Đáp án:
\(m = 44{\text{ gam}}\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
MgO \hfill \\
F{e_2}{O_3} \hfill \\
CuO \hfill \\
\end{gathered} \right. + HCl\xrightarrow{{}}\left\{ \begin{gathered}
MgC{l_2} \hfill \\
FeC{l_3} \hfill \\
CuC{l_2} \hfill \\
\end{gathered} \right. + NaOH\xrightarrow{{}}\left\{ \begin{gathered}
Mg{(OH)_2} \hfill \\
Fe{(OH)_3} \hfill \\
Cu{(OH)_2} \hfill \\
\end{gathered} \right.\)
Ta có:
\({m_{HCl}} = 400.14,6\% = 58,4{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{58,4}}{{36,5}} = 1,6{\text{ mol}}\)
Ta có:
\({n_{HCl}} = {n_{NaCl}} = {n_{NaOH}} = 1,6{\text{ mol}}\)
\({n_{OH}} = {n_{NaOH}} = 1,6{\text{ mol}}\)
\({m_{kt}} = {m_{kl}} + {m_{OH}} = {m_{kl}} + 1,6.17 = 58,4{\text{ gam}}\)
\({m_{kl}} = 58,4 - 1,6.17 = 31,2{\text{ gam}}\)
\({m_{{H_2}O}} = \frac{1}{2}{n_{HCl}} = 0,8{\text{ mol = }}{{\text{n}}_O}\)
\( \to m = {m_{kl}} + {m_O} = 31,2 + 0,8.16 = 44{\text{ gam}}\)
