bo de hsg 11.11993
DESCRIPTION
Bo de on HSG Hoa 11TRANSCRIPT
LUYN HSG S 1
Cu 1:
1. Cho bit hin tng, vit phng trnh phn ng minh ho khi cho t t n d dung dch Na2CO3 vo dung dch AlCl3?
2. Tnh pH ca dung dch CH3COONa nng 0,1M bit CH3COOH c Ka=10-4,74 .
Cu 2: Cho t t tng git dung dch A cha 0,4 mol HCl vo dung dch B cha 0,3 mol Na2CO3 v 0,15 mol KHCO3 thu c dung dch D v V lit CO2 (ktc)
a. Tnh V?
b. Tnh khi lng kt ta thu c khi cho dung dch Ca(OH)2 d vo dung dch D?
Cu 3: C hai bnh in phn (1) v bnh in phn (2). Trong bnh 1 ng dd (1) l NaOH c th tch 38 ml nng CM = 0,5. Trong bnh 2 cha dung dch gm 2 mui Cu(NO3)2 v NaCl tng khi lng cht tan 258,2 gam. Mc ni tip bnh (1) v bnh (2). in phn cho n khi bnh (2) va c kh thot ra c hai in cc th dng li. Ly dung dch sau phn ng :
+ bnh (1): nh lng xc nh thy nng NaOH sau in phn l 0,95M.
+ bnh (2) em phn ng vi lng d bt Fe. Hi sau phn ng khi lng bt Fe b tan ra l m gam, v thot ra mt kh duy nht l NO c th tch x (lt) c o iu kin tiu chun. Tnh m v x?Cu 4:
1. Bng cc phn ng ha hc chng minh s c mt ca cc ion sau trong cng mt dung dch: Fe3+, NH4+, NO3- ?
2. Trnh by cch phn bit ba dd sau trong cc l ring bit mt nhn: NaCl, AlCl3, CaCl2 m ch dng 1 thuc th?
Cu 5: t chy hon ton mt cht hu c A (dng hi) ch thu c CO2 v H2O vi t l mol tng ng l 1:2. Tm CTPT ca A?
Cu 6: Cho m1 gam hn hp gm Mg, Al vo m2 gam dung dch HNO3 24%. Sau khi cc kim loi tan ht c 8,96 lt ( ktc) hn hp kh X gm NO, N2O, N2 bay ra ( ktc) v dung dch A. Thm mt lng va O2 vo X, sau phn ng thu c hn hp kh Y. Dn Y t t qua dung dch NaOH d c 4,48 lt hn hp kh Z i ra ( ktc). T khi ca Z i vi H2 bng 20. Nu cho dung dch NaOH vo A c lng kt ta ln nht thu c 62,2 gam kt ta. Tnh m1, m2. Bit lng HNO3 ly d 20% so vi lng cn thit.Cu 7: 1/ Cho a gam Fe hon tan trong dd HCl, sau p c cn c 3,1 gam cht rn. Nu cho a gam Fe v b gam Mg cng vo dd HCl nh trn th thu c 3,34 gam cht rn v 448 ml kh hiro ktc. Tm a, b?
2/ Vit p xy ra khi sc H2S vo dd FeCl3; dd CuCl2; dd H2SO4 c?
Cu 8: Hn hp kh A ktc gm hai olefin. t chy ht 7 th tch kh A cn 31 th tch oxi.
1/ Tm CTPT ca hai olefin bit rn olefin cha nhiu cacbon hn chim khong 40 50% th tch A?
2/ Tnh %KL mi olefin
3/ Trn 4,074 lt A vi V lt hiro ri un nng vi Ni. Hn hp kh sau p cho qua t t dd nc brom thy nc brom nht mu v khi lng bnh tng 2,8933 gam. Tnh th tch hiro dng v tnh khi lng phn t trung bnh ca hh ankan thu c. Bit cc kh o ktc, cc p xy ra hon ton v hiu sut p ca hai olefin nh nhau.
Cu 9: Ha tan ht hn hp FeS v FeCO3 bng dung dch HNO3 c hn hp kh A gm 2 kh X v Y c t khi so vi H2 bng 22,805.
1/ Tnh %m mi mui trong hn hp ban u?
2/ Lm lnh A c hn hp B gm X, Y, Z c t khi so vi H2 bng 30,61. Tinh %X b ime ha?
LUYN HSG S 2
Cu 1: Vit cng thc cu to ca: axit nitric; cation amoni; sunfur; hiropeoxit; amoni nitrit v nitr ioxit
Cu 2: Hp cht A c cc nguyn t C, H, O, N vi phn trm khi lng ca C = 20% v N = 46,67%. Phn t A ch cha hai nguyn t nit.
1/ Tm CTPT v CTCT ca A?
2/ Vit phn ng theo s :
3/ Bit rng:
a/ Cho Cu vo dung dch X cha NaNO3 v H2SO4 long thy c (E) bay ra v dung dch ha xanh.
b/ Cho Zn vo dung dch Y cha NaNO3 v NaOH long thy c (D) bay ra v dung dch
Vit p xy ra?
Cu 3: Chia 9,76 gam hn hp X gm Cu v oxit ca st lm hai phn bng nhau.
1/ Ha tan hon ton phn th nht vo dung dch HNO3 thu c dung dch A v 1,12 lt(ktc) hn hp kh B (NO v NO2) c t khi i vi hyr bng 19,8. C cn dung dch A thu c 14,78 gam hn hp mui khan. Xc nh cng thc phn t ca oxit ca st. Tnh khi lng mi cht trong hn hp ban u ?
2/ Cho phn hai vo 100ml dung dch HCl 0,8M. Sau khi phn ng kt thc thu c dung dch C v cht rn D.
a. Tnh nng mol/l dung dch C. Bit rng th tch dung dch khng i.
b. Ha tan hon ton D trong dung dch HNO3 thu c V lt (ktc) kh khng mu v ha nu trong khng kh. Tnh V. Cu 4: Cho hirocacbon X tc dng vi dung dch brom d c dn xut tetrabrom cha 75,8% brom (theo khi lng). Khi cng brom (1:1) thu c mt cp ng phn cis trans.
1/ Xc nh cng thc phn t, cng thc cu to v gi tn X.
2/ Vit phng trnh phn ng ca X vi:
a/ dd KMnO4 (trong mi trng H2SO4)
b/ dd AgNO3/ NH3 (hoc Ag2O/ NH3)
c/ Na kim loi (trong ete)
d/ H2O (xt l Hg2+/ H+)
d/ HBr theo t l mol 1:2.
Cu 5: Hn hp kh A gm hiro v mt olefin 90,2oc v 1atm c t l th tch 1:1. Cho hn hp A qua ng Ni nung nng thu dc hn hp kh B c t khi so vi H2 l 23,2.
1. Xc nh cng thc phn t c th c ca olefin trn.
2. T olefin ny, ngi ta cc th iu ch c isooctan dung lm cht t cho ng c qua hai phn ng. Xc nh cng thc cu to ng ca olefin. Vit cc phng trnh phn ng.
Cu 6: Ha tan ht m(g) hn hp gm FeS2 v Cu2S vo H2SO4 c nng thu c dd A v kh SO2. Hp th ht SO2 vo 1 lt dd KOH 1M thu c dd B. Cho lng dd A tc dng vi mt lng d dd NH3, ly kt ta nung n khi lng khng i c 3,2g cht rn. Cho dd NaOH d vo lng dd A. Ly kt ta nung n khi lng khng i sau thi H2 (d) i qua cht rn cn li sau khi phn ng hon ton thu c 1,62g hi H2O.
a/ Tnh m
b/ Tnh s gam cc mui c trong dung dch B
1
Cu 1: 1/ Hin tng: Si bt kh v kt ta trng, dng keo xut hin ri t t tan
Gii thch: - Mi trng ca dung dch Na2CO3 l mi trng ba z, mi trng ca dung dch AlCl3 l mi trng axit. Khi cho t t dung dch Na2CO3 vo mt mi trng axit, bt kh xut hin, kt ta xut hin nhng l Al(OH)3. 3Na2CO3 + 3H2O + 2AlCl3 2Al(OH)3 + 3CO2 + 6NaCl
Sau, tip tc cho vo th kt ta tan dn theo phn ng:
Na2CO3 + Al(OH)3 ( NaAlO2 + NaHCO3 + H2O
2/ pH = 8,87.
Cu 2: V=2,24 lit v m = 35gCu 3: 16,8 v 4,48Cu 4: 1/ Chng minh c Fe3+ v NH4+ bng dung dch NaOH
- C kta nu (c Fe3+ Fe3+ + 3OH- Fe(OH)3- C kh khai bay ra lm xanh giy qu tm m ( c NH4+ :
NH4+ + OH- NH3 + H2O
- Chng minh c NO3- bng Cu v H2SO4 c thng qua kh mu nu hoc kh khng mu ha nu trong khng kh: Cu + 4H+ + 2NO3- Cu2+ + 2NO2 + 2H2O
hoc3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O v 2NO + O2 2NO2
2/ Dng dd Na2CCO3 th:
+ AlCl3 c kt ta v kh bay ra, nu Na2CO3 d th kt ta tan
+ CaCl2 c kt ta
+ Khng c hin tng l NaClCu 5: CH4 v CH4O
Cu 6: m1 = 23,1 g v m2 = 913,5 gam
Cu 7: Gi s Fe p ht vi HCl => 3,1 gam cht rn ch c FeCl2 ng vi 0,0244 mol => s mol HCl > 0,0244 mol => khi p vi Mg v Fe th lng hiro > 0,0244 mol iu ny tri vi gi thit => Fe d khi p vi HCl => khi HCl p vi Mg v Fe th kim loi cng dT tnh c: Fe = 0,03 mol v Mg = 0,01 mol => a = 1,68 gam v b = 0,24 gam.
Cu 8: 1/ = 2,95 => phi c C2H4. Bin lun suy ra olefin cn li l C4H8.
2/ C2H4 = 35,5%; C4H8 = 64,5%.3/ = 43,33 vC; V = 3,136 lt.
Cu 9: 1/%FeS=20,87% 2/63,33% 2Cu 2: A l ureCu 3:1/ Fe3O4(4,64 g) v Cu (5,12 g) 2/ FeCl2 = 0,3M v CuCl2 = 0,1M
3/ V= 0,448 lt.
Cu 4: : phenyl axetilen (Khi phn ng vi Na th Na y H linh ng lin kt ba ra thnh H2, khi phn ng vi nc c xeton ).
Cu 5: 1/ n = 4, 5, 6.
2/ isobutilen
Cu 6: a/ m = 14,4g
b/ K2SO3 = 39,5 g v KHSO3 =60 g
LUYN HSG S 3Cu 1: Gii thch ti sao
1/ NF3 khng c tnh baz nh NH3.
2/ SnCl2 l cht rn cn SnCl4 l cht lng si 114,10C.
3/ NO2 d dng nh hp cn CO2 v ClO2 th khng th.
4/ Khi cho hn hp KIO3 v KI vo dung dch AlCl3 li c kt ta trng keo xut hin.
5/ in phn dung dch KCl khng mng ngn un nng th thu c kali clorat.
Cu 2: Cho hai kim loi X v Y
1/ Oxi ha ht p gam X th c 1,25p gam oxit. Ha tan mui cacbonat ca Y bng dung dch H2SO4 9,8% va thu c dung dch mui sunfat nng 14,18%. Tm X v Y?
2/ Ha tan a gam hn hp X v Y trong Y chim 30% khi lng bng 50 ml dung dch HNO3 63% (d=1,38 g/ml) khuy u hn hp ti khi phn ng hon ton th thu c cht rn A nng 0,75a gam, dung dch B v 7,3248 lt hn hp kh NO2 v NO 54,60C v 1 atm. C cn B c bao nhiu gam mui khan?
Cu 3: Hn hp A gm mt kim loi R ha tr I v kim loi X ha tr II. Ha tan 4,08 gam A vo dung dch c HNO3 v H2SO4 thu c 1,5 gam hn hp kh B gm N2O v kh D c VB = 560 ml(ktc).
1/ Tnh khi lng mui khan thu c?2/ Nu ha tan 1,02 gam hn hp A theo t l mol X : R = 3:2 th c 179,2 ml ktc hn hp kh Z gm NO v SO2 c t khi so vi hiro = 23,5. Tm R v X bit t l khi lng nguyn t tng ng l 27:16? Tnh khi lng R v X?Cu 4: t chy ht a mol hirocacbon A ri cho sn phm chy hp th ht vo nc vi trong d th c 4 gam kt ta. Lc tch kt ta cn li bnh nc vi trong th thy khi lng gim 1,376 gam.
1/ Tm CTPT ca A?
2/ Clo ha a mol A bng cch chiu sang 3000C th sau phn ng thu c mt hn hp B gm 4 ng phn cha clo. Bit dB/hiro < 93. v hiu sut phn ng t 100%, t s kh nng phn ng ca cc nguyn t H cacbon bc I : II : III = 1 ; 3,3 : 4,4. Tnh s mol cc ng phn trong hn hp B?
Cu 5: 1/ Cho axetilen phn ng vi Br2 trong CCl4 th c ti a my sn phm?
2/ Cho etilen vo dung dch cha HCl, NaCl, KI, CH3OH th thu c nhng sn phm g? Gi tn chng?3/ Chia hh hai ancol no, n chc mch h, lin tip lm hai phn bng nhau.
+ Phn 1 cho p vi Na d c 0,2 mol hiro
+ Phn 2 un nng vi H2SO4 c c 7,704 gam hn hp 3 ete. Tham gia p ete ha c 50% lng ancol c KLPT nh v 40% lng ancol c KLPT ln.
Tm CTPT ca hai ancol
CTTQ ca ancol no, n chc mch h l CnH2n+1OH; p ca hai ancol to 3 ete nh sau:CnH2n+1OH +CnH2n+1OH CnH2n+1O CnH2n+1 + H2O
CmH2m+1OH +CmH2m+1OH CmH2m+1OCmH2m+1 + H2O
CnH2n+1OH +CmH2m+1OH CnH2n+1OCmH2m+1 + H2O
Cu 6: Cho V lt CO2 54,60C v 2,4 atm hp th hon ton vo 200 ml dd hh KOH 1M v Ba(OH)2 0,75M thu c 23,64 gam kt ta. Tnh V?
Cu 7: Nhit phn hon ton R(NO3)2 (vi R l kim loi) thu c 8 gam mt oxit kim loi v 5,04 lt hn hp kh X gm NO2 v O2 (o ktc). Khi lng ca hn hp kh X l 10 gam. Xc nh cng thc ca mui R(NO3)2 ?
Cu 8: Ha tan hh X gm Cu v Fe2O3 trong 400 ml dung dch HCl a (M) thu c dung dch Y v cn li 1,0 gam Cu khng tan. Nhng thanh Mg vo dung dch Y, sau khi phn ng hon ton nhc thanh Mg ra thy khi lng tng thm 4,0 gam so vi khi lng thanh Mg ban u v c 1,12 lt kh H2 (o ktc) thot ra (gi thit ton b lng kim loi thot ra u bm ht vo thanh Mg). Tnh khi lng Cu trong X v gi tr ca a?p s:
Cu 1: 1/ NH3 c tnh baz l do trn N c i e t do khi gp H+ th i e ny kt hp vi H+ => NH3 c tnh baz; trong NF3 th i e t do b gi cht v F c m in ln ht i e ny v pha n do NF3 khng c kh nng cho e kt hp vi H+ hay n khng c tnh baz.
2/Do SnCl2 l hp cht ion cn hp cht SnCl4 l hp cht cng ha tr3/ V N trong NO2 c 1e c thn nn NO2 d kt hp vi nhau e ny c ghp i v th m NO2 d nh hp(hai phn t kt hp vi nhau)
+ CO2 v ClO2 khng c e c thn nn khng nh hp c.4/ Trong dd AlCl3 c phn ng thy phn sau: Al3+ + 3H2O Al(OH)3 + 3H+ (1)
+ Bnh thng th lng Al(OH)3 to ra rt t. Khi thm KI v KIO3 vo th c p: IO3- + 5I-+6H+ 6I2+ 3H2O.Nh c p ny m [H+] gim i lm cho cn bng (1) dch chuyn sang phi lm lng Al(OH)3 tng ln v xut hin kt ta
5/ V KOH v Cl2 to ra s p vi nhau theo p: 3Cl2 + 6KOH KClO3 + 5KCl + 3H2OCu 2: 1/ X l Cu; Y l Fe
2/ 37,575 gam Fe(NO3)2.
Cu 3: 1/ Ta c: = 60 => B c MB > 60 => B l SO2. Ta d dng tnh c: N2O =0,005; SO2 = 0,02 mol
tng s mol e nhn = 0,005.8 + 0,02.2 = 0,08 mol
+ Nu ch to mui nitrat th khi lng mui = 4,08 + 0,08.62 = 9,04 gam
+ Nu ch to mui sunfat th khi lng mui = 4,08 + 0,08.96/2 = 7,92 gam
=>7,92 gam < kl mui< 9,04 gam
2/ 2/ Ta c NO = SO2 = 0,004 mol. Gi 3x va 2x ln lt l s mol ca X v R ta c:
3x.2 + 2x.1 = 0,004.3+0,004.2 => x = 0,0025 mol. t 27a v 16a ln lt l KLNT ca R v X ta c:
2.0,0025.27a+3.0,0025.16a = 1,02 => a = 4 => R = 27a = 108 v X l 64.=> X l Cu(0,48 gam) v R l Ag (0,54 gam)Cu 4:1/ petan
2/ Cc cht trong B l ng phn ca nhau v c dng C5H12-mClm. V dB/hiro < 93 nn m m = 1, 2, 3. Mt khc A( petan) + Cl2 4 sn phm nn gi tr ca m ch c th l 1 do A l isopentan v B c 4 ng phn ng vi CTPT l C5H11Cl.
S mol CH2Cl CH(CH3) CH2 CH3 = 2,4.10-3 mol.
S mol CH3 CCl(CH3) CH2 CH3 = 1,76.10-3 mol.
S mol CH3 CH(CH3) CHCl CH3 = 2,64.10-3 mol.
S mol CH3 CH(CH3) CH2 CH2Cl = 1,2.10-3 mol.
Cu 5: 1/ ti a 3 sn phm: CHBr2-CHBr2; CHBr=CHBr(cis v trans)2/ Da vo c ch p suy ra sn phm l: CH3-CH2-Cl, CH3-CH2-I, CH3-CH2-OH, CH3-CH2-OCH3.
3/etanol v propanol
Cu 6: Ta thy s mol kt ta < Ba(OH)2 nn c hai trng hp xy ra, trong trng hp CO2 p vi dd c hai baz th ta gii bng phng trnh ion thu gn.
CO2 + 2OH- CO32-+ H2O; CO2 + OH- HCO3- sau : Ba2+ + CO32- BaCO3KQ: V = 1,343 hoc 4,253 ltCu 7: Fe(NO3)2.Cu 8: 4,2 gam v 1M. LUYN HSG S 4Cu 1: 1/ a/ Mt oxit ca nit c cng thc NOx, trong nit chim 30,43% v khi lng. Xc nh NOx. Vit phng trnh phn ng ca NOx vi dd NaOH va di dng phn t v ion rt gn? dd sau p c mi trng g?
b/Cho cn bng: N2O2x 2NOx (kh khng mu) (kh nu )
Cho hn hp gm 46g N2O2x v 13,8 gam NOx vo mt bnh kn th tch 10 lt n khi hn hp t trng thi cn bng th p sut trong bnh 1,015 ln p sut ban u, bit nhit khng i bng 27,30C.
Tnh hng s cn bng Kc, Kp ca phn ng.
Khi lm lnh bnh hn hp phn nh nn 00c ta thy mu nu nht dn, vy phn ng thun nhit hay ta nhit? So snh hng s cn bng ca phn ng trn 00C v 27,3 0C?
2/ Gii thch tnh axit baz ca cc dung dch sau: NH4ClO4, NaHS, NaClO4, K2Cr2O7, Fe(NO3)3, (CH3COO)2Mg.Cu 2: 1/ Tnh pH ca cc dd sau:
a/ Dd H2SO4 0,1M. Bit pKa = 2b/ Dd CH3COONa 0,4M. Bit Ka(CH3COOH) = 1,8.10-5.
2/ in li ca axit HA 2M l 0,95%
a/ Tnh hng s phn li ca HA.
b/ Nu pha long 10ml dd axit trn thnh 100ml th in li ca HA l bao nhiu? Tnh pH ca dd lc ny? C nhn xt g v in li khi pha long axit ny? Cu 3:1. Hon thnh cc phong trnh phn ng sau dng phn t v dng ion rt gn:
a/ Cl2 + dd FeSO4
EMBED Equation.DSMT4
e/ Fe + KNO3 + HCl b/ NaHCO3 + dd Ba(OH)2 d
f/ KI + FeCl3 c/ Al + NaNO3 + dd NaOH
g/ O3 + dd KI d/ FeS2 + dd HCl
h/ I2 + Na2S2O3 2/ Cn bng cc phn ng sau theo phng php thng bng electron:
a/ K2SO3 + KMnO4 + KHSO4 K2SO4 + MnSO4 ...
b/. P + NH4ClO4 H3PO4 + N2 + Cl2 +
c. Ha tan mt mui cacbonat ca kim loi M bng dd HNO3 thu c dd v hn hp hai kh NO, CO2.
3/ Cn bng phn ng sau theo phng php ion electron v hon thnh phn ng dng phn t.
FeS + H+ + NO-3 SO42- + N2Ox
Cu 4: 1. Cho m gam hn hp cng s mol FeS2 v Fe3O4 phn ng va vi 500 ml dung dch HNO3 un nng c dung dch A v 14,336 lt hn hp kh B (ktc) gm NO2 v NO c t khi so vi hiro l 19. Tnh m v nng mol/l ca dung dch HNO3 ?2/ Kh 3,48 gam mt oxit ca kim loi M cn 1,344 lt hiro. Ton b lng kim loi M cho phn ng vi dung dch HCl d c 1,008 lt hiro. Tm M v oxit ca n?(cc kh o ktc)Cu 5:1/ Cho hn hp X gm 8,4 gam Fe v 6,4 gam Cu vo dung dch HNO3. Sau khi phn ng hon ton thu c 3,36 lt kh NO (l sn phm kh duy nht, o ktc). Tnh khi lng mui thu c sau phn ng?2/ Khi nung butan vi xc tc thch hp thu c hn hp T gm CH4, C3H6, C2H4, C2H6, C4H8, H2 v C4H10 d. t chy hon ton hn hp T thu c 8,96 lt CO2 (o ktc) v 9,0 gam H2O. Mt khc, hn hp T lm mt mu va ht 12 gam Br2 trong dung dch nc brom. Tnh hiu sut phn ng nung butan?3/ Hp th hon ton 4,48 lt CO2 (ktc) vo 200ml dung dch cha Na2CO3 0,5M v NaOH 0,75M thu c dung dch X. Cho dd BaCl2 d vo dd X. Tnh khi lng kt ta thu c?4/ Ha tan 2m (gam) kim loi M bng dung dch HNO3 c, nng, d hay ha tan m (gam) hp cht X (hp cht ca M vi lu hunh) cng trong dung dch HNO3 c, nng, d th cng thu c kh NO2 (sn phm kh duy nht) c th tch bng nhau cng iu kin nhit v p sut. Gi s nguyn t lu hunh ch b oxi ha ln mc cao nht. Tm M, X?p n 4Cu 1: 1/ a. NOx l NO2. 2NaOH + 2NaOH NaNO2 + NaNO3 + H2O, mi trng bazb/ N2O4 2NO2 (1)mol b: 0,5 0,3
mol p: x 2x
mol cb: 0,5-x 0,3+2x
+ p sut ban u l: P = = 1,97 atm.
+ p sut khi cn bng l: P = 1,97.1,015 = 2 atm.
Tng s mol kh khi cn bng l: n = = 0,8122 mol => 0,5-x + 0,3+2x = 0,8122
x = 0,0122 mol
+ trng thi cn bng th: PNO2 = 0,8 atm v PN2O4 = =1,2 atm.
KP = = 0,533 atm.
+ Tnh KC: [NO2] = (0,3+2x)/10 = 0,03244 mol/l v [N2O4] = 0,04878 mol/l ( KC = = 0,0216 atm.
+ Khi lm lnh mu nu ca hh kh nht dn nn cn bng (1) dch chuyn sang tri c ngha l p thun thu nhit. V K nhit cng cao th cng ln nn K 00C nh hn.2/ + NH4ClO4: HClO4 l axit mnh nht trong cc axit nn ClO4- trung tnh do amoni peclorat c tnh axit v ion amoni phn li ra H+: NH4+ NH3 + H+.
+ NaHS: lng tnh v HS- phn li v thy phn ra H+ v OH-: HS- H+ + S2- v HS- + H2O H2S + OH-+ NaClO4 trung tnh v cc ion trong mui ny khng cho cng khng nhn H+.
+ Fe(NO3)3 c tnh axit v Fe3+ thy phn cho H+: Fe3+ + H2O Fe(OH)2+ + H+.
+ K2Cr2O7 c mi trng axit v: Cr2O72- + H2O 2CrO42- + 2H+.
+ (CH3COO)2Mg lng tnh v: CH3COO- + H2O CH3-COOH + OH-.
V: Mg2+ + H2O Mg(OH)+ + H+.Cu 2: 1/ 1/ Ta c: H2SO4 H+ + HSO4-.
Mol/l: 0,1 0,1 0,1
HSO4- H+ + SO42-.
Mol/l b: 0,1 0,1 0
Mol/l p: x x x
Mol/l cb: 0,1-x 0,1+x x
=> Ka = = 10-2 => x = 0,00844 => [H+] = 0,1 + x = 0,10844 => pH = 0,965
b/ Ta c: CH3COO- + H2O CH3COOH + OH- c Kb = 10-14/Ka = 5,55.10-10.
Mol/l b: 0,4 0 0
Mol/l pli: x x x
Mol/l cb:0,4-x x x
=> Kb = = 5,55.10-10 => x = 1,5.10-5 = [OH-] => [H+] = 6,67.10-10 => pH = 9,1762/ a/ ta c: nng HA phn li bng = 2.0,95/100 = 0,019 mol/l
HA H+ + A-.
Mol/l b: 2 0 0
Mol/l pli: 0,019 0,019 0,019
Mol/l cb: 1,981 0,019 0,019
Ka = = 1,82.10-4.
b/ Nu pha long 10 ml thnh 100 ml th nng ban u gim 10 ln v = 0,2 M. Do :
HA H+ + A-.
Mol/l b: 0,2 0 0
Mol/l pli: x x x
Mol/l cb: 0,2 x x x
Ka = = 1,82.10-4 => x = 5,943.10-3 =>
pH = -lgx= 2,226.
+ NX: khi pha long th in li ca tt c cc cht u tngCu 3: 1/ a/ 3Cl2 + 6FeSO4
EMBED Equation.DSMT4 2FeCl3 + 2Fe2(SO4)3.
b/ 2NaHCO3 + Ba(OH)2 Na2CO3 + BaCO3 + 2H2O
c/ 8Al + 3NaNO3 + 5NaOH + 2H2O 8NaAlO2 + 3NH3.d/ FeS2 + 2HCl FeCl2 + S + H2O2/ a. 5K2SO3 + 2KMnO4 +6KHSO4 9K2SO4 + 2MnSO4 + 3H2O
b/ 8P + 10NH4ClO4 8H3PO4 +5N2 + 5Cl2 + 8H2OTa c NH4ClO4 c c cht cho v nhn e th ta phi tnh xem l c phn t NH4ClO4 l cho hay nhn.
c. 3M2(CO3)x + (8y-2x)HNO3 6M(NO3)y + 2(y-x)NO + 3xCO2 + (4y-x)H2O
3. Ta c:
(12-2x)FeS + 18NO3- + (14x+12)H+ (12-2x)Fe3+ + (12-2x)SO42- + 9N2Ox + (7x+6)H2O
Cu 4: 1/ Ta c NO = NO2 = 0,32 mol. Gi x l s mol ca FeS2 v Fe3O4. Theo nh lut bo ton e ta c:
15x + x = 0,32.3 + 0,32.1 => x = 0,08 mol => m = 28,16 gam.
+ tnh s mol ca HNO3 ta p dng LBT nguyn t cho nit ta c s
FeS2 + Fe3O4 + HNO3 Fe(NO3)3 + NO + NO2 + H2OV s mol N trc p = s mol N sau p nn: = 3.(0,08+0,08.3)+ 0,32+0,32 = 1,6 mol ( CM = 3,2 M2/ p dng LBTKL ( mM = 2,52 gam, oxit l Fe3O4.Cu 5: 1/ 41,1gam.2/ 75%3/ 9,85 gam4/ Cu v Cu2S LUYN HSG S 5
Cu 1: 1/ Th no l nguyn tc y cc i gia cc cp e ha tr ? Da vo nguyn tc hy d on cu trc hnh hc ca cc cht sau: BeF2, BCl3, H2O, H3O+, NH4+, NO3-, NO2-, SO42-.2/ Hy v vc t m men lng cc ca cc lin kt trong cc cht sau: HF, H2O, NH3, CH4, BeF2.
Cu 2:
1/Dung dch NH3 1M c in li bng 4%
a/ Tnh pH ca dung dch
b/ pH ca dd thay i nh th no khi thm vo dd: amoniclorua; axit clohiric; natri hiroxit. c/ in li ca dung dch NH3 thay i nh th no khi: pha long dd; thm vo amoni nitrat; thm vo dd HNO3; thm vo dd KOH.
2/ Tnh th tch dung dch Ba(OH)2 0,025M cn cho vo 100 ml dung dch gm HCl v HNO3 c pH = 1 c dung dch c pH = 2?
Cu 3: 1/Hon thnh p: . Cho cc cht trn u cha photpho v c 3 phn ng oxi ha kh?2/ T cc cht v c cn thit vit phn ng iu ch vinyl axetat.
3/ Tm mui X bit X tha mn cc phn ng sau:
a/ natri alumilat + X 3 mui v nc
b/ amoni cacbonat + X cacbonic + hai mui v nc.
Cu 4: A, B v C l olefin hoc parafin kh ktc. Hh X cha A, B,C trong c hai cht c s mol bng nhau. Trong bnh kn dung tch khng i 11,2 lt ng oxi 00C v 0,6 atm. Sau khi thm m gam hh X vo bnh th p sut = 0,88 atm v nhit = 27,30C. Bt tia la in t chy ht hirocacbon v gi nhit 136,50C p sut trong bnh l P sn phm chy c 4,14 gam nc v 6,16 gam cacbonic.
1/ Tnh P?
2/ Xc nh CTPT vCTCT ca A, B, C bit nu ly tt c olefin trong 2 mol hh em trng hp th khng thu c qu 0,5 gam polime.Cu 5: 1. So snh bn knh ca: Na+, Mg2+; O2-; F-; Al, Na, Mg, Al3+? Tm ion c bn knh nh nht trong cc ion: Li+, Na+, K+, Be2+, Mg2+. 2/ Ha tan 0,775 gam n cht trong HNO3 c hh kh khi lng tng l 5,75 gam v mt dung dch hai axit c oxi vi hm lng oxi l ln nht. trung ha dd hai axit ny cn 0,1 mol NaOH.
a/ Xc nh %V hh kh thu c 900C bit t khi hh so vi hiro l 38,33.
b/ Tm n cht cho v tnh t s s mol ca hai axit
c/ Vit CTCT ca cc axit v ch r trng thi lai ha ca nguyn t trung tm?
3. ha tan 9,18 gam bt Al nguyn cht cn dng dung dch axit A thu c mt kh X v dung dch mui Y . tc dng hon ton vi dung dch mui Y to thnh dung dch mui mi trong sut th cn 290 gam dung dch NaOH 20% . Xc nh axt A?
Cu 6: Hn hp A gm 6 gam cht X l mt axit no, n chc v 0,1 mol Y thuc dy ng ng ca axit lactic. t chy hon ton A cn 0,5 mol O2 thu c 0,5 mol CO2. Tm CTCT, tn v %KL mi cht trong X v Y?
Cu 7: Hh X gm 0,1 mol CuO v 0,05 mol Fe3O4. Cho X p vi hiro nung nng mt thi gian c 18 gam cht rn Y. Cho Y p vi HNO3 c nng d. Tnh s mol HNO3 p? Cu 8: Cho lung kh CO i qua m gam Fe2O3 nung nng thu c 2,428 gam hh rn A gm: Fe, FeO, v Fe2O3 d. Trong A khi lng ca FeO gp 1,35 ln khi lng ca Fe2O3. Khi ho tan A trong 130 ml dd H2SO4 0,1M thu c 0,224 lt kh H2 ktc. Cht rn cn d sau khi phn ng l Fe. Tnh khi lng Fe d v m?p s: Cu 1: 1/ Nguyn tc y cc i gia cc cp e ha tr: cc cp e ha tr trong mt cht s c phn b sao cho chng cch xa nhau nht(hay ni cch khc v n y nhau mnh nht nn n xa nhau nht)
BeF2BCl3H2OH3O+NH4+NO3-NO2-SO42-
ThngTam gicV ngcT dinT dinTam gicV ngcT din
2/ + M men lng cc ca lin kt A-X: l vc t ni gia A v X c hng t nguyn t c m in nh n nguyn t c m in ln. VD: m men lng cc ca HF c biu din l: HF. lin kt A-A c m men lng cc bng 0 v hai nguyn t c cng m in.
+ Ngoi ra ngi ta cn qui c: nu nguyn t trung tm c i e t do th i e t do ny cng to ra mt m men lng cc c hng t nguyn t trung tm ra ngoi.
+ M men lng cc ca phn t: l 1 vct c t hp t nhng vect i din cho cc lin kt trong phn t v vect ca cp e khng lin kt ca nguyn t trung tm.
+ V m men lng cc:
Cu 2: 1/ 12,6 2/ 0,15 lit Cu 3: 1/ A l P; B l Ca3P2; C l PH3; D l P2O5; E l H3PO4.2/ canxi cacbonat canxi oxit canxi cacbua axetilen etanal axit axetic vinyl axetat
p to vinyl axetat: CH3-COOH + CHCH CH3-COOCH=CH2.3/ X l KHSO4 v mui hirosufat c tnh axit nh mt axit mnh. P xy ra nh sau:
2NaAlO2 + 8KHSO4 Na2SO4 + 4K2SO4 + Al2(SO4)3 + 4H2O
(NH4)2CO3 + 2KHSO4 (NH4)2SO4 + K2SO4 + CO2 + H2OCu 4: 1/ P = 1,245 atm2/metan, etan v but 2 en.
Cu 5: 1/ Ta thy Na+, Mg2+; O2-; F-; Al3+ cng c 10 e v cng c 2 lp tc l lp v nh nhau. Mt khc ta thy in tch trong nhn ca: Al3+ = 13+ > Mg2+= 12+ > Na+= 11+ > F-= 9+ > O2- = 8+.
( Lc ht trong nhn ca Al3+ > Mg2+ > Na+ > F- > O2-. T suy ra bn knh ca
Al3+ < Mg2+ < Na+ < F- < O2-.
* So snh Li+, Na+, K+ th Li+ c bn knh nh nht v trong mt phn nhm R gim dn khi Z tng, tng t th Be2+ c bn knh nh hn Mg2+. Do tm ion c bn knh min th ta ch cn so snh Li+ v Be2+. Ta thy s e v ngoi ca hai ion ny bng nhau m Z ca Be2+ ln hn nn sc ht ca nhn vi v mnh hn lm cho R nh hn. KL: Be2+ c bn knh min 2/ a/ V =76,6 => phi c N2O4 ( phi c NO2. D tnh c s mol NO2 = 0,025 v N2O4 = 0,05 mol.
b/ S mol e nhn = 0,025 + 2.0,05 = 0,125 mol. Gi n l s oxi ha max ca n cht cn tm ta c:
0,125 => M = 6,2n => ch c n = 5 v M = 31 l ph hp => n cht cho l Photpho.
c/ P: P + 5HNO3 H3PO4 + 5NO2 + H2O. Sau 2NO2 N2O4.
Do hai axit l H3PO4 v HNO3 d
.
Trng thi lai ha tng ng l sp3 v sp2. T l mol = 1:13/ D thy s mol Al3+ = Al( Tnh c s mol NaOH ti a p vi Al3+ nh hn s mol gi thit cho ( phi c mt sp to ra p c vi NaOH. Trong cc axit th ch c HNO3 to ra c NH4NO3.Cu 6: axit axetic v axit HO-C2H4-COOH(c 2 CTCT vi tn gi l axit 2-hiroxipropanoic v axit 3-hiroxipropanoic)
Cu 7: 0,9 mol
Cu 8: Fe d = 1,652 gam(nu lm ra 1,68 gam l sai); m = 3,4 g
LUYN HSG S 6
Cu 1: Nguyn t A c electron sau cui c cc s lng t: n=2, l=1, m=+1, ms= +1/2.
a/ Vit cu hnh electron, xc nh v tr ca nguyn t A trong h thng tun hon.
b/ Vit CTCT ca cc cht: AH3, A2H4, AH4AlCl3, AO2. Nu trng thi lai ha ca A trong cc cht trn.Cu 2: Cho H2S qua dung dch cha Cd2+ 1.10-3M, Zn2+1.10-2M cho n bo ha (CH2S=0,1M).
a/ C kt ta CdS v ZnS tch ra khng
b/ Nu c th kt ta no s xut hin trc .c/ Khi kt ta th 2 xut hin th nng ca cation kim loi th nht bng bao nhiu?Cho pTZnS = 23,8; pT CdS = 26,1. Coi s to phc hirxo(s thy phn) khng xy ra.
Bit trong dung dch bo ha H2S thH2S H+ + HS- k1= 10-7 v HS- H+ + S2- k2= 10-12,92Cu 3: Cho 3,2g Cu vo a g dung dch H2SO4 95% thu c V1 lt kh X ; phn cn li x l tip bng b g dung dch HNO3 80% thu c V2 lt kh Y. Sau hai ln x l lng Cu cn li l 1,28 g . Bit V1+ V2 = 896cm3 v cc th tch o ktc.
1/ Ly a g dung dch H2SO4 95% trn vi b g dung dch HNO3 80% ri pha long vi nc ti 20 ln thu c dung dch A. Cho 3,2g Cu vo dung dch A. Tnh th tch kh thot ra V3.
2/ Trn V1 lt kh X vi V2 lt kh Y c hn hp Z . Cho kh Z li t t qua dung dch BaCl2 d. Tnh lng kt ta to thnh. Gi s cc phn ng xy ra hon ton.Cu 4: a/ Mt ankin A c CTPT C6H10 c ng phn quang hc. Hir ha hon ton A thu c A. Hi A c ng phn quang hc khng.
b/ Mt Ankin B cng c CTPT l C6H10. B tc dng vi H2 c Ni xc tc cho cht 2 mtylpentan. B khng tc dng vi dung dch AgNO3 trong NH3, B tc dng vi H2O/HgSO4 to cht C6H12O (B,). Xc nh B, B v gi tn B, B.
c/ Hir ha B c xc tc Pd v un nng c cht C. C tc dng vi H2SO4 ri thy phn to thnh cht D . Hy xc nh C v D , bit C v D l cc sn phm chnh.
d/ Tch nc cht D vi H2SO4 c xc tc v un nng. Hy vit c ch phn ng v nu sn phm chnh.
Cu 5: Mt hn hp c khi lng 7,6g gm 0,05 mol mt hircacbon mch thng A v 0,05 mol mt ankin B (ktc). t chy hon ton hn hp trn ri cho ton b sn phm chy hp th ht trong dung dch Ba(OH)2 d thu c 108,35 gam kt ta.
1/ A thuc loi hirocacbon no?
2/ Vit CTPT, CTCT ca A v B bit chng hn km nhau mt cacbon.Cu 6: Cho 6,4 gam oxi v a gamm hh kh A gm hai hirocacbon vo bnh kn dung tch 10 lt. 00C, p sut bnh l 0,4704 atm. Sau khi t chy hon ton hirocacbon gi bnh 1270C, p sut trong bnh l P. Sn phm chy thu c c 0,324 gam nc v 0,528 gam CO2. Tm CTCT v % th tch mi cht trong A?
Cu 7: Ha tan hon ton hn hp MgCl2, FeCl3, CuCl2 vo nc c dd A. Cho t t H2S vo A cho n d th thu c kt ta to ra nh hn 2,51 ln kt ta to ra khi cho dd Na2S d vo A. Nu thay FeCl3 trong A bng FeCl2 vi khi lng nh nhau th t l khi lng kt ta l 3,36. Vit p v tnh % khi lng mi mui trong A?
Cu 8: Dung dch A gm FeCl3, AlCl3, NH4Cl, CuCl2 nng u xp x l 0,1M
1/ Dung dch A c mi trng axit, baz hay trung tnh?
2/ Sc H2S li chm vo A bo ha c kt ta B v dd C. Cho bit cc cht c trong B v C ri vit p xy ra?
p s:
Cu 1: a/ A l nit.
b/ CTCT l:
Cu 2: a/+ c CdS th phi c: [Cd2+].[S2-] > 10-26,1 => [S2-] > 10-26,1/10-3 > 10-23,1 (1)+ c ZnS th phi c: [Zn2+].[S2-] > 10-23,8 => [S2-] > 10-23,8/10-2 > 10-21,8 (2)+ Ta phi tnh [S2-] trong dd bit c kt ta khng.
+ Ta c: H2S H+ + HS-.
Cb: 0,1 0 0
Cphn li: x x x
Ccb: 0,1-x x x
=>k1 = = 10-7 => x = 10-4M.
+ Do ta c:
HS- H+ + S2- k2= 10-12,92Cb: 10-4 10-4 0
Cpli: y y y
Ccb: 10-4-y 10-4+y y
=> =10-12,92 => y = 1,2.10-13 = [S2-] (3)+ So snh (1,2,3) ta c kt lun:
( c kt ta to thnh
( kta ta CdS to ra trc.
c/ Khi ZnS bt u kt ta th: [S2-] = 10-21,8 mol/l => [Cd2+] = 10-4,3 mol/l.Cu 3: 1/ Ta d thy X l SO2; Y l NO2 vi s mol l x v y ta c: x + y = 0,04 (1)
+ p dng LBT e ta c: 2x + y = =0,06 (2)
+ T (1,2) suy ra: x = y = 0,02 mol.
+ P: Cu + 2H2SO4 CuSO4 + SO2 + 2H2O
v Cu + 4HNO3 Cu(NO2)2 + 2NO2 + 2H2O
S mol H2SO4 = 0,04 mol v HNO3 = 0,04 mol
+ Khi pha long ti 20 ln th c hai axit u th long. Do Cu khng p vi H2SO4 long m ch c p sau: 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O
Mol b: 0,05 0,12 0,04 0
Mol p: 0,045 0,12 0,03 0,03
Mol cn: 0,005 0 0,01 0,03
Th tch NO l: V3 = 0,672 lt.
2/ Khi trn X vi Y th c p: NO2 + SO2 NO + SO3 => Z c 0,02 mol NO v 0,02 mol SO3. Khi cho Z vo dd BaCl2 d th: SO3 + H2O H2SO4 sau : BaCl2 + H2SO4 BaSO4 + 2HCl.
Do tnh c khi lng kt ta l 4,66 gam.Cu 4:a/ A l 3 metylpent 1 in. A khng c ng phn quang hc
b/ B l 4 metylpent 2 in
c/ C l ng phn cis.
Cu 5: 1/ ta c BaCO3 = CO2 = 0,55 mol. t CTPT ca A v B ln lt l: CxHy(0,05 mol)v CnH2n-2(0,05 mol) ta c h:
y = 2x => A l anken.
2/ Ta c n+x = 11
+ TH1: n = x+1 th n = 6 v x = 5 => A l C6H12 v B l C5H8(3CTCT).
+ TH2: x = n+1 th n = 5 v x = 6 => A l C5H10 v B l C6H10(6CTCT).
Cu 6: Gii theo pp trung bnh c suy ra phi c CH4 v CnH2. D dng tnh c n = 2 v %metan = 80%.
Cu 7: + t x, y, z ln lt l s mol ca MgCl2, FeCl3, CuCl2.
+ P xy ra khi cho H2S p vi A: 2FeCl3 + H2S 2FeCl2 + S + 2HCl (1)
mol: y y/2
CuCl2 + H2S CuS + 2HCl (2)
Mol: z z
khi lng kt ta l: 16y + 96z (*)
+ P xy ra khi cho Na2S p vi A:
2FeCl3 + 3Na2S 2FeS + S + 6NaCl (3)
mol: y y y/2
CuCl2 + Na2S CuS + 2NaCl (4)
Mol: z z
MgCl2 + Na2S + H2O Mg(OH)2 + 2NaCl + H2S (5)Mol: x x
khi lng kt ta l: 58x +104y +96z (*)
+ T (*) v (*) ta c: 58x + 104y + 96z = 2,51(16y + 96z) (I)+ Khi thay FeCl2 vi khi lng bng FeCl3 th s mol FeCl2 = 162,5y/127.
Khi p vi H2S th ch c p (2) do khi lng kt ta = 96z (**)
Khi p vi Na2S th ch c p (4, 5) v p: FeCl2 + Na2S FeS + 2NaCl do khi lng kt ta = 58x + 96z + 112,6y (***)
+ T (**) v (***) ta c: 58x + 96z + 112,6y = 3,36.96z (II)+ T (I, II) ta c: z = 1,6735y; x = 4,596y.
+ Vy khi lng hh mui = 95x + 162,5y + 135z = 825y.
=> % khi lng ca MgCl2...Cu 8: 1/ A c mi trng axit v cc cation c p thy phn v phn li to ra H+ theo phng trnh:
Fe3+ + H2O Fe(OH)2++ H+;Al3+ + H2O Al(OH)2++ H+;NH4+ NH3 + H+2/ p: 2FeCl3 + H2S 2FeCl2+ S + 2HCl.
&CuCl2 + H2S CuS + 2HCl
+ Kt ta B gm c S, CuS.
+ Dung dch C gm: NH4Cl; AlCl3; HCl, FeCl2. LUYN HSG S 7 Cu 1: Ha tan hon ton m gam hn hp gm 3 kim loi bng dung dch HNO3 thu c V(1) kh D ( ktc) gm NO2 v NO. T khi hi cu D so vi hir bng 18,2.
a/ Tnh tng s gam mui khan to thnh theo m v V. Bit rng khng sinh ra mui NH4NO3.
b/ Tnh th tch ti thiu dd HNO3 37,8% ( d=1,242g/ml) dung nu cho V=1,12(1)Cu 2: 1. Hon thnh s chuyn ha sau:
2/ Cho dung dch A trn p vi CO2 thu c hn hp gm hai mui X v Y. un nng X v Y phn hy hon ton c sn phm gm hn hp kh G v hi nc trong CO2 chim 30% v th tch. Tnh t l mol ca X v Y trong hn hp?
Cu 3: Cho 3,87 gam hh A gm Mg v Al vo 250 ml dd B gm HCl 1M v H2SO4 0,5M thu c dd C v 4,365 lt hiro ktc. Tnh khi lng mui trong C?Cu 4: a/ Isopren c th trng hp thnh 4 loi polime kiu cis 1,4; trans 1,4; trans 1,2; trans 2,4. Hy vit CTCT mi loi mt on mch gm ba mt xch?
b/ Ho tan hon ton 14,6 gam hn hp X gm Al v Sn bng dung dch HCl (d), thu c 5,6 lt kh H2 ( ktc). Th tch kh O2 ( ktc) cn phn ng hon ton vi 14,6 gam hn hp X l
Cu 5: 1/ A, B, C, D v E l nm hirocacbon u c CTPT l C9H12. Khi un nng vi dd thuc tm trong H2SO4 d th A v B u cho sp c CTPT l C9H6O6; C cho C8H6O4.
+ Khi un nng vi brom c xt Fe th A cho 1 sp monobrom; B v C u cho 2 sp monobrom.
+ D v E u p vi AgNO3/NH3 v vi dd HgSO4/H2SO4 un nng sinh ra cc hp cht M v N u c CTPT l C9H14O. Ozon phn M cho nonan-2,3,8-trion cn N cho 2-axetyl-3-metylhexaial.
Tm CTCT ca nm hirocacbon trn v vit p xy ra.
2/ un nng m gam hp cht X cha C, H, O mch thng vi dd cha 8 gam NaOH n p hon ton. Trung ha va ht NaOH d bng 40 ml dd HCl 1M. Lm bay hi hh sau khi trung ha mt cch cn thn c 7,36 gam hh hai ancol n chc v 15,14 gam hh gm hai mui khan. Tm CTCT v tn X?
Cu 6: Hn hp bt A gm 3 kim loi Mg, Zn, Al. Khi ho tan ht 7,539g A vo 1lt dung dch HNO3 thu c 1lt dung dch B v hn hp kh D gm NO v N2O. Thu kh D vo bnh dung tch 3,20lt c cha sn N2 00C v 0,23atm th nhit trong bnh tng ln n 27,30C, p sut tng ln n 1,10atm, khi lng bnh tng thm 3,720g. Nu cho 7,539g A vo 1lt dung dch KOH 2M th sau khi kt thc phn ng khi lng dung dch tng thm 5,718g. Tnh thnh phn phn trm khi lng mi kim loi trong A.Cho Mg = 24,3; Zn = 65,38; Al = 26,98.Cu 7: Cho 43 gam hn hp BaCl2 v CaCl2 vo 1 lt dung dch hn hp Na2CO3 0,1M v (NH4)2CO3 0,25 M. Sau phn ng thu c 39,7 gam kt ta A v dung dch B. 1/ Tnh% khi lng ca mi cht trong hh kt ta.
2/ un nng B v thm vo 200 ml dd Ba(OH)2 0,5M th sau p khi lng dd gim bao nhiu gam?
Cu 8: Ha tan 7,82 g XNO3 vo nc thu c dd A. in phn dd A vi in cc tr. Nu thi gian p l t giy th thu c kim loi ti catot v 0,1792 lt kh ktc ti anot. Nu thi gian p l 2t giy th thu c 0,56 lt kh ktc. Tm X, t bit I = 1,93A?
Cu 9: Cho 43,6 gam hh X gm FeCO3 v FeS p vi HNO3 c nng d thu c 33,6 lt hh kh ktc.
1/ Tnh khi lng mi cht trong X?
2/ Cho hh kh trn vo dd NaOH d th thu c bao nhiu gam mui?
S: 1/ FeS = 8,8 gam v FeCO3 = 34,8 gam
2/ 3 mui NaNO3 + NaNO2 + Na2CO3 = Cu 10: Cho hh gm 2 gam Fe v 3 gam Cu p vi dd HNO3 th thu c 448 ml kh NO duy nht ktc. Tnh khi lng mui to thnh?
S: 5,4 gam Fe(NO3)2. LUYN HSG S 8Cu 1: So snh th tch NO sinh ra duy nht trong hai th nghim sau:
TN1: Cho 6,4 gam Cu vo120 ml dung dch HNO3 1M(long).
TN2: Cho 6,4 gam Cu vo 120 ml dung dch cha HNO3 1M(long) v H2SO4 0,5M.
C cn dung dch TN2 thu c bao nhiu mol mui khanCu 2: Hn hp A gm 1,792 lt hai anken kh ng ng lin tip 00C v 2,5 atm.
1/ Cho hn hp A trn qua nc brom d thy khi lng dung dch nc brom tng 9,8 gam. Tm CTPT v %V mi anken?
2/ Cho A phn ng vi ozon ri thy phn sn phm c mt bt km c dung dch B. Cho B phn ng vi dung dch AgNO3/NH3 d c 1 mol Ag. Tm CTCT v tn hai anken?
Cu 3: Dung dch A cha 2 axt HCl 1M v HNO3 0,5M. Thm t t Mg kim loi vo 100 ml dd A cho ti khi kh ngng thot ra thu c dd B ch cha cc mui ca Mg v 0,963 lt hn hp D gm 3 kh khng mu cn nng 0,772 gam. Trn kh D vi 1 lt O2 sau khi phn ng hon ton cho kh thu c i t t qua dd NaOH d c 1,291 lt kh khng b hp th.
1/ D gm cc kh g. Bit rng trong D c 2 kh c s mol bng nhau v khi p vi NaOH ch c mt kh p to hai mui?(cc kh o ktc)2/ Vit cc ptp ho tan Mg di dng ion3/ Tnh nng cc ion trong dung dch B v tnh khi lng Mg b ha tan?
Cu 4: Cho hn hp G dng bt gm Al, Fe, Cu. Ha tan 23,4 gam G bng mt lng d dung dch H2SO4 c, nng, thu c 15,12 lt kh SO2. Cho 23,4 gam G vo bnh A cha 850 ml dung dch H2SO4 1M (long) d, sau khi phn ng hon ton thu c kh B. Dn t t ton b lng kh B vo ng cha bt CuO d nung nng, thy khi lng cht rn trong ng gim 7,2 gam so vi ban u.
1) Vit p xy ra v tnh thnh phn phn trm theo khi lng ca mi cht trong hn hp G.
2) Cho dd cha m gam mui NaNO3 vo bnh A sau phn ng gia G vi dd H2SO4 long trn, thy thot ra V lt kh NO (sp kh duy nht). Tnh gi tr nh nht ca m V l ln nht. Cc th tch kh o ktcCu 5: T mt loi tinh du ngi ta tch c cht A trong A c 78,95%C, 10,52%H cn li l oxi. T khi ca A so vi hiro l 76. A c phn ng trng gng khi b oxi ha mnh A cho hn hp sn phm l axeton, axit oxalic v axit levulic (CH3COCH2CH2COOH). Vit CTCT ca A bit A phn ng vi brom trong dung mi cacbon tetraclorua theo t l 1:1 c hai dn xut ibromCu 6: Cho 6,5 gam axit hu c X p ht vi NaHCO3 c 2,464 lt CO2 27,30C v 1 atm. Mt khc xc nh MX ngi ta ha tan 22,572 gam X vo 100 gam dung mi Y thy nhit ng c ca dd gim 2,240C. Bit rng c 1 mol X trong 1000 gam dung mi th nhit ng c ca dd gim 1,290C.
Tm CTCT ca X bit X c p cis-trans? Cu 7: Nu hin tng v gii thch hin tng xy ra trong cc TN sau:
1/ cho benzen vo ng nghim cha nc brom, lc k ri yn
2/ Cho brom lng vo ng nghim cha benzen, lc k ri yn
3/ Cho thm bt st vo ng nghim th nghim 2 ri un nh
Cu 8: Ho tan hon ton hn hp X gm Zn, FeCO3, Ag bng lng d dung dch HNO3, thu c hn hp kh A gm 2 hp cht kh c t khi i vi H2 bng 19,2 v dung dch B. Cho B tc dng ht vi dung dch NaOH d to kt ta. Lc kt ta em nung nhit cao n khi lng khng i thu c 5,64 gam cht rn. Tnh khi lng hn hp X. Bit trong X khi lng FeCO3 bng khi lng Zn; mi cht trong X khi tc dng vi dung dch HNO3 trn ch cho 1 sn phm kh.
Cu 9: 1/ Cho 11,2 gam Fe vo 800 ml dd HNO3 1,2 M thu c kh NO duy nht v dd Y. Hi Y ha tan c ti a bao nhiu gam Cu?2/ Cho 61,2 gam hn hp X gm Cu v Fe3O4 tc dng vi dung dch HNO3 long, un nng v khuy u. Sau khi cc phn ng xy ra hon ton, thu c 3,36 lt kh NO (sn phm kh duy nht, ktc), dung dch Y v cn li 2,4 gam kim loi. C cn dd Y, thu c m gam mui khan. Tnh m?Cu 10: Cho 1,572 gam hh A gm Al, Fe, Cu + 40 ml dd CuSO4 1M dd B + hh D gm hai kim loi. B + dd NaOH to kt ta max, nung kt ta ny trong kk n khi lng khng i th thu c 1,82 gam hai oxit. Cho D p vi AgNO3 d th thu c Ag c khi lng ln hn khi lng ca D l 7,876 gam. Tnh khi lng mi cht trong A?S 8Cu 1: Gii bng phng trnh ion thu gn sau: 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O
TN1: 0,03 mol NO TN2: 0,06 mol NO, 0,03 mol Cu(NO3)2 v 0,06 mol CuSO4.
Cu 2: propen v but-2-en.
Cu 3: 1/ ba kh l hiro, NO v N2O ( tm N2O phi xt 3 TH trong trng hp hiro v NO c s mol bng nhau tha mn) 3/ [Cl-] = 1M; [NO3-] = 0,24M, [Mg2+] = 0,62M; Mg = 1,488 gam.
Cu 4: 1/ Nhm: = 23,08 (%), St:= 35,90 (%), ng:41,02 (%) 2/ m = 0,15.85 = 12,75 gam(c hai phn ng ca Cu v Fe2+ vi H+ v NO3-).
Cu 5: A l (CH3)2C=CH-CH2-CH2-C(CH3)=CH-CHO.Cu 6: + Ta c CO2 = 0,1 mol.
+ Tm MX: p dng nh lut Raun(SGKNC11): t ( M = 130 vC.+ t CT ca axit l: R(COOH)n ta c:
R(COOH)n + nNaHCO3 R(COONa)n + nCO2 + nH2O
Mol: 0,1/n 0,1
ta c h:= n = 2 v R = 40(C3H4)
Do CTPT ca X l C3H4(COOH)2. V c p cis-trans nn X c CTCT l: HOOC-C(CH3)=CH-COOH.Cu 7: Ghi nh + Benzen khng tan v nh hn nc. + Benzen ha tan brom tt hn nc
1/ Khng c p xy ra. Ban u c hai lp lp trn l benzen; lp di l nc brom. Khi lc k ri yn th brom trong nc s chuyn sang benzen khi vn c hai lp nhng lp trn cha benzen v brom mu vng, lp di cha nc khng mu.
2/ Brom tan dn vo benzen nhng khng c p xy ra 3/ C p v c kh HBr bay ra. Mu ca brom nht dn.Cu 8: = 38,4 vc; A gm 2 hp cht kh, trong 1 l CO2 ( v ban u c FeCO3), kh cn li c M 0,448 ) thu c hh kh X(cc kh o ktc). t chy hon ton X ri cho spc hp th ht vo 450 ml dd Ba(OH)2 0,2M thy to thnh 12,805 gam kt ta. Nu cho hh X qua dd AgNO3 d trong NH3 thy to ra 4,8 gam kt ta vng. Hiu sut cc p = 100%. Tm CTPT; CTCT ca A, B v tnh %V mi hirocacbon trong X?Cu 4: Ha tan ht hh A gm Al v kim loi X ha tr a trong H2SO4 c nng n khi khng cn kh thot ra c dd B v kh C. Kh C b hp th bi NaOH d c 50,4 gam mui. Nu thm vo A mt lng X bng hai ln lng X trong A(gi nguyn Al) ri ha tan ht bng H2SO4 c nng th lng mui trong dd mi tng thm 32 gam so vi lng mui trong B nhng nu gim mt na lng Al trong A(gi nguyn X) th khi ha tan ta c 5,6 lt C ktc
1/ Tnh KLNT ca X bit tng s ht proton; ntron v electron trong X l 93
2/ Tnh % khi lng cc cht trong A
3/ Tnh s mol H2SO4 dng lc u bit rng khi thm t t dd NaOH 2M vo B th lng kt ta bt u khng i khi dng ht 700 ml dd NaOH trn.Cu 5: Xc nh CTCT ca A(cha C, H, O) v vit cc p bit
+ A p vi Na gii phng Na
+ A p c Cu(OH)2 to thnh dd mu xanh lam
+ A c p trng gng
+ t chy 0,1 mol A th thu c khng qu 7 lt kh sp 136,50C v 1 atm.Cu 6: Ha tan 17,4 gam hn hp 3 kim loi Al, Cu, Fe trong dd HCl d thy thot ra 8,96 lt kh(ktc). Nu cho 34,8 gam hh trn p vi dd CuSO4 d ri lc cht rn to ra ha tan bng HNO3 th thot ra 26,88 lt kh (ktc) c t khi so vi oxi = 1,27. Vit cc p v tnh thnh phn hn hp ban u. Cu 7: A l hh kh ktc gm ba hirocacbon X, Y, Z thuc ba dy ng ng, B l hh oxi v ozon c t khi so vi hiro l 19,2. t chy 1 mol A cn 5 mol B c s mol CO2 = H2O. Khi cho 22,4 lt A qua nc brom d thy c 11,2 lt kh bay ra v khi lng bnh nc brom tng 27 gam. Khi cho 22,4 lt A qua dd AgNO3 d trong dd NH3 thy to thnh 32,4 gam kt ta.
1/ Tnh t khi ca A so vi hiro?
2/ Tm CTPT, CTCT ca X, Y, Z?
Cu 8: Ha tan 115,3 gam hn hp X gm MgCO3 v RCO3 (s mol ca RCO3 gp 2,5 ln ca MgCO3) bng 500 ml dd H2SO4 long c dd A, cht rn B v 0,2 mol CO2. C cn A c 12 gam mui khan. Mt khc nung B ti khi lng khng i th c 0,5 mol CO2 v cht rn B1.
1/ Tnh CM ca H2SO4?2/ Tnh tng KL ca B v B1? 3/ Tm R?Cu 9: Hh X gm FeS2 v Cu2S tan ht trong dd HNO3 va thu c dd Y ch gm hai mui sunfat v 5,6 lt hh kh Z gm NO2 v NO c KL ring = 1,7678 g/l ktc. Hy tnh khi lng hh X?
Cu 10: Ha tan ht hh gm FeS v FeCO3 bng dd HNO3 c nng thu c hh kh A gm hai kh X, Y c t khi so vi hiro l 22,805.
1/ Tnh %KL mi cht trong hh ban u?
2/ Lm lnh hh kh A xung nhit thp hn c hh kh B gm ba kh X, Y, Z c t khi so vi hiro l 30,61. Tnh %X b ime ha thnh Z?
P N 9
Cu 1: 1/ Mui trung ha l mui m trong phn t khng cn H c th phn li thnh H+. Mui axit l mui m trong phn t cn H c th phn li thnh H+. VD: CH3COONa l mui c H nhng khng phn li thnh H+ c ( l mui trung ha; NaHCO3 l mui axit v HCO3- H+ + CO32-.+ V H3PO3 l axit hai ln axit nn ch c th phn li ra ti a l 2H+. Do mui Na2HPO4 cn H nhng khng phn li ra c H+ nn n l mui trung ha.
2/ Ta c:: MnO4- + 8H+ + 5e Mn2+ + 4H2O=> E = E0 + = E0 +
= 1,51 + = 1,51-0,0944pH.
+ KMnO4 p c vi HCl th E ca (MnO4-/Mn2+) > (Cl2/2Cl-) = 1,36 V
( pH = 0 th E ca (MnO4-/Mn2+) = 1,51V > 1,36V => p xy ra
( pH = 3 th E ca (MnO4-/Mn2+) = 1,2268V < 1,36V => p khng xy ra
Cu 2: Ta c: CaCO3 = 0,01 mol; CuO = 0,04 mol. Do :
CuO + CO Cu + CO2 .
Mol: x x x
Ca(OH)2 + CO2 CaCO3 + H2O
Mol: x x
x = 0,01 mol ( cht rn cn li c: CuO d = 0,03 mol v Cu = 0,01 mol.
+ Khi p vi 0,08 mol HNO3 th c p sau:
CuO + 2H+ Cu2+ + H2O
Mol: 0,03 0,06 0,03
3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O (1)
mol b: 0,01 0,02 0,08
=> trong p (1) ta tnh theo H+ => NO = 0,02.2/8 = 0,005 mol => V1 = 0,112 lt+ Khi thm vo 1,52/3 mol HCl th:
3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O (2)
mol b: 0,0025 1,52/3 0,075
=> trong p (2) ta tnh theo Cu => NO = 0,0025.2/3 = 0,005/3 mol => V2 = 0,112/3 lt+ Sau p (2) th cn: 0,5 mol H+ + 0,22/3 mol NO3-+ 0,04 mol Cu2++ Cl-. Khi thm 0,5 mol Mg vo th c p sau:
5Mg + 12H+ + 2NO3- 5Mg2+ + N2 + 6H2O
mol b: 0,5 0,5 0,22/3
D thy p ny phi tnh theo NO3- => N2 = 0,11/3 mol
Mg + 2H+ Mg2+ + H2Mol b: 0,95/3 0,06
phi tnh theo H+ => s mol hiro = 0,03 mol
V3 = 22,4(0,03 + 0,11/3)=4,48/3 lt Mg + Cu2+ Mg2+ + Cu
Mol b: 0,86/3 0,04
=> tnh theo Cu2+ => Mg d = 0,74/3 mol v Cu = 0,04 mol. Vy Mg = 5,92 gam v Cu = 2,56 gam1/V1 = 0,112; V2 = 0,112/3; V3 = 4,48/32/ M c 5,92 gam Mg v 2,56 gam Cu.
Cu 3: + V A, B cng dy ng ng hc p vi AgNO3/NH3 nn A, B l ankin.
+ t CTPTTB ca A, B l: (>2)vi s mol l x (x>0,04) ta c:
+ O2 CO2 + (-1)H2O
mol: x x
+ Trng hp 1: khi CO2 p vi Ba(OH)2 ch to ra mt mui
CO2 + Ba(OH)2 BaCO3 + H2O
Mol: x x
=> x = 0,065. D thy x > 0,04 => < 1,625 => loi
+ Trng hp 2: khi CO2 p vi Ba(OH)2 ch to ra hai mui. D dng tnh c x = CO2 = 0,115 mol => < 2,875 => phi c CHCH.
+ Gi CTPT ca ankin cn li l:CnH2n-2 ta d thy X c: (a > 0,02 mol)
+ P vi AgNO3/NH3 ta c:
CHCH + 2AgNO3 + 2NH3 CAgCAg + 2NH4NO3.
Mol: 0,02 0,02
khi lng kt ta = 0,02.240 = 4,8 gam => ankin cn li khng p vi AgNO3/NH3.
+ Mt khc c hai ankin u th kh nn ch c th l C3H4 v C4H6. Ta loi c C3H4 v cht ny p c vi AgNO3/NH3; cn C4H6 c hai CTCT nhng ch c CH3-CC-CH3 tha mn.
+ Tnh th tch ca C4H6 c kt qu l: 0,42 ltCu 4: 1/ Gi x, y ln lt l s mol ca Al v X trong hh A ban u. Ta c:
2Al + 6H2SO4 Al2(SO4)3 +3SO2 + 6H2O
mol: x x/2 3x/2
2X + 2aH2SO4 X2(SO4)a +aSO2 + 2aH2O
mol: y y/2 ay/2
2NaOH + SO2 Na2SO3 + H2O
mol: (3x+ay)/2 (3x+ay)/2
=> 3x + ay = 0,8 (I)+ Khi tng X ln hai ln th lng mui trong B tng thm y mol X2(SO4)a do : 32 = 2y(X+48a) (II)+ Khi gim mt na lng Al th: 3x/4 + ay/2 = 0,25 (III)+ Gii (I, II, III) c: x = 0,2; ay = 0,2 v Xy = 6,4 => X = 32a => X c th l Cu, Mo, Te nhng ch c Cu tha mn tng s ht l 93.
2/ Al = 45,76% v Cu = 54,24%
3/ H2SO4 = 1 molCu 5: HCOOH
Cu 6: Al = 5,4 gam; Fe = 5,6 gam cn li l Cu.
Cu 7: 1/ T khi = 28. 2/ X = butan; Y = buta 1,3 ien hoc buta 1,2 ien; Z = but 1 in.Cu 8: 1/ 0,4M 2/ 199 gam 3/ Ba Cu 9: 4,5 gam Cu 10: 1/ FeS = 20,87%; FeCO3 = 79,13%.2/ 63,35%
LUYN HSG S 10Cu 1: 1/ Cho hai nguyn t X v Y c electron cui cng ng vi b 4 s lng t sau:
X: n = 2; l = 1; m = -1; ms = -1/2. Y: n = 3; l = 1; m = -1; ms = -1/2.
a/ Da vo cu hnh e hy xc nh v tr ca X, Y trong BTH?
b/ Cho bit trng thi lai ha ca nguyn t trung tm v dng hnh hc ca phn t XY3 v ion [YX4]2-. Vit CTCT ca phn t XY3 v ion [YX4]2-?
2/ C 5 hp cht v c: A, B, C, D, E c khi lng phn t tng dn v c cng nguyn t X(c trong qung Lo Cai). Khi cho 5 hp trn ln lt phn ng vi dd NaOH d u thu c dd c cng cht Y. Hy tm cng thc, gi tn cc cht trn v vit phn ng xy ra?
Cu 2: Tnh hiu ng nhit ca p sau: H2SO3(l) H2S(k) + 3/2 O2(k). Bit:
H2O(l) + SO2(k) H2SO3(l) c H1 = -124 KJ
S(r) + O2(k) SO2(k) c H2 = -594 KJ
H2S(k) + O2(k) S(r) + H2O(l) c H3 = -124 KJCu 3: 1/ a. Dd A cha HF 0,1M v NaF 0,1M. Tnh pH ca A cho pKa ca HF = 3,17?
b. Tnh pH ca A trong hai TH sau: + Thm vo 1 lt A 0,01 mol HCl + Thm vo 1 lt A 0,01 mol NaOH
2/ Cho CO2 li qua dd gm Ba(OH)2 0,1M v Sr(OH)2 0,1M. Cht no kt ta trc? Khi mui th hai bt u kt ta th nng ca cation th nht cn li l bao nhiu? C th dng CO2 tch ion Ba2+ v Sr2+ ra khi nhau c khng? Cho: T ca BaCO3 = 8,1.10-9 v SrCO3 = 9,4.10-10. Mt ion c coi l tch hon ton th nng ca n t nht phi = 10-6.
Cu 4: 1/ Khi clo ha isopentan theo t l 1:1 thu c thu c cc dn xut monoclo vi thnh phn nh sau: 1-clo-2-metylbutan: 30%
1-clo-3-metylbutan: 15%
2-clo-3-metylbutan: 33%
2-clo-2-metylbutan: 22%
Hy so snh kh nng th ca cc nguyn t hiro cacbon bc I; II v III?
2/ Khi cho isobutilen vo dd HBr trong nc c ha tan NaCl; CH3OH c th to ra nhng hp cht g?
Cu 5: Cho m1 gam hn hp gm Mg, Al vo m2 gam dd HNO3 24%. Sau khi cc kim loi tan ht c 8,96 lt ( ktc) hn hp kh X gm NO, N2O, N2 bay ra ( ktc) v dd A. Thm mt lng va O2 vo X, sau phn ng thu c hn hp kh Y. Dn Y t t qua dd NaOH d c 4,48 lt hn hp kh Z i ra ( ktc). T khi ca Z i vi H2 bng 20. Nu cho dd NaOH vo A c lng kt ta ln nht thu c 62,2 gam kt ta.
1/ Tnh m1, m2. Bit lng HNO3 ly d 20% so vi lng cn thit.
2/ Tnh nng % cc cht trong dd sau p?
Cu 6: C 6 hirocacbon dng kh A, B, C, D, E v F l ng phn ca nhau. t chy hh A v O2 d, sau khi ngng t nc ri a v iu kin ban u th th tch hn hp kh cn li gim 40% so vi hn hp ban u, tip tc cho kh cn li qua bnh ng dd KOH d th th tch hh gim 4/7.
1/ Tm CTPT ca A?
2/ Tm CTCT; gi tn 6 cht trn ri vit p xy ra bit: khi p vi brom trong CCl4 th A, B, C, D lm mt mu nhanh; E lm mt mu chm cn F khng p. Cc sn phm thu c t B v C vi brom l nhng ng phn lp th ca nhau. Nhit si ca B cao hn C. Khi cho p vi hiro th A, B v C cho cng mt sn phm G.
Cu 7: t chy 1,8 gam cht A ch c C, H v O cn 1,344 lt oxi ktc c CO2 v nc theo t l mol = 1:1.
1/ Tm CTGN ca A?
2/ Khi cho cng mt lng cht A p ht vi Na v NaHCO3 th s mol H2 v s mol CO2 thu c l bng nhau v bng s mol ca A p. Tm CTCT ca cht A(c KLPT nh nht) tha mn iu kin trn v vit p xy ra?Cu 8: C hai th nghim sau:
+ TN1: Cho 3,07 gam hh D gm Fe v Zn vo 200 ml dd HCl. Sau p c cn dd sau p c 5,91 gam b rn
+ TN2: Cho 3,07 gam hh D vo 400 ml dd HCl trn. Sau p c cn dd sau p c 6,62 gam b rn
1/ Xc nh nng ca dd HCl cho?
2/ CMR trong TN2 HCl kim loi vn d?
P N 10Cu 1: 1/ a. X , Y l O v S.
b. SO3 SO42-.
Trng thi lai ha: sp2 sp3.
Dng hnh hc: tam gic u t din
2/ X l photpho c trong qung apatit v photphorit Lo cai
A = HPO3(axit metaphotphoric); B = H3PO4(axit photphoric); C = NaH2PO4(natri ihirophotphat); D = P2O5 (iphotpho pentoxit) hoc Na2HPO4(natri hirophotphat) v E = H4P2O7(axit iphotphoric)
Cu 2: Ta thy c p cho ta phi t hp 3 p nh sau:-(1)-(2)-(3). Do :
H = -H1 H2-H3 = 842 KJCu 3: 1/ a. pH = 3,17
b. pH = 3,08 v 3,254.
2/ SrCO3 kt ta trc. Khi BaCO3 kt ta th [Sr2+] =
Cu 4: 1/c 9 nguyn t C bc I ng vi (30+15)=45% => 1 ngt C bc I = 5%. Tng t ta c: 1 ngt C bc II = 16,5%;1 ngt C bc III = 22% => T l th tng i gia C bc I:II:III = 1:3,3:4,4
2/ Da vo c ch suy ra c 4sp
Cu 5: m1 = 23,1 gam v m2= 913,5 gam; mdd = 922,6 gam => Mg(NO3)2=6,42%; Al(NO3)3=11,54%; HNO3=3,96%;
Cu 6: 1/ t CxHy l cng thc chung ca 6 cht cho. Gi s ban u c 1 mol CxHy v a mol O2 ta c: CxHy + (x+0,25y)O2 xCO2 + 0,5y H2O
Suy ra: Ban u ( sau p:
+ Theo gi thit ta c h: ( x = ( y = 8; x = 4 tha mn.2/ Ghi nh: anken lm mt mu nhanh nc brom; vng 3 cnh ca xicloankan lm mt mu nc brom chm hp. A = but-1-en; B = cis-but-2-en ; C = trans-but-2-en ; D = isobutilen(2-metylpropen) ; E = metylxiclopropan; F = xiclobutan
Cu 8: 1/ t a v b ln lt l s mol Zn v Fe trong 3,07 gam D ta c: 65a + 56b = 3,07 (*). V Zn p trc nn p theo th t: Zn + 2 HCl ZnCl2 + H2. (1) Fe + 2 HCl FeCl2 + H2. (2)
+ Gi s TN1 kim loi ht ( b rn ch c mui ZnCl2 = a mol v FeCl2 = b mol(HCl bay hi khi c cn)( Trong TN2 lng axit tng ln gp i nn kim loi vn ht tc l b rn vn c ZnCl2 = a mol v FeCl2 = b mol ( KL b rn khng i iu ny tri vi gi thit. Vy trong TN1 axit ht; kim loi d
+ Ta c: c 1 mol kim loi to thnh mui clorua th KL tng 71 gam
theo gt th KL tng 5,91 3,07 = 2,84 gam ( S mol kim loi p = 0,04 mol. Theo p ta thy s mol HCl = 2 x s mol kim loi p = 0,08 mol ( CM = 0,08/0,2 = 0,4M2/ Nu hh ch c Zn th s mol s l nh nht v bng 3,07/65 = 0,04723 mol. Theo phn 1 th s mol kim loi p ti a l 0,04 mol ( kim loi lun d._1298316110.unknown
_1298493101.unknown
_1300907548.unknown
_1356629714.unknown
_1360497178.unknown
_1360498228.unknown
_1360670611.unknown
_1360671308.unknown
_1360671426.unknown
_1360670549.unknown
_1360497548.unknown
_1359206724.unknown
_1359206783.unknown
_1359206696.unknown
_1357007317.unknown
_1356629364.unknown
_1356629377.unknown
_1300907651.unknown
_1298831335.unknown
_1299097230.unknown
_1299097243.unknown
_1299097275.unknown
_1299183557.unknown
_1299097236.unknown
_1299067157.unknown
_1299097212.unknown
_1299066663.unknown
_1298831215.unknown
_1298831257.unknown
_1298751849.unknown
_1298318247.unknown
_1298400293.unknown
_1298434793.unknown
_1298435222.unknown
_1298436658.unknown
_1298437544.unknown
_1298435662.unknown
_1298434944.unknown
_1298405912.unknown
_1298378691.unknown
_1298379410.unknown
_1298397956.unknown
_1298400094.unknown
_1298378863.unknown
_1298377740.unknown
_1298317512.unknown
_1298317933.unknown
_1298318079.unknown
_1298317791.unknown
_1298317087.unknown
_1298317320.unknown
_1298316806.unknown
_1264708612.unknown
_1298315171.unknown
_1298315644.unknown
_1298316092.unknown
_1298315388.unknown
_1298201243.unknown
_1298314474.unknown
_1298315056.unknown
_1298314946.unknown
_1298314375.unknown
_1266607452.unknown
_1288063445.unknown
_1288063446.unknown
_1288032621.unknown
_1266607280.unknown
_1264707583.unknown
_1264707755.unknown
_1264708030.unknown
_1264708140.unknown
_1264707599.unknown
_1264704820.unknown
_1264534921.unknown