Cho m gam hỗn hợp X gồm 2 kim loại kiềm thuộc 2 chu kì liên tiếp...

2M + 2H2​O → 2MOH + H2​
Trường hợp 1: AlCl3​ dư
$A{{l}^{3+}}+3\text{O}{{H}^{-}}\to Al{{\left( OH \right)}_{3}}$
${{n}_{O{{H}^{-}}}}=3{{n}_{Al{{\left( OH \right)}_{3}}}}=0,36={{n}_{M}}$
$\underset{0,36}{\mathop{2M}} +\underset{0,5}{\mathop{2HCl}} \to \underset{0,36}{\mathop{2MCl}} +{{H}_{2}}$ $\underset{0,1}{\mathop{2M}} +2{{H}_{2}}O\to \underset{0,1}{\mathop{2MOH}} +{{H}_{2}}$
$\to {{\overline{M}}_{MCl}}=\dfrac{36,45}{0,36}=101,25$
$\to \left\{ \begin{aligned}
& K:x \\
& Rb:y \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x+y=0,36 \\
& 74,5\text{x}+121y=36,45 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x=0,15 \\
& y=0,2 \\
\end{aligned} \right.$
$\to \%{{m}_{K}}=25,55$
Trường hợp 2: AlCl3​ hết
$\underset{0,18}{\mathop{A{{l}^{3+}}}} \underset{\to }{\mathop{+}} \underset{0,54}{\mathop{3\text{O}{{H}^{-}}}} \to \underset{0,18}{\mathop{Al{{\left( OH \right)}_{3}}}} $
$\underset{(0,18-0,12)}{\mathop{Al{{\left( OH \right)}_{3}}}} +\underset{\underset{0,06}{\mathop{{}}} }{\mathop{O{{H}^{-}}}} \to AlO_{2}^{-}+2{{H}_{2}}O$
${{n}_{O{{H}^{-}}}}=0,54+0,06=0,6={{n}_{M}}$
$\underset{0,6}{\mathop{2M}} +\underset{0,5}{\mathop{2HCl}} \underset{\to }{\mathop{\to }} \underset{0,5}{\mathop{2MCl}} +{{H}_{2}}$
${{n}_{M}}>{{n}_{HCl}}$ nên M sẽ phản ứng với nước tạo MOH
$\underset{0,1}{\mathop{2M}} +2{{H}_{2}}O\to \underset{0,1}{\mathop{2MOH}} +{{H}_{2}}$
Ta có: ${{m}_{MCl}}+{{m}_{MOH}}=36,45$
$\to 0,5\left( M+35,5 \right)+0,1\left( M+17 \right)=36,45$
$\to {{\overline{M}}_{M}}=28,3$
$\to X\left\{ \begin{aligned}
& Na:x \\
& K:y \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x+y=0,6 \\
& 23\text{x}+39y+35,5.0,5+0,1.17=36,45 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x=0,4 \\
& y=0,2 \\
\end{aligned} \right.$
$\to \%{{m}_{Na}}=54,12\%$