Đáp án:
\(V = 2,25M\)
Giải thích các bước giải:
Sơ đồ phản ứng:
\(\left\{ \begin{gathered}
Mg \hfill \\
Zn \hfill \\
Al \hfill \\
\end{gathered} \right. + {O_2}\xrightarrow{{}}\left\{ \begin{gathered}
MgO \hfill \\
ZnO \hfill \\
A{l_2}{O_3} \hfill \\
\end{gathered} \right.\xrightarrow{{ + HCl}}\left\{ \begin{gathered}
MgC{l_2} \hfill \\
ZnC{l_2} \hfill \\
AlC{l_3} \hfill \\
\end{gathered} \right. + {H_2}O\)
BTKL: \({m_{kl}} + {m_{{O_2}}} = {m_{oxit}}\)
\( \to {m_{{O_2}}} = 20,3 - 13,1 = 7,2{\text{ gam}}\)
\( \to {n_{{O_2}}} = \frac{{7,2}}{{32}} = 0,225{\text{ mol}}\)
Bảo toàn O: \( \to {n_{{H_2}O}} = 2{n_{{O_2}}} = 0,45{\text{ mol}}\)
Bảo toàn H: \({n_{HCl}} = 2{n_{{H_2}O}} = 0,9{\text{ mol}}\)
\( \to V = \frac{{0,9}}{{0,4}} = 2,25M\)
BTKL: \({m_{oxit}} + {m_{HCl}} = {m_{muối}} + {m_{{H_2}O}}\)
\( \to {m_{muối}} = 20,3 + 0,9.36,5 - 0,45.18 = 45,05{\text{ gam}}\)
