nK = \(\frac{39}{39}\) = 1 (mol)
PTHH: 2K + 2H2O→ 2KOH + H2
nH2 =\(\frac{1}{2}\).nK = 0,5 mol; nKOH = nK = 1 mol;
mdung dịch = mK + mH2O – mH2
=39 + 362 – 0,5.2
= 400 (gam)
\(C\% KOH = \dfrac{{{m_{KOH}}}}{{m\,dd\,sau}}.100\% \) = \(\dfrac{1.56}{400}.100\%\) = 14%.
Đáp án B