\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right);n_{Ba\left(OH\right)_2}=0,3\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Fe_3O_4}=a\left(mol\right)\\n_{Cu\left(p\text{ư}\right)}=b\left(mol\right)\end{matrix}\right.\)
Ta có sơ đồ phản ứng: \(X\left\{{}\begin{matrix}Fe_3O_4:a\\Cu:b+0,05\end{matrix}\right.+HCl\rightarrow\left\{{}\begin{matrix}FeCl_2:3a\\CuCl_2:b\\HCl\end{matrix}\right.+Cu:0,05+H_2O\)
\(\left\{{}\begin{matrix}FeCl_2:3a\\CuCl_2:b\\HCl\end{matrix}\right.\xrightarrow[]{+0,3\left(mol\right)Ba\left(OH\right)_2}18,4\left(g\right)\downarrow\left\{{}\begin{matrix}Fe\left(OH\right)_2:3a\\Cu\left(OH\right)_2:b\end{matrix}\right.+BaCl_2:0,3+H_2O\)
Quá trình oxi hóa - khử:
\(3Fe^{+\dfrac{8}{3}}+2e\rightarrow3Fe^{+2}\)
3a-------->2a
\(Cu^0\rightarrow Cu^{+2}+2e\)
b---------------->2b
BTe: 2a = 2b => a = b
\(\Rightarrow3a.90+98a=18,4\Leftrightarrow a=0,05\left(mol\right)\)
\(\Rightarrow m=0,05.232+\left(0,05+0,05\right).64=18\left(g\right)\)
BTNT Cl: \(n_{HCl}=2n_{BaCl_2}=0,6\left(mol\right)\)
\(\Rightarrow V=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)
