\(FeCl_x+xAgNO_3\rightarrow Fe\left(NO_3\right)_x+xAgCl\downarrow\)
\(n_{AgCl}=\frac{21,525}{143,5}=0,15\left(mol\right)\)
Theo PT \(n_{FeCl_x}=\frac{0,15}{x}\left(mol\right)\)
Ta có: \(n_{FeCl_x}=\frac{190,5.10\%}{56+35,5x}=\frac{0,15}{x}\left(mol\right)\)
=> x=