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k hoch bi dng hsg
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DY HOT NG HO HC CA KIM LOI.
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.
1. PHNG PHP I STrong 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%
2. 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
3. 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:
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 mol
56g 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 6: 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 Cu Ag (cht oxh mnh)
0,1 0,1 ( mol )
Ag C Tnh oxi ho mnh hn Cu 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: 6,72g
4. 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)
5. 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.
6/ 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
7/ 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 SO28/ 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 DCHMt s cng thc tnh cn nh:
Cng thc lin h: C% = Hoc S =
Cng thc tnh nng mol/lit: CM = =
* 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)Cng thc tnh tan: S = . 100Cng 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 // 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) gmct CuSO4(c trong dd CuSO4 8%) l = (g)Ta c phng trnh: + = 89,6Gii 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: 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: 961,75gBi 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 DCHBi 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) = 7,94MBi 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 ) = 2Mb/ 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 DCHLoi 1: Bi ton pha long hay c dc mt dung dch.
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.
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 chtDung 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: 375mlBi 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 = 0,4MNng 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 352,94gBi 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 thVy 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.
Tnh nng mol/lit ca dung dch D.
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:
CM(dd D) = 0,2M
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 = 0,1MVy 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 295,2gBi 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.
Tnh nng mol/l ca dd C.
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.
Tnh nng mol/l ca dung dch (Z).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:
CMdd(Z) = 0,28MNng 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.
Vit phng trnh ho hc xy ra.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 0,8MBi 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 axt 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.
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.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 htmBaSO4 = 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: NO2Bi 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.Tm cng thc qung.
T qung trn hy iu ch 2 kh c tnh kh.
p s:
FeS2H2S v SO2.
Bi 5: Oxit ng c cng thc CuxOy v c mCu : mO = 4 : 1. Tm cng thc oxit.p s: CuOBi 6: Oxit ca kim loi M. Tm cng thc ca oxit trong 2 trng hp sau:
mM : mO = 9 : 8
%M : %O = 7 : 3
p s:
Al2O3 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:
=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 OXITTnh 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 BAZBi 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 ROKhi lng dd RSO4(5,87%) = (MR + 16) + (98 : 4,9).100 = MR + 2016C% = .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: MgOB - BI TON V OXIT AXITBi 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.
V d 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.
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:
(*) = (*)/(**) = (**)/(***)
= 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 < 4,8b/ Vdd hh axit = 0,06 lit.
CHUYN 6:AXIT TC DNG VI KIM LOICch 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
Dy hot ng ho hc
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 25,2g kim loi R trong dung dch axit HCl, sau phn ng thu c 1,008 lit H2 (ktc). Xc nh kim loi R.
p s:Bi 2: Ho tan hon ton 6,5g mt kim loi A cha r ho tr vo dung dch axit HCl, th thu c 2,24 lit H2 (ktc). Xc nh kim loi A.p s: A l Zn.
Bi 3: Cho 10g mt hn hp gm Fe v Cu tc dng vi dung dch axit HCl, th thu c 3,36 lit kh H2 (ktc). Xc nh thnh phn % v khi lng ca mi kim loi trong hn hp u.p s: % Fe = 84%, % Cu = 16%.
Bi 4: Cho 1 hn hp gm Al v Ag phn ng vi dung dch axit H2SO4 thu c 5,6 lt H2 (ktc). Sau phn ng th cn 3g mt cht rn khng tan. Xc nh thnh phn % theo khi lng cu mi kim loi trong hn hp ban u.p s: % Al = 60% v % Ag = 40%.Bi 5: Cho 5,6g Fe tc dng vi 500ml dung dch HNO3 0,8M. Sau phn ng thu c V(lit) hn hp kh A gm N2O v NO2 c t khi so vi H2 l 22,25 v dd B.
a/ Tnh V (ktc)?b/ Tnh nng mol/l ca cc cht c trong dung dch B.
Hng dn:
Theo bi ra ta c:
nFe = 5,6 : 56 = 0,1 mol
nHNO3 = 0,5 . 0,8 = 0,4 mol
Mhh kh = 22,25 . 2 = 44,5
t x, y ln lt l s mol ca kh N2O v NO2.PTHH xy ra:
8Fe + 30HNO3 ----> 8Fe(NO3)3 + 3N2O + 15H2O (1)
8mol 3mol8x/3 x
Fe + 6HNO3 -----> Fe(NO3)3 + 3NO2 + 3H2O (2)
1mol 3moly/3 y
T l th tch cc kh trn l:Gi a l thnh phn % theo th tch ca kh N2O.
Vy (1 a) l thnh phn % ca kh NO2. Ta c: 44a + 46(1 a) = 44,5
a = 0,75 hay % ca kh N2O l 75% v ca kh NO2 l 25%
T phng trnh phn ng kt hp vi t l th tch ta c:
x = 3y (I) ---> y = 0,012 v x = 0,036
8x/3 + y/3 = 0,1 (II) Vy th tch ca cc kh thu c ktc l:VNO = 0,81(lit) v VNO= 0,27(lit)
Theo phng trnh th:
S mol HNO3 (phn ng) = 10nNO + 2n NO= 10.0,036 + 2.0,012 = 0,384 molS mol HNO3 (cn d) = 0,4 0,384 = 0,016 mol
S mol Fe(NO3)3 = nFe = 0,1 mol
Vy nng cc cht trong dung dch l:
CM(Fe(NO3)3) = 0,2MCM(HNO3)d = 0,032M
Bi 6: ho tan 4,48g Fe phi dng bao nhiu ml dung dch hn hp HCl 0,5M v H2SO4 0,75M.
Hng dn: Gi s phi dng V(lit) dung dch hn hp gm HCl 0,5M v H2SO4 0,75MS mol HCl = 0,5V (mol)
S mol H2SO4 = 0,75V (mol)
S mol Fe = 0,08 mol
PTHH xy ra:
Fe + 2HCl ---> FeCl2 + H2Fe + H2SO4 ---> FeSO4 + H2
Theo phng trnh ta c: 0,25V + 0,75V = 0,08---> V = 0,08 : 1 = 0,08 (lit)
Bi 7: ho tan 4,8g Mg phi dng bao nhiu ml dung dch hn hp HCl 1,5M v H2SO4 0,5M.
a/ Tnh th tch dung dch hn hp axit trn cn dng.
b/ Tnh th tch H2 thu c sau phn ng ktc.
p s:
a/ Vhh dd axit = 160ml.b/ Th tch kh H2 l 4,48 lit.
Bi 8: Ho tan 2,8g mt kim loi ho tr (II) bng mt hn hp gm 80ml dung dch axit H2SO4 0,5M v 200ml dung dch axit HCl 0,2M. Dung dch thu c c tnh axit v mun trung ho phi dng 1ml dung dch NaOH 0,2M. Xc nh kim loi ho tr II em phn ng.Hng dn:
Theo bi ra ta c:
S mol ca H2SO4 l 0,04 mol
S mol ca HCl l 0,04 mol
S mol ca NaOH l 0,02 mol
t R l KHHH ca kim loi ho tr IIa, b l s mol ca kim loi R tc dng vi axit H2SO4 v HCl.
Vit cc PTHH xy ra.
Sau khi kim loi tc dng vi kim loi R. S mol ca cc axit cn li l:S mol ca H2SO4 = 0,04 a (mol)S mol ca HCl = 0,04 2b (mol)
Vit cc PTHH trung ho:
T PTP ta c:
S mol NaOH phn ng l: (0,04 2b) + 2(0,04 a) = 0,02
---> (a + b) = 0,1 : 2 = 0,05Vy s mol kim loi R = (a + b) = 0,05 mol
---> MR = 2,8 : 0,05 = 56 v R c ho tr II ---> R l Fe.
Bi 9: Chia 7,22g hn hp A gm Fe v R (R l kim loi c ho tr khng i) thnh 2 phn bng nhau:
Phn 1: Phn ng vi dung dch HCl d, thu c 2,128 lit H2(ktc)Phn 2: Phn ng vi HNO3, thu c 1,972 lit NO(ktc)
a/ Xc nh kim loi R.
b/ Tnh thnh phn % theo khi lng mi kim loi trong hn hp A.
Hng dn:a/ Gi 2x, 2y (mol) l s mol Fe, R c trong hn hp A --> S mol Fe, R trong 1/2 hn hp A l x, y.
Vit cc PTHH xy ra:
Lp cc phng trnh ton hc;
mhh A = 56.2x + 2y.MR (I)nH= x + ny/2 = 0,095 (II)nNO = x + ny/3 = 0,08 (III)
Gii h phng trnh ta c: MR = 9n (vi n l ho tr ca R)Lp bng: Vi n = 3 th MR = 27 l ph hp. Vy R l nhm(Al)
b/ %Fe = 46,54% v %Al = 53,46%.CHUYN 7: AXIT TC DNG VI BAZ(bi ton hn hp axit tc dng vi hn hp baz)* Axit n: HCl, HBr, HI, HNO3. Ta c nH = nA xit
* Axit a: H2SO4, H3PO4, H2SO3. Ta c nH = 2nA xit hoc nH = 3nA xit* Baz n: KOH, NaOH, LiOH. Ta c nOH = 2nBaZ
* Baz a: Ba(OH)2, Ca(OH)2. Ta c nOH = 2nBaZ
PTHH ca phn ng trung ho: H + OH H2O
*Lu : trong mt hn hp m c nhiu phn ng xy ra th phn ng trung ho c u tin xy ra trc.
Cch lm:
Vit cc PTHH xy ra.
t n s nu bi ton l hn hp.Lp phng trnh ton hc
Gii phng trnh ton hc, tm n.
Tnh ton theo yu cu ca bi.
Lu :Khi gp dung dch hn hp cc axit tc dng vi hn hp cc baz th dng phng php t cng thc tng ng cho axit v baz.
t th tch dung dch cn tm l V(lit)
Tm V cn nh: nHX = nMOH.Bi tp:
Cho t t dung dch H2SO4 vo dung dch NaOH th c cc phn ng xy ra:
Phn ng u tin to ra mui trung ho trc.
H2SO4 + 2NaOH Na2SO4 + H2O ( 1 )
Sau khi s mol H2SO4 = s mol NaOH th c phn ng
H2SO4 + NaOH NaHSO4 + H2O ( 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 H2SO4.
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.
Ngc li:Cho t t dung dch NaOH vo dung dch H2SO4 th c cc phn ng xy ra:
Phn ng u tin to ra mui axit trc.
H2SO4 + NaOH NaHSO4 + H2O ( 1 ) !V sau NaOH d + NaHSO4 Na2SO4 + H2O ( 2 ) !Hoc da vo s mol H2SO4 v s mol NaOH hoc s mol Na2SO4 v NaHSO4 to thnh sau phn ng lp cc phng trnh ton hc v gii.
t n x, y ln lt l s mol ca Na2SO4 v NaHSO4 to thnh sau phn ng.Bi tp p dng:
Bi 1: Cn dng bao nhiu ml dung dch KOH 1,5M trung ho 300ml dung dch A cha H2SO4 0,75M v HCl 1,5M.p s: Vdd KOH 1,5M = 0,6(lit)
Bi 2: trung ho 10ml dung dch hn hp axit gm H2SO4 v HCl cn dng 40ml dung dch NaOH 0,5M. Mt khc ly 100ml dung dch axit em trung ho mt lng xt va ri c cn th thu c 13,2g mui khan. Tnh nng mol/l ca mi axt trong dung dch ban u.Hng dn:
t x, y ln lt l nng mol/lit ca axit H2SO4 v axit HCl
Vit PTHH.Lp h phng trnh:
2x + y = 0,02 (I)
142x + 58,5y = 1,32 (II)
Gii phng trnh ta c:Nng ca axit HCl l 0,8M v nng ca axit H2SO4 l 0,6M.
Bi 3: Cn bao nhiu ml dung dch NaOH 0,75M trung ho 400ml hn hp dung dch axit gm H2SO4 0,5M v HCl 1M.p s: VNaOH = 1,07 lit
Bi 4: trung ho 50ml dung dch hn hp axit gm H2SO4 v HCl cn dng 200ml dung dch NaOH 1M. Mt khc ly 100ml dung dch hn hp axit trn em trung ho vi mt lng dung dch NaOH va ri c cn th thu c 24,65g mui khan. Tnh nng mol/l ca mi axit trong dung dch ban u.
p s: Nng ca axit HCl l 3M v nng ca axit H2SO4 l 0,5MBi 5: Mt dung dch A cha HCl v H2SO4 theo t l s mol 3:1, bit 100ml dung dch A c trung ho bi 50ml dung dch NaOH c cha 20g NaOH/lit.a/ Tnh nng mol ca mi axit trong A.
b/ 200ml dung dch A phn ng va vi bao nhiu ml dung dch baz B cha NaOH 0,2M v Ba(OH)2 0,1M.c/ Tnh tng khi lng mui thu c sau phn ng gia 2 dung dch A v B.
Hng dn:
a/ Theo bi ra ta c:
nHCl : nH2SO4 = 3:1
t x l s mol ca H2SO4 (A1), th 3x l s mol ca HCl (A2)
S mol NaOH c trong 1 lt dung dch l:
nNaOH = 20 : 40 = 0,5 ( mol )
Nng mol/lit ca dung dch NaOH l:
CM ( NaOH ) = 0,5 : 1 = 0,5M
S mol NaOH dung trong phn ng trung ho l:
nNaOH = 0,05 * 0,5 = 0,025 mol
PTHH xy ra :
HCl + NaOH NaCl + H2O (1)
3x 3x
H2SO4 + 2NaOH Na2SO4 + 2H2O (2) x 2x
T PTHH 1 v 2 ta c : 3x + 2x = 0,025 5x = 0,025 x = 0,005
Vy nH2SO4 = x = 0,005 mol
nHCl = 3x = 3*0,005 = 0,015 mol
Nng ca cc cht c dung dch A l:
CM ( A1 ) = 0,005 : 0,1 = 0,05M v CM ( A2 ) = 0,015 : 0,1 = 0,15M
b/ t HA l axit i din cho 2 axit cho. Trong 200 ml dung dch A c:
nHA = nHCl + 2nH2SO4 = 0,015*0,2 + 0,05*0,2*2 = 0,05 mol
t MOH l baz i din v V(lit) l th tch ca dung dch B cha 2 baz cho:
nMOH = nNaOH + 2nBa(OH)2 = 0,2 V + 2 * 0,1 V = 0,4 V
PTP trung ho: HA + MOH MA + H2O (3)
Theo PTP ta c nMOH = nHA = 0,05 mol
Vy: 0,4V = 0,05 V = 0,125 lit = 125 ml
c/ Theo kt qu ca cu b ta c:
nNaOH = 0,125 * 0,2 = 0,025 mol v nBa(OH)2 = 0,125 * 0,1 = 0,0125 mol
nHCl = 0,2 * 0,015 = 0,03 mol v nH2SO4 = 0,2 * 0,05 = 0,01 mol
V P trn l phn ng trung ho nn cc cht tham gia phn ng u tc dng ht nn d phn ng no xy ra trc th khi lng mui thu c sau cng vn khng thay i hay n c bo ton.
mhh mui = mSO + mNa + mBa + mCl = 0,01*96 + 0,025*23 + 0,0125*137 + 0,03*35,5 = 0,96 + 1,065 + 0,575 + 1,7125 = 4,3125 gam
Hoc t:
n NaOH = 0,125 * 0,2 = 0,025 mol mNaOH = 0,025 * 40 = 1g
n Ba(OH)2 = 0,125 * 0,1 = 0,0125 mol mBa (OH)= 0,0125 * 171 = 2,1375g
n HCl = 0,2 * 0,015 = 0,03 mol mHCl = 0,03 * 36,5 = 1,095g
n H2SO4 = 0,2 * 0,05 = 0,01 mol mHSO= 0,01 * 98 = 0,98g
p dng l BTKL ta c: mhh mui = mNaOH + mBa (OH)+ mHCl + mHSO- mHO
V s mol: nH2O = nMOH = nHA = 0,05 mol. mHO = 0,05 *18 = 0,9g
Vy ta c: mhh mui = 1 + 2,1375 + 1,095 + 0,98 0,9 = 4,3125 gam.
Bi 6: Tnh nng mol/l ca dung dch H2SO4 v NaOH bit rng:30ml dung dch NaOH c trung ho ht bi 200ml dung dch NaOH v 10ml dung dch KOH 2M.
30ml dung dch NaOH c trung ho ht bi 20ml dung dch H2SO4 v 5ml dung dch HCl 1M.
p s: Nng ca axit H2SO4 l 0,7M v nng ca dung dch NaOH l 1,1M.Bi 7: 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 dung dch HNO3 l 3M v nng dung dch KOH l 1M.Bi 8: Mt dd A cha HNO3 v HCl theo t l 2 : 1 (mol).
a/ Bit rng khi cho 200ml dd A tc dng vi 100ml dd NaOH 1M, th lng axit d trong A tc dng va vi 50ml Ba(OH)2 0,2M. Tnh nng mol/lit ca mi axit trong dd A.
b/ Nu trn 500ml dd A vi 100ml dd B cha NaOH 1M v Ba(OH)2 0,5M. Hi dd thu c c tnh axit hay baz ?
c/ Phi thm vo dd C bao nhiu lit dd A hoc B c c dd D trung ho.
/S: a/ CM [ HCl ] = 0,2M ; CM [ HSO] = 0,4M
b/ dd C c tnh axit, s mol axit d l 0,1 mol.
c/ Phi thm vo dd C vi th tch l 50 ml dd B.
Bi 9: Ho tan 8g hn hp 2 hiroxit kim loi kim nguyn cht thnh 100ml dung dch X.
a/ 100ml dung dch X c trung ho va bi 800ml dung dch axit axtic CH3COOH, cho 14,72g hn hp mui. Tm tng s mol hai hiroxit kim loi kim c trong 8g hn hp. Tm nng mol/l ca dung dch CH3COOH.
b/ Xc nh tn hai kim loi kim bit chng thuc 2 chu k k tip trong bng tun hon. Tm khi lng tng hiroxit trong 8g hn hp.
Hng dn:
Gi A, B l k hiu ca 2 kim loi kim ( cng chnh l k hiu KLNT ).
Gi s MA < MB v R l k hiu chung ca 2 kim loi ---> MA < MR < MB
Trong 8g hn hp c a mol ROH.
a/ Nng mol/l ca CH3COOH = 0,16 : 0,8 = 0,2M
b/ MR = 33 ---> MA = 23(Na) v MB = 39(K)
mNaOH = 2,4g v mKOH = 5,6g.
CHUYN 8:
AXIT TC DNG VI MUI1/ Phn loi axitGm 3 loi axit tc dng vi mui.
a/ Axit loi 1:
Thng gp l HCl, H2SO4long, HBr,..
Phn ng xy ra theo c ch trao i.
b/ Axit loi 2:
L cc axit c tnh oxi ho mnh: HNO3, H2SO4c.Phn ng xy ra theo c ch phn ng oxi ho kh.
c/ Axit loi 3:
L cc axit c tnh kh.
Thng gp l HCl, HI, H2S.
Phn ng xy ra theo c ch phn ng oxi ho kh.
2/ Cng thc phn ng.a/ Cng thc 1:
Mui + Axit ---> Mui mi + Axit mi.
iu kin: Sn phm phi c:
Kt ta.
Hoc c cht bay hi(kh).Hoc cht in li yu hn.
c bit: Cc mui sunfua ca kim loi k t Pb tr v sau khng phn ng vi axit loi 1.
V d: Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2 (k) BaCl2 + H2SO4 ---> BaSO4(r) + 2HCl
b/ Cng thc 2:
Mui + Axit loi 2 ---> Mui + H2O + sn phm kh.
iu kin:Mui phi c tnh kh.
Mui sinh ra sau phn ng th nguyn t kim loi trong mui phi c ho tr cao nht.
Ch : C 2 nhm mui em phn ng.Vi cc mui: CO32-, NO3-, SO42-, Cl- .
+ iu kin: Kim loi trong mui phi l kim loi a ho tr v ho tr ca kim loi trong mui trc phi ng khng cao nht.Vi cc mui: SO32-, S2-, S2-.+ Phn ng lun xy ra theo cng thc trn vi tt c cc kim loi.
c/ Cng thc 3:
Thng gp vi cc mui st(III). Phn ng xy ra theo quy tc 2.(l phn ng oxi ho kh)2FeCl3 + H2S ---> 2FeCl2 + S(r) + 2HCl.Ch :
Bi tp: Cho t t dung dch HCl vo Na2CO3 (hoc K2CO3) th c cc PTHH sau:
Giai on 1 Ch c phn ng.
Na2CO3 + HCl NaHCO3 + NaCl ( 1 ) x (mol) x mol x mol
Giai on 2 Ch c phn ng
NaHCO3 + HCl d NaCl + H2O + CO2 ( 2 ) x x x mol
Hoc ch c mt phn ng khi s mol HCl = 2 ln s mol Na2CO3.
Na2CO3 + 2HCl 2NaCl + H2O + CO2 ( 3 )
i vi K2CO3 cng tng t.
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 Na2CO3.
Nu T 2 th ch c phn ng (3) v c th d HCl.
Nu 1 < T < 2 th c c 2 phn ng (1) v (2) trn hoc c th vit nh sau.
t x l s mol ca Na2CO3 (hoc HCl) tham gia phn ng ( 1 )
Na2CO3 + HCl NaHCO3 + NaCl ( 1 ) x (mol) x mol x mol
Na2CO3 + 2HCl 2NaCl + H2O + CO2 ( 2 ) !
Tnh s mol ca Na2CO3 (hoc HCl) tham gia phn ng(2!)da vo bi ra v qua phn ng(1).
Th d: Cho t t dung dch cha x(mol) HCl vo y (mol) Na2CO3 (hoc K2CO3). Hy bin lun v cho bit cc trng hp c th xy ra vit PTHH , cho bit cht to thnh, cht cn d sau phn ng:
TH 1: x < y
C PTHH: Na2CO3 + HCl NaHCO3 + NaCl
x x x x mol
- Dung dch sau phn ng thu c l: s mol NaHCO3 = NaCl = x (mol)
- Cht cn d l Na2CO3 (y x) mol
TH 2: x = y
C PTHH : Na2CO3 + HCl NaHCO3 + NaCl
x x x x mol
- Dung dch sau phn ng thu c l: NaHCO3 ; NaCl
- C 2 cht tham gia phn ng u ht.
TH 3: y < x < 2y
C 2 PTHH: Na2CO3 + HCl NaHCO3 + NaCl
y y y y mol
sau phn ng (1) dung dch HCl cn d (x y) mol nn tip tc c phn ng
NaHCO3 + HCl NaCl + H2O + CO2 (x y) (x y) (x y) (x y)
- Dung dch thu c sau phn ng l: c x(mol) NaCl v (2y x)mol NaHCO3 cn dTH 4: x = 2y
C PTHH: Na2CO3 + 2HCl 2NaCl + H2O + CO2
y 2y 2y y mol
- Dung dch thu c sau phn ng l: c 2y (mol) NaCl, c 2 cht tham gia phn ng u ht.TH 5: x > 2y
C PTHH: Na2CO3 + 2HCl 2NaCl + H2O + CO2
y 2y 2y y mol
- Dung dch thu c sau phn ng l: c 2y (mol) NaCl v cn d (x 2y) mol HCl.
Bi tp 5: Cho t t dung dch HCl vo hn hp mui gm NaHCO3 v Na2CO3 (hoc KHCO3 v K2CO3) th c cc PTHH sau:
t x, y ln lt l s mol ca Na2CO3 v NaHCO3.
Giai on 1: Ch c Mui trung ho tham gia phn ng.
Na2CO3 + HCl NaHCO3 + NaCl ( 1 ) x (mol) x mol x mol
Giai on 2: Ch c phn ng
NaHCO3 + HCl d NaCl + H2O + CO2 ( 2 ) (x + y) (x + y) (x + y) mol
i vi K2CO3 v KHCO3 cng tng t.
Bi tp: Cho t t dung dch HCl vo hn hp mui gm Na2CO3; K2CO3; NaHCO3 th c cc PTHH sau:
t x, y, z ln lt l s mol ca Na2CO3; NaHCO3 v K2CO3.
Giai on 1: Ch c Na2CO3 v K2CO3 phn ng.
Na2CO3 + HCl NaHCO3 + NaCl ( 1 ) x (mol) x x x
K2CO3 + HCl KHCO3 + KCl ( 2 ) z (mol) z z z
Giai on 2: c cc phn ng
NaHCO3 + HCl d NaCl + H2O + CO2 ( 3 ) (x + y) (x + y) (x + y) mol
KHCO3 + HCl d KCl + H2O + CO2 ( 4 ) z (mol) z z mol
Bi tp: Cho t t dung dch HCl vo dung dch NaAlO2 th c cc PTHH sau.
NaAlO2 + HCl + H2O Al(OH)3 + NaCl ( 1 )Al(OH)3 + 3HCl d AlCl3 + 3H2O ( 2 )NaAlO2 + 4HCl AlCl3 + NaCl + 2H2O ( 3 )Bi tp p dng:Bi 1: Ho tan Na2CO3 vo V(ml) hn hp dung dch axit HCl 0,5M v H2SO4 1,5M th thu c mt dung dch A v 7,84 lit kh B (ktc). C cn dung dch A thu c 48,45g mui khan.a/ Tnh V(ml) hn hp dung dch axit dng?
b/ Tnh khi lng Na2CO3 b ho tan.Hng dn:
Gi s phi dng V(lit) dung dch gm HCl 0,5M v H2SO4 1,5M.
Na2CO3 + 2HCl ---> 2NaCl + H2O + CO20,25V 0,5V 0,5V 0,25V (mol)Na2CO3 + H2SO4 ---> Na2SO4 + H2O + CO2
1,5V 1,5V 1,5V 1,5V (mol)Theo bi ra ta c:
S mol CO2 = 0,25V + 1,5V = 7,84 : 22,4 = 0,35 (mol) (I)
Khi lng mui thu c: 58,5.0,5V + 142.1,5V = 48,45 (g) (II)V = 0,2 (l) = 200ml.
S mol Na2CO3 = s mol CO2 = 0,35 mol
Vy khi lng Na2CO3 b ho tan:
mNa2CO3 = 0,35 . 106 = 37,1g.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: Ho tan CaCO3 vo 100ml hn hp dung dch gm axit HCl v axit H2SO4 th thu c dung dch A v 5,6 lit kh B (ktc), c cn dung dch A th thu c 32,7g mui khan.
a/ Tnh nng mol/l mi axit trong hn hp dung dch ban u.
b/ Tnh khi lng CaCO3 dng.
Bi 4: Cho 4,2g mui cacbonat ca kim loi ho tr II. Ho tan vo dung dch HCl d, th c kh thot ra. Ton b lng kh c hp th vo 100ml dung dch Ba(OH)2 0,46M thu c 8,274g kt ta. Tm cng thc ca mui v kim loi ho tr II.
p s:
TH1 khi Ba(OH)2 d, th cng thc ca mui l: CaCO3 v kim loi ho tr II l Ca.
TH2 khi Ba(OH)2 thiu, th cng thc ca mui l MgCO3 v kim loi ho tr II l Mg.
Bi 5: Cho 1,16g mui cacbonat ca kim loi R tc dng ht vi HNO3, thu c 0,448 lit hn hp G gm 2 kh c t khi hi so vi hiro bng 22,5. Xc nh cng thc mui (bit th tch cc kh o ktc).Hng dn:Hn hp G gm c kh CO2 v kh cn li l kh X.
C dhh G/ H= 22,5 --> MTB ca hh G = 22,5 . 2 = 45
M MCO= 44 < 45 ---> Mkh X > 45. nhn thy trong cc kh ch c NO2 v SO2 c khi lng phn t ln hn 45. Trong trng hp ny kh X ch c th l NO2.t a, b ln lt l s mol ca CO2 v NO2.
Ta c h nhh G = a + b = 0,02 a = 0,01 MTB hh G = = 45 b = 0,01
PTHH:
R2(CO3)n + (4m 2n)HNO3 ---> 2R(NO3)m + (2m 2n)NO2 + nCO2 + (2m n)H2O.2MR + 60n 2m 2n 1,16g 0,01 mol
Theo PTHH ta c: = ----> MR = 116m 146nLp bng: iu kin 1 n m 4 n12233
m32334
MR56
Ch c cp nghim n = 2, m = 3 --> MR = 56 l ph hp. Vy R l FeCTHH: FeCO3Bi 6: Cho 5,25g mui cacbonat ca kim loi M tc dng ht vi HNO3, thu c 0,336 lit kh NO v V lit CO2. Xc nh cng thc mui v tnh V. (bit th tch cc kh c o ktc)p s: Gii tng t bi 3 ---> CTHH l FeCO3 Bi 7: Ho tan 2,84 gam hn hp 2 mui CaCO3 v MgCO3 bng dung dch HCl d thu c 0,672 lt kh CO2 (ktc). Tnh thnh phn % s mol mi mui trong hn hp.
Bi gii
Cc PTHH xy ra:
CaCO3 + 2HCl CaCl2 + CO2 + H2O (1)MgCO3 + 2HCl MgCl2 + CO2 + H2O (2)T (1) v (2) nhh = nCO = = 0,03 (mol)
Gi x l thnh phn % s mol ca CaCO3 trong hn hp th (1 - x) l thnh phn % s mol ca MgCO3.
Ta c 2 mui = 100x + 84(1 - x) = x = 0,67
% s mol CaCO3 = 67% ; % s mol MgCO3 = 100 - 67 = 33%.
Bi 8: Ho tan 174 gam hn hp gm 2 mui cacbonat v sunfit ca cng mt kim loi kim vo dung dch HCl d. Ton b kh thot ra c hp th ti thiu bi 500 ml dung dch KOH 3M.
a/ Xc nh kim loi kim.
b/ Xc nh % s mol mi mui trong hn hp ban u.
Bi gii
cc PTHH xy ra:
M2CO3 + 2HCl 2MCl + CO2 + H2O (1)M2SO3 + 2HCl 2MCl + SO2 + H2O (2)Ton b kh CO2 v SO2 hp th mt lng ti thiu KOH sn phm l mui axit.
CO2 + KOH KHCO3 (3)SO2 + KOH KHSO3 (4)T (1), (2), (3) v (4)
suy ra: n 2 mui = n 2 kh = nKOH = = 1,5 (mol)
2 mui = = 116 (g/mol) 2M + 60 < < 2M + 80
18 < M < 28, v M l kim loi kim, vy M = 23 l Na.
b/ Nhn thy 2 mui = = 116 (g/mol).
% nNaCO = nNaSO = 50%.
CHUYN 9: DUNG DCH BAZ TC DNG VI MUI.Bi tp: Cho t t dung dch NaOH (hoc KOH) hay Ba(OH)2 (hoc Ca(OH)2) vo dung dch AlCl3 th c cc PTHH sau.
3NaOH + AlCl3 Al(OH)3 + 3NaCl ( 1 )NaOH d + Al(OH)3 NaAlO2 + 2H2O ( 2 )4NaOH + AlCl3 NaAlO2 + 3NaCl + 2H2O ( 3 ) v:
3Ba(OH)2 + 2AlCl3 2Al(OH)3 + 3BaCl2 ( 1 )Ba(OH)2 d + 2Al(OH)3 Ba(AlO2)2 + 4H2O ( 2 )4Ba(OH)2 + 2AlCl3 Ba(AlO2)2 + 3BaCl2 + 4H2O ( 3 )
Ngc li: Cho t t dung dch AlCl3 vo dung dch NaOH (hoc KOH) hay Ba(OH)2 (hoc Ca(OH)2) ch c PTHH sau:AlCl3 + 4NaOH NaAlO2 + 3NaCl + 2H2O
v 2AlCl3 + 4Ba(OH)2 ----> Ba(AlO2)2 + 3BaCl2 + 4H2OBi tp: Cho t t dung dch NaOH (hoc KOH) hay Ba(OH)2 (hoc Ca(OH)2) vo dung dch Al2(SO4)3 th c cc PTHH sau.
6NaOH + Al2(SO4)3 2Al(OH)3 + 3Na2SO4 ( 1 )NaOH d + Al(OH)3 NaAlO2 + 2H2O ( 2 )8NaOH + Al2(SO4)3 2NaAlO2 + 3Na2SO4 + 4H2O ( 3 )V:
3Ba(OH)2 + Al2(SO4)3 2Al(OH)3 + 3BaSO4 ( 1 )Ba(OH)2 d + 2Al(OH)3 Ba(AlO2)2 + 4H2O ( 2 )4Ba(OH)2 + Al2(SO4)3 Ba(AlO2)2 + 3BaSO4 + 4H2O ( 3 )
Ngc li: Cho t t dung dch Al2(SO4)3 vo dung dch NaOH (hoc KOH) hay Ba(OH)2 (hoc Ca(OH)2) th c PTHH no xy ra?Al2(SO4)3 + 8NaOH 2NaAlO2 + 3Na2SO4 + 4H2O (3 )/Al2(SO4)3 + 4Ba(OH)2 Ba(AlO2)2 + 3BaSO4 + 4H2O (3 )// Mt s phn ng c bit:
NaHSO4 (dd) + NaAlO2 + H2O Al(OH)3 + Na2SO4
NaAlO2 + HCl + H2O Al(OH)3 + NaCl
NaAlO2 + CO2 + H2O Al(OH)3 + NaHCO3
Bi tp p dng:
Bi 1: Cho 200ml dung dch NaOH vo 200g dung dch Al2(SO4)3 1,71%. Sau phn ng thu c 0,78g kt ta. Tnh nng mol/l ca dung dch NaOH tham gia phn ng.
p s:
TH1: NaOH thiuS mol NaOH = 3s mol Al(OH)3 = 3. 0,01 = 0,03 mol ---> CM NaOH = 0,15MTH2: NaOH d ---> CM NaOH = 0,35M
Bi 2: Cho 400ml dung dch NaOH 1M vo 160ml dung dch hn hp cha Fe2(SO4)3 0,125M v Al2(SO4)3 0,25M. Sau phn ng tch kt ta em nung n khi lng khng i c cht rn C.
a/ Tnh mrn C.
b/ Tnh nng mol/l ca mui to thnh trong dung dch.
p s:
a/ mrn C = 0,02 . 160 + 0,02 . 102 = 5,24gb/ Nng ca Na2SO4 = 0,18 : 0,56 = 0,32M v nng ca NaAlO2 = 0,07MBi 3: Cho 200g dung dch Ba(OH)2 17,1% vo 500g dung dch hn hp (NH4)2SO4 1,32% v CuSO4 2%. Sau khi kt thc tt c cc phn ng ta thu c kh A, kt ta B v dung dch C.a/ Tnh th tch kh A (ktc)
b/ Ly kt ta B ra sch v nung nhit cao n khi lng khng i th c bao nhiu gam rn?
c/ Tnh nng % ca cc cht trong C.
p s:
a/ Kh A l NH3 c th tch l 2,24 litb/ Khi lng BaSO4 = 0,1125 . 233 = 26,2g v mCuO = 0,0625 . 80 = 5gc/ Khi lng Ba(OH)2 d = 0,0875 . 171 = 14,96gmdd = Tng khi lng cc cht em trn - mkt ta - mkhmdd = 500 + 200 26,21 6,12 1,7 = 666gNng % ca dung dch Ba(OH)2 = 2,25%Bi 4: Cho mt mu Na vo 200ml dung dch AlCl3 thu c 2,8 lit kh (ktc) v mt kt ta A. Nung A n khi lng khng i thu c 2,55 gam cht rn. Tnh nng mol/l ca dung dch AlCl3 .
Hng dn:mrn: Al2O3 --> s mol ca Al2O3 = 0,025 mol ---> s mol Al(OH)3 = 0,05 mol
s mol NaOH = 2s mol H2 = 0,25 mol.
TH1: NaOH thiu, ch c phn ng.
3NaOH + AlCl3 ---> Al(OH)3 + 3NaCl
Khng xy ra v s mol Al(OH)3 to ra trong phn ng > s mol Al(OH)3 cho.TH2: NaOH d, c 2 phn ng xy ra.
3NaOH + AlCl3 ---> Al(OH)3 + 3NaCl
0,15 0,05 0,05 mol
4NaOH + AlCl3 ---> NaAlO2 + 3NaCl + H2O(0,25 0,15) 0,025Tng s mol AlCl3 phn ng 2 phng trnh l 0,075 mol
----> Nng ca AlCl3 = 0,375M
Bi 6: Cho 200ml dung dch NaOH x(M) tc dng vi 120 ml dung dch AlCl3 1M, sau cng thu c 7,8g kt ta. Tnh tr s x?
p s:
TH1: Nng AlCl3 = 1,5M
TH2: Nng AlCl3 = 1,9M
Bi 7: Cho 9,2g Na vo 1
