\(n_{HCl}=10.0,05=0,5\left(mol\right)\)
PTHH: Fe3O4 + 8HCl --> FeCl2 + 2FeCl3 + 4H2O
0,0625<--0,5---->0,0625-->0,125
=> \(m_{Fe_3O_4}=0,0625.232=14,5\left(g\right)\)
\(\left\{{}\begin{matrix}C_{M\left(FeCl_2\right)}=\dfrac{0,0625}{10}=0,00625M\\C_{M\left(FeCl_3\right)}=\dfrac{0,125}{10}=0,0125M\end{matrix}\right.\)