phaûn öùng nhieät luyeän
DESCRIPTION
Baøi 8:. phaûn öùng nhieät luyeän. GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN). GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN). Caàn nhôù. Coâng thöùc vieát phaûn öùng nhieät luyeän. H 2. H 2 O. t o. CO. CO 2. - PowerPoint PPT PresentationTRANSCRIPT
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GV. NGUYEÃN TAÁN TRUNG(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN) GV. NGUYEÃN TAÁN TRUNGGV. NGUYEÃN TAÁN TRUNG(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)
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Caàn nhôù
Coâng thöùc vieát phaûn öùng nhieät luyeän
Oxit KL A +H2COAlC
to
KL A
H2O
+ CO2
Al2O3
CO2;CO
Ñieàu kieän KL A phaûi ñöùng sau Al trong daõy hoaït ñoäng hoaù hoïc BeâKeâtoâp
(K, Na, Ca, Mg, , Mn, Zn, Cr, Fe, …)AlVí duï:
CuO + CO toCu + CO2
MgO + CO to
Khoâng pöù ( vì Mg ñöùng tröôùc Al)
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Baøi taäp aùp duïng 1Khöû heát 6,4 gam MxOy , thaáy caàn 2,688 lit CO (ñkc)Tìm coâng thöùc cuûa oxit ?
Pöù: MxOy + CO to
M + CO2 (1)xy y
GiaûinCO = 2,688/ 22,4 = 0,12 (mol)
6,4gam 0,12mol
y
Theo (1) coù:Mx + 16 y
6,4=
y
0,12 M = 37,33. y/x=18,67.18,67. 2y/x 2y/x
2y/x
MVôùi 2y/x laø hoaù trò M
1 2 3
18,67 37,33 56
Choïn: 2y/x = 3 M = 56 M : FeM : Fe oxit: FeFe22OO33
(Mx +16y)
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Daãn CO dö qua oáng söù nung Daãn CO dö qua oáng söù nung noùng chöùa 21,6 g hoãn hôïp: MgO, noùng chöùa 21,6 g hoãn hôïp: MgO, FeFe33OO44 . Sau pöù thu ñöôïc m gam . Sau pöù thu ñöôïc m gam
raén vaø hh khí. Daãn heát khí vaøo raén vaø hh khí. Daãn heát khí vaøo dd Ca(OH)dd Ca(OH)22 dö , thaáy coù 14 gam dö , thaáy coù 14 gam keát tuûa. Tính m? keát tuûa. Tính m?
Aùp duïng 2: (ÑHKTCN-2000)
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to
CO2CO
m g raén
m = ? ddCa(OH)ddCa(OH)22 dö dö
14 gam keát tuûagam keát tuûa
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to
CO2CO
m g raén
m = ? ddCa(OH)ddCa(OH)22 dö dö
14 gam keát tuûagam keát tuûa
soá mol COsoá mol CO2 2 = haèng soá= haèng soá
Caàn thaáy : CO CO khoâng pöù vôùi ddCa(OH)ddCa(OH)22
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Tính löôïng COCO22:
CO2CO
ddCa(OH)ddCa(OH)22 dö dö
14 gam keát tuûagam keát tuûa
CO2 + Ca(OH)2 CaCO3 + H2O (1)
Theo ñeà ta coù keát tuûa laøTheo ñeà ta coù keát tuûa laø: CaCO3
soá mol keát tuûa soá mol keát tuûa CaCO3 baèng 14/100 = 0,14 Ta coù phaûn öùng taïo keát tuûa:Ta coù phaûn öùng taïo keát tuûa:
0,14 mol0,14 mol
Vaäy: soá mol CO2 baèng 0,14 mol0,14 mol
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to CO2
m g raén
m = ?
0,14 mol0,14 mol
Mg ñöùng tröôùc Al, neân MgO khoâng pö vaø Hieäu suaát pöù ñaït 100%, neân Fe3O4
Chuyeån heát thaønh Fe
MgOFe
Sai soùt cuûa thí sinh : MgO pöù thaønh Mg
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to CO2
m g raén
m = ?
0,14 mol0,14 mol
MgOFe
mMgO
mFe
m Fe3O4
nFepöpöùù
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to CO2
m g raén
m = ?
0,14 mol0,14 mol
MgOFe
mMgO
mFe
m Fe3O4
nFepöpöùù
Fe3O4 + 4 CO 3 Fe + 4 CO2 (2)
Theo ñeà ta coù Pöù:
0,14 mol0,105 mol0,035 mol
Theo (2) m Fe3O4 = 8,12
mFe= 5,88
mMgO = 21,6 – 8,12
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to CO2
m g raén
m = ?
0,14 mol0,14 mol
MgOFe
Fe3O4 + 4 CO 3 Fe + 4 CO2 (2)
Theo ñeà ta coù Pöù:
0,14 mol0,105 mol0,035 mol
Toùm laïi ta coù:m MgO = 13,48
mFe= 5,88 m = 13,48+5,8 8
Vaäy: m = 19,36 gam
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Toùm taét aùp duïng 2:
21,6 gam
MgO
Fe3O4
+ CO (dö)
to CO2
m gam raén
m = ? 0,14 mol0,14 mol
hhA
Neáu thí sinh kheùo nhìn, thì seõ thaáy: Baøi naøy coøn 2 caùch giaûi nhanh hôn nhieàu !
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21,6 gam
Theo ñeà ta deã daøng thaáy baøi toaùn treân coù 4 thnaøh phaàn
Toùm taét aùp duïng 2:MgO
Fe3O4
+ CO (dö)
to CO2
m gam raén
m = ? 0,14 mol0,14 mol
hhA
ÑLBTKL
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21,6 gam
Toùm taét aùp duïng 2:MgO
Fe3O4
+ CO (dö)
to CO2
m gam raén
m = ? 0,14 mol0,14 mol
hhA
Theo ñeà ta coù sô ñoà hôïp thöùc:
hhA + CO Raén + CO2 (1)0,14 mol0,14 mol
Theo (1), ÑLBTKL coù:m hhA + m CO m Raén +m CO2
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21,6 gam
Toùm taét aùp duïng 2:MgO
Fe3O4
+ CO (dö)
to CO2
m gam raén
m = ? 0,14 mol0,14 mol
hhA
hhA + CO Raén + CO2 (1)0,14 mol0,14 mol
Theo (1), ÑLBTKL coù:m hhA + m CO m Raén +m CO2
m Raén = 21,6 + 0,14.28 –0,14. 44 = 19, 36 g
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Daãn CO dö qua oáng söù nung Daãn CO dö qua oáng söù nung noùng chöùa 21,6 g hoãn hôïp: CuO, noùng chöùa 21,6 g hoãn hôïp: CuO, FeFe22OO33 . Sau moät thôøi gian thu . Sau moät thôøi gian thu
ñöôïc m gam raén vaø hh khí. ñöôïc m gam raén vaø hh khí. Daãn heát khí vaøo dd Ca(OH)Daãn heát khí vaøo dd Ca(OH)22 dö , dö , thaáy coù 14 gam keát tuûa. thaáy coù 14 gam keát tuûa. Tính m? Tính m?
Aùp duïng 3:
Sau moät thôøi gianSau moät thôøi gian
Hieäu suaát thöôøng < 100%
ÑLBTKL