Đáp án đúng: B
0,03
Muối gồm:
$\left\{ \begin{array}{l}\text{FeS}{{\text{O}}_{\text{4}}}\text{:}\,\,\text{x}\,\,\text{mol}\\\text{F}{{\text{e}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{2}}}\text{:}\,\text{y}\,\,\text{mol}\end{array} \right.\Rightarrow \text{152x}\,\text{+}\,\text{400y}\,\text{=}\,\text{5,04}\left( \text{1} \right)$
Số mol e nhường$\text{=}\,\text{2x}\,\text{+}\,\text{6y}\Leftrightarrow $ số mol$\text{S}{{\text{O}}_{\text{2}}}\,\text{=}\,\text{x}\,\text{+}\,\text{3y}$
⇒ Số mol H2SO4 phản ứng$=2x+6y\left( mol \right)$
Theo bài ra ta có:$\frac{{{\text{n}}_{\text{Fe}}}}{{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}}}\text{=}\frac{\text{3}}{\text{7}}\Rightarrow \frac{\text{x}\,\text{+}\,\text{2y}}{\text{2x}\,\text{+}\,\text{6y}}\text{=}\frac{\text{3}}{\text{7}}\Rightarrow \text{x}\,\text{-}\,\text{4y}\,\text{=}\,\text{0}\left( \text{2} \right)$
Giải (1) và (2) ta được :$\left\{ \begin{array}{l}\text{x}\,\text{=}\,\text{0,02}\\\text{y}\,\text{=}\,\text{0,005}\end{array} \right.\Rightarrow \text{a}\,\text{=}\,\text{x}\,\text{+}\,\text{2y}\,\text{=}\,\text{0,03}\left( \text{mol} \right)$
