DANH MC CC K HIU V CH VIT TT Cng thc cu to: CTCT Dung dch: dd iu kin tiu chun (0oC v 1 atm): ktc

Khi lng: m

Kim loi: KL

Nguyn t khi: NTK

Nng mol/l ~ nng mol ~ CM

Nng phn trm: C%

Th tch: V

Phng trnh ha hc: PTHH S mol: n

Kt ta:

Kh thot ra:

Khi lng mol: M Hn hp: hh Nhit : t K hiu ha hc: KHHH

PHN I. M U

I. L do chn ti

Ha hc l khoa hc nghin cu cc cht, s bin i v ng dng ca chng. Nhng khng ch hc trn l thuyt m cn phi vn dng vo gii thch cc hin tng ca i sng hng ngy v cn phi gii quyt cc bi ton c lin quan. Vic lm cc bi tp Ha hc khng ch gip cng c kin thc m cn gip cho hc sinh thm hng th vi mn hc hn. c bit i vi cc em hc kh, gii mun lm nhiu bi tp nng cao k nng ca mnh hn na.

Trong chng trnh THCS ni dung mn Ha hc bao gm cc phn v cht nguyn t - phn t; phn ng ha hc; mol v tnh ton ha hc; oxi khng kh; hiro nc; dung dch (lp 8). Cc loi hp cht v c; kim loi; phi kim s lc v bng tun hon cc nguyn t ha hc; hirocacbon nhin liu; dn xut ca hirocacbon polime (lp 9). Trong cc ni dung trn ti mun i su v phn kim loi, c th l cc dng bi tp c lin quan n phn ny vi mc ch tm hiu v xy dng thnh h thng cc dng bi tp nng cao chuyn v kim loi THCS. Cc dng bi tp v kim loi rt hay v phong ph, nhng nu ch lm cc bi tp trong sch gio khoa v sch bi tp thi th ta s khng khai thc ht c cc dng bi tp v ci hay ca n, chnh v vy ti mun chn ti: Tuyn chn h thng bi tp v c chuyn kim loi THCS.

II. Mc ch, nhim v nghin cu

II.1. Mc ch nghin cuNghin cu xy dng thnh h thng cc bi tp nng cao v kim loi trong chng trnh Ha hc ca THCS theo hng pht huy tnh tch cc, sng to v rn luyn nhng kh nng tip cn vi cc bi tp nng cao t hnh thnh k nng tnh ton khi gii cc bi tp khng ch THCS m cn phc v cho qu trnh hc sau ny vi cp cao hn.II.2. Nhim v nghin cu

- Nghin cu cc ni dung tnh cht ca kim loi trn c s tm hiu cc dng bi tp c lin quan n nhng tnh cht .

- a ra cc dng bi tp c bn v nng cao nhng trng tm l cc dng bi tp nng cao v phn kim loi nm trong chng trnh Ha hc THCS.

- Su tm, tm kim cc dng bi tp kh xy dng thnh h thng bi tp nng cao.

- Tng hp v su tm cc phng php gii chi tit v c th.

III. i tng v phng php nghin cu

III.1. i tng nghin cu

Cc dng bi tp nng cao v kim loi trong chng trnh Ha hc THCS.

III.2. Phng php nghin cu

- Su tm cc bi tp nng cao v kim loi.

- Phn loi thnh cc dng khc nhau, sau nu ra cc bi tp c hng dn gii c th.PHN II. NI DUNG Chng 1. C s v tng quan

I. V tr cc bi tp kim loi trong chng trnh SGK Ha hc ca THCS.1. Ha hc lp 8

nh lut bo ton khi lng

Phng trnh ha hc Tnh theo cng thc ha hc v phng trnh ha hc 2. Ha hc lp 9

Chng 1: Cc loi hp cht v c

Chng 2: Kim loi

II. Cc dng bi tp c bn

1. Dng bi tp nh tnh c tnh thc t.2. Bi tp lp cng thc ca mt cht v c v xc nh nguyn t kim loi.3. Bi ton tnh theo cng thc ha hc.4. Bi ton tnh theo phng trnh ha hc.Chng 2. Ni dung nghin cu. Xy dng h thng bi tp nng cao trong chng trnh Ha hc ca THCS.

1. Bi tp l thuyt

Gii thch hin tng v vit phng trnh ha hc

iu ch kim loi

Phn bit v nhn bit kim loi

Tinh ch v tch hn hp thnh cht nguyn cht

2. Bi tp tnh ton

-Bi ton xc nh tn kim loi v cng thc cc hp cht ca chng

- Bi ton hn hp

Bi ton v lng cht d.

Bi ton tng gim khi lng.

Bi ton bin lun

A. BI TP L THUYT I. Gii thch hin tng v vit PTHH

Dng bi ny yu cu ngi hc sinh phi nm r tnh cht ca cc kim loi v c k nng thnh tho trong vic nhn bit hin tng ca phn ng ha hc t gii thch v vit PTHH.

V d: Ha tan Fe bng HCl v sc kh Cl2 i qua hoc cho KOH vo dung dch v lu ngoi khng kh. Gii thch hin tng v vit PTHH. Hng dn giiKhi cho Fe tc dng vi HCl thy c kh thot ra: Fe + 2HCl FeCl2 + H2

2FeCl2 + Cl2 2FeCl3 dung dch chuyn mu vng.

FeCl2 + 2KOH Fe(OH)2 + 2KCl c kt ta trng xanh.

4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3 kt ta chuyn mu nu .

Bi tp vn dng Bi 1. Dung dch M c cha CuSO4 v FeSO4a. Cho Al vo dung dch M, sau phn ng c dung dch N cha 3 mui tan.

b. Cho Al vo dung dch M, sau phn ng c dung dch N cha 2 mui tan.c. Cho Al vo dung dch M, sau phn ng c dung dch N cha 1 mui tan.

Gii thch mi trng hp bng phng trnh phn ng. II. iu ch kim loi v hp cht ca chngThc cht y l kiu bi tp thc hin qu trnh bin ha nhng ch cho bit cht u v cht cui. Hc sinh phi suy ngh v la chn con ng ng nht v ngn nht thc hin (v cht c iu ch c phi tinh khit v v nguyn tc nu i bng con ng di hn nhng khng sai th vn gii quyt c yu cu ca bi nhng s mt nhiu thi gian vit phng trnh dng n mt cch khng cn thit).1. S phn ng: Dng bi ny thng bao gm mt chui phn ng ha hc yu cu phi nm c tnh cht ha hc ca tng cht trong chui phn ng v vit PTHH hon thnh chui phn ng

V d: Vit phng trnh phn ng hon thnh s sau: FeCl2 FeSO4 Fe(NO3)2 Fe(OH)2 Fe Fe2O3 FeCl3 Fe2(SO4)3 Fe(NO3)3 Fe(OH)3 Hng dn giiFe + 2HCl FeCl2 + H2FeCl2 + Ag2SO4 FeSO4 + 2AgCl

FeSO4 + Ba(NO3)2 Fe(NO3)2 + BaSO4

Fe(NO3)2 + 2KOH Fe(OH)2 + 2KNO3 tFe(OH)2 + O2 Fe2O3 + H2O

2Fe + 3Cl2 2FeCl32FeCl3 + 3Ag2SO4 Fe2(SO4)3 + 6AgCl

Fe2(SO4)3 + 3Ba(NO3)2 2Fe(NO3)3 + 3BaSO4

Fe(NO3)3 + 3KOH Fe(OH)3 + 3KNO3

t2Fe(OH)3 Fe2O3 + H2O2FeCl2 (lc nht)+ Cl2 2FeCl3(vng nu)2FeCl3 + Fe 3FeCl210FeSO4 + 2KMnO4 + 8H2SO4 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

Fe2(SO4)3 + Fe 3FeSO44Fe(NO3)2 + O2 + 4HNO3 4Fe(NO3)3 + 2H2O

2Fe(NO3)3 + Cu 2Fe(NO3)2 + Cu(NO3)24Fe(OH)2(trng xanh) + O2 + 2H2O 4Fe(OH)3(nu ) 2. in cht v hon thnh PTHHDng bi ny mi phn ng u bit c cht tham gia hoc cht to thnh bi ch yu cu in vo nhng ch trng sao cho thch hp hon thnh PTHH.

V d: Hon thnh cc phn ng sau:Fe2(SO4)3 + ? Fe(NO3)3 + ? AlCl3 + ? Al2(SO4)3 + ? Al2O3 + KHSO4 ? + ? + ? KHCO3 + Ca(OH)2 ? + ? + ?

NaCl + ? NaOH + ?

Ca(HCO3)2 + ? CaCO3 + ?

Hng dn giiFe2(SO4)3 + 3Ba(NO3)2 2Fe(NO3)3 + 3BaSO4

2AlCl3 + 3Ag2SO4 Al2(SO4)3 + 6 AgCl

Al2O3 + 6KHSO4 Al2(SO4)3 + 3K2SO4 + 3H2O

KHCO3 + Ca(OH)2 CaCO3 + KOH + H2O

in phn c vch ngn:

2NaCl + 2H2O 2NaOH + H2 + Cl2

Ca(HCO3)2 +K2CO3 CaCO3 + 2KHCO3

3. iu ch mt cht t nhiu cht bng nhiu cch:Dng ny ta phi s dng nhiu cch khc nhau iu ch ra mt cht. lm c dng ny th hc sinh cng cn phi nm r tnh cht khng ch ring v kim loi m cn cc hp cht khc lin quan n v i hi cht iu ch c phi tinh khit.

V d 1: Vit cc PT phn ng ch ra:

- 4 cch iu ch Al(OH)3 - 6 cch iu ch FeCl2, Hng dn gii - 4 cch iu ch Al(OH)3: + Kim loi + H2O

+ Oxit kim loi + H2O

+ in phn dung dch mui clorua (c vch ngn)

+ Mui + kim

+ Thy phn mui

+ Mui + axit

AlCl3 + 3NaOH Al(OH)3 + 3NaOH

2AlCl3 + 6 H2O 2Al(OH)3 + 3H2 + 3Cl2

Al4C3 + 12H2O 4Al(OH)3 + 3CH4NaAlO2 + HCl + H2O Al(OH)3 + NaCl

- 6 cch iu ch FeCl2:

Fe + 2HCl FeCl2 + H2Fe + CuCl2 FeCl2 + Cu

FeO + 2HCl FeCl2 + H2O

Fe(OH)2 + 2HCl FeCl2 + 2H2O

FeS + 2HCl FeCl2 + H2S

FeCO3 + 2HCl FeCl2 + CO2 + H2O

2FeCl3 + Fe 3FeCl2FeBr2 + Cl2 FeCl2 + Br2V d 2: Nu cch iu ch Na2CO3 Na ; Al(NO3)3 Al ; FeS2 Fe Hng dn gii+ iu ch Na t Na2CO3

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

C cn dd v in phn nng chy 2NaCl 2Na + Cl2

+ iu ch Al t Al(NO3)3

Al(NO3)3 + 3KOH Al(OH)3 + 3KNO3

2Al(OH)3 Al2O3 + 3H2O

in phn nng chy: 2Al2O3 4Al + 3O2

+ iu ch Fe t Fe2S:

4Fe2S + 11O2 2Fe2O3 + 8SO2

Fe2O3 + 3CO 2Fe + CO2 Bi tp vn dng

Bi 1. Vit phng trnh phn ng biu din cc chuyn ha sau:

a. CuSO4 B C D Cu

b. FeS2 Fe2O3 Fe2(SO4)3 FeCl3 Fe(OH)3Bi 2. Hon thnh s phn ng di y.

Al2O3 Al2(SO4)3 NaAlO2Al Al(OH)3

AlCl3 Al(NO3)3 Al2O3 III. Phn bit v nhn bit cc cht1. L thuyt c bn v thuc th ha hc lp 9 ca THCS(p dng nhn bit v phn bit kim loi)

KL, IonThuc thHin tngGii thch, vit PTHH

Na, KH2OTan + dd trongNa + H2O NaOH + 1/2 H2 K + H2O KOH + 1/2 H2

CaH2OTan + dd cCa + H2O Ca(OH)2 + H2

BaH2OAxit H2SO4

Tan+dd trong trngBa + H2O Ba(OH)2 + H2Ba + H2SO4 BaSO4 + H2

AlAl3+Dd kimDd NH3 dTantrng, khng tan Al + H2O + NaOH NaAlO2 + 3/2H2Al3+ +NH3 + H2O Al(OH)3 + NH4+

Zn2+Dd NH3 d trng sau tanZn2+ + NH3 + H2O Zn(OH)2 + NH4+ Zn(OH)2 + NH3 [Zn(NH3)4](OH)2

Fe Fe2+Fe3+Kh CloDd NaOH

Dd NaOH, NH3Trng xm nu trng xanh ha nu

nu

2Fe(trng xm) + 3Cl2(vng lc) 2FeCl3(nu )Fe2+ + 2OH- Fe(OH)2 (trng xanh)Fe(OH)2 + O2 + H2O Fe(OH)3 (nu )Fe3+ + 3OH- Fe(OH)3

Fe3+ + NH3 + H2O Fe(OH)3 + NH4+

HgHNO3 cTan, kh mu nuHg + 4HNO3 Hg(NO3)2 + 2NO2+ H2O

CuCu2+Cu ()HNO3 cDd NH3 d

AgNO3Tan, dd xanh, kh mu nu xanh sau tan

Tan, dd xanh Cu + 4HNO3 Cu(NO3)2 + NO2 + 2H2O

Cu2+ + NH3 + H2O Cu(OH)2 + NH4+Cu(OH)2 + NH3 [Cu(NH3)4](OH)2

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

Ag

Ag+HNO3 sau cho NaClDd H2S, dd NaOH

Tan, kh mu nu v kt ta trngKt ta enAg + 2HNO3 AgNO3 + NO2 + H2OAgNO3 + NaCl AgCl + NaNO3Ag+ + S2- Ag2S

Ag+ + OH- AgOH

2AgOH Ag2O + H2O

MgMg2+Dd HClDd CO32-Tan, c khtrngMg + 2HCl MgCl2 + H2Mg2+ + CO32- MgCO3

PbPb2+Dd HClDd H2S trngenPb + 2HCl PbCl2 + H2Pb2+ + S2- PbS

Na

K

Ca

Bat trn ngn la v quan st- Mu vng ti

- Mu tm (tm hng)

- Mu da cam

- Mu lc (hi vng)

2. Mt s trng hp nhn bit. Nhn bit bng thuc th t chn.y l loi bi nhn bit m thuc th s dng khng b gh p m c la chn t do. Tuy nhin thuc th la chn phi nhn bit c r tng cht v phi thch hp.V d: C 8 dung dch cha: NaNO3, Mg(NO3)2, Fe(NO3)2, Cu(NO3)2, Na2SO4, MgSO4, FeSO4, CuSO4. Hy nu cc thuc th v trnh by phng n phn bit 8 dung dch ni trn. Hng dn giiThuc th phn bit l: dd BaCl2, dd NaOH. Cch lm nh sau:

- Cho dd BaCl2 vo 8 dung dch s thy 4 dung dch c kt ta l: Na2SO4, MgSO4, FeSO4, CuSO4 (nhm A) cn 4 dung dch khng c hin tng g l: NaNO3, Mg(NO3)2, Fe(NO3)2, Cu(NO3)2 (nhm B). - Trong mi nhm A, B u dng dd NaOH th:

Nhn ra Na2SO4 v NaNO3 khng c hin tng g

Nhn ra CuSO4 v Cu(NO3)2 to kt ta mu xanh:

CuSO4 + 2NaOH Cu(OH)2 + Na2SO4 Xanh

Nhn ra MgSO4 v Mg(NO3)2 to kt ta mu trng:

Mg(NO3)2 + 2NaOH Mg(OH)2 + 2NaNO3 Trng Nhn ra FeSO4 v Fe(NO3)2 to kt ta mu trng hi xanh, sau mt lc kt ta s chuyn thnh mu nu FeSO4 + 2NaOH Fe(OH)2 + Na2SO4 4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3 nu Nhn bit ch bng thuc th qui nhy l dng bi tp bi cho sn mt loi thuc th nht nh v yu cu ch dng thuc th ny nhn bit mt lot cc cht m bi yu cho.V d: Nhn bit cc cht trong mi cp di y ch bng dung dch HCl

a. 4 dung dch : MgSO4, NaOH, BaCl2, NaCl

b. 4 cht rn : NaCl, Na2CO3, BaCO3, BaSO4Hng dn giia. Xt kh nng phn ng ca 4 cht, nhn c ch c MgSO4 to c kt ta vi 2 dung dch khc:

MgSO4 + 2NaOH Mg(OH)2 + Na2SO4 MgSO4 + BaCl2 BaSO4 + MgCl2Suy ra dung dch cn li khng kt ta l NaCl.

- Dng axit HCl ha tan 2 kt ta thy kt ta khng tan l BaSO4 nhn c BaCl2, kt ta tan l Mg(OH)2 + 2HCl MgCl2 + 2H2O th nhn c NaOHb. Ha tan 4 cht rn bng dung dch HCl nhn c BaSO4 khng tan, NaCl tan m khng c kh bay ra. Cn:

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

BaCO3 + 2HCl BaCl2 + CO2 + H2O

- Th ln lt 2 cht rn Na2CO3, BaCO3 vo 2 dung dch va to ra s nhn ra Na2CO3 c kt ta: Na2CO3 + BaCl2 BaCO3 + 2NaCl

Cn li l BaCO3. Khng dng thuc th khc, ch dng cht ca u bi phn bit cc cht cho. Bi tp ny s dng phng php sau:

Da vo mu sc ca cc dung dch. Cc phn ng ha hc c trng ca cc ha cht cn nhn bit. Lp bng nhn bit.V d: C 5 l mt nhn, mi l ng mt trong cc dung dch sau y: NaHSO4, KHCO3, Na2SO3, Mg(HCO3)2, Ba(HCO3)2. Trnh by cch nhn bit tng dung dch ch c dng thm cch un nng.Hng dn gii- un nng cc mu th ng cc ha cht trn, c hai ng nghim cho kt ta v kh bay ln, 3 ng nghim khng cho kt ta.

t Mg(HCO3)2 MgCO3 + CO2 + H2O t Ba(HCO3)2 BaCO3 + CO2 + H2O

- Ly vi git dung dch mt trong hai l ng cc dung dch c kt ta khi un nng trn nh vo cc ng nghim ng cc dung dch khc, mt ng nghim thy c kh bay ln l NaHSO4.

2NaHSO4 + Mg(HCO3)2 Na2SO4 + MgSO4 + 2CO2 + 2H2O

2NaHSO4 + Ba(HCO3)2 Na2SO4 + BaSO4 + 2CO2 + 2H2O

Nh vy cht trong dung dch l no va cho kt ta va c kh bay ln l ng Ba(HCO3)2, l kia l Mg(HCO3)2.- Ly vi git Ba(HCO3)2 bit nh vo hai ng nghim cha cc cht cn li, ng nghim no cho kt ta l Na2SO3 Na2SO3 + Ba(HCO3)2 BaSO3 + 2NaHCO3ng nghim cn li cha dung dch KHCO3. Bi tp vn dng

Bi 1. Hy nhn bit ch bng 2 ha cht n gin t chn:

a. 9 cht rn : Ag2O, BaO, MgO, MnO2, Al2O3, FeO, Fe2O3, CaCO3, CuOb. 6 cht bt: Mg(OH)2, Zn(OH)2, Fe(OH)3, BaCl2, sa, xt n da

c. 3 dung dch: NaCl, HCl, NaNO3, ch bng 2 kim loi.

d. 4 cht bt : Na2CO3, NaCl, BaCO3, BaSO4 ch bng CO2, H2O

Bi 2. C 8 dung dch cha: NaNO3, Mg(NO3)2, Fe(NO3)2, Cu(NO3)2, Na2SO4, MgSO4, FeSO4, CuSO4. Hy nu cc thuc th v trnh by phng n phn bit 8 dung dch ni trn. IV. Tinh ch v tch hn hp thnh cht nguyn cht

* Nguyn tc:

- Bc 1. Chn cht X ch tc dng vi A (m khng tc dng vi B) chuyn A thnh dng A1 kt ta, bay hi, hoc ha tan; tch ra khi B (bng cch lc hoc t tch).

- Bc 2. iu ch li cht A t cht A1S tng qut:

+ X B

A, B + Y A1 (,, tan) A

Nu hn hp A, B u tc dng c vi X th dng cht X' chuyn c A, B thnh A', B' ri tch A', B' thnh 2 cht nguyn cht. Sau tin hnh bc 2 (iu ch li A t A')

V d: Nu phng php tch hn hp gm MgO, Fe2O3, CuO th rn thnh cc cht nguyn cht. Hng dn giiTrc tin ta s kh cc oxit kim loi trn bng hiro nhit cao (ch c oxit kim loi ng sau nhm mi b kh)

Ta c phn ng kh nh sau: CuO + H2 Cu + H2O; Fe2O3 + 3H2 2Fe + 3H2O

Cn li MgO khng b kh. Sau ta cho cc cht thu c tc dng vi axit HCl th Cu khng phn ng v b oxi ha ngoi khng kh to thnh CuO:

2Cu + O2 2CuO. Ta tch c CuO ra khi hn hp.

MgO + 2HCl MgCl2 + H2 ; Fe + 2HCl FeCl2 + H2

Hai mui thu c l MgCl2 v FeCl2 ta cho in phn dung dch th FeCl2 b in phn to thnh Fe, sau Fe b oxi ha thnh Fe2O3 ta tch c Fe2O3

Mui MgCl2 khng b in phn dung dch th ta in phn nng chy to thnh Mg, sau t nng th Mg bc chy trong khng kh to ra MgO:

MgCl2 Mg + Cl2; 2Mg + O2 2MgO Cui cng ta tch c c ba cht trn ra khi hn hp thnh cc cht nguyn cht.

Bi tp vn dng

Bi 1. Qung nhm c Al2O3 ln vi cc tp cht l Fe2O3 v SiO2. Hy nu phn ng nhm tch ring tng oxit ra khi qung nhm. Bi 2. Mt hn hp gm Al, Fe, Cu v Ag. Bng phng php ha hc hy tch ri hon ton cc kim loi ra khi hn hp trn.

B. BI TP TNH TON I. Gii thiu mt s phng php gii Phng php 1. p dng phng php bo ton khi lng v bo ton nguyn t.

- Nguyn tc: Trong cc phn ng ha hc, nguyn t v khi lng ca chng c bo ton.

T suy ra:

tng khi lng cc cht tham gia phn ng bng tng khi lng cht to thnh.

Tng khi lng cc cht trc phn ng bng tng khi lng cc cht sau phn ng.

Phng php 2. p dng khi cho kim loi tc dng vi dd mui.

- Nguyn tc:

Khi bi cho kim loi tc dng vi dung dch mui:

- Kim loi mnh (tr nhng kim loi tc dng vi nc) y kim loi yu ra khi dung dch mui ca kim loi yu.

- Khi cho thanh kim loi vo dung dch mui, sau phn ng khi lng thanh kim loi tng hoc gim:

Vit phng trnh ha hc. Di mi phng trnh ha hc t n s theo s mol cht, sau quy s mol ra khi lng (theo n s trn)

Nu khi lng thanh kim loi tng. Lp phng trnh i s

m kim loi gii phng m kim loi tan = m kim loi tng Nu khi lng thanh kim loi gim:

m kim loi tan m kim loi gii phng = m kim loi gim

- Khi cho thanh kim loi vo dung dch mui, sau khi ly ming kim loi ra th thy khi lng dung dch gim. Ta lp lun nh sau:

m cc cht tham gia = m cht to thnh

m thanh kim loAi + m dd = m' thanh kim loi + m'' dd

Theo nh lut bo tonkhi lng, nu sau phn ng khi lng dung dch nh i bao nhiu c ngha l khi lng dung dch nh i bao nhiu c ngha l khi lng thanh kim loi tng ln by nhiu.

Phng php 3. p dng phng php tng gim khi lng.

- Nguyn tc: So snh khi lng ca cht cn xc nh vi cht m gi thit cho bit lng ca n, t khi lng tng (hay gim) ny, kt hp vi quan h t l mol gia hai cht ny trong phn ng m gii quyt yu cu t ra.

Theo trnh t cc bc sau:

Xc nh mi lin h t l mol gia cht bit (cht A) vi cht cn xc nh (cht B) (c th khng cn thit phi vit phng trnh phn ng, ch lp s phn ng gia hai cht ny nhng phi da vo nh lut bo ton nguyn t xc nh t l mol gia chng).

Xt xem khi chuyn t cht A thnh B (hay ngc li) th khi lng tng ln hay gim xung bao nhiu gam theo t l phn ng v theo cho.

Sau da vo quy tc tam sut, lp phng trnh ton hc gii.

Phng php 4. p dng phng php tng gim th tch.

- Nguyn tc: So snh th tch (hoc s mol) ca cht cn xc nh vi cht m gi thit cho bit lng ca n, t th tch (hoc s mol) tng (hay gim) ny, kt hp vi quan h t l mol gia 2 cht ny trong phn ng m gii quyt yu cu t ra.

Theo trnh t cc bc tng t phng php 3.

Phng php 5. p dng phng php chuyn bi ton hn hp thnh mt cht tng ng (hay phng php s dng i lng trung bnh).

- Nguyn tc: khi trong bi ton xy ra nhiu phn ng nhng cc phn ng cng loi v cng hiu sut phn ng th ta thay hn hp nhiu cht thnh mt cht tng ng. Lc : lng (s mol, khi lng hay th tch) ca cht tng ng bng ca hn hp.

Phng php 6. Phng php p dng s ng cho.

Khi pha trn 2 dung cng loi nng , cng loi cht tan th c th dng phng php ng cho:

Trn m1 gam dung dch c nng C1% vi m2 gam dung dch c nng C2% th thu c dung dch mi c nng C%:

m1 gam dd C1 C2 - C

m1 C2 - C

C => =

m 2 C1 - C

m2 gam dd 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 gam dd C1 C2 - C

V1 C2 - C

C => =

V2 C1 - C

V2 gam dd C2 C1 C S ng cho cn c th p dng trong vic tnh khi lng ring D

V1 lt dd D1 D2 - D

V1 D2 - D

D => =

V2 D1 - D

V2 lt dd D2 D1 - D

(Vi gi thit V = V1 + V2)

II. Mt s dng bi tp nng cao

Dng 1. Bi ton xc nh tn kim loi v hp cht ca chngV d 1: Ha tan hn hp X gm 11,2 gam kim loi M v 69,6 gam oxit MxOy ca kim loi trong 2 lt dung dch HCl, thu c dung dch A v 4,48 lt kh H2 (ktc). Nu cng ha tan hn hp X trong 2 lt dung dch HNO3 th thu c dung ch B v 6,72 lt kh NO (ktc). a. Tm kim loi M.

b. Tm cng thc oxit ca kim loi .

Hng dn gii a. Ta c PTP sau: M + nHCl MCln + n/2H2

amol na/2 mol

MxOy + 2yHCl xMCl2y + yH2O

S mol H2 = 0,2 mol => na/2 = 0,2 => na = 0,4 => a = 0,4/n (vi a l s mol ca kim loi M cn tm). Ta c 0,4/n. M = 11,2 => M = 28nBin lun tm M

n123

M28 (loi)56 (nhn)84 (loi)

Vi M = 56 tha mn, vy kim loi cn tm l Fe; nFe = 0,2 mol

b. Gi cng thc oxit MxOy l FexOy

Ta c s mol ca kh NO l: nNO = 0,3 (mol)

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O (1) 0,2mol 0,2mol S mol NO phn ng (2) l: nNO = 0,3 0,2 = 0,1 (mol) 3FexOy + (12x-2y)HNO3 3xFe(NO3)3 + (3x-2y)NO + (6x-y)H2O (2)

3mol (3x-2y)mol

69,6 0,1mol56x + 16y Ta c: 69,6/(56x+16y) = (3x-2y)/0,1 => 64x = 48y => x = 3; y = 4 Vy cng thc oxit FexOy cn tm l Fe3O4 V d 2: Kh hon ton 4,06 gam 1 oxit kim loi bng CO nhit cao thnh kim loi. Dn ton b kh sinh ra vo bnh ng dd Ca(OH)2 d, thy to thnh 7 gam kt ta. Nu ly lng kim loi to thnh (m gam) ha tan ht vo dd HCl d th thu c 1,176 lt H2 (ktc). Tm cng thc ca oxit kim loi v khi lng ca kim loi.

Hng dn giiTheo bi ta c s sau:

Kh Ca(OH)2

CO CO2 CaCO3

Oxit KL

7

Theo s trn => nCO = nCO2 = nCaCO3 = = 0,07 (mol)

100

p dng nh lut bo ton khi lng ta c: moxit + mCO = mKL + mCO2

mKL = moxit + mCO - mCO2

= 4,06 + (0,07. 28) (0,07. 44) = 2,94 (gam)

Gi tn kim loi trn l M, ha tr l a, ta c:

M + aHCl MCla + a/2 H2 1,176

nH2 = = 0,0525 (mol)

22,4 0,0525 0,105

nM = =

(mol)

a/2 a

2,94

= > MM = = 28a

0,105/a Bin lun:

a123

M 28 (loi) 56 (nhn)84 (loi)

Vy nghim ph hp l: a = 2 => M = 56 (Fe) 0,105

nFe = = 0,0525 (mol)

2

Khi lng ca kim loi st l: mFe = 0,0525. 56 = 2,94 (gam)

t cng thc ca oxit st cn tm l FexOy ta c PT:

FexOy + CO xFe + yCO2

Ta c t l sau:

nFe x 0,0525 3

= = = => x = 3, y = 4

nCO2 y 0,07 4

Vy cng thc ca oxit cn tm l Fe3O4. Bi tp vn dng Bi 1. Kh m gam 1 oxit st cha bit bng CO nng, d n hon ton thu c Fe v kh A. Ha tan ht lng Fe trn bng HCl d thot ra 1,68 lt H2 (ktc). Hp th ton b kh A bng Ca(OH)2 d thu c 10 gam kt ta. Tm cng thc oxit.p s: Fe3O4

Bi 2. em kh hon ton 4 gam hn hp CuO v oxit st FexOy bng kh CO nhit cao, sau khi phn ng thu c 2,88 gam cht rn, em ha tan cht rn ny vo 400 ml dung dch HCl (va ) th c 0,896 lt kh bay ra ( ktc).

a. Tnh khi lng hn hp ban u.

b. Xc nh cng thc phn t ca oxit st.

p s: a. mCuO = 0,8 gam, mFexOy = 3,2 gam.

b. Fe2O3 Dng 2. Bi ton hn hp

V d 1: Cho lung kh CO i qua ng s ng 0,04 mol hn hp A gm FeO v Fe2O3 t nng. Sau khi kt thc th nghim thu c B gm 4 cht nng 4,784 gam. Khi i ra khi ng s cho hp th vo dung dch Ba(OH)2 d thu c 9,602 gam kt ta. Mt khc ha tan cht rn B bng dd HCl d thy thot ra 0,6272 lt kh H2 (ktc). a. Tnh % khi lng oxit trong hn hp A b. Tnh % khi lng cc cht trong B, bit trong B

c nFe3O4 = 1/3 nFeO v nFe2O3 Hng dn: p dng bo ton khi lng v bo ton nguyn tHng dn giia. Ta c 0,04 mol hh A (FeO v Fe2O3) + CO 4,784 hh B + CO2 CO2 + Ba(OH)2 d BaCO3 + H2O

0,046 mol 0,046 mol

nCO2 = nBaCO3 = 0,046 mol v nCOp = nCO2 = 0,046 molp dng nh lut bo ton khi lng, ta c:

mA + mCO = mCO2 + mB

=> mA = 4,784 + 0,046.44 0,046.28 = 5,52 (gam)t nFeO = x mol, nFe2O3 = y mol trong hn hp B, ta c: x + y = 0,04 x = 0,01 mol 72x + 160y = 5,52 y = 0,03 mol 0,01.72

=> %mFeO = x 100% = 13,04%

5,52

%mFe2O3 = 100 13,04 = 86,96%

b. Ta c: nH2 = 0,028 mol Fe + 2HCl FeCl2 + H2

Theo phng trnh ta c: nFe = nH2 = 0,028 mol

theo cu a, c 0,01 mol FeO, 0,03 mol Fe2O3 => tng s mol st ban u bng 0,07 mol.Hn hp B gm: 0,028 mol Fe, a mol FeO, b mol Fe2O3, c mol Fe3O4

p dng nh lut bo ton nguyn t st ta c:

a + 2b + 3c = 0,07 0,028 = 0,042 (1)Li c: a/3 + b/3 c = 0 (2) 72a + 160b + 232c = 4,784 (0,028. 56) = 3,216 (3)T (1), (2) v (3) ta c h PT: a + 2b + 3c = 0,042

a/3 + b/3 c = 0

72a + 160b + 232c = 3,216

Gii h ta c nghim: a = 0,012; b = 6.10-3; c = 6.10-3

Ta c phn trm khi lng cc cht trong B nh sau:

%mFeO = 18,1%; %mFe2O3 = 20%; %mFe3O4 = 29,1%%mFe = 100 (18,1 + 20 + 29,1) = 32,8%V d 2: Hn hp 3 kim loi Fe, Al, Cu nng 17,4 gam. Nu ha tan hn hp bng axit H2SO4 long d th thot ra 8,96 dm3 H2 ( ktc). Cn nu ha tan hn hp bng axit c nng, d th thot ra 12,32 dm3 SO2 ( ktc). Tnh khi lng mi kim loi ban u. Hng dn giiCu khng tan trong H2SO4 long, ch c Fe v Al tan c trong axit long Fe + H2SO4 FeSO4 + H2

2Al + 3H2SO4 Al2(SO4)3 + 3H2

H2SO4 c nng ha tan c 3 kim loi:

2Fe + 6H2SO4 Fe2(SO4)3 + 3SO2 + 6H2O

2Al + 6H2SO4 Al2(SO4)3 + 3SO2 + 6H2O

Cu + 2H2SO4 CuSO4 + SO2 + 2H2O

S mol H2 = 0,4; s mol SO2 = 0,55

Gi s mol ca Fe, Al, Cu ln lt l x, y, z ta c:

56x + 27y + 64z = 17,4H 3 phng trnh : x + 1,5y = 0,4

1,5x + 1,5y + z = 0,55 Gii h phng trnh cho ta c nghim l: x = 0,1; y = 0,2; z = 0,1 Khi lng ca st ban u l: mFe = 0,1. 56 = 5,6 (gam)

Khi lng ca nhm ban u l: mAl = 0,2. 27 = 5,4 (gam)

Khi lng ca ng ban u l: mCu = 0,1. 64 = 6,4 (gam)

Bi tp vn dng

Bi 1. Hn hp 3 kim loi Fe, Al, Cu. Ha tan a gam hn hp bng axit H2SO4 c, nng va th thot ra 15,68 dm3 SO2 (ktc) v nhn c dung dch X. Chia i X, mt na em c cn nhn c 45,1 gam mui khan, cn mt na thm NaOH d ri lc kt ta nung trong khng kh n lng khng i cn nng 12 gam. Tm a v khi lng mi kim loi. p s: a = 23 gam; mFe = 11,2 gam, mAl = 5,4 gam, mCuBi 2. ng cha 4,72 gam hn hp Fe, FeO, Fe2O3 c t nng ri cho dng H2 i qua n d. Sau phn ng trong ng cn li 3,92 gam Fe. Nu cho 4,72 gam hn hp u vo dung dch CuSO4 lc k v phn ng hon ton, lc ly cht rn, lm kh cn nng 4,96 gam. Tnh khi lng tng cht trong hn hp. p s: mFe = 1,68gam ; mFeO = 1,44 gam; mFe2O3 = 1,6 gam. Dng 3. Bi ton tng gim khi lng

C s l thuyt Phn ng trao i CaCO3 + 2HNO3 Ca(NO3)2 + H2O + CO2

a mol a mol

tng khi lng mui = lng NO3- - lng CO32- = 124a 60a = 64a

tng khi lng dung dch = lng CaCO3 lng CO2

Phn ng th Fe + CuSO4 FeSO4 + Cu

a mol a mol a mol a mol

tng khi lng kim loi = gim khi lng dung dch = 64a 56a = 8a

Phn ng ha hp 2Cu + O2 2CuO

tng khi lng kim loi = khi lng O2 phn ng Phn ng phn tchCaCO3 CaO + CO2

gim khi lng CaCO3 = khi lng CO2 V d 1: Cho 80 gam bt ng vo 200 ml dung dch AgNO3, sau mt thi gian phn ng lc c dung dch A v 95,2 gam cht rn B. Cho tip 80 gam bt ch vo dung dch A, phn ng xong lc B tch c dung dch D ch cha mt mui duy nht v 67,05 gam cht rn E. Cho 40 gam bt kim loi R (ha tr 2) vo 1/10 dung dch D, sau phn ng hon ton lc tch c 44,575 gam cht rn E. Tnh nng mol/l ca dung dch AgNO3 v xc nh kim loi R.

Hng dn gii Cu + 2AgNO3 Cu(NO3)2 + 2Ag

x 2x x 2x

95,2 80

S mol x = = 0,1 (mol)

216 64

Pb + Cu(NO3)2 Pb(NO3)2 + Cu

0,1 0,1 0,1 0,1

Theo phng trnh nu ch xy ra phn ng ny th gim khi lng kim loi (do mt Pb = 207 v to Cu = 64) l: (207 64). 0,1 = 14,3 (gam) > 80 67,05 = 12,95 (gam).

Chng t trong dung dch vn cn mui AgNO3 d c phn ng :

Pb + 2AgNO3 Pb(NO3)2 + 2Ag

y 2y y 2y

Phn ng ny lm tng lng = (216 207).y

Vy ta c : (216 207). y = 14,3 12,95 = 1,35 => y = 0,15

S mol AgNO3 ban u 2x + 2y = 0,5 (mol)

0,5

=> Nng mol ca AgNO3 = = 0,4M

0,2

Dung dch D cha Pb(NO3)2 nn nD = 0,1 + 0,15 = 0,25 (mol)

Cho kim loi vo 1/10 dung dch D ta c :

R + Pb(NO3)2 R(NO3)2 + Pb

0,025 0,025 0,025 0,025

tng kim loi = (207 R). 0,025 = 44,575 40 = 4,575 (gam) => R = 24 (Mg)

Vy kim loi cn tm l Magie ~ Mg Bi tp vn dngBi 1. Hai thanh kim loi ging nhau (u to bi cng nguyn t R ha tr II) v c cng khi lng. Th thanh th nht vo dung dch Cu(NO3)2 v thanh th hai vo dung dch Pb(NO3)2. Sau mt thi gian, khi s mol 2 mui phn ng bng nhau ly 2 thanh kim loi ra khi dung dch thy khi lng thanh th nht gim i 0,2%, cn khi lng thanh th hai tng thm 28,4%. Tm nguyn t R. p s: R l km (Zn).Bi 2. C 100 ml mui nitrat ca 1 kim loi ha tr II (dung dch A). Th vo A mt thanh Pb kim loi, sau 1 thi gian khi lng Pb khng i th ly n ra khi dung dch thy khi lng ca n gim i 28,6 gam. Dung dch cn li c th tip vo mt thanh Fe nng 100 gam. Khi lng Fe khng i na th ly ra khi dung dch, thm kh cn nng 130,2 gam. Hi cng thc ca mui ban u v nng mol ca dung dch A. p s: cng thc mui l: Cu(NO3)2, nng ca dd A l 2M. Dng 4. Bi ton v lng cht dGi thit ca dng bi ny c c im l trn mt phng trnh phn ng cho bit lng ca hai cht c mt trn phng trnh m theo l ch cn bit lng ca mt cht l suy ra lng cht cn li.

Phn loi:

Hn hp kim loi tc dng vi 1 axit:

Phi gi nh l khi lng kim loi cho ch c mt kim loi t tnh lng axit dng cho mi trng hp v suy ra khong gii hn ca lng axit cn. Nu d kin cho lng axit ln hn khong gii hn th axit d v kim loi ht, nu lng axit nh hn khong gii hn th kim loi d.

Hn hp kim loi tc dng vi hn hp axit:

Cng phi gi nh lng kim loi ch c mt kim loi cn vi axit phi tnh s mol nguyn t hiro trong axit sau cng xc nh khong gii hn nh bi trn.

Mt kim loi tc dng vi 1 dung dch axit nhng vi lng khc nhau trong nhng th nghim khc nhau:

So snh lng axit ca hai th nghim v lng hiro gii phng hai th nghim t suy ra c mt th nghim d axit v mt th nghim ht axit.

Kim loi mnh y kim loi yu ra khi mui ca kim loi yu:

Cn phi ch xem kim loi no tc dng ht trc. Theo quy lut kim loi mnh y kim loi yu ra khi mui ca kim loi yu th kim loi no mnh hn s ht trc v ty thuc vo cc d kin cn li ta bin lun.V d 1: Cho H2SO4 long, d tc dng vi hn hp gm Mg v Fe thu c 2,016 lt kh ktc. Nu hn hp kim loi ny tc dng vi dd FeSO4 d th khi lng hn hp trn tng ln 1,68 gam.

a. Vit phng trnh phn ng ha hc

b. Tm khi lng mi kim loi trong hn hp trn

Hng dn gii a. Ta c PTHH:

Mg + H2SO4 MgSO4 + H2 (1)

x mol x mol x mol

Fe + H2SO4 FeSO4 + H2 (2)

y mol y mol y mol

Cho hn hp kim loi trn vo dd FeSO4 d th Mg tc dng ht (Fe khng tc dng) theo phng trnh sau:

Mg + FeSO4 MgSO4 + Fe (3)

x mol x mol

khi lng hn hp tng ln 1,68 gam l khi lng chnh lch gia Fe mi to ra v Mg phn ng.

b. Ta c s mol ca kh H2 l 0,09 mol

theo phng trnh (1) v (2) ta ch phng trnh: x + y = 0,09

56y 24x = 1,68

Gii h phng trnh trn ta c nghim: x = 0,0525; y = 0,0375

Vy khi lng ca 2 kim loi trn l:

mFe = 0,0375. 56 = 2,1 (gam)

mMg = 0,0525. 24 = 1,26 (gam)

V d 2: Cho 3,87g hn hp A gm Mg v Al vo 250ml dung dch X cha axit HCl 1M v H2SO4 0,5M c dung dch B v 4,368 lt H2 ( ktc).

Hy chng minh rng trong dung dch B vn cn d axit.

Hng dn gii Mg + 2HCl MgCl2 + H2

2Al + 6HCl 2AlCl3 + 3H2 Mg + H2SO4 MgSO4 + H2 2Al + 3H2SO4 Al2(SO4)3 + 3H2

4,368

nH2 = = 0,195 mol

22,4

nnguyn t H = 0,195. 2 = 0,39 mol (1)

nHCl = 0,25 mol nnguyn t H = 0,25 mol

nH2SO4 = 0,25. 0,5 = 0,125 mol nnguyn t H = 0,25 mol

nnguyn t H = 0,25 + 0,25 = 0,5 mol (2)

So snh s mol nguyn t (1) v (2) ta thy axit cn d.

V 0,5 mol > 0,39 mol

V d 3: Cho 6,8 gam hn hp Fe v CuO tan trong 100 ml axit HCl thu c dung dch A v thot ra 224 ml kh B ( ktc) v lc c cht rn D nng 2,4 gam. Thm tip HCl d vo hn hp A + D th D tan mt phn, sau thm tip NaOH n d v lc kt ta tch ra nung nng trong khng kh n lng khng i cn nng 6,4 gam. Tnh thnh phn phn trm khi lng Fe v CuO trong hn hp ban u.

Hng dn giiS mol kh H2 = 0,01 (mol).

Cht rn D tan mt phn trong axit HCl d th D cha Cu v Fe:

CuO + 2HCl CuCl2 + H2O

Fe + CuCl2 FeCl2 + Cu

Fe + 2HCl FeCl2 + H2

Thm NaOH :

CuCl2 + NaOH Cu(OH)2 + 2NaCl

FeCl2 + 2NaOH Fe(OH)2 + 2NaCl

4Fe(OH)2 + O2 2Fe2O3 + 4H2O (6,4 gam l lng Fe2O3 + CuO)

Cu(OH)2 CuO + H2O

Gi a, x, y ln lt l s mol ca Cu, Fe, CuO ta c h phng trnh sau:

56x + 80y = 6,8

56(x 0,01 a) + 64a = 2,4

160x + 80(y a) = 6,4

Gii h trn ta c x = 0,05; y = 0,05; a = 0,02

Vy phn trm khi lng cc cht trong hn hp ban u l:

%mFe = [(0,05. 56)/6,8]. 100 = 41,18%

%mCuO = [ (0,05. 80)/6,8]. 100 = 58,82% Bi tp vn dng

Bi 1. Ha tan 2,4 gam Mg v 11,2 gam st vo 100 ml dung dch CuSO4 2M th tch ra cht rn A v nhn c dung dch B. Thm NaOH d vo dung dch B ri lc kt ta tch ra nung n lng khng i trong khng kh thu c a gam cht rn D. Vit phng trnh ha hc, tnh lng cht rn A v lng cht rn D. p s: mA = 18,4 gam, mD = 12 gam.Bi 2. Cho 0,411 gam hn hp kim loi st v nhm vo 250 ml dung dch AgNO3 0,12M. Sau khi cc phn ng xy ra hon ton, thu c cht rn A cn nng 3,324 gam v dung dch B. Cho dung dch B tc dng vi dung dch NaOH d, thu c kt ta trng dn dn ha nu.

a. Vit tt c cc phn ng c th xy ra.

b. Tm khi lng mi kim loi trong 0,411 gam hn hp u. p s: mAl = 0,243 gam, mFe = 0,168 gam.Bi 3. 1,36 gam hn hp gm Mg v Fe c ha tan trong 100 ml dung dch CuSO4. Sau phn ng nhn c dung dch A v 1,84 gam cht rn B gm hai kim loi. Thm NaOH d vo A ri lc kt ta tch ra nung nng trong khng kh n lng khng i nhn c cht rn D gm MgO v Fe2O3 nng 1,2 gam. Tnh lng Fe, Mg ban u. p s: mFe = 1,12 gam, mMg = 0,24 gam. Dng 5. Bi ton bin lun

Bin lun ha tr

Bin lun trng hp

Bin lun so snh

Bin lun bng tr s trung bnh

V d 1: Cho 14,7 gam hn hp 2 kim loi kim tc dng ht vi H2O thu c dung dch B v 5,6 lt H2 ( ktc). Trung ha dung dch B bng HNO3, un cn dung dch c hn hp mui D. Tm khi lng D. Xc nh 2 kim loi kim, bit mui c khi lng mol ln hn chim 44,2% khi lng hai mui trong D.

Hng dn giiTa c s mol ca H2 = 0,25 (mol)

Tng s mol 2 kim loi = 0,5 v bng tng s mol 2 mui nitrat.

Lng mui nitrat = lng kim loi + lng gc NO3-

= 14,7 + (0,5. 62) = 45,7 (gam)

Lng mui c khi lng mol ln hn = 45,7. 0,442 = 20,2 (gam)

Gi khi lng mol trung bnh ca 2 kim loi kim trn l R

45,7

Ta c khi lng mol ca mui nitrat = = 91,4 (gam)

0,5

MRNO3 = 91,4 => MR = 91,4 62 = 29,4

Vy 29,4 < R < 29,4

Vi R > 29,4 ta thy ch c kali (K) l tha mn

Vi R < 29,4 ta thy natri (Na) l tha mn

V d 2: Ha tan 3,82 gam hn hp 2 mui sunfat kim loi A v B c ha tr I v II tng ng vo nc thnh dung dch ri thm mt lng va BaCl2 thy tch ra 6,99 gam kt ta.

- Lc b kt ta, ly nc lc em c cn thu c bao nhiu gam mui khan?

- Tm cng thc 2 mui v khi lng mi mui bit A v B c v tr cng chu k trong bng tun hon.

Hng dn giiLng mui khan = 3,82 (0,03. 96) + (0,03. 71) = 3,07 (gam)

Do thay gc SO42- bng 2Cl- vi s mol = 0,03 tnh t BaSO4. 3,82

Khi lng mol trung bnh 2 mui sunfat = = 127,33 (gam)

0,03

Ta c: B + 96 < 127,33 < 2A + 96 => B < 31,33 < 2A

Do hai kim loi cng chu k nn A (ha tr I) < B (ha tr II)

31,33 > B > A > 15,67

Ta thy kim loi B (ha tr II) < 31,33 ch c Be = 9 v Mg = 24, nhng khng th l Be v kim loi ha tr I ng trc n < 15,67 (tri vi gi thit).

Vy 2 kim loi phi tm l Na v Mg

Nn cng thc 2 mui tng ng l: Na2SO4, MgSO4Gi s mol ca Na v Mg ln lt l x, y

Theo u bi ta ch phng trnh: 142x + 120y = 3,82

x + y = 0,03

Gii h trn ta tm c x = 0,01 v y = 0,02

Khi lng 2 mui trn l:

mNa2SO4 = 0,01. 142 = 1,42 (gam) v mMgSO4 = 0,02. 120 = 2,4 (gam)

V d 3: Trn CuO vi mt oxit kim loi ha tr 2 khng i theo t l mol 1: 2 c hn hp X. Cho 1 lung kh CO nng d i qua 2,4 gam X n phn ng hon ton thu c cht rn Y. ha tan ht Y cn 40 ml dung dch HNO3 2,5M, ch thot ra mt kh NO duy nht v dung dch thu c ch cha mui ca 2 kim loi ni trn. Xc nh kim loi cha bit.

Hng dn giip dng bin lun trng hp:

V CO ch kh c nhng oxit kim loi ng sau Al trong dy hot ng ha hc nn c 2 trng hp xy ra.

Trng hp 1: Kim loi phi tm ng sau Al trong dy

hot ng ha hc v oxit ca n b CO kh.

CuO + CO Cu + CO2 (1)

MO + CO M + CO2 (2)

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O (3)

3M + 8HNO3 3M(NO3)2 + 2NO + 4H2O (4)

Theo bi phn ng theo t l s mol l 1 : 2

Coi s mol CuO = x th MO = 2x v s mol HNO3 = 0,1

Ta c h hai phng trnh : 8 8

80x + (M + 16)2x = 2,4 v . x + . 2x = 0,1

3 3

Gii h cho x = 0,0125 v M = 40 (Canxi Ca)

Trng hp ny khng tha mn v canxi ng trc Al trong dy hot ng ha hc v CaO khng b kh bi CO

Trng hp 2: Kim loi phi tm ng trc Al trong

dy hot ng ha hc v oxit ca n khng b CO kh. Khi khng xy ra phn ng (2) m xy ra phn ng (1) v (3) v phn ng di y:

MO + 2HNO3 M(NO3)2 + H2O

2a 4a

Tng t coi s mol ca CuO = a MO = 2a ta c h 2 phng trnh:

80a + (M + 16)2a = 2,4 v 8/3a + 4a = 0,1

Gii h ta c a = 0,015 v M = 24 (magie Mg) tha mn

Vy kim loi cn tm l magie (Mg)

Bi tp vn dng

Bi 1. Hn hp Q nng 16,6 gam gm Mg, oxit ca kim loi A ha tr III v oxit ca kim loi B ha tr II c ha tan bng HCl d thu c kh X bay ln v dung dch Y. Dn X qua bt CuO nung nng thu c 3,6 gam nc. Lm bay hi ht nc ca dung dch Y thu c 24,2 gam hn hp mui khan. em in phn dung dch Y n khi kim loi B tch ht ra cc m th cc dng thot ra 0,71 gam kh clo.

a. Xc nh 2 kim loi A, B bit B khng tan c trong dung dch HCl, khi lng mol ca B ln hn 2 ln khi lng mol ca A.

b. Tnh % khi lng mi cht trong Q. c. Nu tn v ng dng ca hp kim cha ch yu 3 kim loi trn trong k ngh p s: A l Al, B l Cu; %mMg = 29,92%, %mAl2O3 = 61,44%, %mCuO = 8,64%.Bi 2. Hn hp Al v kim loi ha tr 2 tan trong axit H2SO4 long va thu c dung dch A v c H2 thot ra. Cho A tc dng vi dung dch BaCl2 va thy tch ra 93,2 gam kt ta trng. Lc kt ta ri c cn nc lc thu c 36,2 gam mui kh.

a. Tnh th tch H2 thot ra ktc v khi lng 2 kim loi ban u.

b. Tm kim loi cha bit, nu trong hn hp ban u s mol ca n ln hn 33,33% s mol ca Al p s: a. VH2 = 8,96 lt, khi lng hai kim loi l 7,8 gam; b. kim loi l Mg.PHN III. KT LUN

i chiu vi mc ch v nhim v nghin cu ra, bi tiu lun thu c cc kt qu sau:

1. Su tm v phn loi c cc dng bi tp c bn v nng cao v kim loi THCS. C th:

+ Su tm v xy dng c 9 dng bi tp nng cao v kim loi. + a ra c cc v d p dng c li gii c th ca tng dng bi v cc bi tp vn dng c p s.2. Tng hp c 6 phng php gii bi tp thng dng khi gii bi tp ha hc THCS.PAGE 3