Đáp án:
\(CTHH:FeC{l_3}\)
Giải thích các bước giải:
\(\begin{array}{l}
FeC{l_x} + xAgN{O_3} \to xAgCl + Fe{(N{O_3})_x}\\
{m_{FeC{l_x}}} = \dfrac{{200 \times 16,25}}{{100}} = 32,5g\\
{n_{AgCl}} = \dfrac{m}{M} = \dfrac{{86,1}}{{143,5}} = 0,6mol\\
{n_{FeC{l_x}}} = \dfrac{{{n_{AgCl}}}}{x} = \dfrac{{0,6}}{x}(mol)\\
{M_{FeC{l_x}}} = \dfrac{m}{n} = 32,5:\dfrac{{0,6}}{x} = 54,167x\\
{M_{Fe}} + x{M_{Cl}} = 54,167x\\
\Rightarrow x = \dfrac{{56}}{{54,167 - 35,5}} = 3\\
\Rightarrow CTHH:FeC{l_3}
\end{array}\)