Đáp án:

3,

b, \({m_{{{(C{H_3}COO)}_2}Zn}} = 36,6g\)

c,

\(C{\% _{C{H_3}COOH}}dd = \dfrac{{24}}{{200}} \times 100\%  = 12\% \)

\(C{\% _{{{(C{H_3}COO)}_2}Zn}}dd = \dfrac{{36,6}}{{216,2}} \times 100\%  = 16,93\% \)

Giải thích các bước giải:

3,

\(\begin{array}{l}
ZnO + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}O\\
{n_{ZnO}} = 0,2mol\\
 \to {n_{{{(C{H_3}COO)}_2}Zn}} = {n_{ZnO}} = 0,2mol\\
 \to {m_{{{(C{H_3}COO)}_2}Zn}} = 36,6g\\
{n_{C{H_3}COOH}} = 2{n_{ZnO}} = 0,4mol\\
 \to {m_{C{H_3}COOH}} = 24g\\
 \to C{\% _{C{H_3}COOH}}dd = \dfrac{{24}}{{200}} \times 100\%  = 12\% \\
{m_{dd}} = {m_{ZnO}} + {m_{C{H_3}COOH}}dd = 216,2g\\
 \to C{\% _{{{(C{H_3}COO)}_2}Zn}}dd = \dfrac{{36,6}}{{216,2}} \times 100\%  = 16,93\% 
\end{array}\)