Đáp án:
\(\% {V_{{O_2}}} = 46,15\% \)
Giải thích các bước giải:
Ta có:
\({n_{Mg}} = {n_{Fe}} = \frac{{9,6}}{{56 + 24}} = 0,12{\text{ mol}}\)
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
Mg \hfill \\
Fe \hfill \\
\end{gathered} \right. + \left\{ \begin{gathered}
C{l_2} \hfill \\
{O_2} \hfill \\
\end{gathered} \right.\xrightarrow{{{t^o}}}\left\{ \begin{gathered}
oxit \hfill \\
clorua \hfill \\
\end{gathered} \right.\xrightarrow{{ + 0,36{\text{ mol HCl}}}}\left\{ \begin{gathered}
clorua \hfill \\
{H_2}O \hfill \\
\end{gathered} \right.\)
Bảo toàn H: \({n_{{H_2}O}} = \frac{1}{2}{n_{HCl}} = 0,18{\text{ mol}}\)
Bảo toàn O:
\({n_O} = {n_{{H_2}O}} = 0,18{\text{ mol}} \to {{\text{n}}_{{O_2}}} = \frac{1}{2}{n_O} = 0,09{\text{ mol}}\)
Gọi \({n_{C{l_2}}} = x{\text{ mol}}\)
Bảo toàn nguyên tố Cl:
\({n_{AgCl}} = 2{n_{C{l_2}}} + {n_{HCl}} = 2x + 0,36{\text{ mol}}\)
Bảo toàn e:
\({n_{Ag}} = 2{n_{Mg}} + 3{n_{Fe}} - 4{n_{{O_2}}} - 2{n_{C{l_2}}} = 0,12.2 + 0,12.3 - 0,09.4 - 2x = 0,24 - 2x{\text{ mol}}\)
\( \to 143,5.(2x + 0,36) + 108.(0,24 - 2x) = 85,035 \to x = 0,105{\text{ mol}}\)
\( \to \% {V_{{O_2}}} = \frac{{0,09}}{{0,09 + 0,105}} = 46,15\% \)
