a,
$2NaCl+2H_2O\buildrel{{\text{đpmn}}}\over\longrightarrow 2NaOH+H_2+Cl_2$
b,
$m_{NaCl}=\dfrac{58,5}{58,5}=1(kmol)$
$H=50\%$ nên chỉ $1.50\%=0,5$ kmol $NaCl$ phản ứng
$n_{NaOH}=n_{NaCl}=0,5(kmol)$
$\to m_{NaOH}=0,5.40=20(kg)$
$n_{H_2}=n_{Cl_2}=0,25(kmol)$
$\to V_{\uparrow}=(0,25+0,25).22,4=11,2(m^3)$