Đáp án đúng: B
FeO.
$\begin{array}{l}{{\text{n}}_{\text{Al}\,\text{(Y)}}}\,\text{=}\,\frac{{{\text{n}}_{{{\text{H}}_{\text{2}}}\,\text{(1)}}}}{\text{1}\text{,5}}\,\text{=0}\text{,05}\,\text{mol;}\,{{\text{n}}_{\text{Fe}\,\text{(Y)}}}\,\text{=}\,{{\text{n}}_{{{\text{H}}_{\text{2}}}\,\text{(2)}}}\,\text{-}\,{{\text{n}}_{{{\text{H}}_{\text{2}}}\,\text{(1)}}}\,\text{=}\,\text{0}\text{,15}\,\text{mol}\\\Rightarrow \,{{\text{n}}_{\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}\,\text{=}\,\frac{\text{14}\text{,85}\,\text{-}\,\text{27}\text{.0}\text{,05}\,\text{-}\,\text{56}\text{.0}\text{,15}}{\text{102}}\,\text{=}\,\text{0}\text{,05}\,\Rightarrow \,{{\text{n}}_{\text{O}}}\,\text{=}\,\text{0}\text{,15}\\\text{x}\,\text{:}\,\text{y}\,\text{=}\,{{\text{n}}_{\text{Fe}}}\,\text{:}\,{{\text{n}}_{\text{O}}}\,\text{=}\,\text{1}\,\text{:}\,\text{1}\,\Rightarrow \,\text{FeO}\end{array}$
