- Chọn C.

- Ta có: ${{n}_{F{{e}_{3}}{{O}_{4}}}}=\frac{2,32}{232}=0,01\left( mol \right)$.

- PTHH:   3Fe           +        2${{O}_{2}}$ $\xrightarrow{{{t}^{o}}}$ $F{{e}_{3}}{{O}_{4}}$

              3(mol)                 2(mol)          1(mol)

                 x(mol)          y(mol)            $\leftarrow $  0,01(mol)

$x=\frac{0,1\times 3}{1}=0,3\left( mol \right);y=\frac{0,1\times 2}{1}=0,2\left( mol \right)$.

${{m}_{Fe}}=0,03\times 56=1,68\left( gam \right);{{m}_{{{O}_{2}}}}=0,02\times 32=0,64\left( gam \right)$