Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Zn}} = 31,25\% \\
\% {m_{ZnO}} = 68,75\% \\
b)\\
{m_{{\rm{dd}}C{H_3}COOH}} = 168g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
ZnO + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}O\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,1\,mol\\
\% {m_{Zn}} = \dfrac{{0,1 \times 65}}{{20,8}} \times 100\% = 31,25\% \\
\% {m_{ZnO}} = 100 - 31,25 = 68,75\% \\
b)\\
{n_{ZnO}} = \dfrac{{20,8 - 6,5}}{{81}} \approx 0,18\,mol\\
{n_{C{H_3}COOH}} = (0,1 + 0,18) \times 2 = 0,56\,mol\\
{m_{{\rm{dd}}C{H_3}COOH}} = \dfrac{{0,56 \times 60}}{{20\% }} = 168g
\end{array}\)