mdd = mct + mdm = mNaCl + mH2O = 8 + 152 = 160 (g)
\(n_{NaCl}=\dfrac{m_{NaCl}}{M_{NaCl}}=\dfrac{8}{58,5}=\dfrac{16}{117}mol\)
\(n_{H_2O}=\dfrac{m_{H_2O}}{M_{H_2O}}=\dfrac{152}{18}=\dfrac{76}{9}mol\)
\(2NaCl+2H_2O\rightarrow2NaOH+Cl_2+H_2\)
2 2 2 1 1 ( mol )
\(\dfrac{16}{117}\) < \(\dfrac{76}{9}\) ( mol )
\(\dfrac{16}{117}\) \(\dfrac{16}{117}\) ( mol )
\(m_{NaOH}=n_{NaOH}.M=\dfrac{16}{117}.40=\dfrac{640}{117}g\)