Question

Soda caustica (NaOH) se obtine industrialbprin electroliza solutiei de NaCL, conform reactiei: 2NaCL+2 H2O》》2NaOH+Cl2+H2.
Se folosesc 2 tone de NaCL de puritate 60%.
Calculeaza: a) volumul de H2 in conditii normale;
B) volumul de Cl2 obținut la 27°C si 2 atm.
C) masa solutiei de NaOH, 40% obtinuta

Answer 1

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