Đáp án:
\(CTHH:F{e_2}{O_3}\)
Giải thích các bước giải:
\(\begin{array}{l}
F{e_x}{O_y} + 2yHCl \to xFeC{l_{\frac{{2y}}{x}}} + y{H_2}O\\
{m_{{\rm{dd}}HCl}} = d \times V = 52,14 \times 1,05 = 54,747g\\
{m_{HCl}} = \dfrac{{54,747 \times 10}}{{100}} = 5,4747g\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{5,4747}}{{36,5}} = 0,15mol\\
{n_{F{e_x}{O_y}}} = \dfrac{{{n_{HCl}}}}{{2y}} = \dfrac{{0,075}}{y}mol\\
{M_{F{e_x}{O_y}}} = \frac{m}{n} = 4:\dfrac{{0,075}}{y} = \dfrac{{160y}}{3}\\
y = 3 \Rightarrow {M_{F{e_x}{O_y}}} = 160dvC\\
\Rightarrow x{M_{Fe}} + 3{M_O} = 160\\
\Rightarrow x = \dfrac{{160 - 3 \times 16}}{{56}} = 2\\
CTHH:F{e_2}{O_3}
\end{array}\)
