Hòa tan hết 32 gam hỗn hợp X gồm Fe, FeO và Fe2O3 vào 1 lít dung...

Giải thích:
$\begin{aligned}
& + \left\{ \begin{aligned}
& F\text{e}: a mol \\
& O: b mol \\
\end{aligned} \right\}\xleftarrow{quy đổi}\underbrace{\left\{ \begin{aligned}
& F\text{e} \\
& F{{\text{e}}_{x}}{{O}_{y}} \\
\end{aligned} \right\}}_{X, 32 gam}\xrightarrow[(1)]{1,7 mol HN{{O}_{3}}}\underbrace{\left\{ \begin{aligned}
& F{{\text{e}}^{3+}} \\
& N{{O}_{3}}^{-} \\
& {{H}^{+}} \\
\end{aligned} \right\}}_{Y}\xrightarrow[(2)]{0,2 mol Cu}\underbrace{\left\{ \begin{aligned}
& F{{\text{e}}^{2+}}: a mol \\
& C{{u}^{2+}}: 0,2 mol \\
& N{{O}_{3}}^{-} \\
\end{aligned} \right\}}_{Z} \\
& + {{n}_{NO (1), (2)}}=c mol. \\
& + \left\{ \begin{aligned}
& BTE (1), (2): 2\text{a+0,2}\text{.2}=2b+3c \\
& {{m}_{X}}=56\text{a}+16b=32 \\
& {{n}_{{{H}^{+}}}}=2b+4c=1,7 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=0,5 \\
& b=0,25 \\
& c=0,3 \\
\end{aligned} \right.\Rightarrow V=6,72 l\acute{i}t. \\
\end{aligned}$