^H2 ~* '^H2 ~ O'^^ mol. s; -> V = 0,15.22,4 = 3,36 lit ^ Dap an A. Vi du 7: Hon hop X gom hai axit cacboxylic don chuc. Dot chay hoan toan 0,1 mol X can 0,24 mol O2, thu dugc CO2 va 0,2 mol H2O. Cong thuc hai axit la A. HCOOH va CH3COOH. i B. CH2=CHCOOH va C2H5COOH. io, rm • -f i C. CH3COOH va C2H5COOH. , ; D. CH3COOH va CH2=CHCOOH. * 3iwb fefl loarifi dG Nh|n thay X chua 2 axit cacboxylic don chiic -> chua 2 nguyen tu oxi ^ no = 2ncooH = 2nx —> no = 0,2 mol. Bao toan nguyen to O: no (COOH) + 2 no, = 2 nco, + "HOO ,, ^ ^ ^ ,;,fj fin qP;Uv.4™- ^ -> 0,2 + 2.0,24 = 2 nco2 + 0,2 ncoj = 0/24 mol. • rtljiW + So' nguyen tu H trung binh = —S2.= ^-^'^ = 4 _> Loai B va C (vi cac axit "X 0,1 , chua so'nguyen tu H > 4. + So' nguyen tu C trung binh = = -5^= 2,4 -> Loai A (vi cac axit chua 0,1 so nguyen tuC < 2). -> Dap an D. ^ '••m'r\fl•„,,•• Vi d\ 8 (B-11): Hon hgp X gom vinyl axetat, metyl axetat va etyl fomat. Dot chay hoan toan 3,08 gam X, thu dugc 2,16 gam H2O. Phan tram so mol ciia vinyl axetat trong X la A. 75%. B. 72,08%. C. 27,92%. D. 25%. Cty TNHH MTV DWH Khang Vijt' Lai gidi: „ = ^^= 0,12 mol nH = 0,24 mol. '^'^v, off. r "H2O 18 Hon hgp X: vinyl axetat (C4H6O2), metyl axetat (C3H6O2) va etyl fomat (C3H6O2). Cac em can thay rang cac chat trong X deu chua 6 nguyen tu H -> Khi dot chay 1 mol X, thu dugc 3 mol H2O -> nx = •^riH20 = 0,04 mol. Cac chat trong X deu chua 2 nguyen tu O no = 2nx = 0,08 mol. Bao toan kho'i lugng cac nguyen to' trong hon hgp X, ta c6: ,.;! > mx = mc + mH + mo -> mc = 3,08-0,24.1 -0,08.16 = 1,56gam. - ->• nco2 = '^c = 0,13 mol. ' ' > • Nhan thay, khi dot chay 1 mol moi chat metyl axetat va etyl fomat deu thu dugc nco2 = nH20' "eng voi vinyl axetat thi: n^^^ - n^^o = 1- v / Do v^y: nvinyi axetat = r\QQ^ - nH20 0,13 - 0,12 = 0,01. ' " -> % nvinyi axetat =^^.100% = 25% -¥ Dap an D. , ,bfniJ3»,,,A Vi d\ 9 (A-11): Hon hgp X gom axit axetic, axit fomic va axit oxalic. Khi cho m gam X tac dung voi NaHCOs (du) thi thu dugc 15,68 lit khi CO2 (dktc). Mat khac, dot chay hoan toan m gam X can 8,96 lit khi O2 (dktc), thu dugc 35,2 gam CO2 va v mol H2O. Gia tri cua v la A. 0,2. B.0,3. C.0,6. D.0,8. N/ion xet: Khi cho axit cacboxylic tac dung voi NaHCOa ta luon c6: "C02=r»C00H -> no(X) = 2ncooH = 2nco2 " «^ -> no(X) = 2. —'-— = 2.0,7= 1,4 mol. r,y' 22,4 "^^ • ^ Bao toan nguyen to O: no(X) + 2 no2 = 2 nco2 '^H20 ^14 + 2^ =2. ^ + nH,o ^ 2,2 = l,6 + y y = 0,6mol. ' '22,4 44 ''•••ft ^ dm nang 6n luygn thi d?! hgc 18 chuy6n dj H6a hpc - IMguygn Van Hai Lot gidi: Nhan xet: Khi cho axit cacboxylic tac dyng voi NaHCOa ta luon c6: j ,3 J nc02= "COOH ' ' • 1 344 no(x) = 2ncooH = 2 ticoj no(x) = 2. = 2.0,06= 0,12 mol. i'r'f, • 22,4 Cty TNHH MTV DWH Khang Vi^t Bao toan nguyen to O: no (x) + 2 = 2 ncoj + riHjO 0,12 + 2. 2,016 ^ 2 .4.84 + "Hjo ^ "Hjo = 0/08 mol. 22,4 44 mH20= 0,08.18 = 1,44 gam —> Dap an B. Vi dy 11: Cho m gam hon hg-p X gom hai ancol tac dung v6i Na (du), thu duoc ) 4,48 lit khi H2. Mat khac, dot chay hoan toan m gam X, thu dugc 13,44 lit khi CO2 va 16,2 gam H2O. Cac the tich do 6 dktc. Gia tri ciia m la A. 14,5. B. 15,4. C.12,2. D. 13,8. ( • • ' Ldfigiai: nH2 = 0,2 mol. , " ' • • '-aw N;'"^-" # I nco2 = 0,6->nc=0,6; DHJO^O'^ ->nH=l,8 . Y .,1 4 'ri *rr *M> „i, v Dya tren moi quan h^ nguyen to-nhom chiic thi voi ancol: no (X) = noH -> no(X)= noH" 2nH2 = 0,4 mol. ••!«?» m-nmi mod «;i,ib fob ,xtM Bao toan kho'i lugng: m = m(- + mH+ mo ^ m = 0,6.12 + 1,8.1 + 0,4.16 = 15,4 gam -> Dap an B. • M Vi dy 12: Hon hgp X gom 2 amino axit no (chi c6 nhom chiic -COOH va - NH2 trong phan tu), trong do ti 1^ mo : mN = 80:21. De tac dung vua du voi 7,66 gam X can 100 ml dung djch KOH IM. Cho 7,66 gam X tac dyng vtra du voi dung dich HCl, thu dugc dung dich Y. Co can Y thu dugc m gam muoi khan. Gia tri ciia m la A. 11,31. B. 10,58. C.9,85. D. 9,12. Lai gidi: ^ nKOH=0,lmol. I Nhan xet: '» Tu ti 1§ khoi lugng: A,U,A Q ne cjsGf ^- niN 21 De thay: ncooH = ^KOH = O,lmol. mo 80 no 80/16 10 y = = _ ^ ;_y^, y[ ^ 21/14 3 - . ivl^t khac, cac em luu y dya quan h? nguyen to - nhom chuc: no(X)=2ncoOHno(x)=0,2mol. <•:'•: nN(X)= 0,06 mol. HN = "NH2 = "HCl ^ "HCl = 0,03 mol . ^ ^ Bao toan khoi lugng: , • r , m = mx+ mHci = 7,66 + 0,06.36,5 = 9,85 gam. OrM^' ^" " , _> Dap an C. Vi 13: Dun nong m gam hSn hgp X gom cac este voi 350ml dung dich NaOH 2M, thu dugc dung dich Y chiia muol cua mgt axit cacboxylic don chuc va 13,9 gam hon hgp ancol Z. Cho Z tac dyng voi Na du, thu dugc 4,48 lit khi H2 (dktc). Co can Y, nung nong chat ran thu dugc voi CaO cho deh khi phan ling xay ra hoan toan, thu dugc 4,8 gam mgt chat khi. Gia tri cua m la A. 40,6. B.26,6. C. 30,7. D. 34,5. ' " rAv>r^ itT,'.' ".sn ' Loigidi: .\i .life n = 0,7 mol; n = 0,2 mol. H2 Nhan xet: Cac este tao thanh tu cung mgt axit cacboxylic don chuc. Khi ancol tac dung voi Na: -OH + Na -> -ONa+ -H2 iMv^jndaEv 0,2 fa; Mol: 0,4 <- noH=2nH2 =0,4 mol. M|it khac: n.coo-= "OH = 0,4 mol "RCOONa=0,4mol. -> nwaOH dir = 0,7 - 0,4 = 0,3 mol. Phan ling voi toi xut (NaOH he't, RCOONa con du): RCOONa + NaOH ) RH + Na2C03 Mol: 0,3 <- 0,3 0,3 4,8 M RH- 0,3 -= 16 RH la CH4. Bao toan khoi lugng: m + mwaOH = mRCooNa + mz -> m = 0,4.82 + 18,4 - 0,4.40 = 30,7 gam. —> Dap an C. Ca'm nang 6n luygn thi dgi hpc 18 chuy§n dg H6a hqc - Nguygn Van Hki Vi dy 14: Do't chay hoan toan m gam hon hgp X gom hai ancol, thu dugc 11,2 h't khi C02 (dktc) va 12,6 gam H2O. Mat khac, cho m gam X tac dung voi Na (du), thu dixgc 4,48 h't khi H2 (dktc). Gia tri cua m la m 0 - . ,„ A. 12,4. ; B. 15,2. C. 12,6. D. 13,8. . ^ Laigidi: rico2=^=0,5mol;nH2O=-^=0'7mol. V j ,j rto ^rt, ^ ff, Ta c6: mc = 0,5.12 = 6,0 gam; mH = 0,7.2 = 1,4 gam. ' '' <• H Khi X tac dung voi Na: n"^-:- '< mcj "* '•onn ni'<»'P!' wl- - i'-^ -OH + Na >-ONa+ t riD,b :,:n.b :xmt Mol: 0,4 0,2 .„. no = noH = 2nH2 -> no = 0,4 mol -> mo = 0,4.16 = 6,4 gam. Bao toan kho'i lugng: mx = m^ + mpi + mo = 6,0 + 1,4 + 6,4 = 13,8 ^ j,^ ^ ->Dap an D. ., . Cty TN TV DVVH Khang Vi§t CAC AXIT vo CO mm HIND 1 AXIT CLOHIDRIC: HCl a, Lithuyet + Tfnft dung voi kim loai, bazo, oxit baza, muoi. Vi du: Si n ? f Fe + 2HC1 > FeCh + Hat ''^^^ ''' CaC03+ 2HC1 > CaCh + COat + H2O '^^-'^^'^-^ "'^ + Tin?z fc^""- Tac dung voi cac chat oxi hoa manh: Mn02, KMn04, KCIO3, K2Cr207. Vidy: ,.,^„„ „ ;„,,.,,„/, ^ Mn02 + 4HC1 -^-^ MnCh + CI2 +2H2O " ry^.^ ^^ ^^^^ 2KMn04 + 16HC1 > 2KC1 + 2MnCl2 +5CI2 + 8H2O qfi(} K2Cr207 + 14HC1 2KC1 + 2CrCl3 + 3Cl2 +7H2O COH/ b. Vidumau , ^ •cr^^' rfi '•'i-i ' Vidul: Cho cac phan ling sau: .,;.^d Au. -^iyti^'^^ '-Xji::; (a) HCl +Mn02 > MnCh +CI2 +2H2O. , ,„,f^eH (b) 2HCl + Fe > FeCh +H2. , .v,t X: jfefb gnii^ XiVt. (c) 6HCl + 2A1 > 2AICI3 + 3H2. " ' -W! M i:i ,i/:}c:: i (d) 16HCl + 2KMna > 2KCl + 2MnCl2 + SCh + 8H2O. iM>f §' ?. - Cac phan ling trong do HCl the hi?n tinh oxi hoa la " s f 8 f i A.(b),(c). B.(a),(b). C. (b), (c), (d). D.(a),(d). • Laigidi: Nhan xet: Trong cac phan ung (b) va (c), so' oxi hoa cua hidro giam tu +1 (trong HCl) xuohgO (khi H2). , , -> Dap an A. Vi dv 2: Hoa tan hoan toan 7,6 gam hSn hgp hot FesOA va Cu trong 200ml dung djch HCl 1,2M (loang). Sau khi cac phan ung xay ra hoan toan, thu dugc dung dich X (khong chua axit du). Co can X thu du^c m gam muo'i khan. Gia tri cua m la A. 10,39. B. 14,20. C. 5,16. D. 11,10. Lffigidi: Gpi so mol: Fe304 = a; Cu = b. ' ^ • ' j Theo bai: 232a + 64b = 7,6. • ?-•<.}( • ••: '^^j^ X :,,;. i ^han xet: chi c6 Fe304 phan ung tryc tiep voi axit. f| ,* g .j : Trong X khong con axit du nen Fe304 phan ung vua dii voi HCl: Cac phan ung hoa hpc: Caim nang On luy$n Ihi dgi hgc 18 chuy6n dg H6a hgc - Nguygn VSn Hit i * • Fe304 + 8HC1 > FeCh + 2FeCh + 2H2O Mol: 0,03 <- 0,24 -» 0,03 0,06 -> a = 0,03 -> b = 0,01 mol. Cu + 2FeCl3 >• CuCh + 2FeCl2 J t arHQUii >J > TiX. Mol: 0,01 -> 0,02 ^ 0,01 0,02 m = 0,01.135 + 0,04.162,5+ 0,05.127 = 14,2 gam -> Dap an B. Vi 3: Day gom cac chat deu tac dung dugc voi dung dich HCl loang la A. KNO3, CaC03, Fe(OH)3. ^ B. FeS, BaS04, KOH. , .^.^^^ C. AgNCte, (NH4)2C03, CuS. D. NaHCOs, FeS, CuO. ' ,,, Lai gidi: Loai A vi KNO3 khong tac dung; loai B, C vi BaS04 va CuS khong tan trong dung dich HCl loang. ^ • Dap an D. Cac phuang trinh hoa hoc: ^''^ ^ ' iOnM>i NaHCOa + HCl > NaCl + H2O + CO2 '' FeS + 2HC1 > FeCh + H2S CuO + 2HC1 > CuCh + H2O Vi du 4: Hoa tan hoan toan 8,55 gam hon hop gom Na, K va Ba vao nuoc, thu duoc dung dich X va 1,792 lit khi H2 (dktc). Dung djch Y gom HCl va H2SO4, ti 1$ mol tuong ung la 2:1. Trung hoa dung djch X boi dung dich Y, tong khoi luong cac muoi dugc tao ra la A. 13,81 gam. B. 11,39 gam. C. 15,23 gam. D. 19,07 gam. 1,792 n^, = 0,08 mol. Cac phan ling hoa hgc: ' r, ^ 22,4 ' • : Na + H2O )• Na* + OH" + -Hi 2 8 K + H2O > + OH- + iH2 o n Ba + 2H2O > Ba2- + 20H- + H2 Nhan xet: n^^. = 2nyi^ =0,Id mo\. "H* ° "HCl + 2nH2S04 = 2a + 2.a = 4a mol n^,. = 2a mol; n^ i_ = a mol. Trung hoa X boi Y: H* + OH" > H2O -> 4a = 0,16 -> a = 0,04 mol. Khoi lugng muoi thu dugc = 8,55 + m + m 2- = 8,55 + 0,08.35,5 + 0,04.96 Cl so^ • * = 15,23 gam —> Dap an C. Trong Y: Cty TNHH MTV DVVH Khang Vijt Vi dV 5: Cho 2,13 gam hon hgp X gom Mg, Cu va Al a dang bgt tac dyng hoan toan voi O2 thu dugc hon hgp Y gom cac oxit c6 khoi lugng 3,33 gam. "The tich dung dich HCl 2M vua du de phan ung het voi Y la A. 150 ml. B.SOml. C. 75 ml. D. 90 ml. Lai gidi: 01 bai nay, cac em rJia't thiet phai ap dung bao toan khoi lugng de h'nh khoi lugng oxi tham gia phan ung: mkimiiHii + moxi = moxit —^ moxi = 3,33— 2,13 = 1,2 gam. 12 noo = — =0/0375 mol -> no = 0,075 mol -> n ^ =0,075 mol. ^-^2 32 ^ (0x11) Khi cho oxit bazo tac dung voi axit tao ra nuoc: . .,fm i. i 02- + 2H^ > H2O. , ^ ; ;, Suy ra: n^+= 0,15 mol-> n^ci^ 0,15 mol -> VHCI = 0,075 lit = 75ml Dap an C. Vi d\ 6: Hoa tan hoan toan 7,8 gam K vao 500ml HCl 0,2M, thu dugc khi H2 va dung dich X. Co can X thu dugc m gam chat ran khan. Gia trj ciia m la A. 14,90. B. 13,05. C. 7,45. D. 20,50. •<• , Lai gidi: iiiX = . HK = 0,2 mol; = 0,1 mol; nHci =0,1 mol. ,, ;j ,. K + HCl > KCl + lH2t «o.'HBOH n.: foH ,(> 2 'h-, Mol: 0,1 0,1 0,1 't^»^'- Luu y: K con du se tiep tuc phan ung voi nuoc: >. Z' K + H2O > KOH + iH2t 2 • OH£' - • - Mol: • 0,1 0,1 r<^r , -> m = mKci+ mKOH = 74,5.0,1 + 56.0,1 = 13,05 gam Dap an B. Vi dy 7: Hoa tan hoan toan 8,97 gam kim loai kiem M vao SOOnil dung djch HCl 0,2M, thu dugc khi hidro va dung djch X. Co c?n X thu dugc 14,73 chat ran khan Y. Kim loai kiem M la A.Rb. B. K. C.Na. D.Li. Lai gidi: Jl/ion Ob ^ Cac phan ung hoa hgc: •< •= , I^^/TU' »• .| , M + HCl > MCI + -H2 va M +H2O > MOH + -H2 2 2 N/ian xet: Bao toan khoi lugng my = mi4 + m^i_ + m^^^. m^„. =14,73-8,97-0,1.35,5 = 2,21 gamn^„. = 0,13 mol. OH UH 73 Ca^m nang 6n luy^n thi dgi hqc 18 chuy6n dg H6a hgc - Nguygn VSn Hai 8 97 riM=nHci +nQ^.= 0,23 -> M=-^=39(K)^ Dap an B. Vi dv 8: Cho m gam hon hop X gom Cu, Mg, Fe tac dung voi axit HCl du, thu dugc dung dich Y, 448ml khi (dktc) va 0,64 gam chat ran. Cho dung dich NaOH du vao Y, loc ket tiia va nung trong khong khi toi khoi luong khong doi, thu dugc 1,2 gam chat ran. Gia tri ciia m la , A. 0,80. B. 1,16. C. 1,44. D. 0,84. Loigidi: N^flnxef: mcu= 0,64 gamncu = 0,01. r •. Ggi so mol trong X: Fe = x; Mg = y -> x + y = nH2 = 0,02 mol. ' So do phan ung : Fe FeCh > Fe(OH)2 -^^^ lFe203 Mg -±»£U MgCh > Mg(OH)2 —^ MgO ^ mFe203 + mMgO = 1/2 ^ 80x + 40y = 1,2 -» x = y = 0,01 mol. m = 0,64 + 0,01.56 + 0,01.24 = 1,44 gam. -> Dap an C. ' Luu f. Khi nung ngoai khong khi, Fe(OH)2 chuyen thanh Fe(OH)3 va bi phan hiiy thanh Fe203. Vi du 9: Hoa tan hoan toan hon hgp X gom Fe va Mg bang mot lugng vira dii dung dich HCl 20%, thu dugc dung dich Y. Nong do cua FeCk trong Y la 15,76%. Nong do phan tram cua MgCh trong Y la A. 24,24%. B. 11,79%. C. 28,21%. D. 15,76%. , Loigidi: Fe + 2HC1 ^ FeCh + H2 Mg + 2HC1 > MgCh + H2 K.i i Xet voi 100 gam dung dich Y: -> m^^^ij = 1^,76 gam. mMgci2 + mH20 = 100 -15,76 ^ 84,24 gam. Nhan xet: Lugng H2O trong Y cung chinh la lugng H2O c6 trong dung dich HCl bandau. Mat khac, do nong do HCl bang 20% mH20 = 4mHci ^ mMgCl2 + 4mHci = 84,24. , Bao toan nguyen to'Cl: 31 52 2 1 2nFeci2 + 2nMgCl2 = "HCl "> + ~n^gClj = ^"^HCl mMgCl2 =11'79 ^ C%(MgCl2) = 11,79% -> Dap an B. Cty TNHH MTV DVVH Khang Vift ' do ^'^^ ^^"^ ^ NaC\a KCl voi HiS04 dac, du. jChi thoat ra cho hoa tan vao nuoc thu dugc dung dich Y. Cho bgt Zn du vao Y thu dugc 448ml khi (dktc). Khoi lugng NaCl trong X la ^ 0,585. B. 1,170. C. 1,755. D. 2,340. ' Loigidi: Goi so mol: NaCl = x; KCl = y. Cac phuong trinh phan ung: jsjaCl + H2SO4 —^ NaHS04 + HClt '''' ' jCCl + H2SO4 KHSO4 + HClt Zn + 2HC1 > ZnCh + Hat " >''•'' ^ Bao toan nguyen to: CI" > HCl > -Hi Ta c6: mx = 58,5x + 74,5y = 2,66 va nH2 = 0,5(x + y) = 0,02. _» x = 0,02; y = 0,02 -> mNaci = 0,02.58,5 = 1,17 gam Dap an B. Vi du 11: Hoa tan het m gam hon hgp Mg va MgCOa trong dung dich HCl, thu dugc 4,48 lit hon hgp khi X (dktc). Ti khoi cua X so voi H2 la 11,5. Gia tri cvia mla A. 13,2. B. 10,8. C. 19,2. D. 9,0. Loigidi: ^-^••«:MfBrtf 6'> Cac phuong trinh phan ung: ' •' i A Mg + 2HC1 )• MgCh + H2 Mol: a a •t^|,\V, MgCOs + 2HC1 > MgCh + CQ2 + H2O ''"^ Mol: b b • 'iliiA:: Theo bai: M = 11,5.2 = 23 va a + b = 0,2 mol. , ,^ , Ap dung cong thuc cua phuong phap duong cheo, ta c6: , "H2 44 -23 2- 23 =- — =- -> a = b = 0,1 mol 1 b 1 ^C02 m = m^g + mMgcos " " ^^'^ 8^ ^ ^ ^' 2. AXIT SUNFURIC: H2SO4 Li thuyet Dung d/clz H2SO4/oflng: Tinh axit manh '1,1 , ^ Fe + H2SO4 > FeS04 + H2t , j , FeS + H2SO4 > FeS04 + H2St , "*" Dwn^djc^ H2SO4 (fflc: Tinh oxi hoa manh Ngoai tinh axit manh, axit sunfuric dac con the hi^n tinh oxi hoa m^nh, tac diing dugc voi nhieu kim loai, hgp chat, : ff;-!;! » {'"-l C^m nang 6n luy$n thi dgi hpc 18 chuySn dg H6a hpc - NguySn VSn HJi Cu + 2H2S04(^flc) ——> CuS04 +SO2 + 2H2O :/> 2Fe + 6H2SO4 (dac) Fe2(S04)3 + 3SO2 + 6H2O ',' ' 2FeO + 4H2SO4 (dac) > Fe2(S04)3 + SO2 +4H2O 2Fe304 + IOH2SO4 (dac) > 3Fe2(S04)3 + SO2 + IOH2O Luu y: Cac kim loai Al, Fe, Cr khong tac dung vdi Ji2S04 dac, nguQi. Dieu che So do: Quang pirit FeS2 hoac S ) SO2 —^ SO3 —^ ' .• ,.H, »,I '• ,. H2S04.nS03 > (n+1) H2SO4 Cac phan ung: S + O2 > SO2 4FeS2 + IIO2 —^ 2Fe203+ 8SO2 •• • "ID :6i fiB'lii^H 2SO2 + O2 2S03 ,-xm:o:.fii , H2SO4 + nSOs > H2S04.nS03 (Oleum) ^ H2S04.nS03+ nH20 > (n+l)H2S04 b. Vi dv mau: Vi dy 1: Cho day cac chat sau: KBr, S, Si02, FeO, Cu va Fe203. So'chat trong day CO the bi oxi hoa boi dung djch axit H2SO4 (dac, nong) la A. 4. B.5. ' G.3.c|f: D.2. Lai gidi: Nhan xet: Axit H2SO4 dac, nong the hien tinh oxi hoa khi tac dung vai chat CO tinh khu (chiia nguyen to' dang 6 muc oxi hoa tha'p). 2KBr + 3H2SO4 2KHS04 + Br2 + SO2 + 2H2O S + 2H2SO4 —^ 3SO2 + 2H2O : ' 2FeO + 4H2SO4 —^ Fe2(S04)3 + SO2 + 4H2O Cu + 2H2SO4 CuS04 + SO2 + 2H2O ^ fH',:^:: Dap an A. Luu y: Khi tac dyng voi H2SO4 d^c, FeaOs the hi^n tinh baza: FeaOs + 3H2SO4 —^ Fe2(S04)3 + 3H2O Vi dy 2: Cho oxit ciia kim loai M (hoa tri 2) tac dung vira dii vai dung dich H2SO410% (loang), thu dug-c dung dich muoi c6 nong dp bang 14,45%. Kim loai Mia ' ' "" A. Mg. : B. Fe. , C. Cu. ' ^ D. Zn. Lcn gidi: Phan ling hoa hpc: MO + H2SO4 > MSO4 + H2O * Cty TNHH MTV DWH Khang ViSt Gia thie't dung djch ban dau chiia 1 mol H2SO4 (tiic chiia 98 gam H2SO4) 100 _> Khoi lupng dung djch H2SO4 = 98. — = 980 gam. ^^^^^•^ -=0,1445 ->M = 56 (Fe)-^ Dap an B. 980 + M + 16 Vi dy 3: Nung hon hop X gom a mol Fe va 0,015 mol Cu trong khong khi mpt thoi gian, thu dupe 6,32 gam chat ran Y. Hoa tan hoan toan Y bang dung dich H2SO4 dac nong (du), thu dupe 0,672 lit khi SO2 (san pham khu duy nha't 6 dktc). Gia tri ciia a la A. 0,04. B.0,05. C. 0,07. D. 0,06. Lm gtat: ^ 0 672 -bVT' ".'^ ncri^=—^ =0,03 mol. , , ,, , i. , Bao toan khoi lupng: mo2 = my " "^x = 6,32 - 56a - 0,96 = 5,36 - 56a. Nhan xet: ne'u dya theo phuong trinh phan ung se rat dai va kho giai. Cach 1:6 day, cac em can su dyng so do phan ung: .tysj.j, : Fe,Cu(l) > Y (2) > Fe^3Cu^M3) Xet su trao doi electron 6 cac giai doan: (1) -> (3): Fe -3e > Fe*-^ nenhuong = 3 np^ = 3a OsH »• ; Cu -2e > Cu*^ nenhiKmg=2ncu =0,03 "^s: (1) (2): O2 +4e > 2Ct^ ^ nenhan = 4no2 = ^^^^^^=0,67-7a (2) (3): S** +2e > SO2 nenh,^n=2nso2 =0,06. - Bao toan electron: 3a + 0,03 = 0,67 - 7a + 0,06 ^ a = 0,07 mol -> Dap an C. CachZ: Qui doi Y thanh: Fe (a mol); Cu (0,015 mol) va O (b mol). Bao toan khoi lupng: 56a + 16b = 6,32 - 0,015.64 = 5,36. Bao toan electron: 3a + 2.0,015 = 2b + 2.0,03 a = 0,07; b = 0,09 , ,, . Dap an C. Vi dy 4 (B-12): Cho cac chat rieng bi^t sau: FeS04, AgNOa, Na2S03, H2S, HI, Fe304, Fe203 tac dung voi dung dich H2SO4 dac, nong. So truang hpp xay ra phan ung oxi hoa - khu la - ? Rf 'A. 6. . B.3. C.4. , DJfjfi Lai gidi: Nhan xef: Phan ung oxi hoa-khu xay ra khi H2SO4 tac dung vai chat c6 tinh khu (chiia nguyen to dang 0 muc oxi hoa tha'p). 2FeS04 + 2H2SO4 Fe2(S04)3 + SO2 + 2H2O 77 im nang 6n luy^n thi dgi hoc 18 chuygn dg H6a hqc - Nguygn Van Hi\ HlS + 3H2SO4 ' > 4SO2 + 4H2O , 8HI + H2SO4 ^'-^ 4I2 + H2S + 4H2O f-SnO'w! ;oi*:^f ' 2Fe304 + IOH2SO4 3Fe2(S04)3 + SO2 + IOH2O , 1/ Dap an C. Luu y: ^eiOj, AgNOs xay ra phan ung trao doi, Na2SQj khong tr.c dung: Fe203 + 3H2SO4 Fe2(S04)3 + 3H2O 2AgN03 + H2SO4 Ag2S04i + 2HN03 i d\ 5: Cho 6,08 gam hon hop gom Li, Na va Ba vao nuoc (du), thu duoc dung dich X va 1,344 h't khi H2 (dktc). Dung dich Y gom HCl IM va H2SO4 0,5M. Trung hoa dung dich X boi dung dich Y, tong khoi lugng cac muoi dugctaorala , : A. 10,38 gam. B. 11,09 gam. C. 11,80 gam. D. 9,42 gam. Lbi gidi: 1 344 "H2 = = 0,06 mol. y ,,. Cac phan ung hoa hoc: Li + H2O > + OH- + -H2 2 ^, ^ ^ 1 • *• iiJ Na + H2O > Na^ + OH" + -H2 Ba + 2H2O > Ba2+ + 20H- + H2 Nhan xet: n^^. = 2nH2 = 0,12 mol. ' ''' Mat khac, nong do HCl gap 2 Ian H2SO4 trong cung mot the ti'ch thi "Ha = 2nH2S04 • Goi nHci = 2a -» nH2S04 = a- -> Trong Y: n + = 4amol;n o_ = amol;n_. = 2amol. . , ,, , H ,,.• ,,.',„ SO4 CI ii ion'ji Trung hoa X boi Y: H* + OH" > H2O ^y^:-yyh 4a = 0,12 a = 0,03 mol. \V Khoi luong muoi thu dug-c = 6,08 + m . + m 2- ' ' CI SO^ ; ,1- X i V. ',K . = 6,08 + a06.35,5 + 0,03.96 = 11,09 gam. ^ Dap an B. , 1 dvi 6: Hoa tan hoan toan 5,28 gam hon hgp bpt Fe304 va Cu trong 80ml dung dich H2SO4 IM (loang, vua dii). Sau khi cac phan ung xay ra hoan toan, thu duoc dung djch X. Co can X thu du(?c m gam muoi khan. Gia trj ciia m la A. 8,64. B.7,68. C. 15,68. D. 11,68. Cty TNHH IVITV DVVH Khang Vigt Lcngiai: Gpi so mol: Fe304 = a; Cu = b. Theo bai: 232a + 64b = 5,28. ;i; jV/ian xet: chi CO Fe304 phan ling true Hep voi axit. ji b uAi Trong X khong con axit du nen Fe304 phan ling vua du voi H2SO4: Cac phan ung hoa hpc: Fe304 + 4H2SO4 > FeS04 + Fe2(S04)3 + 4H2O • • ' : Mol: 0,02 <- 0,08 ^ 0,02 0,02 a = 0,02 ^ b = 0,01 mol. "' ~ ' ' CuS04 + 2FeS04 •' -> 0,01 0,02 ' ' ' m = 0,01.160 + 0,01.400 + 0,04.152 = 11,68 gam Dap an D. n.' Vi 7: Cho 3,68 gam hon hop gom Al, Mg va Zn tac dung voi mot luong vua du dung dich H2SO410%, thu dxxgc dung dich X va 2,24 lit khi H2 (dktc). Khoi lugng dung dich X la A. 101,68 gam. B. 88,20 gam. C. 101,48 gam. D. 97,80 gam. Lcngiai: Bao toan nguyen to hidro: nH2S04 = "H2 = 0,10 mol. ''^ " '' ' -> Khoi luong dung dich H2SO4 = 0,10.98. ^ = 98 gam. Cu + Fe2(S04)3 Mol: 0,01 -> 0,01 10 Bao toan khoi luong: 3,68 + 98 = mx + 0,10.2 ^ mx = 101,48 -> Dap an C. Luu y: Bai nay cac em di quen tru khoi luong khi H2 bay ra, va chi tinh: mx = 3,68 + 98 = 101,68 va se chgn nham dap an A! Vi 8: Hoa tan hoan toan 6,44 gam hon hgp X gom Al, Fe va Zn bang mot luong vua du dung dich H2SO4 loang, thu dugc 2,688 lit H2 (dktc) va dung dich chua m gam muoi. Gia tri cua m la ' " A. 19,04. B. 20,54. C. 17,96. D. 14,50. Lai gidi: Cach 1: Bao toan nguyen to H: nH2S04 = "H2 " 0,12 mol. Bao toan khoi luong: 6,44 + 0,12.98 = m + 0,12.2 -> m = 17,96 gam -> Dap an C. Cach 2: Ta c6: nM = nH2S04 = "HJ = 0,12 mol. Neu 1 mol kirn loai M > MSO4 thi khoi lugng tang 96 gam . . ; 0,12 tang0,12.96. , Vay m = 6,44 + ai2.96 = 17,96. I- ' -> Dap an C. • „., (^„j,,4i<);/:H:H , ).^- ' •'fm-E ' o [...]... 2: Nhan thay de tao dugc khi N2O thi kim loai phai kha m^nh (nhu Mg, Al, Zn) -> Loai D -> Dap an C Vi dv 2: Hoa tan hoan toan hon hgp gom 0,04 mol FeS2 va a mol CU2S vao axit HlSf03 (vua dii), thu dugc dung dich X (chi chua hai muoi sunfat) va V lit khi duy nhat NO2 (dktc) Gia tri ciia V la A 13,44 B 17,92 C 20,16 D 22,40 Cty TNHH MTV DWH Khang Vi$t im nang 6n luy^n thi d?i hpc 18 chuySn dS H6a hpc... trong h o n h g p ban d a u la A 5,40 gam B 2,70 gam C 1,35 gam D 4,05 gam Ca'm nang 6n luygn thi d^i hgc 18 chuyfin d6 H6a hgc - NguySn van Hai Cty TNHH MTV DVVH Khang Vi§t Bai 4: Dot nong 2,80 gam hon hgp X gom Cu, Zn va M g trong khi oxi d u , thu dugc 4,08 gam hon hgp oxit Y De hoa tan het Y can toi thieu V m l dung djch H2SO4IM Gia tri ciia V la A 80 r?S C I B 60 C 100 g D 120 ^e;/Bai 5: Hoa... do la A H2SO4.2SO3 B H2SO4.3SO3 C H2SO4.4SO3 D H2SO4.SO3 4 Bai 13: Cho m gam hon hgp X gom A l , Cu vao dung dich HCl (du), sau khi ket thiic phan ung sinh ra 3,36 lit khi (a dktc) Neu cho m gam hon hgp X tren vao mot lugng d u axit nitric (dac, ngugi), sau khi ket thiic phan ling sinh ra 6,72 lit khi NO2 (san pham khu duy nhat, a dktc) Gia tri cua m la A 11.5- 'yimn 12,3 C 10,5 D 15,6 Bai 14: Hoa tan... luy^n thi dgi hpc 18 chuy6n ai Hda hpc - Nguyjn Van Hai —> a = 0,2.78 = 15,6 gam - > Loai phuang an B va C Nhan thay: Khi cho 700ml dung dich HCl vao X, toan bp N a A 1 0 2 se chuyen het thanh ket tua Al(OH)3, va sau do bi hoa tan mpt phan trong HCl: > A l ( O H ) 3 i + NaCl NaAlOz + H C l + H2O Vi 5: Cho 500ml dung djch Ba(OH)2 0,1M vao V lit dung dich Al2(S04)3 0,1M; sau k h i cac phan ung ket thiic... l ; Al(OH)3 = 2a - (0,l-6a) = (8a-0,l) mol - > a = 0,015 mol - > V = 0,15 lit -> Dap an D dung dich Y Them t i i t u dung djch K O H I M vao Y, khi het 100ml thi bat > A l ( O H ) 3 i + 3NaCl dau xua't hi^n ket tiia; khi het 400ml hoac 800ml thi deu thu dupe lugng ket chat CO tinh chat ludng ti'nh la B.l Ba(OH)2 + 2Al(OH)3 > Ba(A102)2 + 4H2O Tnmng hop 1: Chi xay ra phan ung (1) —> Ba(OH)2 het (1) Vi... Y thu dugc ket tua Loai A Dap an B Ill Cty Jh: nang On luygn thi dgi hpc 18 chuy8n 66 H6a h(?c - NguySn Van HSi 'TV DWH Khang Vi$t L o t gidi: d v 2 (B-09): H o n h a p M gom hai muoi NaX va NaY (X, Y la hai nguyen to c6 nHCi=0'03-l 6,03 gam M vao nuac roi cho tac dung voi dung dich A g N O s (du), thu t^han xet: K h i nho tung giot H C l thi Na2C03 cQng nhan H C l mgt each " t u dugc 8,61 gam ket tua... = 1,08 gam D a p an B ^ ^ ^'^^ " * 115 C a m nang 6 n luy^n thi dai h o c 18 chuyen de H o a hgc - Nguy§n V a n HSi Cty T N H H M T V D V V H Khang Vigt V i dv 4 (B-09): Nhiing mot thanh Fe nang 100 gam vao 100ml dung dich gom Cu(N03)2 0,2M v a AgNOa 0,2M Sau mot thoi gian lay thanh k i m loai ra, rua sach lam idio can duoc 101,72 gam (gia thiet cac kim loai tao thanh deu bam het vao thanh sat) Kho'i... O 4 0,5M va NaNOs 0,2M Sau khi cac phan ung n grsxay ra hoan toan, thu dxxgc dung dich X va khi N O (san pham k h u duy nhat) C h o V m l dung dich N a O H I M vao dung dich X thi thu dugc lugng ke't tua Ion nhat G i a tri toi thieu cua V la A 360 B.240 C 400 D 120 H m Mol: C.4,8 Lai gidi: D 8,0 iiv/^u.y, *>'i ^ > , = 0,24 mol; n , , ^ _ = 0,04 mol NO3 Fe + 4 H * + NO3> Fe3*+ N O + 2H2O 0,04 0,16...C^m nang 6n luygn thi dai hgc 18 chuy6n H6a hpc - Nguygn Van H&\ V i du 9: H o a tan hoan toan 2,81 gam hon hop X gom Fe203, FeO, C u O can 50ml axit H2SO4IM (loang) Khir hoan toan 2,81 gam X bang khi C O (nung nong) thu... gam Cu phan ung voi 80ml dung dich chua H N O 3 I M va H2SO4 0,5M, sau phan ung thu dugc V m l khi NO (dktc) Gia tri ciia V la A 896 B 560 C 448 D 336 Q1 Cty TNHH MTV DVVH Khang Vigt dm nang On luygn thi dai hpc 18 chuySn dg H6a hpc - Nguygn van Hai Bai 21: Cho m gam hot Cu kim loai vao 200ml dung djch HNOa 2M, c6 khi NO thoat ra De hoa tan het hot Cu, can them tiep 100ml dung dich HCl 0,8M, dong thoi . -> no = 2a - Bao toan nguyen to O: no (on) + 2 no2 = 2 nQQ^ + VXH^Q 2V2 ^ 2Vi _ 2V2 22 ,4 2^ 22 ,4 -> 2a - + ^=-^ = ^-^ + a Vi = 2V2 - 11,2a on 22 ,4 22 ,4 22 ,4 ->. tha'p). 2KBr + 3H2SO4 2KHS04 + Br2 + SO2 + 2H2O S + 2H2SO4 —^ 3SO2 + 2H2O : ' 2FeO + 4H2SO4 —^ Fe2(S04)3 + SO2 + 4H2O Cu + 2H2SO4 CuS04 + SO2 + 2H2O ^ fH',:^::. Cu + 2H2S04(^flc) ——> CuS04 +SO2 + 2H2O :/> 2Fe + 6H2SO4 (dac) Fe2(S04)3 + 3SO2 + 6H2O ',' ' 2FeO + 4H2SO4 (dac) > Fe2(S04)3 + SO2 +4H2O 2Fe304