$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$b,n_{Na}=\dfrac{18,4}{23}=0,8mol$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,4mol.$
$⇒V_{H_2}=0,4.22,4=8,96l.$
$c,Theo$ $pt:$ $n_{NaOH}=n_{Na}=0,8mol.$
$⇒m_{NaOH}=0,8.40=32g.$
$⇒C\%_{NaOH}=\dfrac{32}{200}.100\%=16\%$
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