Em xem lại đề bài nhé
\(\begin{array}{l}
{n_{{H_2}S{O_4}}} = \frac{{200.31,85\% }}{{98}} = 0,65\\
{H_2}S{O_4} + \mathop {MgO}\limits_x \to {H_2}O + MgS{O_4}\\
\mathop {F{e_2}{O_3}}\limits_y {\rm{ }} + {\rm{ }}3{H_2}S{O_4}{\rm{ }} \to {\rm{ }}F{e_2}{(S{O_4})_3}{\rm{ }} + {\rm{ }}3{H_2}O\\
\left. \begin{array}{l}
{m_{hh}} = {m_{MgO}} + {m_{F{e_2}{O_3}}} = 40x + 160y = 3,2\\
{n_{{H_2}S{O_4}}} = {n_{MgO}} + 3{n_{F{e_2}{O_3}}} = x + 3y = 0,65
\end{array} \right\} => \left\{ \begin{array}{l}
x = 2,36\\
y = - 0,57
\end{array} \right.
\end{array}\)