Đáp án đúng: B
19,24g.
$\begin{array}{l}{{n}_{Fe}}=\frac{5,6}{56}=0,1;\,{{n}_{A{{g}^{+}}}}=0,22\\Fe\,\,+\,\,2A{{g}^{+}}\xrightarrow{{}}F{{e}^{2+}}+2Ag\\0,1\to 0,2\,\,\,\to \,\,\,\,\,\,\,\,0,1\\F{{e}^{2+}}\,\,\,+\,\,\,A{{g}^{+}}\xrightarrow{{}}F{{e}^{3+}}+Ag\\0,02\,\leftarrow \,0,02\,\,\xrightarrow{{}}\,0,02\end{array}$
Vậy trong dung dịch X có 0,02 mol$Fe{{(N{{O}_{3}})}_{3}}$ và 0,08 mol$Fe{{(N{{O}_{3}})}_{2}}$
${{m}_{Fe{{(N{{O}_{3}})}_{3}}}}+{{m}_{Fe{{(N{{O}_{3}})}_{2}}}}=0,02.242+0,08.180=19,24\,gam$