47 CCChhhuuuyyynnn 111 NNNGGGUUUYYYNNN TTT --- NNNGGGUUUYYYNNN TTT HHHOOO HHHCCC I. KIN THC C BN: 1/ NT l ht v cng nh, trung ho v in v t to mi cht. NT gm ht nhn mang in tch + v v to bi electron (e)mang in tch (-)2/ Ht nhn to bi prton (p) mang in tch (+) v ntron (n) komang in. Nhng NT cng loi c cng s p trong ht nhn. Khi lng ht nhn = khi lng NT 3/Bit trong NT s p = s e. Electron lun chuyn ng v sp xp thnh tng lp. Nh e m NT c kh nng lin kt c vi nhau.Nguyn t c th ln kt c vi nhau nh e lp ngoi cng. STT ca lp :123 S e ti a :2e8e18e 4/ Nguyn t ho hc l tp hp nhng nguyn t c cng s p trong ht nhn. Vy: s P l s c trng cho mt nguyn t ho hc. - Quan h gia s p v s n:psn s 1,5p (ng vi 83 nguyn t u) - Khi lng tng i ca 1 nguyn t (nguyn t khi) NTK = s n +s p5/ Cch biu din nguyn t: Mi nguyn t c biu din bng mt hay hai ch ci, ch ci u c vit dng hoa, ch ci th hai (nu c) vit thng Mi k hiu ch mt nguyn t ca nguyn t . Vd: K hiu Na biu din {nguyn t natri ,mt nguyn t natri}6/ Mt n v cacbon (vC) = 1/12khi lng ca mt nguyn t C mC = 19,9206.10-27kg Khi lng tuyt i ca mt nguyn t (tnh theo gam) + mT = m e + mp + mn

+ mP~ mn~

1VC~ 19,9206.10-27kg/12~1,66005.10-27kg~1.67.10- 24g, + me~9.11.10 -28 g 7/ Nguyn t khi l khi lng ca 1 nguyn t tnh bng n v C (vit tt l v.C). Mi nguyn t c mt NTK ring.Khi lng 1 nguyn t = khi lng 1vc.NTK (1VC = 112KL ca NT(C) (MC = 1.9926.10- 23g) = 1121.9926.10- 23g= 1.66.10- 24 g) II. BI TP: 1. Trong phn ng ho hc cho bit: a) Ht vi m no c bo ton, ht no c th b chia nh ra? b) Nguyn t c b chia nh khng? c) V sao c s bin i phn t ny thnh phn t khc? V sao c s bin i cht ny thnh cht khc trong phn ng ha hc? 2. 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) 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 tkhi ca X,Y, Z .tn nguyn t, k hiu ho hc ca nguyn t ?5. NTK ca nguyn tC bng 3/4 NTK ca nguyn t O, NTK ca nguyn tO bng 1/2 NTK S. Tnh khi lng ca nguyn t O.(p s:O= 16, S=32) 6. Nguyn t oxi c 8 p trong ht nhn. Cho bit thnh phn ht nhn ca 3 nguyn t X,Y,Z theo bng sau: 47 Nguyn tHt nhn X8p , 8 n Y8p , 9n Z8p , 10 n Nhng nguyn t ny thuc cng mt nguyn t no ? v sao ? 7. Nguyn t st c 26p, 30n, 26e a. Tnh khi lng nguyn t st. b. Tnh khi lng e trong 1Kg st. 8. 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 X. 10. 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 815 s ht mang in. Xc nh nguyn t X thuc nguyn t no ? v s cu to nguyn t X? 12. Trong 1 tp hp cc phn t ng sunfat (CuSO4) c khi lng 160000 vC. Cho bit tp hp c bao nhiu nguyn t mi loi. 13. 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? 14. 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 . 15. Nguyn t Z c tng s ht bng 58 v c nguyn t khi < 40. HiZ thuc nguyn t ho hc no. V s cu to nguyn t ca nguyn t Z ? Cho bit Z l g (kim loi hayphi kim ?)(p s :Z thuc nguyn t Kali ( K )) Hng dn gii : bi 2p + n = 58 n = 58 2p ( 1 ) Mt khc :psn s 1,5p ( 2 ) ps58 2ps 1,5p gii ra c 16,5 sp s19,3(p: nguyn) Vy p c th nhn cc gi tr : 17,18,19 p171819 n242220 NTK = n + p414039 Vy nguyn t Z thuc nguyn t Kali ( K ) 16. 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 kk tip v cng mt phn nhm chnh trong bng tun hon. Tng s in tch ht nhn l 32. CCChhhuuuyyynnn 222 CCCHHHTTT VVV SSS BBBIIINNN III CCCHHHTTT I. KIN THC CBN: 1/Hin tng vt l l s bin i hnh dng hay trng thi ca cht. 2/Hin tng ho hc: l s bin i cht ny thnh cht khc. 3/n cht: l nhng cht c to nn t mt nguyn t ho hc t mt nguyn t hh c th to nhiu n cht khc nhau. 4/Hp cht : l nhng cht c to nn t hai nguyn t ho hc tr ln. 5/Phn t: l ht gm1 s nguyn t lin kt vi nhau v th hin y tnh cht ho hc ca cht. 6/Phn t khi :- L khi lng ca nguyn t tnh bng n v cacbon. 47 - PTK bng tng cc nguyn t khi c trong phn t. 7/Trng thi ca cht:Tu iu kin mt cht c th tn ti trng thi lng, rn, hi. II. BI TP: 1. Khi un nng, ng b phn hu bin i thnh than v nc. Nh vy, phn t ung do nguyn t no to nn? ng l n cht hay hp cht. 2. a) Khi nh dim c la bt chy, hin tng l hin tng g? b) Trong cc hin tng sau y, hin tng no l hin tng ha hc: trng b thi; mc ha tan vo nc; ty mu vi xanh thnh trng. 3. Em hycho bit nhng phng php vt l thng dng dng tch cc cht ra khi mt hn hp. Em hy cho bit hn hp gm nhng cht no th p dng c cc phng php . Cho v d minh ha. 4. Phn t ca mt cht A gm hai nguyn t, nguyn t X lin kt vi mt nguyn t oxi v nng hn phn t hiro 31 ln. a) A l n cht hay hp cht b) Tnh phn t khi ca A c) Tnh nguyn t khi ca X. Cho bit tn v k hiu ca nguyn t. CCChhhuuuyyynnn 333 LLLPPP CCCNNNGGG TTTHHHCCC HHHOOO HHHCCC II. LP CTHH DA VO THNH PHN PHN T: Cht (Do nguyn t to nn) n chtHp 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..) OxitAxitBazMui ( 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:- t CTTQ: AxBy (Bao gm: ( M2Oy , HxA, M(OH)y , MxAy) - Vn dng Qui tc ha tr i vi A,B: a.x = b.y xy= ba (ti gin)thay x= a, y = b vo CT chung ta c CTHH cn lp.(B c th l nhm nguyn t:gc axt,nhm OH) V d:Lp CTHH ca hp cht nhm oxt a b Gii:CTHHcdngchungAlxOy TabithatrcaAl=III,O=II a.x = b.y III.x= II. y xy= IIIII thay x= 2, y = 3 ta c CTHH l: Al2O3 47 * 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.Choccnguynt:Na,C,S,O,H.Hyvitcccngthchohccacchpchtvccth c to thnh cc nguyn t trn? 3.Choccnguynt:Ca,C,S,O,H.Hyvitcccngthchohccacchpchtvccth c to thnh cc nguyn t trn? 2.Lp CTHH hp cht khi bit thnh phn khi lng nguyn t . a. 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: ..A AB BM x mM y m= -Tm c t l : ..A BB Am M x ay m M b= = (t l cc s nguyn dng, ti gin) - Thay x= a, y = b Vit thnh CTHH. V d: Lp CTHH ca st v oxi, bit c 7 phn khi lng st th kt hp vi 3 phn khi lng oxi. Gii:- t cng thc tng qut: FexOy - Ta c t l khi lng cc nguyn t: ..MFe xMO y=mFemO= 73 - Tm c t l : xy=..mFe MOmO MFe= 7.163.56 = 112168= 23

- Thay x = 2, y = 3 Vit thnh CTHH.Fe2O3 * Bi tp vn dng: 1: Lp CTHH ca st v oxi, bit c 7 phn khi lng st th kt hp vi 3 phn khi lng oxi. 5:Phntkhicangsunfatl160vC.TrongcmtnguyntCucnguyntkhil64, mt nguyn t S c nguyn t khi l 32, cn li l nguyn t oxi. Cng thc phn ca hp cht l nh th no? 6: Xc nh cng thc phn t ca CuxOy, bit t l khi lng gia ng v oxi trong oxitl 4 : 1?8:Phntkhicangoxit(cthnhphngmngvoxi)vngsunfatctl1/2.Bitkhi lng ca phn t ng sunfatl 160 vC. Xc nh cng thc phn t ng oxit? 9. Mt nhm oxit c t s khi lng ca 2 nguyn t nhm v oxi bng 4,5:4. Cng thc ho hc ca nhm oxit l g? b. 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: y MBx MA..= BA%% -Rt ra t l x: y = MAA %: MBB % (ti gin) 47 -Vit thnh CTHH n gin: (AaBb )n = MAxBy n = MAxByMAaBb nhn n vo h s a,b ca cng thc AaBb ta c CTHH cn lp. V d: Mt hp cht kh Y c phn t khi l 58 vC, cu to t 2 nguyn t C v H trong nguyn t C chim 82,76% khi lng ca hp cht. Tm cng thc phn t ca hp cht. Gii: - t cng thc tng qut: CxHy -Ta c t l khi lng cc nguyn t: ..MC xMH y= %%CH -Rt ra t l x: y = %CMC: %HMH = 82,7612: 17,241 = 1:2-Thay x= 1,y = 2 vo CxHy ta c CTHH n gin: CH2 -Theo bi ra ta c : (CH2 )n = 58 n = 5814 = 5 Ta c CTHH cn lp : C5H10 * Bi tp vn dng: 1:HpchtXcphntkhibng62vC.Trongphntcahpchtnguyntoxichim25,8% theo khi lng, cn li l nguyn t Na. S nguyn tca nguyn t O v Na trong phn t hp cht l bao nhiu ? 2: Mt hp cht X c thnh phn % v khi lng l :40%Ca, 12%C v 48% O . Xc nh CTHH ca X. Bit khi lng mol ca X l 100g. 3: Tm cng thc ho hcca cc hp cht sau. a) Mt cht lng d bay hi, thnh phn t c 23,8% C, 5,9%H, 70,3%Clv 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 ClTrong 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: Xc nh cng thc ca cc hp cht sau: a)Hp cht to thnh bi magie v oxi c phn t khi l 40, trong phn trm v khi lng ca chng ln lt l 60% v 40%. b)Hp cht to thnh bi lu hunh v oxi c phn t khi l 64, trong phn trm v khi lng ca oxi l 50%. c)Hpchtcang,luhunhvoxicphntkhil160,cphntrmcangvlu hunh ln lt l 40% v 20%. d)Hpchttothnhbistvoxickhilngphntl160,trongphntrmvkhi lng ca oxi l 70%. e)Hp cht ca ng v oxi c phn t khi l 114, phn trm v khi lng ca ng l 88,89%. f)Hpchtcacanxivcacboncphntkhil64,phntrmvkhilngcacacbonl 37,5%. g)A c khi lng mol phn t l 58,5g; thnh phn % v khi lng nguyn t: 60,68% Cl cn li l Na. h)B c khi lng mol phn t l 106g; thnh phn % v khi lng ca cc nguyn t: 43,4% Na; 11,3% C cn li l ca O. i)Cckhilngmolphntl101g;thnhphnphntrmvkhilngccnguynt: 38,61% K; 13,86% N cn li l O. j)D c khi lng mol phn t l 126g; thnh phn % v khi lng ca cc nguyn t: 36,508% Na; 25,4% S cn li l O. k)Ec 24,68% K; 34,81% Mn; 40,51%O. E nng hn NaNO3 1,86 ln. l)F cha 5,88% v khi lng l H cn li l ca S. F nng hn kh hiro 17 ln. m)G c 3,7% H; 44,44% C; 51,86% O. G c khi lng mol phn t bng Al. n)H c 28,57% Mg; 14,285% C; 57,145% O. Khi lng mol phn t ca H l 84g. 47 6. Phn t canxi cacbonat c phn t khi l 100 vC , trong nguyn t canxi chim 40% khi lng, nguyntcacbonchim12%khilng.Khilngcnliloxi.Xcnhcngthcphntca hp cht canxi cacbonat? 7: Trong hp cht XHn c cha 17,65% l hidro. Bit hp cht ny c t khi so vi kh Metan CH4 l 1,0625. X l nguyn t no ? c. 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: y MBx MA..= BA%% - Rt ra t l x: y =MAA %: MBB % (ti gin) - Vit thnh CTHH. Vd:HyxcnhcngthchpchtAbitthnhphn%vkhilngccnguyntl: 40%Cu. 20%S v 40% O. Gii:- t cng thc tng qut: CuxSyOz - Rt ra t l x: y:z= %CuMCu: %SMs : %OMo = 4064: 2032 : 4016= = 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) 3: Hai nguyn t X kt hp vi 1 nguyn t O to ra phn t oxit. Trong phn t, nguyn t oxi chim 25,8% v khi lng. Hi nguyn t X l nguyn t no? 4:MtnguyntMkthpvi3nguyntHtothnhhpchtvihydro.Trongphnt,khi lng H chim 17,65%. Hi nguyn t M l g? 5: Hai nguyn t Y kt hp vi 3 nguyn t O to ra phn t oxit. Trong phn t, nguyn t oxi chim 30% v khi lng. Hi nguyn t X l nguyn t no? 6. Mt hp cht c thnh phn gm 2 nguyn t C v O. Thnh phn ca hp cht c 42,6% l nguyn t C, cn li l nguyn t oxi. Xc nh v t l s nguyn tca C v s nguyn toxi trong hp cht. d.Binlungitrkhilngmol(M)theohatr(x,y)tmNTKhocPTK..bitthnhphn%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: y MBx MA..=BA%% Rt ra t l : ..MBMA= x By A. %. % .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 73%ca R trong hp cht .Gii: Gi % R = a% % O = 73a% Gi ho tr ca R l n CTTQ ca C l:R2On Ta c:2 : n = Ra% : 16% 7 / 3 aR =6112n 47 V n l ha tr 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 hpcho 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 : ..MBMA= x mBy mA... 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 bng37. Gii:Gi ho tr ca M l n CTTQ ca C l:M2On Ta c: ..MBMA= x mBy mA... 16. MA= 2 . 3. 7 y . MA=6112n V n l ha tr 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.Oxitcakimloimchotrthpcha22,56%oxi,cnoxitcakimloimchotrcao cha 50,48%. Tnh nguyn t khi ca kim loi . 2. 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:Nu A :B =8 : 9th 89A nB n= = Theo : t s nguyn t khi ca 2 kim loi l 89AB =nn 89A nB n= = ( n e z+ ) V A, B u c KLNT khng qu 30 vC nn:9ns30 n s3 Ta c bng bin lun sau : n123 A81624 B91827 Suy ra hai kim loi l Mg v Al Chhhuuuyyynnn 444 MMMOOOLLL TTTNNNHHH TTTOOONNN HHHAAA HHHCCC I. KIN THC C BN: 1/ Mol l lng cht cha 6.1023 nguyn t hay phn t. K hiu: n, n v: mol. 2/ Cng thc tnh s mol: - Da theo s nguyn t: AnN=

(Trong : A l s nguyn t hay phn t, N = 6.1023). 47 - Da theo khi lng cht: mnM= (Trong : m l khi lng cht (g), M l KLNT (KLPT). - Da theo th tch ca cht kh iu kin chun: 22, 4Vn =(Trong : V l th tch ca cht kh) 3/ Trong cng iu kin nhit v p sut, cc cht kh u c th tch bng nhau v bng 22,4 lt. II. BI TP: 1. Hy xc nh cng thc cc hp cht sau: a) 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. b) 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. c) Hp cht D bit: 0,2 mol hp cht D c cha 9,2g Na, 2,4g C v 9,6g O. 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. Mui n gm 2 nguyn t ho hc l Na v ClTrong Na chim39,3% theo khi lng. Hy tm cng thc ho hc ca mui n ,bit phn t khi ca n gp 29,25 ln PT Khu m st Tri Cau (Thi Nguyn) c mt loi qung st. Khi phn tch mu qung ny ngi ta nhn thy c 2,8 gam st. Trong mu qung trn, khi lng Fe2O3 ng vi hm lng st ni trn l: A. 6 gam B. 8 gam C. 4 gam D. 3 gam 4. Xc nh cng thc phn t ca CuxOy, bit t l khi lng gia ng v oxi trong oxit l 4 : 1. Vit phng trnh phn ng iu ch ng v ng sunfat t CuxOy (cc ha cht khc t chn). 5. Trong phng th nghim c cc kim loi km v magi, cc dung dch axit sunfuric long H2SO4 v axit clohiric HCl. Mun iu ch c 1,12 lt kh hiro (ktc) phi dng kim loi no, axit no ch cn mt lng nh nht. A.Mg v H2SO4 B.Mgv HCl C.Zn vH2SO4 D.Zn v HCl 6. a)Tm cng thc ca oxit st trong c Fe chim 70% khi lng. b) Kh hon ton 2,4 gam hn hp CuO v FexOy cng s mol nh nhau bng hiro c 1,76 gam kim loi. Ho tan kim loi bng dung dchHCl d thy thot ra 0,488 lt H2 (ktc). Xc nh cng thc ca oxit st. CCChhhuuuyyynnn 555 TTTNNNHHH TTTHHHEEEOOO PPPHHHNNNGGG TTTRRRNNNHHH HHHOOO HHHCCC I. L THUYT: 1.Dng1:Tnhkhilng(hocthtchkh,ktc)cachtnykhibit(hocthtch)ca1cht khc trong phng trnh phn ng. 2. Dng 2: Cho bit khi lng ca 2 cht tham gia, tm khi lng cht to thnh. 3. Dng 3: Tnh theo nhiu phn ng. II. BI TP: Bi 1: Cho 8,4 gam st tc dng vi mt lng dung dchHCl va . Dn ton b lng kh sinh ra qua 16 gam ng (II) oxit nng. 47 a) Tnh th tch kh hiro sinh ra (ktc) b) Tnh khi lng kim loi ng thu c sau phn ng. Bi 2:Khi t, than chy theo s sau: Cacbon+ oxi kh cacbon ioxit a) Vit v cn bng phng trnh phn ng. b) Cho bit khi lng cacbon tc dng bng 9 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 6 kg, khi lng kh cacbonic thu c bng 22 kg, hy tnh khi lng oxi phn ng. p s: b) 33 kg c) 16 kg III. CC DNG BI TP HA HC C BN: 1. Bi tp dng thiu tha: * Cch gii: Trng hp c 2 cht tham gia phn ng u cho bit s mol (hoc khi lng, th tch) ca 2 cht, th lu c th c mt cht d. Khi tnh s mol (hoc khi lng, th tch) cht to thnh phi tnh theo lng cht thiu. * Cch gii chung :- Vit v cn bngPTHH: - 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) * Bi tp vn dng: 1. Khi t, than chy theo s sau : Cacbon+oxi kh cacbon ioxit a) 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. 2.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 p? 3. 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) 4. Dng 6,72 lt kh H2 (ktc) kh 20g St (III) oxit. a.Vit PTHH ca p? 47 b.Tnh khi lng oxit st t thu c? 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 vi14,6 gam dung dch axit clohiric HCl nguyncht. a. Vit phng trnh phn ng xy ra. b. Cht no cn dsau 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? 7. Cho 1 dd c cha 10g NaOH tc dng vi 1 dd c cha 10g HNO3. Hy cho bit dung dch sau phn ng s lm giy qu tm chuyn sang mu g? 8.Trn1ddchatan0,2molCuCl2vi1ddchatan20gNaOH.Lchnhpccchtsaup, c kt ta v nc lc. Nung kt ta n khi khi lng ko i. a) Tnh khi lng cht rn thu c sau khi nung. b) Tnh khi lng cc cht tan c trong nc lc. 2. Bi tp dng hn hp: * Cch gii: Cho m (g) hn hp A (gm M, M) + B. Cc cht trong hn hp A u phn ng hon ton vi lng cht B. * Cch gii chung :- Vit v cn bngPTHH Xy ra. - Tnh s mol cc cht trong qu trnh phn ng theo cc d kin ca bi. - Da vo PTHH, cc d kin bi ton, lp h phng trnh bc nht 2 n (hoc 3 n). Tm lng cc cht trong hn hp hay lng cc cht sn phm theo yu cu . * Bi tp vn dng:1. Cho 5,1 gam hn hp Al v Mgvo dung dch H2SO4 long, d thu c 5,6 lt kh H2 (ktc). Tnh khi lng mi kim loi ban u. Bit phn ng xy ra hon ton. 2. Cho 2,8 gam st tc dng vi14,6 gam dung dch axit clohiric HCl nguyncht. 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)Numunchophnngxyrahontonthphidngthmchtkiamtlnglbaonhiu?p s: b) 8, 4 gam; c) 3,36 lt;d) 8, 4 gam st. 3. Cho hn hp CuO v Fe2O3 tc dng vi H2 nhit thch hp. Hi nu thu c 26,4 gam hn hpng v st, trong khi lng ng gp 1,2 ln khi lng st th cn tt c bao nhiu lt kh hiro. p s: 12,23 lt. 4. Cho mt hn hp cha 4,6 gam natri v 3,9 gam kali tc dng vi nc. a) Vit phng trnh phn ng. b) Tnh th tch kh hiro thu c (ktc) c) Dung dch sau phn ng lm qu tm bin i mu nh th no? p s:b)3,36 lt;c) mu xanh 5. C mt hn hp gm 60% Fe2O3 v 40% CuO. Ngi ta dng H2 (d) kh 20 gam hn hp . 47 a) Tnh khi lng st v khi lng ng thu c sau phn ng. b) Tnh s mol H2 tham gia phn ng. 6. Trong phng th nghim c cc kim loi km v magi, cc dung dch axit sunfuric long H2SO4 v axit clohiric HCl. Mun iu ch c 1,12 lt kh hiro (ktc) phi dng kim loi no, axit no ch cn mt lng nh nht. A.Mg v H2SO4 B.Mgv HCl C.Zn vH2SO4 D.Zn v HCl 7.Cho60,5gamhnhpgmhaikimloikmZnvstFetcdngvidungdchaxitclohiric. Thnh phn phn trm v khi lng ca st chim 46,289% khi lng hn hp.Tnh a) Khi lng mi cht trong hn hp. b) Th tch kh H2 (ktc) sinh ra khi cho hn hp 2 kim loi trn tc dng vi dung dch axit clohiric. c) Khi lng cc mui to thnh. p s: a) 28 gam Fe v 32,5 gam km b) 22,4 lt c)2FeClm = 63,5gamv2ZnClm = 68 gam 8. Kh 15,2 gam hn hp gm Fe2O3 v FeO bng H2 nhit cao thu c st kim loi. ha tan ht lng st ny cn 0,4 mol HCl. a.Tnh % v khi lng ca mi oxitc trong hn hp ban u ? b.Tnh th tch H2 thu c ( ktc)? 9.Cho19,46gamhnhpgmMg,AlvZntrongkhilngcaMagiebngkhilngca nhmtc dng vi dung dch HCl to thnh 16, 352 lt kh H2 thot ra ( ktc).Tnh % v khi lng ca tng kim loi c trong hn hp ? 10. Cho 46,1 (g) hn hp Mg, Fe, Zn phn ng vi dung dch HCl th thu c 17,92 lt H2 (ktc). Tnh thnh phn phn trm v khi lng cc kim loi trong hn hp. Bit rng th tch kh H2 do st to ra gp i th tch H2 do Mg to ra. 11. Ha tan 7,8g hn hp gm Al, Mg bng dd HCl d. Sau phn ng, khi lng dd axit tng thm 7g. Xc nh khi lng mi kim loi trong hn hp u. 12. Ha tan 4,5g hp kim nhm magie trong dd H2SO4 long, d, thy c 5,04 lt kh H2 bay ra (ktc).Tnh thnh phn % khi lng ca cc kim loi trong hp kim. 13.Hatan6ghpkimCu-Fe-Altrongaxitclohidricdthtothnh3,024litH2(ktc)vcnli 1,86g kim loi khng tan. Xc nh thnh phn phn trm khi lng cc kim loi trong hp kim. 14. t chy hon ton 125,6g hn hp FeS2 v ZnS thu c 102,4g SO2. Tnh thnh phn phn trm mi mui trong hn hp u. 15. Cho bt than d vo hn hp 2 oxit Fe2O3 v CuO un nng phn ng xy ra hon ton thu c 2g hn hp kim loi v 2,24 lit kh (ktc). Tnh khi lng hn hp 2 oxit ban u. 16.unnng16,8gambtstv6,4gambtluhunh(khngckhngkh)thucchtrnA. HotanAbngHCldthotrakhB.ChokhBichmquadungdchPb(NO3)2tchrakttaD mu en. Cc phn ng u xy ra 100%. a)Vit phng trnh phn ng cho bit A, B, D l g? b)Tnh th tch kh B (ktc) v khi lng kt ta D. Cn bao nhiu th tch O2 (ktc) t hon ton kh B. 17. Cho 1,84g hn hp mui ACO3 v BCO3 tc dng ht vi dd HCl thu c 0,672 lit CO2 (ktc) v dd X. Tnh khi lng mui trong dd X? 18. Phn hy bng nhit 14,2g hn hp CaCO3 v MgCO3 ta thu c 6,6g CO2 ( ktc). Tnh thnh phn % cc cht trong hn hp. 19. Cho 38,2g hn hp Na2CO3 v K2CO3 vo dd HCl. Dn lng kh sinh ra qua nc vi trong c d th thu c 30g kt ta. 47 Tnh khi lng mi mui trong hn hp u. 20. Cho 0,325g hn hp gm NaCl v KCl c ha tan vo nc. Sau cho dd AgNO3 vo dd trn, ta c1ktta;sykttankhilngkoithycannng0,717g.Tnhthnhphn%cccht trong hn hp. 21. A l mt hn hp bt gm Ba, Mg, Al. -Ly m gam A cho vo nc ti khi ht phn ng thy thot ra 6,94 lt H2 (ktc). -Ly m gam A cho vo dung dch xt d ti ht phn ng thy thot ra 6,72 lt H2 (ktc). -LymgamAhotanbngmtlngvadungdchaxitHClcmtdungdchv 9,184 lt H2 (ktc). Hy tnh m v % khi lng cc kim loi trong A. 22. 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. 3. Bi tp c hiu sut phn ng: * Cch gii: Cch 1: Da vo lng cht thiu tham gia phn ng H = 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 phm H = 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 thitH = 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:Bi1:Nung1kgvicha80%CaCO3thuc112dm3CO2(ktc).Tnhhiusutphnhu CaCO3. Bi 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 + O2 HmlngAl2O3trongqungboxitl40%.cc4tnnhmnguynchtcnbaonhiutn qung. Bit H ca qu trnh sn xut l 90% Bi 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 tcAl + O2 Bi 4: Ngi 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. Bi 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. p s: 89,28% 47 Bi 6: C th iu chbao nhiu kg nhm t 1tn qung boxit c cha 95% nhm oxit, bit hiu sut phn ng l 98%. p s: 493 kg Bi 7: Khi cho kh SO3 tc dng vi nc cho ta dung dch H2SO4. Tnh lng H2SO4 iu ch c khi cho 40 kg SO3tc dng vi nc. Bit hiu sut phn ng l 95%. p s: 46,55 kg Bi 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: A. O,352 tn B. 0,478 tn C. 0,504 tnD. 0,616 tn Hy gii thch s la chn? Gi s hiu sut nung vi t 100%. 4. Tp cht v lng dng d trong phn ng: a. Tp cht: Tp cht l cht c ln trong nguyn liu ban u nhng l cht khng tham gia phn ng. V v phi tnh ra lng nguyn cht (tinh khit) trc khi thc hin tnh ton theo phng trnh phn ng. * Ghi ch: tinh khit = 100% - % tp cht Hoc tinh khit = khi lng cht tinh khit.100% Khi lng ko tinh khit Bi 1: Nung 200g vi c ln tp cht c vi sng CaO v CO2 .Tnh khi lng vi sng thu c nu H = 80% Bi 2: t chy 6,5 g lu hunh khng tinh khit trong kh oxi d c 4,48l kh SO2 ktc. a) Vit PTHH xy ra. b) Tnh tinh khit ca mu lu hunh trn? Bi 3: Ngi ta iu ch vi sng bng cch nung vi( CaCO3) .Tnh lng vi sng thu c t 1 tn vi cha 10% tp cht.Bi 4: 1 nng trng ngi ta dng mui ngm nc CuSO4.5H2O bn rung. Ngi ta bn 25kg mui trn 1ha t >Lng Cu c a v t l bao nhiu ( vi lng phn bn trn). Bit rng mui cha 5% tp cht. ( S 6,08 kg) b. Lng dng d trong phn ng: Lng ly d 1 cht nhm thc hn phn ng hon ton 1 cht khc. Lng ny khng a vo phn ng nn khi tnh lng cn dng phi tnh tng lng cho phn ng + lng ly d. Th d: Tnh th tch dung dch HCl 2M cn dng ho tan ht 10,8g Al, bit dng d 5% so vi lng phn ng. Gii: - 10,80, 427Almoln= =2Al + 6HCl 2AlCl3 + 3H2 0,4mol1,2mol -1, 2HClmoln= Vdd HCl (p) = 1,2/2 = 0,6lit V dd HCl(d) = 0,6.5/100 = 0,03 lit -----> Vdd HCl dng = Vp + Vd = 0,6 + 0,03 = 0,63 lit Bi tp. Trong phng th nghim cn iu ch 5,6 lt kh O2 (ktc). Hi phi dng baonhiu gam KClO3? Bit rng kh oxi thu c sau phn ng b haoht 10%) 47 CCChhhuuuyyynnn 666OOOXXXIII--- HHHIIIRRROOO VVV HHHPPP CCCHHHTTT VVV CCC Bi 1: C 4 bnh ng ring cc kh sau: khng kh, kh oxi, kh hiro, khcacbonic. Bng cch no nhn bit cc cht kh trong mi bnh. Gii thch v vit cc phng trnh phn ng (nu c). Bi 2:Vit phng trnh ha hc biu din s chy trong oxi ca cc n cht: cacbon, photpho, hiro, nhm, magi, lu hunh . Hy gi tn cc sn phm. Bi 3: Vit cc phng trnh phn ng ln lt xy ra theo s : C ) 1 ( CO2 ) 2 ( CaCO3 ) 3 ( CaO ) 4 ( Ca(OH)2 sn xut vi trong l vi ngi ta thng sp xp mt lp than, mt lp vi, sau t l. C nhng phn ng ha hc no xy ra trong l vi? Phn ng no l phn ng to nhit; phn ng no l phn ng thu nhit; phn ng no l phn ng phn hu; phn ng no l phn ng ha hp? Bi 4: T cc ha cht: Zn, nc, khng kh v lu hunh hy iu ch 3 oxit, 2 axit v 2 mui. Vit cc phng trnh phn ng. Bi 5.C 4 l mt nhn ng bn cht bt mu trng gm: Na2O, MgO, CaO, P2O5.Dng thuc th no nhn bit cc cht trn? A. dng nc v dung dch axit H2SO4 B. dng dung dch axit H2SO4 v phenolphthalein C. dng nc v giy qu tm. D. khng c cht no kh c Bi 6. iu ch kh oxi, ngi ta nung KClO3 . Sau mt thi gian nung ta thu c 168,2 gam cht rn v 53,76 lt kh O2(ktc). a)Vit phng trnh phn ng xy ra khi nung KClO3. b)Tnh khi lng KClO3 ban u em nung. c) Tnh % khi lng mol KClO3 b nhit phn. p s:b)245 gam. c) 80% Bi 7. C 3 l ng cc ha cht rn, mu trng ring bit nhng khng c nhn : Na2O, MgO, P2O5. Hy dng cc phng php ha hc nhn bit 3 cht trn. Vit cc phng trnh phn ng xy ra. Bi 8. Ly cng mt lng KClO3 v KMnO4 iu ch kh O2. Cht no chonhiu kh oxi hn? a) Vit phng trnh phn ng v gii thch. b) Nu iu ch cng mt th tch kh oxi th dng cht no kinh t hn? Bit rnggi ca KMnO4 l 30.000/kg v KClO3 l 96.000/kg. p s:11.760 (KClO3) v 14.220 (KMnO4) Bi 9. Hy lp cc phng trnh ha hc theo s phn ng sau: a)St (III) oxit+nhm nhm oxit + st b) Nhm oxit +cacbon nhm cacbua + kh cacbon oxit c) Hiro sunfua+ oxi kh sunfur + nc 47 d) ng (II) hiroxit ng (II) oxit + nc e) Natri oxit+cacbon ioxit Natri cacbonat. Trong cc phn ng trn, phn ng no l phn ng oxi ha kh? Xc nh cht oxi ha, cht kh, s oxi ha, s kh. Bi10.C4chtrn dngbtlAl,Cu,Fe2O3 vCuO.Nuchdngthucthldungdchaxit HClcthnhnbitc4chttrnckhng?Mthintngvvitphngtrnhphnng (nu c). Bi 11. a) C 3 l ng ring r cc cht bt mu trng: Na2O, MgO, P2O5. Hy nu phng php ha hc nhn bit 3 cht . Vit cc phng trnh phn ngxy ra. b)C3ngnghimngringr3chtlngtrongsut,khngmul3dungdchNaCl,HCl, Na2CO3. Khng dng thm mt cht no khc (k c qu tm), lm th no nhn bit ra tng cht. Bi 12.Cho 2,8 gam st tc dng vi14,6 gam dung dch axit clohiric HCl nguyncht. 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? p s: b) 8, 4 gam; c) 3,36 lt;d) 8, 4 gam st. Bi 13.Hon thnh phng trnh ha hc ca nhng phn ng gia cc cht sau: a)Al + O2 ..... b) H2 + Fe3O4 .... + ... c) P + O2 .....d)KClO3 .... + ..... e) S + O2 .....f)PbO + H2 ....+.... Bi 14. Trong phng th nghim c cc kim loi km v magi, cc dung dch axit sunfuric long H2SO4 v axit clohiric HCl. Mun iu ch c 1,12 lt kh hiro (ktc) phi dng kim loi no, axit no ch cn mt lng nh nht. A.Mg v H2SO4 B.Mgv HCl C.Zn vH2SO4 D.Zn v HCl p s: B Bi 15. a ) Hy nu phng php nhn bit cc kh: cacbon ioxit, oxi,nit v hiro b) Trnh by phng php ha hc tch ring tng kh oxi v kh cacbonic ra khi hn hp. Vit cc phng trnh phn ng. Theo em thu c kh CO2

c th cho CaCO3 tc dng vi dung dchaxit HCl c khng? Nu khng th ti sao? Bi 16.a) T nhng ha cht cho sn: KMnO4, Fe, dung dch CuSO4, dung dch H2SO4 long, hy vit cc phng trnh ha hc iu ch cc cht theo s chuyn ha sau: Cu CuO Cu 47 a) Khi in phn nc thu c 2 th tch kh H2 v 1 th tch kh O2(cng iu kin nhit , p sut). T kt qu ny em hy chng minh cng thc ha hc ca nc. Bi 17.Cho cc cht nhm., st, oxi, ng sunfat, nc, axit clohiric. Hy iu ch ng (II) oxit, nhm clorua ( bng hai phng php) v st (II) clorua. Vit cc phng trnh phn ng. Bi 18.C 6 l mt nhn ng cc dung dchcc cht sau: HCl; H2SO4; BaCl2; NaCl; NaOH; Ba(OH)2 Hy chn mt thuc th nhn bit cc dung dch trn, A. qu tmB. dung dch phenolphthalein C. dung dch AgNO3 D. tt c u sai CCChhhuuuyyynnn 888 ddduuunnnggg ddd ccchhh Lu khi lm bi tp:1. S chuyn i gia nng phn trm v nng mol -Cng thc chuyn t nng % sang nng CM. d l khi lng ring ca dung dch g/ml M l phn t khi ca cht tan

-Chuynt nng mol (M) sang nng %.

2. Chuyn i gia khi lng dung dch v th tch dung dch. -Th tch ca cht rn v cht lng:DmV = Trong d l khi lng ring:d(g/cm3) c m (g) v V (cm3) hay ml. d(kg/dm3) c m (kg) v V (dm3) hay lit. 3. Pha trn dung dch a) Phng php ng cho Khi pha trn 2 dung dch c cng loi nng ( CM hay C%), cng loi cht tan th c th dng phng php ng cho. -Trnm1gamdungdchcnngC1%vim2gamdungdchcnngC2%ththuc dung dch mi c nng C%. m1 gam dung dch C1 C2-C C C CC Cmm=1221 1000 .%.Md cCM = dC MCM1000 .%=47 m2 gam dung dch C2 C1 - C-Trn V1 ml dung dch c nng C1 mol vi V2 ml dung dch c nng C2 mol th thu c dung dch mi c nng C mol v gi s c th tch V1+V2 ml: V1 ml dung dch C1C2-C C C CC CVV=1221

V2 ml dung dch C2C1 - C -S ng cho cn c th p dng trong vic tnh khi lng ring D V1 lt dung dch D1 D2-D DD DD DVV=1221

V2 lt dung dch D2 D1 - D (Vi gi thitV= V1+ V2 ) b) Dng phng trnh pha trn: m1C1 + m2C2 = (m1 + m2).C Trong :m1 v m2 l s gam dung dch th nht v dung dch th hai. C1 v C2 l nng % dung dch th nht v dung dch th hai. C l nng dung dch mi to thnh sau khi pha trn m1 (C1 -C)= m2 ( C -C2) C1>C> C2 T phng trnh trn ta rt ra: C CC Cmm=1221 Khi pha trn dung dch, cn ch : -Cxyraphnnggiaccchttanhocgiachttanvidungmi?Nuccnphnbit cht em ha tan vi cht tan. V d: Cho Na2O hay SO3 ha tan vo nc, ta c cc phng trnh sau: Na2O +H2O 2NaOH SO3 +H2O H2SO4 -Khi cht tan phn ng vi dung mi, phi tnh nng ca sn phm ch khng phi tnh nng ca cht tan . V d: Cn thm bao nhiu gam SO3 vo 100 gam dung dch H2SO4 10% c dung dch H2SO4 20%. Hng dn cch gii: Gi s x l s mol SO3 cho thm voPhng trnh: SO3 +H2O H2SO4 x molx mol 4 2SO Hm to thnh l 98x; 3SOmcho thm vo l 80x C% dung dch mi:10020100 8098 10=++xx 47 Gii ra ta cmol x41050= 3SOmthm vo 9,756 gam Cng c th gii theo phng trnh pha trn nh nu trn. 4. Tnh nng cc cht trong trng hp cc cht tan c phn ng vi nhau. a) Vit phng trnh phn ng ha hc xy ra bit cht to thnh sau phn ng. b) Tnh s mol (hoc khi lng) ca cc cht sau phn ng. c) Tnh khi lng hoc th tch dung dch sau phn ng. Cch tnh khi lng sau phn ng: -Nu cht to thnh khng c cht bay hi hoc kt ta m dd sau phn ng =mcc cht tham gia Nu cht to thnh c cht bay hi hay kt ta m dd sau phn ng=mcc cht tham gia- m kh m dd sau phn ng=mcc cht tham gia - m ktta hoc:m dd sau phn ng=mcc cht tham gia - m ktta - mkh Ch : Trng hp c 2 cht tham gia phn ng u cho bit s mol (hoc khi lng) ca 2 cht, th lu c th c mt cht d. Khi tnh s mol (hoc khi lng) cht to thnh phi tnh theo lng cht khngd. d) Nu u bi yu cu tnh nng phn trm cc cht sau phn ng, nn tnhkhi lng cht trong phn ng theo s mol, sau t s mol qui rakhilng tnh nng phn trm. 5. S chuyn t tan sang nng phn trm v ngc li -Chuynttansangnngphntrm:Davonhnghatan,ttnhkhilng dung dch suy ra s gam cht tan trong 100 gam dung dch. -Chuyn t nng phn trm sang tan: T nh ngha nng phn trm, suy ra khi lng nc, khi lng cht tan, t tnh 100 gam nc cha bao nhiu gam cht tan. Biuthclinhgiatan(S)vnngphntrmcachttantrongdungdchboha:C% = % 100100+ SS 6. Bi ton v khi lng cht kt tinh Khi lng cht kt tinh ch tnh khi cht tan vt qu bo ha ca dung dch 1.Khi gp dng bi ton lm bay hi c gam nc t dung dch c nng a% c dung dch mi c nng b%. Hy xc nh khi lng ca dung dch ban u ( bit b% > a%). Gp dng bi ton ny ta nn gii nh sau: - Gi s khi lng ca dung dch ban u l m gam. - Lp c phng trnh khi lng cht tan trc v sau phn ng theo m, c, a, b. 47 + Trc phn ng: 100m a + Sau phn ng: 100) ( c m b -Do ch c nc bay hi cn khi lng cht tan khng thay i Ta c phng trnh: Khi lng cht tan:100) (100c m b m a = T phng trnh trn ta c: a bbcm= (gam) B. Cu hi v Bi tp1. Ho tan 25,5 gam NaCl vo 80 gam nc 200C c dung dch A. Hi dung dch A bo ha hay cha? Bit tan ca NaCl 200C l 38 gam. 2. Khi lm lnh 600 gam dung dch bo ha NaCl t 900C xung 100C th c bao nhiu gam mui NaCl tch ra. Bit rng tan ca NaCl 900C l 50 gam v 100C l 35 gam. 3.Mtdungdchccha26,5gamNaCltrong75gamH2O200C.Hyxcnhlngdungdch NaCl ni trn l bo ha hay cha bo ha? Bit rng tan ca NaCl trong nc 200C l 36 gam. 4. Ha tan 7,18 gam mui NaCl vo 20 gam nc 200C th c dung dch bo ha. tan ca NaCl nhit l : A. 35 gam B.35,9 gamC. 53,85 gam D. 71,8 gam Hy chn phng n ng. a) Vit phng trnh phn ng xy ra v tnh nng mol/l ca dung dch A. b) Tnh th tch dung dch H2SO4 20% (d =1,14 g/ml) cn trung ha dung dch A. c) Tnh nng mol/l ca dung dch thu c sau khi trung ha. 5.a) Ha tan 4 gam NaCl trong 80 gam H2O. Tnh nng phn trm ca dung dch. b) Chuyn sang nng phn trm dung dch NaOH 2M c khi lng ring d = 1,08 g/ml. c)CnbaonhiugamNaOHphachc3ltdungdchNaOH10%.Bitkhilngringca dung dch l 1,115 g/ml. 6. Dung dch H2SO4 c nng 0,2 M (dung dch A). Dung dch H2SO4 c nng 0,5M (dung dch B). a) Nu trn A v B theo t l th tch VA: VB=2 : 3 c dung dch C. Hy xc nh nng mol ca dung dch C. b) Phi trn A v B theo t l no v th tch c ddch H2SO4 c nng 0,3 M. 7. ng sunfat tan vo trong nc to thnh dung dch c mu xanh l, mu xanh cng m nu nng dung dch cng cao. C 4 dung dch c pha ch nh sau (th tch dung dch c coi l bng th tch nc). A.dung dch 1: 100 ml H2O v 2,4 gam CuSO4 B.dung dch 2: 300 ml H2O v 6,4 gam CuSO4 C. dung dch 3: 200 ml H2O v 3,2 gam CuSO4 D. dung dch 4: 400 ml H2O v 8,0 gam CuSO4 47 Hi dung dch no c mu xanh m nht? A. dung dch 1B. Dung dch 2 C. Dung dch 3 D. Dung dch 4 8. Ho tan 5,72 gam Na2CO3.10 H2O (Sa tinh th) vo 44,28 ml nc. Nng phn trm ca dung dch thu c l: A. 4,24 % B. 5,24 % C. 6,5 %D. 5% Hy gii thch s la chn. 9. Ha tan 25 gam CaCl2.6H2O trong 300ml H2O. Dung dch c D l 1,08 g/ml a) Nng phn trm ca dung dch CaCl2 l: A. 4% B. 3,8%C. 3,9 % D. Tt c u sai b) Nng mol ca dung dch CaCl2 l: A. 0,37M B. 0,38MC. 0,39MD. 0,45M Hy chn p s ng. 10.a) Phi ly bao nhiu ml dung dch H2SO4 96%(D =1,84 g/ml) trong c 2,45 gam H2SO4? b)Oxihahonton5,6ltkhSO2(ktc)votrong57,2mldungdchH2SO460%(D=1,5g/ml). Tnh nng % ca dung dch axit thu c 11. Tnh khi lng mui natri clorua c th tan trong 830 gam nc 250C. Bitrng nhit ny tan ca NaCl l 36,2 gam. p s: 300,46 gam 12. Xc nh tan camui Na2CO3 trong nc 180C. Bit rng nhit ny 53 gam Na2CO3 ha tan trong 250 gam nc th c dung dch bo ha. p s: 21,2 gam 13. Ha tan m gam SO3 vo 500 ml dung dch H2SO4 24,5% (D = 1,2 g/ml) thu c dung dch H2SO4 49%. Tnh m? p s: m = 200 gam 14. Lm bay hi 300 gam nc ra khi 700 gam dung dch mui 12% nhn thy c 5 gam mui tch ra khi dung dch bo ha. Hy xc nh nng phn trm ca dung dch mui bo ha trong iu kin th nghim trn. p s: 20% 15. a) tan ca mui n NaCl 200C l 36 gam. Xc nh nng phn trm ca dung dch bo ha nhit trn. b) Dung dch bo ha mui NaNO3 100C l 44,44%. Tnh tan ca NaNO3. p s:a) 26,47%b)80 gam 16.Trn50mldungdchHNO3nngxmol/lvi150mldungdchBa(OH)20,2mol/lthuc dung dch A. Cho mu qu tm vo dung dch A thy qutm chuyn mu xanh. Them t t 100 ml dung dch HCl 0,1mol/l vo dungdch A th thy qu tm tr li mu tm. Tnh nng x mol/l. p s:x = 1 mol/l 17. Ha tan 155 gam natri oxit vo 145 gam nc to thnh dung dch c tnh kim. - Vit phng trnh phn ng xy ra. - Tnh nng % dung dch thu c. p s:66,67% 47 18. Ha tan 25 gam cht X vo 100 gam nc, dung dch c khi lng ring l 1,143g/ml. Nng phn trm v th tch dung dch ln lt l: A.30% v 100 mlB. 25% v 80 ml C.35% v 90 mlD.20% v 109,4 ml Hy chn p s ng? p s:D ng 19. Ha tan hon ton 6,66 gam tinh th Al2(SO4)3. xH2O vo nc thnh dung dch A. Ly 1/10 dung dch A cho tc dng vi dung dch BaCl2 d th thu c 0,699 gam kt ta. Hy xc nh cng thc ca tinh th mui sunfat nhm ngmnc trn. p s: Al2(SO4)3.18H2O 20.C 250 gam dung dch NaOH 6% (dung dch A). a)CnphitrnthmvodungdchAbaonhiugamdungdchNaOH10%cdungdch NaOH 8%? b) Cn ha tan bao nhiu gam NaOH vo dung dch A c dung dch NaOH8%? c)LmbayhincdungdchA,ngitacngthucdungdchNaOH8%.Tnhkhilng ncbay hi? p s:a)250 gamb) 10,87 gamc) 62,5 gam21. a) Cn ly bao nhiu ml dung dch c nng 36 % ( D=1,16 g/ ml) pha 5 lt dung dch axit HClc nng 0,5 mol/l? b) Cho bt nhm d vo 200 ml dung dch axit HCl 1 mol/l ta thu c khH2bay ra. - Vit phng trnh phn ng v tnh th tch kh H2 thot ra ktc. - Dn ton b khhiro thot ra trn cho i qua ng ng bt ng oxit d nung nng th thu c 5,67 gam ng. Vit phng trnh phn ng v tnh hiu sut ca phn ng ny? p s: a)213 ml b) 2,24 lt hiu sut : 90%. 22. Trn ln 50 gam dung dch NaOH 10% vi 450 gam dung dch NaOH 25 %. a) Tnh nng sau khi trn. b) Tnh th tch dung dch sau khi trn bit t khi dung dch ny l 1,05. p s: a) 23,5 %b) 0,4762 lt 23.Trn150gamdungdchNaOH10%vo460gamdungdchNaOHx%tothnhdungdch 6%. x c gi tr l: A. 4,7 B. 4,65 C. 4,71D. 6 Hy chnp sng? p s: A ng. 24.a) Cn thm bao nhiu gam nc vo 500 gam dung dch NaCl 12% c dung dch 8%. b) Phi pha thm nc vo dung dch H2SO4 50% thu c mt dung dch H2SO4 20%. Tnh t l v khi lng nc v lng dung dch axit phidng? c) Cn ly bao nhiu gam tinh th CuSO4. 5 H2O v bao nhiu gam dung dch CuSO4 4% iu ch 500 gam dung dch CuSO4 8%? p s: a) 250 g b) 23 c) 466,67 gam 47 25. Bit tan ca mui KCl 200C l 34 gam.Mt dung dch KCl nngc cha 50gam KCl trong 130 gam nc c lm lnh v nhit 200C.Hy cho bit: a) C bao nhiu gam KCl tan trong dung dch b)c bao nhiu gam KCl tch ra khi dung dch. p s: a) 44,2 gamb) 5,8 gam 26.a)Lmbayhi75mlnctdungdchH2SO4cnng20%cdungdcmicnng 25%.Hy xc nh khi lng ca dung dch ban u. Bit khi lng ring ca nc D = 1 g/ml. b) Xc nh khi lng NaCl kt tinh tr li khi lm lnh 548 gam dung dchmui n bo ha 500C xung 00C. Bit tan ca NaCl 500C l 37 gam v 00C l 35 gam. p s: a) 375 gam b) 8 gam 27.Ho tan NaOH rn vo nc to thnh hai dung dch A v dung dch B vi nng phn trm ca dung dch A gp 3 ln nng phn trm ca dung dch B. Nu em pha trnhai dung dch A v dung dch B theo t l khi lngmA: mB = 5 : 2 th thu c dung dchC c nng phn trm l 20%. Nng phn trm ca hai dung dch A v dung dch B ln lt l: A. 24,7% v 8,24% B.24% v 8% C.27% v 9 % D. 30% v 10% Hy chn phng n ng. p s: A ng. 28.a)Ha tan 24,4 gam BaCl2. xH2O vo 175,6 gam H2O thu c dung dch 10,4%. Tnh x. b) C cn t t 200 ml dung dch CuSO4 0,2M thu c 10 gam tinh th CuSO4. yH2O. Tnh y. 47 n tp ha hcDng I : Vit PTHH gia cc cht v c 1.Vit PTHH biu din cc phn ng ho hc cc th nghim sau :a.Nh vi git axit clohidric vo vi b.Cho mt t diphotpho pentoxit vo dd kali hidroxitc.Nhng thanh st vo dd ng (II) sunfatd.Hp th N2O5 vo H2O2.Cho cc oxit sau : K2O, SO2, BaO, Fe3O4, N2O5, FeO, Fe2O3. Vit PTHH (nu c) ca cc oxit ny ln lt tc dng vi H2O, H2SO4, KOH, HCl3.Vit PTP :a.Kim loi M ho tr n tan trong dd HClb.MgCO3 + HNO3 c.Al+ H2SO4 (long)d.FexOy+ HCl e. Fe + Cl2 f.Cl2 + NaOH4.ChottbtCuvoddHNO3c.Lcuthykhmunubayra,saukhkhng mu b ho nu trong khng kh, cui cng kh ngng thot ra. GT hin tng, vit PTHH xy ra5.C nhng baz sau : Fe(OH)3, Ca(OH)2, KOH, Mg(OH)2, Cu(OH)2

a.Baz no b nhit phn hu ? b.Tc dng c vi dd H2SO4

c.i mu dd phenolphtalein ? 6.Hy m t hin tng quan st c, vit pthh khi th l Al vo nhng dd sau :a.dd H2SO4 2 Mb.dd NaOH dc.dd CuCl2

Dng II. S chuyn ho 1. Vit PTHH theo s sau :MgSO4

SO2 H2SO4MgCl2

HCl 2.Tm cc ch ci A,B,C,D,E thch hp, vit PTHH xy ra (1) A+ Cl2B (2) B + Al (d) AlCl3+A (3) A +O2C (4) C+ H2SO4 D+E+H2O 3. Chn cc cht A,B,C,D thch hp, vit PTHH xy ra A B CuSO4CuCl2Cu(NO3)2 A BC C 4. Hon thnh cc phng trnh di y :47 a. Na2SO4+X1 BaSO4 + Y1

Ca(HCO3)2+X2 CaCO3 + Y2

CuSO4+X3CuS+Y3

MgCl2 +X4Mg3(PO4)2+ Y4

b. A +B CaCO3 +NaClC + DZnS +KNO3 E+ F Ca3(PO4)2 + NaNO3 G+HBaSO4 +MgCl2 c. KHS + AH2S + HCl+BCO2 + CaSO3 +CSO2 + H2SO4 +D BaSO4+ CO2 +. 5. Vit cc PTP theo cc s bin ho sau : Fe2(SO4)2 Fe(OH)3 Cu CuCl2

FeCl3CuSO4 6. Vit cc PTP theo s bin ho+X A+Y Fe2O3FeCl2 +ZB +T trong A,B,X,Y,Z,T l cc cht khc nhau 7.Vit cc PTP theo s hai chiu sau :

SSO2 H2SO4 CuSO4

K2SO3 8.Cho s bin ho :a.A1A2A3

Fe(OH)3 Fe(OH)3 B1B2 B3 Tm cng thc ca cc cht ng vi cc cht A1,, A2, ..vit PTP theo s b. A1 A2 A3 CaCO3CaCO3 CaCO3

B1 B2 B3 .. +X,t0 c. A +Y,t0 + B+E AFeD C +Z,t0 ABit rng : A+HCl D+ C+H2O 47 Dng III. Nhn bit cc cht v c 1.Ch c dng mt thuc th t chn, hy nhn bit dd cc cht ng trong cc l ring r: FeSO4 ; Fe2(SO4)3 ; MgCl2 ; AlCl3 ; CuCl2 ; NaOH2.Dng mt thuc th nhn bit cc dd : Na2CO3 ; NaCl ; Na2S ; Ba(NO3)2

3.Bng pp ho hc nhn bit cc kh ng trong cc l mt nhn : CO2 ; NH3 ; O2 ; N2

4.5 bnh cha 5 kh : N2 ; O2 ; CO2 ; H2 ; CH4. Trnh by pp ho hc nhn ra tng kh5.C 5 dd : HCl ; NaOH ; Na2CO3 ; BaCl2 ; NaCl. Cho php s dng qu tm nhn bit cc dd (bit Na2CO3 cng lm xanh qu tm)6.Ch c s dng dd HCl ; H2O nu pp nhn bit 5 gi bt trng cha cc cht : KNO3 ; K2CO3 ; K2SO4 ; BaCO3 ; BaSO4

7.c 5 cht rn : Fe ; Cu ; Al ; CuO ; FeO. Dng pp ho hc nhn bit tng cht8.5lmtnhn,milchamttrongccchtbtmuenhocxmxmsau:FeS; Ag2O ; CuO ; MnO2 ; FeO. ch dng ng nghim, n cn, vmt dd thuc th nhn bit9.C5ddbmtnhngmccchtsau:H2SO4;Na2SO4;NaOH;BaCl2;MgCl2.Ch dng thm phenol phtalein nu cch xc nh tng dd 10. Ch dng 1 thuc th l kim loi hy nhn bit cc l cha cc dd : Ba(OH)2 ; HNO3 c, ngui ; AgNO3

Dng IV: Tch cc cht v c 1.Trnh by pp ho hc tch ring tng kim loi ra khi hh cha : Ag ; Al ; Fe 2. Tch ring dd tng cht ra khi hh dd : AlCl3 ; FeCl3 ; BaCl2

3. iu ch cht nguyn cht : a. NaCl c ln mt t tp cht l Na2CO3. Lm th na c NaCl nguyn cht ? b. N2 ln cc tp cht : CO ; CO2 ; H2 v hi nc c, C hh 3 oxit : SiO2 ; Al2O3 ; Fe2O3. Trnh by pp ho hc ly tng cht dng nguyn cht4.Mtloithungnblntpchtlcckimloisau:Fe;Zn;Pb;Sn.cthdngdd Hg(NO3)2 ly c Hg tinh khit. em hy nu pp lm v vit PTP5. Bng pp hh tch ring a. Bt Fe ra khi hh : Fe, Cu, CaO b. Tch ring tng cht khi hh : Fe, Fe2O3, Cu (khi lng bo ton) Dng V : Tnh theo phng trnh ho hc, xc nh CT oxit baz 1.Ho tan 16,2 gam ZnO vo 400gam dd HNO3 15% thu c dd A a. Tnh khi lng axit phn ngb. Tnh khi lng mui km to thnh c. Tnh C% cc cht trong dd A2.Hotan10,8gamAltcdngvavi600gamddHClthucddXvVltkh KTC a. Tnh V b. Tnh khi lng mui nhm thu cc. Tnh CM ca dd HCl3. Cho 325 gam dd FeCl3 5% vo 112 gam dd KOH 25%47 a. Cht no tha sau phn ng b. Tnh khi lng cht kt ta thu cc. Tnh C% cc cht trong dd sau phn ng4. Ho tan 8,9 gam hh Mg, Zn vo lng va dd H2SO4 0,2M thu c dd A v 4,48 lt kh ktc a. Tnh % theo khi lng 2 kim loib. Tnh th tch dd axit dng5. Cho 16,8 lt CO2 ktc hp th hon ton vo 600ml dd NaOH 2M thu c dd A a. Tnh khi lng mui thu c trong dd A b. Cho BaCl2 d vo dd A th thu c bao nhiu gam kt ta6. Nhng mt ming Al c khi lng 10 gam vo 500 ml dd CuSO4 0,4M. Sau thi gian phn ng ly ming Al ra, cn nng 11,38 gam a. Tnh m Cu bm vo Alb.TnhCMccchttrongddsauphnng(coiVkhng i)7. Cho 20 gam Al vo 400 ml dd CuCl2 0,5 M. Khi nng dd CuCl2 gim 25% th ly ming Al ra, cn nng bao nhiu gam ?8. ho tan 3,9 gam kim loi X cn dng V ml dd HCl v c 1,344 lt H2 ktc. Mt khc, hotan 3,2 gam oxit ca kim loi Y cng dng va Vml dd HCl trn. Hi X,Y l cc kim loi g ?9. Cho 34,8 gam Fe3O4 tc dng vi 455,2 gam ddHCl20%d thu c dd A. Tnh C% cc cht tan c trong dd A10. Cho 16 gam FexOy tc dng vi lng va 300 ml dd HCl 2 M. Xc nh CT oxit st11. Ho tan 8 gam oxit lim loi ho tr 2 cn 14,6 gam HCl nguyn cht. Tm CT oxit12. Ho tan 20,4 gam oxit kim loi A (ho tr 3) bng 300 ml dd H2SO4 va th thu c 68,4 gam mui khana. Tm CTHH ca oxit trn b. Tnh CM ca dd axit13. ho tan 64 gam mt oxit kim loi (ho tr 3) cn va 800 ml dd HNO3 3Ma. Tm CT oxit b. Tnh CM dd mui sau phn ng14. Ha tan 5 gam vi nguyn cht trong 40 ml dd HCl. Sau phn ng phi dng 20ml dd NaOH trung ho axit d. Mt khc, c 50 ml dd HCl phn ng va vi 150 ml dd NaOH. Tnh CM ca 2 dd15. Cho mt lng bt st vo dd va dd H2 SO4 1 M thu c dd A v kh B. Cho ton b dd A phn ng vi 250 ml dd KOH va . Lc kt ta ri nung n khi lng khng i thu c 20 gam cht rna. Tnh m Fe dng b. Tnh V kh ktcc. Tnh V ml dd axit d. Tnh CM dd KOH2: 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. Dng VI : Bi tp v kim loi 1. Cho cc kim loi Al, Fe, Cu, Ag. Nhng KL tc dng no tc dng c vi axit sunfuric long? dd AgNO3 ? dd NaOH ? dd H2SO4 c k thng v un nng ?. Vit cc PTHH xy ra2. Cho cc cp cht sau : a. Zn + AgCl ; Cu+ Fe(NO3)2 (dd) ; Ag + Cu(NO3)2 (dd) ; Ni + dd CuCl2 ; Al + dd AgNO3

3. Ho tan 5,5 gam hh 2 kim loi Al, Fe trong 500 ml dd HCl va thu c 4,48 lt kh ktca. Tnh % khi lng 2 kim loib. Tnh CM dd HCl47 4.Hotan20gamhhgmAg,Zn,MgtrongddH2SO40,5M(va)thuc6,72ltH2 ktc v 8,7 gam kim loi khng tan a. Tnh % khi lng mi KL b. Tnh V ml dd H2SO4

5. Nhng 594 gamAl vo dd AgNO3 2M.Sau thi gian khi lng thanh Al tng 5% so vi ban u a. Tm m Al phn ngb. Tnh m Ag thu cc. Tnh m mui Al to ra 6. NgmmtmingFe vo 320 gamdd CuSO4 10%. Sau khittc Cu bm htvo Fe, khi lng ming Fe tng 8%. Xc nh khi lng ming Fe ban u7.Cho19,6gammtKlhotrIIphnnghontonvi140mlddAgNO3thuc75,6 gam Aga. X KL b. Tnh CM dd AgNO3 c. Tnh CM dd sau phn ng (coi V khng i) 8.t chy hon ton 41,1 gam kim loi A (ho tr II) bng lng kh clo va , ho tan sp vo nc thu c ddB, cho ddB phn ng vi dd AgNO3 d, thy c 86,1 gam kt ta trng xut hina. Tm A b. Tnh VCLO ktc c. Tnh m mui to thnh9. Ho tan 13 gam kim loi A (ho tr II) bng dd HCl 2M va c dd B.Cho B phn ng vi dd AgNO3 d c 57,4 gam kt taa. Vit PTHH b. Tm A c. Tnh V dd HCl dng10.Ho tan 11,7 gam kim loi X (ho tr I) vo 120,6 gam H2O th thuc 132 gam dd Aa. Tm X b. Tnh C% dd A11. Ho tan 9 gam kim loi B (ho tr III) vo dd HCl d thu c kh C. Dn ton b C sinh ra i qua bt CuO t nng va c 32 gam cht rna. Vit PThh b. Tnh V kh C ktc c. Tm B 12. t chy ht 4,48 g KLA ho tr III bng kh Clo va , ho tan sp vo nc thu c dd B, B+ dd KOH d c kt ta C v dd D. Lc kt ta, nung nhit cao c cht rn E (m = 6,4 g). X A v cho bit thnh phn dd D. Dng VII. Bi tp v phi kim 1.T cc cht : NaCl, H2O, MnO2, HCl, KMnO4. Hy vit ptp iu ch kh clo2.T cc cht : CaCO3, Na2CO3, NaHCO3, HCl.Vit pthh iu ch kh CO2

3.Nung 30 gam vi ( tinh khit 80%) ti phn ng hon ton, kh sinh ra hp th vo 200 gam dd NaOH 5%. Sau phn ng thu c nhng mui no ? bao nhiu gam ?4.Cho 50 gam CaCO3 tc dng vi dd HCl 0,5M (d), kh sinh ra cho vo bnh cha 500ml dd KOH 2M n phta.Tnh V dd HCl, bit th nghim ly d 20% so vi lng cn thitb.Tnh CM mui sinh ra khi hp th kh trong dd kim 5.Muinobnhitphn:Na2CO3,NaHCO3,K2CO3,BaCO3,Ba(HCO3)2,KMnO4.Vitcc pthh xy ra6. Cho cc s sau : a. ABCDA l kh mu vng lc, c. D l kh khng mu, khng chy v khng duy tr s sng.Vit cc pthh, tm A,B,C,D b.X Y ZT Tm X,Y,Z,T. vit pthh. bit X l kh mu vng lc, c. T l oxit baz, rn nng chy nhit cao7.Vit 8 phn ng khc nhau iu ch CO2

47 8.Vit CTHH ca cc oxit ca C, P, Sm embit.trong s oxit no l oxit axit, vit CT axit tng ng v PTHH khi cho axit tc dng vi KOH d9.cho dng CO i qua ng ng CuO nung nng, kh ra cho hp th ht vo dd nc vi trong d thu c 16 gam kt taa.Tnh % CuO b khb.Nu ho tan cht rn cn li trong ng bngdd HNO3 c th c bao nhiu lt NO2 bay ra10.Tin hnh f 5 lt dd NaCl 2M (d = 1,2 g/ml) theo phn ng : f, mnx, c tr 2NaCl+2 H2O 2 NaOH+H2+ Cl2

Sau khi anot thot ra 89,6 lt Cl2 ktc th ngng f, H2O bay hi khng ng k. tnh C% cht tan trong dd sau in phn. B - Bi tp 3.1 Trong cc nhm cht sau, nhm no ton l phi kim. a. Cl2, O2, N2, Pb, Cb. O2, N2, S, P, I2 c. Br2, S, Ni, N2, Pd. Cl2, O2, N2, Pb, C p n:b ng. 3.2 Trong cc nhm cht phi kim sau, nhm no ton l phi kim tn ti trng thi kh trong iu kin thng: a. Cl2, O2, N2, Br2, C b. O2, N2, Cl2, Br2, I2 c. Br2, S, F2, N2, Pd. Cl2, O2, N2, F2 p n:d ng. 3.3 Trong khng kh thnh phn chnh l O2 v N2 c ln mt s kh c l Cl2 v H2S. C th cho hn hp kh ny li qua dung dch no trong cc dung dch sau loi b cc kh c. a. Dung dch NaOHb. Dung dch H2SO4 c. Ncd. Dung dch CuSO4

p n:a ng. 3.4 Kh O2 c ln mt s kh l CO2 v SO2. C th cho hn hp kh ny li qua dung dch no trong cc dung dch sau loi b cc kh c. a. Dung dch CaCl2b. Dung dch Ca(OH)2 c. Dung dch Ca(NO3)2d. Ncp n:b ng. 3.5 Khi iu ch kh SO3 bng phn ng: Na2SO3+ H2SO4 Na2SO4 + SO2 +H2O c th thu kh SO2 bng phng php: a. Di ch nc b. Di ch dung dch Ca(OH)2

c. Di ch khng khd. C a v c u ng p n:d ng. 3.6 O3 (ozon) l: 47 a. Mt dng th hnh ca oxib. L hp cht ca oxi c. Cch vit khc ca O2 d. C a v c u ng p n:d ng. 3.7 Cho s cc phn ng sau: A+ O2 C toB B+ O2 tc xc , C toC C+H2ODD+BaCl2 E+ + F A l cht no trong s cc cht sau: a. Cb. S c. Cl2d. Br2

p n:b ng. 3.8 C ba l ng ba kh ring bit l clo, hiroclorua v O2. C th dng mt cht no trong s cc cht sau ng thi nhn bit c c ba kh: a. Giy qu tm tm tb. Dung dich NaOHc. Dung dch CaCl2d. Dung dich H2SO4

p n:a ng. 3.9 C ba l ng ba dung dch ring bit l BaCl2, Ca(HCO3)2 v MgSO4 b mt nhn. C th dng mt cht no trong s cc cht sau ng thi nhn bit c c ba dung dch: a. Dung dch Ba(OH)2b. Dung dich NaOHc. Dung dch FeCl3d. Dung dich H2SO4

p n:d ng. 3.10 Trong nhng cp cht sau 1. H2SO4 v Na2CO3 2. Na2CO3 v NaCl 3. MgCO3 v CaCl24. Na2CO3 v BaCl2 nhng cp cht no c th phn ng c vi nhau: a. Cp (1) v cp (2) b. Cp (3) v cp (4)c. Cp (2) v cp (3) d. Cp (1) v cp (4) p n:d ng. 3.11 Trong nhng cp cht sau 1. Cl2 v O22. Cl2 v Cu 3. S v O24. Cl2 v Br2 nhng cp cht no c th phn ng c vi nhau: a. Cp (1) v cp (2) b. Cp (3) v cp (4)c. Cp (2) v cp (3) d. Cp (1) v cp (4) p n:c ng. 3.12 Hon thnh phng trnh s phn ng sau: A+ O2 C toB B+ O2 tc xc , C toC C+H2ODD+NaOHE+ H2OE+BaCl2 G+ + F Trong B, C l cc oxit axit, E l mt mui tan. Gii 47 Cc phng trnh phn ng: S+ O2 C toSO2 2SO2+ O2 tc xc , C to2SO3 SO3+H2OH2SO4 H2SO4+2NaOH Na2SO4+ 2H2ONa2SO4+BaCl2BaSO4+ + 2NaCl3.13 Mt cht kh c cng thc phn t l X2. Kh l kh g? Bit rng 1,0 lt kh iu kin tiu chun cn nng 3,1696 gam. Vit cc phng trnh phn ng (nu c) ca kh X2 vi cc cht sau: H2, O2, Cu, dung dch NaOH v nc. Gii: - Mt mol kh iu kin tiu chun chim th tch l 22,4 lt, nn khi lng mol phn t ca kh l: M = 2MX = 22,4. 3,1696 = 71 MX = 35,5 vy nguyn t X l Clo v kh X c cng thc phn t l Cl2. - Cc phng trnh phn ng ca Cl2 vi cc cht cho: +Cl2 +H22HCl +Cl2 +O2khng phn ng+Cl2 +CuCuCl2 +Cl2 +2NaOHNaCl + NaClO+ H2O +Cl2 +H2OHCl + HClO3.14 Cho 1,12 lt kh Cl2 (o ktc) tc dng vi H2 d, hp th ton b sn phm vo nc thu c 100,0 ml dung dch A. Tnh nng mol/l ca dung dch A. Gii - S mol kh Cl2 l: 2Cln = 4 2212 1,,=0,05 mol - Phn ng vi kh H2 d: Cl2 +H22HCl(1) Theo phng trnh phn ng (1) H2 d nn s mol kh HCl sinh ra: nHCl = 22Cln = 2.0,05 = 0,1 mol - Kh HCl tan hon ton vo nc to thnh dung dch axit HCl. - Nng dung dch HCl thu c: CHCl = ltmol1 01 0,,=1,0 mol/l (hay 1,0 M) 3.15 Cho 3,36 lt kh Cl2 (o ktc) tc dng vi H2 d, hp th ton b sn phm vo 100,0 gam nc thu c dung dch B. Tnh nng % ca dung dch B. Gii - S mol kh Cl2 l: 2Cln = 4 2236 3,,= 0,15 mol - Phn ng vi kh H2 d: Cl2 +H22HCl(1) Theo phng trnh phn ng (1) H2 d nn s mol kh HCl sinh ra: 47 nHCl = 22Cln = 2.0,15 = 0,3 mol - Kh HCl tan hon ton vo nc to thnh dung dch axit HCl. - Khi lng dung dch axit HCl thu c: mdung dch HCl = mHCl+O H2m = 36,5.0,3+100,0 = 110,95 gam - N ng % HCl trong dung dch B l: C%HCl =% .,, . ,10095 1103 0 5 36= 9,87% 3.16 Cho 2,40 gam Mg kim loi phn ng hon ton vi V lt kh X2 (o ktc) theo phng trnh phn ng sau: X2 +MgMgX2 Khi lng MgX2 thu c l 9,50 gam. Hy cho bit X2 l kh g? v tnh th tch V ca kh X2 phn ng vi Mg trn. Gii - S mol ca Mg kim loi:nHCl = 2440 2,= 0,10 mol - Phng trnh phn ng:X2 +MgMgX2(1) Theo phng trnh phn ng (1): nMg = 2Xn = 2MgXn = 0,10 mol - Khi lng mol phn t ca MgX2: 2MgXM = 10 050 9,,= 95 2MgXM = MMg +2MX = 95 MX = 35,5 vy nguyn t X l Clo v kh X c cng thc phn t l Cl2. - Th tch kh Cl2 phn ng vi Mg:

2ClV = 22,4.0,10 = 2,24 lt 3.17 Mt mui clorua kim loi cha 79,78% clo theo khi lng. Xc nh cng thc phn t ca mui. Gii - Trong cc hp cht mui clorua, clo c ho tr I.- Gi cng thc phn t ca mui l MCln, trong n l ho tr ca kim loi M.- % khi lng ca M trong hp cht l: 100% - 79,78% = 20,22% Ta c: % ,% ,Mn ,m %m %MCl22 2078 79 5 35= = M = 9n Ch c cp n = 3 v M = 27 (Al) l ph hp.Vy cng thc phn t ca mui l AlCl3. 3.18 Mt mui c cng thc phn t l FeX2 trong Fe chim 44,1% theo khi lng. Xc nh cng thc phn t ca mui v vit 3 phng trnh phn ng trc tip to thnh mui FeX2. Gii - % khi lng ca X trong hp cht l: 100% - 44,1% = 55,9% Ta c: 47 % ,% , M .MM .m %m %XFeXFeX1 449 55562 2= = = MX = 35,5 Vy X l nguyn t Clo, cng thc phn t ca mui l FeCl2. - Ba phng trnh phn ng trc tip to thnh FeCl2 l: Fe+2HCl FeCl2+ H2(1) Fe+CuCl2FeCl2+ Cu(2) FeSO4+BaCl2 FeCl2+ BaSO4+(3) 3.19 Mt mui c cng thc phn t l FeX3. Cho dung dch cha 1,30 gam FeX3 tc dng vi lng d dung dch AgNO3 thu c 3,444 gam kt ta. Xc nh cng thc phn t ca mui v vit 2 phng trnh phn ng trc tip to thnh mui FeX3. Gii - Phng trnh phn ng: FeX3+3AgNO3 Fe(NO3)3+3AgX+(1) - Gi x l s mol ca FeX3, theo phng trnh phn ng (1) th s mol ca AgX l 3xmol. - Ta c h phng trnh: 3FeXm = (56 + 3MX).x = 1,30 gam AgXm= (108 + MX) .3x = 3,444 gam MX = 35,5 v x = 0,008 mol. Vy nguyn t X l Clo v mui l FeCl3. - Hai phng trnh phn ng trc tip to thnh FeCl3 l: 2Fe+3Cl2 2FeCl3 (1) Fe2(SO4)3+3BaCl2 2FeCl3 + 3BaSO4+(2) 3.20 Ho tan 18,4 gam hn hp hai kim loi ho tr II v III bng dung dch axit HCl d thu c dung dch A v kh B. Chia kh B lm hai phn bng nhau. t chy hon ton mt phn thu c 4,5gam nc. a. Hi khi c cn dung dch A thu c bao nhiu gam mui khan? b.emphn2chophnnghontonvikhclorichosnphmhpthvo200,0mldungdch NaOH 20% (d = 1,20 gam/ml). Tnh nng % ca cc cht trong dung dch thu c. Gii: Gi kim loi ho tr II l X c s mol trong 18,4 gam hn hp l x mol.Gi kim loi ho tr III l Y c s mol trong 18,4 gam hn hp l y mol.Phng trnh phn ng: X+2HClXCl2+H2(1) 2Y+6HCl2YCl3+3H2(2) Dung dch A cha XCl2, YCl3 v HCl c th d, kh B l H2. t chy mt na kh B; 2H2+O2 ot2H2O(3) a. Theo cc phng trnh phn ng t (1) - (3): mol , y x n,y x n nH H O H5 023185 42321212 2 2= |.|

\|+ = = |.|

\|+ = =S mol HCl tham gia phn ng:mol n n 0 1232 22, y xH HCl= |.|

\|+ = =47 Theonhlutbotonkhilng,khiccndungdchAlngmuithucl:gam m m m 9 53 5 0 2 0 1 5 36 4 183 2khan mui, , . , . , ,YCl XCl= + = + =b. Phn 2 tc dng vi clo: H2+Cl2 ot2HCl(4) Hp th HCl vo dung dch NaOH: HCl + NaOHNaCl+H2O(5) S mol HCl:mol , y xnnHHCl5 023222=|.|

\| + = = S mol NaOH:mol ,% .% . , . ,nNaOH2 1100 4020 2 1 0 200= =nHCl < nNaOH NaOH d Trong dung dch thu c gm NaOH d v NaCl c s mol:nNaOH d = 1,2 - 0,5 = 0,7 mol v nNaCl = nHCl = 0,5 mol Khi lng dung dch thu c: mdd = 200,0.1,2 + 36,5.0,5 = 258,25 gam Nng cc cht trong dung dch: % , % .,, .% C% , % .,, . ,% CNaOHNaCl84 10 10025 2587 0 4033 11 10025 2585 0 5 58= == = 3.21 Tnh th tch kh clo thu c iu kin tiu chun khi un nng nh 1,58 gam KMnO4 vi dung dch axit clohiric c d. Gii - S mol ca KMnO4: 4KMnOn =15858 1,= 0,010 mol - Phng trnh phn ng:2KMnO4 +16HCl ot2KCl +2MnCl2+5Cl2+8H2O(1) - Theo phng trnh phn ng (1) s mol ca Cl2 sinh ra: 2Cln = 254KMnOn = 0,025 mol - Th tch kh Cl2 thu c: 2ClV = 22,4.0,025 = 0,56 lt 3.22 Tnh th tch kh clo thu c iu kin tiu chun khiun nng nh 2,61 gam MnO2 vi dung dch axit clohiric c d. Lng clo ny phn ng ht bao nhiu gam st kim loi.Gii - S mol ca MnO2: 2MnOn =8761 2,= 0,030 mol - Phng trnh phn ng:MnO2 +4HCl otMnCl2+Cl2+2H2O(1) - Theo phng trnh phn ng (1) s mol ca Cl2 sinh ra:47 2Cln = 2MnOn = 0,030 mol - Th tch kh Cl2 thu c: 2ClV = 22,4.0,030 = 0,672 lt - Phn ng vi Fe: 3Cl2 + 2Fe2FeCl3 (2) nFe = 322Cln = 0,02 mol - Khi lng st tham gia phn ng: mFe = 56.0,02 = 1,12 gam3.23 in phn c mng ngn dung dch NaCl bo ho bng dng in mt chiu thu c 33,6 lt kh clo iu kin tiu chun. Tnh khi lng mui dung dch nc Gia - ven thu c khi cho lng kh clo ny phn ng hon ton vi 200,0 gam dung dch NaOH 60%. Gii - Phng trnh phn ng in phn:2NaCl(dd bo ho) + 2H2O 2NaOH + Cl2 +H2 (1) - S mol ca Cl2 thu c: 2Cln = 4 226 33,, = 1,5 mol - S mol ca NaOH c trong 200,0 gam dung dch: nNaOH = % .% . ,100 4060 0 200= 3,0 mol - Phn ng ca clo vi NaOH: Cl2 + 2NaOHNaCl+NaClO +H2O(2) - S mol NaOH gp hai ln s mol Cl2 nn phn ng va . - Khi lng dung dch nc Gia - ven thu c: m = mdung dch NaOH +2Clm = 200,0 + 71.0,15 = 3,6,5 gam 3.24TinhnhinphncmngngndungdchNaClbohobngdnginmtchiuthuc 33,6m3khcloiukintiuchun.TnhkhilngmuiNaCleminphn,vtnhkhi lng NaOH thu c trong qu trnh in phn. Bit hiu sut thu hi kh clo l 95%. Gii - S mol ca Cl2 thu c: 2Cln = 4 226 33,,.103 = 1,5.103 mol - Phng trnh phn ng in phn:2NaCl(dd bo ho) + 2H2O2NaOH + Cl2 +H2(1) - S mol ca NaCl em in phn v s mol NaOH thu c: nNaCl = nNaOH = 22Cln = 1.1,5.103 = 3.103 mol - Khi lng NaCl cn dng: mNaCl = 3.103.58,5. %%95100=184,74.103 gam = 184,74 kg - Khi lng NaOH tc dng: in phn c mng ngn in phn c mng ngn 47 mNaOH = 3.103.40. %%95100=126,32.103 gam = 126,32 kg 3.25 Hon thnh cc phng trnh phn ng theo s bin ho sau: Cl2 2H HCl 2) OH ( CaCaCl2 NaCl NaCl NaClCaCO3 Na O H2NaOH 2CO Na2CO3 Gii Cc phng trnh phn ng: 2NaCl 2Na +Cl2 (1) 2Na +Cl2 2NaCl(2) H2+Cl2 ot2HCl(3) 2Na+2H2O2NaOH+H2 (4) HCl + NaOHNaCl+H2O(5) 2HCl + Ca(OH)2CaCl2+2H2O(6) CO2 + 2NaOHNa2CO3+H2O(7) Na2CO3 +CaCl2CaCO3 + 2NaCl(8) 3.26 Kim cng l: a. Hp cht ca cacbon vi kim loib. L hp cht ca cacbon vi phi kim c. Mt dng th hnh ca cacbon d. C a v b u ng p n:c ng. 3.27 Chn cu ng trong cc cu sau: a. Cc dng th hnh ng ca cacbon l: kim cng, than ch v than g. b. Cc dng th hnh ng ca cacbon l: kim cng, than ch v cacbon v nh hnh. c. Cc dng th hnh ng ca cacbon l: kim cng, than ch v than hot tnh. d. Cc dng th hnh ng ca cacbon l: kim cng, than ch v than . p n:b ng. 3.28 Kh nng hp ph cao l c tnh ca: a. Than b. Kim cng c. Than chd. Than hot tnh p n:d ng. 3.29 Trong cc phn ng ho hc sau: C+O2 otCO2+ Q(1) C+2CuO otCO2+ 2Cu(2) cacbon lun l: a. Cht oxi hob. Cht kh c. L cht oxi ho v cht khd. Khng l cht oxi ho v cht kh p n:b ng. 3.30 Cacbon oxit (CO) l: a. Oxit axitb. Oxit baz c. Oxit trung tnhd. Oxit lng tnh in phn nng chy 47 p n:c ng. 3.31 Trong cc phn ng ho hc sau: 2CO+O2 ot2CO2+ Q(1) CO+CuO otCO2+ Cu(2) cacbon oxit lun l: a. Cht oxi hob. Khng l cht oxi ho v cht kh c. L cht oxi ho v cht khd. Cht khp n:d ng. 3.32 Cacbon ioxit (hay cn gi l anhirit cacbonic, kh cacbonic: CO2) l: a. Oxit axitb. Oxit baz c. Oxit trung tnhd. Oxit lng tnh p n:a ng. 3.33 Nguyn t R to thnh vi hiro mt hp cht c cng thc phn t RH4. R l nguyn t no trong cc nguyn t sau: a. Sb. Sic. Cd. P p n:a ng. 3.34 Hp th ton b 2,24 lt kh CO2 (o ktc) vo 100,0 ml dung dch NaOH 1,5 M. Dung dch thu c cha nhng mui no? a. NaHCO3b. Na2CO3 c. NaHCO3 v Na2CO3d. Phn ng khng to mui p n:c ng. 3.35 Mt vin than t ong c khi lng 350,0 gam cha 60% cacbon theo khi lng. Tnh nhit lng to ra khi t chy hon ton mt vin than ny. Bit khit chy 1 mol cacbon sinh ra lng nhit l 394 kJ. Gii - Phn ng chy: C+O2 otCO2+ Q - S mol cacbon c trong mt vin than t ong l: nC = % .% .100 1260 350 = 17,5 mol - Lng nhit to ra khi t chy hon ton mt vin than t ong l: Q = 17,5.394 = 6895 kJ 3.36 Tnh th tch kh CO cn ly iu kin tiu chun kh ht 8,0 gam CuO. Bit rng hiu sut phn ng kh l 80%. Gii - S mol CuO: nCuO = 800 8, = 0,10 mol - Phn ng kh CuO CO+CuO otCO2+ Cu - Theo phng trnh phn ng s mol CO bng s mol CuO: nCO = nCuO = 0,10 mol - Th tch CO cn ly: 47 nCO =%% . , . ,80100 4 22 10 0= 2,80 lt 3.37Dn 22,4 lt hn hp kh A gm CO v CO2 qua dung dch NaOH d thy c 1,12 lt kh thot ra. Tnh % theo th tch v % theo khi lng ca hn hp kh A. Bit cc th tch u o iu kin tiu chun. Gii - Gi s mol kh CO trong hn hp A l x mol. - Gi s mol kh CO2 trong hn hp A l y mol. - Khi cho hn hp kh A qua dung dch NaOH: CO2 + 2NaOHNa2CO3+H2O - Kh i ra khi dung dch l CO - Ta c cc phng trnh: nA = nCO+2COn = x+y = 4 224 22,,= 1,0 mol nCO = x= 4 2224 2,,= 0,10 mol2COn =y = 0,90 mol - % theo th tch cc kh trong hn hp A: % % .,,% .CO90 1000 19 01002= =+=y xy%n% % .,,% .CO10 1000 11 0100 = =+=y xx%n- % theo khi lng cc kh trong hn hp A: % , % ., . , ., .% .CO4 93 1009 0 44 1 0 289 0 4410044 28442=+=+=y xy%m% , % ., . , ., .% .CO6 6 1009 0 44 1 0 281 0 2810044 2828=+=+=y xx%m3.38Dntt16,8ltkhCO2vo600,0mldungdchCa(OH)21,0M.Tnhkhilngkttathu c.Gii - S mol kh CO2: 2COn =4 228 16,,= 0,75 mol. - S mol Ca(OH)2 trong dung dch: 2) OH ( Can = 0,6.1,0 = 0,60 mol. - S mol kh CO2 ln hn s mol Ca(OH)2 nn to thnh 2 mui: CO2+Ca(OH)2 CaCO3++H2O(1) CO2+CaCO3 + H2OCa(HCO3)2(2) - Gi s mol mui CaCO3 l x mol. - Gi s mol mui Ca(HCO3)2 y mol. - Ta c cc phng trnh: 2) OH ( Can =x+y = 0,60 mol 2COn = x+2y = 0,75 mol. 47 3CaCOn =x = 0,45 mol - Khi lng kt ta CaCO3:m = 100.0,45 = 45,0 gam3.39Hn hp kh A gm CO v CO2 v kh X. Xc nh kh X c trong hn hp bit rng trong hn hp kh A kh CO c s mol gp 3 ln s mol kh CO2 v hn hp kh A c khi lng mol trung bnh l 32. Gii - Gi s hn hp A c tng s mol kh l 1,0 mol. Gi s mol CO2 trong hn hp l x mol, khi s mol CO l 3x v s mol kh X l 1,0 - 4x. - Khi lng mol trung bnh ca hn hp:

14 0 1 3 28 44 ) , ( .Ax M x xMX + += = 32 324 0 14 0 1 324 0 1128 32=== xxxxMX,) , (, X l kh c khi lng mol l 32 ch c th l O2. 3.40 Dn t t 6,72 lt kh CO2 vo 300,0 ml dung dch NaOH 1,20 M.a. Tnh tng khi lng cc mui trong dung dch thu c.b. Tnh khi lng kt ta khi cho BaCl2 d vo dung dch sau lhi hp th CO2. Gii a- S mol kh CO2: 2COn =4 2272 6,,= 0,30 mol. - S mol NaOH trong dung dch: nNaOH = 0,3.1,20 = 0,36 mol. 2COn