Hòa tan hoàn toàn 18,94 gam hỗn hợp X gồm Al, Zn, MgO, Cu(NO3)2 cần dùng hết 430 ml dung dịch H2SO4

* Hướng dẫn giải

$\begin{array}{l}+\left\{\begin{array}{l}{\overline{M}}_{Khí}=\mathrm{10,482}\\ 1khíhoánâu\end{array}\right.\Rightarrow khílà\left\{\begin{array}{l}NO\\ H_2\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}+n_{H_2}=\mathrm{0,19}\\ 30n_{NO}+2n_{H_2}=\mathrm{0,19.10,842}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=\mathrm{0,06}\\ n_{H_2}=\mathrm{0,13}\end{array}\right.\\ +BTKL\Rightarrow n_{H_{2}O}=\frac{\mathrm{18,94}+\mathrm{0,43.98}-\mathrm{0,19.10,842}-\mathrm{54,34}}{18}=\mathrm{0,26}\Rightarrow n_{N{H_4}^{+}}=\frac{\mathrm{0,43.2}-\mathrm{0,13.2}-\mathrm{0,26.2}}{4}=\mathrm{0,02}\\ \Rightarrow n_{Cu{\left(N{O}_3\right)}_2}=\frac{\mathrm{0,06}+\mathrm{0,02}}{2}=\mathrm{0,04.}\\ +n_{H^{+}}=2n_{H_2}+4n_{NO}+10n_{N{H_4}^{+}}+2n_{O^{2-}}\Rightarrow n_{O^{2-}}=\frac{\mathrm{0,43.2}-\mathrm{2.0,13}-\mathrm{0,06.4}-\mathrm{0,02.10}}{2}=\mathrm{0,08}\Rightarrow n_{MgO}=\mathrm{0,08.}\\ +\left\{\begin{array}{l}27n_{Al}+65n_{Zn}=\mathrm{18,94}-\mathrm{0,04.188}-\mathrm{0,08.40}\\ 3n_{Al}+2n_{Zn}=\mathrm{0,06.3}+\mathrm{0,02.8}+\mathrm{0,13.2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,16}\\ y=\mathrm{0,06}\end{array}\right.\Rightarrow \boxed{\%Al=\mathrm{22,8}\%}\end{array}$