32 chuyen de boi duong hsg hoa 89.doc
TRANSCRIPT
Tng hp kin thc c bn ho hc 8
Tng hp kin thc c bn ho hc 8
Cc khi nim:
1. Vt th, cht.
Vt th: L ton b nhng g xung quanh chng ta v trong khng gian. Vt th gm 2 loi: Vt th t nhin v vt th nhn to
Cht: l nguyn liu cu to nn vt th. Cht c khp mi ni, u c vt th l c cht.
Mi cht c nhng tnh cht nht nh. Bao gm tnh cht vt l v tnh cht ho hc.
Tnh cht vt l: Trng thi (R,L,K), mu sc, mi v, tnh tan, tnh dn in, dn nhit, nhit si (t0s), nhit nng chy (t0nc), khi lng ring (d) Tnh cht ho hc: L kh nng b bin i thnh cht khc: Kh nng chy, n, tc dng vi cht khc2. Hn hp v cht tinh khit.
Hn hp l 2 hay nhiu cht trn li vi nhau. Mi cht trong hn hp c gi l 1 cht thnh phn.
Hn hp gm c 2 loi: hn hp ng nht v hn hp khng ng nht
Tnh cht ca hn hp: Hn hp c tnh cht khng n nh, thay i ph thuc vo khi lng v s lng cht thnh phn.
Cht tinh khit l cht khng c ln cht no khc. Cht tinh khit c tnh cht nht nh, khng thay i.
Khi tch ring cc cht ra khi hn hp ta thu c cc cht tinh khit. tch ring cc cht ra khi hn hp ngi ta c th s dng cc phng php vt l v ho hc: tch, chit, gn, lc, cho bay hi, chng ct, dng cc phn ng ho hc3. Nguyn t.
a. nh ngha: L ht v cng nh, trung ho v in, cu to nn cc cht
b. Cu to: gm 2 phn
Ht nhn: to bi 2 loi ht: Proton v Ntron
Proton: Mang in tch +1, c khi lng 1 vC, k hiu: P
Ntron: Khng mang in, c khi lng 1 vC, k hiu: N
V: cu to t cc lp Electron
Electron: Mang in tch -1, c khi lng khng ng k, k hiu: e
Trong nguyn t, cc e chuyn ng rt nhanh v sp xp thnh tng lp t trong ra.
+ Lp 1: c ti a 2e
+ Lp 2,3,4 tm thi c ti a 8eKhi lng nguyn t = s P + s N + s e = s P + s N (v e c khi lng rt nh)
4. Nguyn t ho hc.
L tp hp nhng nguyn t cng loi, c cng s P trong ht nhn
Nhng nguyn t c cng s P nhng s N khc nhau gi l ng v ca nhau
5. Ho tr.
L con s biu th kh nng lin kt ca nguyn t hay nhm nguyn t
Quy tc ho tr:
ta c: a.x = b.y
(vi a, b ln lt l ho tr ca nguyn t A v B)
So snh n cht v hp cht
n chthp cht
VDSt, ng, oxi, nit, than chNc, mui n, ng
K/NL nhng cht do 1 nguyn t ho hc cu to nnL nhng cht do 2 hay nhiu nguyn t ho hc cu to nn
Phn loiGm 2 loi: Kim loi v phi kim.Gm 2 loi: hp cht v c v hp cht hu c
Phn t
(ht i din)- Gm 1 nguyn t: kim loi v phi kim rn
- Gm cc nguyn t cng loi: Phi kim lng v kh- Gm cc nguyn t khc loi thuc cc nguyn t ho hc khc nhau
CTHH- Kim loi v phi kim rn:
CTHH ( KHHH (A)
- Phi kim lng v kh:
CTHH = KHHH + ch s (Ax)CTHH = KHHH ca cc nguyn t + cc ch s tng ng
AxBy
So snh nguyn t v phn t
nguyn tphn t
nh nghaL ht v cng nh, trung ho v in, cu to nn cc chtL ht v cng nh, i din cho cht v mang y tnh cht ca cht
S bin i trong phn ng ho hc.Nguyn t c bo ton trong cc phn ng ho hc.Lin kt gia cc nguyn t trong phn t thay i lm cho phn t ny bin i thnh phn t khc
Khi lngNguyn t khi (NTK) cho bit nng nh khc nhau gia cc nguyn t v l i lng c trng cho mi nguyn t
NTK l khi lng ca nguyn t tnh bng n v CacbonPhn t khi (PTK) l khi lng ca 1 phn t tnh bng n v Cacbon
PTK = tng khi lng cc nguyn t c trong phn t.
p dng quy tc ho tr
1. Tnh ho tr ca 1 nguyn t
Gi ho tr ca nguyn t cn tm (l a)
p dng QTHT: a.x = b.y ( a = b.y/x
Tr li
2. Lp CTHH ca hp cht.
Gi cng thc chung cn lp
p dng QTHT: a.x = b.y (
Tr li.
*** C th dng quy tc cho lp nhanh 1 CTHH: Trong CTHH, ho tr ca nguyn t ny l ch s ca nguyn t kia.
Lu : Khi cc ho tr cha ti gin th cn ti gin trc
6. Phn ng ho hc.
L qu trnh bin i cht ny thnh cht khc.
Cht b bin i gi l cht tham gia, cht c to thnh gi l sn phm
c biu din bng s :
A + B ( C + D c l: A tc dng vi B to thnh C v D
A + B ( C c l A kt hp vi B to thnh C
A ( C + D c l A b phn hu thnh C v D
Ngoi ra c th chia axit thnh axit mnh v axit yu
Axit mnhAxit trung bnh
Axit yu
Axit rt yu
Oxitaxitbazmui
nh nghaL hp cht ca oxi vi 1 nguyn t khcL hp cht m phn t gm 1 hay nhiu nguyn t H lin kt vi gc axitL hp cht m phn t gm 1 nguyn t kim loi lin kt vi 1 hay nhiu nhm OHL hp cht m phn t gm kim loi lin kt vi gc axit.
CTHHGi nguyn t trong oxit l A ho tr n. CTHH l:
- A2On nu n l
- AOn/2 nu n chnGi gc axit l B c ho tr n.
CTHH l: HnBGi kim loi l M c ho tr n
CTHH l: M(OH)nGi kim loi l M, gc axit l B
CTHH l: MxBy
Tn giTn oxit = Tn nguyn t + oxit
Lu : Km theo ho tr ca kim loi khi kim loi c nhiu ho tr.
Khi phi kim c nhiu ho tr th km tip u ng.- Axit khng c oxi: Axit + tn phi kim + hidric
- Axit c t oxi: Axit + tn phi kim + (r)
- Axit c nhiu oxi: Axit + tn phi kim + ic (ric)Tn baz = Tn kim loi + hidroxit
Lu : Km theo ho tr ca kim loi khi kim loi c nhiu ho tr.Tn mui = tn kim loi + tn gc axit
Lu : Km theo ho tr ca kim loi khi kim loi c nhiu ho tr.
TCHH1. Tc dng vi nc
- Oxit axit tc dng vi nc to thnh dd Axit
- Oxit baz tc dng vi nc to thnh dd Baz
2. Oxax + dd Baz to thnh mui v nc
3. Oxbz + dd Axit to thnh mui v nc
4. Oxax + Oxbz to thnh mui1. Lm qu tm ( hng
2. Tc dng vi Baz ( Mui v nc
3. Tc dng vi oxit baz ( mui v nc
4. Tc dng vi kim loi ( mui v Hidro
5. Tc dng vi mui ( mui mi v axit mi1. Tc dng vi axit ( mui v nc
2. dd Kim lm i mu cht ch th
- Lm qu tm ( xanh
- Lm dd phenolphtalein khng mu ( hng
3. dd Kim tc dng vi oxax ( mui v nc
4. dd Kim + dd mui ( Mui + Baz
5. Baz khng tan b nhit phn ( oxit + nc1. Tc dng vi axit ( mui mi + axit mi
2. dd mui + dd Kim ( mui mi + baz mi
3. dd mui + Kim loi ( Mui mi + kim loi mi
4. dd mui + dd mui ( 2 mui mi
5. Mt s mui b nhit phn
Lu - Oxit lng tnh c th tc dng vi c dd axit v dd - HNO3, H2SO4 c c cc tnh cht ring- Baz lng tnh c th tc dng vi c dd axit v - Mui axit c th phn ng nh 1 axit
Tnh cht ho hc ca cc hp cht v c
Mi quan h gia cc loi hp cht v c
Cc phng trnh ho hc minh ho thng gp
4Al + 3O2 ( 2Al2O3CuO + H2 Cu + H2O
Fe2O3 + 3CO 2Fe + 3CO2S + O2 ( SO2CaO + H2O ( Ca(OH)2Cu(OH)2 CuO + H2O
CaO + 2HCl ( CaCl2 + H2O
CaO + CO2 ( CaCO3Na2CO3 + Ca(OH)2 ( CaCO3( + 2NaOH
NaOH + HCl ( NaCl + H2O
2NaOH + CO2 ( Na2CO3 + H2O
BaCl2 + Na2SO4 ( BaSO4( + 2NaCl
SO3 + H2O ( H2SO4P2O5 + 3H2O ( 2H3PO4P2O5 + 6NaOH ( 2Na3PO4 + 3H2O
N2O5 + Na2O ( 2NaNO3BaCl2 + H2SO4 ( BaSO4( + 2HCl
2HCl + Fe ( FeCl2 + H22HCl + Ba(OH)2 ( BaCl2 + 2H2O
6HCl + Fe2O3 ( 2FeCl3 + 3H2O
2HCl + CaCO3 ( CaCl2 + 2H2O
iu ch cc hp cht v c
`
Tnh cht ho hc ca kim loi
Dy hot ng ho hc ca kim loi.
K, Na, Mg, Al, Zn, Fe, Pb, (H), Cu, Ag, Au
(Khi No May A Zp St Phi Hi Cc Bc Vng) ngha:
KBaCaNaMgAlZnFeNiSnPbHCuAgHgAuPt
+ O2: nhit thng
nhit cao
Kh phn ng
KBaCaNaMgAlZnFeNiSnPbHCuAgHgAuPt
Tc dng vi nc
Khng tc dng vi nc nhit thng
KBaCaNaMgAlZnFeNiSnPbHCuAgHgAuPt
Tc dng vi cc axit thng thng gii phng HidroKhng tc dng.
KBaCaNaMgAlZnFeNiSnPbHCuAgHgAuPt
Kim loi ng trc y kim loi ng sau ra khi mui
KBaCaNaMgAlZnFeNiSnPbHCuAgHgAuPt
H2, CO khng kh c oxit
kh c oxit cc kim loi ny nhit cao
Ch :
Cc kim loi ng trc Mg phn ng vi nc nhit thng to thnh dd Kim v gii phng kh Hidro.
Tr Au v Pt, cc kim loi khc u c th tc dng vi HNO3 v H2SO4 c nhng khng gii phng Hidro.
So snh tnh cht ho hc ca nhm v st
* Ging:
- u c cc tnh cht chung ca kim loi.
- u khng tc dng vi HNO3 v H2SO4 c ngui
* Khc:Tnh chtAl (NTK = 27)Fe (NTK = 56)
Tnh cht
vt l- Kim loi mu trng, c nh kim, nh, dn in nhit tt.
- t0nc = 6600C
- L kim loi nh, d dt mng, do.- Kim loi mu trng xm, c nh kim, dn in nhit km hn Nhm.
- t0nc = 15390C
- L kim loi nng, do nn d rn.
Tc dng vi
phi kim2Al + 3Cl2 2AlCl32Al + 3S Al2S32Fe + 3Cl2 2FeCl3Fe + S FeS
Tc dng vi
axit2Al + 6HCl ( 2AlCl3 + 3H2Fe + 2HCl ( FeCl2 + H2
Tc dng vi
dd mui2Al + 3FeSO4 ( Al2(SO4)3 + 3FeFe + 2AgNO3 ( Fe(NO3)2 + 2Ag
Tc dng vi
dd Kim2Al + 2NaOH + H2O
( 2NaAlO2 + 3H2Khng phn ng
Hp cht- Al2O3 c tnh lng tnh
Al2O3 + 6HCl ( 2AlCl3 + 3H2O
Al2O3+ 2NaOH(2NaAlO2 + H2O
- Al(OH)3 kt ta dng keo, l hp cht lng tnh
- FeO, Fe2O3 v Fe3O4 u l cc oxit baz
Fe(OH)2 mu trng xanh
Fe(OH)3 mu nu
Kt lun- Nhm l kim loi lng tnh, c th tc dng vi c dd Axit v dd Kim. Trong cc phn ng ho hc, Nhm th hin ho tr III- St th hin 2 ho tr: II, III
+ Tc dng vi axit thng thng, vi phi kim yu, vi dd mui: II
+ Tc dng vi H2SO4 c nng, dd HNO3, vi phi kim mnh: III
Gang v thp
GangThp
/N- Gang l hp kim ca St vi Cacbon v 1 s nguyn t khc nh Mn, Si, S (%C=2(5%)- Thp l hp kim ca St vi Cacbon v 1 s nguyn t khc (%C 2Al2O3 (r)Phn ng khng c s thay i s oxi ho.
BaO (r) + H2O (l) ----> Ba(OH)2 (dd)
2/ Phn ng phn hu.
- c im ca phn ng: C th xy ra s thay i s oxi ho hoc khng.
V d:
Phn ng c s thay i s oxi ho.
2KClO3 (r) -------> 2KCl (r) + 3O2 (k)
Phn ng khng c s thay i s oxi ho.
CaCO3 (r) -----> CaO (r) + CO2 (k)
II/ Phn ng c s thay i s oxi ho.
1/ Phn ng th.
c im ca phn ng: Nguyn t ca n cht thay th mt hay nhiu nguyn t ca mt nguyn t trong hp cht.
V d:
Zn (r) + 2HCl (dd) ----> ZnCl2 (dd) + H2 (k)
2/ Phn ng oxi ho - kh.
c im ca phn ng: Xy ra ng thi s oxi ho v s kh. hay xy ra ng thi s nhng electron v s nhn electron.
V d:
CuO (r) + H2 (k) ------> Cu (r) + H2O (h)
Trong :
H2 l cht kh (Cht nhng e cho cht khc)
CuO l cht oxi ho (Cht nhn e ca cht khc)
T H2 -----> H2O c gi l s oxi ho. (S chim oxi ca cht khc)
T CuO ----> Cu c gi l s kh. (S nhng oxi cho cht khc)
III/ Phn ng khng c thay i s oxi ho.
1/ Phn ng gia axit v baz.
c im ca phn ng: Sn phm thu c l mui v nc.
V d:
2NaOH (dd) + H2SO4 (dd) ----> Na2SO4 (dd) + 2H2O (l)
NaOH (dd) + H2SO4 (dd) ----> NaHSO4 (dd) + H2O (l)
Cu(OH)2 (r) + 2HCl (dd) ----> CuCl2 (dd) + 2H2O (l)
Trong :Phn ng trung ho (2 cht tham gia trng thi dung dch).
c im ca phn ng: l s tc dng gia axit v baz vi lng va .
Sn phm ca phn ng l mui trung ho v nc.
V d:
NaOH (dd) + HCl (dd) ----> NaCl (dd) + H2O (l)
2/ Phn ng ga axit v mui.
c im ca phn ng: Sn phm thu c phi c t nht mt cht khng tan hoc mt cht kh hoc mt cht in li yu.
V d:
Na2CO3 (r) + 2HCl (dd) ----> 2NaCl (dd) + H2O (l) + CO2 (k)
BaCl2 (dd) + H2SO4 (dd) -----> BaSO4 (r) + 2HCl (dd)
Lu : BaSO4 l cht khng tan k c trong mi trng axit.
3/ Phn ng gia baz v mui.
c im ca phn ng:
+ Cht tham gia phi trng thi dung dch (tan c trong nc)
+ Cht to thnh (Sn phm thu c) phi c t nht mt cht khng tan hoc mt cht kh hoc mt cht in li yu.
+ Ch cc mui kim loi m oxit hay hiroxit c tnh cht lng tnh phn ng vi dung dch baz mnh.
V d:
2NaOH (dd) + CuCl2 (dd) ----> 2NaCl (dd) + Cu(OH)2 (r)
Ba(OH)2 (dd) + Na2SO4 (dd) ---> BaSO4 (r) + 2NaOH (dd)
NH4Cl (dd) + NaOH (dd) ---> NaCl (dd) + NH3 (k) + H2O (l)
AlCl3 (dd) + 3NaOH (dd) ----> 3NaCl (dd) + Al(OH)3 (r)
Al(OH)3 (r) + NaOH (dd) ---> NaAlO2 (dd) + H2O (l)
4/ Phn ng gia 2 mui vi nhau.
c im ca phn ng:
+ Cht tham gia phi trng thi dung dch (tan c trong nc)
+ Cht to thnh (Sn phm thu c) phi c t nht mt cht khng tan hoc mt cht kh hoc mt cht in li yu.
V d:
NaCl (dd) + AgNO3 (dd) ----> AgCl (r) + NaNO3 (dd)
BaCl2 (dd) + Na2SO4 (dd) ----> BaSO4 (r) + 2NaCl (dd)
2FeCl3 (dd) + 3H2O (l) + 3Na2CO3 (dd) ----> 2Fe(OH)3 (r) + 3CO2 (k) + 6NaCl (dd)
gii thiu 1 s phng php
cn bng phng trnh ho hc.1/ Cn bng phng trnh theo phng php i s.
V d: Cn bng phng trnh phn ng
P2O5 + H2O -> H3PO4a cc h s x, y, z vo phng trnh ta c:
- Cn c vo s nguyn t P ta c: 2x = z
(1)
- Cn c vo s nguyn t O ta c: 5x + y = z (2)
- Cn c vo s nguyn t H ta c: 2y = 3z (3)
Thay (1) vo (3) ta c: 2y = 3z = 6x => y = = 3x
Nu x = 1 th y = 3 v z = 2x = 2.1 = 2
=> Phng trnh dng cn bng nh sau: P2O5 + 3H2O -> 2H3PO4
V d: Cn bng phng trnh phn ng.Al + HNO3 (long) ----> Al(NO3)3 + NO + H2O
Bc 1: t h s bng cc n s a, b, c, d trc cc cht tham gia v cht to thnh (Nu 2 cht m trng nhau th dng 1 n)
Ta c.
a Al + b HNO3 ----> a Al(NO3)3 + c NO + b/2 H2O.
Bc 2: Lp phng trnh ton hc vi tng loi nguyn t c s thay i v s nguyn t 2 v.
Ta nhn thy ch c N v O l c s thay i.
N: b = 3a + c (I)
O: 3b = 9a + c + b/2 (II)
Bc 3: Gii phng trnh ton hc tm h s
Thay (I) vo (II) ta c.
3(3a + c) = 9a + c + b/2
2c = b/2 ----> b = 4c ---> b = 4 v c = 1. Thay vo (I) ---> a = 1.
Bc 4: Thay h s va tm c vo phng trnh v hon thnh phng trnh.
Al + 4 HNO3 ----> Al(NO3)3 + NO + 2 H2O
Bc 5: Kim tra li phng trnh va hon thnh.
2/ Cn bng theo phng php electron.
V d:
Cu + HNO3 (c) -----> Cu(NO3)2 + NO2 + H2O
Bc 1: Vit PTP xc nh s thay i s oxi ho ca nguyn t.
Ban u: Cu0 ----> Cu+ 2 Trong cht sau phn ng Cu(NO3)2 Ban u: N+ 5 (HNO3) ----> N+ 4 Trong cht sau phn ng NO2 Bc 2: Xc nh s oxi ho ca cc nguyn t thay i.
Cu0 ----> Cu+ 2N+ 5 ----> N+ 4Bc 3: Vit cc qu trnh oxi ho v qu trnh kh.
Cu0 2e ----> Cu+ 2 N+ 5 + 1e ----> N+ 4Bc 4: Tm bi chung cn bng s oxi ho.
1 Cu0 2e ----> Cu+ 22 N+ 5 + 1e ----> N+ 4Bc 5: a h s vo phng trnh, kim tra, cn bng phn khng oxi ho - kh v hon thnh PTHH.
Cu + 2HNO3 (c) -----> Cu(NO3)2 + 2NO2 + H2O
+ 2HNO3 (c) ----->
Cu + 4HNO3 (c) -----> Cu(NO3)2 + 2NO2 + 2H2O
3/ Cn bng theo phng php bn phn ng ( Hay ion - electron)
Theo phng php ny th cc bc 1 v 2 ging nh phng php electron.
Bc 3: Vit cc bn phn ng oxi ho v bn phn ng kh theo nguyn tc:
+ Cc dng oxi ho v dng kh ca cc cht oxi ho, cht kh nu thuc cht in li mnh th vit di dng ion. Cn cht in li yu, khng in li, cht rn, cht kh th vit di dng phn t (hoc nguyn t). i vi bn phn ng oxi ho th vit s e nhn bn tri cn bn phn ng th vit s e cho bn phi.
Bc 4: Cn bng s e cho nhn v cng hai bn phn ng ta c phng trnh phn ng dng ion.
Mun chuyn phng trnh phn ng dng ion thnh dng phn t ta cng 2 v nhng lng tng ng nh nhau ion tri du (Cation v anion) b tr in tch.
Ch : cn bng khi lng ca na phn ng.
Mi trng axit hoc trung tnh th ly oxi trong H2O.
Bc 5: Hon thnh phng trnh.
Mt s phn ng ho hc thng gp.
Cn nm vng iu kin xy ra phn ng trao i trong dung dch.
Gm cc phn ng:
1/ Axit + Baz Mui + H2O
2/ Axit + Mui Mui mi + Axt mi
3/ Dung dch Mui + Dung dch Baz Mui mi + Baz mi
4/ 2 Dung dch Mui tc dng vi nhau 2 Mui mi
iu kin xy ra phn ng trao i l: Sn phm thu c phi c t nht mt cht khng tan hoc mt cht kh hoc phi c H2O v cc cht tham gia phi theo yu cu ca tng phn ng.
Tnh tan ca mt s mui v baz.
Hu ht cc mui clo rua u tan ( tr mui AgCl , PbCl2 )
Tt c cc mui nit rat u tan.
Tt c cc mui ca kim loi kim u tan.
Hu ht cc baz khng tan ( tr cc baz ca kim loi kim, Ba(OH)2 v Ca(OH)2 tan t.* Na2CO3 , NaHCO3 ( K2CO3 , KHCO3 ) v cc mui cacbonat ca Ca, Mg, Ba u tc dng c vi a xt.
NaHCO3 + NaHSO4 Na2SO4 + H2O + CO2
Na2CO3 + NaHSO4 Khng xy ra
NaHCO3 + NaOH Na2CO3 + H2O
Na2CO3 + NaOH Khng xy ra
2NaHCO3 Na2CO3 + H2O + CO2NaHCO3 + Ba(OH)2 BaCO3 + NaOH + H2O 2NaHCO3 + 2KOH Na2CO3 + K2CO3 + 2H2O
Na2CO3 + Ba(OH)2 BaCO3 + 2NaOH
Ba(HCO3)2 + Ba(OH)2 2BaCO3 + 2H2O
Ca(HCO3)2 + Ba(OH)2 BaCO3 + CaCO3 + 2H2O
NaHCO3 + BaCl2 khng xy ra
Na2CO3 + BaCl2 BaCO3 + 2NaClBa(HCO3)2 + BaCl2 khng xy ra
Ca(HCO3)2 + CaCl2 khng xy ra NaHSO3 + NaHSO4 Na2SO4 + H2O + SO2
Na2SO3 + H2SO4 Na2SO4 + H2O + SO22NaHSO3 + H2SO4 Na2SO4 + 2H2O + 2SO2
Na2SO3 + 2NaHSO4 2Na2SO4 + H2O + SO2
2KOH + 2NaHSO4 Na2SO4 + K2SO4 + H2O
(NH4)2CO3 + 2NaHSO4 Na2SO4 + (NH4)2SO4 + H2O + CO2Fe + CuSO4 FeSO4 + Cu
Cu + Fe SO4 khng xy ra Cu + Fe2(SO4)3 2FeSO4 + CuSO4
Fe + Fe2(SO4)3 3FeSO4
2FeCl2 + Cl2 2FeCl3
Bng tnh tan trong nc ca cc axit baz - muiNhm hiroxit v gc axitHiro v cc kim loi
H
IK
INa
IAg
IMg
IICa
IIBa
IIZn
IIHg
IIPb
IICu
IIFe
IIFe
IIIAl
III
- OHtt-kitk-kkkkk
- Clt/bttktttttitttt
- NO3t/bttttttttttttt
- CH3COOt/bttttttttttt-t
= St/bttk-ttkkkkkk
= SO3t/bttkkkkkkkkk-
= SO4t/kbttitikt-ktttt
= CO3t/bttkkkkk-k-k-
= SiO3k/kbttkkkkkkkk
= PO4t/kbttkkkkkkkkkkk
t : hp cht khng tan c trong nc .
k: hp cht khng tan
i: hp cht t tan.
b: hp cht bay hi hoc d bi phn hu thnh kh bay ln.
kb : hp cht khng bay hi.
Vch ngang - " :hp cht khng tn ti hoc b phn hu trong nc.
Mt s PTHH cn lu :
V d: Ho tan m( gam ) MxOy vo dung dch axit (HCl, H2SO4, HNO3)
Ta c PTHH cn bng nh sau: lu 2y/x l ho tr ca kim loi M
MxOy + 2yHCl xMCl2y/x + yH2O
2MxOy + 2yH2SO4 xM2(SO4)2y/x + 2yH2O
MxOy + 2yHNO3 xM(NO3)2y/x + yH2O
VD: Ho tan m( gam ) kim loi M vo dung dch a xit (HCl, H2SO4)
Ta c PTHH cn bng nh sau: lu x l ho tr ca kim loi M2M + 2xHCl 2MClx + xH2
p dng:
Fe + 2HCl FeCl2 + H2
2Al + 2*3 HCl 2AlCl3 + 3H2
6
2M + xH2SO4 M2(SO4)x + xH2p dng:
Fe + H2SO4 FeSO4 + H22Al + 3H2SO4 Al2(SO4)3 + 3H2Cc phn ng iu ch mt s kim loi:
i vi mt s kim loi nh Na, K, Ca, Mg th dng phng php in phn nng chy cc mui Clorua.
PTHH chung: 2MClx (r ) 2M(r ) + Cl2( k )(i vi cc kim loi ho tr II th nh n gin phn h s)
i vi nhm th dng phng php in phn nng chy Al2O3, khi c cht xc tc Criolit(3NaF.AlF3) , PTHH: 2Al2O3 (r ) 4Al ( r ) + 3 O2 (k )
i vi cc kim loi nh Fe , Pb , Cu th c th dng cc phng php sau:
- Dng H2: FexOy + yH2 xFe + yH2O ( h )- Dng C: 2FexOy + yC(r ) 2xFe + yCO2 ( k )
- Dng CO: FexOy + yCO (k ) xFe + yCO2 ( k ) - Dng Al( nhit nhm ): 3FexOy + 2yAl (r ) 3xFe + yAl2O3 ( k )- PTP nhit phn st hir xit:
4xFe(OH)2y/x + (3x 2y) O2 2xFe2O3 + 4y H2O
Mt s phn ng nhit phn ca mt s mui
1/ Mui nitrat
Nu M l kim loi ng trc Mg (Theo dy hot ng ho hc)
2M(NO3)x 2M(NO2)x + xO2(Vi nhng kim loi ho tr II th nh n gin phn h s )
Nu M l kim loi k t Mg n Cu (Theo dy hot ng ho hc)
4M(NO3)x 2M2Ox + 4xNO2 + xO2
(Vi nhng kim loi ho tr II th nh n gin phn h s )
Nu M l kim loi ng sau Cu (Theo dy hot ng ho hc)
2M(NO3)x 2M + 2NO2 + xO2
(Vi nhng kim loi ho tr II th nh n gin phn h s)
2/ Mui cacbonat
- Mui trung ho: M2(CO3)x (r) M2Ox (r) + xCO2(k)(Vi nhng kim loi ho tr II th nh n gin phn h s)
- Mui cacbonat axit: 2M(HCO3)x(r) M2(CO3)x(r) + xH2O( h ) + xCO2(k)(Vi nhng kim loi ho tr II th nh n gin phn h s)
3/ Mui amoni
NH4Cl NH3 (k) + HCl ( k )NH4HCO3 NH3 (k) + H2O ( h ) + CO2(k)NH4NO3 N2O (k) + H2O ( h )NH4NO2 N2 (k) + 2H2O ( h )(NH4)2CO3 2NH3 (k) + H2O ( h ) + CO2(k)2(NH4)2SO4 4NH3 (k) + 2H2O ( h ) + 2SO2 ( k ) + O2(k)
Bi 1: Vit cc phng trnh ho hc biu din cc phn ng ho hc cc th nghim sau:
a) Nh vi git axit clohiric vo vi.
b) Ho tan canxi oxit vo nc.
c) Cho mt t bt iphotpho pentaoxit vo dung dch kali hirxit.
d) Nhng mt thanh st vo dung dch ng(II) sunfat.
e) Cho mt mu nhm vo dung dch axit sunfuric long.
f) Nung mt t st(III) hirxit trong ng nghim.
g) Dn kh cacbonic vo dung dch nc vi trong n d.
h) Cho mt t natri kim loi vo nc.
Bi 2: C nhng baz sau: Fe(OH)3, Ca(OH)2, KOH, Mg(OH)2. Hy cho bit nhng baz no:
a) B nhit phn hu?
b) Tc dng c vi dung dch H2SO4?
c) i mu dung dch phenolphtalein t khng mu thnh mu hng?
Bi 3: Cho cc cht sau: canxi oxit, kh sunfur, axit clohiric, bari hirxit, magi cacbonat, bari clorua, iphotpho penta oxit. Cht no tc dng c vi nhau tng i mt. Hy vit cc phng trnh ho hc ca phn ng.
Hng dn: Lp bng thy c cc cp cht tc dng c vi nhau r hn.
Bi 4: Cho cc oxit sau: K2O, SO2, BaO, Fe3O4, N2O5. Vit phng trnh ho hc(nu c) ca cc oxit ny ln lt tc dng vi nc, axit sunfuric, dung dch kali hiroxit.
Bi 5: Cho mt lng kh CO d i vo ng thu tinh t nng c cha hn hp bt gm: CuO, K2O, Fe2O3 (u ng thu tinh cn li b hn kn). Vit tt c cc phng trnh ho hc xy ra.
Bi 6: Nu hin tng v vit PTHH minh ho
a/ Cho Na vo dung dch Al2(SO4)3
b/ Cho K vo dung dch FeSO4
c/ Ho tan Fe3O4 vo dung dch H2SO4 long.
d/ Nung nng Al vi Fe2O3 to ra hn hp Al2O3 v FexOy.
PTHH tng qut:
3x Fe2O3 + ( 6x 4y ) Al 6 FexOy + ( 3x 2y ) Al2O3Bi 7: Cho th nghim
MnO2 + HCl Kh A
Na2SO3 + H2SO4 ( l ) Kh BFeS + HCl Kh CNH4HCO3 + NaOHd Kh DNa2CO3 + H2SO4 ( l ) Kh Ea. Hon thnh cc PTHH v xc nh cc kh A, B, C, D, E.
b. Cho A tc dng C, B tc dng vi dung dch A, B tc dung vi C, A tc dung dch NaOH iu kin thng, E tc dng dung dch NaOH. Vit cc PTHH xy ra.
Bi 8: Nu hin tng xy ra, gii thch v vit PTHH minh ho khi:
1/ Sc t t n d CO2 vo dung dch nc vi trong; dung dch NaAlO2.
2/ Cho t t dung dch axit HCl vo dung dch Na2CO3.
3/ Cho Na vo dung dch MgCl2, NH4Cl.
4/ Cho Na vo dung dch CuSO4, Cu(NO3)2.
5/ Cho Ba vo dung dch Na2CO3, (NH4)2CO3, Na2SO4.
6/ Cho Fe vo dung dch AgNO3 d
7/ Cho t t n d dung dch NaOH vo dung dch AlCl3, Al2(SO4)3.
8/ Cho Cu ( hoc Fe ) vo dung dch FeCl3.
9/ Cho t t n d bt Fe vo hn hp dung dch gm AgNO3 v Cu(NO3)2.
10/ Sc t t NH3 vo dung dch AlCl3Mt s phng php
gii ton ho hc thng dng.
1. Phng php s hc
Gii cc php tnh Ho hc cp II ph thng, thng thng s dng phng php s hc: l cc php tnh da vo s ph thuc t l gia cc i lng v cc php tnh phn trm. C s ca cc tnh ton Ho hc l nh lut thnh phn khng i c p dng cho cc php tnh theo CTHH v nh lut bo ton khi lng cc cht p dng cho c php tnh theo PTHH. Trong phng php s hc ngi ta phn bit mt s phng php tnh sau y:
a. Phng php t l.
im ch yu ca phng php ny l lp c t l thc v sau l p dng cch tnh ton theo tnh cht ca t l thc tc l tnh cc trung t bng tch cc ngoi t.
Th d: Tnh khi lng ccbon ixit CO2 trong c 3 g cacbon.
Bi gii
1mol CO2 = 44g
Lp t l thc: 44g CO2
c 12g C
xg
3g C
44 : x = 12 : 3
=> x =
Vy, khi lng cacbon ixit l 11g
Th d 2: C bao nhiu gam ng iu ch c khi cho tng tc 16g ng sunfat vi mt lng st cn thit.
Bi gii
Phng trnh Ho hc: CuSO4 + Fe - > FeSO4 + Cu
160g
64g
16g
xg
=> x =
Vy iu ch c 6,4g ng.
b. Phng php tnh theo t s hp thc.
Dng c bn ca php tnh ny tnh theo PTHH tc l tm khi lng ca mt trong nhng cht tham gia hoc to thnh phn ng theo khi lng ca mt trong nhng cht khc nhau. Phng php tm t s hp thc gia khi lng cc cht trong phn ng c pht biu nh sau:
T s khi lng cc cht trong mi phn ng Ho hc th bng t s ca tch cc khi lng mol cc cht vi cc h s trong phng trnh phn ng. C th biu th di dng ton hc nh sau:
Trong : m1 v m2 l khi lng cc cht, M1, M2 l khi lng mol cc cht cn n1, n2 l h s ca PTHH.
Vy khi tnh khi lng ca mt cht tham gia phn ng Ho hc theo khi lng ca mt cht khc cn s dng nhng t s hp thc tm c theo PTHH nh th no ? minh ho ta xt mt s th d sau:
Th d 1: Cn bao nhiu gam Ptat n da cho phn ng vi 10g st III clorua ?
Bi gii
PTHH FeCL3 + 3KOH -> Fe(OH)3 + 3KCL
10g ?
Tnh t s hp thc gia khi lng Kali hirxit v st II clorua
MKOH = (39 + 16 + 1) = 56g
* Tm khi lng KOH: m
Th d 2: Cn bao nhiu gam st III chorua cho tng tc vi kalihirxit thu c 2,5g Kaliclorua?
Bi gii
PTHH FeCl3 + 3 KOH - > Fe(OH)3 + 3KCl
Tnh t s hp thc gia khi lng FeCl3 v Kaliclorua
; MKCL 74,5g
* Tnh khi lng FeCl3:
c. Phng php tnh theo tha s hp thc.
Hng s c tnh ra t t l hp thc gi l tha s hp thc v biu th bng ch ci f. Tha s hp thc c tnh sn v c trong bng tra cu chuyn mn.
Vic tnh theo tha s hp thc cng cho cng kt qu nh php tnh theo t s hp thc nhng c tnh n gin hn nh cc bng tra cu c sn.
Th d: Theo th d 2 trn th tha s hp thc l:
f =
=>
Vy, khi lng FeCl3 l 1,86g
2. Phng php i s
Trong cc phng php gii cc bi ton Ho hc phng php i s cng thng c s dng. Phng php ny c u im tit kim c thi gian, khi gii cc bi ton tng hp, tng i kh gii bng cc phng php khc. Phng php i s c dng gii cc bi ton Ho hc sau:a. Gii bi ton lp CTHH bng phng php i s.
Th d: t chy mt hn hp 300ml hirocacbon v amoniac trong oxi c d. Sau khi chy hon ton, th tch kh thu c l 1250ml. Sau khi lm ngng t hi nc, th tch gim cn 550ml. Sau khi cho tc dng vi dung dch kim cn 250ml trong c 100ml nit. Th tch ca tt c cc kh o trong iu kin nh nhau. Lp cng thc ca hirocacbon
Bi gii
Khi t chy hn hp hirocacbon v amoniac trong oxi phn ng xy ra theo phng trnh sau:
4NH3 + 3O2 -> 2N2 + 6H2O
(1)
CxHy + (x + O2 -> xCO2 + H2O (2)
Theo d kin bi ton, sau khi t chy amoniac th to thnh 100ml nit. Theo PTHH (1) sau khi t chy hon ton amoniac ta thu c th tch nit nh hn 2 ln th tch amoniac trong hn hp ban u, vy th tch amonac khi cha c phn ng l 100. 2 = 200ml. Do th tch hiro ccbon khi cha c phn ng l 300 - 200 = 100ml. Sau khi t chy hn hp to thnh (550 - 250) = 300ml, cacbonnic v (1250 - 550 - 300) = 400ml hi nc.
T ta c s phn ng:
CxHy + (x + ) O2 -> xCO2 + H2O
100ml
300ml 400ml
Theo nh lut Avogaro, c th thay th t l th tch cc cht kh tham gia v to thnh trong phn ng bng t l s phn t hay s mol ca chng.
CxHy + 5O2 -> 3CO2 + 4 H2O
=> x = 3; y = 8
Vy CTHH ca hydrocacbon l C3H8
b. Gii bi ton tm thnh phn ca hn hp bng phng php i s.
Th d: Ho tan trong nc 0,325g mt hn hp gm 2 mui Natriclorua v Kaliclorua. Thm vo dung dch ny mt dung dch bc Nitrat ly d - Kt ta bc clorua thu c c khi lng l 0,717g. Tnh thnh phn phn trm ca mi cht trong hn hp.
Bi gii
Gi MNaCl l x v mKcl l y ta c phng trnh i s:
x + y = 0,35 (1)
PTHH: NaCl + AgNO3 -> AgCl ( + NaNO3KCl + AgNO3 -> AgCl ( + KNO3
Da vo 2 PTHH ta tm c khi lng ca AgCl trong mi phn ng:
mAgCl = x .= x . = x . 2,444
mAgCl = y .= y . = y . 1,919
=> mAgCl = 2,444x + 1,919y = 0,717
(2)
T (1) v (2) => h phng trnh
Gii h phng trnh ta c: x = 0,178
y = 0,147
=> % NaCl = .100% = 54,76%
% KCl = 100% - % NaCl = 100% - 54,76% = 45,24%.
Vy trong hn hp: NaCl chim 54,76%, KCl chim 45,24%
3. Phng php p dng nh lut bo ton nguyn t v khi lng.
a/ Nguyn tc:
Trong phn ng ho hc, cc 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 cc cht to thnh.
+ Tng khi lng cc cht trc phn ng bng tng khi lng cc cht sau phn ng.
b/ Phm vi p dng:
Trong cc bi ton xy ra nhiu phn ng, lc ny i khi khng cn thit phi vit cc phng trnh phn ng v ch cn lp s phn ng thy mi quan h t l mol gia cc cht cn xc nh v nhng cht m cho.
Bi 1. Cho mt lung kh clo d tc dng vi 9,2g kim loi sinh ra 23,4g mui kim loi ho tr I. Hy xc nh kim loi ho tr I v mui kim loi .
Hng dn gii:
t M l KHHH ca kim loi ho tr I.
PTHH: 2M + Cl2 2MCl
2M(g) (2M + 71)g
9,2g 23,4g
ta c: 23,4 x 2M = 9,2(2M + 71)
suy ra: M = 23.
Kim loi c khi lng nguyn t bng 23 l Na.
Vy mui thu c l: NaCl
Bi 2: Ho tan hon ton 3,22g hn hp X gm Fe, Mg v Zn bng mt lng va dung dch H2SO4 long, thu c 1,344 lit hiro ( ktc) v dung dch cha m gam mui. Tnh m?
Hng dn gii:
PTHH chung: M + H2SO4 MSO4 + H2 nHSO = nH= = 0,06 mol
p dng nh lut BTKL ta c:
mMui = mX + m HSO- m H= 3,22 + 98 * 0,06 - 2 * 0,06 = 8,98g
Bi 3: C 2 l st khi lng bng nhau v bng 11,2g. Mt l cho tc dng ht vi kh clo, mt l ngm trong dung dch HCl d. Tnh khi lng st clorua thu c.
Hng dn gii:
PTHH:
2Fe + 3Cl2 2FeCl3 (1)
Fe + 2HCl FeCl2 + H2 (2)
Theo phng trnh (1,2) ta c:
nFeCl = nFe = = 0,2mol nFeCl = nFe = = 0,2mol
S mol mui thu c hai phn ng trn bng nhau nhng khi lng mol phn t ca FeCl3 ln hn nn khi lng ln hn.
mFeCl= 127 * 0,2 = 25,4g mFeCl= 162,5 * 0,2 = 32,5g
Bi 4: Ho tan hn hp 2 mui Cacbonnat kim loi ho tr 2 v 3 bng dung dch HCl d thu c dung dch A v 0,672 lt kh (ktc).
Hi c cn dung dch A thu c bao nhiu gam mui khc nhau?
Bi gii:
Bi 1: Gi 2 kim loi ho tr II v III ln lt l X v Y ta c phng trnh phn ng:
XCO3 + 2HCl -> XCl2 + CO2 + H2O
(1)
Y2(CO3)3 + 6HCl -> 2YCl3 + 3CO2 + 3H2O (2).
S mol CO2 thot ra (ktc) phng trnh 1 v 2 l:
Theo phng trnh phn ng 1 v 2 ta thy s mol CO2 bng s mol H2O.
v
Nh vy khi lng HCl phn ng l:
mHCl = 0,06 . 36,5 = 2,19 gam
Gi x l khi lng mui khan ()
Theo nh lut bo ton khi lng ta c:
10 + 2,19 = x + 44 . 0,03 + 18. 0,03
=> x = 10,33 gam
Bi ton 2: Cho 7,8 gam hn hp kim loi Al v Mg tc dng vi HCl thu c 8,96 lt H2 ( ktc). Hi khi c cn dung dch thu c bao nhiu gam mui khan.
Bi gii: Ta c phng trnh phn ng nh sau:
Mg + 2HCl -> MgCl2 + H2(2Al + 6HCl -> 2AlCl3 + 3H2(S mol H2 thu c l:
Theo (1, 2) ta thy s mol HCL gp 2 ln s mol H2Nn: S mol tham gia phn ng l:
n HCl = 2 . 0,4 = 0,8 mol
S mol (s mol nguyn t) to ra mui cng chnh bng s mol HCl bng 0,8 mol. Vy khi lng Clo tham gia phn ng:
mCl = 35,5 . 0,8 = 28,4 gam
Vy khi lng mui khan thu c l:
7,8 + 28,4 = 36,2 gam
4. Phng php tng, gim khi lng.
a/ 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 2 cht ny m gii quyt yu cu t ra.
b/ Phm v s dng:
i vi cc bi ton phn ng xy ra thuc phn ng phn hu, phn ng gia kim loi mnh, khng tan trong nc y kim loi yu ra khi dung sch mui phn ng, ...c bit khi cha bit r phn ng xy ra l hon ton hay khng th vic s dng phng php ny cng n gin ho cc bi ton hn.
Bi 1: Nhng mt thanh st v mt thanh km vo cng mt cc cha 500 ml dung dch CuSO4. Sau mt thi gian ly hai thanh kim loi ra khi cc th mi thanh c thm Cu bm vo, khi lng dung dch trong cc b gim mt 0,22g. Trong dung dch sau phn ng, nng mol ca ZnSO4 gp 2,5 ln nng mol ca FeSO4. Thm dung dch NaOH d vo cc, lc ly kt ta ri nung ngoi khng kh n khi lng khng i , thu c 14,5g cht rn. S gam Cu bm trn mi thanh kim loi v nng mol ca dung dch CuSO4 ban u l bao nhiu?
Hng dn gii:
PTHH
Fe + CuSO4 FeSO4 + Cu ( 1 )Zn + CuSO4 ZnSO4 + Cu ( 2 )Gi a l s mol ca FeSO4
V th tch dung dch xem nh khng thay i. Do t l v nng mol ca cc cht trong dung dch cng chnh l t l v s mol.
Theo bi ra: CM ZnSO = 2,5 CM FeSONn ta c: nZnSO= 2,5 nFeSO
Khi lng thanh st tng: (64 - 56)a = 8a (g)
Khi lng thanh km gim: (65 - 64)2,5a = 2,5a (g)
Khi lng ca hai thanh kim loi tng: 8a - 2,5a = 5,5a (g)
M thc t bi cho l: 0,22g
Ta c: 5,5a = 0,22 a = 0,04 (mol)
Vy khi lng Cu bm trn thanh st l: 64 * 0,04 = 2,56 (g)
v khi lng Cu bm trn thanh km l: 64 * 2,5 * 0,04 = 6,4 (g)
Dung dch sau phn ng 1 v 2 c: FeSO4, ZnSO4 v CuSO4 (nu c)
Ta c s phn ng:
NaOH d t, kk
FeSO4 Fe(OH)2 Fe2O3 a a (mol)
mFeO = 160 x 0,04 x = 3,2 (g)
NaOH d t
CuSO4 Cu(OH)2 CuO
b b b (mol)
mCuO = 80b = 14,5 - 3,2 = 11,3 (g) b = 0,14125 (mol)
Vy nCuSO ban u = a + 2,5a + b = 0,28125 (mol)
CM CuSO = = 0,5625 M
Bi 2: Nhng mt thanh st nng 8 gam vo 500 ml dung dch CuSO4 2M. Sau mt thi gian ly l st ra cn li thy nng 8,8 gam. Xem th tch dung dch khng thay i th nng mol/lit ca CuSO4 trong dung dch sau phn ng l bao nhiu?
Hng dn gii:
S mol CuSO4 ban u l: 0,5 x 2 = 1 (mol)
PTHH
Fe + CuSO4 FeSO4 + Cu ( 1 )
1 mol 1 mol56g 64g lm thanh st tng thm 64 - 56 = 8 gam
M theo bi cho, ta thy khi lng thanh st tng l: 8,8 - 8 = 0,8 gam
Vy c = 0,1 mol Fe tham gia phn ng, th cng c 0,1 mol CuSO4 tham gia phn ng.
S mol CuSO4 cn d : 1 - 0,1 = 0,9 mol
Ta c CM CuSO = = 1,8 M
Bi 3: Dn V lit CO2 (ktc) vo dung dch cha 3,7 gam Ca(OH)2. Sau phn ng thu c 4 gam kt ta. Tnh V?
Hng dn gii:
Theo bi ra ta c:
S mol ca Ca(OH)2 = = 0,05 mol
S mol ca CaCO3 = = 0,04 mol
PTHH
CO2 + Ca(OH)2 CaCO3 + H2O
Nu CO2 khng d:
Ta c s mol CO2 = s mol CaCO3 = 0,04 mol
Vy V(ktc) = 0,04 * 22,4 = 0,896 lt
Nu CO2 d:
CO2 + Ca(OH)2 CaCO3 + H2O
0,05 0,05 mol 0,05
CO2 + CaCO3 + H2O Ca(HCO3)20,01(0,05 - 0,04) mol
Vy tng s mol CO2 tham gia phn ng l: 0,05 + 0,01 = 0,06 mol
V(ktc) = 22,4 * 0,06 = 1,344 lt
Bi 4: Ho tan 20gam hn hp hai mui cacbonat kim loi ho tr 1 v 2 bng dung dch HCl d thu c dung dch X v 4,48 lt kh ( ktc) tnh khi lng mui khan thu c dung dch X.Bi gii: Gi kim loi ho tr 1 v 2 ln lt l A v B ta c phng trnh phn ng sau:
A2CO3 + 2HCl -> 2ACl + CO2( + H2O (1)
BCO3 + 2HCl -> BCl2 + CO2( + H2O (2)
S mol kh CO2 ( ktc) thu c 1 v 2 l:
Theo (1) v (2) ta nhn thy c 1 mol CO2 bay ra tc l c 1 mol mui cacbonnat chuyn thnh mui Clorua v khi lng tng thm 11 gam (gc CO3 l 60g chuyn thnh gc Cl2 c khi lng 71 gam).
Vy c 0,2 mol kh bay ra th khi lng mui tng l:
0,2 . 11 = 2,2 gam
Vy tng khi lng mui Clorua khan thu c l:
M(Mui khan) = 20 + 2,2 = 22,2 (gam)
Bi 5: Ho tan 10gam hn hp 2 mui Cacbonnat kim loi ho tr 2 v 3 bng dung dch HCl d thu c dung dch A v 0,672 lt kh (ktc).
Hi c cn dung dch A thu c bao nhiu gam mui khc nhau?
Bi gii
Mt bi ton ho hc thng l phi c phn ng ho hc xy ra m c phn ng ho hc th phi vit phng trnh ho hc l iu khng th thiu.
Vy ta gi hai kim loi c ho tr 2 v 3 ln lt l X v Y, ta c phn ng:
XCO3 + 2HCl -> XCl2 + CO2 + H2O
(1)
Y2(CO3)3 + 6HCl -> 2YCl3 + 3CO2 + 3H2O (2).
S mol cht kh to ra chng trnh (1) v (2) l:
= 0,03 mol
Theo phn ng (1, 2) ta thy c 1 mol CO2 bay ra tc l c 1 mol mui Cacbonnat chuyn thnh mui clorua v khi lng tng 71 - 60 = 11 (gam) ( ).S mol kh CO2 bay ra l 0,03 mol do khi lng mui khan tng ln:
11 . 0,03 = 0,33 (gam).
Vy khi lng mui khan thu c sau khi c cn dung dch.
m (mui khan) = 10 + 0,33 = 10,33 (gam).
Bi 6: Ho tan 20gam hn hp hai mui cacbonat kim loi ho tr 1 v 2 bng dung dch HCl d thu c dung dch X v 4,48 lt kh ( ktc) tnh khi lng mui khan thu c dung dch X.
Bi gii: Gi kim loi ho tr 1 v 2 ln lt l A v B ta c phng trnh phn ng sau:
A2CO3 + 2HCl -> 2ACl + CO2( + H2O (1)
BCO3 + 2HCl -> BCl2 + CO2( + H2O (2)
S mol kh CO2 ( ktc) thu c 1 v 2 l:
Theo (1) v (2) ta nhn thy c 1 mol CO2 bay ra tc l c 1 mol mui cacbonnat chuyn thnh mui Clorua v khi lng tng thm 11 gam (gc CO3 l 60g chuyn thnh gc Cl2 c khi lng 71 gam).
Vy c 0,2 mol kh bay ra th khi lng mui tng l:
0,2 . 11 = 2,2 gam
Vy tng khi lng mui Clorua khan thu c l:
M(Mui khan) = 20 + 2,2 = 22,2 (gam)
Bi 1: Nhng mt thanh kim loi M ho tr II vo 0,5 lit dd CuSO4 0,2M. Sau mt thi gian phn ng, khi lng thanh M tng ln 0,40g trong khi nng CuSO4 cn li l 0,1M.
a/ Xc nh kim loi M.
b/ Ly m(g) kim loi M cho vo 1 lit dd cha AgNO3 v Cu(NO3)2 , nng mi mui l 0,1M. Sau phn ng ta thu c cht rn A khi lng 15,28g v dd B. Tnh m(g)?
Hng dn gii:
a/ theo bi ra ta c PTHH .
M + CuSO4 MSO4 + Cu (1)
S mol CuSO4 tham gia phn ng (1) l: 0,5 ( 0,2 0,1 ) = 0,05 mol
tng khi lng ca M l:
mtng = mkl gp - mkl tan = 0,05 (64 M) = 0,40
gii ra: M = 56 , vy M l Fe
b/ ta ch bit s mol ca AgNO3 v s mol ca Cu(NO3)2. Nhng khng bit s mol ca Fe
(cht kh Fe Cu2+ Ag+ (cht oxh mnh)
0,1 0,1 ( mol )
Ag+ C Tnh oxi ho mnh hn Cu2+ nn mui AgNO3 tham gia phn ng vi Fe trc.
PTHH:
Fe + 2AgNO3 Fe(NO3)2 + 2Ag (1)
Fe + Cu(NO3)2 Fe(NO3)2 + Cu (2)
Ta c 2 mc so snh:
- Nu va xong phn ng (1): Ag kt ta ht, Fe tan ht, Cu(NO3)2 cha phn ng. Cht rn A l Ag th ta c: mA = 0,1 x 108 = 10,8 g
- Nu va xong c phn ng (1) v (2) th khi cht rn A gm: 0,1 mol Ag v 0,1 mol Cu
mA = 0,1 ( 108 + 64 ) = 17,2 g
theo cho mA = 15,28 g ta c: 10,8 < 15,28 < 17,2
vy AgNO3 phn ng ht, Cu(NO3)2 phn ng mt phn v Fe tan ht.
mCu to ra = mA mAg = 15,28 10,80 = 4,48 g. Vy s mol ca Cu = 0,07 mol.
Tng s mol Fe tham gia c 2 phn ng l: 0,05 ( p 1 ) + 0,07 ( p 2 ) = 0,12 mol
Khi lng Fe ban u l: 6,72g5. Phng php lm gim n s.
Bi ton 1: (Xt li bi ton nu phng php th nht)
Ho tan hn hp 20 gam hai mui cacbonnat kim loi ho tr I v II bng dung dch HCl d thu c dung dch M v 4,48 lt CO2 ( ktc) tnh khi lng mun to thnh trong dung dch M.
Bi gii
Gi A v B ln lt l kim loi ho tr I v II. Ta c phng trnh phn ng sau:
A2CO3 + 2HCl -> 2ACl + H2O + CO2( (1)
BCO3 + 2HCl -> BCl2 + H2O + CO2( (2)
S mol kh thu c phn ng (1) v (2) l:
Gi a v b ln lt l s mol ca A2CO3 v BCO3 ta c phng trnh i s sau:
(2A + 60)a + (B + 60)b = 20 (3)
Theo phng trnh phn ng (1) s mol ACl thu c 2a (mol)
Theo phng trnh phn ng (2) s mol BCl2 thu c l b (mol)
Nu gi s mui khan thu c l x ta c phng trnh:
(A + 35.5) 2a + (B + 71)b = x (4)
Cng theo phn ng (1, 2) ta c:
a + b =
(5)
T phng trnh (3, 4) (Ly phng trnh (4) tr (5)) ta c:
11 (a + b) = x - 20 (6)
Thay a + b t (5) vo (6) ta c:
11 . 0,2 = x - 20
=> x = 22,2 gam
Bi ton 2: Ho tan hon ton 5 gam hn hp 2 kim loi bng dung dch HCl thu c dung dch A v kh B, c cn dung dch A thu c 5,71 gam mui khan tnh th tch kh B ktc.
Bi gii: Gi X, Y l cc kim loi; m, n l ho tr, x, y l s mol tng ng, s nguyn t khi l P, Q ta c:
2X + 2n HCl => 2XCln = nH2( (I)
2Y + 2m HCl -> 2YClm + mH2( (II).
Ta c: xP + y Q = 5 (1)
x(P + 35,5n) + y(Q + 35,5m) = 5,71 (2)
Ly phng trnh (2) tr phng trnh (1) ta c:
x(P + 35,5n) + y(Q + 35,5m)- xP - yQ = 0,71
=> 35,5 (nx + my) = 0,71
Theo I v II:
=> th tch: V = nx + my = (lt)
6. Phng php dng bi ton cht tng ng.
a/ Nguyn tc:
Khi trong bi ton xy ra nhiu phn ng nhng cc phn ng cng loi v cng hiu sut th ta thay hn hp nhiu cht thnh 1 cht tng ng. Lc lng (s mol, khi lng hay th tch) ca cht tng ng bng lng ca hn hp.
b/ Phm vi s dng:
Trong v c, phng php ny p dng khi hn hp nhiu kim loi hot ng hay nhiu oxit kim loi, hn hp mui cacbonat, ... hoc khi hn hp kim loi phn ng vi nc.
Bi 1: Mt hn hp 2 kim loi kim A, B thuc 2 chu k k tip nhau trong bng h thng tun hon c khi lng l 8,5 gam. Hn hp ny tan ht trong nc d cho ra 3,36 lit kh H2 (ktc). Tm hai kim loi A, B v khi lng ca mi kim loi.
Hng dn gii:
PTHH
2A + 2H2O 2AOH + H2 (1)
2B + 2H2O 2BOH + H2 (2)
t a = nA , b = nB
ta c: a + b = 2 = 0,3 (mol) (I)
trung bnh: = = 28,33
Ta thy 23 < = 28,33 < 39
Gi s MA < MB th A l Na, B l K hoc ngc li.
mA + mB = 23a + 39b = 8,5 (II)
T (I, II) ta tnh c: a = 0,2 mol, b = 0,1 mol.
Vy mNa = 0,2 * 23 = 4,6 g, mK = 0,1 * 39 = 3,9 g.
Bi 2: Ho tan 115,3 g hn hp gm MgCO3 v RCO3 bng 500ml dung dch H2SO4 long ta thu c dung dch A, cht rn B v 4,48 lt CO2 (ktc). C cn dung dch A th thu c 12g mui khan. Mt khc em nung cht rn B ti khi lng khng i th thu c 11,2 lt CO2 (ktc) v cht rn B1. Tnh nng mol/lit ca dung dch H2SO4 long dng, khi lng ca B, B1 v khi lng nguyn t ca R. Bit trong hn hp u s mol ca RCO3 gp 2,5 ln s mol ca MgCO3.
Hng dn gii:
Thay hn hp MgCO3 v RCO3 bng cht tng ng CO3PTHH
CO3 + H2SO4 SO4 + CO2 + H2O (1)
0,2 0,2 0,2 0,2
S mol CO2 thu c l: nCO = = 0,2 (mol)
Vy nHSO = nCO = 0,2 (mol)
CM HSO = = 0,4 M
Rn B l CO3 d:
CO3 O + CO2 (2)
0,5 0,5 0,5
Theo phn ng (1): t 1 mol CO3 to ra 1 mol SO4 khi lng tng 36 gam.
p dng nh lut bo ton khi lng ta c:
115,3 = mB + mmui tan - 7,2
Vy mB = 110,5 g
Theo phn ng (2): t B chuyn thnh B1, khi lng gim l:
mCO = 0,5 * 44 = 22 g.
Vy mB = mB - mCO = 110,5 - 22 = 88,5 g
Tng s mol CO3 l: 0,2 + 0,5 = 0,7 mol
Ta c + 60 = 164,71 = 104,71
V trong hn hp u s mol ca RCO3 gp 2,5 ln s mol ca MgCO3.
Nn 104,71 = R = 137
Vy R l Ba.
Bi 3: ho tan hon ton 28,4 gam hn hp 2 mui cacbonat ca 2 kim loi thuc phn nhm chnh nhm II cn dng 300ml dung dch HCl aM v to ra 6,72 lit kh (ktc). Sau phn ng, c cn dung dch thu c m(g) mui khan. Tnh gi tr a, m v xc nh 2 kim loi trn.
Hng dn gii:
nCO = = 0,3 (mol)
Thay hn hp bng CO3
CO3 + 2HCl Cl2 + CO2 + H2O (1)
0,3 0,6 0,3 0,3
Theo t l phn ng ta c:
nHCl = 2 nCO = 2 * 0,3 = 0,6 mol
CM HCl = = 2M
S mol ca CO3 = nCO = 0,3 (mol)
Nn + 60 = = 94,67
= 34,67
Gi A, B l KHHH ca 2 kim loi thuc phn nhm chnh nhm II, MA < MB
ta c: MA < = 34,67 < MB tho mn ta thy 24 < = 34,67 < 40.
Vy hai kim loi thuc phn nhm chnh nhm II l: Mg v Ca.
Khi lng mui khan thu c sau khi c cn l: m = (34,67 + 71)* 0,3 = 31,7 gam.
7/ Phng php bo ton s mol nguyn t.
a/ Nguyn tc p dng:
Trong mi qu trnh bin i ho hc: S mol mi nguyn t trong cc cht c bo ton.
b/ V d: Cho 10,4g hn hp bt Fe v Mg (c t l s mol 1:2) ho tan va ht trong 600ml dung dch HNO3 x(M), thu c 3,36 lit hn hp 2 kh N2O v NO. Bit hn hp kh c t khi d = 1,195. Xc nh tr s x?
Hng dn gii:
Theo bi ra ta c:
nFe : nMg = 1 : 2 (I) v 56nFe + 24nMg = 10,4 (II)
Gii phng trnh ta c: nFe = 0,1 v nMg = 0,2
S phn ng.
Fe, Mg + HNO3 ------> Fe(NO3)3 , Mg(NO3)2 + N2O, NO + H2O
0,1 v 0,2 x 0,1 0,2 a v b (mol)
Ta c:
a + b = = 0,15 v = 1,195 ---> a = 0,05 mol v b = 0,1 mol
S mol HNO3 phn ng bng:
nHNO= nN = 3nFe(NO) + 2nMg(NO)+ 2nNO + nNO
= 3.0,1 + 2.0,2 + 2.0,05 + 0,1 = 0,9 mol
Nng mol/lit ca dung dch HNO3:
x(M) = .1000 = 1,5M
8/ Phng php lp lun kh nng.a/ Nguyn tc p dng:
Khi gii cc bi ton ho hc theo phng php i s, nu s phng trnh ton hc thit lp c t hn s n s cha bit cn tm th phi bin lun ---> Bng cch: Chn 1 n s lm chun ri tch cc n s cn li. Nn a v phng trnh ton hc 2 n, trong c 1 n c gii hn (tt nhin nu c 2 n c gii hn th cng tt). Sau c th thit lp bng bin thin hay d vo cc iu kin khc chn cc gi tr hp l.
b/ V d:
Bi 1: Ho tan 3,06g oxit MxOy bng dung dich HNO3 d sau c cn th thu c 5,22g mui khan. Hy xc nh kim loi M bit n ch c mt ho tr duy nht.
Hng dn gii:
PTHH: MxOy + 2yHNO3 -----> xM(NO3)2y/x + yH2O
T PTP ta c t l:
= ---> M = 68,5.2y/x
Trong : t 2y/x = n l ho tr ca kim loi. Vy M = 68,5.n (*)
Cho n cc gi tr 1, 2, 3, 4. T (*) ---> M = 137 v n =2 l ph hp.
Do M l Ba, ho tr II.
Bi 2: A, B l 2 cht kh iu kin thng, A l hp cht ca nguyn t X vi oxi (trong oxi chim 50% khi lng), cn B l hp cht ca nguyn t Y vi hir (trong hiro chim 25% khi lng). T khi ca A so vi B bng 4. Xc nh cng thc phn t A, B. Bit trong 1 phn t A ch c mt nguyn t X, 1 phn t B ch c mt nguyn t Y.
Hng dn gii:
t CTPT A l XOn, MA = X + 16n = 16n + 16n = 32n.
t CTPT A l YOm, MB = Y + m = 3m + m = 4m.
d = = = 4 ---> m = 2n.
iu kin tho mn: 0 < n, m < 4, u nguyn v m phi l s chn.
Vy m ch c th l 2 hay 4.
Nu m = 2 th Y = 6 (loi, khng c nguyn t no tho)
Nu m = 4 th Y = 12 (l cacbon) ---> B l CH4v n = 2 th X = 32 (l lu hunh) ---> A l SO29/ Phng php gii hn mt i lng.
a/ Nguyn tc p dng: Da vo cc i lng c gii hn, chng hn:
KLPTTB (), ho tr trung bnh, s nguyn t trung bnh, ....
Hiu sut: 0(%) < H < 100(%)
S mol cht tham gia: 0 < n(mol) < S mol cht ban u,...
suy ra quan h vi i lng cn tm. Bng cch:
Tm s thay i gi tr min v max ca 1 i lng no dn n gii hn cn tm.
Gi s thnh phn hn hp (X,Y) ch cha X hay Y suy ra gi tr min v max ca i lng cn tm.
b/ V d:
Bi 1: Cho 6,2g hn hp 2 kim loi kim thuc 2 chu k lin tip trong bng tun hon phn ng vi H2O d, thu c 2,24 lit kh (ktc) v dung dch A.
a/ Tnh thnh phn % v khi lng tng kim loi trong hn hp ban u.
Hng dn:
a/ t R l KHHH chung cho 2 kim loi kim cho
MR l khi lng trung bnh ca 2 kim loi kim A v B, gi s MA < MB
---.> MA < MR < MB .
Vit PTHH xy ra:
Theo phng trnh phn ng:
nR = 2nH= 0,2 mol. ----> MR = 6,2 : 0,2 = 31
Theo ra: 2 kim loi ny thuc 2 chu k lin tip, nn 2 kim loi l:
A l Na(23) v B l K(39)
Bi 2:
a/ Cho 13,8 gam (A) l mui cacbonat ca kim loi kim vo 110ml dung dch HCl 2M. Sau phn ng thy cn axit trong dung dch thu c v th tch kh thot ra V1 vt qu 2016ml. Vit phng trnh phn ng, tm (A) v tnh V1 (ktc).
b/ Ho tan 13,8g (A) trn vo nc. Va khuy va thm tng git dung dch HCl 1M cho ti 180ml dung dch axit, thu c V2 lit kh. Vit phng trnh phn ng xy ra v tnh V2 (ktc).
Hng dn:
a/ M2CO3 + 2HCl ---> 2MCl + H2O + CO2
Theo PTHH ta c:
S mol M2CO3 = s mol CO2 > 2,016 : 22,4 = 0,09 mol
---> Khi lng mol M2CO3 < 13,8 : 0,09 = 153,33 (I)
Mt khc: S mol M2CO3 phn ng = 1/2 s mol HCl < 1/2. 0,11.2 = 0,11 mol
---> Khi lng mol M2CO3 = 13,8 : 0,11 = 125,45 (II)
T (I, II) --> 125,45 < M2CO3 < 153,33 ---> 32,5 < M < 46,5 v M l kim loi kim
---> M l Kali (K)
Vy s mol CO2 = s mol K2CO3 = 13,8 : 138 = 0,1 mol ---> VCO = 2,24 (lit)
b/ Gii tng t: ---> V2 = 1,792 (lit)
Bi 3: Cho 28,1g qung lmt gm MgCO3; BaCO3 (%MgCO3 = a%) vo dung dch HCl d thu c V (lt) CO2 ( ktc).
a/ Xc nh V (lt).
Hng dn:
a/ Theo bi ra ta c PTHH:
MgCO3 + 2HCl MgCl2 + H2O + CO2 (1)
x(mol) x(mol)
BaCO3 + 2HCl BaCl2 + H2O + CO2 (2)
y(mol) y(mol)
CO2 + Ca(OH)2 CaCO3 + H2O (3)
0,2(mol) 0,2(mol) 0,2(mol)
CO2 + CaCO3 + H2O Ca(HCO3)2 (4)
Gi s hn hp ch c MgCO3.Vy mBaCO3 = 0
S mol: nMgCO3 = = 0,3345 (mol)
Nu hn hp ch ton l BaCO3 th mMgCO3 = 0
S mol: nBaCO3 = = 0,143 (mol)
Theo PT (1) v (2) ta c s mol CO2 gii phng l:
0,143 (mol) nCO2 0,3345 (mol)
Vy th tch kh CO2 thu c ktc l: 3,2 (lt) VCO 7,49 (lt)
Chuyn 2:
tan - nng dung dch
Mt s cng thc tnh cn nh:
Cng thc tnh tan: Stcht = . 100
Cng thc tnh nng %: C% = . 100%
mdd = mdm + mct Hoc mdd = Vdd (ml) . D(g/ml)
* Mi lin h gia tan ca mt cht v nng phn trm dung dch bo ho ca cht mt nhit xc nh.
C 100g dm ho tan c Sg cht tan to thnh (100+S)g dung dch bo ho.
Vy: x(g) // y(g) // 100g //
Cng thc lin h: C% = Hoc S =
Cng thc tnh nng mol/lit: CM = =
EMBED Equation.3 * Mi lin h gia nng % v nng mol/lit.
Cng thc lin h: C% = Hoc CM =
Trong :
mct l khi lng cht tan( n v: gam)
mdm l khi lng dung mi( n v: gam)
mdd l khi lng dung dch( n v: gam)
V l th tch dung dch( n v: lit hoc mililit)
D l khi lng ring ca dung dch( n v: gam/mililit)
M l khi lng mol ca cht( n v: gam)
S l tan ca 1 cht mt nhit xc nh( n v: gam)
C% l nng % ca 1 cht trong dung dch( n v: %)
CM l nng mol/lit ca 1 cht trong dung dch( n v: mol/lit hay M)
Dng 1: Ton tan
Phn dng 1: Bi ton lin quan gia tan ca mt cht v nng phn trm dung dch bo ho ca cht .
Bi 1: 400C, tan ca K2SO4 l 15. Hy tnh nng phn trm ca dung dch K2SO4 bo ho nhit ny?
p s: C% = 13,04%
Bi 2: Tnh tan ca Na2SO4 100C v nng phn trm ca dung dch bo ho Na2SO4 nhit ny. Bit rng 100C khi ho tan 7,2g Na2SO4 vo 80g H2O th c dung dch bo ho Na2SO4.
p s: S = 9g v C% = 8,257%
Phn dng 2: Bi ton tnh lng tinh th ngm nc cn cho thm vo dung dch cho sn.
Cch lm:
Dng nh lut bo ton khi lng tnh:
* Khi lng dung dch to thnh = khi lng tinh th + khi lng dung dch ban u.
* Khi lng cht tan trong dung dch to thnh = khi lng cht tan trong tinh th + khi lng cht tan trong dung dch ban u.
* Cc bi ton loi ny thng cho tinh th cn ly v dung dch cho sn c cha cng loi cht tan.
Bi tp p dng:
Bi 1: Tnh lng tinh th CuSO4.5H2O cn dng iu ch 500ml dung dch CuSO4 8%(D = 1,1g/ml).
p s: Khi lng tinh th CuSO4.5H2O cn ly l: 68,75g
Bi 2: iu ch 560g dung dch CuSO4 16% cn phi ly bao nhiu gam dung dch CuSO4 8% v bao nhiu gam tinh th CuSO4.5H2O.
Hng dn
* Cch 1:
Trong 560g dung dch CuSO4 16% c cha.
mct CuSO4(c trong dd CuSO4 16%) = = = 89,6(g)
t mCuSO4.5H2O = x(g)
1mol(hay 250g) CuSO4.5H2O cha 160g CuSO4
Vy x(g) // cha = (g)
mdd CuSO4 8% c trong dung dch CuSO4 16% l (560 x) g
mct CuSO4(c trong dd CuSO4 8%) l = (g)
Ta c phng trnh: + = 89,6
Gii phng trnh c: x = 80.
Vy cn ly 80g tinh th CuSO4.5H2O v 480g dd CuSO4 8% pha ch thnh 560g dd CuSO4 16%.
* Cch 2: Gii h phng trnh bc nht 2 n.
* Cch 3: Tnh ton theo s ng cho.
Lu : Lng CuSO4 c th coi nh dd CuSO4 64%(v c 250g CuSO4.5H2O th c cha 160g CuSO4). Vy C%(CuSO4) = .100% = 64%.
Phn dng 3: bi ton tnh lng cht tan tch ra hay thm vo khi thay i nhit mt dung dch bo ho cho sn.
Cch lm:
Bc 1: Tnh khi lng cht tan v khi lng dung mi c trong dung dch bo ho t1(0c)
Bc 2: t a(g) l khi lng cht tan A cn thm hay tch ra khi dung dch ban u, sau khi thay i nhit t t1(0c) sang t2(0c) vi t1(0c) khc t2(0c).
Bc 3: Tnh khi lng cht tan v khi lng dung mi c trong dung dch bo ho t2(0c).
Bc 4: p dng cng thc tnh tan hay nng % dung dch bo ho(C% ddbh) tm a.
Lu : Nu yu cu tnh lng tinh th ngm nc tch ra hay cn thm vo do thay i nhit dung dch bo ho cho sn, bc 2 ta phi t n s l s mol(n)Bi 1: 120C c 1335g dung dch CuSO4 bo ho. un nng dung dch ln n 900C. Hi phi thm vo dung dch bao nhiu gam CuSO4 c dung dch bo ho nhit ny.
Bit 120C, tan ca CuSO4 l 33,5 v 900C l 80.
p s: Khi lng CuSO4 cn thm vo dung dch l 465g.
Bi 2: 850C c 1877g dung dch bo ho CuSO4. Lm lnh dung dch xung cn 250C. Hi c bao nhiu gam CuSO4.5H2O tch khi dung dch. Bit tan ca CuSO4 850C l 87,7 v 250C l 40.
p s: Lng CuSO4.5H2O tch khi dung dch l: 961,75g
Bi 3: Cho 0,2 mol CuO tan trong H2SO4 20% un nng, sau lm ngui dung dch n 100C. Tnh khi lng tinh th CuSO4.5H2O tch khi dung dch, bit rng tan ca CuSO4 100C l 17,4g/100g H2O.
p s: Lng CuSO4.5H2O tch khi dung dch l: 30,7g
Dng 2: Ton nng dung dch
Bi 1: Cho 50ml dung dch HNO3 40% c khi lng ring l 1,25g/ml. Hy:
a/ Tm khi lng dung dch HNO3 40%?
b/ Tm khi lng HNO3?
c/ Tm nng mol/l ca dung dch HNO3 40%?
p s:
a/ mdd = 62,5g
b/ mHNO = 25g
c/ CM(HNO) = 7,94M
Bi 2: Hy tnh nng mol/l ca dung dch thu c trong mi trng hp sau:
a/ Ho tan 20g NaOH vo 250g nc. Cho bit DHO = 1g/ml, coi nh th tch dung dch khng i.
b/ Ho tan 26,88 lt kh hiro clorua HCl (ktc) vo 500ml nc thnh dung dch axit HCl. Coi nh th dung dch khng i.
c/ Ho tan 28,6g Na2CO3.10H2O vo mt lng nc va thnh 200ml dung dch Na2CO3.
p s:
a/ CM( NaOH ) = 2M
b/ CM( HCl ) = 2,4M
c/ CM(Na2CO3) = 0,5M
Bi 3: Cho 2,3g Na tan ht trong 47,8ml nc thu c dung dch NaOH v c kh H2 thot ra . Tnh nng % ca dung dch NaOH?
p s: C%(NaOH) = 8%
chuyn 3:
pha trn dung dch
Loi 1: Bi ton pha long hay c dc mt dung dch.a) c im ca bi ton:
Khi pha long, nng dung dch gim. Cn c dc, nng dung dch tng.
D pha long hay c c, khi lng cht tan lun lun khng thay i.
b) Cch lm:
C th p dng cng thc pha long hay c c
TH1: V khi lng cht tan khng i d pha long hay c c nn.
mdd(1).C%(1) = mdd(2).C%(2) TH2: V s mol cht tan khng i d pha long hay c dc nn.
Vdd(1). CM (1) = Vdd(2). CM (2)
Nu gp bi ton bi ton: Cho thm H2O hay cht tan nguyn cht (A) vo 1 dung dch (A) c nng % cho trc, c th p dng quy tc ng cho gii. Khi c th xem:
- H2O l dung dch c nng O%
- Cht tan (A) nguyn cht cho thm l dung dch nng 100%
+ TH1: Thm H2O
Dung dch u C1(%) C2(%) - O
C2(%) =
H2O O(%) C1(%) C2(%)
+ TH1: Thm cht tan (A) nguyn cht
Dung dch u C1(%) 100 - C2(%)
C2(%) =
Cht tan (A) 100(%) C1(%) C2(%)
Lu : T l hiu s nng nhn c ng bng s phn khi lng dung dch u( hay H2O, hoc cht tan A nguyn cht) cn ly t cng hng ngang.
Bi ton p dng:
Bi 1: Phi thm bao nhiu gam H2O vo 200g dung dch KOH 20% c dung dch KOH 16%.
p s: mH2O(cn thm) = 50g
Bi 2: C 30g dung dch NaCl 20%. Tnh nng % dung dch thu c khi:
Pha thm 20g H2O
C c dung dch ch cn 25g.
p s: 12% v 24%
Bi 3: Tnh s ml H2O cn thm vo 2 lit dung dch NaOH 1M thu c dung dch mi c nng 0,1M.
p s: 18 lit
Bi 4: Tnh s ml H2O cn thm vo 250ml dung dch NaOH1,25M to thnh dung dch 0,5M. Gi s s ho tan khng lm thay i ng k th tch dung dch.
p s: 375ml
Bi 5: Tnh s ml dung dch NaOH 2,5%(D = 1,03g/ml) iu ch c t 80ml dung dch NaOH 35%(D = 1,38g/ml).
p s: 1500ml
Bi 6: Lm bay hi 500ml dung dch HNO3 20%(D = 1,20g/ml) ch cn 300g dung dch. Tnh nng % ca dung dch ny.
p s: C% = 40%
Loi 2:Bi ton ho tan mt ho cht vo nc hay vo mt dung dch cho sn.
a/ c im bi ton:
Ho cht em ho tan c th l cht kh, cht lng hay cht rn.
S ho tan c th gy ra hay khng gy ra phn ng ho hc gia cht em ho tan vi H2O hoc cht tan trong dung dch cho sn.
b/ Cch lm:
Bc 1: Xc nh dung dch sau cng (sau khi ho tan ho cht) c cha cht no:
Cn lu xem c phn ng gia cht em ho tan vi H2O hay cht tan trong dung dch cho sn khng? Sn phm phn ng(nu c) gm nhng cht tan no? Nh rng: c bao nhiu loi cht tan trong dung dch th c by nhiu nng .
. Nu cht tan c phn ng ho hc vi dung mi, ta phi tnh nng ca sn phm phn ng ch khng c tnh nng ca cht tan .
Bc 2: Xc nh lng cht tan(khi lng hay s mol) c cha trong dung dch sau cng.
. Lng cht tan(sau phn ng nu c) gm: sn phm phn ng v cc cht tc dng cn d.
. Lng sn phm phn ng(nu c) tnh theo ptt phi da vo cht tc dng ht(lng cho ), tuyt i khng c da vo lng cht tc dng cho d (cn tha sau phn ng)
Bc 3: Xc nh lng dung dch mi (khi lng hay th tch)
. tnh th tch dung dch mi c 2 trng hp (tu theo bi)
Nu khng cho bit khi lng ring dung dch mi(Dddm)
+ Khi ho tan 1 cht kh hay 1 cht rn vo 1 cht lng c th coi:
Th tch dung dch mi = Th tch cht lng
+ Khi ho tan 1 cht lng vo 1 cht lng khc, phi gi s s pha trn khng lm thy i ng k th tch cht lng, tnh:
Th tch dung dch mi = Tng th tch cc cht lng ban u.
Nu cho bit khi lng ring dung dch mi(Dddm)
Th tch dung dch mi: Vddm =
mddm: l khi lng dung dch mi
+ tnh khi lng dung dch mi
mddm = Tng khi lng(trc phn ng) khi lng kt ta(hoc kh bay ln) nu c.
Bi tp p dng:
Bi 1: Cho 14,84g tinh th Na2CO3 vo bnh cha 500ml dung dch HCl 0,4M c dung dch B. Tnh nng mol/lit cc cht trong dung dch B.
p s: Nng ca NaCl l: CM = 0,4M
Nng ca Na2CO3 cn d l: CM = 0,08M
Bi 2: Ho tan 5,6lit kh HCl ( ktc) vo 0,1lit H2O to thnh dung dch HCl. Tnh nng mol/lit v nng % ca dung dch thu c.
p s:
CM = 2,5M
C% = 8,36%
Bi 3: Cho 200g SO3 vo 1 lt dung dch H2SO4 17%(D = 1,12g/ml) c dung dch A. Tnh nng % dung dch A.
p s: C% = 32,985%
Bi 4: xc nh lng SO3 v lng dung dch H2SO4 49% cn ly pha thnh 450g dung dch H2SO4 83,3%.
p s:
Khi lng SO3 cn ly l: 210g
Khi lng dung dch H2SO4 49% cn ly l 240g
Bi 5: Xc nh khi lng dung dch KOH 7,93% cn ly khi ho tan vo 47g K2O th thu c dung dch 21%.
p s: Khi lng dung dch KOH 7,93% cn ly l 352,94g
Bi 6: Cho 6,9g Na v 9,3g Na2O vo nc, c dung dch A(NaOH 8%). Hi phi ly thm bao nhiu gam NaOH c tinh khit 80%(tan hon ton) cho vo c dung dch 15%?
p s: - Khi lng NaOH c tinh khit 80% cn ly l 32,3g
Loi 3: Bi ton pha trn hai hay nhiu dung dch.
a/ c im bi ton.
Khi pha trn 2 hay nhiu dung dch vi nhau c th xy ra hay khng xy ra phn ng ho hc gia cht tan ca cc dung dch ban u.
b/ Cch lm:
TH1: Khi trn khng xy ra phn ng ho hc(thng gp bi ton pha trn cc dung dch cha cng loi ho cht)
Nguyn tc chung gii l theo phng php i s, lp h 2 phng trnh ton hc (1 theo cht tan v 1 theo dung dch)
Cc bc gii:
Bc 1: Xc nh dung dch sau trn c cha cht tan no.
Bc 2: Xc nh lng cht tan(mct) c trong dung dch mi(ddm)
Bc 3: Xc nh khi lng(mddm) hay th tch(Vddm) dung dch mi.
mddm = Tng khi lng( cc dung dch em trn )
+ Nu bit khi lng ring dung dch mi(Dddm)
Vddm =
+ Nu khng bit khi lng ring dung dch mi: Phi gi s s hao ht th tch do s pha trn dung dch l khng ng k, c.
Vddm = Tng th tch cc cht lng ban u em trn
+ Nu pha trn cc dung dch cng loi cht tan, cng loi nng , c th gii bng quy tc ng cho.
m1(g) dd C1(%) C2 C3
C3(%)
m2(g) dd C2(%) C3 C1
( Gi s: C1< C3 < C2 ) v s hao ht th tch do s pha trn cc dd l khng ng k.
=
+ Nu khng bit nng % m li bit nng mol/lit (CM) th p dng s :
V1(l) dd C1(M) C2 C3
C3(M)
V2(g) dd C2(M) C3 C1
( Gi s: C1< C3 < C2 )
=
+ Nu khng bit nng % v nng mol/lit m li bit khi lng ring (D) th p dng s :
V1(l) dd D1(g/ml) D2 D3
D3(g/ml)
V2(l) dd D2(g/ml) D3 D1
(Gi s: D1< D3 < D2) v s hao ht th tch do s pha trn cc dd l khng ng k.
=
TH2: Khi trn c xy ra phn ng ho hc cng gii qua 3 bc tng t bi ton loi 2 (Ho tan mt cht vo mt dung dch cho sn). Tuy nhin, cn lu .
bc 1: Phi xc nh cng thc cht tan mi, s lng cht tan mi. Cn ch kh nng c cht d(do cht tan ban u khng tc dng ht) khi tnh ton.
bc 3: Khi xc nh lng dung dch mi (mddm hay Vddm)
Tac: mddm = Tng khi lng cc cht em trng khi lng cht kt ta hoc cht kh xut hin trong phn ng.
Th tch dung dch mi tnh nh trng hp 1 loi bi ton ny.
Th d: p dng phng php ng cho.
Mt bi ton thng c nhiu cch gii nhng nu bi ton no c th s dng c phng php ng cho gii th s lm bi ton n gin hn rt nhiu.
Bi ton 1: Cn bao nhiu gam tinh th CuSO4 . 5H2O ho vo bao nhiu gam dung dch CuSO4 4% iu ch c 500 gam dung dch CuSO4 8%.
Bi gii: Gii Bng phng php thng thng:
Khi lng CuSO4 c trong 500g dung dch bng:
(1)
Gi x l khi lng tinh th CuSO4 . 5 H2O cn ly th: (500 - x) l khi lng dung dch CuSO4 4% cn ly:
Khi lng CuSO4 c trong tinh th CuSO4 . 5H2O bng:
(2)
Khi lng CuSO4 c trong tinh th CuSO4 4% l:
(3)
T (1), (2) v (3) ta c:
=> 0,64x + 20 - 0,04x = 40.
Gii ra ta c:
X = 33,33g tinh th
Vy khi lng dung dch CuSO4 4% cn ly l:
500 - 33,33 gam = 466,67 gam.
+ Gii theo phng php ng cho
Gi x l s gam tinh th CuSO4 . 5 H2O cn ly v (500 - x) l s gam dung dch cn ly ta c s ng cho nh sau:
=>
Gii ra ta tm c: x = 33,33 gam.
Bi ton 2: Trn 500gam dung dch NaOH 3% vi 300 gam dung dch NaOH 10% th thu c dung dch c nng bao nhiu%.
Bi gii: Ta c s ng cho:
=>
Gii ra ta c: C = 5,625%
Vy dung dch thu c c nng 5,625%.
Bi ton 3: Cn trn 2 dung dch NaOH 3% v dung dch NaOH 10% theo t l khi lng bao nhiu thu c dung dch NaOH 8%.
Bi gii:
Gi m1; m2 ln lt l khi lng ca cc dung dch cn ly. Ta c s ng cho sau:
=>
Vy t l khi lng cn ly l:
(
Bi ton p dng:
Bi 1: Cn pha ch theo t l no v khi lng gia 2 dung dch KNO3 c nng % tng ng l 45% v 15% c mt dung dch KNO3 c nng 20%.
p s: Phi ly 1 phn khi lng dung dch c nng d 45% v 5 phn khi lng dung dch c nng 15% trn vi nhau.
Bi 2: Trn V1(l) dung dch A(cha 9,125g HCl) vi V2(l) dung dch B(cha 5,475g HCl) c 2(l) dung dch D.
Coi th tch dung dch D = Tng th tch dung dch A v dung dch B.
a) Tnh nng mol/lit ca dung dch D.
b) Tnh nng mol/lit ca dung dch A, dung dch B (Bit hiu nng mol/lit ca dung dch A tr nng mol/lit dung dch B l 0,4mol/l)
p s:
a) CM(dd D) = 0,2M
b) t nng mol/l ca dung dch A l x, dung dch B l y ta c:
x y = 0,4 (I)
V th tch: Vdd D = Vdd A + Vdd B = + = 2 (II)
Gii h phng trnh ta c: x = 0,5M, y = 0,1M
Vy nng mol/l ca dung dch A l 0,5M v ca dung dch B l 0,1M.
Bi 3: Hi phi ly 2 dung dch NaOH 15% v 27,5% mi dung dch bao nhiu gam trn vo nhau c 500ml dung dch NaOH 21,5%, D = 1,23g/ml?
p s: Dung dch NaOH 27,5% cn ly l 319,8g v dung dch NaOH 15% cn ly l 295,2g
Bi 4: Trn ln 150ml dung dch H2SO4 2M vo 200g dung dch H2SO4 5M( D = 1,29g/ml ). Tnh nng mol/l ca dung dch H2SO4 nhn c.
p s: Nng H2SO4 sau khi trn l 3,5M
Bi 5: Trn 1/3 (l) dung dch HCl (dd A) vi 2/3 (l) dung dch HCl (dd B) c 1(l) dung dch HCl mi (dd C). Ly 1/10 (l) dd C tc dng vi dung dch AgNO3 d th thu c 8,61g kt ta.
a) Tnh nng mol/l ca dd C.
b) Tnh nng mol/l ca dd A v dd B. Bit nng mol/l dd A = 4 nng d mol/l dd B.
p s: Nng mol/l ca dd B l 0,3M v ca dd A l 1,2M.
Bi 6: Trn 200ml dung dch HNO3 (dd X) vi 300ml dung dch HNO3 (dd Y) c dung dch (Z). Bit rng dung dch (Z) tc dng va vi 7g CaCO3.
a) Tnh nng mol/l ca dung dch (Z).
b) Ngi ta c th iu ch dung dch (X) t dung dch (Y) bng cch thm H2O vo dung dch (Y) theo t l th tch: VHO : Vdd(Y) = 3:1.
Tnh nng mol/l dung dch (X) v dung dch (Y)? Bit s pha trn khng lm thay i ng k th tch dung dch.
p s:
a) CMdd(Z) = 0,28M
b) Nng mol/l ca dung dch (X) l 0,1M v ca dung dch (Y) l 0,4M.
Bi 7: trung ho 50ml dung dch NaOH 1,2M cn V(ml) dung dch H2SO4 30% (D = 1,222g/ml). Tnh V?
p s: Th tch dung dch H2SO4 30% cn ly l 8,02 ml.
Bi 8: Cho 25g dung dch NaOH 4% tc dng vi 51g dung dch H2SO4 0,2M, c khi lng ring D = 1,02 g/ml. Tnh nng % cc cht sau phn ng.
p s:
Nng % ca dung dch Na2SO4 l 1,87%
Nng % ca dung dch NaOH (d) l 0,26%
Bi 9:Trn ln 100ml dung dch NaHSO4 1M vi 100ml dung dch NaOH 2M c dung dch A.
a) Vit phng trnh ho hc xy ra.
b) C cn dung dch A th thu c hn hp nhng cht no? Tnh khi lng ca mi cht.
p s: b) Khi lng cc cht sau khi c cn.
Khi lng mui Na2SO4 l 14,2g
Khi lng NaOH(cn d) l 4 g
Bi 10: Khi trung ho 100ml dung dch ca 2 axit H2SO4 v HCl bng dung dch NaOH, ri c cn th thu c 13,2g mui khan. Bit rng c trung ho 10 ml dung dch 2 axit ny th cn va 40ml dung dch NaOH 0,5M. Tnh nng mol/l ca mi axit trong dung dch ban u.
p s: Nng mol/l ca axit H2SO4 l 0,6M v ca axit HCl l 0,8M
Bi 11: Tnh nng mol/l ca dung dch H2SO4 v dung dch NaOH bit rng:
C 30ml dung dch H2SO4 c trung ho ht bi 20ml dung dch NaOH v 10ml dung dch KOH 2M.Ngc li: 30ml dung dch NaOH c trung ho ht bi 20ml dung dch H2SO4 v 5ml dung dch HCl 1M.
p s: Nng mol/l ca dd H2SO4 l 0,7M v ca dd NaOH l 1,1M.
Hng dn gii bi ton nng bng phng php i s:
Th d: Tnh nng ban u ca dung dch H2SO4 v dung dch NaOH bit rng:
- Nu 3 lt dung dch NaOH vo 2 lt dung dch H2SO4 th sau phn ng dung dch c tnh kim vi nng 0,1M.
- Nu 2 lt dung dch NaOH vo 3 lt dung dch H2SO4 th sau phn ng dung dch c tnh axit vi nng 0,2M.
Bi gii
PTHH: 2NaOH + H2SO4 -> Na2SO4 + 2H2O
Gi nng dung dch xt l x v nng dung dch axit l y th:
* Trong trng hp th nht lng kim cn li trong dung dch l
0,1 . 5 = 0,5mol.
Lng kim tham gia phn ng l: 3x - 0,5 (mol)
Lng axt b trung ho l: 2y (mol)
Theo PTP s mol xt ln hn 2 ln H2SO4Vy 3x - 0,5 = 2y.2 = 4y hay 3x - 4y = 0,5 (1)
* Trong trng hp th 2 th lng a xt d l 0,2.5 = 1mol
Lng a xt b trung ho l 3y - 1 (mol)
Lng xt tham gia phn ng l 2x (mol). Cng lp lun nh trn ta c:
3y - 1 = . 2x = x hay 3y - x = 1 (2)
T (1) v (2) ta c h phng trnh bc nht:
Gii h phng trnh ny ta c x = 1,1 v y = 0,7.
Vy, nng ban u ca dung dch H2SO4 l 0,7M ca dung dch NaOH l 1,1M.Bi 12: Tnh nng mol/l ca dung dch NaOH v dung dch H2SO4. Bit nu ly 60ml dung dch NaOH th trung ho hon ton 20ml dung dch H2SO4. Nu ly 20ml dung dch H2SO4 tc dng vi 2,5g CaCO3 th mun trung ho lng axit cn d phi dng ht 10ml dung dch NaOH trn.
p s: Nng mol/l ca dd H2SO4 l 1,5M v ca dd NaOH l 1,0M.
Bi 13: Tnh nng mol/l ca dung dch HNO3 v dung dch KOH. Bit
20ml dung dch HNO3 c trung ho ht bi 60ml dung dch KOH.
20ml dung dch HNO3 sau khi tc dng ht vi 2g CuO th c trung ho ht bi 10ml dung dch KOH.
p s: Nng ca dung dch HNO3 l 3M v ca dung dch KOH l 1M.
Bi 14: C 2 dung dch H2SO4 l A v B.
a) Nu 2 dung dch A v B c trn ln theo t l khi lng 7:3 th thu c dung dch C c nng 29%. Tnh nng % ca dd A v dd B. Bit nng dd B bng 2,5 ln nng dd A.
b) Ly 50ml dd C (D = 1,27g/ml) cho phn ng vi 200ml dd BaCl2 1M. Tnh khi lng kt ta v nng mol/l ca dd E cn li sau khi tch ht kt ta, gi s th tch dd thay i khng ng k.
Hng dn:
a/ Gi s c 100g dd C. c 100g dd C ny cn em trn 70g dd A nng x% v 30g dd B nng y%. V nng % dd C l 29% nn ta c phng trnh:
mH2SO4(trong dd C) = + = 29 (I)
Theo bi ra th: y = 2,5x (II)
Gii h (I, II) c: x% = 20% v y% = 50%
b/ nH2SO4( trong 50ml dd C ) = = = 0,1879 mol
nBaCl2 = 0,2 mol > nH2SO4. Vy axit phn ng ht
mBaSO4 = 0,1879 . 233 = 43,78g
Dung dch cn li sau khi tch ht kt ta c cha 0,3758 mol HCl v 0,2 0,1879 = 0,0121 mol BaCl2 cn d.
Vy nng ca dd HCl l 1,5M v ca dd BaCl2 l 0,0484M
Bi 15: Trn dd A cha NaOH v dd B cha Ba(OH)2 theo th tch bng nhau c dd C. Trung ho 100ml dd C cn ht 35ml dd H2SO4 2M v thu c 9,32g kt ta. Tnh nng mol/l ca cc dd A v B. Cn trn bao nhiu ml dd B vi 20ml dd A ho tan va ht 1,08g bt Al.
p s: nH2SO4 = 0,07 mol; nNaOH = 0,06 mol; nBa(OH)2 = 0,04 mol.
CM(NaOH) = 1,2M; CM(Ba(OH)) = 0,8M.
Cn trn 20ml dd NaOH v 10ml dd Ba(OH)2 ho tan ht 1,08g bt nhm.
Chuyn 4:
Xc nh cng thc ho hcPhng php 1: Xc nh cng thc ho hc da trn biu thc i s.
* Cch gii:
Bc 1: t cng thc tng qut.
Bc 2: Lp phng trnh(T biu thc i s)
Bc 3: Gii phng trnh -> Kt lun
Cc biu thc i s thng gp.
Cho bit % ca mt nguyn t.
Cho bit t l khi lng hoc t l %(theo khi lng cc nguyn t).
Cc cng thc bin i.
Cng thc tnh % ca nguyn t trong hp cht.
CTTQ AxBy AxBy%A = .100% --> =
Cng thc tnh khi lng ca nguyn t trong hp cht.
CTTQ AxBy AxBy mA = nAB.MA.x --> =
Lu :
xc nh nguyn t kim loi hoc phi kim trong hp cht c th phi lp bng xt ho tr ng vi nguyn t khi ca kim loi hoc phi kim .
Ho tr ca kim loi (n): 1 n 4, vi n nguyn. Ring kim loi Fe phi xt thm ho tr 8/3.
Ho tr ca phi kim (n): 1 n 7, vi n nguyn.
Trong oxit ca phi kim th s nguyn t phi kim trong oxit khng qu 2 nguyn t.
Bi tp p dng:
Bi 1: Mt oxit nit(A) c cng thc NOx v c %N = 30,43%. Tm cng thc ca (A).
p s: NO2
Bi 2: Mt oxit st c %Fe = 72,41%. Tm cng thc ca oxit.
p s: Fe3O4 Bi 3: Mt oxit ca kim loi M c %M = 63,218. Tm cng thc oxit.
p s: MnO2Bi 4: Mt qung st c cha 46,67% Fe, cn li l S.
a) Tm cng thc qung.
b) T qung trn hy iu ch 2 kh c tnh kh.
p s:
a) FeS2b) H2S v SO2.
Bi 5: Oxit ng c cng thc CuxOy v c mCu : mO = 4 : 1. Tm cng thc oxit.
p s: CuO
Bi 6: Oxit ca kim loi M. Tm cng thc ca oxit trong 2 trng hp sau:
a) mM : mO = 9 : 8
b) %M : %O = 7 : 3
p s:
a) Al2O3
b) Fe2O3
Bi 7: Mt oxit (A) ca nit c t khi hi ca A so vi khng kh l 1,59. Tm cng thc oxit A.
p s: NO2Bi 8: Mt oxit ca phi kim (X) c t khi hi ca (X) so vi hiro bng 22. Tm cng thc (X).
p s:
TH1: CO2TH2: N2O
Phng php 2: Xc nh cng thc da trn phn ng.
Cch gii:
Bc 1: t CTTQ
Bc 2: Vit PTHH.
Bc 3: Lp phng trnh ton hc da vo cc n s theo cch t.
Bc 4: Gii phng trnh ton hc.
Mt s gi :
Vi cc bi ton c mt phn ng, khi lp phng trnh ta nn p dng nh lut t l.
Tng qut:
C PTHH: aA + bB -------> qC + pD (1)
Chun b: a b.MB q.22,4
cho: nA p nB p VC (l ) ktc
Theo(1) ta c:
= =
Bi tp p dng:
Bi 1: t chy hon ton 1gam nguyn t R. Cn 0,7 lit oxi(ktc), thu c hp cht X. Tm cng thc R, X.
p s: R l S v X l SO2
Bi 2: Kh ht 3,48 gam mt oxit ca kim loi R cn 1,344 lit H2 (ktc). Tm cng thc oxit.
y l phn ng nhit luyn.
Tng qut:
Oxit kim loi A + (H2, CO, Al, C) ---> Kim loi A + (H2O, CO2, Al2O3, CO hoc
CO2)
iu kin: Kim loi A l kim loi ng sau nhm.
p s: Fe3O4Bi 3: Nung ht 9,4 gam M(NO3)n thu c 4 gam M2On. Tm cng thc mui nitrat
Hng dn:
Phn ng nhit phn mui nitrat.
Cng thc chung:
-----M: ng trc Mg---> M(NO2)n (r) + O2(k)M(NO3)3(r) -----t------ -----M: ( t Mg --> Cu)---> M2On (r) + O2(k) + NO2(k)
-----M: ng sau Cu------> M(r) + O2(k) + NO2(k)
p s: Cu(NO3)2.
Bi 4: Nung ht 3,6 gam M(NO3)n thu c 1,6 gam cht rn khng tan trong nc. Tm cng thc mui nitrat em nung.
Hng dn: Theo ra, cht rn c th l kim loi hoc oxit kim loi. Gii bi ton theo 2 trng hp.
Ch :
TH: Rn l oxit kim loi.
Phn ng: 2M(NO3)n (r) ----t----> M2Om (r) + 2nO2(k) + O2(k)
Hoc 4M(NO3)n (r) ----t----> 2M2Om (r) + 4nO2(k) + (2n m)O2(k)
iu kin: 1 n m 3, vi n, m nguyn dng.(n, m l ho tr ca M )
p s: Fe(NO3)2Bi 5: t chy hon ton 6,8 gam mt hp cht v c A ch thu c 4,48 lt SO2(ktc) v 3,6 gam H2O. Tm cng thc ca cht A.
p s: H2S
Bi 6: Ho tan hon ton 7,2g mt kim loi (A) ho tr II bng dung dch HCl, thu c 6,72 lit H2 (ktc). Tm kim loi A.
p s: A l Mg
Bi 7: Cho 12,8g mt kim loi R ho tr II tc dng vi clo va th thu c 27g mui clorua. Tm kim loi R.
p s: R l Cu
Bi 8: Cho 10g st clorua(cha bit ho tr ca st ) tc dng vi dung dch AgNO3 th thu c 22,6g AgCl(r) (khng tan). Hy xc nh cng thc ca mui st clorua.
p s: FeCl2
Bi 9: Ho tan hon ton 7,56g mt kim loi R cha r ho tr vo dung dch axit HCl, th thu c 9,408 lit H2 (ktc). Tm kim loi R.
p s: R l Al
Bi 10: Ho tan hon ton 8,9g hn hp 2 kim loi A v B c cng ho tr II v c t l mol l 1 : 1 bng dung dch HCl dng d thu c 4,48 lit H2(ktc). Hi A, B l cc kim loi no trong s cc kim loi sau y: ( Mg, Ca, Ba, Fe, Zn, Be )
p s:A v B l Mg v Zn.
Bi 11: Ho tan hon ton 5,6g mt kim loi ho tr II bng dd HCl thu c 2,24 lit H2(ktc). Tm kim loi trn.
p s: Fe
Bi 12: Cho 4,48g mt oxit ca kim loi ho tr tc dng ht 7,84g axit H2SO4. Xc nh cng thc ca oxit trn.
p s: CaO
Bi 13: ho tan 9,6g mt hn hp ng mol (cng s mol) ca 2 oxit kim loi c ho tr II cn 14,6g axit HCl. Xc nh cng thc ca 2 oxit trn. Bit kim loi ho tr II c th l Be, Mg, Ca, Fe, Zn, Ba.
p s: MgO v CaO
Bi 14: Ho tan hon ton 6,5g mt kim loi A cha r ho tr vo dung dch HCl th thu c 2,24 lit H2(ktc). Tm kim loi A.
p s: A l Zn
Bi 15: C mt oxit st cha r cng thc, chia oxit ny lm 2 phn bng nhau.
a/ ho tan ht phn 1 cn dng 150ml dung dch HCl 1,5M.
b/ Cho lung kh H2 d i qua phn 2 nung nng, phn ng xong thu c 4,2g st.
Tm cng thc ca oxit st ni trn.
p s: Fe2O3
Bi 16: Kh hon ton 4,06g mt oxit kim loi bng CO nhit cao thnh kim loi. Dn ton b kh sinh ra vo bnh ng nc vi trong d, thy to thnh 7g kt ta. Nu ly lng kim loi sinh ra ho tan ht vo dung dch HCl d th thu c 1,176 lit kh H2 (ktc). Xc nh cng thc oxit kim loi.
Hng dn:
Gi cng thc oxit l MxOy = amol. Ta c a(Mx +16y) = 4,06
MxOy + yCO -----> xM + yCO2 a ay ax ay (mol)
CO2 + Ca(OH)2 ----> CaCO3 + H2O
ay ay ay (mol)
Ta c ay = s mol CaCO3 = 0,07 mol.---> Khi lng kim loi = M.ax = 2,94g.
2M + 2nHCl ----> 2MCln + nH2 ax 0,5nax (molTa c: 0,5nax = 1,176 :22,4=0,0525molhaynax=0,105Lptl:
EMBED Equation.3 =28.Vy M = 28n ---> Ch c gi tr n = 2 v M = 56 l ph hp. Vy M l Fe. Thay n = 2 ---> ax = 0,0525.
Ta c: = = = ----> x = 3 v y = 4. Vy cng thc oxit l Fe3O4.
Chuyn 5:
Bi ton v oxit v hn hp oxit
Tnh cht:
Oxit baz tc dng vi dung dch axit.
Oxit axit tc dng vi dung dch baz.
Oxit lng tnh va tc dng vi dung dch axit, va tc dng dung dch baz.
Oxit trung tnh: Khng tc dng c vi dung dch axit v dung dch baz.
Cch lm:
Bc 1: t CTTQ
Bc 2: Vit PTHH.
Bc 3: Lp phng trnh ton hc da vo cc n s theo cch t.
Bc 4: Gii phng trnh ton hc.
Bc 5: Tnh ton theo yu cu ca bi.
A - Ton oxit baz
Bi tp p dng:
Bi 1: Cho 4,48g mt oxit ca kim loi ho tr tc dng ht 7,84g axit H2SO4. Xc nh cng thc ca oxit trn.
p s: CaO
Bi 2: Ho tan hon ton 1 gam oxit ca kim loi R cn dng 25ml dung dch hn hp gm axit H2SO4 0,25M v axit HCl 1M. Tm cng thc ca oxit trn.
p s: Fe2O3Bi 3: C mt oxit st cha r cng thc, chia oxit ny lm 2 phn bng nhau.
a/ ho tan ht phn 1 cn dng150ml dung dch HCl 1,5M.
b/ Cho lung kh H2 d i qua phn 2 nung nng, phn ng xong thu c 4,2g st.
Tm cng thc ca oxit st ni trn.
p s: Fe2O3Bi 4: Ho tan hon ton 20,4g oxit kim loi A, ho tr III trong 300ml dung dch axit H2SO4 th thu c 68,4g mui khan. Tm cng thc ca oxit trn.
p s:
Bi 5: ho tan hon ton 64g oxit ca kim loi ho tr III cn va 800ml dung dch axit HNO3 3M. Tm cng thc ca oxit trn.
p s:
Bi 6: Khi ho tan mt lng ca mt oxit kim loi ho tr II vo mt lng va dung dch axit H2SO4 4,9%, ngi ta thu c mt dung dch mui c nng 5,78%. Xc nh cng thc ca oxit trn.
Hng dn:
t cng thc ca oxit l RO
PTHH: RO + H2SO4 ----> RSO4 + H2O
(MR + 16) 98g (MR + 96)g
Gi s ho tan 1 mol (hay MR + 16)g RO
Khi lng dd RSO4(5,87%) = (MR + 16) + (98 : 4,9).100 = MR + 2016
C% = .100% = 5,87%
Gii phng trnh ta c: MR = 24, kim loi ho tr II l Mg.p s: MgO
Bi 7: Ho tan hon ton mt oxit kim loi ho tr II bng dung dch H2SO4 14% va th thu c mt dung dch mui c nng 16,2%. Xc nh cng thc ca oxit trn.
p s: MgO
B - bi ton v oxit axit
Bi tp 1: Cho t t kh CO2 (SO2) vo dung dch NaOH(hoc KOH) th c cc PTHH xy ra:
CO2 + 2NaOH Na2CO3 + H2O ( 1 )Sau khi s mol CO2 = s mol NaOH th c phn ng.
CO2 + NaOH NaHCO3 ( 2 )
Hng gii: xt t l s mol vit PTHH xy ra.
t T =
Nu T 1 th ch c phn ng ( 2 ) v c th d CO2.
Nu T 2 th ch c phn ng ( 1 ) v c th d NaOH.
Nu 1 < T < 2 th c c 2 phn ng ( 1 ) v ( 2 ) trn hoc c th vit nh sau:
CO2 + NaOH NaHCO3 ( 1 ) / tnh theo s mol ca CO2.V sau : NaOH d + NaHCO3 Na2CO3 + H2O ( 2 ) /Hoc da vo s mol CO2 v s mol NaOH hoc s mol Na2CO3 v NaHCO3 to thnh sau phn ng lp cc phng trnh ton hc v gii.
t n x,y ln lt l s mol ca Na2CO3 v NaHCO3 to thnh sau phn ng.
Bi tp p dng:
1/ Cho 1,68 lit CO2 (ktc) sc vo bnh ng dd KOH d. Tnh nng mol/lit ca mui thu c sau phn ng. Bit rng th tch dd l 250 ml.
2/ Cho 11,2 lit CO2 vo 500ml dd NaOH 25% (d = 1,3g/ml). Tnh nng mol/lit ca dd mui to thnh.
3/ Dn 448 ml CO2 (ktc) sc vo bnh cha 100ml dd KOH 0,25M. Tnh khi lng mui to thnh.Bi tp 2: Cho t t kh CO2 (SO2) vo dung dch Ca(OH)2 (hoc Ba(OH)2) th c cc phn ng xy ra:
Phn ng u tin to ra mui trung ho trc.
CO2 + Ca(OH)2 CaCO3 + H2O ( 1 )Sau khi s mol CO2 = 2 ln s mol ca Ca(OH)2 th c phn ng
2CO2 + Ca(OH)2 Ca(HCO3)2 ( 2 )Hng gii : xt t l s mol vit PTHH xy ra:
t T =
Nu T 1 th ch c phn ng ( 1 ) v c th d Ca(OH)2.
Nu T 2 th ch c phn ng ( 2 ) v c th d CO2.
Nu 1 < T < 2 th c c 2 phn ng (1) v (2) trn hoc c th vit nh sau:CO2 + Ca(OH)2 CaCO3 + H2O ( 1 ) tnh theo s mol ca Ca(OH)2 .
CO2 d + H2O + CaCO3 Ca(HCO3)2 ( 2 ) !
Hoc da vo s mol CO2 v s mol Ca(OH)2 hoc s mol CaCO3 to thnh sau phn ng lp cc phng trnh ton hc v gii.
t n x, y ln lt l s mol ca CaCO3 v Ca(HCO3)2 to thnh sau phn ng.
Bi tp p dng:
Bi 1: Ho tan 2,8g CaO vo nc ta c dung dch A.
a/ Cho 1,68 lit kh CO2 hp th hon ton vo dung dch A. Hi c bao nhiu gam kt ta to thnh.
b/ Nu cho kh CO2 sc qua dung dch A v sau khi kt thc th nghim thy c 1g kt ta th c bao nhiu lt CO2 tham gia phn ng. ( cc th tch kh o ktc )
p s:
a/ mCaCO3 = 2,5g
b/ TH1: CO2 ht v Ca(OH)2 d. ---> VCO = 0,224 lit
TH2: CO2 d v Ca(OH)2 ht ----> VCO = 2,016 lit
Bi 2:Dn 10 lt hn hp kh gm N2 v CO2 (ktc) sc vo 2 lit dung dch Ca(OH)2 0,02M, thu c 1g kt ta. Hy xc nh % theo th tch ca kh CO2 trong hn hp.
p s:
TH1: CO2 ht v Ca(OH)2 d. ---> VCO = 0,224 lit v % VCO = 2,24%
TH2: CO2 d v Ca(OH)2 ht ----> VCO = 1,568 lit v % VCO = 15,68%
Bi 3: Dn V lit CO2(ktc) vo 200ml dung dch Ca(OH)2 1M, thu c 10g kt ta. Tnh v.
p s:
TH1: CO2 ht v Ca(OH)2 d. ---> VCO = 2,24 lit.
TH2: CO2 d v Ca(OH)2 ht ----> VCO = 6,72 lit.
Bi 4: Cho m(g) kh CO2 sc vo 100ml dung dch Ca(OH)2 0,05M, thu c 0,1g cht khng tan. Tnh m.
p s:
TH1: CO2 ht v Ca(OH)2 d. ---> mCO2 = 0,044g
TH2: CO2 d v Ca(OH)2 ht ----> mCO2 = 0,396g
Bi 5: Phi t bao nhiu gam cacbon khi cho kh CO2 to ra trong phn ng trn tc dng vi 3,4 lit dung dch NaOH 0,5M ta c 2 mui vi mui hiro cacbonat c nng mol bng 1,4 ln nng mol ca mui trung ho.
p s:
V th tch dung dch khng thay i nn t l v nng cng chnh l t l v s mol. ---> mC = 14,4g.
Bi 6: Cho 4,48 lit CO2(ktc) i qua 190,48ml dung dch NaOH 0,02% c khi lng ring l 1,05g/ml. Hy cho bit mui no c to thnh v khi lng lf bao nhiu gam.
p s: Khi lng NaHCO3 to thnh l: 0,001.84 = 0,084g
Bi 7: Thi 2,464 lit kh CO2 vo mt dung dch NaOH th c 9,46g hn hp 2 mui Na2CO3 v NaHCO3. Hy xc nh thnh phn khi lng ca hn hp 2 mui . Nu mun ch thu c mui NaHCO3 th cn thm bao nhiu lt kh cacbonic na.
p s: 8,4g NaHCO3 v 1,06g Na2CO3. Cn thm 0,224 lit CO2.
Bi 8: t chy 12g C v cho ton b kh CO2 to ra tc dng vi mt dung dch NaOH 0,5M. Vi th tch no ca dung dch NaOH 0,5M th xy ra cc trng hp sau:
a/ Ch thu c mui NaHCO3(khng d CO2)?
b/ Ch thu c mui Na2CO3(khng d NaOH)?
c/ Thu c c 2 mui vi nng mol ca NaHCO3 bng 1,5 ln nng mol ca Na2CO3?
Trong trng hp ny phi tip tc thm bao nhiu lit dung dch NaOH 0,5M na c 2 mui c cng nng mol.
p s:
a/ nNaOH = nCO2 = 1mol ---> Vdd NaOH 0,5M = 2 lit.
b/ nNaOH = 2nCO= 2mol ---> Vdd NaOH 0,5M = 4 lit.
c/
t a, b ln lt l s mol ca mui NaHCO3 v Na2CO3.
Theo PTHH ta c:
nCO2 = a + b = 1mol (I)
V nng mol NaHCO3 bng 1,5 ln nng mol Na2CO3 nn.
= 1,5 ---> a = 1,5b (II)
Gii h phng trnh (I, II) ta c: a = 0,6 mol, b = 0,4 mol
nNaOH = a + 2b = 0,6 + 2 x 0,4 = 1,4 mol ---> Vdd NaOH 0,5M = 2,8 lit.Gi x l s mol NaOH cn thm v khi ch xy ra phn ng.
NaHCO3 + NaOH ---> Na2CO3 + H2O
x(mol) x(mol) x(mol)
nNaHCO3 (cn li) = (0,6 x) mol
nNa2CO3 (sau cng) = (0,4 + x) mol
V bi cho nng mol 2 mui bng nhau nn s mol 2 mui phi bng nhau.
(0,6 x) = (0,4 + x) ---> x = 0,1 mol NaOH
Vy s lit dung dch NaOH cn thm l: Vdd NaOH 0,5M = 0,2 lit.
Bi 9: Sc x(lit) CO2 (ktc) vo 400ml dung dch Ba(OH)2 0,5M th thu c 4,925g kt ta. Tnh x.
p s:
TH1: CO2 ht v Ca(OH)2 d. ---> VCO = 0,56 lit.
TH2: CO2 d v Ca(OH)2 ht ----> VCO = 8,4 lit.
C - Ton hn hp oxit.
Cc bi ton vn dng s mol trung bnh v xc nh khong s mol ca cht.
1/ i vi cht kh. (hn hp gm c 2 kh)
Khi lng trung bnh ca 1 lit hn hp kh ktc:
MTB =
Khi lng trung bnh ca 1 mol hn hp kh ktc:
MTB =
Hoc: MTB = (n l tng s mol kh trong hn hp)Hoc: MTB = (x1l % ca kh th nht)Hoc: MTB = dhh/kh x . Mx2/ i vi cht rn, lng. MTB ca hh =
Tnh cht 1:
MTB ca hh c gi tr ph thuc vo thnh phn v lng cc cht thnh phn trong hn hp.Tnh cht 2:
MTB ca hh lun nm trong khong khi lng mol phn t ca cc cht thnh phn nh nht v ln nht.
Mmin < nhh < MmaxTnh cht 3:
Hn hp 2 cht A, B c MA < MB v c thnh phn % theo s mol l a(%) v b(%)
Th khong xc nh s mol ca hn hp l.
< nhh <
Gi s A hoc B c % = 100% v cht kia c % = 0 hoc ngc li.
Lu :
- Vi bi ton hn hp 2 cht A, B (cha bit s mol) cng tc dng vi 1 hoc c 2 cht X, Y ( bit s mol). bit sau phn ng ht A, B hay X, Y cha. C th gi thit hn hp A, B ch cha 1 cht A hoc B
- Vi MA < MB nu hn hp ch cha A th:
nA = > nhh =
Nh vy nu X, Y tc dng vi A m cn d, th X, Y s c d tc dng ht vi hn hp A, B
Vi MA < MB, nu hn hp ch cha B th:
nB = < nhh =
Nh vy nu X, Y tc dng cha vi B th cng khng tc dng ht vi hn hp A, B.
Ngha l sau phn ng X, Y ht, cn A, B d.3/ Khi lng mol trung bnh ca mt hn hp ()
Khi lng mol trung bnh (KLMTB) ca mt hn hp l khi lng ca 1 mol hn hp .
= = (*)
Trong :
mhh l tng s gam ca hn hp.
nhh l tng s mol ca hn hp.
M1, M2, ..., Mi l khi lng mol ca cc cht trong hn hp.
n1, n2, ..., ni l s mol tng ng ca cc cht.
Tnh cht: Mmin < < Mmaxi vi cht kh v th tch t l vi s mol nn (*) c vit li nh sau:
= (**)
T (*) v (**) d dng suy ra:
= M1x1 + M2x2 + ... + Mixi (***)
Trong : x1, x2, ..., xi l thnh phn phn trm (%) s mol hoc th tch (nu hn hp kh) tng ng ca cc cht v c ly theo s thp phn, ngha l: 100% ng vi x = 1.
50% ng vi x = 0,5.
Ch : Nu hn hp ch gm c hai cht c khi lng mol tng ng M1 v M2 th cc cng thc (*), (**) v (***) c vit di dng:
(*) = (*)/(**) = (**)/(***)
EMBED Equation.3 = M1x + M2(1 - x) (***)/ Trong : n1, V1, x l s mol, th tch, thnh phn % v s mol hoc th tch (hn hp kh) ca cht th nht M1. n gin trong tnh ton thng thng ngi ta chn M1 > M2.
Nhn xt: Nu s mol (hoc th tch) hai cht bng nhau th = v ngc li.
Bi tp p dng:
Bi 1: Ho tan 4,88g hn hp A gm MgO v FeO trong 200ml dung dch H2SO4 0,45M(long) th phn ng va , thu c dung dch B.
a/ Tnh khi lng mi oxit c trong hn hp A.
b/ tc dng va vi 2 mui trong dung dch B cn dng V(lit) dung dch NaOH 0,2M, thu c kt ta gm 2 hirxit kim loi. Lc ly kt ta, em nung trong khng kh n khi lng khng i thu c m gam cht rn khan(phn ng hon ton). Tnh V v m.
p s:
a/ mMgO = 2g v mFeO = 2,88g
b/ Vdd NaOH 0,2M = 0,9 lit v mrn = 5,2g.
Bi 2: ho tan 9,6g mt hn hp ng mol (cng s mol) ca 2 oxit kim loi c ho tr II cn 14,6g axit HCl. Xc nh cng thc ca 2 oxit trn. Bit kim loi ho tr II c th l Be, Mg, Ca, Fe, Zn, Ba.
p s: MgO v CaO
Bi 3: Kh 9,6g mt hn hp gm Fe2O3 v FeO bng H2 nhit cao, ngi ta thu c Fe v 2,88g H2O.
a/ Vit cc PTHH xy ra.
b/ Xc nh thnh phn % ca 2 oxit trong hn hp.
c/ Tnh th tch H2(ktc) cn dng kh ht lng oxit trn.
p s:
b/ % Fe2O3 = 57,14% v % FeO = 42,86%
c/ VH = 3,584 lit
Bi 4: Cho X v Y l 2 oxit ca cng mt kim loi M. Bit khi ho tan cng mt lng oxit X nh nhau n hon ton trong HNO3 v HCl ri c cn dung dch th thu c nhng lng mui nitrat v clorua ca kim loi M c cng ho tr. Ngoi ra, khi lng mui nitrat khan ln hn khi lng mui clorua khan mt lng bng 99,38% khi lng oxit em ho tan trong mi axit. Phn t khi ca oxit Y bng 45% phn t khi ca oxit X. Xc nh cc oxit X, Y.
p s:
Bi 5: Kh 2,4g hn hp gm CuO v Fe2O3 bng H2 nhit cao th thu c 1,76g hn hp 2 kim loi. em hn hp 2 kim loi ho tan bng dd axit HCl th thu c V(lit) kh H2.
a/ Xc nh % v khi lng ca mi oxit trong hn hp.
b/ Tnh V ( ktc).
p s:
a/ % CuO = 33,33% ; % Fe2O3 = 66,67%
b/ VH = 0,896 lit.
Bi 6: Ho tan 26,2g hn hp Al2O3 v CuO th cn phi dng va 250ml dung dch H2SO4 2M. Xc nh % khi lng mi cht trong hn hp.
p s: % Al2O3 = 38,93% v % CuO = 61,07%.
Bi 7: Cho hn hp A gm 16g Fe2O3 v 6,4g CuO vo 160ml dung dch H2SO4 2M. Sau phn ng thy cn m gam rn khng tan.
a/ Tnh m.
b/ Tnh th tch dung dch hn hp gm axit HCl 1M v axit H2SO4 0,5M cn dng phn ng ht hn hp A.
p s:
a/ 3,2 < m < 4,8
b/ Vdd hh axit = 0,06 lit.
Chuyn 6:
Axit tc dng vi kim loi
Cch lm:
1/ Phn loi axit:
Axit loi 1: Tt c cc axit trn( HCl, H2SO4long, HBr,...), tr HNO3 v H2SO4 c.
Axit loi 2: HNO3 v H2SO4 c.
2/ Cng thc phn ng: gm 2 cng thc.
Cng thc 1: Kim loi phn ng vi axit loi 1.
Kim loi + Axit loi 1 ----> Mui + H2iu kin:
Kim loi l kim loi ng trc H trong dy hot ng ho hc Bktp.
Dy hot ng ho hc Bktp.
K, Na, Ba, Ca, Mg, Al, Zn, Fe, Ni, Sn, Pb, H, Cu, Hg, Ag, Pt, Au.
c im:
Mui thu c c ho tr thp(i vi kim loi c nhiu ho tr)
Th d: Fe + 2HCl ----> FeCl2 + H2 Cu + HCl ----> Khng phn ng.
Cng thc 2: Kim loi phn ng vi axit loi 2:
Kim loi + Axit loi 2 -----> Mui + H2O + Sn phm kh.
c im:
Phn ng xy ra vi tt c cc kim loi (tr Au, Pt).
Mui c ho tr cao nht(i vi kim loi a ho tr)
Bi tp p dng:
Bi 1: Ho tan ht 25,2