Đáp án:
a, \({V_{{H_2}}} = 2,24l\)
b, \({m_{{{(C{H_3}COO)}_2}Zn}} = \)14,64g
Giải thích các bước giải:
\(\begin{array}{l}
2C{H_3}COOH + Zn \to {(C{H_3}COO)_2}Zn + {H_2}\\
{n_{Zn}} = 0,1mol\\
\to {n_{{H_2}}} = {n_{Zn}} = 0,1mol \to {V_{{H_2}}} = 2,24l\\
{n_{{{(C{H_3}COO)}_2}Zn}} = {n_{Zn}} = 0,1mol\\
\to {m_{{{(C{H_3}COO)}_2}Zn}} = 18,3g\\
\to {m_{{{(C{H_3}COO)}_2}Zn}}thực tế= \dfrac{{{m_{{{(C{H_3}COO)}_2}Zn}} \times H}}{{100}} = \dfrac{{18,3 \times 80}}{{100}} = 14,64g
\end{array}\)
