Đáp án:
a, \(C{M_{C{H_3}COOH}} = \dfrac{{0,1}}{{0,025}} = 4M\)
b, \({V_{{H_2}}} = 1,12l\)
c, \({V_{{O_2}}} = 0,56l\)
Giải thích các bước giải:
\(\begin{array}{l}
2C{H_3}COOH + Mg \to {(C{H_3}COO)_2}Mg + {H_2}\\
{n_{{{(C{H_3}COO)}_2}Mg}} = 0,05mol\\
\to {n_{C{H_3}COOH}} = 2{n_{{{(C{H_3}COO)}_2}Mg}} = 0,1mol\\
\to C{M_{C{H_3}COOH}} = \dfrac{{0,1}}{{0,025}} = 4M\\
{n_{{H_2}}} = {n_{{{(C{H_3}COO)}_2}Mg}} = 0,05mol\\
\to {V_{{H_2}}} = 1,12l\\
{H_2} + \dfrac{1}{2}{O_2} \to {H_2}O\\
{n_{{O_2}}} = \dfrac{1}{2}{n_{{H_2}}} = 0,025mol\\
\to {V_{{O_2}}} = 0,56l
\end{array}\)
