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phaûn öùng nhieät luyeän

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.

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phaûn öùng nhieät luyeän

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  1. 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)

  2. Caàn nhôù Coâng thöùc vieát phaûn öùng nhieät luyeän • H2 • H2O to • CO • CO2 Oxit KL A KL A + + • Al2O3 • Al • C • CO2;CO • Ñieàu kieän • KL A phaûi ñöùng sau Al • trong daõy hoaït ñoäng hoaù hoïc BeâKeâtoâp Al (K, Na, Ca, Mg, , Mn, Zn, Cr, Fe, …) • Ví duï: to Cu + CO2 CuO + CO  to MgO + CO  Khoâng pöù ( vì Mg ñöùng tröôùc Al)

  3. Baøi taäp aùp duïng 1 Khöû heát 6,4 gam MxOy , thaáy caàn 2,688 lit CO (ñkc) Tìm coâng thöùc cuûa oxit ? • Giaûi nCO = 2,688/ 22,4 = 0,12 (mol) to y M + CO2 (1) x MxOy + CO  y Pöù: (Mx +16y) y 0,12mol 6,4gam y Theo (1) coù: Mx + 16 y = 6,4 0,12 18,67. 2y/x =  M = 37,33. y/x 3 2 1 2y/x 18,67 37,33 56 M Vôùi 2y/x laø hoaù trò M  oxit:Fe2O3  M : Fe  M = 56 Choïn: 2y/x = 3

  4. Aùp duïng 2:(ÑHKTCN-2000) Daãn CO dö qua oáng söù nung noùng chöùa 21,6 g hoãn hôïp: MgO, Fe3O4 . Sau pöù thu ñöôïc m gam raén vaø hh khí. Daãn heát khí vaøo dd Ca(OH)2 dö , thaáy coù 14 gam keát tuûa. Tính m?

  5. MgO + CO (dö) m g raén Fe3O4 CO2 CO ddCa(OH)2 dö to 14 gam keát tuûa • Toùm taét aùp duïng 2: 21,6 gam m = ?

  6. Toùm taét aùp duïng 2: MgO + CO (dö) m g raén Fe3O4 CO2 CO ddCa(OH)2 dö to 21,6 gam 14 gam keát tuûa m = ? soá mol CO2 = haèng soá Caàn thaáy : CO khoâng pöù vôùi ddCa(OH)2

  7. Tính löôïng CO2: CO2 CO ddCa(OH)2 dö 14 gam keát tuûa Theo ñeà ta coù keát tuûa laø: CaCO3 soá mol keát tuûa CaCO3 baèng 14/100 = 0,14 Ta coù phaûn öùng taïo keát tuûa: CO2 + Ca(OH)2  CaCO3+ H2O (1) 0,14 mol 0,14 mol Vaäy: soá mol CO2 baèng0,14 mol

  8. Toùm taét aùp duïng 2: MgO 21,6 gam + CO (dö) CO2 m g raén Fe3O4 to m = ? MgO Fe 0,14 mol Mg ñöùng tröôùc Al, neân MgO khoâng pö vaø Hieäu suaát pöù ñaït 100%, neân Fe3O4 Chuyeån heát thaønh Fe Sai soùt cuûa thí sinh : MgO pöù thaønh Mg

  9. Toùm taét aùp duïng 2: MgO Fe MgO 0,14 mol 21,6 gam + CO (dö) CO2 m g raén Fe3O4 to m = ? pöù mMgO mFe m Fe3O4 nFe

  10. Toùm taét aùp duïng 2: MgO Fe MgO 0,14 mol 21,6 gam + CO (dö) CO2 m g raén Fe3O4 to m = ? pöù mMgO mFe m Fe3O4 = 8,12 mFe= 5,88 m Fe3O4 nFe Theo ñeà ta coù Pöù: Fe3O4 + 4 CO  3 Fe + 4 CO2 (2) 0,035 mol 0,105 mol 0,14 mol  mMgO = 21,6 – 8,12 Theo (2) 

  11. Toùm taét aùp duïng 2: MgO Fe MgO 0,14 mol 21,6 gam + CO (dö) CO2 m g raén Fe3O4 to m = ? mMgO = 13,48 mFe= 5,88 Theo ñeà ta coù Pöù: Fe3O4 + 4 CO  3 Fe + 4 CO2 (2) 0,035 mol 0,105 mol 0,14 mol Toùm laïi ta coù:  m = 13,48+5,8 8 Vaäy: m = 19,36 gam

  12. Toùm taét aùp duïng 2: hhA MgO 0,14 mol 21,6 gam + CO (dö) CO2 m gam raén Fe3O4 m = ? to Neáu thí sinh kheùo nhìn, thì seõ thaáy: Baøi naøy coøn 2 caùch giaûi nhanh hôn nhieàu !

  13. Toùm taét aùp duïng 2: hhA MgO 0,14 mol + CO (dö) CO2 m gam raén Fe3O4 m = ? to ÑLBTKL 21,6 gam Theo ñeà ta deã daøng thaáy baøi toaùn treân coù 4 thnaøh phaàn

  14. Toùm taét aùp duïng 2: hhA MgO 0,14 mol 21,6 gam + CO (dö) CO2 m gam raén Fe3O4 m = ? to Theo ñeà ta coù sô ñoà hôïp thöùc: hhA + CO  Raén + CO2 (1) 0,14 mol 0,14 mol Theo (1), ÑLBTKL coù: m hhA + m CO  m Raén +m CO2

  15. Toùm taét aùp duïng 2: hhA MgO 0,14 mol 21,6 gam + CO (dö) CO2 m gam raén Fe3O4 m = ? to hhA + CO  Raén + CO2 (1) 0,14 mol 0,14 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

  16. Aùp duïng 3: Daãn CO dö qua oáng söù nung noùng chöùa 21,6 g hoãn hôïp: CuO, Fe2O3 . Sau moät thôøi gian thu ñöôïc m gam raén vaø hh khí. Daãn heát khí vaøo dd Ca(OH)2 dö , thaáy coù 14 gam keát tuûa. Tính m? Sau moät thôøi gian Hieäu suaát thöôøng < 100% ÑLBTKL

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