$m_{Na}=17.54,117\%=9,2g$
$⇒n_{Na}=9,2/23=0,4mol$
$⇒m_K=17-9,2=7,8g$
$⇒n_K=7,8/39=0,2mol$
$2Na+2H_2O\to 2NaOH+H_2↑(1)$
$2K+2H_2O\to 2KOH+H_2↑(2)$
$a/n_{H_2}=n_{H_2(1)}+n_{H_2(2)}=0,2+0,1=0,3mol$
$⇒V_{H_2}=0,3.22,4=6,72l$
$b/Na_2O+H_2O\to 2NaOH$
$K_2O+H_2O\to 2KOH$
$n_{NaOH}=n_{Na}=0,4mol⇒n_{Na_2O}=1/2.n_{NaOH}=1/2.0,4=0,2mol⇒m_{Na_2O}=0,2.62=12,4g$
$n_{KOH}=n_K=0,2⇒n_{K_2O}=1/2.n_{KOH}=1/2.0,2=0,1mol⇒m_{KOH}=0,1.56=5,6g$