$n_{H_2}=3,36/22,4=0,15mol$

$a/2Na+2H_2O\to 2NaOH+H_2$

$2K+H_2O\to 2KOH+H_2$

b/Gọi $n_{Na}=a;n_K=b$

Ta có :

$m_{hh}=23a+39b=10,1$

$n_{H_2}=0,5a+0,5b=0,15$

Ta có hpt :

$\left\{\begin{matrix}
23a+39b=10,1 & \\ 
0,5a+0,5b=0,15 & 
\end{matrix}\right.$

$⇔\left\{\begin{matrix}
a=0,1 & \\ 
b=0,2 & 
\end{matrix}\right.$

$⇒\%m_{Na}=\dfrac{23.0,1.100\%}{10,1}=22,77\%$

$⇒\%m_K=100\%-22,77\%=77,23\%$

$b/n_{NaOH}=n_{Na}=0,1mol⇒m_{NaOH}=0,1.40=4g$

$n_{KOH}=n_K=0,2mol⇒m_{KOH}=0,2.56=11,2g$

Đáp án:

Giải thích các bước giải:

$a,PTPƯ:$

$2K+2H_2O→2KOH+H_2↑$

$2Na+2H_2O→2NaOH+H_2↑$

$b,n_{H_2}\dfrac{3,36}{22,4}=0,15mol.$

$\text{Gọi}$ $n_{Na}$ $là$ $a,$ $n_{K}$ $là$ $b.$

$\text{Theo đề bài ta có hệ pt:}$ $\left \{ {{23a+39b=10,1} \atop {0,5a+0,5b=0,15}} \right.$ $⇒\left \{ {{a=0,1} \atop {b=0,2}} \right.$

$⇒\%m_{Na}=\dfrac{0,1.23}{10,1}.100\%=22,78\%$

$⇒\%m_{K}=100\%-22,78\%=77,22\%$

$c,Theo$ $pt1:$ $n_{KOH}=n_{K}=0,2mol.$

$⇒m_{KOH}=0,2.56=11,2g.$

$Theo$ $pt2:$ $n_{NaOH}=n_{Na}=0,1mol.$

$⇒m_{NaOH}=0,1.40=4g.$

chúc bạn học tốt!