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Probleme chimie

 
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sHoTeR

[þig]



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(since 30-04-2018 22:10)
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Post Posted: 23-10-2013, 16:19:51 | Translate post to: ... (Click for more languages)

1) Se ard 19,2 fier . Aflati masa de oxigen necesara si numarul de moli Fe3O4 .
2) Se descompun 42,5 g KClO3 de puritate 80% . Determinati masa de KCl obtinuta .
3) Se dau 630g acid azotic de concentratie= 30% ce reactioneaza cu Mg.
Se cere masa de sare obtinuta.
4) 184g de sodiu reactioneaza cu apa obtinanduse o solutie de soda caustica 25% . Ce masa de soda a reactionat (ms = ?).
5 rep celui care imi face astea . Puppy Dog Eyes
PS : Daca sunt bune maresc la 10.
Problemele astea sunt valabile pana maine la 7:30



"Usually, when you are the absolute best.
You get hated the most"

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Fough

[Nutella Lover]



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(since 13-12-2019 11:16)
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Post Posted: 24-10-2013, 14:53:44 | Translate post to: ... (Click for more languages)

ehm..sper si eu ca e 7 dupa-masa..n-am mai dat pe aici -

ok...1)...la prima vedere par gresite datele..adica din principiu 19.2/56 iti da numar irational de moli de fier...dar o las asa....
3Fe+2 O2-->Fe3O4...scriem pe sus si jos...masa de Fe3O4=19.2*242(masa ferofericului)/168(masa celor 3 moli de fier)=27.65 g
moli=27.65/272=0.11425 (rezultat care il consider o ineptie din datele problemei)

2)la fel...rezultate ciudate....KClO3--->KCl+3/2 O2.
masa pura de clorat=42.5*4/5=34g
facem sus/jos/plm....masa de Kcl=34*74.5/122.5=20.67 g

3)630g..concentratie=30%...avem masa HNO3=630*3/10=189 grame
Reactia: 2HNO3 + Mg--->Mg(NO3)2 +H2

masa=189*148/(2*63)=222 g

4)Na + H2O --->NaOH + 1/2 H2;

184/23-->8 moli Na intra in reactie -->8 moli soda --->320 grame soda pura...puritate=25%--->masa soda=320*4=1280 grame.


Sper sa fie ok rezolvarile si date la timp Thinking ...pentru cine mai are probleme de chimie/info/mate ...activati-mi casuta (fough/Fatboss)



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sHoTeR

[þig]



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Post Posted: 27-10-2013, 13:39:15 | Translate post to: ... (Click for more languages)

Nu au fost la timp , dar merg si asa , iti dau cum am promis .
PS : Le-am copiat in clasa -.



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