Cho 31,6 gam hỗn hợp X gồm Fe và Fe3O4 tan hết trong dung dịch...

Phương pháp: Tính được ${{n}_{{{H}_{2}}}}$
Đặt ${{n}_{HCl}}=x\text{ mol}$
BTNT "H": ${{n}_{{{H}_{2}}O}}=({{n}_{HCl}}-2{{n}_{{{H}_{2}}}})/2$
$31,5(g)\text{ X}\left\{ \begin{aligned}
& F\text{e} \\
& F{{\text{e}}_{3}}{{O}_{4}} \\
\end{aligned} \right.+HCl:x\to {{H}_{2}}:0,1+60,7(g)\text{ Y}\left\{ \begin{aligned}
& F\text{e}C{{l}_{2}} \\
& F\text{e}C{{l}_{3}} \\
\end{aligned} \right.+{{H}_{2}}O:\dfrac{x-0,2}{2}$
BTKL: ${{m}_{X}}+{{m}_{HCl}}={{m}_{{{H}_{2}}}}+{{m}_{Y}}+{{m}_{{{H}_{2}}O}}\Rightarrow x\Rightarrow {{n}_{{{H}_{2}}O}}$
BTNT "O": ${{n}_{F{{\text{e}}_{3}}{{O}_{4}}}}={{n}_{{{H}_{2}}O}}:4\Rightarrow {{m}_{F{{\text{e}}_{3}}{{O}_{4}}}}$
Hướng dẫn giải:
${{n}_{{{H}_{2}}}}=0,1\text{ mol}$
Đặt ${{n}_{HCl}}=x\text{ mol}$
BTNT "H": ${{n}_{{{H}_{2}}O}}=({{n}_{HCl}}-2{{n}_{{{H}_{2}}}})/2=(x-0,2)/2\text{ (mol)}$
$31,6(g)\text{ X}\left\{ \begin{aligned}
& F\text{e} \\
& F{{\text{e}}_{3}}{{O}_{4}} \\
\end{aligned} \right.+HCl:x\to {{H}_{2}}:0,1+60,7(g)\text{ Y}\left\{ \begin{aligned}
& F\text{e}C{{l}_{2}} \\
& F\text{e}C{{l}_{3}} \\
\end{aligned} \right.+{{H}_{2}}O:\dfrac{x-0,2}{2}$
BTKL: ${{m}_{X}}+{{m}_{HCl}}={{m}_{{{H}_{2}}}}+{{m}_{Y}}+{{m}_{{{H}_{2}}O}}$
$\Rightarrow 31,6+36,5\text{x}=0,1.2+60,7+18(x-0,2)/2\Rightarrow x=1$
$\Rightarrow {{n}_{{{H}_{2}}O}}=0,4\text{ mol}$
BTNT "O": ${{n}_{F{{\text{e}}_{3}}{{O}_{4}}}}={{n}_{{{H}_{2}}O}}:4=0,1\text{ mol}$
$\Rightarrow {{m}_{F{{\text{e}}_{3}}{{O}_{4}}}}=0,1.232=23,2\text{ gam}$