gv. nguyeÃn taÁn trung (trung taâm luyeän thi chaát … · cho 21,84g kali kim loaïi vaøo...
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![Page 1: GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát … · Cho 21,84g kali kim loaïi vaøo 200g moät dung dòch chöùa Fe 2(SO 4) 3 5% , FeSO 4 3,04% vaø Al 2(SO 4) 3 8,55%](https://reader033.vdocuments.pub/reader033/viewer/2022041901/5e6067fb4cf51b6ebc24388e/html5/thumbnails/1.jpg)
GV. NGUYEÃN TAÁN TRUNG(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)
![Page 2: GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát … · Cho 21,84g kali kim loaïi vaøo 200g moät dung dòch chöùa Fe 2(SO 4) 3 5% , FeSO 4 3,04% vaø Al 2(SO 4) 3 8,55%](https://reader033.vdocuments.pub/reader033/viewer/2022041901/5e6067fb4cf51b6ebc24388e/html5/thumbnails/2.jpg)
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Coù 3 coâng thöùc vieát phaûn öùng
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Coâng thöùc 1: Kim loai tan trong H2O
Bazô + H2 (1)KL + H2O →Bazô + Muoái → Bazô môùi+ Muoái môùi (2)
Saûn phaåm cuûa (2) phaûi coù:Chaát keát tuûaChaát bay hôi
Chaát khoù ñieän ly hôn
Muoái pöù: Tan hoaëc ít tan
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NaOH + CuSO4 → Cu(OH)2↓ + Na2SO4
2Na + 2 H2O + CuSO4 → Cu(OH)2↓ + Na2SO4 + H2↑
Ví duï 1:Cho Na phaûn öùng vôùi dung dòchCuSO4. Vieát phöông trình phaûn öùng.
Na + H2O → NaOH + H2↑2 2 2
2
Coâng thöùc 1: Kim loai tan trong H2OBazô + H2 (1)KL + H2O →
Bazô + Muoái → Bazô môùi+ Muoái môùi (2)
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Ví duï 2: (ÑH Noâng Nghieäp 1 – 1997)Cho 21,84g kali kim loaïi vaøo 200g moät dung dòch
chöùa Fe2(SO4)3 5% , FeSO4 3,04% vaø Al2(SO4)3 8,55% veà khoái löôïng.Sau phaûn öùng, loïc taùch, thu ñöôïc keáttuûa A vaø dung dòch B. Nung keát tuûa A trong khoângkhí ñeán khoái löôïng khoâng ñoåi.
1. Vieát phöông trình caùc phaûn öùng hoaù hoïc ñaõ xaûyra.
2. Tính khoái löôïng chaát raén thu ñöôïc sau khi nungkeát tuûa A.3. Tính noàng ñoä phaàn traêm khoái löôïng caùc chaáttaïo thaønh trong dung dòch B.
Fe=56, K=39, S=32, Al=27, O=16, H=1.
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Giaûi:
Soá mol K= 21,8439
=0,56 mol
Soá mol Fe2(SO4)3=5x200100x400
= 0,025 mol
Soá mol FeSO4=3,04x200100x152
Soá mol Al2(SO4)3= 100x3428,55x200
= 0,04 mol
= 0,05 mol
1. Caùc phaûn öùng
Caùc phaûn öùng treân ñöôïc xaùc ñònh chính xaùc nhôø ñònh löôïng sau:Al(OH)3 + KOH = KAlO2 + 2 H2O
Coù theå coù theâm :
K + H2O = KOH + H2↑Fe2(SO4)3 + 6KOH = 2 Fe(OH)3↓ + 3 K2SO4
FeSO4 + 2KOH = Fe(OH)2↓ + K2SO4Al2(SO4)3 + 6 KOH = 2Al(OH)3↓ + 3K2SO4
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Theo ñeà ta co ù Caùc phaûn öùng:
Fe2(SO4)3+6 KOH = 2Fe(OH)3↓ + 3K2SO4 (2)
0,04 0,04 0,040,08
0,025 0,15 0,05 0,075FeSO4 + 2 KOH = Fe(OH)2↓ + K2SO4 (3)
Al2(SO4)3 + 6 KOH = 2Al(OH)3↓ +3 K2SO4 (4)0,05 0,3 0,1 0,15
K + H2O = KOH + H2↑ (1)½0,56 0,56 0,28 molmol
molmol
molmol
molmol(2), (3), (4) ⇒ Soá Σmol KOH pöù = 0,53 mol
⇒ Soá mol KOH dö =0,56 – 0,53 = 0,03 mol
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Al(OH)3 + KOH = KAlO2 + 2 H2O (5)0,03 0,03 0,03 mol
Theo (4), (5) ⇒ Soá mol Al(OH)3 dö = 0,1 – 0,03 = 0,07 mol (*)2. Khi nung keát tuûa A:
0,04 0,02 mol
0,05 0,025 mol
0,07 0,035 mol
2Fe(OH)2 + O2 = Fe2O3 + 2 H2O (6)to½
2Fe(OH)3 = Fe2O3 + 3 H2O (7)to
2Al(OH)3 = Al2O3 + 3H2O (8)to
(2), (3), (4),(*) ⇒Caùc pöù nung keât tuûa taïo raén
Vì sau (4) coøn KOH, neân coù theâm pöù:
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Theo (6), (7), (8) Theo (6), (7), (8) tata cocoùù KhoKhoááii llööôôïïngng chachaáátt raraéénn sausau khikhi nungnung::
Dung dòch B coù: K2SO4 =206,87
174x0,256 =22,29%x100
KAlO2 = 206,8798 x 0,03 x100 =1,42%
mddB = 200+ mK – m ↑ - m↓ = 206,87 gamH2
10,77 gam.
Fe2O3 = 160 x 0,045 = 7,2 gam.Al2O3 = 102 x 0,035 = 3,57 gam.
3.T3.Tíính nh nonoààngng ññooää phaphaàànn traêmtraêm khokhoááii llööôôïïngng cacaùùcc chachaáátt tataïïoothathaøønhnh trongtrong dung dung dòchdòch B.B.
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Ví duï 3: : ((ÑÑH H ThuyThuyûû LôLôïïii –– 1997)1997)Cho 9,2 gam Na vaøo 160 ml dung
dòch coù khoái löôïng rieâng laø 1,25 g/ml chöùaFe2(SO4)3 vôùi noàng ñoä töông öùng laø0,125M vaø 0,25M. Sau phaûn öùng ngöôøi tataùch keát tuûa ra vaø ñem nung ñeán khoáilöôïng khoâng ñoåi.
1. Tính khoái löôïng caùc chaát raén thuñöôïc sau khi nung.
2. Tính noàng ñoä phaàn traêm cuûa caùcmuoái taïo thaønh trong dung dòch.
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9,2Soá mol Na =23
=0,4 mol
Soá mol Fe2(SO4)3 = 0,16x0,125 = 0,02 mol
Soá mol Al2(SO4)3 = 0,16x0,25 = 0,04 molKhoái löôïng 160 ml dd = 160x1,25g/ml = 200 g
0,4 0,4 0,4 0,2 mol
0,040,02 0,12 0,06 mol
Giaûi:
Na + H2O → NaOH + 1/2 H2↑ (1)
Fe2(SO4)3+ 6NaOH → 2Fe(OH)3↓ + 3 Na2SO4 (2)
Caùc phaûn öùng:
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Al2(SO4)3+6NaOH→ 2 Al(OH)3↓ + 3 Na2SO4 (3)0,04 0,24 0,08 0,12 mol
Soá mol NaOH coøn dö = 0,4 – (0,12+0,24)=0,04 mol
VVìì cocoøønn ddöö NaOHNaOH , , neânneân Al(OH)Al(OH)3 3 bòbò tan tan theotheo::Al(OH)3 + NaOH → NaAlO2 + 2 H2O (4)0,04 0,04 0,04 mol
Theo (2), (3), (4) Theo (2), (3), (4) KeKeáátt tutuûûaa thuthu ñöñöôôïïcc gogoààmm::Fe(OH)3: 0,04 molAl(OH)3:0,08 - 0,04 = 0,04 mol
CaCaùùcc phaphaûûnn öùöùngng nungnung kekeáátt tutuûûaa
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0,04 0,02 mol
0,04 0,02 mol1. 1. KhoKhoááii llööôôïïngng chachaáátt raraéénn sausau khikhi nungnung::
KhoKhoááii llööôôïïngng FeFe22OO33 = 160x0,02=3,2 = 160x0,02=3,2 gamgam.KhoKhoááii llööôôïïngng AlAl22OO33 = 102x0,02=2,04 = 102x0,02=2,04 gamgam..
5,24 gam.2. 2. NoNoààngng ññooää % % cacaùùcc muomuoááii trongtrong dung dung dòchdòch::
Khoái löôïng Na2SO4 = 142x0,18=25,56 gam.Khoái löôïng NaAlO2 = 82x0,04=3,28 gam.
2Fe(OH)3 = Fe2O3 + 3 H2Oto
2Al(OH)3 = Al2O3 + 3 H2Oto
(5)
(6)
Theo (5), (6) Theo (5), (6) tata suysuy rara ñöñöôôïïcc:
Theo (2), (3), (4) Theo (2), (3), (4) tata suysuy rara ñöñöôôïïcc khokhoááii llööôôïïngng cacaùùcc muomuoááii:
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VaVaääyy khokhoááii llööôôïïngng dung dung dòchdòch luluùùcc sausau::(9,2 + 200) –(0,4 + 4,28 + 3,12) = 201,4 gam.
201,4C% Na2SO4=
25,56 100 =12,71%
201,4C% NaAlO2=
3,28 100 =1,63%
Tính khoái löôïng dung dòch:
Theo (1), (2), (3), Theo (1), (2), (3), tata ttíínhnh ñöñöôôïïcc khokhoááii llööôôïïngng cacaùùcc chachaáátt::Khoái löôïng H2↑ = 2x0,2 =0,4 gam.Khoái löôïng Fe(OH)3 ↓ = 107x0,04=4,28 gam.Khoái löôïng Al(OH)3 ↓ = 78x0,04=3,12 gam.
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CoângCoâng ththöùöùcc 22: Kim loaïi khoâng tan trong nöôùc.
(cô cheá kim loaïi ñaåy kim loaïi ra khoûi muoái )
KLA + Muoái KLB → KLB + Muoái KLA
ÑÑieieààuu kiekieäänn::
KLA khoâng tan trong nöôùcKL A ñöùng tröôùc KL B ( trong
daõy hoaït ñoäng hoaù hoïc Beâkeâoâp)Muoái :Tan
Ví duï: Zn + CuSO4 → Cu + ZnSO4
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Giaûi:
3,78 gam boät Al phaûn öùng vöøañuû vôùi dung dòch muoái XCl3 taïothaønh dung dòch Y. Khoái löôïng chaáttan trong dung dòch Y giaûm 4,06 gam so vôùi dung dòch XCl3 . XaÙcñònh coâng thöùc cuûa muoái XCL3.
Phaûn öùng:Al + XCl3 → AlCl3 + X (1)
Ví duï 4:(:(ÑÑHQGTP.HCM HQGTP.HCM –– 1998)1998)
3,7827 X+3.35,5 27 +3.35,5
mC.tan giaûm: 4,06 gmC.tan giaûm: X-27 g
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Theo (1) coù:
3,78
27
4,06
X-27=
⇒ X = 56
⇒ X : Fe
⇒ XCl3 : Fe Cl3
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Coâng thöùc 3: khi gaëp saétPöù xaûy ra theo qui taéc α
Oh2 + Kh1 Kh2Oh1+→
TQ:
Fe2+
Fe
Cu2+
Cu
Ag+
Ag
Fe3+
Fe2+
I2
2I-
Daõy ñieän hoaù:
a. Cu+ Fe(NO3)3b. Fe + Fe(NO3)3
Oh1
Kh2Kh1
Oh2
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Ví duï 5: Vieát caùc phaûn öùnga. Cu+ Fe(NO3)3
Cu +2Fe3+ Cu2+2Fe2+ +→
TQ:
Fe2+
Fe
Cu2+
Cu
Ag+
Ag
Fe3+
Fe2+
I2
2I-
Daõy ñieän hoaù:
Cu2+
Fe2+Cu
Fe3+
a. Cu+ Fe(NO3)3b. Fe + Fe(NO3)3
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Ví duï 5: Vieát caùc phaûn öùngb. Fe + Fe(NO3)3
Cu +2Fe3+ 3Fe2+→
TQ:
Fe2+
Fe
Cu2+
Cu
Ag+
Ag
Fe3+
Fe2+
I2
2I-
Daõy ñieän hoaù:
Fe2+
Fe2+Fe
Fe3+
a. Cu+ Fe(NO3)3b. Fe + Fe(NO3)3
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Cho 6,4 gam Cu phaûn öùng ôùi300 gam dd Fe(NO3)3 24,2% thu ñöôïc dd A coù khoái löôïngrieâng baèng 1,446 g/ml. Tínhnoàng ñoä mol/l cuûa dd A.
Aùp duïng 6:
Soá mol Cu = 0,1 (mol)= 0,1 (mol)Soá mol Fe(NOFe(NO33))3 3 = 0,3 (mol)= 0,3 (mol)
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Bñ:
Pöù:Sau: 0,1
0,2 0,10,20,2
0,1
0,1
0
2Fe(NO3)3 + Cu → 2Fe(NO3)2 + Cu(NO3)2 (1)0,3 0,1 0 0 (mol)
(mol)
(mol)
- Theo ñeà ta coù pöù:
Theo (1) ta coù:mdd= 6,4 + 300 =306,4g
⇒Vdd= 306,41,532
Vaäy
= 200 (ml) = 0,2 (lít)
:[Fe(NO3)2]= 1(M) [Cu(NO3)2]= 0,5(M)