Đáp án đúng: A
173,2 gam.
Phân tích: $\displaystyle \underbrace{\text{Fe, F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{,Cu,CuO}}_{\text{33,2 gam}}\text{ +}\,\,\text{HCl }\left( \text{1mol} \right)\text{ }$
$\displaystyle \Rightarrow \left\{ \begin{array}{l}\text{1,6}\,\text{gam}\left( \text{Cu} \right)\text{ }\\{{\text{H}}_{\text{2}}}\text{ 0,1mol }\\\text{dd Y}\xrightarrow{\text{AgN}{{\text{O}}_{\text{3}}}}\left\{ \begin{array}{l}\text{m gam}\downarrow \\\text{0,025 mol NO}\end{array} \right.\end{array} \right.\text{ }$
$\displaystyle {{\text{n}}_{\text{Cu du}}}\text{ =}\frac{\text{1,6}}{\text{64}}\text{=0,025 mol;}$$\displaystyle {{\text{n}}_{{{\text{H}}_{\text{2}}}}}\text{ = }{{\text{n}}_{\text{Fe}}}\text{ = 0,1}\left( \text{mol} \right)\text{ }$
Đặt số mol của Fe3O4 là a (mol) và số mol CuO là b (mol)
Ta có:
$\displaystyle \begin{array}{l}\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\text{+ 8HCl }\xrightarrow{{}}\text{ FeC}{{\text{l}}_{\text{2}}}\text{+2FeC}{{\text{l}}_{\text{3}}}\\\,\,\,\,\,\text{a }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\,\,\,\,\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ 8a }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\,\,\,\,\,\,\,\,\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ a }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\,\,\,\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ 2a}\end{array}$
$\displaystyle \begin{array}{l}\text{Cu +2FeC}{{\text{l}}_{\text{3}}}\xrightarrow{{}}\text{ CuC}{{\text{l}}_{\text{2}}}\text{+ 2FeC}{{\text{l}}_{\text{3}}}\\\,\text{a }\leftarrow \text{ 2a}\end{array}$
Suy ra trong 33,2g hỗn hợp X có
$\displaystyle {{\text{n}}_{\text{Cu}}}\text{= a+ 0,025 }\left( \text{mol} \right)$
$\displaystyle \Rightarrow \text{ 33,2 = 0,1}\text{.56+ 232}\text{. a +}\left( \text{a + 0,025} \right)\text{. 64 + 80b}$
$\displaystyle \Rightarrow \text{ 296a +80b = 26}\,\,\,\,\,\,\,\left( \text{1} \right)$
Vì khi cho AgNO3 dư vào dung dịch Y thấy thoát ra khí NO nên HCl chắc chắn còn dư
$\displaystyle \Rightarrow \text{ }{{\text{n}}_{\text{HCl du}}}\,\text{=}\,\text{1-2}{{\text{n}}_{{{\text{H}}_{\text{2}}}}}\text{-8}{{\text{n}}_{\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}}}\text{-2}{{\text{n}}_{\text{CuO}}}\,\text{=}\,\text{1-0,2-8a-2b}\,\text{=}\,\text{0,8-8a-2b}$
Ta có:
$\displaystyle \begin{array}{l}\text{3F}{{\text{e}}^{\text{2+}}}\text{+ 4}{{\text{H}}^{\text{+}}}\,\,\,\,\text{+ }\,\,\text{N}{{\text{O}}_{\text{3}}}^{\text{-}}\xrightarrow{{}}\text{ 3F}{{\text{e}}^{\text{3+}}}\text{+NO + 4}{{\text{H}}_{\text{2}}}\text{O}\\\,\,\,\,\,\,\,\,\,\,\,\,\,0,075\text{ }\,\,\,\leftarrow \text{ }0,1~~~~~~~~~~~~~~\leftarrow ~~~~~~~~~~\,\,\,\,\,\,\,\,\,\,~0,025\end{array}$
$\displaystyle \Rightarrow \text{ }{{\text{n}}_{\text{HCl}\,\,\text{du}}}\text{ = 0,1}\left( \text{mol} \right)\text{ }\Rightarrow \text{ 8a+ 2b = 0,7 }\,\,\,\text{ }\left( \text{2} \right)$
$\displaystyle \left( \text{1} \right)\text{;}\,\left( \text{2} \right)\Rightarrow \text{a}\,\text{=}\frac{\text{1}}{\text{12}}\left( \text{mol} \right)\text{; b}\,\text{=}\frac{\text{1}}{\text{60}}\left( \text{mol} \right)\text{ }$
Suy ra dung dịch Y gồm
$\left\{ \begin{array}{l}\text{HCl du 0,1 mol}\\\text{FeC}{{\text{l}}_{\text{2}}}\text{ 0,35 mol}\\\text{ CuC}{{\text{l}}_{\text{2}}}\text{ 0,1 mol }\end{array} \right.$
Từ phương trình (I), ta thấy FeCl2 dư 0,275 mol nên ta có thêm phản ứng :
$\displaystyle \text{A}{{\text{g}}^{\text{+}}}\text{+ F}{{\text{e}}^{\text{2+}}}\xrightarrow{{}}\text{ F}{{\text{e}}^{\text{3+}}}\text{+ Ag}\downarrow $$\displaystyle \Rightarrow \text{ }{{\text{n}}_{\text{Ag}}}\text{=0,275 mol; }{{\text{n}}_{\text{AgCl}}}\text{= }{{\text{n}}_{\text{C}{{\text{l}}^{\text{-}}}}}\text{=}{{\text{n}}_{\text{HCl}}}_{\,\text{bd}}\text{= 1 mol}$
$\displaystyle \text{m=0,275}\text{.108+1}\text{.}\left( \text{108+35,5} \right)\text{=173,2 }\left( \text{gam} \right)$
Vậy m=173,2 gam.
