a,
$2NaCl+2H_2O\buildrel{{\text{đpmn}}}\over\longrightarrow 2NaOH+H_2+Cl_2$
b,
$n_{NaCl}=\dfrac{58,5}{58,5}=1(kmol)$
Theo PTHH:
$n_{NaOH}=n_{NaCl}=1(kmol)$
$\to m_{NaOH}=1.40.50\%=20kg$
$n_{H_2}=n_{Cl_2}=\dfrac{n_{NaCl}}{2}=0,5(kmol)$
$\to V_{\uparrow}=(0,5+0,5).22,4.50\%=11,2m^3$