Hòa tan hoàn toàn 15,2 gam hỗn hợp X gồm FeO (0,02 mol); Fe(NO3)2; FeCO3; Cu (a gam)

* Hướng dẫn giải

$$\begin{gathered} Đặt x, y, z là số mol NO; N{O}_2; C{O}_2 \\ {m}_Y=30x+46y+44z=\frac{2.64\left(x+y+z\right)}{3}\left(1\right) \\ {n}_{Fe{\left(N{O}_3\right)}_2}=\frac{x+y}{2};n_{FeC{O}_3}=z \\ {m}_X=\mathrm{0,02.72}+\frac{180\left(x+y\right)}{2}+116z+a=\mathrm{15,2}\left(2\right) \\ {n}_{HCl}=4a+2y+2z+\mathrm{0,02.2} \\ \to m muối =\mathrm{0,02.56}+\frac{56\left(x+y\right)}{2}+56z+a+\mathrm{35,5}\left(4x+2y+2z+\mathrm{0,02.2}\right)=\mathrm{18,88}\left(3\right) \\ Bảo toàn electron: \\ \mathrm{0,02}+\frac{x+y}{2}+z+\frac{2a}{64}=3x+y+n_{Ag} \\ \to n_{Ag}=\frac{a}{32}-\mathrm{2,5}x-\mathrm{0,5}y+z+\mathrm{0,02} \\ m\downarrow =\mathrm{143,5}\left(4x+2y+2z+\mathrm{0,02.2}\right)+108\left(\frac{a}{32}-\mathrm{2,5}x-\mathrm{0,5}y+z+\mathrm{0,02}\right)=\mathrm{50,24}\left(4\right) \\ \left(1\right)\left(2\right)\left(3\right)\left(4\right)\to x=\mathrm{0,02};y=\mathrm{0,06};z=\mathrm{0,04};a=\mathrm{1,92} \\ m=\mathrm{0,02.56}+\frac{56\left(x+y\right)}{2}+56z+a+17\left(4x+2y+2z+\mathrm{0,02.2}\right)=\mathrm{12,96} \\ \to m+a=\mathrm{14,88} \end{gathered}$$