Đáp án đúng: C
3b > 8a.
$\begin{array}{l}\,\,\,\text{Fe}\,\,\,\text{+}\,\,\,\,\text{4}{{\text{H}}^{\text{+}}}\,\text{+}\,\text{NO}_{\text{3}}^{\text{-}}\,\xrightarrow{{}}\,\text{F}{{\text{e}}^{\text{3+}}}\,\text{+}\,\text{NO}\,\text{+}\,\text{2}{{\text{H}}_{\text{2}}}\text{O}\\\text{0,25b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0,25b}\\\,\,\,\text{2F}{{\text{e}}^{\text{3+}}}\text{+}\,\,\,\,\,\,\,\,\text{Fe}\xrightarrow{{}}\,\text{3F}{{\text{e}}^{\text{2+}}}\\\text{0,25b}\,\,\to \,\,\,\text{0,125b}\end{array}$
Để dung dịch X hòa tan được Cu suy ra axit còn dư hoặc$\text{F}{{\text{e}}^{\text{3+}}}$ chưa phản ứng hết nên Fe phải hết.
⇒0,25b + 0,125b > a ⇒ 3b > 8a.
