Đáp án đúng: A
9,28 gam.
$\displaystyle \begin{array}{l}\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{+ 4CO }\xrightarrow{{{\text{t}}^{\text{o}}}}\text{3Fe + 4C}{{\text{O}}_{\text{2}}}\\\,\,\,\,\,\text{x }\,\,\,\,\,\,\,\,\,\,\,\,\,\text{4x }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{4x }\left( \text{mol} \right)\end{array}$ Hỗn hợp khí CO (0,2 – 4x) mol; CO2 : 4x mol Tỉ khối với He bằng 10,2 ⇒28.(0,2-4x) + 44.4x = 10,2.4.0,2 ⇒ x = 0,04 mol ⇒ mFe3O4 = 9,28 gam.