$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$b,n_{Na}=\dfrac{4,6}{23}=0,2mol.$
$Theo$ $pt:$ $n_{NaOH}=n_{Na}=0,2mol.$
$⇒m_{NaOH}=0,2.40=8g.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$c,m_{ddNaOH}=4,6+100-(0,1.2)=104,4g.$
$⇒C\%_{NaOH}=\dfrac{m_{NaOH}}{m_{ddNaOH}}.100\%=\dfrac{8}{104,4}.100\%=7,6\%$
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