\(7,68\left\{{}\begin{matrix}Fe\\Fe_3O_4\\Fe_2O_3\end{matrix}\right.+HCl_{\left(0,26mol\right)}\rightarrow X\rightarrow NaOH_{Dư_{\left(cr:m\left(g\right)\right)=?}}\)
Chất rắn là Fe2O3
\(n_{O\left(oxit\right)}=\frac{1}{2}n_{HCl}=0,13\left(mol\right)\)
\(m_{O\left(oxit\right)}=0,13.16=2,08\left(g\right)\)
\(m_{Fe.trong.hh.bđ}=7,68-2,08=5,6\left(g\right)\)
\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
\(2Fe^O\rightarrow Fe^{+3}\)
0,1_______0,05__
\(n_{Fe2O3}=0,05\left(mol\right)\)
\(\Rightarrow m_{Fe2O3}=0,05.160=8\left(g\right)\)
