Ta có nK = \(\dfrac{78}{39}\) = 2 ( mol )
K + H2O \(\rightarrow\) KOH + \(\dfrac{1}{2}\)H2
2.....................2...........1
=> mKOH = 56 . 2 = 112 ( gam )
Ta có Mdung dịch = Mtham gia - MH2
= 78 + 923 - 1 . 2
= 999
=> C%KOH = \(\dfrac{112}{999}\times100\approx11,211\%\)