Cate grame de sodiu trebuie sa se adauge peste 200 g apa astfel incat sa se obtina o solutie de concentratie 7,68%. Clasa IX chimie
Răspunsuri la întrebare
Răspuns:
-calculez md ,masa de NaOH,din c%
c=mdx100/ms, unde ms= md+m,apa
md= cxms/100
ms= md+ m,apa= md+(200-m,apa intrata in reactie))
md= 7,68(md+200-m,apa reactie)/100
100md=7,68md-7,68m,apa+1536
92,32md= 1536-7,68m,apa
md= 16,6-0,08m,apa
-deduc m,apa si m,Na din ecuatia chimica
23g.......18g..........40g
Na + H2O----> NaOH + 1/2H2
x,,,,,..,,y............(.16,6-0,08m,apa)....
y= m,apa=18(16,6-0,08m,apa)/40
m,apa=0,45(16,6-0,08m,apa)=>m,apa+0,036m,apa= 7,47-->m,apa= 7,2g
x= 23(16,6-0,08m,apa)/40----> mNa= 9,22g
plicație:
sodiu cand se introduce in apa formeaza prin reactie NaOH, unde el va fi md
n g b g md g
2Na + 2H2O --> 2NaOH + H2O
2x23 2x18 2x40
stim ca c% = mdx100/ms
unde md = masa de NaOH formata
ms = m.apa initiala - b (masa de appa consumata din reactie) + md
c% = 7,68%
=> 7,68 = mdx100/ms
=> 100md = 7,68(200-b+md)
b = 2x18xmd/2x40 = 0,45md
=> 100md = 7,68(200-0,45md+md)
=> 13,021md = 200+0,55md
=> md = 15,70 g NaOH format
=> n = 2x23xmd/2x40
= 9,026 g Na