Cho 36,24 gam hỗn hợp X gồm Fe, Fe3O4,

Chọn đáp án B.

Ta có: $n_{AgCl}=n_{HCl}=\mathrm{1,15}mol\Rightarrow n_{Ag}=\frac{m_{\downarrow }-\mathrm{143,5}{n}_{AgCl}}{108}=\mathrm{0,075}mol$

$\stackrel{BT:e}{\to }{n}_{F{e}^{2+}}=n_{Ag}+3n_{NO}=\mathrm{0,15}mol$$\stackrel{BTDT}{\to }{n}_{F{e}^{3+}}=\frac{n_{C{l}^{-}}-n_{H^{+}}-2n_{F{e}^{2+}}}{3}=\mathrm{0,25}mol$

$\stackrel{BT:N}{\to }{n}_{Fe{\left(N{O}_3\right)}_2}=\frac{n_{N{O}_2}+n_{NO}-n_{HN{O}_3}}{2}=\mathrm{0,06}mol$

Ta có hệ:$\left\{\begin{array}{l}56n_{Fe}+232n_{F{e}_3{O}_4}=m_X-180n_{Fe{\left(N{O}_3\right)}_2}=\mathrm{25,44}\\ \stackrel{BT:Fe}{\to }{n}_{Fe}+3n_{F{e}_3{O}_4}=n_{F{e}^{2+}}+n_{F{e}^{3+}}-n_{Fe{\left(N{O}_3\right)}_2}=\mathrm{0,34}\end{array}\right.\to \left\{\begin{array}{l}{n}_{Fe}=\mathrm{0,04}mol\\ n_{F{e}_3{O}_4}=\mathrm{0,1}mol\end{array}\right.$$\Rightarrow \%n_{Fe}=20\%.$