Đổi 6,72m3 = 6720dm3 = 6720 lít
Ta có: \(n_{Cl_2}=\dfrac{6720}{71}\left(mol\right)\)
PTHH: \(2NaCl+2H_2O\xrightarrow[có.màng.ngăn]{điện.phân}Cl_2+H_2+2NaOH\)
Theo PT: \(n_{NaCl}=2.n_{Cl_2}=2.\dfrac{6720}{71}=\dfrac{13440}{71}\left(mol\right)\)
\(\Rightarrow m_{NaCl}=\dfrac{13440}{71}.58,5=11073,80282\left(g\right)\)