cho 3.2 g hon hop gom mgo va fe203 tac dung voi 200 g dd h2s04 nong do 31,85% a)tinh khoi luong moi axIT BAN DAU B)TINH C% MUOI THU DC SAU PHAN UNG

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