Đáp án đúng: C
9,15.
nFe = 0,02 ; nHCl = 0,06 ⇒ nH+ dư = 0,02 mol
$\displaystyle \begin{array}{l}3F{{e}^{2+}}~~+~~4{{H}^{+}}+\text{ }N{{O}_{3}}^{-}\xrightarrow{{}}3F{{e}^{3+}}+\text{ }NO\text{ }+\text{ }2{{H}_{2}}O\\0,015\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,02\end{array}$
$\displaystyle \begin{array}{l}F{{e}^{2+}}+\text{ }A{{g}^{+}}\xrightarrow{{}}F{{e}^{3+}}+\text{ }Ag\\0,005\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,005\end{array}$
$\displaystyle \Rightarrow {{m}_{Ag}}=\text{ }0,54\text{ }gam$
$\displaystyle A{{g}^{+}}+\text{ }C{{l}^{-}}\xrightarrow{{}}AgCl~~$
$\displaystyle \Rightarrow {{m}_{AgCl}}\text{ }=\text{ }8,61~~\Rightarrow m\text{ }=\text{ }9,15$