Hướng dẫn:

${{m}_{Fe}}={{m}_{gang}}.97%=500.97%=485\left( kg \right)$ 

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{485}{56}=8,66(kmol)$

${{n}_{F{{e}_{2}}{{O}_{3}}}}=\frac{{{n}_{Fe}}}{2}=\frac{8,66}{2}=4,33\left( kmol \right)$ 

$\Rightarrow {{m}_{F{{e}_{2}}{{O}_{3}}}}={{n}_{F{{e}_{2}}{{O}_{3}}}}.{{M}_{F{{e}_{2}}{{O}_{3}}}}=4,33.160=692,8\left( kg \right)$ 

$\Rightarrow {{m}_{quang}}=\frac{{{m}_{F{{e}_{2}}{{O}_{3}}}}}{50%}=\frac{692,8}{50%}=1385,6\left( kg \right)$ 

  Chọn đáp án D