$PTPƯ:2K+2H_2O\xrightarrow{} 2KOH+H_2↑$
$n_{K}=\dfrac{3,9}{39}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{K}=0,05mol.$
$Theo$ $pt:$ $n_{KOH}=n_{K}=0,1mol.$
$⇒m_{KOH}=0,1.56=5,6g.$
$⇒m_{dd\ spư}=m_{K}+m_{H_2O}-m_{H_2}=3,9+96,2-(0,05.2)=100g.$
$⇒C\%_{KOH}=\dfrac{5,6}{100}.100\%=5,6\%$
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