Cho m gam hỗn hợp E gồm NaHCO3, Fe2O3, ZnO, MgCO3 tác dụng vừa đủ...

$\left\{ \begin{aligned}
& NaHC{{O}_{3}} \\
& F{{e}_{2}}{{O}_{3}} \\
& MgC{{O}_{3}} \\
& \underbrace{ZnO}_{m\left( g \right)} \\
\end{aligned} \right.+{{H}_{2}}S{{O}_{4}}\to \left\{ \begin{aligned}
& \left\{ \begin{aligned}
& ZnS{{O}_{4}} \\
& N{{a}_{2}}S{{O}_{4}} \\
& F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}} \\
& MgS{{O}_{4}} \\
\end{aligned} \right.\xrightarrow{+BaC{{l}_{2}}\text{ du}}\underset{0,6\text{ mol}}{\mathop{BaS{{O}_{4}}}} \\
& C{{O}_{2}} \\
& {{H}_{2}}O:193,08\left( g \right) \\
\end{aligned} \right.$
BT nguyên tố $S\to {{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{BaS{{O}_{4}}}}=0,6\text{ mol}$
BTKL $\to {{m}_{E}}+{{m}_{dd{{\text{H}}_{2}}S{{O}_{4}}}}=$ mmuối​ $+{{m}_{{{H}_{2}}O}}+{{m}_{C{{O}_{2}}}}$
$\to {{n}_{C{{O}_{2}}}}=0,22$ mol
$\Rightarrow {{n}_{{{H}_{2}}O}}$ ​sinh ra từ phản ứng​ $=\dfrac{193,08-{{m}_{{{H}_{2}}O\text{ trong dd }{{\text{H}}_{2}}S{{O}_{4}}}}}{18}=0,66$
BT nguyên tố $H\to {{n}_{NaHC{{O}_{3}}}}=2{{n}_{{{H}_{2}}O\text{ sinh ra}}}-2{{n}_{{{H}_{2}}S{{O}_{4}}}}=0,12\text{ mol}$
E gồm $\left\{ \begin{aligned}
& NaHC{{O}_{3}}:0,12 \\
& MgC{{O}_{3}}:0,1 \\
& F{{e}_{2}}{{O}_{3}}:x \\
& ZnO:y \\
\end{aligned} \right.\xrightarrow{+{{H}_{2}}S{{O}_{4}}}\left\{ \begin{aligned}
& N{{a}_{2}}S{{O}_{4}}:0,06 \\
& MgS{{O}_{4}}:0,1 \\
& F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}:x \\
& ZnS{{O}_{4}}:y \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& BTNT\text{ S }\to 3x+y+0,1+0,06=0,6 \\
& \overline{{{M}_{E}}}=\dfrac{0,22.84+160x+81y}{0,22+x+y}=94,96 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=0,08 \\
& y=0,2 \\
\end{aligned} \right.$
$\Rightarrow \%{{m}_{F{{e}_{2}}{{O}_{3}}}}=\dfrac{0,08.160}{94,96.\left( 0,22+0,08+0,2 \right)}.100=26,96\%$