Hòa tan hết 11,02 gam hỗn hợp X

Chọn đáp án A

Giải thích:

$\begin{array}{l}+n_T=\frac{\mathrm{2,688}}{\mathrm{22,4}}=\mathrm{0,12}\Rightarrow n_{C{O}_2}=\mathrm{0,05};n_{H_2}=\mathrm{0,02};n_{NO}=\mathrm{0,05}\Rightarrow n_{FeC{O}_3}=n_{C{O}_2}=\mathrm{0,05.}\\ +n_{H^{+}trongZ}=n_{HCl}-2n_{C{O}_2}-2n_{H_2}-4n_{NO}=\mathrm{0,06}\Rightarrow Z+AgN{O}_3\stackrel{}{\to }{n}_{NO}=\frac{\mathrm{0,06}}{4}=\mathrm{0,015.}\end{array}$

$\left.\begin{array}{l}+Tchua{H}_2\Rightarrow N{O_3}^{-}het\\ +n_{Al}=x;n_{Fe{\left(N{O}_3\right)}_2}=y;n_{KN{O}_3}=z\end{array}\right\}\Rightarrow Z\stackrel{\mathrm{0,45}molNaOH}{\to }dd\left\{\begin{array}{l}{K}^{+}:zmol\\ N{a}^{+}:\mathrm{0,45}mol\\ C{l}^{-}:\mathrm{0,4}mol\\ Al{O_2}^{-}:xmol\end{array}\right\}(*)$

$\begin{array}{l}\Rightarrow \left\{\begin{array}{l}BTNTcho(*):\mathrm{0,45}+z=\mathrm{0,4}+x\\ m_X=27x+180y+\mathrm{0,05.116}=\mathrm{11,02}\\ BTNTN:2y+z=\mathrm{0,05}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,06}\\ y=\mathrm{0,02}\\ z=\mathrm{0,01}\end{array}\right.\\ +BTEchotoanboquatrình:3\underset{\mathrm{0,06}}{\underbrace{n_{Al}}}+\underset{\mathrm{0,02}}{\underbrace{n_{Fe{\left(N{O}_3\right)}_2}}}+\underset{\mathrm{0,05}}{\underbrace{n_{FeC{O}_3}}}=2\underset{\mathrm{0,02}}{\underbrace{n_{H_2}}}+3\underset{\mathrm{0,065}}{\underbrace{n_{NO}}}+\underset{?}{\underbrace{n_{A{g}^{+}pö}}}\Rightarrow n_{A{g}^{+}pö}=\mathrm{0,015}\\ \Rightarrow m_{kettua}=m_{Ag}+m_{AgCl}=\mathrm{0,015.108}+\mathrm{0,4.143,5}=\mathrm{59,02}gam.\end{array}$