Trộn 8,1 gam Al với 35,2 gam hỗn

Chọn đáp án B.

Dung dịch Z chứa Al³⁺ (0,3 mol), Fe²⁺, Fe³⁺, H⁺ dư, Cl^–.

Kết tủa gồm AgCl và Ag trong đó:$n_{AgCl}=n_{C{l}^{-}}=\mathrm{1,9}mol\Rightarrow n_{Ag}=\mathrm{0,075}mol$

 $\stackrel{BT:e}{\to }{n}_{F{e}^{2+}}=3n_{NO}+n_{Ag}=\mathrm{0,15}mol$và$n_{H^{+}}=4n_{NO}=\mathrm{0,1}mol$

$\stackrel{BTDT\left(Z\right)}{\to }3n_{F{e}^{3+}}+3n_{A{l}^{3+}}+2n_{F{e}^{2+}}+n_{H^{+}}=n_{C{l}^{-}}\Rightarrow n_{F{e}^{3+}}=\mathrm{0,2}mol$

$\stackrel{BT:H}{\to }{n}_{H_{2}O}=\frac{\mathrm{1,9}+\mathrm{0,15}-\mathrm{0,1}}{2}=\mathrm{0,975}mol\stackrel{BTKL}{\to }{m}_T=\mathrm{9,3}gam$

$\left\{\begin{array}{l}{n}_{NO}+n_{N_{2}O}=\mathrm{0,275}\\ 30n_{NO}+44n_{N_{2}O}=\mathrm{9,3}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{NO}=\mathrm{0,2}mol\\ n_{N_{2}O}=\mathrm{0,075}mol\end{array}\right.\stackrel{BT:N}{\to }{n}_{Fe{\left(N{O}_3\right)}_2}=\mathrm{0,1}mol\Rightarrow \%m_{Fe{\left(N{O}_3\right)}_2}=\mathrm{41,57}\%$