Hòa tan hoàn toàn 15,74 gam hỗn hợp X chứa Na, K, Ca và Al trong...

Giải thích:
$\begin{aligned}
& + Số mol phản ứng: \underbrace{\left\{ \begin{aligned}
& Na, K \\
& Ca, Al \\
\end{aligned} \right\}}_{15,74 gam}+{{H}_{2}}O\xrightarrow{{}}\underbrace{\left\{ \begin{aligned}
& {{K}^{+}}, N{{a}^{+}},C{{a}^{2+}} \\
& Al{{O}_{2}}^{-}: x mol \\
& O{{H}^{-}}: y mol \\
\end{aligned} \right\}}_{26,04 gam}+\underbrace{{{H}_{2}}\uparrow }_{0,43 mol} \\
& (Na,K,Ca)+{{H}_{2}}O\xrightarrow{{}}\left\{ \begin{aligned}
& {{K}^{+}}, N{{a}^{+}},C{{a}^{2+}} \\
& O{{H}^{-}}:(x+y) mol \\
\end{aligned} \right.+\underbrace{{{H}_{2}}}_{\dfrac{x+y}{2} mol} \\
& Al+O{{H}^{-}}+{{H}_{2}}O\xrightarrow{{}}AlO_{2}^{-}+\dfrac{3}{2}{{H}_{2}} \\
& x \leftarrow x \to 3x/2 mol \\
& + \left\{ \begin{aligned}
& {{m}_{({{O}^{2-}}, O{{H}^{-}})}}=26,04-15,74=10,3 \\
& \dfrac{x+y}{2}+\dfrac{3x}{2}={{n}_{{{H}_{2}}}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 32x+17y=10,3 \\
& 2x+0,5y=0,43 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=0,12 \\
& y=0,38 \\
\end{aligned} \right. \\
& \Rightarrow \%{{m}_{Al}}=\dfrac{0,12.27}{15,74}=20,58\% \\
\end{aligned}$