Question

Se supun electrolizei 400g NaCl=puritate 58,5%
Se cere:a.)mNa=?
b.)Volum Cl2

Na-23 Cl,35,5

Answer 1

puritatatea = m pură/m impură 
m pură = 58,5*400/100 = 234 g NaCl
M NaCl = 23+35,5 = 58,5
V c.n. = 22,4 l
M Na = 23
 NaCl = Na + 1/2 Cl2 
58,5 g NaCl...........23 g Na.......1/2*22,4 l Cl2
234 g NaCl...............x.........................y
x = 92 g Na
y = 44,8 l Cl2