Giải thích các bước giải:
4,
\(\begin{array}{l}
Mg + 2C{H_3}COOH \to {(C{H_3}COO)_2}Mg + {H_2}\\
{n_{Mg}} = 0,1mol\\
\to {n_{{H_2}}} = {n_{Mg}} = 0,1mol\\
\to {V_{{H_2}}} = 2,24l\\
\to {n_{C{H_3}COOH}} = {n_{Mg}} = 0,1mol\\
\to {m_{C{H_3}COOH}} = 6g
\end{array}\)
5,
\(\begin{array}{l}
Zn + 2C{H_3}COOH \to {(C{H_3}COO)_2}Zn + {H_2}\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Zn}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Zn}} = 6,5g\\
\to {n_{{{(C{H_3}COO)}_2}Zn}} = {n_{Zn}} = 0,1mol\\
\to {m_{{{(C{H_3}COO)}_2}Zn}} = 18,3g
\end{array}\)
6,
\(\begin{array}{l}
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{B{r_2}}} = 0,1mol\\
\to {n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,1mol\\
\to {V_{{C_2}{H_4}}} = 2,24l\\
\to {n_{{C_2}{H_4}B{r_2}}} = {n_{B{r_2}}} = 0,1mol\\
\to {m_{{C_2}{H_4}B{r_2}}} = 18,8g
\end{array}\)
