Đáp án đúng: B
22,40
Vì Cu nên Fe3+ chuyển về Fe2+
$\displaystyle \begin{array}{l}\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{+ 6HCl }\!\!~\!\!\text{ }\xrightarrow{{}}\text{ 2 FeC}{{\text{l}}_{\text{3}}}\text{+ 3}{{\text{H}}_{\text{2}}}\text{O}\\\text{Cu + 2F}{{\text{e}}^{\text{3+}}}\xrightarrow{{}}\text{C}{{\text{u}}^{\text{2+}}}\text{+ 2F}{{\text{e}}^{\text{3+}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{2x}\\\text{F}{{\text{e}}^{\text{2+}}}\text{+ A}{{\text{g}}^{\text{+}}}\xrightarrow{{}}\text{F}{{\text{e}}^{\text{3+}}}\text{+ Ag}\\\text{A}{{\text{g}}^{\text{+}}}\text{+ C}{{\text{l}}^{\text{-}}}\text{ }\!\!~\!\!\text{ }\xrightarrow{{}}\text{AgCl}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{6x}\end{array}$
m kết tủa = mAg + AgCl = 108.2x + 143,5.6x = 86,16 => x = 0,08 mol
Chất rắn sau phản ứng còn lại là 0,2 m gam => lượng phản ứng là 0,5 m gam
=> 0,8m = mFe2O3 + mCu( pư) = 0,08( 160 + 64) => m = 22,4g
