Hỗn hợp X gồm Cu2O, FeO và kim loại

Chọn đáp án A

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}C{u}_{2}O:xmol\\ FeO:ymol\\ M:\mathrm{0,5}(x+y)mol\end{array}\right\}\stackrel{\mathrm{2,1}molHN{O}_3}{\to }\left\{\begin{array}{l}C{u}^{2+},F{e}^{3+}\\ M^{n+},N{O_3}^{-}\mathrm{,...}\end{array}\right\}+NO\uparrow +H_{2}O\\ +BTKL:n_{H_{2}O}=\frac{48+\mathrm{2,1.63}-\mathrm{157,2}-\mathrm{0,2.30}}{18}=\mathrm{0,95}mol\Rightarrow n_{N{H_4}^{+}}=\frac{\mathrm{2,1}-\mathrm{0,95.2}}{4}=\mathrm{0,05}mol.\end{array}$

$\begin{array}{l}+n_{H^{+}}=4n_{NO}+10n_{N{H_4}^{+}}+2n_{O^{2-}}\Rightarrow n_{O^{2-}}=\frac{\mathrm{2,1}-\mathrm{0,2.4}-\mathrm{0,05.10}}{2}=\mathrm{0,4}mol\Rightarrow n_M=\mathrm{0,2}mol.\\ +\left\{\begin{array}{l}BTE:2x+y+\mathrm{0,2}n=\mathrm{0,2.3}+\mathrm{0,05.8}=1\\ m_X=144x+72y+\mathrm{0,2}M=48\end{array}\right.\Rightarrow \left\{\begin{array}{l}2x+y=1-\mathrm{0,2}n\\ 72(2x+y)+\mathrm{0,2}M=48\end{array}\right.\Rightarrow 72(1-\mathrm{0,2}n)+\mathrm{0,2}M=48\\ \Rightarrow \mathrm{0,2}M=\mathrm{14,4}n-24\Rightarrow n=2;M=24\left(Mg\right)\Rightarrow \%Mg=10\%.\end{array}$