${{n}_{HCl}}=0,1.0,8=0,08\,\,mol$

${{n}_{NaOH}}=0,8:40=0,02\text{ }mol$

$A+{{H}_{2}}O\xrightarrow{{}}AOH+\frac{1}{2}{{H}_{2}}$ (1)

$AOH+HCl\xrightarrow{{}}ACl+{{H}_{2}}O$ (2)

$NaOH+HCl\xrightarrow{{}}NaCl+{{H}_{2}}O$ (3)

0,02 mol 0,02 mol 0,02 mol.

${{m}_{NaCl\,(3)}}\text{ }=\text{ }0,02.58,5=1,17g$

${{m}_{ACl}}=5,4-1,17=4,23g$

${{n}_{ACl}}={{n}_{AOH}}=0,08-0,02=0,06\text{ }mol$

$\overline{{{M}_{ACl}}}=\frac{{{m}_{ACl}}}{{{n}_{ACl}}}=\frac{4,23}{0,06}=70,5\Rightarrow \overline{{{M}_{A}}}=70,5-35,5=35\Rightarrow A$ chứa $Na\left( 23 \right)$ và $K\left( 39 \right)$

Đặt ${{n}_{Na}}=x\,\,mol$, ${{n}_{K}}=y\,\,mol$, ta có: $\left\{ \begin{align}& x+y=0,06 \\ & 58,5x+74,5y=4,23 \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align}& x=0,015 \\ & y=0,045 \\ \end{align} \right.$

${{m}_{Na}}=0,015.23=0,345$g

${{m}_{K}}=0,045.39=1,755$g