Đáp án:

 a) 6,72l

b) 9,4 g và 12,4 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
mNa = 17 \times 54,117\%  = 9,2g\\
mK = 17 - 9,2 = 7,8g\\
nNa = \dfrac{{9,2}}{{23}} = 0,4\,mol\\
nK = \dfrac{{7,8}}{{39}} = 0,2\,mol\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
n{H_2} = \dfrac{{0,4}}{2} + \frac{{0,2}}{2} = 0,3\,mol\\
 \Rightarrow V{H_2} = 0,3 \times 22,4 = 6,72l\\
b)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{K_2}O + {H_2}O \to 2KOH\\
nN{a_2}O = \dfrac{{0,4}}{2} = 0,2\,mol\\
 \Rightarrow mN{a_2}O = 0,2 \times 62 = 12,4g\\
n{K_2}O = \dfrac{{0,2}}{2} = 0,1\,mol\\
 \Rightarrow m{K_2}O = 0,1 \times 94 = 9,4g
\end{array}\)

$m_{Na}=17.54,117\%=9,2g$

$⇒n_{Na}=9,2/23=0,4mol$

$⇒m_K=17-9,2=7,8g$

$⇒n_K=7,8/39=0,2mol$

$2Na+2H_2O\to 2NaOH+H_2↑(1)$

$2K+2H_2O\to 2KOH+H_2↑(2)$

$a/n_{H_2}=n_{H_2(1)}+n_{H_2(2)}=0,2+0,1=0,3mol$

$⇒V_{H_2}=0,3.22,4=6,72l$

$b/Na_2O+H_2O\to 2NaOH$

$K_2O+H_2O\to 2KOH$

$n_{NaOH}=n_{Na}=0,4mol⇒n_{Na_2O}=1/2.n_{NaOH}=1/2.0,4=0,2mol⇒m_{Na_2O}=0,2.62=12,4g$

$n_{KOH}=n_K=0,2⇒n_{K_2O}=1/2.n_{KOH}=1/2.0,2=0,1mol⇒m_{KOH}=0,1.56=5,6g$