CATOD(-) 2H⁺(din H2O) +2e---> H2
ANOD (+) 2Cl ⁻(din NaCl) -2e----> Cl2
2NaCl + 2H2O---> H2+Cl2+2NaOH
n,NaCl= m/M= 585g/58,5g/mol= 10mol
din ecuatie, deduc: 2mol NaCl........1molH2
10mol................x= 5molH2 in conditii normale(1atm si 273Kelvin)
pV= nRT---> V= 5X0,082X300.........................CALCULEAZA !!!!!