CC DNG BI TP BI DNG HSGChuyn 1:Bi tp v nguyn t, nguyn t ha hc 1/
Nguyn t (NT):- Ht v cng nh , trung ha v in, to nn cc cht.Cu to: +
Ht nhn mang in tch (+)(Gm: Proton(p) mang in tch (+) v ntron khng
mang in ). Khi lng ht nhn c coi l khi lng nguyn t.+ V nguyn t cha 1
hay nhiu electron (e) mang in tch (-). Electron chuyn ng rt nhanh
quanh htnhn v sp xp theo lp (th tsp xp (e) ti a trong tng lp t
trong ra ngoi:STT ca lp : 12 3 S e ti a : 2e 8e 18e Trong nguyn t:-
S p = s e = s in tch ht nhn =s th t ca nguyn t trong bng h thng tun
hon cc nguyn t ha hc - Quan h gia s p v s n:p n 1,5p ( ng vi 83
nguyn t )- Khi lng tng i ca 1 nguyn t ( nguyn t khi )NTK = s n +s p
- Khi lng tuyt i ca mt nguyn t ( tnh theo gam ) + mT = m e + mp +
mn + mP mn
1VC 1.67.10- 24g, + me9.11.10 -28 gNguyn t c th ln kt c vi nhau
nh e lp ngoi cng.2/ Nguyn t ha hc (NTHH): l tp hp nhng nguyn t cng
loi c cng s p trong ht nhn.- S p l s c trng ca mt NTHH.- Mi NTHH c
biu din bng mt hay hai ch ci. Ch ci u vit di dng in hoa ch ci th
hai l ch thng. l KHHH- Nguyn t khi l khi lng ca nguyn ttnh bng VC.
Mi nguyn t c mt NTK ring. Khi lng 1 nguyn t = khi lng 1vc.NTKNTK =
1khoiluongmotnguyentukhoiluong dvc
ma Nguyn t = a.m 1vc .NTK (1VC = 112KL ca NT(C) (MC = 1.9926.10-
23g) = 1121.9926.10- 23g= 1.66.10- 24 g)* Bi tp vn dng:1. Bit nguyn
t C c khi lng bng 1.9926.10- 23 g. Tnh khi lng bng gam ca nguyn t
Natri. Bit NTK Na = 23. (p s: 38.2.10- 24 g)2.NTK ca nguyn 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= 32,S=16)3. Bit rng 4 nguyn t Mage nng bng 3 nguyn
t nguyn t X. Xc nh tn,KHHH ca nguyn t X. (p s:O= 32)4.Nguyn t X nng
gp hai ln nguyn t oxi .b)nguyn t Y nh hn nguyn t Magie 0,5 ln .c)
nguyn t Z nng hn nguyn t Natri l 17 vc .Hy tnh nguyn tkhi ca X,Y, Z
.tn nguyn t, k hiu ho hc ca nguyn t ? 5.Nguyn t M c s n nhiu hn s p
l 1 v s ht mang in nhiu hn s ht khng mang in l 10. Hy xc nh M l
nguyn t no?6.Tng s ht p, e, n trong nguyn t l 28, trong s ht khng
mang in chim xp x 35% .Tnh s ht mi loa .V s cu to nguyn t .7.Nguyn
t st c 26p, 30n, 26ea.Tnh khi lng nguyn t stb.Tnh khi lng e trong
1Kg st8.Nguyn t X c tng cc ht l 52 trong s ht mang in nhiu hn s ht
khng mang in l 16 ht.a)Hy xc nh s p, s n v s e trong nguyn t X.b) V
s nguyn t X.c) Hy vit tn, k hiu ho hc v nguyn t khi ca nguyn t
X.19. Mt nguyn t X c tng s ht e, p, n l 34. S ht mang in nhiu hn s
ht khng mang in l 10. Tm tn nguyn t X. V s cu to ca nguyn t X v ion
c to ra t nguyn t X10.Tm tn nguyn t Y c tng s ht trong nguyn t l
13. Tnh khi lng bng gam ca nguyn t.11. Mt nguyn t X c tng s ht l
46, s ht khng mang in bng 815 s ht mang in. Xc nh nguyn t X thuc
nguyn t no ? v s cu to nguyn t X ?12.Nguyn t Z c tng s ht bng 58 v
c nguyn t khi < 40 . 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 ))H ng d ngii: bi 2p + n = 58 n = 58 2p ( 1
)Mt khc :p n 1,5p ( 2 )p 58 2p1,5p gii ra c 16,5 p 19,3( p : nguyn
)Vy p c th nhn cc gi tr : 17,18,19P 17 18 19N 24 22 20NTK = n + p
41 40 39 Vy nguyn t Z thuc nguyn t Kali ( K )13.Tm 2 nguyn t A, B
trong cc trng hp sau y :a) Bit A, B ng k tip trong mt chu k ca bng
tun hon v c tng s in tch ht nhn l 25.b) A, B thuc 2 chu k k tip v
cng mt phn nhm chnh trong bng tun hon. Tng s in tch ht nhn l 32.14:
Trong 1 tp hp cc phn t ng sunfat (CuSO4) c khi lng 160000 vC. Cho
bit tp hp c bao nhiu nguyn t mi loi.3. S to thnh ion (dnh cho HSG
lp 9) t cu trc bo ha (8e lp ngoi cng hoc 2e i vi H) th cc nguyn t c
th nhng hoc nhn thm electron to ra nhng phn mang in - gi l ion* Kim
loi v Hiro : nhng e to ion dng ( cation)Mne M n +
(Ca2e Ca 2 +
)* Cc phi kim nhn e to ionm (anion)X + ne X n- ( Cl+1e Cl 1-)*
Bi tp vn dng: 1.Hp cht X c to thnh t cation M+v anion Y2- . Mi ion
u do 5 nguyn t ca 2 nguyn t to nn. Tng s proton trong M+ l 11 cn
tng s electron trong Y2- l 50.Xc nh CTPT ca hp cht X v gi tn ? ng
dng ca cht ny trong nng nghip . Bit rng 2 nguyn t trong Y2- thuc
cng phn nhm trong 2 chu k lin tip ca bng tun hon cc ng.t.Hng dn gii
:t CTTQ ca hp chtXl M2YGi s ionM+ gm 2 nguyn t A, B: ionM+ dng :
AxBy+ c :x+y = 5 ( 1 )x.pA +y.pB = 11( 2) Gi s ionY 2- gm 2 nguyn t
R, Q : ionY2- dng : R xQy2- c : x + y = 5 (3) xpR + y.pQ = 48 (4 )
do s e > s p l 2 T ( 1 ) v (2) ta c s proton trung bnh ca A v B
: 112, 25p 1 trong AxBy+ c 1 nguyn t c p < 2,2( H hoc He ) v 1
nguyn t c p > 2,2 V He khng to hp cht ( do tr ) nnnguyn t c p
< 2,2l H( gi s l B )T ( 1 ) v ( 2) ta c :x.pA +(5 x ).1 = 11 pA
=61Apx + (1 x < 5 )X 1 2 34pA7(N) 4(B) 3(Li)2,5 (loi)ion M+ NH4+
khng xc nh ion Tng t: s proton trung bnh ca R v Q l :489, 65p c 1
nguyn t c s p < 9,6 ( gi s l R )V Q v R lin tip trong nhm nn :
pQ = pR + 8( 5 )T (3) ,(4) , ( 5) ta c :xpR + (5- x)( pR + 8)= 48
5pR 8x = 8 8 8 '5Rxp+ 2x 1 2 34pR3,2 4,8 6,48(O ) pQ khng xc nh
ion16 ( S )Vy CTPT ca hp cht X l (NH4 )2SO4Chuyn 2:Bi tp v cng thc
ha hcA. Tnh theo CTHH:1: Tnh TP% ccnguyn t theo khi lng.* Cch
gii:CTHH c dng AxBy- Tm khi lng mol ca hp cht. MAxBy = x.MA + y.
MB- Tm s mol nguyn t mi nguyn t trong 1 mol hp cht : x, y (ch s s
nguyn t ca cc nguyn t trong CTHH)- Tnh thnh phn % mi nguyn t theo
cng thc: %A = .100%mAMAxBy= ..100%x MAMAxByV d:Tm TP % ca S v O
trong hp cht SO2- Tm khi lng mol ca hp cht : MSO2 = 1.MS + 2. MO =
1.32 + 2.16 = 64(g)- Trong 1 molSO2 c 1 mol nguyn t S (32g), 2 mol
nguyn t O (64g)- Tnh thnh phn %: %S = 2.100%mSMSO= 1.3264.100% =
50%%O = 2.100%mOMSO= 2.1664.100% = 50% (hay 100%- 50% = 50%)* Bi tp
vn dng:1 : Tnh thnh phn % theo khi lng cc nguyn t trong cc hp cht
:a/ H2Ob/ H2SO4 c/ Ca3(PO4)22: Tnh thnh phn phn trm v khi lng ca cc
nguyn t c trong cc hp cht sau:a) CO; FeS2; MgCl2; Cu2O; CO2; C2H4;
C6H6.b) FeO; Fe3O4; Fe2O3; Fe(OH)2; Fe(OH)3.c) CuSO4; CaCO3; K3PO4;
H2SO4. HNO3; Na2CO3.d) Zn(OH)2; Al2(SO4)3; Fe(NO3)3. (NH4)2SO4;
Fe2(SO4)3.3:Trongcchpchtsau,hpcht nochmlngFecaonht:FeO;Fe2O3;Fe3O4;
Fe(OH)3;FeCl2 ;Fe SO4.5H2O ?4: Trong cc loi phn bn sau, loi phn bn
no c hm lng N cao nht: NH4NO3; NH4Cl; (NH4)2SO4; KNO3; (NH2)2CO?2:
Tm khi lng nguyn t trong mt lng hp cht.* Cch gii:CTHH c dng AxBy-
Tnh khi lng mol ca hp cht. MAxBy = x.MA + y. MB- Tm khi lng mol ca
tng nguyn t trong 1 mol hp cht: mA = x.MA , mB = y. MB- Tnh khi lng
tng nguyn t trong lng hp cht cho.mA = . mA mAxByMAxBy = . . x MA
mAxByMAxBy , mB = . mB mAxByMAxBy = . . y MB mAxByMAxByV d: Tm khi
lng ca Cc bon trong 22g CO2Gii:- Tnh khi lng mol ca hp cht. MCO2 =
1.Mc + 2. MO = 1.12 + 2. 16 = 44(g)- Tm khi lng mol ca tng nguyn t
trong 1 mol hp cht:mC = 1.Mc = 1.12 = 12 (g)- Tnh khi lng tng nguyn
t trong lng hp cht cho.mC = . 22mC mCOMCO = 1.12.2244 =6(g)* Bi tp
vn dng: 1: Tnh khoi lng moi nguyen to co trong cac lng chat sau:
a)26g BaCl2; 8g Fe2O3; 4,4g CO2; 7,56g MnCl2; 5,6g NO.b)12,6g HNO3;
6,36g Na2CO3; 24g CuSO4; 105,4g AgNO3; 6g CaCO3. c) 37,8g Zn(NO3)2;
10,74g Fe3(PO4)2; 34,2g Al2(SO4)3; 75,6g Zn(NO3)2.2: Mt ngi lm vn
dng 500g (NH4)2SO4 bn rau. Tnh khi lng N bn cho rau?B/ Lp CTHH da
vo Cu to nguyn t:Kin thc c bn phn 1* Bi tp vn dng:3 Vy ion Y2-l
SO42-1.Hp cht A c cng thc dng MXy trong M chim 46,67% v khi lng. M
l kim loi, X l phi kim c 3 lp e trong nguyn t. Ht nhn M c n p = 4.
Ht nhn X c n= p ( n, p, n, p l s ntron v proton ca nguyn tM v X ).
Tng s proton trong MXy l 58. Xc nh cc nguyn t M v X (p s : M c p =
26 ( Fe ), X c s proton = 16 ( S ) )2. Nguyn t A c n p = 1, nguyn t
B cn=p. Trong phn t AyB c tng s proton l 30, khi lng ca nguyn t A
chim 74,19% .Tm tn ca nguyn t A, B v vit CTHH ca hp cht AyB ? Vit
PTHH xy ra khi cho AyB v nc ri bm t t kh CO2 vo dung dch thu c3.Tng
s ht tronghp cht AB2 = 64. S ht mang in trong ht nhn nguyn t A nhiu
hn s ht mang in trong ht nhn nguyn t B l 8. Vit cng thc phn t hp
cht trn.Hng d n bi1 :Nguyn t M c :n p = 4n = 4 + p NTK = n + p = 4
+ 2p Nguyn t X c : n = p NTK = 2pTrong MXy c 46,67% khi lng l M nn
ta c :
4 2 46, 67 7.2 ' 53, 33 8py p+ (1) Mt khc :p + y.p = 58yp =58 p
( 2) Thay ( 2) vo (1) ta c :4 + 2p= 78. 2 (58 p )gii ra p = 26v yp
= 32M c p = 26 ( Fe ) X tha mn hm s :p = 32y ( 1y 3)y 1 23P
32(loi)16 10,6 ( loi)Vy X c s proton = 16 ( S ) C/ Lp CTHH da vo
thnh phn phn t, CTHH tng qut : Cht (Do nguyn t to nn) n chtHpcht(Do
1 ng.t to nn)(Do 2 ng.t tr ln to nn) CTHH: AX AxBy + x=1 (gm cc n
cht kim loi, S, C, Si..)(Qui tc ha tr: a.x = b.y)+ x= 2(gm : O2,
H2,, Cl2,, N2, Br2 , I2..) Oxit Axit Baz Mui ( M2Oy)( HxA ) (
M(OH)y ) (MxAy) 1. Lp CTHH hp cht khi bit thnh phn nguyn t v bit ha
tr ca chngCch gii:- CTHH c dng chung : AxBy (Bao gm: ( M2Oy , HxA,
M(OH)y , MxAy)Vn dng Qui tc ha tr i vi hp cht 2 nguyn t A, B (B c
th l nhm nguyn t:gc axt,nhm OH): a.x = b.yxy= ba (ti gin) thay x=
a, y = b vo CT chung ta c CTHH cn lp. V dLp CTHH ca hp cht nhm oxt
a b4Gii:CTHHcdngchungAlxOyTabit hatrcaAl=III,O=II a.x = b.yIII.x=
II. y xy= IIIII thay x= 2, y = 3 ta c CTHH l: Al2O3* Bi tp vn
dng:1.Lp cng thc ha hc hp cht c to bi ln lt t cc nguyn t Na, Ca, Al
vi (=O,; -Cl; = S; - OH; = SO4 ; - NO3 ; =SO3 ; = CO3 ; - HS; -
HSO3;- HSO4; - HCO3;=HPO4 ;-H2PO4 )
2. Cho cc nguyn t: Na, C, S, O, H. Hy vit cc cng thc ho hc ca cc
hp cht v c c th c to thnh cc nguyn t trn?3. Cho cc nguyn t: Ca, C,
S, O, H. Hy vit cc cng thc ho hc ca cc hp cht v c c th c to thnh cc
nguyn t trn?2. Lp CTHH hp cht khi bit thnh phn khi lng nguyn t . 1:
Bit t l khi lng cc nguyn t trong hp cht.Cch gii:-t cng thc tng qut:
AxBy- Ta c t l khi lng cc nguyn t: ..MA xMB y =mAmB-Tm c t l : xy
=..mA MBmB MA= ab (t l cc s nguyn dng, ti gin)- Thay x= a, y = b -
Vit thnh CTHH.V d:: Lap CTHH cua sat va oxi, biet c 7 phan khoi lng
sat th ket hp vi 3 phan khoi lng oxi.Gii:- t cng thc tng qut:
FexOy- Ta c t l khi lng cc nguyn t: ..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: Lap CTHH cua sat va oxi, biet
c 7 phan khoilng sat thket hp vi3 phan khoi lng oxi.2: Hp cht B (hp
cht kh ) bit t l v khi lng cc nguyn t to thnh: mC : mH = 6:1, mt lt
kh B (ktc) nng 1,25g.3: Hp cht C, bit t l v khi lng cc nguyn t l :
mCa : mN : mO = 10:7:24 v 0,2 mol hp cht C nng 32,8 gam.4: Hp cht D
bit: 0,2 mol hp cht D c cha 9,2g Na, 2,4g C v 9,6g O5: Phan t khoi
cua ong sunfat la 160 vC. Trong o co mot nguyen t Cu co nguyen t
khoi la 64, mot nguyen t S co nguyen t khoi la 32, con lai la
nguyen t oxi. Cong thc phan cua hp chat la nh the nao?6:Xc nh cng
thc phn t ca CuxOy, bit t l khi lng gia ng v oxi trong oxit l 4 :
1? 7: Trong 1 tap hp cac phan t ong sunfat (CuSO4) co khoi lng
160000 vC. Cho biet tap hp o co bao nhieu nguyen t moi loai.8 :
Phan t khoi cua ong oxit (co thanh phan gom ong va oxi)va ong
sunfat co t le 1/2.Biet khoilng cua phan t ong sunfat la 160 vC.Xac
nh cong thc phan t ong oxit?9. Mot nhom oxit co t so khoi lng cua 2
nguyen to nhom va oxi bang 4,5:4. Cong thc hoa hoc cua nhom oxit o
la g?2. Bitkhi lng cc nguyn t trong mt lng hp cht, Bit phn t khi hp
cht hoc ch a bit PTK(bi ton t chy) Bi ton c dng : tm (g)AxByCzt chy
m(g) cc hp cht cha A,B,C+Trng hp bit PTKTm c CTHH ng+Trng hp cha
bit PTKTm c CTHH n ginCch gii: - Tm mA, mB, mC trongm(g) cc hp cht
cha cc nguyn t A,B,C.+ Nu (mA + m B) = m (g)AxByCz Trong h/c khng c
nguyn t CT : x : y =MAmA : MBmB = a:b (t l cc s nguyn dng, ti gin)
CTHH: AaBb+ Nu (mA + m B) m (g)AxByCz Trong h/c c nguyn t C 5 m C =
m (g)AxByCz- (mA + m B)T : x : y : z =MAmA : MBmB : mcMc = a:b:c (t
l cc s nguyn dng, ti gin) CTHH: AaBbCcCch gii khc: Da vo phng trnh
phn ng chy tng qut CxHy +020 042 2 2HyxCyx +
,_
+CxHy0z +020 02 42 2 2HyxCz yx +
,_
+- Lp t l s mol theo PTHH v s mol theo d kin bi ton suy ra x, y,
z.V d: t chy 4,5 g hp cht hu c A. Bit A cha C, H, 0 v thu c 9,9g kh
C02 v 5,4g H20. Lp cng thc phn t ca A. Bit kh lng phn t A bng
60.Gii: - Theo bi ra:mol nA075 , 0605 , 4 ,mol nC225 , 0449 , 920
,mol nH3 , 0184 , 502 - Phng trnh phn ng :CxHy0z +020 02 42 2
2HyxCz yx + ,_
+1mol . ,_
+2 4z yx(mol). x (mol)) (2molySuy ra : 82 . 3 , 0 075 , 013225 ,
0 075 , 01 yyxxMt khc;MC3H80z= 60 Hay : 36 + 8 + 16z =60 > z = 1
Vy cng thc ca A l C3H80 * Bi tp vn dng:+ Trng hp cha bit PTKTm c
CTHH n gin1: t chy hon ton 13,6g hp cht A,th thu c 25,6g SO2 v
7,2gH2O. Xc nh cng thc ca A2:ot chayhoantoanmgamchat Acandunghet
5,824dm3O2(ktc). San pham co CO2 va H2O c chia oi. Phan 1 cho i qua
P2O5 thay lng P2O5 tang 1,8 gam. Phan 2 cho i qua CaO thay lng CaO
tang 5,32 gam. Tm m va cong thc n gian A. Tm cong thc phan t A va
biet A the kh (k thng) co so C 4.3: t chy hon ton 13,6g hp cht A,
th thu c 25,6 g S02 v 7,2g H20. Xc nh cng thc A +Trng hp bit PTKTm
c CTHH ng1: t chy hon ton 4,5g hp cht hu c A .Bit A cha C, H, O v
thu c 9,9g kh CO2 v 5,4g H2O. lp cng thc phn t ca A. Bit phn t khi
A l 60.2: t chy hon ton 7,5g hyroccbon A ta thu c 22g CO2 v 13,5g
H2O. Bit t khi hI so vi hyr bng 15. Lp cng thc phn t ca A.3: t chy
hon ton 0,3g hp cht hu c A . Bit A cha C, H, O v thu c 224cm3 kh
CO2 (ktc) v 0,18g H2O. lp cng thc phn t ca A.Bit t khi ca A i vi
hiro bng 30.64: t chy 2,25g hp cht hu c A cha C, H, O phi cn 3,08
lt oxy (ktc) v thu c VH2O =5\4 VCO2.Bit t khi hi ca A i vi H2 l 45.
Xc nh cng thc ca A5: Hiro A l cht lng , c t khi hi so vi khng kh
bng 27. t chy A thu c CO2 v H2O theo t l khi lng 4,9 :1 . tm cng
thc ca AS:A la C4H103: Bit thnh phn phn trm v khi lng cc nguyn t,
cho bit NTK, phn t khi.Cch gii: - Tnh khi lng tng nguyn t trong 1
mol hp cht.- Tnh s mol nguyn t tng nguyn t trong 1 mol hp cht.- Vit
thnh CTHH.Hoc: - t cng thc tng qut: AxBy- Ta c t l khi lng cc nguyn
t: y MBx MA..= BA%%- Rt ra t l x: y = MAA %: MBB % (ti gin)- 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.Vi d. Mot hp chat kh Y co phan t
khoi la 58 vC, cau tao t 2 nguyen to C
vaHtrongonguyentoCchiem82,76%khoi lngcuahpchat. Tmcong thc phan t
cua hp chat. 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 : C5H8* Bi
tp vn dng:1: Hp cht X c phn t khi bng 62 vC. Trong phn t ca hp cht
nguyn t oxi chim 25,8% theo khilng, cn lil nguyn t Na. S nguyn t ca
nguyn t O v Na trong phn t hp cht l bao nhiu ?2: Mot hp chat X co
thanh phan % ve khoi lng la :40%Ca, 12%C va 48% O . Xac nh CTHH cua
X. Biet khoi lng mol cua X la 100g.3:Tm cng thc ho 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: Xac nh cong thc cua cac hp
chat sau:a) Hp chat tao thanh bi magie va oxi co phan t khoi la 40,
trong o phan tram ve khoi lng cua chung lan lt la 60% va 40%.b) Hp
chat tao thanh bi lu huynh va oxi co phan t khoi la 64, trong o
phan tram ve khoi lng cua oxi la 50%.c) Hp chat cua ong, lu huynh
va oxi co phan t khoi la 160, co phan tram cua ong va lu huynh lan
lt la 40% va 20%.d) Hp chat tao thanh bi sat va oxi co khoi lng
phan t la 160, trong o phan tram ve khoi lng cua oxi la 70%.e) Hp
chat cua ong va oxi co phan t khoi la 114, phan tram ve khoi lng
cua ong la 88,89%.f) Hp chat cua canxi va cacbon co phan t khoi la
64, phan tram ve khoi lng cua cacbon la 37,5%.g) A co khoi lng mol
phan t la 58,5g; thanh phan % ve khoi lng nguyen to: 60,68% Cl con
lai la Na.7h) B co khoilng molphan t la 106g;thanh phan % ve
khoilng cua cac nguyen to: 43,4% Na; 11,3% C con lai la cua O.i)
Ccokhoi lngmol phantla101g; thanhphanphantramvekhoi lng cac nguyen
to: 38,61% K; 13,86% N con lai la O.j) D co khoilng molphan t la
126g;thanh phan % ve khoilng cua cac nguyen to: 36,508% Na; 25,4% S
con lai la O.k) Eco 24,68% K; 34,81% Mn; 40,51%O. E nang hn NaNO3
1,86 lan.l) F cha 5,88% ve khoi lng la H con lai la cua S. F nang
hn kh hiro 17 lan.m) G co 3,7% H; 44,44% C; 51,86% O. G co khoi lng
mol phan t bang Al.n) H co 28,57% Mg; 14,285% C; 57,145% O. Khoi
lng mol phan t cua H la 84g.6 . Phan t canxi cacbonat co phan t
khoi la 100 vC , trong o nguyen t canxi chiem 40% khoi lng, nguyen
to cacbon chiem 12% khoi lng. Khoi lng con lai la oxi. Xac nh cong
thc phan t cua hp chat canxi cacbonat?7.Mot hpchat co phan t khoi
bang 62 vC. trongphant cua hpchat nguyen to oxi chiem 25,8% theo
khoi lng, con lai la nguyen to Na. Xac nh ve t le so nguyen tcua O
va so nguyen tNa trong hp chat.8 .Trong hp chat XHn co cha 17,65%
la hidro. Biet hp chat nay co tkhoi so vi kh Metan CH4 la 1,0625. X
la nguyen to nao ?4: Bit thnhphnphntrmvkhi lngccnguyntmbi
khngchobit 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.V d: Hy xc nh cng thc hp cht A bit
thnh phn % v khi lng cc nguyn t l: 40%Cu. 20%S v 40% O.Gii:- t cng
thc tng qut: CuxSyOz - Rt ra t l x: y:z= %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)2. Nung 2,45
gam mt cht ha hc A thy thot ra 672 ml kh O2 (ktc). Phn rn cn li cha
52,35% kali v 47,65% clo (v khi lng). Tm cng thc ha hc ca A.3 . Hai
nguyen t X ket hp vi 1 nguyen t O tao ra phan t oxit. Trong phan t,
nguyen t oxi chiem 25,8% ve khoi lng. Hoi nguyen to X la nguyen to
nao?4. Mot nguyen t M ket hp vi 3 nguyen tH tao thanh hp chat
vihyro. Trong phan t, khoi lng H chiem 17,65%. Hoi nguyen to M la
g?5 . Hai nguyen t Y ket hp vi 3 nguyen t O tao ra phan t oxit.
Trong phan t, nguyen t oxi chiem 30% ve khoi lng. Hoi nguyen to X
la nguyen to nao?6. Mot hp chat co thanh phan gom 2 nguyen to C va
O. Thanh phan cua hp chat co 42,6% la nguyen to C, con lai la
nguyen to oxi. Xac nh ve t le so nguyen tcua C va so nguyen toxi
trong hp chat.7 .Lp cng thc phn t ca A .Bit em nung 4,9 gam mt mui
v c Ath thu c 1344 ml kh O2 ( ktc), phn cht rn cn li cha 52,35% K v
47,65% Cl. H ng dn gii: n2O= 4 , 22344 , 1 = 0,06 (mol) m2O = 0,06
. 32 =1,92 (g) p dng LBT khi lng ta c: m cht rn = 4,9 1,92 = 2,98
(g) m K = 10098 , 2 35 , 52 =1,56 (g) n K = 3956 , 1 = 0,04 (mol)
mCl = 2,98 1,56 = 1,42 (g) n Cl = 5 , 3542 , 1 = 0,04 (mol) Gi cng
thc tng qut ca B l: KxClyOzta c:8x : y : z=0,04 : 0,04 : 0,06 2 = 1
: 1 : 3V i vi hp cht v c ch s ca cc nguyn t l ti gin nn cng thc ho
hc ca A l KClO3. 5: Bin lun gi tr khilng mol(M) theo ha tr(x,y) tm
NTK hoc PTK..bit thnh phn % v khi lng hoc t l khi lng cc nguyn t. +
Tr ng 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:Bl 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 a R =6112n V n l ht ca nguyn t nn n phi nguyn dng,
ta c bng sau: n I II III IVR 18,637,356 76,4loi loi Fe loi Vy cng
thc phn t ca C l Fe2O3.+ Tr ng 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 ht ca nguyn t nn n phi nguyn dng, ta c bng sau: n I II III IVM
18,637,356 76,4loi loi Fe loi Vy cng thc phn t ca C l Fe2O3.* Bi tp
vn dng:1. Oxit cua kim loai mc hoa tr thap cha 22,56% oxi, con oxit
cua kim loai o mc hoa tr cao cha 50,48%. Tnh nguyen t khoi cua kim
loai o.2. Co mot hon hp gom 2 kim loai A va B co t le khoi lng
nguyen t 8:9. Biet khoi lng nguyen t cua A, B eu khong qua 30 vC.
Tm 2 kim loai *Giai: Neu A :B =8 : 9th 89A nB n '
9Theo e : t so nguyen t khoi cua 2 kim loai la 89AB nen 89A nB n
' ( n z+ )V A, B eu co KLNT khong qua 30 vC nen:9n 30 n 3 Ta co
bang bien luan sau :n 1 2 3A 8 16 24B 9 18 27Suy ra hai kim loai la
Mg va AlD/ Lp CTHH hp cht kh da vo t khiCch gii chung:- Theo cng
thc tnh t khi cc cht kh: d A/B = MBMA- Tm khi lng mol (M) cht cn tm
NTK,PTK ca cht Xc nh CTHH.V d : Cho 2 kh A v B c cng thc ln lt l
NxOy v NyOx . t khi hi i vi Hyro ln lt l: d A/H2 = 22, d B/A =
1,045. Xc nh CTHHca A v BGii: Theo bi ra ta c: - d NxOy/H2 = 2
MHMA= 2MA= 22 MA =MNxOy = 2.22 = 4414x+ 16y = 44(1)- d NyOx/NxOy =
MAMB= 44MB = 1,045 MB = MNyOx = 44.1,045 = 45,9814y+ 16x = 45,98
(2)gi tr tha mn k bi ton: x = 2 , y= 1 A = N2O , B = NO2* Bi tp vn
dng:1.Cho 2 cht kh AOx c TP% O = 50% v BHy c TP% H = 25% . bit
dAOx/BHy = 4. Xc nh CTHH ca 2 kh trn.2.Mt oxit ca Nit c cng thc
NxOy. Bit khi lng ca Nit trong phn t chim 30,4%. ngoira c 1,15 gam
oxit ny chim th tch l 0,28 lt (ktc).Xc nh CTHH ca oxit trn.3. C 3
hirocacbon A, B, CA: CxH2x+2B : Cx' H2x'C : Cx' H2x'- 2Bit d B/A =
1,4 ; d A/C = 0,75 . Xc nh CTHH ca A, B, C.E/Lpcng thc ho hc hp cht
da vo ph ng trnh phn ng ho hc: 1.Dng ton c bn 1: Tm nguyn t hay hp
cht ca nguyn t trong trng hp cho bit ha tr ca nguyn t, khi bi ton
cho bit lng cht(hay lng hp cht ca nguyn t cn tm) v lng mt cht khc(c
th cho bng gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh)
trong mt phn ng ha hc.Cch gii chung:Bi ton c dng : a M + bBcC + d
D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)- t cng thc
cht cho theo bi ton : - Gi a l s mol, A l NTK hay PTK ca cht cn tm.
- Vit phng trnh phn ng, t s mol a vo phng trnh v tnh s mol cc cht c
lin quan theo a v A. -Lp phng trnh, giitm khi lng mol (M(g)) cht cn
tm NTK,PTK ca cht Xc nh nguyn t hay hp cht ca nguyn t cn tm. L u :
Lng cht khc trong phn ng ha hc c th cho nhng dng sau:1. Cho dng trc
tip bng: gam, mol.V d1: Cho 7,2g mt kim loi ho tr II phn ng hon ton
vi dung dch HCl, thu c 0,3 mol H2 iu kin tiu chun. Xc nh tn kim loi
dng. 10Gii: - Gi CTHH ca kim loi l : Mt x l s mol , A l NTK ca kim
loi dng phn ng . Ta c Phng trnh phn ng:M + 2HCl > MCl2 + H21mol
1mol x (mol) x (mol)Suy ra ta c h s :m M = x . A = 7,2 (g) (1)nM =
n H2 = x= 0,3 (mol) (2)Th (2) vo (1) ta c A = 7, 20, 3 = 24(g) NTK
ca A = 24.Vy A l kim loi Mg2/ Cho dng gin tip bng: V(ktc)V d2: Cho
7,2g mt kim loi ho tr II phn ng hon ton vi dung dch HCl, thu c 6,72
lt H2 iu kin tiu chun. Xc nh tn kim loi dng. GiiTm : nH2 = 6, 7222,
4 = 0,3 (mol) Bi ton quay v v d 1 * Cho 7,2g mt kim loi ho tr II
phn ng hon ton vi dung dch HCl, thu c 0,3 mol H2 iu kin tiu chun.
Xc nh tn kim loi dng. (gii nh v d 1)3/ Cho dng gin tip bng: mdd,
C%V d 3: Cho 7,2g mt kim loi ho tr II phn ng hon ton 100g dung dch
HCl 21,9%. Xc nh tn kim loi dng. Gii t x l s mol, A l NTK ca kim
loi dng phn ng . p dng : C % = .100% mctmdd m HCl =. %100%mdd c
=100.21, 9100 = 21,9 (g) n HCl = mM =21, 936, 5 = 0,6 (mol)*Tr v bi
ton cho dng trc tip: Cho 7,2g mt kim loi ho tr II phn ng hon ton
0,6 mol HCl . Xc nh tn kim loi dng. Ta c Phng trnh phn ng:M +
2HCl>MCl2 +H21mol 2mol x (mol) 2x (mol) Suy ra ta c h s : m A =
x . A = 7,2 (g) (1)nHCl = 2x= 0,6 (mol) x= 0,3 (mol) (2)Th (2) vo
(1) ta c A = 7, 20, 3 = 24(g) NTK ca A = 24.Vy A l kim loi Mg4/ Cho
dng gin tip bng: Vdd, CMV d 4 : Cho 7,2g mt kim loi ho tr II phn ng
hon ton 100 ml dung dch HCl 6 M. Xc nh tn kim loi dng. Gii11Tmn
HCl= ? p dng : CM = nV n HCl= CM.V = 6.0,1 = 0,6 (mol) *Tr v bi ton
cho dng trc tip: Cho 7,2g mt kim loi ho tr II phn ng hon ton 0,6
mol HCl. Xc nh tn kim loi dng. (Gii nh v d 3)5/ Cho dng gin tip bng
: mdd, CM ,d (g/ml)V d 5 : Cho 7,2g mt kim loi ho tr II phn ng hon
ton 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim loi dng.
Gii- Tm Vdd (da vo mdd, d (g/ml)):td= mV Vdd H Cl = md = 1201, 2 =
100 (ml) =0,1(l) - Tmn HCl= ? p dng : CM = nV n HCl= CM. V = 6. 0,1
= 0,6 (mol) *Tr v bi ton cho dng trc tip: Cho 7,2g mt kim loi ho tr
II phn ng hon ton 0,6 mol HCl. Xc nh tn kim loi dng. (Gii nh v d
3)6/ Cho dng gin tip bng: Vdd, C%, d (g/ml)V d 6 : Cho 7,2g mt kim
loi ho tr II phn ng hon ton 83,3 ml dung dch HCl 21,9 % ( d= 1,2
g/ml). Xc nh tn kim loi dng. Gii - Tm m dd (da vo Vdd, d
(g/ml)):td= mV mdd H Cl = V.d= 83,3 . 1,2 = 100 (g) dd HCl.p dng :
C % = .100% mctmdd m HCl =. %100%mdd c =100.21, 9100 = 21,9 (g) n
HCl = mM =21, 936, 5 = 0,6 (mol)*Tr v bi ton cho dng trc tip: Cho
7,2g mt kim loi ho tr II phn ng hon ton 0,6 mol HCl. Xc nh tn kim
loi dng. (Gii nh v d 3)Vn dng 6 dng ton trn: Ta c th thit lp c 6 bi
ton lp CTHH ca mt hp cht khi bit thnh phn nguyn t, bit ha tr vi lng
HCl cho 6 dng trn. Bi 1 :Cho 12 g mt Oxt kim loi ho tr II phn ng
hon ton vi 0,6 mol HCl . Xc nh tn kim loi dng. Gii- Gi CTHH ca oxit
l:MOt x l s mol , A l PTK ca o xt dng phn ng . Ta c Phng trnh phn
ng:MO +2HCl >MCl2 + H2O1mol1mol x (mol)2x (mol)Suy ra ta c h s :
m MO = x . A =12(g) (1)12nHCl = 2x= 21, 936, 5 = 0,6(mol) x= 0,6:2
= 0,3 (mol) (2)Th (2) vo (1) ta c A = 120, 3 = 40(g) MM = MMO - MO
= 40 16 = 24 (g) NTK ca M = 24.Vy M l kim loi Mg CTHH ca o xt l
MgOBi 2:Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 21,9 g
HCl . Xc nh tn kim loi dng. Bi 3: Cho 12 g mt Oxt kim loi ho tr II
phn ng hon ton 100g dung dch HCl21,9%. Xc nh tn kim loi dng. Bi 4 :
Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton 100 ml dung dch HCl
6 M. Xc nh tn kim loi dng. Bi 5 : Cho 12 g mt Oxt kim loi ho tr II
phn ng hon ton 120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim
loi dng. Bi 6:Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton 83,3
ml dung dch HCl 21,9 % ( d= 1,2 g/ml). Xc nh tn kim loi dng. 2. Dng
ton c bn 2: Tm nguyn t hay hp cht ca nguyn t trong trng hp ch a bit
ha tr ca nguyn t, khi bi ton cho bit lng cht(hay lng hp cht ca
nguyn t cn tm) v lng mt cht khc(c th cho bng gam, mol, V(ktc) , cc
i lng v nng dd, tan, t khi cht kh) trong mt phn ng ha hc,.Cch gii
chung:Bi ton c dng : a M + bB cC + d D(Trong cc cht M, B, C, D :c
th l mt n cht hay 1 hp cht)- t cng thc cht cho theo bi ton : - Gi a
l s mol, A l NTK hay PTK, x, y.... l ha tr ca nguyn t ca chthy hp
cht ca nguyn t cn tm. - Vit phng trnh phn ng, t s mol a vo phng
trnh v tnh s mol cc cht c lin quan theo a v A. -Lp phng trnh, bin
lun gi tr khi lng mol (M(g)) theo ha tr (x,y) ca nguyn t cn tm ( 1,
x y 5) t NTK,PTK ca cht Xc nh nguyn t hay hp cht ca nguyn t cn tm.
V d 1.2: Cho 7,2g mt kim loi cha r ha tr, phn ng hon ton vi 0,6
HCl. Xc nh tn kim loi dng. Gii:- Gi CTHH kim loi l : M- Gi x l s
mol, A l NTK ca kim loi M, n l ha tr ca kim loi MTa c Phng trnh phn
ng:2M +2nHCl >2MCln + nH22(mol ) 2n(mol) x (mol) nx (mol)Suy ra
ta c h s : m M = x . A =7,2(g)(1) nHCl = xn=0,6(mol) x= 0,6:n (2)Th
(2) vo (1) ta c A = 7, 2.0, 6n = 12.n Vn phi nguyn dng, ta c bng
sau: 13n I II IIIA 12 24 36loi Mg loiA = 24 (g) NTK ca kim loi = 24
Kim loi l MgT ta c th thit lp c 6 bi ton(phn dng c bn 1) v 6 bi ton
(phn dng c bn 2) vi lng HCl cho 6 dng trn Bi 1.1 :Cho 12 g mt Oxt
kim loi ho tr II phn ng hon ton vi 0,6 mol HCl . Xc nh tn kim loi
dng. Gii- Gi CTHH ca oxit l:MOt x l s mol , A l PTK ca o xt dng phn
ng . Ta c Phng trnh phn ng:MO +2HCl >MCl2 + H2O1mol1mol x
(mol)2x (mol)Suy ra ta c h s : m MO = x . A =12(g)(1)nHCl = 2x= 21,
936, 5 = 0,6(mol) x= 0,6:2 = 0,3 (mol)(2)Th (2) vo (1) ta c A =
120, 3 = 40(g) MM = MMO - MO = 40 16 = 24 (g) NTK ca M = 24.Vy M l
kim loi Mg CTHH ca o xt l MgOBi 2.1:Cho 12 g mt Oxt kim loi ho tr
II phn ng hon ton vi 21,9 g HCl . Xc nh tn kim loi dng. Bi 3.1 :
Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 100g dung dch
HCl 21,9%. Xc nh tn kim loi dng. Bi 4.1 : Cho 12 g mt Oxt kim loi
ho tr II phn ng hon ton vi 100 ml dung dch HCl 6 M. Xc nh tn kim
loi dng. Bi 5.1 :Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi
120 g dung dch HCl 6 M ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 6.1
:Cho 12 g mt Oxt kim loi ho tr II phn ng hon ton vi 120 ml dung dch
HCl21,9 % ( d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 7.2: Cho 7,2g mt
kim loi cha r ha tr, phn ng hon ton vi 0,6 molHCl. Xc nh tn kim loi
dng. Bi 8.2:ho 7,2g mt kim loi cha r ha tr , phn ng hon ton vi 21,9
g HCl . Xc nh tn kim loi dng. Bi 9.2: Cho 7,2g mt kim loi cha r ha
tr , phn ng hon ton vi 100g dung dch HCl 21,9%. Xc nh tn kim loi
dng. Bi 10.2: Cho 7,2g mt kim loi cha r ha tr , phn ng hon ton vi
100 ml dung dch HCl 6 M. Xc nh tn kim loi dng. Bi 11.2 : Cho 7,2g
mt kim loi cha r ha tr , phn ng hon ton vi 120 g dung dch HCl 6 M (
d= 1,2 g/ml). Xc nh tn kim loi dng. Bi 12.2 : Cho 7,2g mt kim loi
cha r ha tr ,phn ng hon ton vi 83,3 ml dung dch HCl 21,9 % ( d= 1,2
g/ml). Xc nh tn kim loi dng. 14Bi 13: Cho 7,22 gam hon hp X gom Fe
va kim loai M co hoa tr khong oi. Chia hon hp thanh 2 phan bang
nhau. - Hoa tan het phan 1 trong dung dch HCl, c 2,128 lt H2.- Hoa
tan het phan 2 trong dung dch HNO3, c 1,792 lt kh NO duy nhat.Xac
nh kim loai M va % khoi lng moi kim loai trong hon hp X.ap so:M
(Al) va %Fe = 77,56% ; %Al = 22,44%Bi 14:Kh 3,48 gam mt oxit kim
loi M cn dng 1,344 lt kh hiro ( ktc). Ton b l-ng kim loi thu c tc
dng vi dung dch HCl d cho 1,008 lt kh hiro ktc.Tm kim loi M v oxit
ca n .(CTHH oxit : Fe3O4)Mt sdng bi ton bin lun v lp CTHH (Dnh cho
HSG K9) DNG:BIN LUN THEO N S TRONG GII PHNG TRNHBi 1 : Ha tan mt
kim loi cha bit ha tr trong 500ml dd HCl th thy thot ra 11,2 dm3 H2
( KTC). Phi trung ha axit d bng 100ml dd Ca(OH)2 1M. Sau c cn dung
dch thu c th thy cn li 55,6 gam mui khan. Tm nng M ca dung dch axit
dng; xc nh tn ca kim loi dng.Gii : Gi s kim loi l R c ha tr l x 1
x, nguyn 3 s mol Ca(OH)2 = 0,11 = 0,1 mols mol H2 = 11,2 : 22,4 =
0,5 molCc PTP: 2R + 2xHCl 2RClx+ xH2 (1)1/x (mol) 1 1/x 0,5Ca(OH)2+
2HClCaCl2+ 2H2O(2)0,1 0,2 0,1t cc phng trnh phn ng (1) v (2) suy
ra:nHCl = 1 + 0,2 = 1,2 mol nng M ca dung dch HCl : CM = 1,2 : 0,5
= 2,4 Mtheo cc PTP ta c : 55, 6 (0,1 111) 44,5xRClm gam ta c : 1x (
R + 35,5x ) = 44,5 R =9x X 1 2 3R 918 27Vy kim loi tho mn u bi l
nhm Al ( 27, ha tr III )Bi 2: Khi lm ngui 1026,4 gam dung dch bo ha
R2SO4.nH2O ( trong R l kim loi kim v n nguyn, tha iu kin 7< n
< 12 ) t 800Cxung 100C th c 395,4 gam tinh th R2SO4.nH2O tch ra
khi dung dch.Tm cng thc phn t ca Hirat ni trn. Bit tan ca R2SO4
800C v 100C ln lt l 28,3 gam v 9 gam.Gii: S( 800C) = 28,3 gam trong
128,3 gam ddbh c 28,3g R2SO4 v 100g H2OVy : 1026,4gam ddbh 226,4 g
R2SO4 v 800 gam H2O.Khi lng dung dch bo ho ti thi im
100C:1026,4395,4= 631 gam 100C, S(R2SO4 ) = 9 gam, nn suy ra: 109
gam ddbh c cha9 gam R2SO4 vy 631 gam ddbh c khi lng R2SO4 l : 631
952,1109gamkhi lng R2SO4 khan c trong phn hirat b tch ra : 226,4
52,1 = 174,3 gamV s mol hirat = s mol mui khan nn : 395, 4 174, 32
96 18 2 96 R n R+ + +442,2R-3137,4x +21206,4 = 0 R =7,1n 48 cho R l
kim loi kim ,7 < n< 12 , nnguyn ta c bng bin lun:n 8 9 10 11R
8,8 18,6 23 30,1Kt qu ph hp l n = 10 , kim loi l Na cng thc hirat l
Na2SO4.10H2ODNG:BIN LUN THEO TRNG HP Bi1:Hn hp A gm CuO v mt oxit
ca kim loi ha tr II( khng i ) c t l mol 15 1: 2. Cho kh H2 d i qua
2,4 gam hn hp A nung nng th thu c hn hp rn B. ha tan ht rn B cn dng
ng 80 ml dung dch HNO3 1,25M v thu c kh NO duy nht.Xc nh cng thc ha
hc ca oxit kim loi. Bit rng cc phn ng xy ra hon ton. Gii: t CTTQ ca
oxit kim loi l RO.Gi a, 2a ln lt l s mol CuO v RO c trong 2,4 gam
hn hp AV H2 ch kh c nhng oxit kim loi ng sau Al trong dy BKTp nn c
2 kh nng xy ra:- R l kim loi ng sau Al :Cc PTP xy ra:CuO + H2Cu+H2O
a aRO + H2R+H2O2a 2a3Cu + 8HNO3 3Cu(NO3)2+2NO + 4H2Oa83a3R + 8HNO3
3R(NO3)2+2NO + 4H2O 2a163a Theo bi:8 160, 0125 0, 08 1, 25 0,13
340( )80 ( 16)2 2, 4a aaR Caa R a + ' '+ + Khng nhn Ca v kt qu tri
vi gi thit R ng sau Al- Vy R phi l kim loi ng trc Al CuO + H2Cu+H2O
a a3Cu + 8HNO3 3Cu(NO3)2+2NO + 4H2Oa83aRO + 2HNO3 R(NO3)2+ 2H2O 2a
4aTheo bi : 80, 015 4 0,1324( )80 ( 16).2 2, 4aa aR Mga R a + ' '+
+ Trng hp ny tho mn vi gi thit nn oxit l:MgO.Bi2 : Khi cho a (mol )
mt kim loi R tan va ht trong dung dch cha a (mol ) H2SO4 th thu c
1,56 gam mui v mt kh A. Hp th hon ton kh A vo trong 45ml dd NaOH
0,2M th thy to thnh 0,608 gam mui. Hy xc nh kim loi dng. Gii :Gin l
ha tr ca kim loi R .V cha r nng ca H2SO4 nnc th xy ra 3 phn ng:2R +
nH2SO4 R2 (SO4 )n+ nH2 (1)2R + 2nH2SO4 R2 (SO4 )n+ nSO2 +2nH2O
(2)2R + 5nH2SO4 4R2 (SO4 )n+ nH2S +4nH2O (3)kh A tc dng c vi NaOH
nn khng th l H2 P (1) khng ph hp.V s mol R = s mol H2SO4 = a , nn
:Nu xy ra ( 2) th : 2n = 2 n =1 ( hp l )Nu xy ra ( 3) th : 5n = 2 n
=25( v l )Vy kim loi R ha tr I v kh A l SO22R + 2H2SO4 R2 SO4 + SO2
+2H2Oa(mol) a2a2aGi s SO2 tc dng vi NaOH to ra 2 mui NaHSO3 ,
Na2SO3SO2+ NaOH NaHSO3 t:x (mol)x xSO2+ 2NaOH Na2SO3+ H2O16y (mol)
2y ytheo ta c : 2 0, 2 0, 045 0, 009104 126 0, 608x yx y+ '+ gii h
phng trnh c 0, 0010, 004xy 'Vy gi thit phn ng to 2 mui l ng.Ta c:s
mol R2SO4 = s mol SO2= x+y = 0,005(mol)Khi lng ca R2SO4 :(2R+ 96)
0,005=1,56 R =108. Vy kim loi dng l Ag.DNG: BIN LUN SO SNHBi 1:C mt
hn hp gm 2 kim loi A v B c t l khi lng nguyn t 8:9. Bit khi lng
nguyn t ca A, B u khng qu 30 vC. Tm 2 kim loi Gii:Theo : t s nguyn
t khi ca 2 kim loi l 89AB nn 89A nB n ' ( n z+ )V A, B u c KLNT
khng qu 30 vC nn:9n 30 n 3 Ta c bng bin lun sau :n 1 2 3A 8 16 24B
9 18 27Suy ra hai kim loi l Mg v AlBi 2 :Ha tan 8,7 gam mt hn hp gm
K v mt kim loi Mthuc phn nhm chnh nhm II trong dung dch HCl d th
thy c 5,6 dm3 H2 ( KTC). Ha tan ring 9 gam kim loi M trong dung dch
HCl d th th tch kh H2 sinh ra cha n 11 lt ( KTC). Hy xc nh kim loi
M.Gii:t a, b ln lt l s mol ca mi kim loi K, M trong hn hpTh nghim
1:2K + 2HCl 2KCl + H2 a a/2M + 2HCl MCl2+ H2 b b s mol H2= 5,60, 25
2 0, 52 22, 4ab a b + + Th nghim 2:M + 2HCl MCl2+ H2 9/M(mol)
9/MTheo bi: 9 1122, 4 M 18,3(1) Mt khc:39 . 8, 7 39(0, 5 2 ) 8, 72
0, 5 0, 5 2a b M b bMa b a b+ + ' '+ b = 10,878 M V0 < b Z la 8.
Xac nh kim loai Y va Z.Chuyn III.Bi tp v phng trnh ha hc ha hc a.Lp
ph ng trnh ha hc: Cch gii chung:- Vit s ca phn ng (gm CTHH ca cc
cht p v sn phm).- Cn bng s nguyn t ca mi nguyn t (bng cch chn cc h
s thch hp in vo trc cc CTHH).- Vit PTHH.Lu : Khi chn h s cn bng:+
Khi gp nhm nguyn t -> Cn bng nguyn c nhm.+ Thng cn bng nguyn t c
s nguyn t l cao nht bng cch nhn cho 2,4+ Mt nguyn t thay i s nguyn
t 2 v PT, ta chn h s bng cch ly BSCNN ca 2 s trn chia cho s nguyn t
ca nguyn t .V d:?K + ?O2-> ?K2OGii:4K + O2-> 2K2O+ Khi gp mt
s phng trnh phc tp cn phi dng phng php cn bng theo phng php i s:V d
1:Cn bng PTHH sau :FeS2+ O2-> Fe2O3+ SO2 Gii:- t cc h s:aFeS2+
bO2-> cFe2O3+ dSO219 - Tnh s nguyn t cc nguyn t trc v sau phn ng
theo cc h s trong PTHH: Ta c: + S nguyn t Fe:a = 2c + S nguyn t S
:2a = d+ S nguyn t O : 2b = 3c + 2dt a = 1 c = 1/2, d = 2, b = 3/2
+ 2.2 = 11/2Thay a, b, c, d vo PT:aFeS2+ bO2->cFe2O3+ dSO2FeS2+
11/2O2->1/2Fe2O3+2SO2
Hay: 2FeS2+ 11O2 -> Fe2O3+4SO2
V d 2 Cn bng PTHH sau: FexOy+H2Fe +H2O Gii:- t cc h s: a FexOy+
b H2cFe + d H2O - Tnh s nguyn t cc nguyn t trc v sau phn ng theo cc
h s trong PTHH: Ta c: + S nguyn t Fe: a.x = c + S nguyn t O : a.y =
d+ S nguyn t H : 2b = 2dt a = 1 c = x, d = b = yThay a, b, c, d vo
PT: FexOy+ y H2 xFe + y H2O
* Bi tp vn dng:1 . Hay chon CTHH va he so thch hp at vao nhng
cho co dau hoi trong cac PTP sau e c PTP ung :a/ ?Na+ ?2Na2Ob/ 2HgO
? Hg+?c/ ? H2 + ? 2H2O d/ 2Al+6HCl ?AlCl3+?2: Hoan thanh cacs o PHH
sau e c PTHH ung:a/ CaCO3 + HCl ------> CaCl2 + CO2 + H2 b/
C2H2+ O2 ---------> CO2 +H2O c/ Al +H2SO4 -------->Al2(SO4)3
+ H2 d/ KHCO3+ Ba(OH)2 ------->BaCO3 + K2CO3 + H2O e/ NaHS+
KOH------> Na2S+ K2S + H2O f/ Fe(OH)2+ O2+ H2O
------>Fe(OH)33:ot chay khaxetylen (C2H2) trong kh oxi sinh ra
kh cacbonic va hi nc .Dan honhpkh vaodungdchncvoi trong(Ca(OH)2)th
thucchatkettua canxicacbonat (CaCO3) .Viet cac PTP xay ra . 4: Hon
thnh cc PTHH cho cc p sau:Na2O+ H2O ->NaOH.BaO + H2O ->
Ba(OH)2CO2+ H2O ->H2CO3N2O5+ H2O ->HNO3P2O5+ H2O ->
H3PO4NO2+ O2+ H2O-> HNO3SO2+ Br2+ H2O ->H2SO4+ HBrK2O +
P2O5-> K3PO4Na2O+ N2O5-> NaNO3Fe2O3+H2SO4-> Fe2(SO4)3+
H2OFe3O4+ HCl -> FeCl2+ FeCl3+ H2OKOH+ FeSO4->
Fe(OH)2+K2SO4Fe(OH)2+ O2-> Fe2O3+ H2O.KNO3->KNO2+
O2AgNO3->Ag + O2+ NO2Fe + Cl2-> FeClnFeS2+ O2-> Fe2O3+
SO2FeS + O2-> Fe2O3+ SO2FexOy+ O2-> Fe2O3Cu + O2+ HCl
->CuCl2+ H2OFe3O4+ C ->Fe + CO220Fe2O3+ H2->Fe + H2O.
FexOy+ Al ->Fe + Al2O3Fe + Cl2-> FeCl3CO + O2->CO25. Hon
thnh cc phng trnh ha hc sau: FexOy+H2SO4 Fe 2(SO4) 2y / x+
H2OFexOy+H2 Fe+H2OAl(NO3)3 Al2O3+NO2 +O2KMnO4 + HCl Cl2+ KCl
+MnCl2+H2OFe 3O4 + Al Fe+ Al2O3FeS2 + O2---->Fe2O3 + SO2KOH +
Al2(SO4)3---->K2SO4 + Al(OH)3FeO + HNO3---->Fe(NO3)3 + NO +
H2OFexOy + CO---->FeO + CO26. Hon thnh chui bin ho sau:P2O5
H3PO4 H2KClO3O2Na2O NaOHH2O H2H2O KOH7:Hon thnh s chuyn ho sau (ghi
r iu kin phn ng) v cho bit cc phn ng trn thuc loi no?. KMnO47
KOHH2OO2 Fe3O4FeH2H2O8 H2SO4 KClO3B: Tnh theo ph ng trnh ha hc Cch
gii chung:- Vit v cn bngPTHH.- Tnh s mol ca cht bi cho.- Da vo
PTHH, tm s mol cc cht m bi yu cu.- Tnh ton theo yu cu ca bi (khi
lng, th tch cht kh)1.Dng ton c bn: Cho bit lng mt cht (c th cho bng
gam, mol, V(ktc) , cc i lng v nng dd, tan, t khi cht kh), tm lng cc
cht cn li trong mt phn ng ha hc.Cch gii :Bi ton c dng : a M + b B c
C + d D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)- Tnh
s mol ca cht bi cho.- Da vo PTHH, tm s mol cc cht m bi yu cu.- Tnh
ton theo yu cu ca bi * Tr ng hp 1: Cho dng trc tip bng : gam, mol.
V d1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi
lngkim loi dng. Gii: Ta c Phng trnh phn ng:Mg+ 2HCl
>MgCl2+H21mol 2mol x (mol) 0,6 (mol)x = 0,6. 1 / 2 = 0,3 (mol)
mMg = n.M = 0,3. 24 = 7,2 (g) *Tr ng hp 2: Cho dng gin tip bng :
V(ktc) V d2:Chokimloi Mgphnnghontonvi dungdchHCl. thuc6,72ltkh
(ktc) . Xc nh khi lngkim loi dng. Gii214 3 5 6Tm : nH2 = 6, 7222, 4
= 0,3 (mol) Ta c Phng trnh phn ng:Mg+ 2HCl >MgCl2+H21mol 1mol x
(mol)0,3 (mol)x = 0,3. 1 / 1 = 0,3 (mol) mMg = n.M = 0,3. 24 = 7,2
(g)*Tr ng hp 3: Cho dng gin tip bng : mdd, c% V d 3: Cho kim loi Mg
phn ng hon ton vi 100g dung dch HCl 21,9%. Xc nh khi lng kim loi
dng. Gii Ta phi tmn HCl phn ng ? p dng : C % = .100% mctmdd m HCl
=. %100%mdd c =100.21, 9100 = 21,9 (g) n HCl = mM =21, 936, 5 = 0,6
(mol)*Tr v bi ton 1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl.
Xc nh khi lng kim loi dng. (Gii nh v d 1)*Tr ng hp 4:Cho dng gin
tip bng : Vdd, C MV d 4 : Cho kim loi Mg phn ng hon ton vi 100 ml
dung dch HCl 6 M. Xc nh khilng kim loi dng. Gii: Tmn HCl= ? p dng :
CM = nV n HCl= CM.V = 6.0,1 = 0,6 (mol)*Tr v bi ton 1: Cho kim loi
Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi dng.(Gii nh
v d 1)*Tr ng hp 5: Cho dng gin tip bng : mdd, C M ,d (g/ml)V d 5 :
Cho kim loi Mg phn ng hon ton vi 120 g dung dch HCl 6 M ( d= 1,2
g/ml). Xc nh khi lng kim loi dng. Gii: Tmn HCl= ? - Tm Vdd (da vo
mdd, d (g/ml)):td= mV
Vdd H Cl = md = 1201, 2 = 100 (ml) =0,1(l) - Tmn HCl= ? p dng :
CM = nV n HCl= CM. V = 6. 0,1 = 0,6 (mol) *Tr v bi ton 1: Cho kim
loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng kim loi
dng.(Gii nh v d 1)*Tr ng hp 6:Cho dng gin tip bng : Vdd, C%, d
(g/ml) V d 6 : Cho kim loi Mg phn ng hon ton vi 83,3 ml dung dch
HCl 21,9 % ( d= 1,2 g/ml). Xc nh khi lng kim loi dng. Gii: Tmn HCl=
? - Tm m dd (da vo Vdd, d (g/ml)):td= mV mdd H Cl = V.d= 83,3 . 1,2
= 100 (g) dd HCl.22p dng : C % = .100% mctmdd m HCl =. %100%mdd c
=100.21, 9100 = 21,9 (g) n HCl = mM =21, 936, 5 = 0,6 (mol)*Tr v bi
ton 1: Cho kim loi Mg phn ng hon ton vi 0,6 mol HCl. Xc nh khi lng
kim loi dng.(Gii nh v d 1)Vn dng 6 dng ton trn:Ta c th thit lp c 9
bi ton tm cc i lng lin quan n nng dung dch( C%, CM., mdd, Vdd, khi
lng ring ca dd(d(g/ml)) ca cht phn ng).1. Cho 7,2 g kim loi Mg phn
ng hon ton vi 100g dung dch HCl . Xc nh nng % dd HClcn dng. 2. Cho
7,2 g kim loi Mg phn ng hon ton vidung dch HCl 21,9% . Xc nh khi
lng dd HClcn dng. 3: Cho 7,2 g kim loi Mg phn ng hon ton vi 100 ml
dung dch HCl .Xc nh nng Mol/ lt dd HCl cn dng. 4. Cho 7,2 g kim loi
Mg phn ng hon ton vi dung dch HCl 6M .Xc nh th tchdd HCl cn dng. 5.
Cho 7,2 g kim loi Mg phn ng hon ton vi dung dch HCl 6 M ( d = 1,2
g/ml). Xc nh khi lng dd HCl cn dng. 6.Cho 7,2 g kim loi Mg phn ng
hon ton vi 120g dung dch HCl ( d = 1,2 g/ml). Xc nh nng Mol/ltdd
HCl cn dng. 7.Cho 7,2 g kim loi Mg phn ng hon ton vi dung dch HCl
21,9%( d = 1,2 g/ml). Xc nh th tch dd HCl cn dng. 8. Cho 7,2 g kim
loi Mg phn ng hon ton vi 120 g dung dch HCl 6 M . Xc nh khi l-ng
ring dd HCl cn dng. 9.Cho 7,2 g kim loi Mg phn ng hon ton vi 83,3
ml dung dch HCl 21,9% . Xc nh khi lng ring dd HCl cn dng. 2.Dng ton
tha thiu:1. Tr ng hp ch c 2 cht phn ng: PTHH c dng: a M + b B c C +
d D(Trong cc cht M, B, C, D :c th l mt n cht hay 1 hp cht)* Cho bit
lng 2 cht trong phn ng (c th cho bng gam, mol, V(ktc) , cc i lng v
nng dd, tan, t khi cht kh), tm lng cc cht cn li trong mt phn ng ha
hc.Cch gii chung : - Vit v cn 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)V d: Khi t, than
chy theo s sau : Cacbon+ oxi kh cacbon ioxita) Vit v cn bng phng
trnh phn ng.b) Cho bit khi lng cacbon tc dng bng 18 kg, khi lng oxi
tc dng bng 24 kg. Hy tnh khi lng kh cacbon ioxit to thnh.c) Nu khi
lng cacbon tc dng bng 8 kg, khi lng kh cacbonic thu c bng 22 kg, hy
tnh khi lng cacbon cn d v khi lng oxi phn ng.Gii:23a. PTHH:C +
O2t0CO2b. S mol C: nC = 18.000 : 12 = 1500 mol.- S mol O2: nO2 =
24.000 : 32= 750 mol.Theo PTHH, ta c t s: 1nC= 11500 = 1500 > 12
nO= 1750 = 750.=> O2 p ht, C d.- Theo pthh: nCO2 = nO2 = 750
mol.- Vy khi lng CO2 to thnh: mCO2 = 750. 44 = 33.000gam = 33kg.c.
S mol CO2: nCO2 = 22.000 : 44 = 500 mol. - Theo PTHH: nC = nO2 =
nCO2 = 500 mol.- Khi lng C tham gia p: mC = 500. 12 = 6.000g =
6kg.=> Khi lng C cn d: 8 6 = 2kg.- Khi lng O2 tham gia p: mO2 =
500 . 32 = 16000g = 16kg.* Bi tp vn dng: 1: Cho 22,4g Fe tc dng vi
dd long c cha 24,5g axit sulfuric.a. Tnh s mol mi cht ban u v cho
bit cht d trong p?b. Tnh khi lng cht cn d sau p?c. Tnh th tch kh
hidro thu c ktc?d. Tnh khi lng mui thu c sau p2 : Cho dd cha 58,8g
H2SO4 tc dng vi 61,2g Al2O3.a. Tnh s mol mi cht ban u ca hai cht
p?b. Sau p cht no d, d bao nhiu gam?c. Tnh khi lng mui nhm sunfat
to thnh?(bitH2SO4 + Al2O3 Al2(SO4)3+ H2O )3: Dng 6,72 lt kh H2
(ktc) kh 20g St (III) oxit.a. Vit PTHH ca p?b. Tnh khi lng oxit st
t thu c?4: Cho 31g Natri oxit vo 27g nc.a. Tnh khi lng NaOH thu
c?b. Tnh nng % ca dd thu c sau p?5: Cho 4,05g kim loi Al vo dd
H2SO4, sa p thu c 3,36 lt kh ktc.a. Tnh khi lng Al p?b. Tnh khi lng
mui thu c v khi lng axit p?c. ha tan ht lng Al cn d cn phi dng them
bao nhiu gam axit?6 .Cho 2,8 gam st tc dng 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 phidng thm cht kia mt lng l bao
nhiu?2.Tr ng hp c nhiu cht phn ng: * Cho bit lng mt hn hp nhiu cht
phn ng vi mt lng cht phn ng khc (c th cho bng gam, mol, V(ktc) , cc
i lng v nng dd, tan, t khi cht kh), tm l-ng cc cht cn li trong qu
trnh phn ng ha hc.Bi ton c dng : cho hn hp A( gm M, M) phn ng vi B
chng minh hh A ht hay B ht:Cch gii chung : - Vit v cn bngPTHH:PTHH
c dng: a M +b Bc C +d Da M + bB c C + dD(Trong cc cht M, M, B, C,
D, C, D: c th l mt n cht hay 1 hp cht)- Tnh s mol cahn hp v s mol
cc cht trong qu trnh phn ng . Bin lun lng hn hp hay lng cht phn ng
vi hh theo cc d kin ca bi tonlin quan n lng hh hay cht phn ng , xc
nh lng hh ht hay cht phn ng vi hh ht - Da vo PTHH, tm lng cc cht cn
litheo lng cht p ht.V d: Cho 3,78 gam hon hp gom Mg va Al tac dung
vi 0,5 mol HCl24a. Chng minh rang sau phan ng vi Mg va Al , axit
van con d ?b. Neu phan ng tren lam thoat ra 4,368 lt kh H2 (ktc) .
Hay tnh so gam Mg va Al a dung ban au?Gii: a. Ta c PTHH:2Al + 6 HCl
2 AlCl3 + 3 H2 (1)x (mol)3x 3.2xMg + 2 HCl MgCl2+H2(2) y (mol)2y
yGi s lng hn hp ht :- Theo bi ra : 27x + 24y = 3,78 > 24 (x+y)
3, 7824 = 0,16> x +y(3)- Theo PT (1) (2) n HCl = 3x + 2y
