{"id":493,"date":"2020-04-24T11:45:33","date_gmt":"2020-04-24T09:45:33","guid":{"rendered":"https:\/\/feb.ch.bme.hu\/?page_id=493"},"modified":"2020-04-28T11:50:17","modified_gmt":"2020-04-28T09:50:17","slug":"feb-es-feladatok-sztochiometria","status":"publish","type":"page","link":"https:\/\/feb.ch.bme.hu\/index.php\/szakmai-napok\/sztochiometria\/feb-es-feladatok-sztochiometria\/","title":{"rendered":"Szt\u00f6chiometria – FEB-es feladatok"},"content":{"rendered":"

Szt\u00f6chiometria – FEB-es feladatok<\/h1>\n

Vissza a szt\u00f6chiometria t\u00e9mak\u00f6rre!<\/a><\/p>\n

3\/f-1<\/strong> Oxid\u00e1ci\u00f3s sz\u00e1mok alapj\u00e1n rendezd a k\u00f6vetkez\u0151 egyenleteket!<\/p>\n

    \n
  1. FeCl3<\/sub> + SnCl2<\/sub> = FeCl2<\/sub> + SnCl4<\/sub><\/li>\n
  2. MnO4<\/sub>–<\/sup> + Cl– <\/sup>+ H+<\/sup> = Mn2+<\/sup>+ Cl2<\/sub> + H2<\/sub>O<\/li>\n
  3. HgCl2<\/sub> + SnCl2<\/sub> = Hg2<\/sub>Cl2<\/sub> + SnCl4<\/sub><\/li>\n
  4. P4<\/sub> + HNO3<\/sub> = H3<\/sub>PO4<\/sub> + NO<\/li>\n
  5. Cr2<\/sub>O3<\/sub> + KNO3<\/sub> + KOH = K2<\/sub>CrO4<\/sub> + KNO2<\/sub><\/li>\n
  6. Fe2+<\/sup> + NO3<\/sub>–<\/sup> + H+<\/sup> = Fe3+<\/sup> +\u00a0 NO<\/li>\n
  7. MnSO4<\/sub> + NaOH + KNO3<\/sub> = Na2<\/sub>MnO4<\/sub> + Na2<\/sub>SO4<\/sub> + KNO2<\/sub><\/li>\n
  8. KClO3<\/sub> +\u00a0 KI\u00a0 +\u00a0 H2<\/sub>SO4<\/sub> = K2<\/sub>SO4<\/sub>\u00a0 +\u00a0 KCl\u00a0 +\u00a0 I2<\/sub><\/li>\n
  9. Na2<\/sub>TeO3<\/sub> +\u00a0 NaI\u00a0 +\u00a0 HCl = NaCl\u00a0 +\u00a0 Te\u00a0 +\u00a0 H2<\/sub>O\u00a0 +\u00a0 I2<\/sub><\/li>\n
  10. H2<\/sub>O2 <\/sub>+ KMnO4<\/sub> + HCl = MnCl2<\/sub> + H2<\/sub>O + O2<\/sub> + KCl<\/li>\n
  11. Cr2<\/sub>O3 <\/sub>+ Na2<\/sub>CO3<\/sub> + NaNO3<\/sub> = Na2<\/sub>CrO4<\/sub> + NaNO2<\/sub> + CO2<\/sub><\/li>\n
  12. MnO +\u00a0 PbO2<\/sub>\u00a0 +\u00a0 HNO3<\/sub> = HMnO4<\/sub>\u00a0 +\u00a0 Pb(NO3<\/sub>)2<\/sub>\u00a0 +\u00a0 H2<\/sub>O<\/li>\n
  13. K2<\/sub>Cr2<\/sub>O7<\/sub> + KI + H2<\/sub>SO4<\/sub> = Cr2<\/sub>(SO4<\/sub>)3<\/sub> + I2<\/sub> + K2<\/sub>SO4<\/sub>\u00a0 + H2<\/sub>O<\/li>\n
  14. I2<\/sub> + SO2<\/sub> + H2<\/sub>O = I–<\/sup> + SO4<\/sub>2-<\/sup> + H+<\/sup><\/li>\n
  15. KMnO4<\/sub> + NaNO2<\/sub> + H2<\/sub>SO4<\/sub> = K2<\/sub>SO4<\/sub> + MnSO4<\/sub> + NaNO3<\/sub><\/li>\n
  16. Bi2<\/sub>S3<\/sub> + NO3<\/sub>–<\/sup> = Bi3+<\/sup> + NO + S<\/li>\n
  17. Na2<\/sub>CrO4<\/sub> + KI + H2<\/sub>SO4<\/sub> = Cr2<\/sub>(SO4<\/sub>)3<\/sub> + I2<\/sub> + Na2<\/sub>SO4<\/sub> + K2<\/sub>SO4<\/sub><\/li>\n
  18. NO3<\/sub>\u2013<\/sup> + Zn + OH\u2013<\/sup> = NH3<\/sub> + [Zn(OH)4<\/sub>]2\u2013<\/sup><\/li>\n
  19. KMnO4<\/sub> + C6<\/sub>H12<\/sub>O6<\/sub> + H2<\/sub>SO4<\/sub> = K2<\/sub>SO4<\/sub> + MnSO4<\/sub> + CO2<\/sub><\/li>\n
  20. Mn(NO3<\/sub>)2<\/sub> + PbO2<\/sub> + HNO3<\/sub> = HMnO4<\/sub> + Pb(NO3<\/sub>)2<\/sub> + H2<\/sub><\/li>\n
  21. KMnO4<\/sub> + MnSO4 + ZnO = MnO2<\/sub> + K2<\/sub>SO4<\/sub> + ZnSO4<\/sub><\/li>\n
  22. Fe + HNO3 (h\u00edg) <\/sub>\u00a0= Fe(NO3<\/sub>)3<\/sub> + NO + H2<\/sub>O<\/li>\n
  23. (NH4<\/sub>)2<\/sub>Cr2<\/sub>O7<\/sub> = Cr2<\/sub>O3<\/sub> + N2<\/sub> + H2<\/sub>O<\/li>\n
  24. BaO2<\/sub> + H2<\/sub>SO4<\/sub> = H2<\/sub>O2<\/sub> + BaSO4<\/sub><\/li>\n
  25. FeS +\u00a0 HNO3<\/sub> = Fe(NO3<\/sub>)3<\/sub>\u00a0 +\u00a0 S\u00a0 +\u00a0 NO<\/li>\n
  26. Cu2+<\/sup> + I–<\/sup> = CuI + I2<\/sub><\/li>\n
  27. CuS + HNO3<\/sub> = Cu(NO3<\/sub>)2<\/sub> + NO2<\/sub> + H2<\/sub>SO4<\/sub><\/li>\n
  28. Sn + HNO3<\/sub> = Sn(NO3<\/sub>)2<\/sub> + NH4<\/sub>NO3<\/sub> + H2<\/sub>O<\/li>\n
  29. Fe3<\/sub>O4<\/sub> + HCl = FeCl3<\/sub> + FeCl2<\/sub> + H2<\/sub>O<\/li>\n
  30. KClO3<\/sub> + H2<\/sub>SO4<\/sub> = K2<\/sub>SO4<\/sub> + KClO4<\/sub> + ClO2<\/sub><\/li>\n
  31. BrO3<\/sub>\u2011<\/sup> + Br –<\/sup> + H+<\/sup> = Br2<\/sub> + H2<\/sub>O<\/li>\n
  32. P2<\/sub>H4<\/sub> = PH3<\/sub> + P4<\/sub>H2<\/sub><\/li>\n
  33. S + NaOH = Na2<\/sub>S + Na2<\/sub>S2<\/sub>O3<\/sub><\/li>\n
  34. NO2<\/sub> + NaOH = NaNO3<\/sub> + NaNO2<\/sub> + H2<\/sub>O<\/li>\n
  35. KClO3<\/sub> = KCl + KClO4<\/sub><\/li>\n
  36. As2<\/sub>S3<\/sub> + NH4<\/sub>OH + H2<\/sub>O2<\/sub> = (NH4<\/sub>)3<\/sub>AsO4<\/sub> + (NH4<\/sub>)2<\/sub>SO4<\/sub> + H2<\/sub>O<\/li>\n
  37. AgNO3<\/sub> + AsH3<\/sub> + H2<\/sub>O = Ag + H3<\/sub>AsO4<\/sub> + HNO3<\/sub><\/li>\n
  38. FeS2<\/sub> + O2<\/sub> = Fe2<\/sub>O3<\/sub> + SO2<\/sub><\/li>\n<\/ol>\n

     <\/p>\n

    3\/f-2 <\/strong>40 t\u00f6meg % rezet tartalmaz\u00f3 r\u00e9z-cink \u00f6tv\u00f6zet 5,00 g-j\u00e1t f\u00f6l\u00f6s mennyis\u00e9g\u0171 s\u00f3savoldattal reag\u00e1ltatjuk. Mekkora t\u00e9rfogat\u00fa standard\u00e1llapot\u00fa g\u00e1z fejl\u0151dik?<\/p>\n

    M (Cu) = 63,5 g\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M (Zn) = 65,4 g<\/p>\n

     <\/p>\n

    3\/f-3<\/strong> Magn\u00e9zium-, kalcium- \u00e9s b\u00e1rium- karbon\u00e1tb\u00f3l \u00e1ll\u00f3 kever\u00e9k 1,1200 g-j\u00e1t 520 \u00b0C-ra hev\u00edtj\u00fck. Ekkor csak a magn\u00e9zium- karbon\u00e1t bomlik. A hev\u00edt\u00e9s sor\u00e1n 35,3 cm3<\/sup> standard\u00e1llapot\u00fa sz\u00e9n- dioxid g\u00e1z fejl\u0151dik. A visszamaradt anyagot 20,00 cm3<\/sup> 2,000 mol\/dm3<\/sup><\/strong> \u2013es s\u00f3savval reag\u00e1ltatjuk, \u00e9s a sz\u00e9n- dioxid elt\u00e1vol\u00edt\u00e1sa ut\u00e1n 100 cm3<\/sup>– re eg\u00e9sz\u00edtj\u00fck ki az oldatot. Ebb\u0151l a t\u00f6rzsoldatb\u00f3l \u00a010,00 cm3<\/sup> \u2013t kim\u00e9rve 0,1150 mol\/dm3<\/sup>– es n\u00e1trium- hidroxid- oldattal titr\u00e1ljuk a savfelesleget, a fogy\u00e1s 19,40 cm3<\/sup>. Sz\u00e1m\u00edtsd ki a porever\u00e9k n\/n%-os \u00e9s m\/m%-os \u00f6sszet\u00e9t\u00e9l\u00e9t!<\/p>\n

     <\/p>\n

    3\/f-4<\/strong> Ismeretlen t\u00f6meg\u0171 BaCl2 <\/sub>.<\/sup> xH2<\/sub>O krist\u00e1lyt 50 ml desztill\u00e1lt v\u00edzhez (\u03c1=1 g\/cm3<\/sup>) adagoltunk\u00a0\u00a0 20 \u00b0C-on. Ezen a h\u0151m\u00e9rs\u00e9kleten nem old\u00f3dott fel az \u00f6sszes krist\u00e1ly, a tel\u00edtett oldattal BaCl2<\/sub> .<\/sup> 2 H2<\/sub>O tart egyens\u00falyt. Az oldatot lesz\u0171rt\u00fck, a szil\u00e1rd anyag t\u00f6meg\u00e9t megm\u00e9rt\u00fck: 1,28 g volt. Az oldatot 0 \u00b0C-ra h\u0171t\u00f6tt\u00fck, ekkor \u00fajabb krist\u00e1lykiv\u00e1l\u00e1s (BaCl2 <\/sub>.<\/sup> 2 H2<\/sub>O) k\u00f6vetkezett be. Sz\u0171r\u00e9s ut\u00e1n az \u00f6sszes szil\u00e1rd anyagb\u00f3l 100 ml t\u00f6rzsoldatot k\u00e9sz\u00edtett\u00fcnk desztill\u00e1lt v\u00edzzel, melynek 10-10 cm3<\/sup>-es r\u00e9szleteit AgNO3<\/sub>– oldattal (c=0,2 mol\/dm3<\/sup>, f=0,997) titr\u00e1ltuk, az \u00e1tlagfogy\u00e1s 18,15 ml volt.<\/p>\n

    Mekkora volt az ismeretlen minta t\u00f6mege?<\/strong><\/p>\n

    H\u00e1ny krist\u00e1lyv\u00edzzel krist\u00e1lyosodott a kiindul\u00e1si anyag?<\/strong><\/p>\n

    Oldhat\u00f3s\u00e1gok: 20 \u00b0C: 35,7 g BaCl2<\/sub>\/100 g v\u00edz, 0 \u00b0C: 30,7 g BaCl2<\/sub>\/ 100 g v\u00edz<\/p>\n

     <\/p>\n

    3\/f-5<\/strong> 20 \u00b0C-on tel\u00edtett n\u00e1trium- karbon\u00e1t-oldat 14,69 g-j\u00e1hoz pontosan 25,00 cm3<\/sup> 3,0025 mol\/dm3<\/sup>– es s\u00f3savat m\u00e9r\u00fcnk. A reakci\u00f3 befejezt\u00e9vel- az \u00f6sszes sz\u00e9n- dioxid elt\u00e1voz\u00e1sa ut\u00e1n- 250,0 cm3<\/sup> t\u00f6rzsoldatot k\u00e9sz\u00edt\u00fcnk bel\u0151le, majd ebb\u0151l 10-10 cm3<\/sup>-t k\u00f6r\u00fclbel\u00fcl 0,1 mol\/dm3<\/sup> koncentr\u00e1ci\u00f3j\u00fa NaOH- oldattal titr\u00e1lunk, az \u00e1tlagfogy\u00e1s 10,50 cm3<\/sup>. A NaOH- oldat pontos koncentr\u00e1ci\u00f3j\u00e1nak meghat\u00e1roz\u00e1s\u00e1ra a 3,0025 mol\/dm3<\/sup>\u2013es s\u00f3sav 10,00 cm3<\/sup>\u2013\u00e9b\u0151l 250 cm3<\/sup> t\u00f6rzsoldatot k\u00e9sz\u00edt\u00fcnk, melynek 10 cm3<\/sup>-et megtitr\u00e1ljuk a k\u00f6r\u00fclbel\u00fcl 0,1 mol\/dm3<\/sup>-es NaOH- oldattal, az \u00e1tlagfogy\u00e1s 12,12 cm3<\/sup>.<\/p>\n

    Hat\u00e1rozd meg a NaOH- oldat pontos koncentr\u00e1ci\u00f3j\u00e1t! Mekkora t\u00f6meg\u0171 n\u00e1trium- karbon\u00e1tot tartalmazott a kiadott minta? Milyen m\/m%-os a 20 \u00b0C-on tel\u00edtett oldat? 100 g v\u00edz h\u00e1ny g n\u00e1trium- karbon\u00e1tot old 20 \u00b0C-on?<\/strong><\/p>\n

     <\/p>\n

    3\/f-6 <\/strong>10 mol HCl-t 1000 g v\u00edzben oldunk. Az oldat 1000 g-hoz 200 g Ba(OH)2<\/sub>-t adunk.<\/p>\n

    Mi lesz a reakci\u00f3 ut\u00e1n az elegy m\u00f3lsz\u00e1zal\u00e9kos \u00f6sszet\u00e9tele?<\/strong><\/p>\n

    MBa(OH)2<\/sub>=171 g\/mol\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0MHCl<\/sub>=36,5 g\/mol<\/p>\n

     <\/p>\n

    3\/f-7<\/strong> R\u00f3ka az oroszl\u00e1nyi uszoda s\u00f3savtartalm\u00e1t szeretn\u00e9 meghat\u00e1rozni titr\u00e1l\u00e1ssal, ami az al\u00e1bbi kieg\u00e9sz\u00edtend\u0151 egyenlet szerint j\u00e1tsz\u00f3dik le:<\/p>\n

    K2<\/sub>Cr2<\/sub>O7<\/sub> +\u00a0\u00a0\u00a0 HCl =\u00a0\u00a0\u00a0 KCl +\u00a0\u00a0\u00a0 CrCl3 <\/sub>\u00a0+\u00a0\u00a0\u00a0 Cl2<\/sub><\/p>\n

    a) Az uszoda viz\u00e9nek 10,0 cm3<\/sup>-es r\u00e9szlet\u00e9re 11,5 cm3<\/sup>, 0,02 mol\/dm3<\/sup>-es K2<\/sub>Cr2<\/sub>O7<\/sub>-oldat fogyott. Milyen koncentr\u00e1ci\u00f3j\u00fa a v\u00edz s\u00f3savra n\u00e9zve?<\/strong><\/p>\n

    b) M\u00e1snapra v\u00e1ltozott a s\u00f3savtartalom, ez\u00e9rt \u00fajra meg kellett titr\u00e1lni. Viszont most Soma gonosz volt, \u00e9s megl\u00f6kte R\u00f3ka kez\u00e9t titr\u00e1l\u00e1s k\u00f6zben. \u00cdgy a 10 cm3<\/sup>-es mint\u00e1hoz 20,0 cm3 <\/sup>0,02 mol\/dm3 <\/sup>K2<\/sub>Cr2<\/sub>O7<\/sub> oldatot adott. A felesleg meghat\u00e1roz\u00e1s\u00e1hoz a dikrom\u00e1tionokat KI hozz\u00e1ad\u00e1s\u00e1val reduk\u00e1lta, majd a k\u00e9pz\u0151d\u00f6tt j\u00f3d mennyis\u00e9g\u00e9t Na2<\/sub>S2<\/sub>O3<\/sub> seg\u00edts\u00e9g\u00e9vel hat\u00e1rozta meg. A tioszulf\u00e1tos titr\u00e1l\u00e1s sor\u00e1n 8,5 cm3<\/sup> 0,1 mol\/dm3<\/sup>-es Na2<\/sub>S2<\/sub>O3<\/sub>-oldat fogyott. Mennyi volt a s\u00f3sav koncentr\u00e1ci\u00f3ja a m\u00e1sodik napon?<\/strong><\/p>\n

    Cr2<\/sub>O7<\/sub>2-<\/sup> +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I–<\/sup>+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 H+\u00a0 <\/sup>=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Cr3+<\/sup>+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I2<\/sub>+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 H2<\/sub>O<\/p>\n

    I2<\/sub> +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Na2<\/sub>S2<\/sub>O3<\/sub> =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 NaI +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Na2<\/sub>S4<\/sub>O6<\/sub><\/p>\n

    Az egyenleteket rendezni kell!<\/p>\n

     <\/p>\n

    3\/f-8 <\/strong>Az oxig\u00e9n fejleszt\u00e9shez G\u00e1bor az al\u00e1bbi rendezend\u0151 reakci\u00f3egyenlet alapj\u00e1n hajtott v\u00e9gre reakci\u00f3t(a sz\u00fcks\u00e9ges v\u00edz nem elfelejtend\u0151):<\/p>\n

    KMnO4 <\/sub>\u00a0\u00ad\u00ad+ \u00a0\u00a0\u00a0\u00a0\u00a0 H2<\/sub>O2<\/sub>\u00a0\u00a0 \u00ad\u00ad+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 H2<\/sub>SO4<\/sub>=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 MnSO4<\/sub>\u00ad\u00ad\u00a0 +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 K2<\/sub>SO4<\/sub>\u00a0 \u00ad\u00ad+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 O2<\/sub><\/p>\n

    H\u00e1ny dm3<\/sup> 0 <\/strong>\u00b0<\/strong>C-os atmoszf\u00e9ra nyom\u00e1s\u00fa O2<\/sub> g\u00e1zt siker\u00fclt G\u00e1bornak fejlesztenie, ha csak 201,3 g permangan\u00e1tb\u00f3l indultunk ki?<\/strong><\/p>\n

    A robbant\u00e1shoz 2 l-es palack \u00e1ll rendelkez\u00e9sre \u00e9s tudjuk, hogy a l\u00e9gk\u00f6ri nyom\u00e1s 5x-\u00f6s\u00e9t b\u00edrja ki. 17 \u00baC-on h\u00e1ny g KMnO4 <\/sub>\u2013t kell haszn\u00e1lnunk, hogy m\u00e9g \u00e9ppen ne robbanjon fel G\u00e1bor kez\u00e9ben a palack?<\/strong><\/p>\n

    M(KMnO4<\/sub>)=158 g\/mol M(H2<\/sub>SO4<\/sub>)=98 g\/mol\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M(H2<\/sub>O2<\/sub>)=34 g\/mol<\/p>\n

     <\/p>\n

    3\/f-9<\/strong> Vir\u00e1g \u00e9s Kata a gyerekekkel s\u00f3skaf\u0151zel\u00e9ket k\u00e9sz\u00edt, \u00e9s k\u00edv\u00e1ncsiak a szert\u00e1rban tal\u00e1lhat\u00f3 s\u00f3ska-aroma (H2<\/sub>C2<\/sub>O4 <\/sub>\u2219 x H2<\/sub>O) krist\u00e1lyv\u00edztartalm\u00e1ra. Ez\u00e9rt a szil\u00e1rd porb\u00f3l bem\u00e9rtek 0,6021 g-ot \u00e9s 100,0 ml oldatot k\u00e9sz\u00edtettek bel\u0151le. Ennek 10,0 ml-es r\u00e9szlet\u00e9t 0,0200 M-os f=1,068 faktor\u00fa KMnO4<\/sub> m\u00e9r\u0151oldattal titr\u00e1lt\u00e1k, az \u00e1tlagos fogy\u00e1s 10,10 ml lett. <\/strong><\/p>\n

    H\u00e1ny mol krist\u00e1lyvizet tartalmaz a szert\u00e1rban l\u00e9v\u0151 ox\u00e1lsav?<\/strong><\/p>\n

    A kieg\u00e9sz\u00edtend\u0151 reakci\u00f3egyenlet:<\/p>\n

    (COOH)2<\/sub>\u00a0+\u00a0\u00a0KMnO4<\/sub>\u00a0+\u00a0\u00a0H2<\/sub>SO4<\/sub>\u00a0=\u00a0CO2<\/sub>\u00a0+\u00a0\u00a0MnSO4<\/sub>\u00a0+ K2<\/sub>SO4<\/sub><\/p>\n

    M(H2<\/sub>C2<\/sub>O4<\/sub>)=90,03 g\/mol<\/p>\n

     <\/p>\n

    3\/f-10<\/strong> Zsolti az esti k\u00eds\u00e9rletez\u00e9s sor\u00e1n cinket \u00e9s alum\u00edniumot 2:3 t\u00f6megar\u00e1nyban \u00f6nt\u00f6tte \u00f6ssze \u00e9s a keletkez\u0151 kever\u00e9k 5,00 g-j\u00e1t NaOH-oldattal reag\u00e1ltatta. Ezen k\u00edv\u00fcl tudjuk, hogy Zsolti 174 cm magas \u00e9s 80 kg \u00e9s 20 \u00f3ra 13 perckor hajtotta v\u00e9gre a k\u00eds\u00e9rletet. Seg\u00edtsetek Zsoltinak \u00e9s adj\u00e1tok meg, hogy mekkora t\u00e9rfogat\u00fa standard \u00e1llapot\u00fa g\u00e1z fejl\u0151d\u00f6tt<\/strong>!<\/p>\n

     <\/p>\n

    3\/f-11<\/strong> A FEB t\u00e1boros H\u00e1t Izs\u00e1k huncotkod\u00f3 vegy\u00e9sz technikus egy \u00f3vatlan pillanatban beszabadul a k\u00e9mia szert\u00e1rba. R\u00f6vid ideje volt a kipender\u00edt\u00e9sig, de gyorsan tal\u00e1lt 5 g Fe-at \u00e9s 60%-os ( \u03c1 =1,51 g\/cm3<\/sup>)\u00a0 H2<\/sub>SO4 <\/sub>\u00a0oldatot. A vasat feloldotta a k\u00e9nsavban, mik\u00f6zben 5% H2<\/sub>SO4 <\/sub>\u00a0felesleget alkalmazott.<\/p>\n

    Fe + H2<\/sub>SO4<\/sub> = FeSO4<\/sub> + H2<\/sub><\/p>\n

    a) Mekkora az alkalmazott H2<\/sub>SO4<\/sub> -oldat molkoncentr\u00e1ci\u00f3ja ( mol\/dm3<\/sup>) ?<\/strong><\/p>\n

    b) Az old\u00e1sn\u00e1l h\u00e1ny dm3<\/sup> H2<\/sub>SO4<\/sub> -oldatot haszn\u00e1lt fel?<\/strong><\/p>\n

    Mekkora az oldatban (az old\u00e1s ut\u00e1n) a FeSO4<\/sub> molt\u00f6rtje ha felt\u00e9telezz\u00fck, hogy a H2<\/sub> g\u00e1z teljesen elt\u00e1vozott?<\/strong> (Figyelem a H2<\/sub>SO4<\/sub>-et 5% feleslegben alkalmaztuk!) M(Fe) = 56 g\/mol<\/p>\n

     <\/p>\n

    3\/f-12 <\/strong>18,6 g t\u00f6meg\u0171, magn\u00e9zium-kloridb\u00f3l (MgCl2<\/sub>), f\u00e9m magn\u00e9ziumb\u00f3l (Mg) \u00e9s magn\u00e9zium-karbon\u00e1tb\u00f3l (MgCO3<\/sub>) \u00e1ll\u00f3 kever\u00e9khez h\u00edg k\u00e9nsavat adunk, \u00e9s ekkor 5,08 liter (dm3<\/sup>) t\u00e9rfogat\u00fa, 0 \u00b0C h\u0151m\u00e9rs\u00e9klet\u0171, 101,325 kPa nyom\u00e1s\u00fa g\u00e1zkever\u00e9k keletkezik.<\/p>\n

    Mennyi a kever\u00e9k kl\u00f3rtartalma, ha \u00f6sszes magn\u00e9zium-tartalma 45,71 t\u00f6meg%? Milyen a kever\u00e9k t\u00f6meg \u00e9s mol%-os \u00f6sszet\u00e9tele?<\/strong><\/p>\n

    (Relat\u00edv atomt\u00f6megek: Mg: 24; Cl: 35,5; C: 12; H: 1; O: 16.)<\/p>\n

     <\/p>\n

    3\/f-13<\/strong> 200 kg 88,4%-os NaIO3<\/sub>-b\u00f3l h\u00e1ny kg tiszta j\u00f3d nyerhet\u0151<\/strong>, a kieg\u00e9sz\u00edtend\u0151 reakci\u00f3egyenlet alapj\u00e1n, ha a kitermel\u00e9s 91%-os.<\/p>\n

    NaIO3<\/sub> + NaHSO3<\/sub> + Na2<\/sub>CO3<\/sub> = I2<\/sub> + Na2<\/sub>SO4<\/sub> + CO2<\/sub><\/p>\n

     <\/p>\n

    3\/f-14<\/strong> Zn-Cu \u00f6tv\u00f6zet 2 grammj\u00e1t s\u00f3savban oldjuk. H\u00e1ny t\u00f6meg% rezet tartalmaz az \u00f6tv\u00f6zet, ha 6%-os vesztes\u00e9g mellett 375,0 cm3<\/sup> standard \u00e1llapot\u00fa H2<\/sub> g\u00e1z keletkezik?<\/strong> (A Zn m\u00f3lt\u00f6mege 65,34 g\/m\u00f3l, a Cu m\u00f3lt\u00f6mege 63,5 g\/m\u00f3l.)<\/p>\n

     <\/p>\n

    3\/f-15<\/strong> Fenol \u00e9s p-nitrofenol kever\u00e9k\u00e9nek \u00f6sszet\u00e9tel\u00e9t hat\u00e1rozzuk meg az al\u00e1bbi elj\u00e1r\u00e1ssal. A kever\u00e9k 220 mg-j\u00e1b\u00f3l t\u00f6rzsoldatot k\u00e9sz\u00edt\u00fcnk egy 100 cm3<\/sup>-es m\u00e9r\u0151lombikban. Ebb\u0151l 10,00-10,00 cm3<\/sup>-eket m\u00e9r\u00fcnk ki k\u00e9tjel\u0171 pipetta \u00e9s Griffin-labda seg\u00edts\u00e9g\u00e9vel csiszolatos Erlenmeyer-lombikokba. Minden r\u00e9szlethez 0,8 g szil\u00e1rd KBr-ot \u00e9s 20 cm3<\/sup> 2 M k\u00e9nsavat adunk, valamint 20 cm3<\/sup> 1\/60 M-os KBrO3<\/sub>-oldatot pipett\u00e1zunk. A lombikokat ledug\u00f3zzuk, v\u00e1runk 20 percet (elmegy\u00fcnk hadij\u00e1t\u00e9kozni), majd 1-1 g KI-ot adunk az oldatokhoz. \u00dajra lez\u00e1rjuk a lombikokat \u00e9s v\u00e1runk 15 percet. V\u00e9g\u00fcl 0,1 M-os 0,937-es faktor\u00fa n\u00e1trium-tioszulf\u00e1t-m\u00e9r\u0151oldattal megtitr\u00e1ljuk, a v\u00e9gpont el\u0151tt 10 csepp kem\u00e9ny\u00edt\u0151 indik\u00e1tort adva az oldatokhoz. A kapott fogy\u00e1sok: 11,25 cm3<\/sup>, 11,30 cm3<\/sup> \u00e9s 11,25 cm3<\/sup>.<\/p>\n

    Mi a kever\u00e9k \u00f6sszet\u00e9tele?<\/strong><\/p>\n

    KBrO3<\/sub> + KBr + H2<\/sub>SO4<\/sub> = Br2<\/sub> + H2<\/sub>O + K2<\/sub>SO4<\/sub><\/p>\n

    \"\"<\/p>\n

    Br2<\/sub> + I–<\/sup> = Br–<\/sup> + I2<\/sub><\/p>\n

    I2<\/sub> + S2<\/sub>O3<\/sub>2-<\/sup> \u2192 I\u2013<\/sup> + S4<\/sub>O6<\/sub>2-<\/sup><\/p>\n

     <\/p>\n

    3\/f-16<\/strong> Ismeretlen koncentr\u00e1ci\u00f3j\u00fa K2<\/sub>CrO4<\/sub>-oldat 200 ml-b\u0151l kivesz\u00fcnk 10,00 cm3<\/sup>-es r\u00e9szleteket \u00e9s ezekhez adunk feleslegben vett KI-ot, majd az oldatot megsavany\u00edtjuk. A keletkezett I2<\/sub>-ot 0,200\u00a0mol\/dm3<\/sup>-es tioszulf\u00e1ttal (S2<\/sub>O3<\/sub>2-<\/sup>) titr\u00e1ljuk. A fogy\u00e1sok \u00e1tlaga 24,00 cm3<\/sup>.<\/p>\n

    a) Mekkora a kiindul\u00e1si 200 ml-es oldat koncentr\u00e1ci\u00f3ja K2<\/sub>CrO4<\/sub>-ra n\u00e9zve?<\/strong><\/p>\n

    b) Mennyi az oldott anyag t\u00f6mege?<\/strong><\/p>\n

    A rendezend\u0151 egyenletek:<\/p>\n

    CrO4<\/sub>2-<\/sup> + I–<\/sup> + H+<\/sup> = I2<\/sub> + Cr3+<\/sup> + H2<\/sub>O<\/p>\n

    S2<\/sub>O3<\/sub>2-<\/sup> + I2<\/sub> = I–<\/sup> + S4<\/sub>O6<\/sub>2-<\/sup><\/p>\n

     <\/p>\n

    3\/f-17<\/strong> Fenolt (C6<\/sub>H6<\/sub>O) hat\u00e1rozunk meg. 15 cm3<\/sup> mint\u00e1b\u00f3l 100 cm3<\/sup> t\u00f6rzsoldatot k\u00e9sz\u00edt\u00fcnk. Az ebb\u0151l vett 10 cm3<\/sup>-es mint\u00e1ba 1 g KBrO3<\/sub>-ot \u00e9s felesleges mennyis\u00e9g\u0171 KBr-ot adunk. A keletkezett br\u00f3m a fenollal reag\u00e1l, ezt k\u00f6vet\u0151en feleslegben KI-ot adunk az oldatba. A keletkezett j\u00f3dot 22,5 cm3<\/sup> 0,02 M Na2<\/sub>S2<\/sub>O3<\/sub> oldattal tudtuk megtitr\u00e1lni.<\/p>\n

    Mennyi volt a fenol molarit\u00e1sa (m\u00f3l\/dm3<\/sup>) az eredeti oldatban? <\/strong><\/p>\n

    (Ar(C)=12, Ar(O)=16, Ar(H)=1)<\/p>\n

    \"\"<\/p>\n

     <\/p>\n

    3\/f-18<\/strong> 10g Na2<\/sub>CO3<\/sub> \u00e9s NaHCO3<\/sub> szil\u00e1rd porkever\u00e9ket s\u00f3savval kezelnek. Feleslegben haszn\u00e1lt s\u00f3sav hat\u00e1s\u00e1ra a lej\u00e1tsz\u00f3dott reakci\u00f3 ut\u00e1n az elegyet sz\u00e1razra p\u00e1rolj\u00e1k (csak NaCl marad). Az \u00edgy keletkezett term\u00e9k t\u00f6mege pontosan annyi, mint a kiindul\u00e1si kever\u00e9k t\u00f6mege volt. Sz\u00e1m\u00edtsuk ki a kiindul\u00e1si kever\u00e9k t\u00f6meg %-os \u00f6sszet\u00e9tel\u00e9t!<\/p>\n

    Milyen m\u00f3lar\u00e1nyban kevert\u00e9k \u00f6ssze a k\u00e9t vegy\u00fcletet?<\/strong><\/p>\n

    Na2<\/sub>CO3<\/sub> + 2 HCl = 2 NaCl + CO2<\/sub> + H2<\/sub>O<\/p>\n

    NaHCO3<\/sub> + HCl = NaCl + CO2<\/sub> + H2<\/sub>O<\/p>\n

    M(Na2<\/sub>CO3<\/sub>) = 106 g\/mol\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M(NaHCO3<\/sub>) = 84 g\/mol.<\/p>\n

     <\/p>\n

    3\/f-19<\/strong> 1 liter 0,5 m\u00f3los HBr oldat elk\u00e9sz\u00edt\u00e9s\u00e9hez h\u00e1ny g 20% szennyez\u00e9st tartalmaz\u00f3 br\u00f3m sz\u00fcks\u00e9ges<\/strong>, ha a reakci\u00f3 sor\u00e1n\u00a0 a vesztes\u00e9g 20%? A lej\u00e1tsz\u00f3d\u00f3 reakci\u00f3k (az egyenleteket rendezd!):<\/p>\n

    P + Br3<\/sub> = PBr3<\/sub><\/p>\n

    PBr3<\/sub> + H2<\/sub>O = H3<\/sub>PO3<\/sub> + HBr<\/p>\n

     <\/p>\n

    3\/f-20<\/strong> Egy cink- magn\u00e9zium porkever\u00e9k h\u00edg s\u00f3savb\u00f3l 8,82 dm3<\/sup> standard \u00e1llapot\u00fa g\u00e1zt fejleszt. Az el\u0151z\u0151 mint\u00e1val azonos t\u00f6meg\u0171 cink-aluminium porkever\u00e9k h\u00edg s\u00f3savb\u00f3l 10,878 dm3<\/sup> standard \u00e1llapot\u00fa g\u00e1zt fejleszt. A k\u00e9t mint\u00e1ban a cink t\u00f6mege azonos volt.<\/p>\n

    H\u00e1ny gramm volt a mint\u00e1k t\u00f6mege, \u00e9s egy mint\u00e1ban h\u00e1ny gramm cink volt?<\/strong><\/p>\n

    MZn<\/sub>=65,4 g\/mol\u00a0\u00a0\u00a0 MMg<\/sub>=24,3 g\/mol\u00a0\u00a0\u00a0\u00a0 MAl<\/sub>=27 g\/mol<\/p>\n

     <\/p>\n

    Vissza a szt\u00f6chiometria t\u00e9mak\u00f6rre!<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

    Szt\u00f6chiometria – FEB-es feladatok Vissza a szt\u00f6chiometria t\u00e9mak\u00f6rre! 3\/f-1 Oxid\u00e1ci\u00f3s sz\u00e1mok alapj\u00e1n rendezd a k\u00f6vetkez\u0151 egyenleteket! FeCl3 + SnCl2 = FeCl2 + SnCl4 MnO4– + Cl– + H+ = Mn2++ Cl2 + H2O HgCl2 + SnCl2 = Hg2Cl2 + SnCl4 P4 + HNO3 = H3PO4 + NO Cr2O3 + KNO3 + KOH = K2CrO4 + KNO2 Fe2+ + NO3– + H+ = Fe3+ +\u00a0 NO MnSO4 + NaOH + KNO3 = Na2MnO4 + Na2SO4 + KNO2 KClO3 +\u00a0 KI\u00a0 +\u00a0 H2SO4 = K2SO4\u00a0 +\u00a0 KCl\u00a0 +\u00a0 I2 Na2TeO3 +\u00a0 NaI\u00a0 +\u00a0 HCl = NaCl\u00a0 +\u00a0 Te\u00a0 +\u00a0 H2O\u00a0 +\u00a0 I2 H2O2 + KMnO4 + HCl = MnCl2 + H2O + O2 + KCl Cr2O3 + Na2CO3 + NaNO3 = Na2CrO4 + NaNO2 + CO2 MnO +\u00a0 PbO2\u00a0 +\u00a0 HNO3 = HMnO4\u00a0 +\u00a0 Pb(NO3)2\u00a0 +\u00a0 H2O K2Cr2O7 + KI + H2SO4 = Cr2(SO4)3 + I2 + K2SO4\u00a0 + H2O I2 +…<\/p>\n

    B\u0151vebben<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":294,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_eb_attr":"","footnotes":""},"class_list":["post-493","page","type-page","status-publish"],"featured_image_src":null,"featured_image_src_square":null,"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/pages\/493","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/comments?post=493"}],"version-history":[{"count":4,"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/pages\/493\/revisions"}],"predecessor-version":[{"id":604,"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/pages\/493\/revisions\/604"}],"up":[{"embeddable":true,"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/pages\/294"}],"wp:attachment":[{"href":"https:\/\/feb.ch.bme.hu\/index.php\/wp-json\/wp\/v2\/media?parent=493"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}