$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$b,n_{Na}=\dfrac{6,9}{23}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,15mol.$
$⇒m_{H_2}=0,15.2=0,3g.$
$c,Theo$ $pt:$ $n_{NaOH}=n_{Na}=0,3mol.$
$⇒m_{NaOH}=0,3.40=12g.$
$⇒m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+108,9-0,3=115,5g.$
$⇒C\%_{NaOH}=\dfrac{12}{115,5}.100\%=10,39\%$
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