a,
$2Na+2H_2O\to 2NaOH+H_2$
$2K+2H_2O\to 2KOH+H_2$
$NaOH+HCl\to NaCl+H_2O$
$KOH+HCl\to KCl+H_2O$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
Gọi x, y là số mol Na, K.
$\Rightarrow 0,5x+0,5y=0,1$ (1)
Ta có $n_{NaCl}=n_{Na}=x; n_{KCl}=n_K=y (mol)$
$\Rightarrow 58,5x+74,5y=13,3$ (2)
$(1)(2)\Rightarrow x=y=0,1$
$\%m_{Na}=\dfrac{0,1.23.100}{0,1.23+0,1.39}=37,1\%$
$\%m_K=62,9\%$
b,
$2H_2+O_2\buildrel{{t^o}}\over\to 2H_2O$
$\Rightarrow n_{O_2}=0,05(mol)$
$V_{O_2}=0,05.22,4=1,12l$