Đáp án đúng: A
22,7g.
$\begin{array}{l}\text{Trong}\,\text{12}\,\text{gam}\,\text{X:}\\{{\text{m}}_{\text{Fe}}}\,\text{=}\,\text{12}\text{.46,67}{\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}\,\text{=}\,\text{5,6}\,\,\text{gam}\,\Rightarrow \,{{\text{n}}_{\text{Fe}}}\,\text{=}\,\frac{\text{5,6}}{\text{56}}\,\text{=}\,\text{0,1}\,\text{mol}\\\Rightarrow \,{{\text{m}}_{\text{Cu}}}\,\text{=}\,\text{12}\,\text{-}\,\text{5,6}\,\,\text{=}\,\text{6,4}\,\text{gam}\,\,\Rightarrow {{\text{n}}_{\text{Cu}}}\text{=0,1}\,\text{mol}\end{array}$
Sau phản ứng còn lại 4,8 gam kim loại chưa tan (Cu)
⇒ dung dịch Y có muối$\text{Fe(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{2}}$ và$\text{Cu(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$
${{\text{n}}_{\text{Fe(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{2}}}}\,\text{=}\,{{\text{n}}_{\text{Fe}}}\,\text{=}\,\text{0,1}\,\text{mol}$
${{\text{n}}_{\text{Cu(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}}}\,\text{=}\,\text{0,1}\,\text{-}\,\frac{\text{4,8}}{\text{64}}\,\text{=}\,\text{0,025}\,\text{mol}$
$\text{m}\,\text{=}\,\text{0,1}\text{.180}\,\text{+}\,\text{0,025}\text{.188}\,\text{=}\,\text{22,7}\,\text{gam}$
