46 g m g md n moli
2Na + 2H2O ---> 2NaOH + H2
2x23 2x18 2x40 1
=> m = 36 g apa consumata
m.apa ramasa = 436 - m = 400 g
md = 80 g NaOH
n = 1 mol H2
c% = mdx100/ms
= 80x100/400 = 20% sol NaOH
80 g md m a
NaOH + HCl --> NaCl + H2O
40 36,5 58,5 18
=> md = 73 g HCl
=> ms = mdx100/c%
= 73x100/10 = 730 g sol HCl
m.apa sol HCl = 730-73 = 657 g apa
=> m = 117 g NaCl
a = 36 g apa nou formata
=> m.apa totala = 400+657+36 = 1093 g
c%.finala = 117x100/1093
= 10,71%
