Hòa tan hoàn toàn 15,74 gam hỗn hợp

Chọn đáp án C

Giải thích:

$\begin{array}{l}Sodophanung:\underset{\mathrm{15,74}gam}{\underbrace{\left\{\begin{array}{l}Na,K\\ Ca,Al\end{array}\right\}}}+H_{2}O\stackrel{}{\to }\underset{\mathrm{26,04}gam}{\underbrace{\left\{\begin{array}{l}{K}^{+},N{a}^{+},C{a}^{2+}\\ Al{O_2}^{-}:xmol\\ O{H}^{-}:ymol\end{array}\right\}}}+\underset{\mathrm{0,43}mol}{\underbrace{H_2\uparrow }}\\ (Na,K,Ca)+H_{2}O\stackrel{}{\to }\left\{\begin{array}{l}{K}^{+},N{a}^{+},C{a}^{2+}\\ O{H}^{-}:(x+y)mol\end{array}\right.+\underset{\frac{x+y}{2}mol}{\underbrace{H_2}}\end{array}$

$\begin{array}{l}Al+O{H}^{-}+H_{2}O\stackrel{}{\to }Al{O}_2^{-}+\frac{3}{2}{H}_2\\ x\leftarrow x\to 3x/2mol\\ +\left\{\begin{array}{l}{m}_{(O^{2-},O{H}^{-})}=\mathrm{26,04}-\mathrm{15,74}=\mathrm{10,3}\\ \frac{x+y}{2}+\frac{3x}{2}=n_{H_2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}32x+17y=\mathrm{10,3}\\ 2x+\mathrm{0,5}y=\mathrm{0,43}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,12}\\ y=\mathrm{0,38}\end{array}\right.\\ \Rightarrow \%m_{Al}=\frac{\mathrm{0,12.27}}{\mathrm{15,74}}=\mathrm{20,58}\%\end{array}$