Hòa tan hết 11,02 gam hỗn hợp X gồm FeCO3, Fe(NO3)2 và Al vào dung

* Hướng dẫn giải

Chọn đáp án A

$\begin{array}{l}+ n_T=\frac{2,688}{22,4}=0,12\Rightarrow n_{C{O}_2}=0,05; n_{H_2}=0,02; n_{NO}=0,05\Rightarrow n_{FeC{O}_3}=n_{C{O}_2}=0,05.\\ + n_{H^{+} trong Z}=n_{HCl}-2n_{C{O}_2}-2n_{H_2}-4n_{NO}=0,06\Rightarrow Z+AgN{O}_3\to n_{NO}=\frac{0,06}{4}=0,015\\ \left.\begin{array}{l}+ T ch¨òua H_2\Rightarrow N{O_3}^{-} ~na~o héat\\ + n_{Al}=x; n_{Fe{\left(N{O}_3\right)}_2}=y; n_{KN{O}_3}=z\end{array}\right\}\Rightarrow Z\stackrel{0,45 mol NaOH}{\to }dd \left\{\begin{array}{l}{K}^{+}: z mol\\ N{a}^{+}: 0,45 mol\\ C{l}^{-}: 0,4 mol\\ Al{O_2}^{-}: x mol\end{array}\right\} (*)\\ \Rightarrow \left\{\begin{array}{l}BT~NT cho (*): 0,45+z=0,4+x\\ m_X=27x+180y+0,05.116=11,02\\ BTNT N: 2y+z=0,05\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,06\\ y=0,02\\ z=0,01\end{array}\right.\\ \begin{array}{l}+ BTE cho toa\backslash on bo¨a quàu tr̀inh: 3\underset{0,06}{\underbrace{n_{Al}}}+\underset{0,02}{\underbrace{n_{Fe{\left(N{O}_3\right)}_2}}}+\underset{0,05}{\underbrace{n_{FeC{O}_3}}}=2\underset{0,02}{\underbrace{n_{H_2}}}+3\underset{0,065}{\underbrace{n_{NO}}}+\underset{?}{\underbrace{n_{A{g}^{+} p¨o}}}\Rightarrow n_{A{g}^{+} p¨o}=0,015\\ \Rightarrow m_{kéat tu^ua}=m_{Ag}+m_{AgCl}=0,015.108+0,4.143,5=59,02 gam.\end{array}\end{array}$