a)
$2K + 2H_2O \to 2KOH + H_2$
b)
$n_K = \dfrac{0,78}{39} = 0,02(mol)$
Theo PTHH : $n_{H_2} = \dfrac{1}{2}n_K = 0,01(mol)$
$V = 0,01.22,4 = 0,224(lít)$
c) $m_{dd\ sau\ pư} = m_K + m_{H_2O} - m_{H_2} = 0,78 + 200 - 0,01.2 = 200,76(gam)$
$C\%_{KOH} = \dfrac{0,02.56}{200,76}.100\% = 0,56\%$