giao an day them mon hoa hoc 8 hay
DESCRIPTION
Mon Hoa Hoc 8 HaTRANSCRIPT
Cc dng bi tp ha hc chng trnh lp 8-THCS
Cc dng bi tp ha hc chng trnh lp 8-THCSCHUYN 1: BI TP V NGUYN T- NGUYN T HA HC1/ Nguyn t (NT):- Ht v cng nh , trung ha v in, to nn cc cht.Cu to: + Ht nhn mang in tch (+)(Gm: Proton(p) mang in tch (+) v ntron khng mang in ). Khi lng ht nhn c coi l khi lng nguyn t. + V nguyn t cha 1 hay nhiu electron (e) mang in tch (-). Electron chuyn ng rt nhanh quanh ht nhn v sp xp theo lp (th t sp xp (e) ti a trong tng lp t trong ra ngoi: STT ca lp :123 S e ti a :2e8e18eTrong nguyn t:- S p = s e = s in tch ht nhn = s th t ca nguyn t trong bng h thng tun hon cc nguyn t ha hc - Quan h gia s p v s n : p n 1,5p ( ng vi 83 nguyn t )- Khi lng tng i ca 1 nguyn t ( nguyn t khi )NTK = s n + s p - Khi lng tuyt i ca mt nguyn t ( tnh theo gam ) + mT = m e + mp + mn
+ mP mn 1VC 1.67.10- 24 g,
+ me 9.11.10 -28 gNguyn t c th ln kt c vi nhau nh e lp ngoi cng.2/ Nguyn t ha hc (NTHH): l tp hp nhng nguyn t cng loi c cng s p trong ht nhn.- S p l s c trng ca mt NTHH.- Mi NTHH c biu din bng mt hay hai ch ci. Ch ci u vit di dng in hoa ch ci th hai l ch thng. l KHHH- Nguyn t khi l khi lng ca nguyn t tnh bng VC. Mi nguyn t c mt NTK ring. Khi lng 1 nguyn t = khi lng 1vc.NTK
NTK = m a Nguyn t = a.m 1vc .NTK
(1VC = KL ca NT(C) (MC = 1.9926.10- 23 g) = 1.9926.10- 23 g= 1.66.10- 24 g)* Bi tp vn dng:1. Bit nguyn t C c khi lng bng 1.9926.10- 23 g. Tnh khi lng bng gam ca nguyn t Natri. Bit NTK Na = 23. (p s: 38.2.10- 24 g)2.NTK ca nguyn t C bng 3/4 NTK ca nguyn t O, NTK ca nguyn t O bng 1/2 NTK S. Tnh khi lng ca nguyn t O. (p s:O= 32,S=16)3. Bit rng 4 nguyn t Mage nng bng 3 nguyn t nguyn t X. Xc nh tn,KHHH ca nguyn t X. (p s:O= 32)4.Nguyn t X nng gp hai ln nguyn t oxi .b)nguyn t Y nh hn nguyn t Magie 0,5 ln .c) nguyn t Z nng hn nguyn t Natri l 17 vc .Hy tnh nguyn t khi ca X,Y, Z .tn nguyn t, k hiu ho hc ca nguyn t ? 5.Nguyn t M c s n nhiu hn s p l 1 v s ht mang in nhiu hn s ht khng mang in l 10. Hy xc nh M l nguyn t no?6.Tng s ht p, e, n trong nguyn t l 28, trong s ht khng mang in chim xp x 35% .Tnh s ht mi loa .V s cu to nguyn t .7.Nguyn t st c 26p, 30n, 26ea.Tnh khi lng nguyn t stb.Tnh khi lng e trong 1Kg st8.Nguyn t X c tng cc ht l 52 trong s ht mang in nhiu hn s ht khng mang in l 16 ht.a)Hy xc nh s p, s n v s e trong nguyn t X.b) V s nguyn t X.c) Hy vit tn, k hiu ho hc v nguyn t khi ca nguyn t X.9. Mt nguyn t X c tng s ht e, p, n l 34. S ht mang in nhiu hn s ht khng mang in l 10. Tm tn nguyn t X. V s cu to ca nguyn t X v ion c to ra t nguyn t X10.Tm tn nguyn t Y c tng s ht trong nguyn t l 13. Tnh khi lng bng gam ca nguyn t.11. Mt nguyn t X c tng s ht l 46, s ht khng mang in bng s ht mang in. Xc nh nguyn t X thuc nguyn t no ? v s cu to nguyn t X ?12.Nguyn t Z c tng s ht bng 58 v c nguyn t khi < 40 . Hi Z thuc nguyn t ho hc no. V s cu to nguyn t ca nguyn t Z ? Cho bit Z l g ( kim loi hay phi kim ? ) (p s :Z thuc nguyn t Kali ( K ))Hng dngii : bi 2p + n = 58 n = 58 2p ( 1 )Mt khc : p n 1,5p ( 2 ) p 58 2p 1,5p gii ra c 16,5 p 19,3 ( p : nguyn )Vy p c th nhn cc gi tr : 17,18,19p171819
n242220
NTK = n + p414039
Vy nguyn t Z thuc nguyn t Kali ( K )13.Tm 2 nguyn t A, B trong cc trng hp sau y :a) Bit A, B ng k tip trong mt chu k ca bng tun hon v c tng s in tch ht nhn l 25.b) A, B thuc 2 chu k k tip v cng mt phn nhm chnh trong bng tun hon. Tng s in tch ht nhn l 32. 14: Trong 1 tap hp cac phan t ong sunfat (CuSO4) co khoi lng 160000 vC. Cho biet tap hp o co bao nhieu nguyen t moi loai.
3. S to thnh ion (dnh cho HSG lp 9) t cu trc bo ha ( 8e lp ngoi cng hoc 2e i vi H ) th cc nguyn t c th nhng hoc nhn thm electron to ra nhng phn mang in - gi l ion* Kim loi v Hiro : nhng e to ion dng ( cation)M ne M n + (Ca 2e Ca 2 + )* Cc phi kim nhn e to ion m (anion)X+ ne X n- ( Cl + 1e Cl 1- )* Bi tp vn dng: 1.Hp cht X c to thnh t cation M+ v anion Y2- . Mi ion u do 5 nguyn t ca 2 nguyn t to nn. Tng s proton trong M+ l 11 cn tng s electron trong Y2- l 50.Xc nh CTPT ca hp cht X v gi tn ? ng dng ca cht ny trong nng nghip . Bit rng 2 nguyn t trong Y2- thuc cng phn nhm trong 2 chu k lin tip ca bng tun hon cc ng.t.Hng dn gii :t CTTQ ca hp cht X l M2YGi s ion M+ gm 2 nguyn t A, B : ion M+ dng: AxBy+ c: x + y = 5( 1 )x.pA + y.pB = 11 ( 2) Gi s ion Y 2- gm 2 nguyn t R, Q: ion Y2- dng: R xQy2- c: x + y = 5(3) xpR + y.pQ = 48(4 ) do s e > s p l 2 T ( 1 ) v (2) ta c s proton trung bnh ca A v B : 1 trong AxBy+ c 1 nguyn t c p < 2,2 ( H hoc He ) v 1 nguyn t c p > 2,2 V He khng to hp cht ( do tr ) nn nguyn t c p < 2,2 l H ( gi s l B )T ( 1 ) v ( 2) ta c: x.pA + (5 x ).1 = 11 pA = ( 1 x < 5 )x1 2 3 4
pA7(N) 4(B) 3(Li) 2,5 (loi)
ion M+ NH4+ khng xc nh ion
Tng t: s proton trung bnh ca R v Q l : c 1 nguyn t c s p < 9,6 ( gi s l R )V Q v R lin tip trong nhm nn : pQ = pR + 8 ( 5 )T (3) ,(4) , ( 5) ta c: xpR + (5- x)( pR + 8) = 48 5pR 8x = 8 x1 2 3 4 Vy ion Y2- l SO42-
pR3,2 4,8 6,4 8 ( O )
pQ khng xc nh ion 16 ( S )
Vy CTPT ca hp cht X l (NH4 )2SO4
Chuyn II. Bi tp v cng thc ha hc : a.Tnh theo CTHH:1: Tm TP% cc nguyn t theo khi lng.* Cch gii: CTHH c dng AxBy- Tm khi lng mol ca hp cht. MAxBy = x.MA + y. MB- Tm s mol nguyn t mi nguyn t trong 1 mol hp cht : x, y (ch s s nguyn t ca cc nguyn t trong CTHH)
- Tnh thnh phn % mi nguyn t theo cng thc: %A = =
V d: Tm TP % ca S v O trong hp cht SO2- Tm khi lng mol ca hp cht : MSO2 = 1.MS + 2. MO = 1.32 + 2.16 = 64(g)- Trong 1 mol SO2 c 1 mol nguyn t S (32g), 2 mol nguyn t O (64g)
- Tnh thnh phn %: %S = = = 50%
%O = = = 50% (hay 100%- 50% = 50%)* Bi tp vn dng:1: Tnh thanh phan % theo khoi lng cac nguyen to trong cac hp chat : a/ H2O b/ H2SO4 c/ Ca3(PO4)22: Tnh thanh phan phan tram ve khoi lng cua cac nguyen to co trong cac hp chat sau:a) CO; FeS2; MgCl2; Cu2O; CO2; C2H4; C6H6.b) FeO; Fe3O4; Fe2O3; Fe(OH)2; Fe(OH)3.c) CuSO4; CaCO3; K3PO4; H2SO4. HNO3; Na2CO3.d) Zn(OH)2; Al2(SO4)3; Fe(NO3)3. (NH4)2SO4; Fe2(SO4)3.3: Trong cc hp cht sau, hp cht no c hm lng Fe cao nht: FeO ; Fe2O3 ; Fe3O4 ; Fe(OH)3 ; FeCl2 ; Fe SO4.5H2O ?4: Trong cc loi phn bn sau, loi phn bn no c hm lng N cao nht: NH4NO3; NH4Cl; (NH4)2SO4; KNO3; (NH2)2CO?2: Tm khi lng nguyn t trong mt lng hp cht.* Cch gii: CTHH c dng AxBy- Tnh khi lng mol ca hp cht. MAxBy = x.MA + y. MB- Tm khi lng mol ca tng nguyn t trong 1 mol hp cht: mA = x.MA , mB = y. MB- Tnh khi lng tng nguyn t trong lng hp cht cho.
mA = = , mB = = V d: Tm khi lng ca Cc bon trong 22g CO2Gii:- Tnh khi lng mol ca hp cht. MCO2 = 1.Mc + 2. MO = 1.12 + 2. 16 = 44(g)- Tm khi lng mol ca tng nguyn t trong 1 mol hp cht: mC = 1.Mc = 1.12 = 12 (g)- Tnh khi lng tng nguyn t trong lng hp cht cho.
mC = = = 6(g)* Bi tp vn dng: 1: Tnh khoi lng moi nguyen to co trong cac lng chat sau: a) 26g BaCl2; 8g Fe2O3; 4,4g CO2; 7,56g MnCl2; 5,6g NO.b) 12,6g HNO3; 6,36g Na2CO3; 24g CuSO4; 105,4g AgNO3; 6g CaCO3. c) 37,8g Zn(NO3)2; 10,74g Fe3(PO4)2; 34,2g Al2(SO4)3; 75,6g Zn(NO3)2.2: Mt ngi lm vn dng 500g (NH4)2SO4 bn rau. Tnh khi lng N bn cho rau?B/ Lp CTHH da vo Cu to nguyn t: Kin thc c bn phn 1* Bi tp vn dng:1.Hp cht A c cng thc dng MXy trong M chim 46,67% v khi lng. M l kim loi, X l phi kim c 3 lp e trong nguyn t. Ht nhn M c n p = 4. Ht nhn X c n= p ( n, p, n, p l s ntron v proton ca nguyn t M v X ). Tng s proton trong MXy l 58. Xc nh cc nguyn t M v X (p s : M c p = 26 ( Fe ), X c s proton = 16 ( S ) )2. Nguyn t A c n p = 1, nguyn t B c n=p. Trong phn t AyB c tng s proton l 30, khi lng ca nguyn t A chim 74,19% .Tm tn ca nguyn t A, B v vit CTHH ca hp cht AyB ? Vit PTHH xy ra khi cho AyB v nc ri bm t t kh CO2 vo dung dch thu c3. Tng s ht tronghp cht AB2 = 64. S ht mang in trong ht nhn nguyn t A nhiu hn s ht mang in trong ht nhn nguyn t B l 8. Vit cng thc phn t hp cht trn.Hng dn bi1:Nguyn t M c : n p = 4 n = 4 + p NTK = n + p = 4 + 2p Nguyn t X c : n = p NTK = 2pTrong MXy c 46,67% khi lng l M nn ta c : (1) Mt khc : p + y.p = 58 yp = 58 p ( 2) Thay ( 2) vo (1) ta c:4 + 2p = . 2 (58 p ) gii ra p = 26 v yp = 32M c p = 26 ( Fe ) X tha mn hm s: p = ( 1 y 3 )y12 3
P32(loi)1610,6 ( loi)
Vy X c s proton = 16 ( S )
C/ lp CTHH da vo Thnh phn phn t,CTHH tng qut: Cht (Do nguyn t to nn)
n cht Hp cht (Do 1 ng.t to nn) (Do 2 ng.t tr ln to nn) CTHH: AX AxBy + x=1 (gm cc n cht kim loi, S, C, Si..) (Qui tc ha tr: a.x = b.y)
+ x= 2(gm : O2, H2,, Cl2,, N2, Br2 , I2..)
Oxit Axit Baz Mui ( M2Oy) ( HxA ) ( M(OH)y ) (MxAy) 1.Lp CTHH hp cht khi bit thnh phn nguyn t v bit ha tr ca chng Cch gii: - CTHH c dng chung : AxBy (Bao gm: ( M2Oy , HxA, M(OH)y , MxAy)Vn dng Qui tc ha tr i vi hp cht 2 nguyn t A, B
(B c th l nhm nguyn t:gc axt,nhm OH): a.x = b.y = (ti gin) thay x= a, y = b vo CT chung ta c CTHH cn lp. V d Lp CTHH ca hp cht nhm oxt a b
Gii: CTHH c dng chung AlxOy Ta bit ha tr ca Al=III,O=II a.x = b.y III.x= II. y = thay x= 2, y = 3 ta c CTHH l: Al2O3* Bi tp vn dng:1.Lp cng thc ha hc hp cht c to bi ln lt t cc nguyn t Na, Ca, Al vi (=O,; -Cl; = S; - OH; = SO4 ; - NO3 ; =SO3 ; = CO3 ; - HS; - HSO3 ;- HSO4; - HCO3; =HPO4 ; -H2PO4 ) 2. Cho cc nguyn t: Na, C, S, O, H. Hy vit cc cng thc ho hc ca cc hp cht v c c th c to thnh cc nguyn t trn?3. Cho cc nguyn t: Ca, C, S, O, H. Hy vit cc cng thc ho hc ca cc hp cht v c c th c to thnh cc nguyn t trn?
2.Lp CTHH hp cht khi bit thnh phn khi lng nguyn t . 1: Bit t l khi lng cc nguyn t trong hp cht.Cch gii: - t cng thc tng qut: AxBy
- Ta c t l khi lng cc nguyn t: =
- Tm c t l : = = (t l cc s nguyn dng, ti gin) - Thay x= a, y = b - Vit thnh CTHH.V d:: Lap CTHH cua sat va oxi, biet c 7 phan khoi lng sat th ket hp vi 3 phan khoi lng oxi.Gii: - t cng thc tng qut: FexOy
- Ta c t l khi lng cc nguyn t: = = - Tm c t l : = = = = - Thay x= 2, y = 3 - Vit thnh CTHH. Fe2O3* Bi tp vn dng:1: Lap CTHH cua sat va oxi, biet c 7 phan khoi lng sat th ket hp vi 3 phan khoi lng oxi.2: Hp cht B (hp cht kh ) bit t l v khi lng cc nguyn t to thnh: mC : mH = 6:1, mt lt kh B (ktc) nng 1,25g.3: Hp cht C, bit t l v khi lng cc nguyn t l : mCa : mN : mO = 10:7:24 v 0,2 mol hp cht C nng 32,8 gam.4: Hp cht D bit: 0,2 mol hp cht D c cha 9,2g Na, 2,4g C v 9,6g O5: Phan t khoi cua ong sunfat la 160 vC. Trong o co mot nguyen t Cu co nguyen t khoi la 64, mot nguyen t S co nguyen t khoi la 32, con lai la nguyen t oxi. Cong thc phan cua hp chat la nh the nao?6:Xc nh cng thc phn t ca CuxOy, bit t l khi lng gia ng v oxi trong oxit l 4 : 1? 7: Trong 1 tap hp cac phan t ong sunfat (CuSO4) co khoi lng 160000 vC. Cho biet tap hp o co bao nhieu nguyen t moi loai.8: Phan t khoi cua ong oxit (co thanh phan gom ong va oxi)va ong sunfat co t le 1/2. Biet khoi lng cua phan t ong sunfat la 160 vC. Xac nh cong thc phan t ong oxit?9. Mot nhom oxit co t so khoi lng cua 2 nguyen to nhom va oxi bang 4,5:4. Cong thc hoa hoc cua nhom oxit o la g?
2. Bit khi lng cc nguyn t trong mt lng hp cht, Bit phn t khi hp cht hoc cha bit PTK(bi ton t chy) Bi ton c dng : t m (g)AxByCz t chy m(g) cc hp cht cha A,B,C
+Trng hp bit PTK Tm c CTHH ng
+Trng hp cha bit PTK Tm c CTHH n ginCch gii: - Tm mA, mB, mC trong m(g) cc hp cht cha cc nguyn t A,B,C.
+ Nu (mA + m B) = m (g)AxByCz Trong h/c khng c nguyn t C
T : x : y = : = a:b (t l cc s nguyn dng, ti gin) CTHH: AaBb
+ Nu (mA + m B) m (g)AxByCz Trong h/c c nguyn t C
m C = m (g)AxByCz - (mA + m B)
T : x : y : z = : : = a:b:c (t l cc s nguyn dng, ti gin)
CTHH: AaBbCcCch gii khc: Da vo phng trnh phn ng chy tng qut
CxHy +
CxHy0z + - Lp t l s mol theo PTHH v s mol theo d kin bi ton suy ra x, y, z.V d: t chy 4,5 g hp cht hu c A. Bit A cha C, H, 0 v thu c 9,9g kh C02 v 5,4g H20. Lp cng thc phn t ca A. Bit kh lng phn t A bng 60.Gii:
- Theo bi ra: , , - Phng trnh phn ng :
CxHy0z +
1mol . (mol). x (mol)
Suy ra : Mt khc;MC3H80z = 60 Hay : 36 + 8 + 16z =60 > z = 1 Vy cng thc ca A l C3H80
* Bi tp vn dng:
+Trng hp cha bit PTK Tm c CTHH n gin1: t chy hon ton 13,6g hp cht A,th thu c 25,6g SO2 v 7,2g H2O. Xc nh cng thc ca A
2: ot chay hoan toan m gam chat A can dung het 5,824 dm3 O2 (ktc). San pham co CO2 va H2O c chia oi. Phan 1 cho i qua P2O5 thay lng P2O5 tang 1,8 gam. Phan 2 cho i qua CaO thay lng CaO tang 5,32 gam. Tm m va cong thc n gian A. Tm cong thc phan t A va biet A the kh (k thng) co so C 4.3: t chy hon ton 13,6g hp cht A, th thu c 25,6 g S02 v 7,2g H20. Xc nh cng thc A
+Trng hp bit PTK Tm c CTHH ng1: t chy hon ton 4,5g hp cht hu c A .Bit A cha C, H, O v thu c 9,9g kh CO2 v 5,4g H2O. lp cng thc phn t ca A. Bit phn t khi A l 60.2: t chy hon ton 7,5g hyroccbon A ta thu c 22g CO2 v 13,5g H2O. Bit t khi hI so vi hyr bng 15. Lp cng thc phn t ca A.3: : t chy hon ton 0,3g hp cht hu c A . Bit A cha C, H, O v thu c 224cm3 kh CO2 (ktc) v 0,18g H2O. lp cng thc phn t ca A.Bit t khi ca A i vi hiro bng 30.4:t chy 2,25g hp cht hu c A cha C, H, O phi cn 3,08 lt oxy (ktc) v thu c VH2O =5\4 VCO2 .Bit t khi hi ca A i vi H2 l 45. Xc nh cng thc ca A5: Hyro A l cht lng , c t khi hi so vi khng kh bng 27. t chy A thu c CO2 v H2O theo t l khi lng 4,9 :1 . tm cng thc ca AS: A la C4H103: Bit thnh phn phn trm v khi lng cc nguyn t, cho bit NTK, phn t khi. Cch gii: - Tnh khi lng tng nguyn t trong 1 mol hp cht.- Tnh s mol nguyn t tng nguyn t trong 1 mol hp cht.- Vit thnh CTHH.Hoc: - t cng thc tng qut: AxBy
Ta c t l khi lng cc nguyn t: =
Rt ra t l x: y = : (ti gin)
Vit thnh CTHH n gin: (AaBb )n = MAxBy n =
nhn n vo h s a,b ca cng thc AaBb ta c CTHH cn lp.Vi d. Mot hp chat kh Y co phan t khoi la 58 vC, cau tao t 2 nguyen to C va H trong o nguyen to C chiem 82,76% khoi lng cua hp chat. Tm cong thc phan t cua hp chat. Gii : - t cng thc tng qut: CxHy
Ta c t l khi lng cc nguyn t: =
Rt ra t l x: y = : = : = 1:2 Thay x= 1,y = 2 vo CxHy ta c CTHH n gin: CH2
Theo bi ra ta c : (CH2 )n = 58 n = = 5
Ta c CTHH cn lp : C5H8
* Bi tp vn dng:1: Hp cht X c phn t khi bng 62 vC. Trong phn t ca hp cht nguyn t oxi chim 25,8% theo khi lng, cn li l nguyn t Na. S nguyn t ca nguyn t O v Na trong phn t hp cht l bao nhiu ?2: Mot hp chat X co thanh phan % ve khoi lng la :40%Ca, 12%C va 48% O . Xac nh CTHH cua X. Biet khoi lng mol cua X la 100g.3:Tm cng thc ho hc ca cc hp cht sau. a) Mt cht lng d bay hi, thnh phn t c 23,8% C, 5,9%H, 70,3%Cl v c PTK bng 50,5. b ) Mt hp cht rn mu trng, thnh phn t c 4o% C, 6,7%H, 53,3% O v c PTK bng 180.4:Mui n gm 2 nguyn t ho hc l Na v Cl Trong Na chim 39,3% theo khi lng . Hy tm cng thc ho hc ca mui n, bit phn t khi ca n gp 29,25 ln PTK H2.5: Xac nh cong thc cua cac hp chat sau:a) Hp chat tao thanh bi magie va oxi co phan t khoi la 40, trong o phan tram ve khoi lng cua chung lan lt la 60% va 40%.b) Hp chat tao thanh bi lu huynh va oxi co phan t khoi la 64, trong o phan tram ve khoi lng cua oxi la 50%.c) Hp chat cua ong, lu huynh va oxi co phan t khoi la 160, co phan tram cua ong va lu huynh lan lt la 40% va 20%.d) Hp chat tao thanh bi sat va oxi co khoi lng phan t la 160, trong o phan tram ve khoi lng cua oxi la 70%.e) Hp chat cua ong va oxi co phan t khoi la 114, phan tram ve khoi lng cua ong la 88,89%.f) Hp chat cua canxi va cacbon co phan t khoi la 64, phan tram ve khoi lng cua cacbon la 37,5%.g) A co khoi lng mol phan t la 58,5g; thanh phan % ve khoi lng nguyen to: 60,68% Cl con lai la Na.h) B co khoi lng mol phan t la 106g; thanh phan % ve khoi lng cua cac nguyen to: 43,4% Na; 11,3% C con lai la cua O.i) C co khoi lng mol phan t la 101g; thanh phan phan tram ve khoi lng cac nguyen to: 38,61% K; 13,86% N con lai la O.j) D co khoi lng mol phan t la 126g; thanh phan % ve khoi lng cua cac nguyen to: 36,508% Na; 25,4% S con lai la O.k) E co 24,68% K; 34,81% Mn; 40,51%O. E nang hn NaNO3 1,86 lan.l) F cha 5,88% ve khoi lng la H con lai la cua S. F nang hn kh hiro 17 lan.m) G co 3,7% H; 44,44% C; 51,86% O. G co khoi lng mol phan t bang Al.n) H co 28,57% Mg; 14,285% C; 57,145% O. Khoi lng mol phan t cua H la 84g.6. Phan t canxi cacbonat co phan t khoi la 100 vC , trong o nguyen t canxi chiem 40% khoi lng, nguyen to cacbon chiem 12% khoi lng. Khoi lng con lai la oxi. Xac nh cong thc phan t cua hp chat canxi cacbonat?7. Mot hp chat co phan t khoi bang 62 vC. trong phan t cua hp chat nguyen to oxi chiem 25,8% theo khoi lng, con lai la nguyen to Na. Xac nh ve t le so nguyen t cua O va so nguyen t Na trong hp chat.8: Trong hp chat XHn co cha 17,65% la hidro. Biet hp chat nay co t khoi so vi kh Metan CH4 la 1,0625. X la nguyen to nao ?
4: Bit thnh phn phn trm v khi lng cc nguyn t m bi khng cho bit NTK,phn t khi.Cch gii: - t cng thc tng qut: AxBy
- Ta c t l khi lng cc nguyn t: =
- Rt ra t l x: y = : (ti gin) - Vit thnh CTHH.V d: Hy xc nh cng thc hp cht A bit thnh phn % v khi lng cc nguyn t l: 40%Cu. 20%S v 40% O.Gii: - t cng thc tng qut: CuxSyOz
- Rt ra t l x: y:z = : : = : : = 0.625 : 0.625 : 2.5 = 1:1:4 - Thay x = 1, y = 1, z = 4 vo CTHH CuxSyOz, vit thnh CTHH: CuSO4
* Bi tp vn dng:1: Hai nguyn t X kt hp vi 1 nguyn t oxi to ra phn t oxit . Trong phn t, nguyn t oxi chim 25,8% v khi lng .Tm nguyn t X (s: Na)2:Nung 2,45 gam mt cht ha hc A thy thot ra 672 ml kh O2 (ktc). Phn rn cn li cha 52,35% kali v 47,65% clo (v khi lng). Tm cng thc ha hc ca A.3: Hai nguyen t X ket hp vi 1 nguyen t O tao ra phan t oxit. Trong phan t, nguyen t oxi chiem 25,8% ve khoi lng. Hoi nguyen to X la nguyen to nao?4: Mot nguyen t M ket hp vi 3 nguyen t H tao thanh hp chat vi hyro. Trong phan t, khoi lng H chiem 17,65%. Hoi nguyen to M la g?5: Hai nguyen t Y ket hp vi 3 nguyen t O tao ra phan t oxit. Trong phan t, nguyen t oxi chiem 30% ve khoi lng. Hoi nguyen to X la nguyen to nao?6. Mot hp chat co thanh phan gom 2 nguyen to C va O. Thanh phan cua hp chat co 42,6% la nguyen to C, con lai la nguyen to oxi. Xac nh ve t le so nguyen t cua C va so nguyen t oxi trong hp chat.7: Lp cng thc phn t ca A .Bit em nung 4,9 gam mt mui v c A th thu c 1344 ml kh O2 ( ktc), phn cht rn cn li cha 52,35% K v 47,65% Cl. Hng dn gii:
n = = 0,06 (mol) m = 0,06 . 32 =1,92 (g)
p dng LBT khi lng ta c: m cht rn = 4,9 1,92 = 2,98 (g)
m K = =1,56 (g) n K = = 0,04 (mol)
mCl = 2,98 1,56 = 1,42 (g) n Cl = = 0,04 (mol) Gi cng thc tng qut ca B l: KxClyOz ta c:
x : y : z = 0,04 : 0,04 : 0,06 2 = 1 : 1 : 3V i vi hp cht v c ch s ca cc nguyn t l ti gin nn cng thc ho hc ca A l KClO3.
5: Bin lun gi tr khi lng mol(M) theo ha tr(x,y) tm NTK hoc PTK..bit thnh phn % v khi lng hoc t l khi lng cc nguyn t.+Trng hp cho thnh phn % v khi lng Cch gii: - t cng thc tng qut: AxBy
- Ta c t l khi lng cc nguyn t: =
Rt ra t l : = .Bin lun tm gi tr thch hp MA ,MB theo x, y Vit thnh CTHH.
V d: B l oxit ca mt kim loi R cha r ho tr. Bit thnh phn % v khi lng ca oxi trong hp cht bng % ca R trong hp cht .
Gii: Gi % R = a% % O = a%
Gi ho tr ca R l n CTTQ ca C l: R2On
Ta c: 2 : n = : R = V n l ht ca nguyn t nn n phi nguyn dng, ta c bng sau: nIIIIIIIV
R18,637,35676,4
loiloiFeloi
Vy cng thc phn t ca C l Fe2O3.+Trng hp cho t l v khi lng Cch gii: - t cng thc tng qut: AxBy - Ta c t l khi lng cc nguyn t: MA.x : MB..y = mA : mB
- Tm c t l : = .Bin lun tm gi tr thch hp MA ,MB theo x, y - Vit thnh CTHH. V d:
C l oxit ca mt kim loi M cha r ho tr. Bit t l v khi lng ca M v O bng .Gii:
Gi ho tr ca M l n CTTQ ca C l: M2On
Ta c: = = . MA = V n l ht ca nguyn t nn n phi nguyn dng, ta c bng sau: nIIIIIIIV
M18,637,35676,4
loiloiFeloi
Vy cng thc phn t ca C l Fe2O3.
* Bi tp vn dng:1. oxit cua kim loai mc hoa tr thap cha 22,56% oxi, con oxit cua kim loai o mc hoa tr cao cha 50,48%. Tnh nguyen t khoi cua kim loai o.2. Co mot hon hp gom 2 kim loai A va B co t le khoi lng nguyen t 8:9. Biet khoi lng nguyen t cua A, B eu khong qua 30 vC. Tm 2 kim loai
*Giai:Neu A : B = 8 : 9 th
Theo e : t so nguyen t khoi cua 2 kim loai la nen ( n z+ )V A, B eu co KLNT khong qua 30 vC nen : 9n 30 n 3 Ta co bang bien luan sau :n123
A81624
B91827
Suy ra hai kim loai la Mg va Al
D/ lp CTHH hp cht kh da vo t khi . Cch gii chung:
- Theo cng thc tnh t khi cc cht kh: d A/B =
- Tm khi lng mol (M) cht cn tm NTK,PTK ca cht Xc nh CTHH.V d : Cho 2 kh A v B c cng thc ln lt l NxOy v NyOx . t khi hi i vi Hyro ln lt l: d A/H2 = 22 , d B/A = 1,045. Xc nh CTHHca A v B Gii: Theo bi ra ta c:
- d NxOy/H2 = = = 22 MA = MNxOy = 2.22 = 44 14x+ 16y = 44 (1)
- d NyOx/NxOy = = = 1,045 MB = MNyOx = 44.1,045 = 45,9814y+ 16x = 45,98 (2)
gi tr tha mn k bi ton: x = 2 , y= 1 A = N2O , B = NO2* Bi tp vn dng:1. Cho 2 cht kh AOx c TP% O = 50% v BHy c TP% H = 25% . bit dAOx/BHy = 4. Xc nh CTHH ca 2 kh trn.2. Mt oxit ca Nit c cng thc NxOy. Bit khi lng ca Nit trong phn t chim 30,4%. ngoi ra c 1,15 gam oxit ny chim th tch l 0,28 lt (ktc).Xc nh CTHH ca oxit trn.3. C 3 Hyro ccbon A, B, CA: CxH2x+2B : Cx' H2x'C : Cx' H2x'- 2Bit d B/A = 1,4 ; d A/C = 0,75 . Xc nh CTHH ca A, B, C.
E/Lp cng thc ho hc hp cht da vo phng trnh phn ng ho hc:1.Dng ton c bn 1: Tm nguyn t hay hp cht ca nguyn t trong trng hp cho bit ha tr ca nguyn t, khi bi ton cho bit lng cht (hay lng hp cht ca nguyn t cn tm) v lng mt cht khc (c th cho bng gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh) trong mt phn ng ha hc.
Cch gii chung: Bi ton c dng : a M + bB cC + d D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)- t cng thc cht cho theo bi ton : - Gi a l s mol, A l NTK hay PTK ca cht cn tm. - Vit phng trnh phn ng, t s mol a vo phng trnh v tnh s mol cc cht c lin quan theo a v A.
-Lp phng trnh, gii tm khi lng mol (M(g)) cht cn tm NTK,PTK ca cht Xc nh nguyn t hay hp cht ca nguyn t cn tm. Lu : Lng cht khc trong phn ng ha hc c th cho nhng dng sau: 1.Cho dng trc tip bng : gam, mol.V d1: Cho 7,2g mt kim loi ho tr II phn ng hon ton vi dung dch HCl, thu c 0,3 mol H2 iu kin tiu chun. Xc nh tn kim loi dng. Gii: - Gi CTHH ca kim loi l : M t x l s mol , A l NTK ca kim loi dng phn ng . Ta c Phng trnh phn ng:M + 2HCl > MCl2 + H21mol 1mol x (mol) x (mol)
Suy ra ta c h s : m M = x . A = 7,2 (g)(1)nM = n H2 = x = 0,3 (mol) (2)
Th (2) vo (1) ta c A = = 24(g) NTK ca A = 24.Vy A l kim loi Mg2/ Cho dng gin tip bng : V(ktc)V d2: Cho 7,2g mt kim loi ho tr II phn ng hon ton vi dung dch HCl, thu c 6,72 lt H2 iu kin tiu chun. Xc nh tn kim loi dng. Gii
Tm : nH2 = = 0,3 (mol)
Bi ton quay v v d 1 * Cho 7,2g mt kim loi ho tr II phn ng hon ton vi dung dch HCl, thu c 0,3 mol H2 iu kin tiu chun. Xc nh tn kim loi dng. (gii nh v d 1)3/ Cho dng gin tip bng :mdd, c%V d 3: Cho 7,2g mt kim loi ho tr II phn ng hon ton 100g dung dch HCl 21,9%. Xc nh tn kim loi dng. Gii t x l s mol , A l NTK ca kim loi dng phn ng .
p dng : C % = m HCl = = = 21,9 (g)
n HCl = = = 0,6 (mol)*Tr v bi ton cho dng trc tip: Cho 7,2g mt kim loi ho tr II phn ng hon ton 0,6 mol HCl . Xc nh tn kim loi dng. Ta c Phng trnh phn ng:M + 2HCl > MCl2 + H21mol 2mol x (mol) 2x (mol) Suy ra ta c h s : m A = x . A = 7,2 (g) (1)
nHCl = 2x = 0,6 (mol) x = 0,3 (mol) (2)
Th (2) vo (1) ta c A = = 24(g) NTK ca A = 24.Vy A l kim loi Mg4/ Cho dng gin tip bng : Vdd, CMV d 4 : Cho 7,2g mt kim loi ho tr II phn ng hon ton 100 ml dung dch HCl 6 M. Xc nh tn kim loi dng. Gii
Tm n HCl = ? p dng : CM = n HCl = CM.V = 6.0,1 = 0,6 (mol) *Tr v bi ton cho dng trc tip: Cho 7,2g mt kim loi ho tr II phn ng hon ton 0,6 mol HCl. Xc nh tn kim loi dng. (Gii nh v d 3)5/ Cho dng gin tip bng : mdd, CM ,d (g/ml)V d 5 : Cho 7,2g mt kim loi ho tr II phn ng hon ton 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim loi dng. Gii
- Tm Vdd (da vo mdd, d (g/ml)): t d = Vdd H Cl = = = 100 (ml) =0,1(l)
- Tm n HCl = ? p dng : CM = n HCl = CM. V = 6. 0,1 = 0,6 (mol) *Tr v bi ton cho dng trc tip: Cho 7,2g mt kim loi ho tr II phn ng hon ton 0,6 mol HCl. Xc nh tn kim loi dng. (Gii nh v d 3)6/ Cho dng gin tip bng : Vdd, C%, d (g/ml)V d 6 : Cho 7,2g mt kim loi ho tr II phn ng hon ton 83,3 ml dung dch HCl 21,9 % ( d= 1,2 g/ml). Xc nh tn kim loi dng. Gii
- Tm m dd (da vo Vdd, d (g/ml)): t d = mdd H Cl = V.d = 83,3 . 1,2 = 100 (g) dd HCl.
p dng : C % = m HCl = = = 21,9 (g)
n HCl = = = 0,6 (mol)*Tr v bi ton cho dng trc tip: Cho 7,2g mt kim loi ho tr II phn ng hon ton 0,6 mol HCl. Xc nh tn kim loi dng. (Gii nh v d 3)
Vn dng 6 dng ton trn: Ta c th thit lp c 6 bi ton lp CTHH ca mt hp cht khi bit thnh phn nguyn t, bit ha tr vi lng HCL cho 6 dng trn. Bi 1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 0,6 mol HCl . Xc nh tn kim loi dng. Gii - Gi CTHH ca oxit l: MOt x l s mol , A l PTK ca o xt dng phn ng . Ta c Phng trnh phn ng:MO + 2HCl > MCl2 + H2O1mol 1mol x (mol) 2x (mol)
Suy ra ta c h s : m MO = x . A = 12(g) (1)
nHCl = 2x = = 0,6(mol) x= 0,6:2 = 0,3 (mol) (2)
Th (2) vo (1) ta c A = = 40(g) MM = MMO - MO = 40 16 = 24 (g)
NTK ca M = 24.Vy M l kim loi Mg CTHH ca o xt l MgOBi 2: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 21,9 g HCl . Xc nh tn kim loi dng. Bi 3: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton 100g dung dch HCl 21,9%. Xc nh tn kim loi dng. Bi 4: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton 100 ml dung dch HCl 6 M. Xc nh tn kim loi dng. Bi 5: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 6: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton 83,3 ml dung dch HCl 21,9 % ( d= 1,2 g/ml). Xc nh tn kim loi dng. 2.Dng ton c bn 2: Tm nguyn t hay hp cht ca nguyn t trong trng hp cha bit ha tr ca nguyn t, khi bi ton cho bit lng cht (hay lng hp cht ca nguyn t cn tm) v lng mt cht khc (c th cho bng gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh) trong mt phn ng ha hc,.Cch gii chung: Bi ton c dng : a M + bB cC + d D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)- t cng thc cht cho theo bi ton : - Gi a l s mol, A l NTK hay PTK, x, y.... l ha tr ca nguyn t ca chthy hp cht ca nguyn t cn tm. - Vit phng trnh phn ng, t s mol a vo phng trnh v tnh s mol cc cht c lin quan theo a v A.
-Lp phng trnh, bin lun gi tr khi lng mol (M(g)) theo ha tr (x,y) ca nguyn t cn tm ( 1 5) t NTK,PTK ca cht Xc nh nguyn t hay hp cht ca nguyn t cn tm. V d1.2: Cho 7,2g mt kim loi cha r ha tr, phn ng hon ton vi 0,6 HCl. Xc nh tn kim loi dng. Gii:- Gi CTHH kim loi l : M- Gi x l s mol, A l NTK ca kim loi M, n l ha tr ca kim loi MTa c Phng trnh phn ng:2M + 2nHCl > 2MCln + nH22(mol ) 2n(mol) x (mol) nx (mol)
Suy ra ta c h s : m M = x . A = 7,2(g) (1)
nHCl = xn = 0,6(mol) x= 0,6:n (2)
Th (2) vo (1) ta c A = = 12.n V n phi nguyn dng, ta c bng sau: nIIIIII
A122436
loiMgloi
A = 24 (g) NTK ca kim loi = 24 Kim loi l MgT ta c th thit lp c 6 bi ton (phn dng c bn 1) v 6 bi ton (phn dng c bn 2) vi lng HCL cho 6 dng trn .Bi 1.1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 0,6 mol HCl . Xc nh tn kim loi dng. Gii - Gi CTHH ca oxit l: MOt x l s mol , A l PTK ca o xt dng phn ng . Ta c Phng trnh phn ng:MO + 2HCl > MCl2 + H2O1mol 1mol x (mol) 2x (mol)
Suy ra ta c h s : m MO = x . A = 12(g) (1)
nHCl = 2x = = 0,6(mol) x= 0,6:2 = 0,3 (mol) (2)
Th (2) vo (1) ta c A = = 40(g) MM = MMO - MO = 40 16 = 24 (g)
NTK ca M = 24.Vy M l kim loi Mg CTHH ca o xt l MgOBi 2.1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 21,9 g HCl . Xc nh tn kim loi dng. Bi 3.1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 100g dung dch HCl 21,9%. Xc nh tn kim loi dng. Bi 4.1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 100 ml dung dch HCl 6 M. Xc nh tn kim loi dng. Bi 5.1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 6.1: Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 120 ml dung dch HCl 21,9 % ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 7.2: Cho 7,2g mt kim loi cha r ha tr, phn ng hon ton vi 0,6 mol HCl. Xc nh tn kim loi dng. Bi 8.2:ho 7,2g mt kim loi cha r ha tr , phn ng hon ton vi 21,9 g HCl . Xc nh tn kim loi dng. Bi 9.2: Cho 7,2g mt kim loi cha r ha tr , phn ng hon ton vi 100g dung dch HCl 21,9%. Xc nh tn kim loi dng. Bi 10.2: Cho 7,2g mt kim loi cha r ha tr , phn ng hon ton vi 100 ml dung dch HCl 6 M. Xc nh tn kim loi dng. Bi 11.2: Cho 7,2g mt kim loi cha r ha tr , phn ng hon ton vi 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 12.2: Cho 7,2g mt kim loi cha r ha tr ,phn ng hon ton vi 83,3 ml dung dch HCl 21,9 % ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 13: Cho 7,22 gam hon hp X gom Fe va kim loai M co hoa tr khong oi. Chia hon hp thanh 2 phan bang nhau. Hoa tan het phan 1 trong dung dch HCl, c 2,128 lt H2. Hoa tan het phan 2 trong dung dch HNO3, c 1,792 lt kh NO duy nhat.Xac nh kim loai M va % khoi lng moi kim loai trong hon hp X.ap so: M (Al) va %Fe = 77,56% ; %Al = 22,44%Bi 14: Kh 3,48 gam mt oxit kim loi M cn dng 1,344 lt kh hiro ( ktc). Ton b lng kim loi thu c tc dng vi dung dch HCl d cho 1,008 lt kh hiro ktc.Tm kim loi M v oxit ca n .(CTHH oxit: Fe3O4)Mt s dng bi ton bin lun v lp CTHH (Dnh cho HSG K9)DNG: BIN LUN THEO N S TRONG GII PHNG TRNHBi 1:Ha tan mt kim loi cha bit ha tr trong 500ml dd HCl th thy thot ra 11,2 dm3 H2 ( KTC). Phi trung ha axit d bng 100ml dd Ca(OH)2 1M. Sau c cn dung dch thu c th thy cn li 55,6 gam mui khan. Tm nng M ca dung dch axit dng; xc nh tn ca kim loi dng.Gii : Gi s kim loi l R c ha tr l x 1 x, nguyn 3 s mol Ca(OH)2 = 0,1 1 = 0,1 mols mol H2 = 11,2 : 22,4 = 0,5 molCc PTP: 2R+2xHCl 2RClx +xH2 (1)1/x (mol)11/x0,5Ca(OH)2 +2HCl CaCl2+2H2O (2)0,10,20,1t cc phng trnh phn ng (1) v (2) suy ra:nHCl = 1 + 0,2 = 1,2 mol nng M ca dung dch HCl :CM = 1,2 : 0,5 = 2,4 M
theo cc PTP ta c :
ta c : ( R + 35,5x ) = 44,5 R= 9x x123
R9 1827
Vy kim loi tho mn u bi l nhm Al ( 27, ha tr III )Bi2:Khi lm ngui 1026,4 gam dung dch bo ha R2SO4.nH2O ( trong R l kim loi kim v n nguyn, tha iu kin 7< n < 12 ) t 800C xung 100C th c 395,4 gam tinh th R2SO4.nH2O tch ra khi dung dch.Tm cng thc phn t ca Hirat ni trn. Bit tan ca R2SO4 800C v 100C ln lt l 28,3 gam v 9 gam.Gii:S( 800C) = 28,3 gam trong 128,3 gam ddbh c 28,3g R2SO4 v 100g H2OVy :1026,4gam ddbh 226,4 g R2SO4 v 800 gam H2O.Khi lng dung dch bo ho ti thi im 100C:1026,4 395,4 = 631 gam 100C, S(R2SO4 ) = 9 gam, nn suy ra: 109 gam ddbh c cha 9 gam R2SO4
vy 631 gam ddbh c khi lng R2SO4 l : khi lng R2SO4 khan c trong phn hirat b tch ra :226,4 52,1 = 174,3 gam
V s mol hirat = s mol mui khan nn : 442,2R-3137,4x +21206,4 = 0 R = 7,1n 48 cho R l kim loi kim , 7 < n < 12 , n nguyn ta c bng bin lun:n891011
R8,818,62330,1
Kt qu ph hp l n = 10 , kim loi l Na cng thc hirat l Na2SO4.10H2O
DNG : BIN LUN THEO TRNG HP Bi1:Hn hp A gm CuO v mt oxit ca kim loi ha tr II( khng i ) c t l mol 1: 2. Cho kh H2 d i qua 2,4 gam hn hp A nung nng th thu c hn hp rn B. ha tan ht rn B cn dng ng 80 ml dung dch HNO3 1,25M v thu c kh NO duy nht.Xc nh cng thc ha hc ca oxit kim loi. Bit rng cc phn ng xy ra hon ton. Gii: t CTTQ ca oxit kim loi l RO.Gi a, 2a ln lt l s mol CuO v RO c trong 2,4 gam hn hp AV H2 ch kh c nhng oxit kim loi ng sau Al trong dy BKTp nn c 2 kh nng xy ra:- R l kim loi ng sau Al :Cc PTP xy ra:CuO+H2 Cu + H2O a aRO+H2 R + H2O2a 2a3Cu+8HNO3 3Cu(NO3)2+ 2NO +4H2O
a3R+8HNO3 3R(NO3)2+ 2NO +4H2O
2a
Theo bi:Khng nhn Ca v kt qu tri vi gi thit R ng sau Al- Vy R phi l kim loi ng trc Al CuO+H2 Cu + H2O a a3Cu+8HNO3 3Cu(NO3)2+ 2NO +4H2O
aRO+2HNO3 R(NO3)2+2H2O 2a4a
Theo bi : Trng hp ny tho mn vi gi thit nn oxit l: MgO.Bi2: Khi cho a (mol ) mt kim loi R tan va ht trong dung dch cha a (mol ) H2SO4 th thu c 1,56 gam mui v mt kh A. Hp th hon ton kh A vo trong 45ml dd NaOH 0,2M th thy to thnh 0,608 gam mui. Hy xc nh kim loi dng. Gii:Gi n l ha tr ca kim loi R .V cha r nng ca H2SO4 nn c th xy ra 3 phn ng:2R+nH2SO4 R2 (SO4 )n +nH2 (1) 2R+2nH2SO4 R2 (SO4 )n +nSO2 + 2nH2O(2)2R+5nH2SO4 4R2 (SO4 )n +nH2S + 4nH2O(3)kh A tc dng c vi NaOH nn khng th l H2 P (1) khng ph hp.V s mol R = s mol H2SO4 = a , nn :Nu xy ra ( 2) th : 2n = 2 n =1 ( hp l )
Nu xy ra ( 3) th : 5n = 2 n = ( v l )Vy kim loi R ha tr I v kh A l SO22R+2H2SO4 R2 SO4 +SO2 + 2H2O
a(mol)aGi s SO2 tc dng vi NaOH to ra 2 mui NaHSO3 , Na2SO3SO2+NaOH NaHSO3 t : x (mol) xxSO2+2NaOH Na2SO3 +H2Oy (mol) 2yy
theo ta c : gii h phng trnh c Vy gi thit phn ng to 2 mui l ng.Ta c: s mol R2SO4 = s mol SO2 = x+y = 0,005 (mol)Khi lng ca R2SO4 : (2R+ 96)0,005 = 1,56 R = 108 .Vy kim loi dng l Ag.
DNG:BIN LUN SO SNHBi 1:C mt hn hp gm 2 kim loi A v B c t l khi lng nguyn t 8:9. Bit khi lng nguyn t ca A, B u khng qu 30 vC. Tm 2 kim loi
Gii: Theo : t s nguyn t khi ca 2 kim loi l nn ( n z+ )V A, B u c KLNT khng qu 30 vC nn : 9n 30 n 3 Ta c bng bin lun sau :n123
A81624
B91827
Suy ra hai kim loi l Mg v Al
Bi 2:Ha tan 8,7 gam mt hn hp gm K v mt kim loi M thuc phn nhm chnh nhm II trong dung dch HCl d th thy c 5,6 dm3 H2 ( KTC). Ha tan ring 9 gam kim loi M trong dung dch HCl d th th tch kh H2 sinh ra cha n 11 lt ( KTC). Hy xc nh kim loi M.Gii:t a, b ln lt l s mol ca mi kim loi K, M trong hn hpTh nghim 1:2K+2HCl2KCl+H2 aa/2M+2HClMCl2 + H2 bb
s mol H2 = Th nghim 2:M+2HCl MCl2 + H2 9/M(mol) 9/M
Theo bi: M > 18,3 (1)
Mt khc: b =
V 0 < b < 0,25 nn suy ra ta c : < 0,25 M < 34,8 (2)T (1) v ( 2) ta suy ra kim loi ph hp l Mg
DNGBIN LUN THEO TR S TRUNG BNH( Phng php khi lng mol trung bnh)Bi 1:Cho 8 gam hn hp gm 2 hyroxit ca 2 kim loi kim lin tip vo H2O th c 100 ml dung dch X. Trung ha 10 ml dung dch X trong CH3COOH v c cn dung dch th thu c 1,47 gam mui khan. 90ml dung dch cn li cho tc dng vi dung dch FeClx d th thy to thnh 6,48 gam kt ta.Xc nh 2 kim loi kim v cng thc ca mui st clorua. Gii:t cng thc tng qut ca hn hp hiroxit l ROH, s mol l a (mol)Th nghim 1:
mhh = = 0,8 gam ROH+CH3COOH CH3COOR+H2O(1)1 mol1 mol
suy ra : 33vy c 1kim loi A > 33 v mt kim loi B < 33V 2 kim loi kim lin tip nn kim loi l Na, KC th xc nh tng khi lng (1) : m = 1,47 0,8=0,67 gam
nROH = 0,67: ( 59 17 ) =
ROH = = 50 17 = 33 Th nghim 2:mhh = 8 - 0,8 = 7,2 gamxROH + FeClx Fe(OH)x +xRCl (2)
(+17)x (56+ 17x) 7,2 (g) 6,48 (g)
suy ra ta c: gii ra c x = 2Vy cng thc ha hc ca mui st clorua l FeCl2Bi2: X l hn hp 3,82 gam gm A2SO4 v BSO4 bit khi lng nguyn t ca B hn khi lng nguyn t ca A l1 vC. Cho hn hp vo dung dch BaCl2 va ,thu c 6,99 gam kt ta v mt dung dch Y. a) C cn dung dch Y th thu c bao nhiu gam mui khanb) Xc nh cc kim loi A v B Gii:a)A2SO4 +BaCl2 BaSO4 +2AClBSO4+BaCl2 BaSO4 +BCl2Theo cc PTP :
S mol X = s mol BaCl2 = s mol BaSO4 = Theo nh lut bo ton khi lng ta c:
3,82 + (0,03. 208) 6.99 = 3,07 gam
b) Ta c M1 = 2A + 96 v M2 = A+ 97
Vy : (*)T h bt ng thc ( *) ta tm c :15,5 < A < 30Kim loi ha tr I tho mn iu kin trn l Na (23)Suy ra kim loi ha tr II l Mg ( 24)
* Bi tp vn dng:1.Hoa tan hoan toan 3,78 gam mot kim loai M vao dung dch HCl thu c 4,704 lt kh H2 (ktc) . Xac nh kim loai M ?2. Kh hon ton 16g bt oxit st nguyn cht bng CO nhit cao .Sau phn ng kt thc khi lng cht rn gim 4,8g.Xc nh cng thc ca oxit st dng. 3.Kh hon ton 23,2g mt oxit ca st (cha r ho tr ca st )bng kh CO nhit cao. Sau phn ng thy khi lng cht rn gim i 6,4g so vi ban u . Xc nh cng thc ca oxit st4.C mt oxt st cha r cng thc , chia oxits ny lm 2 phn bng nhau : - ho tan ht phn 1 phi cn 0,225 mol HCl . - Cho mt lung kh H2 d i qua phn 2 nung nng, phn ng xong thu c 4,2g Fe .Tm cng thc ca oxit ni trn5. Cho 4,48g mt oxt kim loi ho tr tc dng ht vi 7,84g axitsunfuric. xc nh cng thc oxt kim loi .6. Cho 16 gam FexOy tc dng vi lng va 0,6 mol HCl. Xc nh CT oxit st 7: Co 1 oxit sat cha biet.- Hoa tan m gam oxit can 0,45 mol HCl .- Kh toan bo m gam oxit bang CO nong, d thu c 8,4 gam sat. Tm cong thc oxit.8: Kh hoan toan 4,06g mot oxit kim loai bang CO nhiet o cao thanh kim loai. Dan toan bo kh sinh ra vao bnh ng Ca(OH)2 d, thay tao thanh 7g ket tua. Neu lay lng kim loai sinh ra hoa tan het vao dung dch HCl d th thu c 1,176 lt kh H2 (ktc). Xac nh cong thc phan t oxit kim loai.9.Hoa tan hoan toan 3,6 gam mot kim loai hoa tr II bang dung dch HCl co 3,36 lt kh H2 thoat ra ktc. Hoi o la kim loai nao ?10. Hoa tan 2,4 gam oxit cua mot kim loai hoa tr II can dung 2,19 gam HCl. Hoi o la oxit cua kim loai nao ?11.Cho 10,8 gam kim loai hoa tri III tac dung vi dung dch HCl d thay tao thanh 53,4 gam muoi . Xac nh ten kim loai o.12. A la oxit cua nit co phan t khoi la 92 co t le so nguyen t N va O la 1 : 2. B la mot oxit khac cua nit. ktc 1 lt kh B nang bang 1 lt kh CO2 . Tm cong thc phan t cua A va B ?13.Hoa tan hoan toan 1,44 gam kim loai hoa tr II bang 7.35g H2SO4. e trung hoa lng axit d can dung 0.03 mol NaOH, Xac nh ten kim loai ? (bi t H2SO4 + NaOH Na2SO4 + H2O )14.Xac nh cong thc phan t cua A, biet rang khi ot chay 1 mol chat A can 6,5 mol oxi thu c 4 mol CO2 va 5 mol nc .15. ot chay m gam chat A can dung 4,48 lt O2 thu c 2,24 lt CO2 va 3,6 gam nc . Tnh m biet the tch cac chat kh eu dc o ktc .16. ot chay 16 gam chat A can 4,48 lt kh oxi (ktc) thu c kh CO2 va hi nc theo t le so mol la 1 : 2 . Tnh khoi lng CO2 va H2O tao thanh ?17.Hoa tan hoan toan 3,78 gam mot kim loai M vao dung dch HCl thu c 4,704 lt kh H2 (ktc) . Xac nh kim loai M ?18.Hoa tan hoan toan hon hp 4 g hai kim loai A, B cung hoa tr II va co t le mol la ! : 1 bang dung dch HCl thu c 2,24 lt kh H2 ( ktc). Hoi A, B la cac kim loai nao trong cac kim loai sau : Mg , Ca , Ba , Zn , Fe , Ni . (Biet : Mg = 24 , Ca= 40 , Ba= 137 , Zn = 65, Fe = 56 , Ni = 58). 19.Nguyen t khoi cua 3 kim loai hoa tr 2 t le vi nhau theo t so la 3 : 5 : 7 . T le so mol cua chung trong hon hp la 4 : 2 : 1 . Sau khi hoa tan 2,32 gam hon hp trong HCl d thu c 1,568 lt H2 ktc . Xac nh 3 kim loai biet chung eu ng trc H2 trong day Beketop (u phn ng c vi HCl ). 20. Kh 3,48 gam mt oxit kim loi M cn dng 1,344 lt kh hiro ktc. Ton b lng kim loi thu c tc dng vi dung dch HCl d cho 1,008 lt kh hiro ktc.Tm kim loi M v oxit ca n .21. Mot hon hp kim loai X gom 2 kim loai Y, Z co t so khoi lng 1 : 1. Trong 44,8g hon hp X, so hieu mol cua A va B la 0,05 mol. Mat khac nguyen t khoi Y > Z la 8. Xac nh kim loai Y va Z.
Chuyn III. Bi tp v phng trnh ha hc ha hc a.Lp phng trnh ha hc:Cch gii chung: - Vit s ca phn ng (gm CTHH ca cc cht p v sn phm).- Cn bng s nguyn t ca mi nguyn t (bng cch chn cc h s thch hp in vo trc cc CTHH).- Vit PTHH.Lu : Khi chn h s cn bng:+ Khi gp nhm nguyn t -> Cn bng nguyn c nhm.+ Thng cn bng nguyn t c s nguyn t l cao nht bng cch nhn cho 2,4+ Mt nguyn t thay i s nguyn t 2 v PT, ta chn h s bng cch ly BSCNN ca 2 s trn chia cho s nguyn t ca nguyn t .V d: ?K+ ?O2-> ?K2OGii: 4K+O2-> 2K2O+ Khi gp mt s phng trnh phc tp cn phi dng phng php cn bng theo phng php i s:V d 1: Cn bng PTHH sau : FeS2+O2-> Fe2O3+SO2 Gii: - t cc h s: aFeS2+bO2-> cFe2O3+dSO2 - Tnh s nguyn t cc nguyn t trc v sau phn ng theo cc h s trong PTHH: Ta c: + S nguyn t Fe: a = 2c + S nguyn t S : 2a = d + S nguyn t O : 2b = 3c + 2dt a = 1 c = 1/2, d = 2, b = 3/2 + 2.2 = 11/2Thay a, b, c, d vo PT: aFeS2 +bO2-> cFe2O3 +dSO2 FeS2 +11/2O2-> 1/2Fe2O3 + 2SO2 Hay: 2FeS2 +11O2 -> Fe2O3 + 4SO2 V d 2 Cn bng PTHH sau: FexOy + H2 Fe + H2O Gii: - t cc h s: a FexOy + b H2 c Fe + d H2O - Tnh s nguyn t cc nguyn t trc v sau phn ng theo cc h s trong PTHH: Ta c: + S nguyn t Fe: a.x = c + S nguyn t O : a.y = d + S nguyn t H : 2b = 2dt a = 1 c = x, d = b = yThay a, b, c, d vo PT: FexOy + y H2 x Fe + y H2O * Bi tp vn dng:1: Hay chon CTHH va he so thch hp at vao nhng cho co dau hoi trong cac PTP sau e c PTP ung : a/ ?Na + ? 2Na2O b/ 2HgO t0 ? Hg + ? c/ ? H2 + ? t0 2H2O d/ 2Al + 6HCl ?AlCl3 + ?2: Hoan thanh cacs o PHH sau e c PTHH ung : a/ CaCO3 + HCl ------> CaCl2 + CO2 + H2 b/ C2H2 + O2 ---------> CO2 + H2O c/ Al + H2SO4 --------> Al2(SO4)3 + H2 d/ KHCO3 + Ba(OH)2 ------->BaCO3 + K2CO3 + H2O e/ NaHS + KOH ------> Na2S + K2S + H2O f/ Fe(OH)2 + O2 + H2O ------> Fe(OH)33: ot chay kh axetylen (C2H2) trong kh oxi sinh ra kh cacbonic va hi nc .Dan hon hp kh vao dung dch nc voi trong ( Ca(OH)2) th thu c chat ket tua canxicacbonat (CaCO3) .Viet cac PTP xay ra . 4: Hon thnh cc PTHH cho cc p sau:Na2O +H2O-> NaOH.BaO+H2O ->Ba(OH)2CO2+H2O-> H2CO3N2O5+H2O -> HNO3P2O5+H2O->H3PO4NO2+O2+H2O-> HNO3SO2+Br2+H2O-> H2SO4+HBrK2O+P2O5-> K3PO4Na2O+N2O5-> NaNO3Fe2O3 + H2SO4-> Fe2(SO4)3+H2OFe3O4+HCl-> FeCl2+FeCl3+H2OKOH +FeSO4-> Fe(OH)2+ K2SO4Fe(OH)2+O2-> Fe2O3+H2O.KNO3 -> KNO2+O2AgNO3 -> Ag+O2+NO2Fe+Cl2-> FeClnFeS2+O2-> Fe2O3+SO2FeS+O2-> Fe2O3+SO2FexOy+O2-> Fe2O3Cu+O2+HCl-> CuCl2+H2OFe3O4+C-> Fe+CO2Fe2O3+H2-> Fe+H2O. FexOy+Al-> Fe+Al2O3Fe+Cl2->FeCl3CO+O2-> CO25. Hon thnh cc phng trnh ha hc sau: FexOy + H2SO4 Fe 2(SO4) 2y / x + H2OFexOy + H2 Fe + H2OAl(NO3)3 Al2O3 + NO2 + O2KMnO4 + HCl Cl2 + KCl + MnCl2 + H2OFe 3O4 + Al Fe + Al2O3FeS2 + O2 ----> Fe2O3 + SO2KOH + Al2(SO4)3 ----> K2SO4 + Al(OH)3FeO + HNO3 ----> Fe(NO3)3 + NO + H2OFexOy + CO ----> FeO + CO26. Hon thnh chui bin ho sau: P2O5 H3PO4 H2
KClO3 O2Na2ONaOH
H2OH2H2OKOH7: Hon thnh s chuyn ho sau (ghi r iu kin phn ng) v cho bit cc phn ng trn thuc loi no?.1
KMnO4 7 KOH4653
H2O O2 Fe3O4 Fe H2 H2O 8 H2SO4
KClO32
B: Tnh theo phng trnh ha hcCch gii chung: - Vit v cn bng PTHH.- Tnh s mol ca cht bi cho.- Da vo PTHH, tm s mol cc cht m bi yu cu.- Tnh ton theo yu cu ca bi (khi lng, th tch cht kh)1.Dng ton c bn : Cho bit lng mt cht (c th cho bng gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh), tm lng cc cht cn li trong mt phn ng ha hc.Cch gii : Bi ton c dng : a M + b B c C + d D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)- Tnh s mol ca cht bi cho.- Da vo PTHH, tm s mol cc cht m bi yu cu.- Tnh ton theo yu cu ca bi * Trng hp 1: Cho dng trc tip bng : gam, mol.V d1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi dng. Gii: Ta c Phng trnh phn ng:Mg + 2HCl > MgCl2 + H21mol 2mol x (mol) 0,6 (mol) x = 0,6. 1 / 2 = 0,3 (mol) mMg = n.M = 0,3. 24 = 7,2 (g)
*Trng hp 2: Cho dng gin tip bng : V(ktc)V d2: Cho kim loi Mg phn ng hon ton vi dung dch HCl. thu c 6,72 lt kh (ktc) . Xc nh khi lng kim loi dng. Gii
Tm : nH2 = = 0,3 (mol) Ta c Phng trnh phn ng:Mg + 2HCl > MgCl2 + H21mol 1mol x (mol) 0,3 (mol) x = 0,3. 1 / 1 = 0,3 (mol) mMg = n.M = 0,3. 24 = 7,2 (g) *Trng hp 3: Cho dng gin tip bng : mdd, c%V d 3: Cho kim loi Mg phn ng hon ton vi 100g dung dch HCl 21,9%. Xc nh khi lng kim loi dng. Gii Ta phi tm n HCl phn ng ?
p dng : C % = m HCl = = = 21,9 (g)
n HCl = = = 0,6 (mol)*Tr v bi ton 1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi dng. (Gii nh v d 1)*Trng hp 4: Cho dng gin tip bng : Vdd, CMV d 4 : Cho kim loi Mg phn ng hon ton vi 100 ml dung dch HCl 6 M. Xc nh khi lng kim loi dng.
Gii: Tm n HCl = ? p dng : CM = n HCl = CM.V = 6.0,1 = 0,6 (mol)*Tr v bi ton 1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi dng.(Gii nh v d 1)*Trng hp 5: Cho dng gin tip bng : mdd, CM ,d (g/ml)V d 5 : Cho kim loi Mg phn ng hon ton vi 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh khi lng kim loi dng. Gii: Tm n HCl = ?
- Tm Vdd (da vo mdd, d (g/ml)): t d = Vdd H Cl = = = 100 (ml) =0,1(l)
- Tm n HCl = ? p dng : CM = n HCl = CM. V = 6. 0,1 = 0,6 (mol) *Tr v bi ton 1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi dng.(Gii nh v d 1)*Trng hp 6: Cho dng gin tip bng : Vdd, C%, d (g/ml)V d 6 : Cho kim loi Mg phn ng hon ton vi 83,3 ml dung dch HCl 21,9 % ( d= 1,2 g/ml). Xc nh khi lng kim loi dng. Gii: Tm n HCl = ?
- Tm m dd (da vo Vdd, d (g/ml)): t d = mdd H Cl = V.d = 83,3 . 1,2 = 100 (g) dd HCl.
p dng : C % = m HCl = = = 21,9 (g)
n HCl = = = 0,6 (mol)*Tr v bi ton 1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi dng.(Gii nh v d 1)Vn dng 6 dng ton trn:Ta c th thit lp c 9 bi ton tm cc i lng lin quan n nng dung dch( C%, CM., mdd, Vdd, khi lng ring ca dd(d(g/ml)) ca cht phn ng).1. Cho 7,2 g kim loi Mg phn ng hon ton vi 100g dung dch HCl . Xc nh nng % dd HCl cn dng. 2. Cho 7,2 g kim loi Mg phn ng hon ton vi dung dch HCl 21,9% . Xc nh khi lng dd HCl cn dng. 3: Cho 7,2 g kim loi Mg phn ng hon ton vi 100 ml dung dch HCl .Xc nh nng Mol/ lt dd HCl cn dng. 4. Cho 7,2 g kim loi Mg phn ng hon ton vi dung dch HCl 6M .Xc nh th tch dd HCl cn dng. 5. Cho 7,2 g kim loi Mg phn ng hon ton vi dung dch HCl 6 M ( d = 1,2 g/ml). Xc nh khi lng dd HCl cn dng. 6. Cho 7,2 g kim loi Mg phn ng hon ton vi 120g dung dch HCl ( d = 1,2 g/ml). Xc nh nng Mol/lt dd HCl cn dng. 7. Cho 7,2 g kim loi Mg phn ng hon ton vi dung dch HCl 21,9%( d = 1,2 g/ml). Xc nh th tch dd HCl cn dng. 8. Cho 7,2 g kim loi Mg phn ng hon ton vi 120 g dung dch HCl 6 M . Xc nh khi lng ring dd HCl cn dng. 9. Cho 7,2 g kim loi Mg phn ng hon ton vi 83,3 ml dung dch HCl 21,9% . Xc nh khi lng ring dd HCl cn dng. 2.Dng ton tha thiu :1. Trng hp ch c 2 cht phn ng : PTHH c dng : a M + b B c C + d D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)* Cho bit lng 2 cht trong phn ng (c th cho bng gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh), tm lng cc cht cn li trong mt phn ng ha hc.Cch gii chung : - Vit v cn bng PTHH:- Tnh s mol ca cht bi cho.- Xc nh lng cht no phn ng ht, cht no d bng cch:- Lp t s : S mol cht A bi cho(>; =; T s ca cht no ln hn -> cht d; t s ca cht no nh hn, cht p ht.- Da vo PTHH, tm s mol cc cht sn phm theo cht p ht.- Tnh ton theo yu cu ca bi (khi lng, th tch cht kh)
V d: Khi t, than chy theo s sau : Cacbon + oxi kh cacbon ioxita) Vit v cn bng phng trnh phn ng.b) Cho bit khi lng cacbon tc dng bng 18 kg, khi lng oxi tc dng bng 24 kg. Hy tnh khi lng kh cacbon ioxit to thnh.c) Nu khi lng cacbon tc dng bng 8 kg, khi lng kh cacbonic thu c bng 22 kg, hy tnh khi lng cacbon cn d v khi lng oxi phn ng.Gii:a. PTHH: C+O2t0CO2b. S mol C: nC = 18.000 : 12 = 1500 mol.- S mol O2: nO2 = 24.000 : 32 = 750 mol.
Theo PTHH, ta c t s: = = 1500 > = = 750.=> O2 p ht, C d.- Theo pthh: nCO2 = nO2 = 750 mol.- Vy khi lng CO2 to thnh: mCO2 = 750. 44 = 33.000gam = 33kg.c. S mol CO2: nCO2 = 22.000 : 44 = 500 mol.- Theo PTHH: nC = nO2 = nCO2 = 500 mol.- Khi lng C tham gia p: mC = 500. 12 = 6.000g = 6kg.=> Khi lng C cn d: 8 6 = 2kg.- Khi lng O2 tham gia p: mO2 = 500 . 32 = 16000g = 16kg.* Bi tp vn dng: 1: Cho 22,4g Fe tc dng vi dd long c cha 24,5g axit sulfuric.a. Tnh s mol mi cht ban u v cho bit cht d trong p?b. Tnh khi lng cht cn d sau p?c. Tnh th tch kh hidro thu c ktc?d. Tnh khi lng mui thu c sau p2: Cho dd cha 58,8g H2SO4 tc dng vi 61,2g Al2O3.a. Tnh s mol mi cht ban u ca hai cht p?b. Sau p cht no d, d bao nhiu gam?c. Tnh khi lng mui nhm sunfat to thnh?(bit H2SO4 + Al2O3 Al2(SO4)3 + H2O )3: Dng 6,72 lt kh H2 (ktc) kh 20g St (III) oxit.a. Vit PTHH ca p?b. Tnh khi lng oxit st t thu c?4: Cho 31g Natri oxit vo 27g nc.a. Tnh khi lng NaOH thu c?b. Tnh nng % ca dd thu c sau p?5: Cho 4,05g kim loi Al vo dd H2SO4, sa p thu c 3,36 lt kh ktc.a. Tnh khi lng Al p?b. Tnh khi lng mui thu c v khi lng axit p?c. ha tan ht lng Al cn d cn phi dng them bao nhiu gam axit?6. Cho 2,8 gam st tc dng vi 14,6 gam dung dch axit clohiric HCl nguyn cht.a. Vit phng trnh phn ng xy ra.b. Cht no cn d sau phn ng v d bao nhiu gam?c. Tnh th tch kh H2 thu c (ktc)?d. Nu mun cho phn ng xy ra hon ton th phi dng thm cht kia mt lng l bao nhiu?2.Trng hp c nhiu cht phn ng : * Cho bit lng mt hn hp nhiu cht phn ng vi mt lng cht phn ng khc (c th cho bng gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh), tm lng cc cht cn li trong qu trnh phn ng ha hc.Bi ton c dng : cho hn hp A( gm M, M) phn ng vi B
chng minh hh A ht hay B ht:Cch gii chung : - Vit v cn bng PTHH:PTHH c dng : a M + b B c C + d D a M + bB c C + dD(Trong cc cht M, M, B, C, D, C, D: c th l mt n cht hay 1 hp cht)- Tnh s mol ca hn hp v s mol cc cht trong qu trnh phn ng . Bin lun lng hn hp hay lng cht phn ng vi hh theo cc d kin ca bi ton lin quan n lng hh hay cht phn ng , xc nh lng hh ht hay cht phn ng vi hh ht - Da vo PTHH, tm lng cc cht cn li theo lng cht p ht.V d: Cho 3,78 gam hon hp gom Mg va Al tac dung vi 0,5 mol HCl a. Chng minh rang sau phan ng vi Mg va Al , axit van con d ?b. Neu phan ng tren lam thoat ra 4,368 lt kh H2 (ktc) . Hay tnh so gam Mg va Al a dung ban au ?Gii: a. Ta c PTHH: 2Al + 6 HCl 2 AlCl3 + 3 H2 (1)
x (mol) 3x Mg + 2 HCl MgCl2 + H2 (2) y (mol) 2y yGi s lng hn hp ht :
Theo bi ra : 27x + 24y = 3,78 > 24 (x+y) = 0,16 > x +y (3) Theo PT (1) (2) n HCl = 3x + 2y < 3 (x +y) (4)Kt hp (3) (4) : 3x + 2y < 3 (x +y) < 3.0,16 = 0,48 Vy : n HCl phn ng = 3x + 2y < 0,48 m bi theo bi ra n HCl = 0,5 (mol) Nn lng hn hp ht, A xt cn d . b. Lng hn hp ht nn ta c PT : 27x + 24y = 3,78 (5)
Theo (1) (2) : n H2 = + y = = 0,195 (6)
Gii h phng trnh:
x = 0,06 (mol) , y = 0,09 (mol)m Al = n. M = 0,06. 27 = 1,62 (g), m Mg = n. M = 0,09. 24 = 2,16 (g),
* Bi tp vn dng:1. Cho 8,4 gam hon hp Zn va Mg tac dung vi 3,65 g HCl a. Chng minh rang sau phan ng axit van con d ?b. Neu thoat ra 4,48 lt kh (ktc) . Hay tnh so gam Mg va Al a dung ban au2. Cho 7,8 gam hon hp Mg va Al tac dung vi 0,5 mol dung dch H2SO4 a. Chng minh rang sau phan ng vi Mg va Al , axit van con d ?b. Neu phan ng tren lam thoat ra 4,368 lt kh H2 (ktc) . Hay tnh % ve khoi lng cua Mg va Al a dung ban au ?3. Ho tan hn hp gm 37,2 gam Zn v Fe trong 1 mol dung dch H2SO4 a. Chng minh rng hn hp tan ht. b. Nu ho tan hn hp trn vi lng gp i vo cng lng axit trn th hn hp c tan ht khng.4. Ho tan hn hp gm Mg v Fe trong dung dch ng 7,3 gam HCl ta thu c 0,18 gam H2. Chng minh sau phn ng vn cn d axit.5. Ngui ta tin hnh 2 th nghim sau:TN1: Cho 2,02 gam hn hp Mg, Zn vo cc ng 200ml dung dch HCl . Sau phn ng un nng cho nc bay hi ht thu c 4,86 gam cht rn.TN2: Cho 2,02 gam hn hp trn vo cc ng 400ml dung dch HCl trn. Sau khi c cn thu c 5,57 gam cht rn.a. Chng minh trong TN1 axit ht, TN2 axit d.b. Tnh th tch kh (ktc) bay ra TN1.c. Tnh s mol HCl tham gia phn ng.d. Tnh s gam mi kim loi6. Cho a gam Fe ho tan trong dung dch HCl (TN1) sau khi c cn dung dch thu c 3,1 gam cht rn. Nu cho a gam Fe v b gam Mg ( TN2) vo dung dch HCl cng vi lng trn th thu c 3,34 gam cht rn . Bit th tch H2 (ktc) thot ra c 2 TN u l 448 ml. Tnh a,b bit rng TN2 Mg hot ng mnh hn Fe. Ch khi Mg phn ng xong th Fe mi phn ng.7. Cho 22 gam hn hp X gm Al v Fe phn ng vi dung dch cha 0,6 mol HCl . Chng minh hn hp X tan ht.8. Cho 3,87 gam hn hp A gm Mg v Al vo 0,25mol HCl v 0,125 mol H2SO4 ta thu c dung dch B v 4,368 lit H2 (ktc) .a. Chng minh trong dung dch vn cn d axit.b. Tnh % cc kim loi trong A.9. Ho tan 7,8 gam hn hp gm Mg v Zn vo dung dch H2SO4. Sau phn ng thu c dung dch A v 2,24 lit kh. Chng minh sau phn ng kim loi vn cn d.10. Hoa tan 13,2 gam hon hp A gom 2 kim loai co cung hoa tr vao 0.6 mol HCl . Co can dung dch sau phan ng thu c 32,7 gam hon hp muoi khan.a. Chng minh hon hp A khong tan het.b. Tnh the tch hiro sinh ra (ktc).3. Dng Ton hn hp :
Bi ton c dng : cho m (g) hn hp A ( gm M, M) phn ng hon ton vi lng cht B Tnh thnh phn % ca hn hp hay lng sn phm.1. Trng hp trong hn hp c mt s cht khng phn ng vi cht cho: cho m (g) hn hp A(gm M, M) + ch c mt cht phn ng hon ton vi lng cht B.Cch gii chung : - Xc nh trong hn hp A (M, M) cht no phn ng vi B. vit v cn bng PTHH.- Tnh s mol cc cht trong qu trnh phn ng theo cc d kin ca bi ton lin quan n lng hh hay lng cht phn ng, xc nh lng cht no trong hn hp phn ng, lng cht khng phn ng.- Da vo PTHH, cc d kin bi ton, tm lng cc cht trong hn hp hay lng cc cht sn phm theo yu cu .V d: Cho 9,1 gam hn hp kim loi Cu v Al phn ng hon ton vi dd HCl, thu c 3,36 lt kh (ktc). Tnh TP % ca hn hp kim loi.Gii: - Cho hn hp kim loi vo HCl ch c Al phn ng theo PT: 2Al + 6 HCl 2 AlCl3 + 3 H2 (1)
x (mol) 3x
- Theo PT: n H2 = = = 0,15 (mol) x = 0,1 (mol)
m Al = n.M = 0,1. 27 = 2,7 (g) m Cu = m hh - m Al = 9,1 - 2,7 = 6,4 (g) * Bi tp vn dng:1. Cho 8 gam hon hp gom Cu va Fe tac dung vi dung dch HCl d tao thanh 1,68 lt kh H2 thoat ra ( ktc ). Tnh % ve khoi lng cua tng kim loai co trong hon hp ?2. Cho hon hp gom Ag va Al tac dung vi dung dch H2SO4 d tao thanh 6,72 lt kh H2 thoat ra ( ktc) va 4,6 g chat ran khong tan. Tnh % ve khoi lng cua tng kim loai co trong hon hp ?2.Trng hp cc cht trong hn hp u tham gia phn ngcho m (g) hn hp A ( gm M, M) + cc cht trong n hp A u phn ng hon ton vi lng cht B.Cch gii chung : - Vit v cn bng PTHH XY RA..- Tnh s mol cc cht trong qu trnh phn ng theo cc d kin ca bi ton lin quan n lng hh hay lng cht phn ng .- Da vo PTHH, cc d kin bi ton, Lp h phng trnh bc nht 1 n( hoc 2 n ). tm lng cc cht trong hn hp hay lng cc cht sn phm theo yu cu .V d. t chy 29,6 gam hn hp kim loi Cu v Fe cn 6,72 lt kh oxi iu kin tiu chun.Tnh khi lng cht rn thu c theo 2 cch.Gii: noxi = 6,72 : 22,4 = 0,3 molmoxi = 0,3 x 32 = 9,6 gamPTP : 2Cu + O2 -> 2CuO (1) x (mol) : x/2 : x 3 Fe + 2O2 -> Fe3O4 (2) y (mol) 2y/3 y/3 Cch 1: p dng LBTKL cho phn ng (1) v (2) ta c : mst + mng + moxi = m oxu = 29,6 + 9,6 = 39,2 gam Cch 2: Gi x,y l s mol ca Cu v Fe trong hn hp ban u (x,y nguyn dng)Theo bi ra ta c: 64x + 56y = 29,6 x/2 + 2y/3 = 0,3 x = 0,2; y = 0,3 khi lng oxit thu c l: 80x + (232y:3 ) = 80 . 0,2 + 232 . 0,1 = 39,2 gam * Bi tp vn dng:1. Kh 15,2 gam hon hp gom Fe2O3 va FeO bang H2 nhiet o cao thu c sat kim loai . e hoa tan het lng sat nay can 0,4 mol HCl.a.Tnh % ve khoi lng cua moi oxit co trong hon hp ban au ?b.Tnh the tch H2 thu c ( ktc)?2. Cho 19,46 gam hon hp gom Mg , Al va Zn trong o khoi lng cua Magie bang khoi lng cua nhom tac dung vi dung dch HCl tao thanh 16, 352 lt kh H2 thoat ra ( ktc ) . Tnh % ve khoi lng cua tng kim loai co trong hon hp ?3. Kh 15,2 gam hon hp gom Fe2O3 va FeO bang H2 nhiet o cao thu c sat kim loai . e hoa tan het lng sat nay can 0,4 mol HCl .a.Tnh % ve khoi lng cua moi oxit co trong hon hp ban au ?b.Tnh the tch H2 thu c ktc ?4. Cho mot luong CO d i qua ong s cha 15,3 gam hon hp gom FeO va ZnO nung nong , thu c mot hon hp chat ran co khoi lng 12, 74 gam . Biet trong ieu kien th nghiem hieu suat cac phan ng eu at 80% .a.Tnh % ve khoi lng cua moi oxit co trong hon hp ban au ?b.e hoa tan hoan toan lng chat ran thu c sau phan ng tren phai dung bao nhieu lt dung dch HCl 2M ?5. Cho luong kh CO i qua ong s ng m gam hon hp gom Fe , FeO , Fe2O3 nung nong . Sau khi ket thuc th nghiem , thu c 64 gam chat ran A va 11,2 lt kh B (ktc) co t khoi hi so vi hiro la 20,4. Tnh m ?6. Cho 11 gam hon hp gom Al va Fe tac dung va u vi dung dch HCl 2M tao thanh 8,96 lt kh H2 thoat ra ktc .a.Tnh % ve khoi lng cua tng kim loai co trong hon hp ?b. Tnh the tch dung dch HCl a tham gia phan ng ?7. Cho 8,8 gam hon hp gom Mg va MgO tac dung va u vi dung dch HCl 14,6% .Co can dung dch sau phan ng thu c 28,5 gam muoi khan.a. Tnh % ve khoi lng cua tng chat co trong hon hp ?b. Tnh khoi lng dung dch HCl a tham gia phan ng ?c. Tnh nong o phan tram cua muoi tao thanh sau phan ng ?8. Cho mot luong CO d i qua ong s cha 15,3 gam hon hp gom FeO va ZnO nung nong , thu c mot hon hp chat ran co khoi lng 12, 74 gam . Biet trong ieu kien th nghiem hieu suat cac phan ng eu at 80%.a. Tnh % ve khoi lng cua moi oxit co trong hon hp ban au ?b. e hoa tan hoan toan lng chat ran thu c sau phan ng tren phai dung bao nhieu lt dung dch HCl 2M ?9. Chia hon hp gom Fe va Fe2O3 lam 2 phan bang nhau.Phan 1 : cho mot luong CO i qua va nung nong thu c 11,2 gam Fe.Phan 2 : ngam trong dung dch HCl . Sau phan ng thu c 2,24 lt kh H2 ktcTnh % ve khoi lng cua moi chat co trong hon hp ban au ?10. Cho 46,1 (g) hon hp Mg, Fe, Zn phan ng vi dung dch HCl th thu c 17,92 lt H2 (ktc). Tnh thanh phan phan tram ve khoi lng cac kim loai trong hon hp. Biet rang the tch kh H2 do sat tao ra gap oi the tch H2 do Mg tao ra.
4. dng ton Tng gim khi lngTrng hp1: Kim loai phan ng vi muoi cua kim loai yeu hn.Cch gii chung : - Goi x (g) la khoi lng cua kim loai manh.- Lap phng trnh hoa hoc.- Da vao d kien e bai va PTHH e tm lng kim loai tham gia.- T o suy ra lng cac chat khac. Lu y: Khi cho mieng kim loai vao dung dch muoi, Sau phan ng thanh kim loai tang hay giam:
- Neu thanh kim loai tang:
- Neu khoi lng thanh kim loai giam:
- Neu e bai cho khoi lng thanh kim loai tang a% hay giam b% th nen at thanh kim loai ban au la m gam. Vay khoi lng thanh kim loai tang a% m hay b% m .* Bi tp vn dng:1: Cho mot la ong co khoi lng la 6 gam vao dung dch AgNO3. Phan ng xong, em la kim loai ra ra nhe, lam kho can c 13,6 gam. Tnh khoi lng ong a phan ng.2. Ngam mot mieng sat vao 320 gam dung dch CuSO4 10%. Sau khi tat ca ong b ay ra khoi dung dch CuSO4 va bam het vao mieng sat, th khoi lng mieng sat tang len 8%. Xac nh khoi lng mieng sat ban au.3.Nhung thanh sat co khoi lng 50 gam vao 400ml dung dch CuSO4. Sau mot thi gian khoi lng thanh sat tang 4%. a. Xac nh lng Cu thoat ra. Gia s ong thoat ra eu bam vao thanh sat.b. Tnh nong o mol/l cua dung dch sat(II) sunfat tao thanh. Gia s the tch dung dch khong thay oi.4. Hai thanh kim loai giong nhau (eu tao bi cung nguyen to R hoa tr II) va co cung khoi lng. Tha thanh th nhat vao dung dch Cu(NO3)2 va thanh thu hai vao dung dch Pb(NO3)2. Sau mot thi gian, khi so mol 2 muoi phan ng bang nhau lay 2 thanh kim loai o ra khoi dung dch thay khoi lng thanh th nhat giam i 0,2%, con khoi lng thanh th hai tang them 28,4%. Tm nguyen to R.5: Cho mot thanh Pb kim loai tac dung va u vi dung dch muoi nitrat cua kim loai hoa tr II, sau mot thi gian khi khoi lng thanh Pb khong oi th lay ra khoi dung dch thay khoi lng no giam i 14,3 gam. Cho thanh sat co khoi lng 50 gam vao dung dch sau phan ng tren, khoi lng thanh sat khong oi na th lay ra khoi dung dch, ra sach, say kho can nang 65,1 gam. Tm ten kim loai hoa tr II.6. Nhung mot thoi sat 100 gam vao dung dch CuSO4 . Sau mot thi gian lay ra ra sach , say kho can nang 101,6 gam . Hoi khoi kim loai o co bao nhieu gam sat , bao nhieu gam ong ? 7.Cho mot ban nhom co khoi lng 60 gam vao dung dch CuSO4. Sau mot thi gian lay ra ra sach, say kho can nang 80,7gam. Tnh khoi lng ong bam vao ban nhom ?8. Ngam mot la ong vao dung dch AgNO3. Sau phan ng khoi lng la ong tang 0,76 gam . Tnh so gam ong a tham gia phan ng ?9. Ngam inh sat vao dung dch CuSO4. Sau mot thi gian lay ra ra sach, say kho can nang hn luc au 0,4 gam a. Tnh khoi lng sat va CuSO4 a tham gia phan ng ?b. Neu khoi lng dung dch CuSO4a dung tren la 210 gam co khoi lng rieng la 1,05 g/ml . Xac nh nong o mol ban au cua dung dch CuSO4 ?10. Cho 333 gam hon hp 3 muoi MgSO4 , CuSO4 va BaSO4 vao nc c dung dch D va mot phan khong tan co khoi lng 233 gam . Nhung thanh nhom vao dung dch D . Sau phan ng khoi lng thanh kim loai tang 11,5 gam . Tnh % ve khoi lng cua moi muoi co trong hon hp tren ?11. Cho ban sat co khoi lng 100 gam vao 2 lt dung dch CuSO4 1M. Sau mot thi gian dung dch CuSO4 co nong o la 0,8 M . Tnh khoi lng ban kim loai , biet rang the tch dung dch xem nh khong oi va khoi lng ong bam hoan toan vao ban sat ?12. Nhung mot la kem vao 500 ml dung dch Pb(NO3)2 2M . Sau mot thi gian khoi lng la kem tang 2,84 gam so vi ban au . a.Tnh lng Pb a bam vao laZn, biet rang lng Pb sinh ra bam hoan toan vao la Zn.b. Tnh mong o M cac muoi co trong dung dch sau khi lay la kem ra, biet rang the tch dung dch xem nh khong oi ? Trng hp2 : Tang giam khoi lng cua chat ket tua hay khoi lng dung dch sau phan nga) Khi gap bai toan cho a gam muoi clorua (cua kim loai Ba, Ca, Mg) tac dung vi dung dch cacbonat tao muoi ket tua co khoi lng b gam. Hay tm cong thc muoi clorua.- Muon tm cong thc muoi clorua phai tm so mol (n) muoi.
o giam khoi lng muoi clorua = a b la do thay Cl2 (M = 71) bang CO3 (M = 60).
Xac nh cong thc phan t muoi: T o xac nh cong thc phan t muoi.b) Khi gap bai toan cho m gam muoi cacbonat cua kim loai hoa tr II tac dung vi H2SO4 loang d thu c n gam muoi sunfat. Hay tm cong thc phan t muoi cacbonat.Muon tm cong thc phan t muoi cacbonat phai tm so mol muoi.
(do thay muoi cacbonat (60) bang muoi sunfat (96)
Xac nh cong thc phan t muoi RCO3: Suy ra cong thc phan t cua RCO3.* Bi tp vn dng:1. Co 100 ml muoi nitrat cua kim loai hoa tr II (dung dch A). Tha vao A mot thanh Pb kim loai, sau mot thi gian khi lng Pb khong oi th lay no ra khoi dung dch thay khoi lng cua no giam i 28,6 gam. Dung dch con lai c tha tiep vao o mot thanh Fe nang 100 gam. Khi lng sat khong oi na th lay ra khoi dung dch, tham kho can nang 130,2 gam. Hoi cong thc cua muoi ban au va nong o mol cua dung dch A.2. Hoa tan muoi nitrat cua mot kim loai hoa tr II vao nc c 200 ml dung dch (A). Cho vao dung dch (A) 200 ml dung dch K3PO4, phan ng xay ra va u, thu c ket tua (B) va dung dch (C). Khoi lng ket tua (B) va khoi lng muoi nitrat trong dung dch (A) khac nhau 3,64 gam.a. Tm nong o mol/l cua dung dch (A) va (C), gia thiet the tch dung dch thay oi do pha tron va the tch ket tua khong ang ke.b. Cho dung dch NaOH (lay d) vao 100 ml dung dch (A) thu c ket tua (D), loc lay ket tua (D) roi em nung en khoi lng khong oi can c 2,4 gam chat ran. Xac nh kim loai trong muoi nitrat.5. Dng ton theo s hp thc hiu sut phn ng Cch 1: Da vo lng cht thiu tham gia phn ngH = Lng thc t phn ng .100% Lng tng s ly- Lng thc t phn ng c tnh qua phng trnh phn ng theo lng sn phm bit.- Lng thc t phn ng < lng tng s ly.- Lng thc t phn ng , lng tng s ly c cng n v.Cch 2: Da vo 1 trong cc cht sn phmH = Lng sn phm thc t thu c .100% Lng sn phm thu theo l thuyt- Lng sn phm thu theo l thuyt c tnh qua phng trnh phn ng theo lng cht tham gia phn ng vi gi thit H = 100%- Lng sn phm thc t thu c thng cho trong bi.- Lng sn phm thc t thu c < Lng sn phm thu theo l thuyt- Lng sn phm thc t thu c v Lng sn phm thu theo l thuyt phi c cng n v o.* Bi tp vn dng:1: Nung 1 kg vi cha 80% CaCO3 thu c 112 dm3 CO2 (ktc) .Tnh hiu sut phn hu CaCO3.2:a) Khi cho kh SO3 hp nc cho ta dung dch H2SO4. Tnh lng H2SO4 iu ch c khi cho 40 Kg SO3 hp nc. Bit Hiu sut phn ng l 95%. b) Ngi ta dng qung boxit sn xut nhm theo s phn ng sau:
Al2O3 in phn nng chy, xc tc Al + O2Hm lng Al2O3 trong qung boxit l 40% . c c 4 tn nhm nguyn cht cn bao nhiu tn qung. Bit H ca qu trnh sn xut l 90%3:C th iuch bao nhiu kg nhm t 1 tn qung bxit c cha 95% nhm oxit, bit hiu sut phn ng l 98%.PT: Al2O3 in phn nng chy, xc tc Al + O2 4Ngi ta dng 490kg than t l chy my. Sau khi l ngui, thy cn 49kg than cha chy.a) Tnh hiu sut ca s chy trn.b) Tnh lng CaCO3 thu c, khi cho ton b kh CO2 vo nc vi trong d.5:Ngi ta iu ch vi sng (CaO) bng cch nung vi (CaCO3). Lng vi sng thu c t 1 tn vi c cha 10% tp cht l 0,45 tn. Tnh hiu sut phn ng.6:C th iu ch bao nhiu kg nhm t 1tn qung boxit c cha 95% nhm oxit, bit hiu sut phn ng l 98%.7:Khi cho kh SO3 tc dng vi nc cho ta dung dch H2SO4. Tnh lng H2SO4 iu ch c khi cho 40 kg SO3 tc dng vi nc. Bit hiu sut phn ng l 95%. 8.Ngi ta iu ch vi sng (CaO) bng cch nung vi CaCO3. Lng vi sng thu c t 1 tn vi c cha 10% tp cht l: Hy gii thch s la chn? Gi s hiu sut nung vi t 100%.9. Tnh khoi lng H2SO4 95% thu c t 60 kg quang pirit neu hieu suat p/ ng la 85% ?10. Dung 150 gam quang pirit cha 20% chat tr ieu che H2SO4. em toan bo lng axit ieu che c hoa tan va u m gam Fe2O3. Tat ca phan ng xay ra hoan toan, hay a. Tnh khoi lng H2SO4 ieu che c ?b. Tnh m ?11. T 1 tan quang pirit cha 90% FeS2 co the ieu che bao nhieu lt H2SO4 am ac 98% (d = 1,84 g/ml) , biet hieu suat trong qua trnh ieu che la 80% ?12. Trong cong nghiep ieu che H2SO4 t FeS2 theo s o sau:
FeS2 SO2 SO3 H2SO4a. Viet phng trnh phan ng va ghi ro ieu kien.b. Tnh lng axit 98% ieu che c t 1 tan quang cha 60% FeS2.Biet hieu suat cua qua trnh la 80%.13. ieu che HNO3 trong cong nghiep theo s o:
NH3 NO NO2 HNO3a. Viet phng trnh phan ng va ghi ro ieu kien.b. Tnh the tch NH3 ( ktc) cha 15% tap chat khong chay can thiet e thu c 10 kg HNO3 31,5%. Biet hieu suat cua qua trnh la 79,356%