Đáp án:
\({V_{dd{{\text{H}}_2}S{O_4}}} = 0,12{\text{ lít}}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
CuO \hfill \\
MgO \hfill \\
F{e_2}{O_3} \hfill \\
A{l_2}{O_3} \hfill \\
\end{gathered} \right. + {H_2}S{O_4}\xrightarrow{{}}\left\{ \begin{gathered}
CuS{O_4} \hfill \\
MgS{O_4} \hfill \\
F{e_2}{(S{O_4})_3} \hfill \\
A{l_2}{(S{O_4})_3} \hfill \\
\end{gathered} \right. + {H_2}O\)
Gọi \({n_{{H_2}S{O_4}}} = {n_{{H_2}O}} = x\)
BTKL:
\({m_A} + {m_{{H_2}S{O_4}}} = {m_{muối}} + {m_{{H_2}O}} \to 3,84 + 98x = 11,52 + 18x \to x = 0,096{\text{ mol}}\)
\( \to {V_{dd{{\text{H}}_2}S{O_4}}} = \frac{{{n_{{H_2}S{O_4}}}}}{{{C_{M{{\text{H}}_2}S{O_4}}}}} = \frac{{0,096}}{{0,8}} = 0,12{\text{ lít}}\)
