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Vickyvichy
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Se dizolva 160g de NaOH in 1440 g apa. Jumatate din solutia astfel obtinuta se trateaza cu 400 g de solutie HCl de concentratie 36,5%(reactia 1). Cealalta jumatate se amesteca cu CuCl2 si se formeaza un precipitat(reactia2). Se cere
a) scrierea ecuatiilor reactiilor ce au loc
b)concentratia solutiei de hidroxid de sodiu
c)masa de solutie luata in exces, in cazul primei reactii
d) compozitia procentuala a amestecului obtinut din prima reactie
e) masa de precipitat care s-a format in a doua reactie

Răspuns :

Mirceavaman
a)Reactii:
NaOH + HCl -> NaCl + H2O
2NaOH + CuCl2 -> 2NaCl + Cu(OH)2
b)
ms=md+mapa=1440+160=1600g solutie
c=md*100/ms=160*100/1600=10%
c)
10/100=mdfolosit/800 => mdfolosit=80g NaOH
40g......36,5g....58,5g...18g
NaOH + HCl -> NaCl + H2O
80g.......x=?......y=?.......z=?
x=80*36,5/40=73g HCl
calculam masa solutiei folosita
36,5/100=73/ms => ms=73*100/36,5=200g solutie HCl
excesul=400-200=200g solutie
d)
am.final: apa, NaCl si HCl nereactionat
macid=c*mexces/100=36,5*200/100=73g acid
mapadinsolutiaacidului=ms-73-73=400-146=254g
mapadinsolutiabazei=800-80=720g
x=80*58,5/40=117g NaCl
y=80*18/40=36g H2O
mapatotala=254+36+720=1010g
mamestec=1010+117+73=1200g
1200g amestec..73g acid..117g sare..1010g apa
..100% amestec....a% acid.......b% sare........c% apa
a=73*100/1200=6,08% acid
b=117*100/1200=9,75% sare
c=1010*100/1200=84,16% apa
e)
masa de NaOH=80g (fiind egala cu cea de la punctul c)
80g...........135g
2NaOH + CuCl2 -> 2NaCl + Cu(OH)2
80g...........d=?
d=80*135/80=135g precipitat

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