0,5 moli m g
Cl2 + 2Na = 2NaCl
2 2x58,5
=> m = 0,5x2x58,5/2
= 29,25 g NaCl la randament 100% (Ct)
la 80% cat s-ar fi obtinut...(Cp)?
=>
(Ct) 29,25 g ....................... 100%
(Cp) g .................................... 80%
=> Cp = Ctx80/100 = 29,25x80/100
= 23,4 g NaCl s-au obtinut practic