Cho 38,55 gam hỗn hợp X gồm Mg

Chọn đáp án D

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}{\overline{M}}_{Khí}=18\\ 1khíhoanâu\end{array}\right.\Rightarrow khíla\left\{\begin{array}{l}NO\\ H_2\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}+n_{H_2}=\mathrm{0,175}\\ 30n_{NO}+2n_{H_2}=\mathrm{0,175.18}=\mathrm{3,15}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=\mathrm{0,1}\\ n_{H_2}=\mathrm{0,075}\end{array}\right.\\ +BTKL\Rightarrow n_{H_{2}O}=\frac{\mathrm{38,55}+\mathrm{0,725.98}-\mathrm{3,15}-\mathrm{96,55}}{18}=\mathrm{0,55}\Rightarrow n_{N{H_4}^{+}}=\frac{\mathrm{0,725.2}-\mathrm{0,075.2}-\mathrm{0,55.2}}{4}=\mathrm{0,05}\\ \Rightarrow BTBT\left(N\right):n_{Fe{\left(N{O}_3\right)}_2}=\frac{\mathrm{0,1}+\mathrm{0,05}}{2}=\mathrm{0,075.}\\ +n_{H^{+}}=2n_{H_2}+4n_{NO}+10n_{N{H_4}^{+}}+2n_{O^{2-}}\Rightarrow n_{O^{2-}}=\frac{\mathrm{0,725.2}-\mathrm{0,075.2}-\mathrm{0,1.4}-\mathrm{0,05.10}}{2}=\mathrm{0,2}\Rightarrow n_{ZnO}=\mathrm{0,2.}\\ +\left\{\begin{array}{l}27n_{Al}+24n_{Mg}=\mathrm{38,55}-\mathrm{0,075.180}-\mathrm{0,2.81}\\ 3n_{Al}+2n_{Mg}=\mathrm{0,1.3}+\mathrm{0,05.8}+\mathrm{0,075.2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,15}\\ y=\mathrm{0,2}\end{array}\right.\Rightarrow \%molAl=24\%gannhatvoi25\%.\end{array}$