Đáp án đúng: C
10 : 3.
$n_{Al}$ ban đầu: x mol$n_{Fe_2O_3}$ ban đầu : y mol
Hỗn hợp sau phản ứng:$Al_2O_3$, Fe (2y mol), Al dư (x-2y mol)Phần 1:$\displaystyle \begin{array}{l}\text{Fe}\xrightarrow{{}}{{\text{H}}_{\text{2}}}\\\text{y}\,\,\to \,\,\,\,\text{y}\\\,\,\,\,\text{Al}\,\,\,\,\xrightarrow{{}}\,\,\,\,\frac{\text{3}}{\text{2}}{{\text{H}}_{\text{2}}}\\\frac{\left( \text{x-2y} \right)}{\text{2}}\,\,\,\to \,\,\frac{\text{3(x}\,\text{-}\,\text{2y)}}{\text{2}\text{.2}}\end{array}$
Phần 2$\displaystyle \begin{array}{l}\,\,\,\,\text{Al}\,\,\,\,\xrightarrow{{}}\,\,\,\,\frac{\text{3}}{\text{2}}{{\text{H}}_{\text{2}}}\\\frac{\left( \text{x-2y} \right)}{\text{2}}\,\,\,\to \,\,\frac{\text{3(x}\,\text{-}\,\text{2y)}}{\text{2}\text{.2}}\end{array}$
$\displaystyle \Rightarrow \,\text{y}\,\text{+}\,\frac{\text{3(x}\,\text{-}\,\text{2y)}}{\text{2}\text{.2}}\text{=2}\text{.}\frac{\text{3(x}\,\text{-}\,\text{2y)}}{\text{2}\text{.2}}$
$\Rightarrow$ x : y = 10 : 3
