Đáp án:
\({m_X} = 19,3{\text{ gam}}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
Cu \hfill \\
Mg \hfill \\
Fe \hfill \\
\end{gathered} \right.\xrightarrow{{HN{O_3}}}\left\{ \begin{gathered}
Cu{(N{O_3})_2} \hfill \\
Mg{(N{O_3})_2} \hfill \\
Fe{(N{O_3})_2} \hfill \\
\end{gathered} \right.\xrightarrow{{{t^o}}}\left\{ \begin{gathered}
CuO \hfill \\
MgO \hfill \\
F{e_2}{O_3} \hfill \\
\end{gathered} \right.\)
\( \to {m_{muối}} = {m_{kl}} + {m_{N{O_3}}} \to {m_{N{O_3}}} = 46,3 - 15,3 = 31{\text{ gam}}\)
\( \to {n_{N{O_3}}} = \frac{{31}}{{62}} = 0,5{\text{ mol}}\)
Nhận thấy:
\({n_{N{O_3}}} = 2{n_{Cu}} + 2{n_{Mg}} + 3{n_{Fe}}\)
\({n_O} = {n_{Cu}} = {n_{Mg}} + \frac{3}{2}{n_{Fe}} = \frac{1}{2}{n_{N{O_3}}} = 0,25{\text{ mol}}\)
\({m_X} = {m_{kl}} + {m_O} = 15,3 + 0,25.16 = 19,3{\text{ gam}}\)
