Đáp án đúng: B
42,12.
$\displaystyle {{\text{n}}_{\text{Fe}}}\text{=0,15}\,\text{mol;}\,{{\text{n}}_{\text{AgN}{{\text{O}}_{\text{3}}}}}\text{=0,39}\,\text{mol}$
$\displaystyle \begin{array}{l}\,\,\text{Fe}\,\,\text{+}\,\,\text{2AgN}{{\text{O}}_{\text{3}}}\xrightarrow{{}}\text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{2}}}\text{+2Ag}\\\text{0,15 }\,\to \,\,\,\text{0,3 }\!\!~\!\!\text{ }\,\,\text{ }\!\!~\!\!\text{ }\,\xrightarrow{{}}\,\text{0,15mol}\end{array}$
$\displaystyle \begin{array}{l}\text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{2}}}\text{+AgN}{{\text{O}}_{\text{3}}}\xrightarrow{{}}\text{Fe}{{\left( \text{N}{{\text{O}}_{\text{3}}} \right)}_{\text{3}}}\text{+Ag}\\\,\,\text{0,09 }\!\!~~~\!\!\text{ }\xleftarrow{{}}\text{0,09 mol}\end{array}$
$\displaystyle \Rightarrow {{\text{n}}_{\text{Ag}}}\text{=}\,\text{0,39}\,\text{mol}$
$\displaystyle \Rightarrow {{\text{m}}_{\text{Ag}}}\text{=42,12}\,\text{gam}$
