Đáp án:
a) 4,48l
b)27,6%
Giải thích các bước giải:
\(\begin{array}{l}
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
{n_{Zn}} = 0,2mol \to {n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
\to {m_{{H_2}}} = 0,2 \times 2 = 0,4g\\
\to {V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
\to {n_{C{H_3}{\rm{COOH}}}} = 2{n_{Zn}} = 0,4mol\\
\to {m_{C{H_3}{\rm{COOH}}}} = 0,4 \times 60 = 24g\\
\to {m_{{\rm{dd}}C{H_3}{\rm{COOH}}}} = \dfrac{{24}}{{20\% }} \times 100\% = 120g\\
\to {n_{{{(C{H_3}{\rm{COO)}}}_2}Zn}} = {n_{Zn}} = 0,2mol\\
\to {m_{{{(C{H_3}{\rm{COO)}}}_2}Zn}} = 0,2 \times 183 = 36,6g\\
\to {m_{{\rm{dd}}}}_{{{(C{H_3}{\rm{COO)}}}_2}Zn} = {m_{Zn}} + {m_{{\rm{dd}}}}_{C{H_3}{\rm{COOH}}} - {m_{{H_2}}} = 132,6g\\
\to C{\% _{{{(C{H_3}{\rm{COO)}}}_2}Zn}} = \dfrac{{36,6}}{{132,6}} \times 100\% = 27,6\%
\end{array}\)
