Ioni prezenti in solutie: Na⁺, H⁺, OH⁻, Cl⁻
Catod(-) H⁺ +1e⁻ ⇒ H·
H· + H· ⇒ H₂↑
Anod(+) Cl⁻ -1e⁺ ⇒ Cl·
Cl· + Cl· ⇒ Cl₂↑
In solutie: Na⁺ + OH⁻ ⇒ NaOH
Reactia globala(totala) : 2NaCl + 2H₂O ⇒ 2NaOH + Cl₂↑ + H₂↑
p (puritatea) = m pur/m impur *100 => m pur = p*mimpur/100 = 60*2000/100 = 1200kg NaCl
n = m/M = 1200/58,5 = 20,512 kmoli = 20512,82 moli NaCl
a) 2 kmoli NaCl....................22,4 m³ H₂
20,512 kmoli.....................x = 229,743 m³ H₂
b) 2 kmoli NaCl................1 kmol Cl₂
20,512 kmoli...................x = 10,256 kmoli
pV = nRT => V = nRT/p = 10,256*0,082*300/2 = 126,153 m³ Cl₂
c) n NaCl = nNaOH = 20,512 kmoli
md NaOH = 20,512*40 = 820,512kg
ms = 100md/C = 2051,282kg sol. NaOH
